Learn Fuzzy Logic → Genetic Algorithms → Particle Swarm Optimization, with the required maths explained just before it is used. The answers stay below the questions.
\r\n| Question | Marks | Attempt rule | Prepare for |
|---|---|---|---|
| Q1(a) | 2 | Compulsory | A short numerical or theory question |
| Q1(b) | 2 | Compulsory | A short numerical or theory question, including swarm intelligence |
| Q2, Q3, Q4 | 3 each | Attempt any two | GA, PSO and fuzzy logic; the mix can include theory and numericals |
Your class update specifically highlights fuzzy union/intersection/difference/complement, GA with supplied operators, two-particle PSO minimisation, fuzzy rules, architecture, advantages and disadvantages. Work GA and PSO through iteration 2. The mock question numbering is a practice arrangement; the actual exam can rearrange the topics.
\r\nFuzzy relation matrices are kept as additional Unit 2 practice after the three main topics. ACO and GA-for-TSP are in the TEE route; they are not mixed into this M2 reading path.
\r\nFull question → Given → To find → Formula and why → numbered calculation steps → final answer. Read one worked example, then write the same calculation in your notebook. Keep its intermediate table: in GA and PSO that table is the input to the next iteration.
\r\nFirst complete fuzzy operations, triangle membership and the Mamdani architecture. Next finish one GA two-generation solution and one PSO two-iteration solution. Then do the remaining variants, theory answers and mocks. Use the formula page for revision after learning the symbols; it is not a substitute for the worked explanations.
\r\nYaad rakho: Fuzzy = degree aur rules. GA = select, cross, mutate. PSO = velocity, position, best memories.
"}]},{"id":"core-maths","title":"1. The maths these three topics use","unit":"Start here","scope":"Short prerequisites, with worked examples. Read once before Fuzzy → GA → PSO.","sections":[{"id":"reading-formulas","title":"Symbols, brackets and iteration labels: how to read a formula","html":"A variable is a name for a number that can change. In f(x)=x²+1, x is the input, f is the name of the scoring rule, and f(x) is its result. An equation says that the expressions on both sides have the same value. A formula tells you how to calculate a quantity.
| Read this | Meaning | Concrete example |
|---|---|---|
| x² | x multiplied by itself | 3² = 3 × 3 = 9 |
| 2x or 2(x) | 2 multiplied by x | For x=3: 2x=6, while x²=9 |
| f(3) | Apply the rule named f to input 3 | If f(x)=x²+1, then f(3)=10 |
| x₁ and x₂ | First and second named entries; the small lower number is a label | If a position is (3,7), x₁=3 and x₂=7 |
| x at time t | A version of x at a particular iteration | Old x=3; after a move x=2. The time label is not squaring. |
| = and ≈ | Exactly equal; approximately equal after rounding | 1/3 ≈ 0.333333 |
| ∈ and { } | Belongs to; a collection of values | x ∈ {2,5} means x may be 2 or 5 |
Evaluate f(x)=2x²+1 at x=−3. [2 marks · Worked example]
The rule is f(x)=2x²+1. The supplied input is x=−3.
f(−3): the output of this rule for input −3.
![\"f(x)=2x^2+1\"](\"math/4ee7030c8a3d9d3a.svg\")
The question supplies the rule. Replace x by −3, square that whole input, multiply the square by 2, and add 1. Parentheses keep the negative input together.
![\"f(-3)=2(-3)^2+1\"](\"math/c18467bab709f205.svg\")
![\"(-3)^2=(-3)\\times(-3)=9\"](\"math/ac40c2845414cf0d.svg\")
![\"2\\times9=18\"](\"math/c271680245ea1d71.svg\")
![\"f(-3)=18+1=19\"](\"math/27fc155eedcc5888.svg\")
Evaluate g(x)=(x−2)/4 at x=10. [2 marks · Worked example]
The rule is g(x)=(x−2)/4. The supplied input is x=10.
g(10): the output of this rule for input 10.
![\"g(x)=\\frac{x-2}{4}\"](\"math/ef554c6170bf393e.svg\")
The fraction bar groups the numerator x−2. Subtract 2 from the input first, then divide the entire difference by 4.
![\"g(10)=\\frac{10-2}{4}\"](\"math/e276417a70627f82.svg\")
![\"10-2=8\"](\"math/bdb801d401c03c68.svg\")
![\"g(10)=8/4=2\"](\"math/40041b1514c36954.svg\")
At iteration 0, x=4. The rule is new x = old x − 1. Find x after two iterations. [2 marks · Worked example]
Initial value x=4 at iteration 0. Every update subtracts 1 from the current value. Perform two updates.
The value after update 1 and after update 2.
![\"x^{(t+1)}=x^{(t)}-1\"](\"math/14d3646dbde001b1.svg\")
t labels the current iteration, and t+1 labels the next. The parenthesised upper label records the step; it is not a power. Apply the supplied rule once, then use that new answer as the input to the second update.
![\"x^{(1)}=4-1=3\"](\"math/6df71b72d0a31e70.svg\")
![\"x^{(2)}=3-1=2\"](\"math/40962d708fa42c89.svg\")
x at iteration 2 is 2. Squaring the initial value would give 4²=16; that is a different operation.
Yaad rakho: Symbol sirf naam hai. Pehle bolo kaunsa number chahiye, phir uski jagah value likho.
"},{"id":"objective","title":"Position, score, minimisation and maximisation","html":"An intelligent system uses information and a decision procedure to choose an output or an action. In M2, you study three different procedures.
| Topic | Input | What the procedure does | Output |
|---|---|---|---|
| Fuzzy logic | Temperature and occupancy | Applies rules such as IF hot AND crowded THEN fast | A fan-speed setting |
| Genetic algorithm (GA) | A population of candidate answers | Selects, combines and changes candidate answers | A good solution and its fitness |
| Particle swarm optimization (PSO) | Positions and velocities of particles | Moves candidate answers using individual and shared experience | A best position and its objective value |
An objective function is the calculation used to score an answer. In minimisation, a smaller score is better; in maximisation, a larger score is better. The letter x names a candidate answer. The notation f(x), read “f of x”, means its score. Parentheses here do not mean f multiplied by x.
For the minimisation problem ![\"f(x)=(x-3)^2\"](\"math/17ba9de444962fb9.svg\")
, find the score at ![\"x=2\"](\"math/feb11b3ddc232396.svg\")
. [2 marks · Practice]
,
The objective value f(x).
Replace x by the supplied number. Square means multiply the result by itself.
![\"x-3=2-3=-1\"](\"math/55d4a2b861701691.svg\")
![\"f(2)=(-1)^2=(-1)\\times(-1)=1\"](\"math/28ec806f80fe7e4c.svg\")
For the minimisation problem , find the score at ![\"x=5\"](\"math/7a95cf78643aae43.svg\")
. [2 marks · Practice]
,
The objective value f(x).
Replace x by the supplied number. Square means multiply the result by itself.
![\"x-3=5-3=2\"](\"math/c893e43cc49d3cae.svg\")
![\"f(5)=(2)^2=(2)\\times(2)=4\"](\"math/4ad5b97ffc47b26c.svg\")
For the minimisation problem , find the score at ![\"x=-1\"](\"math/2854d883b15c7a60.svg\")
. [2 marks · Practice]
,
The objective value f(x).
Replace x by the supplied number. Square means multiply the result by itself.
![\"x-3=(-1)-3=-4\"](\"math/cca2cd1e0a06e8d4.svg\")
![\"f(-1)=(-4)^2=(-4)\\times(-4)=16\"](\"math/dce608d3d6bd8495.svg\")
Comparing these candidates: x = 2 has score 1, x = 5 has score 4, and x = −1 has score 16. For minimisation, x = 2 is the best of these three. A negative position is allowed unless the question gives a restriction; a low score, not a low coordinate, decides the winner.
Yaad rakho: x = answer ki jagah; f(x) = us answer ka score. Min mein chhota score jeetega.
"},{"id":"arithmetic","title":"Fractions, brackets, minimum and maximum","html":"A fraction means division: the top number is the numerator, and the bottom is the denominator. Calculate each bracket first. min selects the smallest listed number; max selects the largest. Neither word means addition.
Evaluate ![\"\\frac{32-25}{35-25}\"](\"math/df0b8ec6b8392714.svg\")
. [2 marks · Practice]
Top: 32 − 25; bottom: 35 − 25.
The quotient.
Subtract separately in the numerator and denominator, then divide.
![\"32-25=7\"](\"math/c65340df2ae7cb64.svg\")
![\"35-25=10\"](\"math/9a5b4bf9b146b2b2.svg\")
![\"\\frac{7}{10}=0.7\"](\"math/6b3a1b0f1231fc45.svg\")
Evaluate ![\"\\frac{35-32}{35-25}\"](\"math/55a4e1a39f439f8c.svg\")
. [2 marks · Practice]
Top: 35 − 32; bottom: 35 − 25.
The quotient.
Subtract separately in the numerator and denominator, then divide.
![\"35-32=3\"](\"math/b8b4baa992603f77.svg\")
![\"\\frac{3}{10}=0.3\"](\"math/df25a279590f4f48.svg\")
Evaluate ![\"\\frac{7-6}{8-6}\"](\"math/5627768e8762bda4.svg\")
. [2 marks · Practice]
Top: 7 − 6; bottom: 8 − 6.
The quotient.
Subtract separately in the numerator and denominator, then divide.
![\"7-6=1\"](\"math/e03d8b2e4fa9e9e0.svg\")
![\"8-6=2\"](\"math/2511936ca0026cfc.svg\")
![\"\\frac{1}{2}=0.5\"](\"math/0ef4a9ef78e1b7d5.svg\")
Find ![\"\\max\\{\\min(0.7,0.2),\\min(0.4,0.8)\\}\"](\"math/f5116e7e9b08a405.svg\")
. [2 marks · Practice]
Two inner min operations followed by one outer max.
One number.
Work from the innermost brackets outward. This exact pattern occurs in fuzzy relation composition.
![\"\\min(0.7,0.2)=0.2\"](\"math/0f24b8a13d9d75d1.svg\")
![\"\\min(0.4,0.8)=0.4\"](\"math/8fd31f9127b73dc1.svg\")
![\"\\max(0.2,0.4)=0.4\"](\"math/06c831ab89f196ab.svg\")
Yaad rakho: Pehle bracket, phir divide. AND ka min; OR ka max.
"},{"id":"probability","title":"Percentages, probabilities and cumulative intervals","html":"A probability is a number between 0 and 1. In these finite examples, 0 means an event has no probability mass and 1 means it occurs with certainty under the model. In a continuous distribution, a single exact value can have probability zero without being logically impossible; probabilities then describe intervals. 25% means 25 out of 100, so it equals 25÷100=0.25. For equally likely outcomes, probability is favourable outcomes divided by all possible outcomes. For GA selection, it is the candidate’s weight divided by the sum of weights.
![\"P(\\text{event})=\\frac{\\text{favourable](\"math/8e42f83b578b6254.svg\")
A bag contains 3 red balls and 7 blue balls. One ball is drawn uniformly at random. Find the probability of red and of blue. [2 marks · Worked example]
Red=3; blue=7. Each ball is equally likely.
Both probabilities and their percentages.
![\"P(\\text{colour})=\\frac{\\text{balls](\"math/6e0b3ead99011455.svg\")
All balls are equally likely, so counting them gives probability. The denominator counts all balls; the numerator counts the requested colour.
![\"3+7=10\"](\"math/90e46c6fe6c61dca.svg\")
![\"P(R)=3/10=0.3=30\\%\"](\"math/ac0dd61435e485bc.svg\")
![\"P(B)=7/10=0.7=70\\%\"](\"math/927c15e53970ce9d.svg\")
A weight need not itself be a probability. Weights 1, 2, 3, 4 add to 10. Dividing every weight by 10 makes their total 1: this is normalisation. Cumulative means “total accumulated so far”. It turns separate shares into boundaries along a line from 0 to 1.
Candidates A, B, C and D have weights 1, 2, 3 and 4. Find probabilities and cumulative probabilities. Which candidate is selected by random draw r=0.55? [2 marks · Worked example]
Weights [1,2,3,4], draw 0.55. Use intervals including their lower endpoint and excluding their upper endpoint.
Four probabilities, their running totals and the selected candidate.
![\"P_i=F_i/(F_A+F_B+F_C+F_D)\"](\"math/791fc41b905c43f5.svg\")
F is weight. P is its share. Add P values in row order for the cumulative boundary.
![\"1+2+3+4=10\"](\"math/45a76e7db8947072.svg\")
| Candidate | Division | Probability |
|---|---|---|
| A | 1÷10 | 0.1 |
| B | 2÷10 | 0.2 |
| C | 3÷10 | 0.3 |
| D | 4÷10 | 0.4 |
| Candidate | Running addition | Interval |
|---|---|---|
| A | 0+0.1=0.1 | 0 ≤ r < 0.1 |
| B | 0.1+0.2=0.3 | 0.1 ≤ r < 0.3 |
| C | 0.3+0.3=0.6 | 0.3 ≤ r < 0.6 |
| D | 0.6+0.4=1 | 0.6 ≤ r ≤ 1 |
0.55 is at least 0.3 and less than 0.6. It lies in C’s interval. The draw is not compared with C’s weight 3.
![\"Each](\"fig/depth_roulette.12804497d2.svg\")
Two unnormalised class scores are 0.12 and 0.08. Convert them to probabilities. [2 marks · Worked example]
Scores A=0.12 and B=0.08; they describe the only two alternatives.
Normalised probabilities.
Divide each score by the sum, preserving their ratio.
![\"0.12+0.08=0.20\"](\"math/f3e0db112ddd0eb5.svg\")
![\"P(A)=0.12/0.20=0.6\"](\"math/fd57d6dd0af66643.svg\")
![\"P(B)=0.08/0.20=0.4\"](\"math/a6f1afe6f9d94092.svg\")
Yaad rakho: Probability = apna hissa ÷ sabka total. Cumulative = ab tak ke saare hisse jodna.
"},{"id":"binary","title":"Binary from the beginning: positions, place values and bits","html":"A position is simply a place occupied by one digit in a number. A digit is the symbol written there. The place value tells us how much one unit in that position is worth.
\r\nStart with an ordinary number you already know: 23. Its two digits are 2 and 3. They do not contribute the same kind of unit.
| LEFT: tens place | RIGHT: ones place | |
|---|---|---|
| Value of this place | 10 | 1 |
| Digit written here | 2 | 3 |
| Contribution | 2 × 10 = 20 | 3 × 1 = 3 |
![\"23=(2\\times](\"math/46c502cb15486e6a.svg\")
The 2 means two tens because of where it sits. In 32, the same digit 2 sits on the right and means only two ones. Changing a digit’s position can change its contribution.
Yaad rakho: Position = digit ki jagah. Place value = us jagah par likhe ONE ki keemat. 23 mein 2 ka contribution 20 hai, kyunki woh tens ki jagah par hai.
Our ordinary number system is decimal: it has ten possible digits, 0 to 9. Its places are ones, tens, hundreds, and so on. Moving one place left multiplies the place value by 10.
\r\nBinary has only two possible digits: 0 and 1. One binary digit is called a bit. Therefore, a 4-bit number means four digit boxes, each holding either 0 or 1.
\r\nIn binary, grouping happens in twos. The rightmost box counts single units. After one single, a second single cannot be written as digit “2”, because binary has no digit 2. Instead, the two singles become one group of two in the next box on the left.
\r\nBinary 10 means one group of two and zero single units.
It represents the ordinary number 2. Read its digits as “one-zero” when discussing binary, so you do not accidentally read it as decimal ten.
Two groups of two make 4, so the next box to the left is the fours place. Two groups of four make 8, so the next box is the eights place. This is why the values double.
| Ordinary number | Binary digits | Meaning |
|---|---|---|
| 0 | 0 | No single units |
| 1 | 1 | One single unit |
| 2 | 10 | One two + zero ones = 2 |
| 3 | 11 | One two + one one = 3 |
| 4 | 100 | One four + zero twos + zero ones = 4 |
Yaad rakho: Decimal mein 10 ka group; binary mein 2 ka group. Do singles = ek pair; do pairs = ek group of four.
First draw four empty boxes. Start with the box furthest to the right. Give that place the value 1. Move left one box at a time, doubling the previous value.
\r\nPlace value: what ONE in this box is worth
← Double each time you move one box LEFT
We built the values by moving right to left: 1 → 2 → 4 → 8. Once the boxes are on paper, reading them in the usual left-to-right direction shows 8, 4, 2, 1. These are the same four places, viewed in the opposite direction.
Position number is not place value. The third box from the right has position number 3, but its place value is 4. “Third” tells you where the box is; “4” tells you what a 1 in it contributes.
If you add a fifth box on the left, its value is 8 × 2 = 16. The rightmost place stays 1; adding a box on the left does not change the existing places.
Yaad rakho: Sabse RIGHT se 1 likho. LEFT jaate jaate double karo: 1, 2, 4, 8, 16. Question ke digits baad mein in boxes mein rakho.
Keep the digit 1 the same and move it into a different place. Put 0 in every other box. Watch how its position changes the number it represents:
| Binary digits | Where the 1 sits | Decimal value |
|---|---|---|
| 0001 | 1st place from the right: ones | 1 × 1 = 1 |
| 0010 | 2nd place from the right: twos | 1 × 2 = 2 |
| 0100 | 3rd place from the right: fours | 1 × 4 = 4 |
| 1000 | 4th place from the right: eights | 1 × 8 = 8 |
In all four rows, the written digit is still 1. Its position decides whether that 1 contributes one, two, four or eight.
\r\nNow put 1 0 1 1 in the four boxes. The values 8, 4, 2 and 1 are labels for the boxes. The written digits tell us which values to include:
\r\nPlace value: what ONE in this box is worth
Digits written in those boxes
Each box contributes
The 0 above does not mean the place value changes from 4 to 0. The place is still worth 4 per unit; the digit says to take zero units of it. Its contribution is therefore 0 × 4 = 0.
The three 1s are the same digit, but they sit in different places. They contribute 8, 2 and 1. Therefore, binary 1011 represents decimal 11.
Yaad rakho: Box ki value fixed. Andar 1 hai toh value lo; 0 hai toh skip karo. Phir li hui values jodo.
Decode means convert the binary digits into an ordinary decimal number. In exam notation, a small 2 below a number means binary; a small 10 means decimal. These small labels identify the number system. They are not multipliers.
![\"(1011)_2=(11)_{10}\"](\"math/2c7d9feaea569d4c.svg\")
This line says: “one-zero-one-one in binary and eleven in decimal represent the same quantity.”
Convert the binary number ![\"(1011)_2\"](\"math/2f07200d2566e6c6.svg\")
to decimal. [2 marks · Practice]
Binary digits: 1011, in this exact order.
The ordinary decimal number represented by these digits.
Each digit tells us how many of that place value to include. Multiply each digit by its place value, then add.
![\"\\text{Number}=\\text{sum](\"math/8e2952f36a427b0a.svg\")
| Place value | 8 | 4 | 2 | 1 |
|---|---|---|---|---|
| Binary digit | 1 | 0 | 1 | 1 |
The digit is 1, so include the value of this place.
![\"1\\times](\"math/c26b14275b7516e9.svg\")
The digit is 0, so this place contributes zero. Its place value still stays the same.
![\"0\\times](\"math/ff395ff0ae2a5dd3.svg\")
The digit is 1, so include the value of this place.
![\"1\\times](\"math/46c6b9d4e71ea84d.svg\")
The digit is 1, so include the value of this place.
![\"1\\times](\"math/37d06720ad44993d.svg\")
![\"8+0+2+1=11\"](\"math/1006619e9b2d19e6.svg\")
![\"(1011)_2=(11)_{10}\"](\"math/9299eb74dd4bc751.svg\")
.Convert the binary number ![\"(0110)_2\"](\"math/e242d7ba8a184724.svg\")
to decimal. [2 marks · Practice]
Binary digits: 0110, in this exact order.
The ordinary decimal number represented by these digits.
Each digit tells us how many of that place value to include. Multiply each digit by its place value, then add.
| Place value | 8 | 4 | 2 | 1 |
|---|---|---|---|---|
| Binary digit | 0 | 1 | 1 | 0 |
The digit is 0, so this place contributes zero. Its place value still stays the same.
![\"0\\times](\"math/c667fae833dbf441.svg\")
The digit is 1, so include the value of this place.
![\"1\\times](\"math/60e4619429362147.svg\")
The digit is 1, so include the value of this place.
The digit is 0, so this place contributes zero. Its place value still stays the same.
![\"0\\times](\"math/cbd4799920019db6.svg\")
![\"0+4+2+0=6\"](\"math/972060f72bffad8b.svg\")
![\"(0110)_2=(6)_{10}\"](\"math/c6cd8f6b20d7ec32.svg\")
.Convert the binary number ![\"(1001)_2\"](\"math/92700f027de744f9.svg\")
to decimal. [2 marks · Practice]
Binary digits: 1001, in this exact order.
The ordinary decimal number represented by these digits.
Each digit tells us how many of that place value to include. Multiply each digit by its place value, then add.
| Place value | 8 | 4 | 2 | 1 |
|---|---|---|---|---|
| Binary digit | 1 | 0 | 0 | 1 |
The digit is 1, so include the value of this place.
The digit is 0, so this place contributes zero. Its place value still stays the same.
The digit is 0, so this place contributes zero. Its place value still stays the same.
![\"0\\times](\"math/b584619dcb7531a8.svg\")
The digit is 1, so include the value of this place.
![\"8+0+0+1=9\"](\"math/64fd3b60b066ebf6.svg\")
![\"(1001)_2=(9)_{10}\"](\"math/d83c62995f78037e.svg\")
.Convert the binary number ![\"(10011)_2\"](\"math/4d6378ea1db035a2.svg\")
to decimal. [2 marks · Practice]
Binary digits: 10011, in this exact order.
The ordinary decimal number represented by these digits.
Each digit tells us how many of that place value to include. Multiply each digit by its place value, then add.
| Place value | 16 | 8 | 4 | 2 | 1 |
|---|---|---|---|---|---|
| Binary digit | 1 | 0 | 0 | 1 | 1 |
The digit is 1, so include the value of this place.
![\"1\\times](\"math/2d5a7498be08cfdd.svg\")
The digit is 0, so this place contributes zero. Its place value still stays the same.
The digit is 0, so this place contributes zero. Its place value still stays the same.
The digit is 1, so include the value of this place.
The digit is 1, so include the value of this place.
![\"16+0+0+2+1=19\"](\"math/4c61d1192553e0c2.svg\")
![\"(10011)_2=(19)_{10}\"](\"math/9557b36b09cf5595.svg\")
.Encode means write an ordinary number using binary digits. The direction of the task is now reversed: the number is known, and you have to choose the 0s and 1s.
\r\nFor four bits, write the places 8, 4, 2, 1. Start with the largest place on the left. Use it only if it does not exceed the amount still needed. Whenever you use a place, subtract it before moving to the next one.
Yaad rakho: Number banana hai. Badi value se start: jitna bacha hai usmein value fit ho toh 1 aur subtract; fit nahi ho toh 0.
Write the decimal number 6 as an unsigned 4-bit binary number. [2 marks · Practice]
Decimal number: 6. Required length: 4 bits, meaning exactly 4 digit boxes.
The binary digits, including any zero at the beginning needed to keep the required length.
At each place, compare its value with the amount still needed. If that amount is at least the place value, write 1 and subtract that value. Otherwise write 0 and keep the remaining amount unchanged.
This works because each place can be used either once (1) or not at all (0). The selected places must add up to the given number.
| Place value | 8 | 4 | 2 | 1 |
|---|---|---|---|---|
| Digit to choose | □ | □ | □ | □ |
Start with the whole number 6 still to make. Work from the largest place on the LEFT to the smallest on the RIGHT.
We still need 6. Taking 8 would be too much, so skip this place. Write 0 in this box.
We used nothing from this place. The remaining amount stays 6.
We still need 6. This is at least 4, so take one 4. Write 1 in this box.
![\"\\text{Remaining}=6-4=2\"](\"math/2a4c16cf6f52dfc6.svg\")
We still need 2. This is at least 2, so take one 2. Write 1 in this box.
![\"\\text{Remaining}=2-2=0\"](\"math/8d90e4663ec0a413.svg\")
We still need 0. Taking 1 would be too much, so skip this place. Write 0 in this box.
We used nothing from this place. The remaining amount stays 0.
The boxes now contain 0 1 1 0. Join them in the same order: 0110.
The remaining amount is 0, so we have made the required number exactly.
![\"0\\times](\"math/c3e3fe9d6dd4079b.svg\")
![\"(6)_{10}=(0110)_2\"](\"math/17dc4cb6ab18cf42.svg\")
. The answer has 4 bits.Write the decimal number 11 as an unsigned 4-bit binary number. [2 marks · Practice]
Decimal number: 11. Required length: 4 bits, meaning exactly 4 digit boxes.
The binary digits, including any zero at the beginning needed to keep the required length.
At each place, compare its value with the amount still needed. If that amount is at least the place value, write 1 and subtract that value. Otherwise write 0 and keep the remaining amount unchanged.
This works because each place can be used either once (1) or not at all (0). The selected places must add up to the given number.
| Place value | 8 | 4 | 2 | 1 |
|---|---|---|---|---|
| Digit to choose | □ | □ | □ | □ |
Start with the whole number 11 still to make. Work from the largest place on the LEFT to the smallest on the RIGHT.
We still need 11. This is at least 8, so take one 8. Write 1 in this box.
![\"\\text{Remaining}=11-8=3\"](\"math/a475ecc5627f41d9.svg\")
We still need 3. Taking 4 would be too much, so skip this place. Write 0 in this box.
We used nothing from this place. The remaining amount stays 3.
We still need 3. This is at least 2, so take one 2. Write 1 in this box.
![\"\\text{Remaining}=3-2=1\"](\"math/21e8a95558e5db30.svg\")
We still need 1. This is at least 1, so take one 1. Write 1 in this box.
![\"\\text{Remaining}=1-1=0\"](\"math/9f9f05b8fd1b9885.svg\")
The boxes now contain 1 0 1 1. Join them in the same order: 1011.
The remaining amount is 0, so we have made the required number exactly.
![\"1\\times](\"math/f63d65f39c03b2d0.svg\")
![\"(11)_{10}=(1011)_2\"](\"math/1b92e7546fc83d0f.svg\")
. The answer has 4 bits.Write the decimal number 19 as an unsigned 5-bit binary number. [2 marks · Practice]
Decimal number: 19. Required length: 5 bits, meaning exactly 5 digit boxes.
The binary digits, including any zero at the beginning needed to keep the required length.
At each place, compare its value with the amount still needed. If that amount is at least the place value, write 1 and subtract that value. Otherwise write 0 and keep the remaining amount unchanged.
This works because each place can be used either once (1) or not at all (0). The selected places must add up to the given number.
| Place value | 16 | 8 | 4 | 2 | 1 |
|---|---|---|---|---|---|
| Digit to choose | □ | □ | □ | □ | □ |
Start with the whole number 19 still to make. Work from the largest place on the LEFT to the smallest on the RIGHT.
We still need 19. This is at least 16, so take one 16. Write 1 in this box.
![\"\\text{Remaining}=19-16=3\"](\"math/401912fa7840eae2.svg\")
We still need 3. Taking 8 would be too much, so skip this place. Write 0 in this box.
We used nothing from this place. The remaining amount stays 3.
We still need 3. Taking 4 would be too much, so skip this place. Write 0 in this box.
We used nothing from this place. The remaining amount stays 3.
We still need 3. This is at least 2, so take one 2. Write 1 in this box.
We still need 1. This is at least 1, so take one 1. Write 1 in this box.
The boxes now contain 1 0 0 1 1. Join them in the same order: 10011.
The remaining amount is 0, so we have made the required number exactly.
![\"1\\times](\"math/684f11c4b4f0e966.svg\")
![\"(19)_{10}=(10011)_2\"](\"math/a03240c4b6abd4c3.svg\")
. The answer has 5 bits.Write the decimal number 0 as an unsigned 4-bit binary number. [2 marks · Practice]
Decimal number: 0. Required length: 4 bits, meaning exactly 4 digit boxes.
The binary digits, including any zero at the beginning needed to keep the required length.
At each place, compare its value with the amount still needed. If that amount is at least the place value, write 1 and subtract that value. Otherwise write 0 and keep the remaining amount unchanged.
This works because each place can be used either once (1) or not at all (0). The selected places must add up to the given number.
| Place value | 8 | 4 | 2 | 1 |
|---|---|---|---|---|
| Digit to choose | □ | □ | □ | □ |
Start with the whole number 0 still to make. Work from the largest place on the LEFT to the smallest on the RIGHT.
We still need 0. Taking 8 would be too much, so skip this place. Write 0 in this box.
We used nothing from this place. The remaining amount stays 0.
We still need 0. Taking 4 would be too much, so skip this place. Write 0 in this box.
We used nothing from this place. The remaining amount stays 0.
We still need 0. Taking 2 would be too much, so skip this place. Write 0 in this box.
We used nothing from this place. The remaining amount stays 0.
We still need 0. Taking 1 would be too much, so skip this place. Write 0 in this box.
We used nothing from this place. The remaining amount stays 0.
The boxes now contain 0 0 0 0. Join them in the same order: 0000.
The remaining amount is 0, so we have made the required number exactly.
![\"0\\times](\"math/edfc0319e8a1b583.svg\")
![\"(0)_{10}=(0000)_2\"](\"math/a5e664b79f15e056.svg\")
. The answer has 4 bits.Leading zeros are zeros at the beginning, on the left. Binary 110 and 0110 both represent decimal 6:
![\"(110)_2=(1\\times4)+(1\\times2)+(0\\times1)=6\"](\"math/d9a164fc63da4660.svg\")
![\"(0110)_2=(0\\times8)+(1\\times4)+(1\\times2)+(0\\times1)=6\"](\"math/c20ea277129e9ff7.svg\")
The added leftmost place contributes zero. It changes the length from three bits to four bits, but not the value. If a GA question requires four bits, write 0110.
This rule is specifically about zeros on the left. Appending a zero on the right shifts the existing digits into places worth twice as much. For example, binary 1100 is 8 + 4 = 12, not 6.
For these GA conversions, unsigned means the bits represent zero or a positive whole number, with no negative sign. The smallest 4-bit value is 0000. The largest is 1111 because all four places are included.
Find the largest unsigned integer that can be represented using 4 bits. Can decimal 19 fit in 4 bits? [2 marks · Practice]
Four places with values 8, 4, 2 and 1. Each digit is either 0 or 1.
The largest possible value and whether 19 can be represented.
A digit contributes the most when it is 1. Set every digit to 1 and add all the place values.
The largest 4-bit string is 1111.
![\"8+4+2+1=15\"](\"math/f3965404fe997578.svg\")
19 is greater than 15, so four bits are insufficient. Add the 16-place on the left. Then 19 = 16 + 2 + 1, giving 10011 in five bits.
The general rule is ![\"\\text{largest](\"math/b2b8f1010a02780c.svg\")
. Here n means the number of bits. The power ![\"2^n\"](\"math/c0a4368527e8571c.svg\")
means multiplying n copies of 2; for four bits, ![\"2^4=2\\times2\\times2\\times2=16\"](\"math/750f21b62900f171.svg\")
, so the maximum is 16 − 1 = 15. There are 16 different values, from 0 through 15, because zero is also one value.
| Bits | Place values from left to right | Unsigned values |
|---|---|---|
| 3 | 4, 2, 1 | 0 to 7 |
| 4 | 8, 4, 2, 1 | 0 to 15 |
| 5 | 16, 8, 4, 2, 1 | 0 to 31 |
If an exam gives a number outside the capacity of the stated bit length, state that inconsistency. Do not silently cut off a digit: removing the 16-place from 10011 would change 19 into 3. Also retain any separate allowed range stated in the question.
A chromosome is the representation of one candidate solution in a genetic algorithm. With the direct unsigned binary encoding used here, decimal 6 is represented by the four-bit chromosome 0110.
\r\nDuring crossover or mutation, the algorithm changes these bits. Afterwards, decode the resulting chromosome to find its new value of x. Then use that x in the given objective or fitness function. The binary string, its decimal value, and its fitness score are different things.
Yaad rakho: Encode: number se bits. Decode: bits se number. GA mein bits badle toh x dobara decode karo, phir uska score nikalo.
"},{"id":"coordinates","title":"Vectors, coordinates, matrices and transpose from the first entry","html":"A scalar is one number, such as 5. A vector is an ordered list, such as (3,7). Order matters: the first number may be horizontal position and the second vertical position. A matrix is a rectangular table of numbers. These words describe how numbers are arranged, not a new kind of arithmetic.
| Arrangement | Example | How to read it |
|---|---|---|
| Scalar | 5 | One value |
| Row vector | [3,7] | One row, two entries |
| Column vector | ![\"\\begin{bmatrix}3\\\\7\\end{bmatrix}\"](\"math/a263cada62354d7d.svg\") | Two rows, one entry |
| Matrix | ![\"\\begin{bmatrix}3&7\\\\4&2\\end{bmatrix}\"](\"math/a9c176e9eb771437.svg\") | Two rows, two columns |
For a matrix A, A₂₁ means the entry in row 2, column 1. The first lower label is the row; the second is the column. Matrix size is written number of rows × number of columns. A 2×3 matrix has two rows containing three entries each.
![\"Read](\"fig/depth_matrix.a8870f4ab6.svg\")
Let A have first row [2,5,8] and second row [1,4,7]. State its size and find A₂₃ and A₁₂. [2 marks · Worked example]
![\"A=\\begin{bmatrix}2&5&8\\\\1&4&7\\end{bmatrix}\"](\"math/01196c6c529af391.svg\")
Matrix dimensions and two entries.
Read the lower labels in row–column order.
There are 2 horizontal rows and 3 vertical columns. Size=2×3.
Row 2 is [1,4,7]. Its third entry is 7.
Row 1 is [2,5,8]. Its second entry is 5.
A particle has position x=(3,7), velocity v=(−1,2), and best position p=(1,8). Find p−x, half of that difference, and x+v. [2 marks · Worked example]
Position x=(3,7); velocity v=(−1,2); personal-best position p=(1,8). First and second coordinates stay in the same order.
The difference p−x, the scaled difference 0.5(p−x), and the updated position x+v.
Match first with first and second with second. A scalar multiplies every coordinate.
![\"p-x=(1-3,\\;8-7)=(-2,1)\"](\"math/162ee1a9497444ec.svg\")
![\"0.5(p-x)=(0.5(-2),\\;0.5(1))=(-1,0.5)\"](\"math/2b6de1af24a02bfa.svg\")
![\"x+v=(3+(-1),\\;7+2)=(2,9)\"](\"math/1d4e7e4fb543eeb3.svg\")
The superscript T in Aᵀ means transpose. It does not mean multiply or raise A to a power. Row 1 of A becomes column 1 of Aᵀ; row 2 becomes column 2. An m×n arrangement becomes n×m.
Find the transpose of A = [[2,5,8],[1,4,7]]. [2 marks · Worked example]
Aᵀ.
Turn each old row into a new column, keeping its order.
[2,5,8] becomes the first column from top to bottom.
[1,4,7] becomes the second column from top to bottom.
![\"A^T=\\begin{bmatrix}2&1\\\\5&4\\\\8&7\\end{bmatrix}\"](\"math/1317eeff1ada1bd8.svg\")
Calculate the dot product of [2,3] and [4,5]. [2 marks · Worked example]
First vector a=[2,3]; second vector b=[4,5]. Each has two entries.
The dot product a·b, a single number obtained by adding the two matching products.
![\"a\\cdot](\"math/b383a742b04996eb.svg\")
Multiply corresponding entries, then add the products. Markov matrix multiplication later repeats this for each destination column.
![\"2\\times4=8\"](\"math/104d4bd59aaafd44.svg\")
![\"3\\times5=15\"](\"math/9c14aa0e11753341.svg\")
![\"8+15=23\"](\"math/bb009310f34186b1.svg\")
Yaad rakho: Vector = order wali list. Matrix = rows-columns ka table. Entry padhte waqt pehle row, phir column.
"},{"id":"graphs","title":"Graph axes, intervals and the fractions inside membership functions","html":"The horizontal axis runs left to right and shows input x. The vertical axis runs bottom to top and shows the output. A point (3,0.4) means: input 3 gives output 0.4. In a membership graph, the vertical output is a degree between 0 and 1.
To read the graph at x=3, start at 3 on the bottom axis, move vertically until you reach the line, then move horizontally to the vertical axis. The height you read is the membership.
![\"Halfway](\"fig/depth_slope.c91a1030aa.svg\")
| Read this | Meaning | Concrete example |
|---|---|---|
| x < 5 | Strictly less than 5 | 4.9 qualifies; 5 does not |
| x ≤ 5 | Less than or equal to 5 | Both 4.9 and 5 qualify |
| 2 ≤ x ≤ 8 | x is between 2 and 8 including both ends | The interval is [2,8] |
| [2,8) | Include 2; exclude 8 | Useful for roulette interval boundaries |
| Piecewise formula | Choose one calculation according to where x lies | Rising branch, falling branch or outside |
A straight membership line rises from (2,0) to (6,1). Find its height at x=3. [2 marks · Worked example]
Left input=2; peak input=6; required input=3.
Membership height.
![\"\\text{height}=\\frac{\\text{distance](\"math/b04bfa99f868280c.svg\")
A straight line changes by the same proportion of its height as the proportion travelled along its width.
![\"3-2=1\"](\"math/c1d7fd7ffca66c17.svg\")
![\"6-2=4\"](\"math/0e4232a9d761c2cf.svg\")
![\"\\mu(3)=\\frac{1}{4}=0.25\"](\"math/07bbe7a121e1fdbb.svg\")
A straight membership line rises from (2,0) to (6,1). Find its height at x=4. [2 marks · Worked example]
Left input=2; peak input=6; required input=4.
Membership height.
A straight line changes by the same proportion of its height as the proportion travelled along its width.
![\"4-2=2\"](\"math/bae97278ddb3a17b.svg\")
![\"\\mu(4)=\\frac{2}{4}=0.5\"](\"math/d97d0314d106677e.svg\")
A straight membership line rises from (10,0) to (30,1). Find its height at x=25. [2 marks · Worked example]
Left input=10; peak input=30; required input=25.
Membership height.
A straight line changes by the same proportion of its height as the proportion travelled along its width.
![\"25-10=15\"](\"math/14c6ec773492ed50.svg\")
![\"30-10=20\"](\"math/04505a3de0e682e9.svg\")
![\"\\mu(25)=\\frac{15}{20}=0.75\"](\"math/5aad4f1600b806e2.svg\")
A membership line falls from (5,1) to (8,0). Find its height at x=7. [2 marks · Worked example]
Peak=5; right zero=8; x=7.
Membership.
![\"\\mu(x)=\\frac{\\text{distance](\"math/a265ac2e0c52f5aa.svg\")
![\"8-7=1\"](\"math/8f69aa364500c9a0.svg\")
![\"8-5=3\"](\"math/11fb664a99823789.svg\")
![\"\\mu(7)=1/3\\approx0.333333\"](\"math/245668271fba02af.svg\")
Yaad rakho: Chadhti line: left se kitna aage? Utarti line: right tak kitna baaki? Dono mein poori width se divide.
"},{"id":"sums-averages","title":"Sum notation, ordinary averages and weighted averages","html":"The capital Greek letter Σ is read “sum”. The lower label says where to start and the upper label where to stop. The small index i is just a counter: first item, second item, third item.
![\"\\sum_{i=1}^{3}a_i=a_1+a_2+a_3\"](\"math/187790b2a1d823b4.svg\")
For a₁=2, a₂=5 and a₃=8, calculate Σaᵢ and their mean. [2 marks · Worked example]
Three numbers: 2,5,8.
Sum and ordinary average.
![\"\\bar](\"math/5c0dc23066626f45.svg\")
The bar means average. Each number has equal weight.
![\"\\sum_{i=1}^3](\"math/73da4a922a419fa3.svg\")
![\"2+5=7\\quad7+8=15\"](\"math/a4c248316325eed6.svg\")
![\"\\bar](\"math/30b09f1f2d830b9a.svg\")
If values 20 and 80 have weights 3 and 1, this is equivalent to averaging [20,20,20,80]. Therefore you divide by the sum of weights, not by the number of distinct values. Weights may also be decimals. Their total need not initially be 1.
![\"\\bar](\"math/b4ce56a94f70502d.svg\")
Find the weighted average of [20, 80] with corresponding weights [3, 1]. [2 marks · Worked example]
| Value | Its weight |
|---|---|
| 20 | 3 |
| 80 | 1 |
The weighted average of the listed values.
![\"\\bar](\"math/91cd48ba0a9e464c.svg\")
zᵢ is a value and wᵢ is its corresponding weight. Multiply each value by its own weight, add, then divide by total weight. Here all weights are nonnegative and their total is positive.
![\"20\\times3=60\"](\"math/6b7da7f3a0b59a78.svg\")
![\"80\\times1=80\"](\"math/5b5619f67f240a03.svg\")
![\"60+80=140\"](\"math/4012d4bc25614db2.svg\")
![\"3+1=4\"](\"math/c37463b701447787.svg\")
![\"\\bar](\"math/dbdc025973fedfed.svg\")
Find the weighted average of [20, 50, 80] with corresponding weights [0.2, 0.6, 0.2]. [2 marks · Worked example]
| Value | Its weight |
|---|---|
| 20 | 0.2 |
| 50 | 0.6 |
| 80 | 0.2 |
The weighted average of the listed values.
zᵢ is a value and wᵢ is its corresponding weight. Multiply each value by its own weight, add, then divide by total weight. Here all weights are nonnegative and their total is positive.
![\"20\\times0.2=4\"](\"math/396c5c156cb14965.svg\")
![\"50\\times0.6=30\"](\"math/f5c2b1ec334e1160.svg\")
![\"80\\times0.2=16\"](\"math/9cd289b939140103.svg\")
![\"4+30+16=50\"](\"math/cd5630515c6da6da.svg\")
![\"0.2+0.6+0.2=1\"](\"math/62f59e75ba76247e.svg\")
![\"\\bar](\"math/21b878898a4a0815.svg\")
Find the weighted average of [65, 85] with corresponding weights [0.3, 0.5]. [2 marks · Worked example]
| Value | Its weight |
|---|---|
| 65 | 0.3 |
| 85 | 0.5 |
The weighted average of the listed values.
zᵢ is a value and wᵢ is its corresponding weight. Multiply each value by its own weight, add, then divide by total weight. Here all weights are nonnegative and their total is positive.
![\"65\\times0.3=19.5\"](\"math/3285138cffe8477f.svg\")
![\"85\\times0.5=42.5\"](\"math/bf9c01dfa8fdf718.svg\")
![\"19.5+42.5=62\"](\"math/6cfaa5029f5a6cb9.svg\")
![\"0.3+0.5=0.8\"](\"math/8108e5ca96fd589b.svg\")
![\"\\bar](\"math/a313fb00fc0904f5.svg\")
![\"\\prod_{i=1}^{3}a_i=a_1a_2a_3\"](\"math/40f044c5860eb614.svg\")
For [0.5,0.2,0.4], first 0.5×0.2=0.1, then 0.1×0.4=0.04. Σ would instead give 1.1. Naive Bayes multiplies likelihood factors; it does not add them.
Yaad rakho: Sigma = sab jodo. Product Pi = sab multiply. Weighted mean = value×weight ke total ko weights ke total se divide.
"},{"id":"powers-exponentials","title":"Powers, roots, absolute values and exp on the calculator","html":"A positive integer power repeats multiplication. 2³=2×2×2=8; it does not mean 2×3. A nonzero number raised to power 0 equals 1. A negative power means a reciprocal: 2⁻²=1/(2²)=1/4. A square root reverses a square for a nonnegative result: √9=3.
| Read this | Meaning | Concrete example |
|---|---|---|
| 2⁰, 2¹, 2², 2³ | Binary place values | 1,2,4,8 |
| |x| | Absolute value: distance from zero | |−3|=3 and |3|=3 |
| e ≈ 2.71828 | A fixed mathematical constant | e⁰=1 |
| exp(u) | Another way of writing eᵘ | exp(−2)=e⁻²≈0.135335 |
| ln(x) | The natural logarithm; the exponent giving x from e | ln(e²)=2; ln(1)=0 |
Calculate exp(−x²/2) at x=-2. [2 marks · Worked example]
x=-2; exp means e raised to the bracketed exponent.
The exponential value.
Calculate the complete exponent first. Then use the calculator’s e-to-the-power function.
![\"(-2)^2=4\"](\"math/4193a49c3f887499.svg\")
![\"u=-\\frac{4}{2}=-2\"](\"math/3d2cdafe17449912.svg\")
![\"e^u=e^{-2}\\approx](\"math/93bf357c53e138fe.svg\")
Calculate exp(−x²/2) at x=0. [2 marks · Worked example]
x=0; exp means e raised to the bracketed exponent.
The exponential value.
Calculate the complete exponent first. Then use the calculator’s e-to-the-power function.
![\"(0)^2=0\"](\"math/cd561d277fcba773.svg\")
![\"u=-\\frac{0}{2}=0\"](\"math/4e1e416350f910b4.svg\")
![\"e^u=e^{0}\\approx](\"math/4263edec66681677.svg\")
Calculate exp(−x²/2) at x=2. [2 marks · Worked example]
x=2; exp means e raised to the bracketed exponent.
The exponential value.
Calculate the complete exponent first. Then use the calculator’s e-to-the-power function.
![\"(2)^2=4\"](\"math/730976d327ad7f5e.svg\")
On an fx-991ES-style calculator, eˣ is normally the shifted function above ln. Enter the complete exponent in brackets. Use the negative-number key for −2 where required. Confirm against e⁰=1 before a longer calculation. The exact key labels depend on the model.
Yaad rakho: exp ka matlab e ki power. Pehle exponent ka poora hisaab, phir eˣ.
"}]},{"id":"fuzzy-core","title":"2. Fuzzy logic, from meaning to full solutions","unit":"Unit 2","scope":"Definitions → membership graphs → operations → rules → defuzzification → complete controller.","sections":[{"id":"membership-meaning","title":"Fuzzy sets: what membership actually means","html":"\r\nSuppose the input is temperature 32°C and the fuzzy label is Hot. A membership function is the rule that converts that temperature into a degree of matching the label. If μHot(32)=0.7, read it aloud as “the membership of 32 degrees in Hot is 0.7”. The lower word Hot names the set; 32 is the input; 0.7 is the output.
The same temperature can have membership 0.3 in Medium and 0.7 in Hot. These are two descriptions of one input. Fuzzy memberships do not generally have to sum to 1. By contrast, probabilities of mutually exclusive, exhaustive outcomes do.
| Read this | Meaning | Concrete example |
|---|---|---|
| Universe U | All input values being considered | Temperatures from 0 to 40°C |
| Crisp value | One definite measured number | 32°C |
| Linguistic term | A word naming a fuzzy category | Low, Medium, High |
| μA(x) | Degree to which input x fits category A | μHot(32)=0.7 |
Yaad rakho: 32 temperature hai; 0.7 us temperature ka Hot label mein degree hai. Dono ko mix mat karo.
Suppose a controller says a room is hot only when its temperature is at least 30°C. At 29.9°C it says “not hot”; at 30°C it says “hot”. That sharp boundary is a crisp rule. Fuzzy logic allows a gradual degree of belonging to the category “Hot”.
\r\nDefinition. A fuzzy set A on a universe U assigns every element x a membership degree μA(x) between 0 and 1. The symbol μ is the Greek letter “mu”; read μA(x) as “the degree to which x belongs to set A”.
![\"\\mu_A:U\\rightarrow[0,1]\"](\"math/d2598c83177cc2bf.svg\")
| Term | Meaning | Temperature example |
|---|---|---|
| Universe U | All allowed input values | Temperatures from 0°C to 40°C |
| Element x | The particular input being considered | 32°C |
| Fuzzy set A | A category described by a membership function | Hot |
| Membership μA(x) | How strongly x fits that category | μHot(32) = 0.7 |
Membership 0 means no belonging; 1 means full belonging; 0.7 means partial belonging under the chosen definition. It does not mean a 70% chance that the measured temperature is 32°C. Nor is it a 70% chance that the sensor is correct. Fuzziness concerns the gradual meaning of “Hot”; probability concerns uncertainty about which event occurs.
| Input | Warm membership | Hot membership | Interpretation |
|---|---|---|---|
| 28 | 0.8 | 0.1 | Fits Warm strongly and Hot weakly |
| 32 | 0.3 | 0.7 | Fits Hot more than Warm |
| 35 | 0 | 1 | Fully Hot in this chosen model |
These memberships need not add to 1. Overlapping categories are allowed. A crisp set is the special case in which membership can only be 0 or 1. A crisp set can also contain infinitely many elements; “crisp” does not mean “finite”.
On U = {20, 25, 30, 35}, the Hot memberships are [0, 0.2, 0.6, 1]. Write the fuzzy set Hot and identify the fully and partly belonging inputs. [2 marks · Practice]
Ordered inputs [20, 25, 30, 35]; memberships [0, 0.2, 0.6, 1].
The set notation and interpretation.
In the common fuzzy-set notation μ/x, the slash separates membership from element; it is not division.
| x | μHot(x) |
|---|---|
| 20 | 0 |
| 25 | 0.2 |
| 30 | 0.6 |
| 35 | 1 |
![\"Hot=\\{0/20,\\;0.2/25,\\;0.6/30,\\;1/35\\}\"](\"math/5a7f23351df5307b.svg\")
35 fully belongs; 25 and 30 partly belong; 20 does not belong.
Distinguish crisp and fuzzy sets. [2 marks · Practice]
| Crisp set | Fuzzy set |
|---|---|
| Membership is 0 or 1 | Membership is any value in [0,1] |
| A sharp boundary separates members | A membership function permits gradual boundaries |
| Example: temperature ≥ 30°C | Example: degree of being Hot increases gradually |
Yaad rakho: μ ka matlab: is category mein kitna fit hota hai. Chance nahi.
"},{"id":"triangle","title":"Triangular membership: choose the side, then calculate","html":"\r\nTri(2,5,8) means membership starts at zero at x=2, rises to 1 at x=5, then falls to zero at x=8. The first three numbers are input positions, not three membership degrees.
| Where is x? | Meaning | Calculation |
|---|---|---|
| x≤2 or x≥8 | Outside, or at a zero-height endpoint | 0 |
| 2<x<5 | Part-way up the rising edge | (x−2)/(5−2) |
| x=5 | At the peak | 1 |
| 5<x<8 | Part-way down the falling edge | (8−x)/(8−5) |
For x=3: distance travelled is 3−2=1; total rising width is 5−2=3; the height is 1÷3. For x=7: distance still remaining is 8−7=1; total falling width is 8−5=3; again the height is 1÷3. Use the branch containing the input before doing any division.
Yaad rakho: a = start, b = peak, c = end. x kahan pada hai, wahi fraction decide karega.
The horizontal axis gives the input x. The vertical axis gives membership μ. To read μ at an input, go vertically up from that x to the line, then horizontally to the membership axis.
A triangle has three parameters: a is the left zero, b is the peak with membership 1, and c is the right zero. We assume a < b < c.
![\"Rising](\"fig/v2_triangle.4eb0b27c9d.svg\")
![\"\\mu_A(x)=\\begin{cases}0,&x\\le](\"math/4010d79d1ea394d4.svg\")
On the rising side, x − a is how far you have climbed across the horizontal base; b − a is the full rising width. Their ratio goes from 0 to 1. On the falling side, c − x is the remaining horizontal distance to the right zero; c − b is the full falling width. That ratio goes from 1 to 0.
A fuzzy set A has triangular membership Tri(2, 5, 8). Find membership at x = 3. [2 marks · Practice]
Left endpoint a = 2; peak b = 5; right endpoint c = 8; input x = 3.
The membership degree μA(x), a number from 0 to 1.
Locate x relative to a, b, c before choosing the branch.
2 < 3 ≤ 5, so membership increases as x moves toward the peak.
Rising side
![\"\\mu_A(3)=\\frac{3-2}{5-2}=\\frac{1}{3}=0.333333333\"](\"math/feefc8b7a33635af.svg\")
The degree of belonging to A is 0.333333333. It is a membership degree, not a probability.
A fuzzy set A has triangular membership Tri(2, 5, 8). Find membership at x = 7. [2 marks · Practice]
Left endpoint a = 2; peak b = 5; right endpoint c = 8; input x = 7.
The membership degree μA(x), a number from 0 to 1.
Locate x relative to a, b, c before choosing the branch.
5 < 7 < 8, so membership decreases as x moves away from the peak.
Falling side
![\"\\mu_A(7)=\\frac{8-7}{8-5}=\\frac{1}{3}=0.333333333\"](\"math/379229dc05ca4e37.svg\")
The degree of belonging to A is 0.333333333. It is a membership degree, not a probability.
A fuzzy set A has triangular membership Tri(2, 5, 8). Find membership at x = 5. [2 marks · Practice]
Left endpoint a = 2; peak b = 5; right endpoint c = 8; input x = 5.
The membership degree μA(x), a number from 0 to 1.
Locate x relative to a, b, c before choosing the branch.
2 < 5 ≤ 5, so membership increases as x moves toward the peak.
Rising side
![\"\\mu_A(5)=\\frac{5-2}{5-2}=\\frac{3}{3}=1\"](\"math/650bc1ad2a8788e3.svg\")
The degree of belonging to A is 1. This is full membership at the peak.
A fuzzy set A has triangular membership Tri(2, 5, 8). Find membership at x = 9. [2 marks · Practice]
Left endpoint a = 2; peak b = 5; right endpoint c = 8; input x = 9.
The membership degree μA(x), a number from 0 to 1.
Locate x relative to a, b, c before choosing the branch.
9 is outside the open support (2, 8) or at an endpoint.
Outside the triangle
![\"\\mu_A(9)=0\"](\"math/857573d5b79c2b73.svg\")
The degree of belonging to A is 0. It is a membership degree, not a probability.
Yaad rakho: Peak se left: x − a. Peak se right: c − x. Neeche usi side ki poori width.
"},{"id":"trapezoid-intuition","title":"Trapezoids, shoulders and intuition-based fuzzification","html":"\r\nTrap(a,b,c,d) has four input positions: start rising at a, reach height 1 at b, stay at height 1 until c, and fall to zero at d. For Trap(0,2,6,8), any input between 2 and 6 has full membership. The rising width is 2−0=2; the falling width is 8−6=2.
| Input | Locate it | Numerator | Denominator | Membership |
|---|---|---|---|---|
| 1 | Rising: 0<1<2 | 1−0=1 | 2−0=2 | 1/2=0.5 |
| 4 | Flat top: 2≤4≤6 | No division | No division | 1 |
| 7 | Falling: 6<7<8 | 8−7=1 | 8−6=2 | 1/2=0.5 |
| 9 | Beyond right endpoint | No division | No division | 0 |
A left shoulder can stay at height 1 for all sufficiently small inputs, then fall. A right shoulder does the reverse. If a diagram uses a vertical edge, read its stated endpoint convention; do not divide by a zero-width edge.
A trapezoid has four parameters a < b ≤ c < d: rise from a to b, stay at 1 between b and c, then fall from c to d. The flat top means a whole interval fully belongs.
![\"From](\"fig/v2_trapezoid.dbfff598ea.svg\")
![\"\\mu(x)=\\begin{cases}0,&x\\le](\"math/8db3f8710aa967fd.svg\")
For Trap(10,20,30,40), calculate μ(15). [2 marks · Practice]
a=10, b=20, c=30, d=40; x=15.
Membership at the input.
Choose the piece of the trapezoid containing x. A plateau needs no division.
x = 15 lies on the rising side.
![\"\\mu(15)=(15-10)/(20-10)=5/10=0.5\"](\"math/dcb555d1584d4cd8.svg\")
For Trap(10,20,30,40), calculate μ(25). [2 marks · Practice]
a=10, b=20, c=30, d=40; x=25.
Membership at the input.
Choose the piece of the trapezoid containing x. A plateau needs no division.
x = 25 lies on the flat top.
![\"\\mu(25)=1\"](\"math/3a0f8e229e9dd73e.svg\")
For Trap(10,20,30,40), calculate μ(35). [2 marks · Practice]
a=10, b=20, c=30, d=40; x=35.
Membership at the input.
Choose the piece of the trapezoid containing x. A plateau needs no division.
x = 35 lies on the falling side.
![\"\\mu(35)=(40-35)/(40-30)=5/10=0.5\"](\"math/b4c061d09e040d15.svg\")
For Trap(10,20,30,40), calculate μ(45). [2 marks · Practice]
a=10, b=20, c=30, d=40; x=45.
Membership at the input.
Choose the piece of the trapezoid containing x. A plateau needs no division.
x = 45 lies on the outside support.
![\"\\mu(45)=0\"](\"math/e052435017d8c2af.svg\")
A left shoulder remains fully true at the low end; a right shoulder remains fully true at the high end. A Gaussian has a smooth rounded peak. In a numerical, use the supplied graph or expression: do not invent breakpoints from a label such as “High”.
\r\nFuzzification converts a measured crisp input into its membership degrees. In the intuition method, a domain expert defines useful linguistic categories and membership functions from experience. For example, the designer chooses the Warm triangle (15,25,35); after that choice, μWarm(32) = (35−32)/(35−25) = 0.3 is a calculation, not another guess.
\r\nYaad rakho: Shape banana design choice; shape milne ke baad membership calculate karni hai.
"},{"id":"set-operations","title":"Union, intersection, complement and difference","html":"\r\nIf A assigns element u₁ a degree 0.7 and B assigns the same element degree 0.4, every operation compares these two degrees. Do not compare u₁ from A with u₂ from B. ∪ means union (“A OR B”); ∩ means intersection (“A AND B”); superscript c means complement (“NOT”).
At one element u, μA(u)=0.7 and μB(u)=0.4. Find union, intersection, complement of B and A\\B. [2 marks · Worked example]
At the same element u, A assigns membership 0.7 and B assigns membership 0.4.
The membership at u in A∪B, A∩B, Bᶜ and A minus B.
Use max for OR, min for AND, and 1 minus membership for NOT. Difference is A AND NOT B.
![\"\\max(0.7,0.4)=0.7\"](\"math/709d5f49fad6a33b.svg\")
![\"\\min(0.7,0.4)=0.4\"](\"math/f5bde40636dee1a9.svg\")
![\"1-0.4=0.6\"](\"math/553cb087421dc8d7.svg\")
![\"\\min(0.7,0.6)=0.6\"](\"math/d73caed266f7eed3.svg\")
Use the standard max–min convention unless the question specifies another t-norm or t-conorm. “T-norm” simply names a fuzzy AND operator; here it is min. “T-conorm” names a fuzzy OR operator; here it is max.
| Operation | Meaning | Rule |
|---|---|---|
| Union A ∪ B | A OR B | Take the larger membership |
| Intersection A ∩ B | A AND B | Take the smaller membership |
| Complement Aᶜ | NOT A | 1 − membership of A |
| Difference A ∖ B | A AND NOT B | min(μA, 1 − μB) |
On U = {u1, u2, u3}, A has memberships [0.2, 0.8, 0.6] and B has memberships [0.7, 0.3, 0.5]. Find A ∪ B, A ∩ B, Bᶜ and A ∖ B using standard max–min fuzzy operations. [3 marks · Practice]
| Element | u1 | u2 | u3 |
|---|---|---|---|
| A | 0.2 | 0.8 | 0.6 |
| B | 0.7 | 0.3 | 0.5 |
Four membership lists in exactly the same element order.
Union is OR → max. Intersection is AND → min. Complement is NOT → 1 minus membership. Difference A ∖ B means A AND NOT B, so first complement B, then intersect with A.
![\"\\mu_{A\\cup](\"math/d5a9710049ed9d75.svg\")
![\"\\mu_{B^c}=1-\\mu_B\\qquad\\mu_{A\\setminus](\"math/a10c281d39aeb04c.svg\")
All operations compare the two memberships of the same u. Do not compare A at u₁ with B at u₂.
![\"\\mu_{B^c}(u_1)=1-0.7=0.3\"](\"math/ad1dcd17cbaf20b9.svg\")
![\"\\mu_{A\\setminus](\"math/3b878090c37dc92f.svg\")
| Element | A | B | Union | Intersection | Bᶜ | A ∖ B |
|---|---|---|---|---|---|---|
| u1 | 0.2 | 0.7 | max(0.2, 0.7) = 0.7 | min(0.2, 0.7) = 0.2 | 1 − 0.7 = 0.3 | min(0.2, 0.3) = 0.2 |
| u2 | 0.8 | 0.3 | max(0.8, 0.3) = 0.8 | min(0.8, 0.3) = 0.3 | 1 − 0.3 = 0.7 | min(0.8, 0.7) = 0.7 |
| u3 | 0.6 | 0.5 | max(0.6, 0.5) = 0.6 | min(0.6, 0.5) = 0.5 | 1 − 0.5 = 0.5 | min(0.6, 0.5) = 0.5 |
On U = {u1, u2, u3}, A has memberships [0, 1, 0.4] and B has memberships [1, 0, 0.4]. Find A ∪ B, A ∩ B, Bᶜ and A ∖ B using standard max–min fuzzy operations. [3 marks · Practice]
| Element | u1 | u2 | u3 |
|---|---|---|---|
| A | 0 | 1 | 0.4 |
| B | 1 | 0 | 0.4 |
Four membership lists in exactly the same element order.
Union is OR → max. Intersection is AND → min. Complement is NOT → 1 minus membership. Difference A ∖ B means A AND NOT B, so first complement B, then intersect with A.
All operations compare the two memberships of the same u. Do not compare A at u₁ with B at u₂.
![\"\\mu_{B^c}(u_1)=1-1=0\"](\"math/2890d976bd930ef0.svg\")
![\"\\mu_{A\\setminus](\"math/478aba1f930ec419.svg\")
| Element | A | B | Union | Intersection | Bᶜ | A ∖ B |
|---|---|---|---|---|---|---|
| u1 | 0 | 1 | max(0, 1) = 1 | min(0, 1) = 0 | 1 − 1 = 0 | min(0, 0) = 0 |
| u2 | 1 | 0 | max(1, 0) = 1 | min(1, 0) = 0 | 1 − 0 = 1 | min(1, 1) = 1 |
| u3 | 0.4 | 0.4 | max(0.4, 0.4) = 0.4 | min(0.4, 0.4) = 0.4 | 1 − 0.4 = 0.6 | min(0.4, 0.6) = 0.4 |
On U = {u1, u2, u3, u4}, A has memberships [0.9, 0.2, 0.5, 0.7] and B has memberships [0.6, 0.8, 0.1, 0.7]. Find A ∪ B, A ∩ B, Bᶜ and A ∖ B using standard max–min fuzzy operations. [3 marks · Practice]
| Element | u1 | u2 | u3 | u4 |
|---|---|---|---|---|
| A | 0.9 | 0.2 | 0.5 | 0.7 |
| B | 0.6 | 0.8 | 0.1 | 0.7 |
Four membership lists in exactly the same element order.
Union is OR → max. Intersection is AND → min. Complement is NOT → 1 minus membership. Difference A ∖ B means A AND NOT B, so first complement B, then intersect with A.
All operations compare the two memberships of the same u. Do not compare A at u₁ with B at u₂.
![\"\\mu_{B^c}(u_1)=1-0.6=0.4\"](\"math/c9c6864aea087c58.svg\")
![\"\\mu_{A\\setminus](\"math/b891d5b3e6b00766.svg\")
| Element | A | B | Union | Intersection | Bᶜ | A ∖ B |
|---|---|---|---|---|---|---|
| u1 | 0.9 | 0.6 | max(0.9, 0.6) = 0.9 | min(0.9, 0.6) = 0.6 | 1 − 0.6 = 0.4 | min(0.9, 0.4) = 0.4 |
| u2 | 0.2 | 0.8 | max(0.2, 0.8) = 0.8 | min(0.2, 0.8) = 0.2 | 1 − 0.8 = 0.2 | min(0.2, 0.2) = 0.2 |
| u3 | 0.5 | 0.1 | max(0.5, 0.1) = 0.5 | min(0.5, 0.1) = 0.1 | 1 − 0.1 = 0.9 | min(0.5, 0.9) = 0.5 |
| u4 | 0.7 | 0.7 | max(0.7, 0.7) = 0.7 | min(0.7, 0.7) = 0.7 | 1 − 0.7 = 0.3 | min(0.7, 0.3) = 0.3 |
Why not subtract A − B? With A=0.2 and B=0.7, raw subtraction gives −0.5, which is not a membership degree. Under this convention, A ∖ B = min(0.2, 0.3) = 0.2. Some alternative fuzzy algebras define other difference operators; always identify the specified convention.
Yaad rakho: Union bada; intersection chhota; complement 1 se ghata; difference = pehle B ka complement, phir min.
"},{"id":"properties","title":"Fuzzy properties, De Morgan laws and α-cuts","html":"\r\n| Read this | Meaning | Concrete example |
|---|---|---|
| Commutative | Changing order changes nothing | max(0.3,0.8)=max(0.8,0.3)=0.8 |
| Associative | Regroup the same operation | max(max(0.2,0.7),0.4)=max(0.2,max(0.7,0.4))=0.7 |
| Idempotent | Combining a membership with itself keeps it | min(0.6,0.6)=0.6 |
| Involution | Complement twice returns the original | 1−(1−0.4)=1−0.6=0.4 |
| De Morgan | NOT swaps AND and OR while complementing each operand | 1−max(a,b)=min(1−a,1−b) |
| Property | Statement for max–min operations |
|---|---|
| Commutative | A ∪ B = B ∪ A; A ∩ B = B ∩ A |
| Associative | (A ∪ B) ∪ C = A ∪ (B ∪ C); likewise for intersection |
| Idempotent | A ∪ A = A; A ∩ A = A |
| Distributive | A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C); and its dual |
| Involution | (Aᶜ)ᶜ = A |
| De Morgan | (A ∪ B)ᶜ = Aᶜ ∩ Bᶜ; (A ∩ B)ᶜ = Aᶜ ∪ Bᶜ |
For fuzzy sets, A ∪ Aᶜ need not equal the entire universe at membership 1; A ∩ Aᶜ need not be the empty set at membership 0.
At an element x, μA(x)=0.4. Calculate μ(A ∪ Aᶜ) and μ(A ∩ Aᶜ). [2 marks · Practice]
μA(x)=0.4; standard complement/max/min.
Whether excluded middle and non-contradiction hold at this element.
Complement first, then use max or min.
![\"\\mu_{A^c}(x)=1-0.4=0.6\"](\"math/54b240c60f89d754.svg\")
![\"\\max(0.4,0.6)=0.6\\ne1\"](\"math/fbe8846f3efa93f5.svg\")
![\"\\min(0.4,0.6)=0.4\\ne0\"](\"math/4fcb041eeb76aeb4.svg\")
Verify the first De Morgan law for μA=0.3 and μB=0.8. [2 marks · Practice]
A=0.3, B=0.8 at the same element.
Both sides of (A ∪ B)ᶜ = Aᶜ ∩ Bᶜ.
Evaluate the operations independently on each side.
![\"1-\\max(0.3,0.8)=1-0.8=0.2\"](\"math/8bcb03e96487eae0.svg\")
![\"\\min(1-0.3,1-0.8)=\\min(0.7,0.2)=0.2\"](\"math/902b79a384f1ef2a.svg\")
For U={a,b,c,d} and memberships A=[0,0.3,1,0.7], find support, core, boundary, height and the 0.7-cut. [2 marks · Worked example]
Universe U={a,b,c,d}. Memberships: A(a)=0, A(b)=0.3, A(c)=1, A(d)=0.7. Cut level α=0.7.
The support, core, boundary and 0.7-cut as sets of elements, and the height as one membership value.
Support: degree>0; core: degree=1; boundary: 0<degree<1; height: largest degree; α-cut: degree≥α.
| Element | Degree | Positive? | Exactly 1? | At least 0.7? |
|---|---|---|---|---|
| a | 0 | No | No | No |
| b | 0.3 | Yes | No | No |
| c | 1 | Yes | Yes | Yes |
| d | 0.7 | Yes | No | Yes |
Support={b,c,d}. Core={c}. Boundary={b,d}. The 0.7-cut={c,d}.
The maximum degree is 1, so the height is 1 and the set is normal.
Find the 0.5-cut of the triangular fuzzy set Tri(2,5,8). [2 marks · Worked example]
Triangle start a=2, peak b=5, end c=8; cut level α=0.5. Retain all inputs whose membership is at least 0.5.
Left and right endpoints of the retained interval.
The triangle rises past 0.5, remains above it, then falls through 0.5. Solve height=0.5 on both slopes.
![\"(x-2)/3=0.5\"](\"math/60bd77c1b7954114.svg\")
![\"x-2=0.5\\times3=1.5\"](\"math/58646302eb803fe5.svg\")
![\"x=1.5+2=3.5\"](\"math/5def47a62d44d4a8.svg\")
![\"(8-x)/3=0.5\"](\"math/101167a6754cbdc0.svg\")
![\"8-x=1.5\"](\"math/35ee639a9616fa80.svg\")
![\"x=8-1.5=6.5\"](\"math/4ae644eed157e001.svg\")
The graph is at or above 0.5 for 3.5≤x≤6.5. Include the endpoints because the condition is ≥.
Support: inputs with μ > 0. Core: inputs with μ = 1. Boundary: inputs with 0 < μ < 1. Height: largest membership (more generally the supremum). A set is normal when its height is 1; it is subnormal when its height is below 1. A crossover point has membership 0.5. An α-cut contains inputs with μ ≥ α.
For Tri(2,5,8), support is (2,8), core is {5}, height is 1, and the 0.5-cut is [3.5,6.5]. Solve (x−2)/3=0.5 → x=3.5 and (8−x)/3=0.5 → x=6.5.
A one-dimensional fuzzy set is convex if every α-cut is an interval (or empty). The membership between two inputs must not dip below the smaller endpoint membership. A single triangle is convex; two separated peaks with a lower valley may not be.
Yaad rakho: Fuzzy A aur NOT A dono partial ho sakte hain. Isliye crisp waali 1 aur 0 identities blindly nahi lagani.
"},{"id":"mamdani-architecture","title":"Mamdani architecture: what every block does","html":"\r\nA controller turns measurements into an action setting. Consider temperature 32°C and occupancy 7 people. The input numbers first become membership degrees. The rules then combine those degrees. Only the final defuzzification returns one fan-speed number.
| Stage | What goes in | Operation | What comes out |
|---|---|---|---|
| Fuzzifier | 32°C and 7 people | Use the supplied membership graphs | Temperature Medium=.3, High=.7; occupancy High=.5 |
| Knowledge base | Chosen labels and their shapes | Store graphs and IF–THEN rules | A reusable definition of how the controller works |
| Inference | Medium=.3 AND High=.5 | Rule strength=min(.3,.5)=.3 | “Fan High” fires at .3 |
| Aggregation | All clipped output sets | Take the largest output degree at each speed | One combined fuzzy output shape |
| Defuzzifier | Combined shape | Apply the specified MoM, centroid or representative method | One numerical speed |
The antecedent is the IF part (“temperature is Medium AND occupancy is High”). The consequent is the THEN part (“fan speed is High”). A rule strength of 0.3 is a membership height; it does not mean the fan runs at 30%.
A fuzzy inference system (FIS) maps inputs to an output by evaluating fuzzy IF–THEN rules. A Mamdani FIS uses fuzzy sets in the rule consequents, combines their inferred output sets, and defuzzifies the result.
\r\n| Block | What it stores or calculates | Concrete meaning |
|---|---|---|
| Fuzzifier | Membership values of the measured inputs | 32°C may be Medium to degree 0.3 and High to degree 0.7 |
| Knowledge base: database | Ranges and membership functions | Medium = Tri(15,25,35) |
| Knowledge base: rule base | IF–THEN rules | IF temperature High AND occupancy High THEN fan Very High |
| Inference engine | Combines conditions and applies consequents | min for AND; clips the output triangle at the rule strength |
| Aggregator | Combines all rule outputs | At every output location, take the maximum clipped membership |
| Defuzzifier | Converts the aggregate to one crisp value | Use centroid or mean of maxima, as requested |
In “IF temperature is High AND occupancy is High THEN fan speed is Very High”, the IF part is the antecedent (condition), and the THEN part is the consequent (output category). The input variable temperature, the label High, and the numerical membership μHigh(32) are different things.
\r\nA strength such as 0.3 says how strongly a rule fires. It is on the membership scale 0–1. Fan speed might be on a 0–100% scale. Defuzzification is the step that produces a number on that output scale.
Explain the architecture of a Mamdani fuzzy inference system. [3 marks · Practice]
A Mamdani FIS contains a fuzzifier, knowledge base, inference engine and defuzzifier. The database specifies membership functions and the rule base stores IF–THEN rules. Crisp inputs are fuzzified. The inference engine calculates rule firing strengths and constructs output fuzzy sets. These sets are aggregated, then defuzzified into a crisp action. For a fan controller, temperature and occupancy are inputs and fan speed is the output.
Yaad rakho: Input ka degree → rule ka strength → output ka shape → final number.
"},{"id":"rules-clipping","title":"Rule firing, clipping and aggregation: four solved cases","html":"\r\nA rule firing at strength 0.3 cannot give its consequent a membership above 0.3. At every candidate speed z, keep the smaller of 0.3 and the original output-set membership. The speed axis is unchanged.
At speeds 40, 60 and 80, output set A has memberships [0.1,0.8,0.2] and set B has [0,0.4,1]. Rule A fires at 0.3 and rule B at 0.5. Clip and max-aggregate the sets. [2 marks · Worked example]
| Speed | Original A membership | Original B membership |
|---|---|---|
| 40 | 0.1 | 0 |
| 60 | 0.8 | 0.4 |
| 80 | 0.2 | 1 |
Rule A strength=0.3; rule B strength=0.5. Compare memberships at the same speed.
The combined membership at each speed.
![\"\\mu_{out}(z)=\\max(\\min(0.3,\\mu_A(z)),\\min(0.5,\\mu_B(z)))\"](\"math/89f0491f916a2f27.svg\")
Inner min limits a rule’s height; outer max keeps the stronger support at that speed.
At 40: min(.3,.1)=.1. At 60: min(.3,.8)=.3. At 80: min(.3,.2)=.2.
At 40: min(.5,0)=0. At 60: min(.5,.4)=.4. At 80: min(.5,1)=.5.
| Speed | Clipped A | Clipped B | max(A,B) |
|---|---|---|---|
| 40 | 0.1 | 0 | 0.1 |
| 60 | 0.3 | 0.4 | 0.4 |
| 80 | 0.2 | 0.5 | 0.5 |
Yaad rakho: Clip = height ki ceiling. Aggregate = har same x par sabse upar wala height.
AND/OR combine conditions within a rule. Implication uses the resulting strength to modify that rule’s output membership function. Aggregation merges different rule outputs. Here we use min-AND, max-OR, min-implication and max-aggregation.
![\"\\alpha_{AND}=\\min(a,b)\\qquad\\alpha_{OR}=\\max(a,b)\"](\"math/e446839a4e5e2d93.svg\")
![\"\\mu_{\\mathrm{rule}}(z)=\\min(\\alpha,\\mu_{\\mathrm{consequent}}(z))\"](\"math/32c896372ab93b2e.svg\")
![\"\\mu_{\\mathrm{aggregate}}(z)=\\max_r\\mu_{\\mathrm{rule}\\](\"math/dd5c7064495bb995.svg\")
α (“alpha”) is the rule firing strength. z names a possible output value. The subscript r labels a rule, not a multiplication.
A rule has two condition degrees 0.7 and 0.5, joined by AND. Find its firing strength. [2 marks · Practice]
a=0.7; b=0.5; connective=AND.
The strength α.
Use the smaller degree because both conditions must hold.
![\"\\alpha=\\min(0.7,0.5)\"](\"math/c7cb0e48bf98cd0f.svg\")
![\"\\alpha=0.5\"](\"math/45f07f09a68169c8.svg\")
A rule has two condition degrees 0.3 and 0, joined by AND. Find its firing strength. [2 marks · Practice]
a=0.3; b=0; connective=AND.
The strength α.
Use the smaller degree because both conditions must hold.
![\"\\alpha=\\min(0.3,0)\"](\"math/7194f00ae81bc1b1.svg\")
![\"\\alpha=0\"](\"math/7b95185b7bc8e8a2.svg\")
A rule has two condition degrees 0.2 and 0.8, joined by OR. Find its firing strength. [2 marks · Practice]
a=0.2; b=0.8; connective=OR.
The strength α.
Use the larger degree because either condition is sufficient.
![\"\\alpha=\\max(0.2,0.8)\"](\"math/a8a03390bc687846.svg\")
![\"\\alpha=0.8\"](\"math/2a9a9be0e9cd69dc.svg\")
Rule 1 fires at 0.3 and its original output membership at z=60 is 0.8. Rule 2 fires at 0.6 and its original membership at z=60 is 0.2. Find the aggregate membership at z=60 using min-implication and max-aggregation. [3 marks · Practice]
Rule 1: α₁=0.3, μ₁(60)=0.8. Rule 2: α₂=0.6, μ₂(60)=0.2.
Membership at the single output location z=60 after implication and aggregation.
Clip each original degree at its rule strength; then take the larger clipped degree.
![\"\\mu_{R1}(60)=\\min(0.3,0.8)=0.3\"](\"math/e2cf40ea63fed461.svg\")
![\"\\mu_{R2}(60)=\\min(0.6,0.2)=0.2\"](\"math/a21d3bc0188016ea.svg\")
![\"\\mu_{out}(60)=\\max(0.3,0.2)=0.3\"](\"math/84156b76989dd58e.svg\")
![\"Above](\"fig/v2_clip.f101b5cfa9.svg\")
Yaad rakho: AND ke liye min; output clip karne ke liye bhi min; rules ko jodne ke liye max. Har min ka role alag hai.
"},{"id":"mom","title":"Mean of maxima: discrete peaks and clipped plateaus","html":"\r\nMaximum membership is a height. Mean of maxima averages the horizontal locations attaining that height. If heights at speeds [20,50,80] are [.2,.6,.6], the maximum height is .6 and the qualifying speeds are 50 and 80; MoM=(50+80)/2=65. Do not average [.6,.6], which would return a degree rather than a speed. A continuous flat maximum from 50 to 80 also has midpoint 65.
Defuzzification converts the aggregate membership function into one crisp output. For MoM, first find the largest membership height, then average the output locations attaining that height. Do not average the membership values themselves.
![\"z^*_{MoM}=\\frac{z_1+\\cdots+z_m}{m}\"](\"math/bfe66044c82b5d4a.svg\")
For a continuous flat maximum from a to b, the mean is its midpoint:
![\"z^*_{MoM}=\\frac{a+b}{2}\"](\"math/5be124f1bdbedfca.svg\")
For the following discrete aggregate, find the mean-of-maxima output. [2 marks · Practice]
| Output z | Membership μ |
|---|---|
| 10 | 0.2 |
| 20 | 0.8 |
| 30 | 0.8 |
| 40 | 0.3 |
MoM crisp output.
Find all locations at the largest height, then average only those locations.
![\"\\mu_{max}=0.8\"](\"math/55721303722e693a.svg\")
20, 30
![\"z^*=\\frac{20+30}{2}=25\"](\"math/caf06fb51932ddd6.svg\")
For the following discrete aggregate, find the mean-of-maxima output. [2 marks · Practice]
| Output z | Membership μ |
|---|---|
| 1 | 0.1 |
| 2 | 0.4 |
| 3 | 0.9 |
| 4 | 0.2 |
MoM crisp output.
Find all locations at the largest height, then average only those locations.
![\"\\mu_{max}=0.9\"](\"math/c286f3dd87b25655.svg\")
3
![\"z^*=\\frac{3}{1}=3\"](\"math/fc739e8ece1c9e47.svg\")
A single triangular output Tri(50,65,80) is clipped at α=0.3. Find its crisp value by mean of maxima. [3 marks · Practice]
a=50, b=65, c=80; clip height α=0.3.
Both plateau endpoints and their mean.
Set the rising and falling membership fractions equal to 0.3. Every point between those intersections has the maximum height.
![\"\\frac{z_L-50}{65-50}=0.3\"](\"math/145de080f101d4ed.svg\")
![\"z_L-50=0.3\\times15=4.5\"](\"math/39a20fc95a439ada.svg\")
![\"z_L=54.5\"](\"math/5ce2529d6b5b0ef5.svg\")
![\"\\frac{80-z_R}{80-65}=0.3\"](\"math/cbfa4bdb9123ba28.svg\")
![\"80-z_R=4.5\"](\"math/9c5f88f119fb8224.svg\")
![\"z_R=75.5\"](\"math/d02ea96dcfb95060.svg\")
![\"z^*=(54.5+75.5)/2=65\"](\"math/fd9133cfe48a7467.svg\")
For several disjoint continuous maximum plateaus, a continuous MoM uses a length-weighted mean over all maximum intervals; simply averaging two outside endpoints is not generally valid. In tabulated questions, average the listed locations at the maximum.
Yaad rakho: Sabse upar ka level dekho; us level par x-values ka mean lo.
"},{"id":"centroid","title":"Centroid: three table sums and one area calculation","html":"\r\nImagine the area under the output graph as a thin sheet of uniform material. The centroid is its horizontal balance point. More area on one side pulls the answer toward that side. In a discrete problem, each membership is a weight. In a continuous problem, the entire area supplies the weights.
The integral sign ∫ means adding very thin strips across a continuous interval. The denominator ∫μ(z)dz is the total area. The numerator ∫zμ(z)dz is the first moment: position multiplied by area, accumulated across the shape. Their ratio has units of the output, such as fan-speed percent.
A non-overlapping aggregate consists of a rectangle of width 4 and height 0.5 centred at z=2, plus a triangle of base 4 and height 0.5 centred horizontally at z=6. Find the area centroid. [2 marks · Worked example]
Rectangle: width 4, height 0.5, horizontal centre 2. Triangle: base 4, height 0.5, horizontal centroid 6. Their areas do not overlap.
The combined horizontal centroid.
![\"z^*=\\frac{A_1c_1+A_2c_2}{A_1+A_2}\"](\"math/97228eac9007fb53.svg\")
Each piece contributes area times its own centre. Use this only for a valid decomposition of the aggregate.
![\"A_1=4\\times0.5=2\"](\"math/925738e6d7bc9738.svg\")
![\"A_2=\\tfrac12\\times4\\times0.5=1\"](\"math/39c5ce3ae9dbf13b.svg\")
![\"A_1c_1=2\\times2=4\"](\"math/3e993e836c87a393.svg\")
![\"A_2c_2=1\\times6=6\"](\"math/c3ef1e042430e3fc.svg\")
![\"A=2+1=3\\qquad](\"math/0582db5cb8a7455e.svg\")
![\"z^*=10/3\\approx3.333333\"](\"math/772602a59fae475c.svg\")
For overlapping clipped sets, first form their max-aggregate. Adding the areas of the original overlapping sets would count the overlap twice.
The centroid uses the whole aggregate membership distribution. For a discrete output table, multiply each output location by its membership, add the products, then divide by total membership.
![\"z^*=\\frac{\\sum_i](\"math/4ca780545740bc9c.svg\")
The index i simply numbers table entries. Σ (“sigma”) means add all listed terms. The denominator must be nonzero.
Find the discrete centroid of the aggregate membership table. [3 marks · Practice]
| z | μ(z) |
|---|---|
| 10 | 0.2 |
| 20 | 0.6 |
| 30 | 0.4 |
One centroid value on the output scale.
This question supplies memberships at discrete output locations; use the discrete centroid, not a peak-only approximation.
![\"z^*=\\frac{\\sum](\"math/0daa1f2e296038c3.svg\")
| z | μ | z × μ |
|---|---|---|
| 10 | 0.2 | 2 |
| 20 | 0.6 | 12 |
| 30 | 0.4 | 12 |
![\"2+12+12=26\"](\"math/1b63f80f68e4f903.svg\")
![\"0.2+0.6+0.4=1.2\"](\"math/762eaecf8eca8b36.svg\")
![\"z^*=\\frac{26}{1.2}=21.666666667\"](\"math/f7410bf10136fdc7.svg\")
Find the discrete centroid of the aggregate membership table. [3 marks · Practice]
| z | μ(z) |
|---|---|
| 0 | 0.5 |
| 5 | 1 |
| 10 | 0.5 |
One centroid value on the output scale.
This question supplies memberships at discrete output locations; use the discrete centroid, not a peak-only approximation.
| z | μ | z × μ |
|---|---|---|
| 0 | 0.5 | 0 |
| 5 | 1 | 5 |
| 10 | 0.5 | 5 |
![\"0+5+5=10\"](\"math/31c26d8043dbf781.svg\")
![\"0.5+1+0.5=2\"](\"math/723c388b267a6d46.svg\")
![\"z^*=\\frac{10}{2}=5\"](\"math/6b4a65f7e66f3cfb.svg\")
Find the discrete centroid of the aggregate membership table. [3 marks · Practice]
| z | μ(z) |
|---|---|
| 20 | 0 |
| 40 | 0.3 |
| 60 | 0.6 |
| 80 | 0.1 |
One centroid value on the output scale.
This question supplies memberships at discrete output locations; use the discrete centroid, not a peak-only approximation.
| z | μ | z × μ |
|---|---|---|
| 20 | 0 | 0 |
| 40 | 0.3 | 12 |
| 60 | 0.6 | 36 |
| 80 | 0.1 | 8 |
![\"0+12+36+8=56\"](\"math/eacf08e4fd9566bf.svg\")
![\"0+0.3+0.6+0.1=1\"](\"math/0cd832e4cf2e8857.svg\")
![\"z^*=\\frac{56}{1}=56\"](\"math/8184446a5afe0a48.svg\")
![\"z^*=\\frac{\\int](\"math/a3227ed1a9289bb5.svg\")
The denominator is the area under the aggregate. The numerator is its first moment about zero: area weighted by horizontal location. For a shape made of non-overlapping rectangles and triangles, compute each part’s area A and its own centroid c, then use ![\"z^*=(\\sum](\"math/033d1d87c88bf6a8.svg\")
. Overlapping regions of a max-aggregate must not be counted twice.
An aggregate is a rectangle of height 0.5 from z=0 to z=4, followed by a right triangle decreasing from height 0.5 at z=4 to 0 at z=6. Find its area centroid. [3 marks · Practice]
Rectangle [0,4], height 0.5; triangle with vertices (4,0),(4,0.5),(6,0).
Centroid of the combined area.
The parts do not overlap except at their boundary. Weight each part’s horizontal centroid by its area.
![\"A_1=4\\times0.5=2\\qquad](\"math/478220c1039e3c18.svg\")
![\"A_2=\\tfrac12\\times2\\times0.5=0.5\"](\"math/931f8a69a034cfb8.svg\")
![\"c_2=(4+4+6)/3=14/3\"](\"math/9b85cc1cdfb06f4d.svg\")
![\"A_1c_1+A_2c_2=2(2)+0.5(14/3)=19/3\"](\"math/f37e6894a3c89e61.svg\")
![\"z^*=(19/3)/(2+0.5)=38/15\\approx2.533333\"](\"math/c0212421c50f4e75.svg\")
Yaad rakho: Centroid = har location × uska weight, total karke total weight se divide.
"},{"id":"weighted-average","title":"Weighted average and centroid are not automatically the same","html":"\r\nThis is the same weighted-average calculation taught in Essential maths. Here a representative output value c is weighted by a rule strength α. Keep the pair together: strength .3 belonging to speed 65 contributes .3×65. The denominator is the total strength. It is not automatically the number of rules, and the representative is not automatically the area-centre of a clipped set.
Some classroom solutions use one representative (often the peak) per output label and weight it by the firing strength. That is a weighted-average or height-based calculation. It is exact for a specified weighted combination of singleton outputs, but is not generally the centroid of a max-aggregated, clipped Mamdani shape.
Always name the method. A table of memberships at actual output locations supports the discrete centroid. A list of category peaks and strengths supports the specified representative-weighted calculation. Full continuous output shapes support area centroid.
Using the weighted-average method, combine output representatives [25, 75] mph with respective strengths [0.1, 0.7]. [3 marks · Practice]
| Output representative | Strength |
|---|---|
| 25 | 0.1 |
| 75 | 0.7 |
One crisp output by the specified weighted-average method.
Each representative is multiplied by its strength; add these products and divide by the total strength. Here α is a strength and z is a representative value. These are not automatically the full area of a clipped Mamdani set.
![\"z^*=\\frac{\\sum_i\\alpha_i](\"math/b26ca7a8925830d0.svg\")
| Strength × value | Product |
|---|---|
| 0.1 × 25 | 2.5 |
| 0.7 × 75 | 52.5 |
![\"2.5+52.5=55\"](\"math/648cd602080c422e.svg\")
![\"0.1+0.7=0.8\"](\"math/e794808140502a26.svg\")
![\"z^*=\\frac{55}{0.8}=68.75\"](\"math/5b92e77608da1213.svg\")
Using the weighted-average method, combine output representatives [40, 65, 85] % fan speed with respective strengths [0, 0.3, 0.5]. [3 marks · Practice]
| Output representative | Strength |
|---|---|
| 40 | 0 |
| 65 | 0.3 |
| 85 | 0.5 |
One crisp output by the specified weighted-average method.
Each representative is multiplied by its strength; add these products and divide by the total strength. Here α is a strength and z is a representative value. These are not automatically the full area of a clipped Mamdani set.
| Strength × value | Product |
|---|---|
| 0 × 40 | 0 |
| 0.3 × 65 | 19.5 |
| 0.5 × 85 | 42.5 |
![\"0+19.5+42.5=62\"](\"math/0bdf70679aafe819.svg\")
![\"0+0.3+0.5=0.8\"](\"math/76b7c63a5ed0de17.svg\")
![\"z^*=\\frac{62}{0.8}=77.5\"](\"math/31b2865a5ddd122a.svg\")
Using the weighted-average method, combine output representatives [25, 40, 60] minutes with respective strengths [0.6, 0.4, 0.2]. [3 marks · Practice]
| Output representative | Strength |
|---|---|
| 25 | 0.6 |
| 40 | 0.4 |
| 60 | 0.2 |
One crisp output by the specified weighted-average method.
Each representative is multiplied by its strength; add these products and divide by the total strength. Here α is a strength and z is a representative value. These are not automatically the full area of a clipped Mamdani set.
| Strength × value | Product |
|---|---|
| 0.6 × 25 | 15 |
| 0.4 × 40 | 16 |
| 0.2 × 60 | 12 |
![\"15+16+12=43\"](\"math/368823bf9efe0162.svg\")
![\"0.6+0.4+0.2=1.2\"](\"math/9f1b10f7784e3b8c.svg\")
![\"z^*=\\frac{43}{1.2}=35.833333333\"](\"math/603521823aee6db6.svg\")
| Method | What it uses | Limitation |
|---|---|---|
| MoM | Locations at maximum membership | Ignores lower portions of the shape; can change abruptly |
| Area centroid | Entire aggregated output area | Needs an area/moment calculation or numerical integration |
| Representative weighted average | Supplied representative values and strengths | Not interchangeable with a clipped Mamdani area centroid |
No method is universally “most accurate”. The correct exam calculation is the method specified by the question. If all memberships or all weights are zero, division by zero is undefined; a controller needs an explicit fallback rule.
Yaad rakho: Formula similar dikhega, par weight kya hai? Membership, area, ya rule strength—pehle woh identify karo.
\r\nThis is a different example from the two-input fan design. With Warm=Tri(20,30,40) and Hot=Tri(30,40,50), temperature 32 gives Warm=(40−32)/10=0.8 and Hot=(32−30)/10=0.2. The slide uses output representatives 50 and 80.
Using the weighted-average method, combine output representatives [50, 80] % fan speed with respective strengths [0.8, 0.2]. [3 marks · Practice]
| Output representative | Strength |
|---|---|
| 50 | 0.8 |
| 80 | 0.2 |
One crisp output by the specified weighted-average method.
Each representative is multiplied by its strength; add these products and divide by the total strength. Here α is a strength and z is a representative value. These are not automatically the full area of a clipped Mamdani set.
| Strength × value | Product |
|---|---|
| 0.8 × 50 | 40 |
| 0.2 × 80 | 16 |
![\"40+16=56\"](\"math/dc4a907872eaaa0b.svg\")
![\"0.8+0.2=1\"](\"math/ffd94fb58f716964.svg\")
There are 3 temperature labels and 3 occupancy labels. Pair each temperature label with each occupancy label: 3×3=9 possible AND rules. At the specified input, any rule containing a zero input membership has strength min(0,other)=0 and contributes no output. The rules with nonzero strength remain; their output labels come from the supplied rule table, not from the magnitude of the strength.
This is the fan-controller design question shared in class: temperature 32°C and occupancy 7. The photos contain inconsistent graph breakpoints and arithmetic. A design question permits a stated design; below, every membership function is specified explicitly. These choices make the solution reproducible and should not be read as a uniquely recoverable answer from the photos.
Design a Mamdani fan controller for temperature 0–40°C and occupancy 0–10, with three input labels each and five fan-speed labels on 0–100%. For T=32°C and occupancy=7, use the membership functions and nine rules below. Calculate all memberships, firing strengths, and active output sets. Find fan speed by (a) mean of maxima and (b) the specified representative-weighted average. [10 marks · Class-question adaptation; full design]
| Variable | Low / Very Low | Medium / Low | High / Medium | High output | Very High output |
|---|---|---|---|---|---|
| Temperature | Tri(0,10,20) | Tri(15,25,35) | Tri(25,35,40) | — | — |
| Occupancy | Tri(0,2,4) | Tri(3,5,7) | Tri(6,8,10) | — | — |
| Fan speed | Tri(0,10,20) | Tri(10,25,40) | Tri(20,40,60) | Tri(50,65,80) | Tri(70,85,100) |
Input T=32°C; occupancy=7. Use min-AND, min-clipping and max-aggregation.
| Temperature | Occupancy | Fan output |
|---|---|---|
| Low | Low | Very Low |
| Low | Medium | Low |
| Low | High | Medium |
| Medium | Low | Low |
| Medium | Medium | Medium |
| Medium | High | High |
| High | Low | Medium |
| High | Medium | High |
| High | High | Very High |
Input memberships → nine rule strengths → clipped output → two named crisp outputs.
Use the appropriate triangle branch for membership. AND takes the smaller degree. Each output is clipped at that strength; rules with the same consequent combine by max. MoM uses only the highest plateau. The requested representative weighted average uses category peaks and strengths.
![\"\\alpha_r=\\min(\\mu_{T,r},\\mu_{O,r})\\quad\\mu_{out}(z)=\\max_r\\min(\\alpha_r,\\mu_{C_r}(z))\"](\"math/2e1abd53c3392e1f.svg\")
![\"At](\"fig/v2_fan_temp.434c9b6635.svg\")
![\"\\mu_L(32)=0\"](\"math/90541f01c8f3e577.svg\")
![\"\\mu_M(32)=\\frac{35-32}{35-25}=\\frac3{10}=0.3\"](\"math/8b417cc7c6fd3b3f.svg\")
![\"\\mu_H(32)=\\frac{32-25}{35-25}=\\frac7{10}=0.7\"](\"math/1456bdcfcdae3dcf.svg\")
![\"At](\"fig/v2_fan_people.1861531dd1.svg\")
![\"\\mu_L(7)=0\\qquad\\mu_M(7)=0\"](\"math/0dc9da58647d7498.svg\")
![\"\\mu_H(7)=\\frac{7-6}{8-6}=\\frac12=0.5\"](\"math/527793f08bd07193.svg\")
| T label / degree | O label / degree | min = strength | Output |
|---|---|---|---|
| Low / 0 | Low / 0 | 0 | Very Low |
| Low / 0 | Medium / 0 | 0 | Low |
| Low / 0 | High / 0.5 | 0 | Medium |
| Medium / 0.3 | Low / 0 | 0 | Low |
| Medium / 0.3 | Medium / 0 | 0 | Medium |
| Medium / 0.3 | High / 0.5 | 0.3 | High |
| High / 0.7 | Low / 0 | 0 | Medium |
| High / 0.7 | Medium / 0 | 0 | High |
| High / 0.7 | High / 0.5 | 0.5 | Very High |
![\"\\mu_{out}(z)=\\max\\{\\min(0.3,\\mu_H(z)),\\min(0.5,\\mu_{VH}(z))\\}\"](\"math/2ead49cb7f0674dc.svg\")
![\"High](\"fig/v2_fan_output.8708d0ea77.svg\")
The maximum is 0.5, belonging to Very High. Its triangle is (70,85,100).
![\"\\frac{z_L-70}{15}=0.5\\Rightarrow](\"math/a4fdc221a6cfa166.svg\")
![\"\\frac{100-z_R}{15}=0.5\\Rightarrow](\"math/fba2b937f7b534a8.svg\")
![\"z^*_{MoM}=\\frac{77.5+92.5}{2}=85\"](\"math/a4e234f2232ff44d.svg\")
![\"0.3\\times65=19.5\"](\"math/e44a237331cdc730.svg\")
![\"0.5\\times85=42.5\"](\"math/dfe0b2ebeb9b2e0c.svg\")
![\"z^*_{WA}=\\frac{19.5+42.5}{0.3+0.5}=\\frac{62}{0.8}=77.5\"](\"math/1cf8861b16222d21.svg\")
How to adapt it in the exam: if breakpoints or output representatives differ, recompute the memberships from those data. The sequence of operations stays unchanged. If the question explicitly asks for the area centroid, integrate the max-aggregated shape; do not silently substitute the weighted average above.
Yaad rakho: Temperature aur people ka degree nikalo → 9 min → active outputs clip → jo method poocha hai woh lagao.
\r\n\r\n\r\nFor the stated Very High triangle (70,85,100) clipped at height 0.5, solve the rising edge: (z−70)/(85−70)=0.5, so z−70=0.5×15=7.5 and z=77.5. On the falling edge, (100−z)/(100−85)=0.5, so 100−z=7.5 and z=92.5. Thus the highest flat interval is [77.5,92.5]. Its midpoint is (77.5+92.5)/2=170/2=85.
Continue with the complete area-centroid calculation for this same fan design: 77.086956522%, with all overlap breakpoints and area/moment arithmetic shown.
"},{"id":"centroid-comparison","title":"The same fan controller: true area centroid versus MoM and weighted average","html":"The preceding class-fan lesson makes the input shapes and nine rules explicit because the photographs do not specify one internally consistent design. Keep that exact stated design here: temperature 32°C, occupancy 7, High output Tri(50,65,80) firing at 0.3 and Very High output Tri(70,85,100) firing at 0.5. This continuation calculates the actual area centroid of their max-aggregate. It does not reinterpret 77.5 as a centroid.
For the stated two-input class fan controller at T=32°C and occupancy=7, calculate the true continuous Mamdani area centroid of High=Tri(50,65,80) clipped at 0.3 and Very High=Tri(70,85,100) clipped at 0.5. Compare it with mean of maxima and the specified representative-weighted average. [6 marks]
Worked teaching continuation of the class-design adaptation; the mark allocation here is for practice.
All other output rules have strength zero. Use min for implication, max for aggregation and speed z on 0–100%. The two active rule strengths were calculated from input memberships: min(0.3,0.5)=0.3 and min(0.7,0.5)=0.5.
Every visible segment of the aggregate, its area and first moment, the centroid, and the differences between the three crisp answers.
![\"z^*_{COA}=\\frac{\\int](\"math/175fa0720f954403.svg\")
Centroid balances all the area under the upper envelope. A straight strip with endpoints (a,u) and (b,v) has the following area A and first moment M; adding only disjoint strips counts overlap once.
![\"A=\\frac{(b-a)(u+v)}2,\\qquad](\"math/10dac72a97306eb6.svg\")
![\"H:\\quad](\"math/6757779ccc7556ad.svg\")
![\"VH:\\quad](\"math/19f28361b506bad3.svg\")
High has a 0.3 plateau from 54.5 to 75.5. Very High has a 0.5 plateau from 77.5 to 92.5.
Very High rises through the High plateau at height 0.3. Solve its rising equation:
![\"(z-70)/15=0.3\\quad\\Rightarrow\\quad](\"math/d00383e24c2ca74a.svg\")
Before 74.5, clipped High supplies the upper envelope. After 74.5, Very High is above it. The descending High edge from 75.5 to 80 lies underneath and contributes no extra aggregate area.
![\"Fan](\"fig/deep_fan_centroid.98bf1c217d.svg\")
Read left to right: (50,0) → (54.5,0.3) → (74.5,0.3) → (77.5,0.5) → (92.5,0.5) → (100,0). Membership is zero outside this support. Each adjacent pair is one straight strip in the table.
![\"A_1=(54.5-50)(0+0.3)/2=0.675\"](\"math/5c3f2b7a4bd76f0b.svg\")
![\"M_1=\\frac{4.5}{6}[(100+54.5)0+(50+109)0.3]=35.775\"](\"math/5570b4038f564dfd.svg\")
The first moment has units of speed multiplied by area. Dividing total moment by total area returns speed.
| a | b | u | v | Area A | First moment M |
|---|---|---|---|---|---|
| 50 | 54.5 | 0 | 0.3 | 0.675 | 35.775 |
| 54.5 | 74.5 | 0.3 | 0.3 | 6 | 387 |
| 74.5 | 77.5 | 0.3 | 0.5 | 1.2 | 91.35 |
| 77.5 | 92.5 | 0.5 | 0.5 | 7.5 | 637.5 |
| 92.5 | 100 | 0.5 | 0 | 1.875 | 178.125 |
![\"\\sum](\"math/1a4fac5b0fc5e065.svg\")
![\"\\sum](\"math/3f79a707c0d7b969.svg\")
![\"z^*_{COA}=1329.75/17.25=1773/23\\approx77.086956522\\%\"](\"math/cf56069827f681bb.svg\")
| Method | Calculation | Answer |
|---|---|---|
| Area centroid | 1329.75 ÷ 17.25 | 77.086956522% |
| Mean of maxima | (77.5+92.5) ÷ 2 | 85% |
| Specified representative weighted average | (0.3×65+0.5×85) ÷ (0.3+0.5) | 77.5% |
MoM uses only the highest plateau. The representative method uses two chosen peaks and two rule strengths. The area centroid uses the full max-aggregate, so these are three legitimate but different calculations.
Boundary coverage of this teaching design. With the ordinary triangles stated in the class adaptation, every temperature degree is zero at T=0 or T=40, and every occupancy degree is zero at occupancy=0 or 10. Then no rule fires and an area centroid is undefined. To make a controller cover the entire declared input range, explicitly redesign outer sets as shoulders or specify an out-of-coverage rule; do not silently divide by zero. The worked T=32, occupancy=7 case has positive area and is unaffected.
Yaad rakho: Centroid mein poore shaded area ka balance chahiye. Peak × strength wala shortcut same calculation nahi hai.
"},{"id":"exam-theory","title":"Fuzzy theory: definitions, architecture, advantages and applications","html":"Explain fuzzy logic with one application. [2 marks · Practice]
Fuzzy logic represents gradual truth or category membership using values between 0 and 1. A temperature controller can use Low, Medium and High temperature sets. It evaluates rules such as IF temperature is High THEN fan speed is High, combines the rule outputs and defuzzifies them into a numerical fan speed. This permits gradual control around category boundaries.
Give the advantages and limitations of a Mamdani fuzzy controller. [3 marks · Practice]
| Advantages | Limitations |
|---|---|
| Human-readable rules express expert knowledge | Membership functions and rules require careful design |
| Can model gradual boundaries without an exact analytic plant model | Rule count can grow rapidly with more inputs and categories |
| Handles overlapping linguistic descriptions | A poorly chosen rule base can produce poor or uncovered outputs |
| Suitable for many control applications | Does not automatically learn or guarantee optimal control |
Describe the design of a fuzzy fan controller. [3 marks · Practice]
Choose temperature and occupancy as inputs and fan speed as output. Define their ranges and Low/Medium/High membership functions. Construct a rule table linking input labels to output labels. Fuzzify measured inputs, use min to obtain AND-rule strengths, clip the consequent sets and max-aggregate them. Apply the specified defuzzification method to obtain a fan speed. Test the design at normal inputs and range boundaries.
| Application | Inputs | Example rule | Output |
|---|---|---|---|
| Washing machine | Dirtiness, load | IF dirt High AND load High THEN wash time Long | Wash duration |
| Loan decision support | Income, credit score | IF income High AND credit Good THEN approval suitability High | Suitability score (not an automatic legal lending decision) |
| Fire warning | Temperature, smoke | IF temperature High OR smoke High THEN risk High | Alarm-risk score |
| Vehicle control | Distance, relative speed | IF distance Small AND closing speed High THEN braking Strong | Braking command |
Yaad rakho: Theory answer mein definition → blocks/steps → ek concrete example. Sirf headings se marks nahi milte.
"},{"id":"smooth-memberships","title":"Gaussian, singleton and bell memberships","html":"\r\nFor a Gaussian membership, c is the input at the centre and σ controls the width. Compute x−c, square it, divide by 2σ², put a minus sign in front, and apply eˣ. The maximum membership is 1 at x=c because the exponent becomes 0. Increasing σ makes the same distance from the centre count as a smaller relative distance, so the curve becomes wider. A Gaussian membership is not the same formula as a normal probability density: there is no 1/(√(2π)σ) multiplier here.
A Gaussian membership function reaches 1 at its centre c and decreases smoothly on both sides. Its width σ must be positive. This is a fuzzy membership curve, not a normalised probability density: there is no 1/(√(2π)σ) factor.
![\"\\mu(x)=\\exp\\left(-\\frac{(x-c)^2}{2\\sigma^2}\\right)\"](\"math/518c1dcbf29cffeb.svg\")
exp(u) means eᵘ. The calculator evaluates it using eˣ; calculate the entire negative exponent first. A squared distance treats equal left and right offsets equally.
For Gaussian membership with c=10 and σ=2, find μ(10). [2 marks · Practice]
x=10; centre c=10; width σ=2.
Membership degree.
Use squared distance from the centre in the Gaussian membership formula.
![\"10-10=0\"](\"math/321dce230aba9ce3.svg\")
![\"(0)^2=0\\qquad2\\sigma^2=2(2^2)=8\"](\"math/2fa85e59ecadded4.svg\")
![\"-0/8=0\"](\"math/7875a0b4018e0854.svg\")
![\"\\mu(10)=e^{0}\\approx1\"](\"math/db4e90ad0923df06.svg\")
For Gaussian membership with c=10 and σ=2, find μ(12). [2 marks · Practice]
x=12; centre c=10; width σ=2.
Membership degree.
Use squared distance from the centre in the Gaussian membership formula.
![\"12-10=2\"](\"math/97fe26072a5dade9.svg\")
![\"(2)^2=4\\qquad2\\sigma^2=2(2^2)=8\"](\"math/0776e93b48581bdc.svg\")
![\"-4/8=-0.5\"](\"math/31d0827fcff0f487.svg\")
![\"\\mu(12)=e^{-0.5}\\approx0.60653066\"](\"math/3045651368f264c4.svg\")
For Gaussian membership with c=10 and σ=2, find μ(14). [2 marks · Practice]
x=14; centre c=10; width σ=2.
Membership degree.
Use squared distance from the centre in the Gaussian membership formula.
![\"14-10=4\"](\"math/83b853a2ad6339ff.svg\")
![\"(4)^2=16\\qquad2\\sigma^2=2(2^2)=8\"](\"math/8237f522c90ae910.svg\")
![\"-16/8=-2\"](\"math/6189f75f7af84207.svg\")
![\"\\mu(14)=e^{-2}\\approx0.135335283\"](\"math/a282bbe0d0c6f2b0.svg\")
A singleton fuzzy set has membership 1 at exactly one specified point and 0 elsewhere. For a singleton at c=50: μ(50)=1, μ(49)=0 and μ(51)=0. A singleton consequent is represented by one numerical output; this is why a specified weighted average of singleton outputs is a legitimate calculation.
\r\n![\"\\mu(x)=\\frac1{1+\\left|\\frac{x-c}{a}\\right|^{2b}}\"](\"math/5710690d4b19691e.svg\")
c is the centre, a>0 controls width and b>0 controls the slope. Absolute value |u| is the nonnegative magnitude: |−2|=2. At x=c membership is 1; at x=c±a membership is 1/2.
For a=2, b=1 and c=10, find generalised-bell membership at x=14. [2 marks · Practice]
a=2, b=1, c=10; x=14.
μ(14).
Subtract the centre, divide by a, take the absolute value, raise to 2b, then use the reciprocal.
![\"|(14-10)/2|=|2|=2\"](\"math/76742f554dcbda62.svg\")
![\"2^{2(1)}=4\"](\"math/de2c89102c7ce99d.svg\")
![\"\\mu(14)=1/(1+4)=0.2\"](\"math/d1a7dee6fdbdc5f8.svg\")
Yaad rakho: Gaussian MF ka centre 1 hota hai. GMM density ke normalising factor ko yahan mat lagao.
"}]},{"id":"ga-core","title":"3. Genetic algorithms","unit":"Unit 3 · GA","scope":"Candidate → fitness → encoding → selection → crossover → mutation → two complete generations.","sections":[{"id":"meaning","title":"GA meaning, biological idea and every basic term","html":"\r\n| Object | Concrete value | Do not confuse it with |
|---|---|---|
| Individual number | Row 2 | The candidate’s value x |
| Candidate value x | 11 | Its chromosome 1011 |
| Chromosome | 1011 | Its fitness |
| Fitness when maximising x² | 11×11=121 | Its row number or bit string |
In this guide, “gene 3” counts the third bit from the left starting at 1. Binary place values were built from the right starting with weight 1. These are two labelling conventions, not contradictory place-value rules. Read the end specified in an exam.
A genetic algorithm (GA) is a population-based optimisation method inspired by biological evolution. It maintains candidate solutions, evaluates their fitness, selects parents, generates children through crossover and mutation, and repeats the process.
In biology, genes carry inherited information. In a GA, an encoding plays the role of inherited information. The analogy motivates the operators; it does not mean that a computer chromosome is literally DNA.
| Term | Meaning | Four-bit example |
|---|---|---|
| Individual | One candidate solution | x=11 |
| Chromosome | The complete encoded candidate | 1011 |
| Gene | One component of an encoding | The third bit |
| Allele | The value at a gene | 0 or 1 for binary encoding |
| Population | The collection of current individuals | [3,5,10,12] |
| Fitness | A score used to favour useful candidates | For max x², fitness of 11 is 121 |
| Generation | One cycle producing a new population | Select → cross → mutate → replace |
| Search space | All candidate solutions permitted | 4-bit integers: 0–15 |
| Elitism | Preserving best individuals unchanged | Copy the best into the next population |
An objective describes what the problem wants. Fitness describes what the selection mechanism favours. For maximising a nonnegative objective, they may be the same. For minimising cost, directly selecting in proportion to cost would favour bad candidates; use a specified transformation or a comparison-based selection method.
For a maximisation GA with objective f(x)=x², find the fitness of x=3. [2 marks · Practice]
x=3; fitness equals the nonnegative objective x².
Fitness, not the encoded chromosome.
Square x to evaluate its objective; binary encoding is a separate representation step.
![\"f(3)=3^2\"](\"math/7cb566d00780f2b1.svg\")
![\"3\\times3=9\"](\"math/40a9936a2387e5b6.svg\")
For a maximisation GA with objective f(x)=x², find the fitness of x=7. [2 marks · Practice]
x=7; fitness equals the nonnegative objective x².
Fitness, not the encoded chromosome.
Square x to evaluate its objective; binary encoding is a separate representation step.
![\"f(7)=7^2\"](\"math/a39be859ebcf3420.svg\")
![\"7\\times7=49\"](\"math/67b20e8d00098635.svg\")
For a maximisation GA with objective f(x)=x², find the fitness of x=12. [2 marks · Practice]
x=12; fitness equals the nonnegative objective x².
Fitness, not the encoded chromosome.
Square x to evaluate its objective; binary encoding is a separate representation step.
![\"f(12)=12^2\"](\"math/40c9415d3dcdd694.svg\")
![\"12\\times12=144\"](\"math/ea37e28a2a0fc9ec.svg\")
Yaad rakho: Chromosome answer ka code; fitness us code wale answer ka score.
"},{"id":"encoding","title":"Encoding: binary, real values and permutations","html":"\r\n![\"x=b_1](\"math/7a27d9903aa92206.svg\")
For four bits, b₁ is the leftmost bit and receives weight 8; b₄ is the rightmost and receives weight 1. In general n−i gives the power belonging to bit i. For 1011, the four terms are 1×8, 0×4, 1×2, 1×1. Add 8+0+2+1=11.
For real-range mapping, d/(2ⁿ−1) is the fraction of the way through the possible codes. Multiplying by U−L travels that fraction of the physical range. Adding L shifts the result to start at the lower limit.
| 3-bit code | Decoded d | Fraction d/7 | Mapped to [−2,5] |
|---|---|---|---|
| 000 | 0 | 0/7=0 | −2+0×7=−2 |
| 011 | 3 | 3/7 | −2+(3/7)×7=1 |
| 111 | 7 | 7/7=1 | −2+1×7=5 |
Binary encoding uses bit strings; real-valued encoding stores real numbers directly; permutation encoding stores an order, such as a TSP city tour. An operator valid for bits may create invalid city tours, so encoding and operators must fit the problem.
| Representation | Candidate | Appropriate use |
|---|---|---|
| Binary | 1011 | An unsigned integer or a coded real value |
| Real-valued | [0.2, 1.7] | Continuous parameters |
| Permutation | [A,C,B,D] | Visit order in TSP |
![\"x=b_1](\"math/f86e986776de93a7.svg\")
b₁ is the leftmost bit, not a coefficient you have to learn. In four bits the weights are 8,4,2,1. The maximum is ![\"2^n-1\"](\"math/9dc91f3c375d809f.svg\")
.
Decode the 4-bit chromosome 0011. [2 marks · Practice]
Bits 0011; place values 8,4,2,1.
Integer x.
Multiply each bit by its matching place value, then add.
| Bit | Place value | Contribution |
|---|---|---|
| 0 | 8 | 0 |
| 0 | 4 | 0 |
| 1 | 2 | 2 |
| 1 | 1 | 1 |
![\"0+0+2+1=3\"](\"math/76153632459f881f.svg\")
Decode the 4-bit chromosome 1010. [2 marks · Practice]
Bits 1010; place values 8,4,2,1.
Integer x.
Multiply each bit by its matching place value, then add.
| Bit | Place value | Contribution |
|---|---|---|
| 1 | 8 | 8 |
| 0 | 4 | 0 |
| 1 | 2 | 2 |
| 0 | 1 | 0 |
![\"8+0+2+0=10\"](\"math/d1e04d751dc741ed.svg\")
Decode the 4-bit chromosome 1111. [2 marks · Practice]
Bits 1111; place values 8,4,2,1.
Integer x.
Multiply each bit by its matching place value, then add.
| Bit | Place value | Contribution |
|---|---|---|
| 1 | 8 | 8 |
| 1 | 4 | 4 |
| 1 | 2 | 2 |
| 1 | 1 | 1 |
If the question maps an n-bit integer d to a real interval [L,U], first decode d, then use:
![\"x=L+\\frac{d}{2^n-1}(U-L)\"](\"math/3013a3fc845c64e5.svg\")
Map the 3-bit string 101 to the real interval [−2,5] using linear binary decoding. [3 marks · Practice]
n=3, L=−2, U=5; string 101.
Real x.
The smallest bit code maps to L and the largest to U; the remaining codes are equally spaced.
![\"x=L+\\frac](\"math/d5a028f7f6ff09e2.svg\")
![\"d=1(4)+0(2)+1(1)=5\"](\"math/5d98008e8ca61fcc.svg\")
![\"U-L=5-(-2)=7\\qquad2^3-1=7\"](\"math/a70472c6ec58ad8d.svg\")
![\"x=-2+(5/7)(7)=3\"](\"math/2d086e3a5cf5f29f.svg\")
Yaad rakho: Pehle bits se integer. Real range mapping explicitly diya ho tabhi next formula lagao.
"},{"id":"selection","title":"Roulette selection, cumulative intervals and alternatives","html":"\r\nBefore the three complete sums below, read the running-total example in Percentages, probabilities and cumulative intervals. For weights [1,2,3,4], candidate C’s own probability is 0.3, but its interval is [0.3,0.6). A draw of 0.55 selects C because it lies inside that interval; it is not required to be less than C’s individual probability.
Roulette-wheel selection gives each candidate a probability proportional to a nonnegative fitness weight. A high weight occupies a larger interval on [0,1]. A supplied random number selects the interval containing it.
![\"P_i=F_i/\\sum_jF_j\"](\"math/2cf98c22ed8b8331.svg\")
Fᵢ is the fitness weight of individual i. The sum must be positive. Probabilities add to 1. Cumulative probability is the running sum and supplies the interval boundaries.
Fitness weights of successive individuals are [1, 2, 3, 4]. Select parents using random numbers [0.05, 0.25, 0.55, 0.9]. [3 marks · Practice]
Fitness weights [1, 2, 3, 4]; draws [0.05, 0.25, 0.55, 0.9].
Parent row number for each draw.
Divide by the fitness sum, form cumulative intervals, then locate each random draw. Use half-open intervals [lower,upper); the final upper endpoint 1 is included.
![\"\\sum](\"math/601b2d3d469972c2.svg\")
| Individual | Fitness | Probability | Cumulative |
|---|---|---|---|
| 1 | 1 | 0.1 | 0.1 |
| 2 | 2 | 0.2 | 0.3 |
| 3 | 3 | 0.3 | 0.6 |
| 4 | 4 | 0.4 | 1 |
| Draw | Selected individual |
|---|---|
| 0.05 | 1 |
| 0.25 | 2 |
| 0.55 | 3 |
| 0.9 | 4 |
Fitness weights of successive individuals are [4, 4, 2]. Select parents using random numbers [0.2, 0.6, 0.85]. [3 marks · Practice]
Fitness weights [4, 4, 2]; draws [0.2, 0.6, 0.85].
Parent row number for each draw.
Divide by the fitness sum, form cumulative intervals, then locate each random draw. Use half-open intervals [lower,upper); the final upper endpoint 1 is included.
![\"\\sum](\"math/a1ef5b129218331e.svg\")
| Individual | Fitness | Probability | Cumulative |
|---|---|---|---|
| 1 | 4 | 0.4 | 0.4 |
| 2 | 4 | 0.4 | 0.8 |
| 3 | 2 | 0.2 | 1 |
| Draw | Selected individual |
|---|---|
| 0.2 | 1 |
| 0.6 | 2 |
| 0.85 | 3 |
Fitness weights of successive individuals are [0, 3, 1]. Select parents using random numbers [0.2, 0.8]. [3 marks · Practice]
Fitness weights [0, 3, 1]; draws [0.2, 0.8].
Parent row number for each draw.
Divide by the fitness sum, form cumulative intervals, then locate each random draw. Use half-open intervals [lower,upper); the final upper endpoint 1 is included.
![\"\\sum](\"math/7f52b89af830db1b.svg\")
| Individual | Fitness | Probability | Cumulative |
|---|---|---|---|
| 1 | 0 | 0 | 0 |
| 2 | 3 | 0.75 | 0.75 |
| 3 | 1 | 0.25 | 1 |
| Draw | Selected individual |
|---|---|
| 0.2 | 2 |
| 0.8 | 3 |
| Method | Procedure | Why it is used |
|---|---|---|
| Tournament | Sample a group and take its best candidate | Only objective comparisons are needed |
| Rank | Sort candidates; assign weights based on rank | Reduces sensitivity to a single huge raw fitness |
| Elitist survival | Preserve top individuals unchanged into the next population | Protects good solutions from destructive variation |
Selection is not automatically “replace the weakest with the fittest”. That is one possible classroom reproduction rule. If a problem gives selected parents, a rank rule, tournament contestants or roulette draws, follow that exact instruction.
Yaad rakho: Probability nikalo → cumulative banao → random number ka interval dekho.
"},{"id":"operators","title":"Crossover and mutation: four complete worked variants","html":"\r\nThe prefix is the part before the cut; the suffix is the part after it. “10 joined with 00 gives 1000” means writing strings next to each other. It is not the arithmetic addition 10+00.
| Operation | Parents / instruction | Construction | Children |
|---|---|---|---|
| One-point crossover | 10|11 and 01|00 | First prefix + second tail; second prefix + first tail | 1000 and 0111 |
| Two-point crossover | 1|01|10 and 0|10|01 | Exchange only the middle blocks | 11010 and 00101 |
| Uniform crossover | A=1010, B=0101, mask=1100 | Child 1 takes A where mask=1 and B where mask=0; child 2 reverses this | 1001 and 0110 |
| Mutation | 1000; flip position 3 from left | The third bit is 0; replace it by 1; keep every other bit | 1010 |
These crossover operators are alternatives. Apply the operator stated in the question; do not apply all three in one generation. Mutation is a separate step after crossover in the worked exam workflow.
In single-point crossover, cut both parents after the specified bit. Child 1 keeps parent 1’s prefix and takes parent 2’s suffix; child 2 does the reverse. A cut after bit 2 in a four-bit string looks like 10|11.
Cross parents 1011 and 0100 after bit 2. Then flip bit 4 of both children, counting from the left. Decode the final children. [3 marks · Practice]
Parents 1011, 0100; cut after bit 2; mutation bit 4.
Children after crossover, after mutation and their decimal values.
Exchange only suffixes. Mutation means 0→1 or 1→0; it does not mean set the bit to 1.
10 | 11 and 01 | 00
10 + 00 → 1000
01 + 11 → 0111
| Crossover child | Bit before | Bit after | Mutated child |
|---|---|---|---|
| 1000 | 0 | 1 | 1001 |
| 0111 | 1 | 0 | 0110 |
| String | Place-value sum | x |
|---|---|---|
| 1001 | 8 + 0 + 0 + 1 | 9 |
| 0110 | 0 + 4 + 2 + 0 | 6 |
Cross parents 0011 and 1100 after bit 1. Then flip bit 2 of both children, counting from the left. Decode the final children. [3 marks · Practice]
Parents 0011, 1100; cut after bit 1; mutation bit 2.
Children after crossover, after mutation and their decimal values.
Exchange only suffixes. Mutation means 0→1 or 1→0; it does not mean set the bit to 1.
0 | 011 and 1 | 100
0 + 100 → 0100
1 + 011 → 1011
| Crossover child | Bit before | Bit after | Mutated child |
|---|---|---|---|
| 0100 | 1 | 0 | 0000 |
| 1011 | 0 | 1 | 1111 |
| String | Place-value sum | x |
|---|---|---|
| 0000 | 0 + 0 + 0 + 0 | 0 |
| 1111 | 8 + 4 + 2 + 1 | 15 |
Cross parents 1110 and 0001 after bit 3. Then flip bit 1 of both children, counting from the left. Decode the final children. [3 marks · Practice]
Parents 1110, 0001; cut after bit 3; mutation bit 1.
Children after crossover, after mutation and their decimal values.
Exchange only suffixes. Mutation means 0→1 or 1→0; it does not mean set the bit to 1.
111 | 0 and 000 | 1
111 + 1 → 1111
000 + 0 → 0000
| Crossover child | Bit before | Bit after | Mutated child |
|---|---|---|---|
| 1111 | 1 | 0 | 0111 |
| 0000 | 0 | 1 | 1000 |
| String | Place-value sum | x |
|---|---|---|
| 0111 | 0 + 4 + 2 + 1 | 7 |
| 1000 | 8 + 0 + 0 + 0 | 8 |
Cross parents 1010 and 1010 after bit 2. Then flip bit 4 of both children, counting from the left. Decode the final children. [3 marks · Practice]
Parents 1010, 1010; cut after bit 2; mutation bit 4.
Children after crossover, after mutation and their decimal values.
Exchange only suffixes. Mutation means 0→1 or 1→0; it does not mean set the bit to 1.
10 | 10 and 10 | 10
10 + 10 → 1010
10 + 10 → 1010
| Crossover child | Bit before | Bit after | Mutated child |
|---|---|---|---|
| 1010 | 0 | 1 | 1011 |
| 1010 | 0 | 1 | 1011 |
| String | Place-value sum | x |
|---|---|---|
| 1011 | 8 + 0 + 2 + 1 | 11 |
| 1011 | 8 + 0 + 2 + 1 | 11 |
Two-point crossover exchanges the segment between two cuts. Uniform crossover uses a given mask to choose, at each position, which parent contributes a gene. Mutation may flip a bit, perturb a real value, or swap two positions in a permutation. These are different operators; apply the one named in the question.
Yaad rakho: Crossover = tail exchange. Mutation = specified bit flip. Flip ka matlab 1 bhi 0 banega.
"},{"id":"algorithm","title":"GA architecture, two-generation flow, constraints and theory","html":"\r\n| Read this | Meaning | Concrete example |
|---|---|---|
| Feasible | Obeys every stated constraint | x≤10: x=9 qualifies; x=12 does not |
| Penalty | A deduction for violating a constraint | For maximisation, subtract 100×excess |
| Elitism | Explicitly preserve a best old individual | Only apply if the question specifies it |
| Replacement | Choose what population the next generation uses | Our full-replacement examples use the new children |
| Stopping condition | A rule ending the repeated process | Stop after two complete generations |
A simple generational GA replaces the population in generations using a chosen encoding, selection rule and variation operators. A more general GA may use steady-state replacement, different encodings, adaptive operators or explicit elitism. Always state the choices used in your worked answer.
\r\nInitial population → first generation’s selection/crossover/mutation → new population → second generation’s selection/crossover/mutation → final population. Stopping after the first children and labelling them iteration 2 is not two full generations.
\r\nStop after a fixed generation budget, after reaching a specified acceptable objective, or after little improvement for a stated number of generations. A finite run does not prove the global optimum.
\r\nA constraint restricts allowed solutions, such as 0≤x≤15 or “visit every city once”. Common approaches are valid-by-design encoding, repairing an invalid child, rejecting it, or using a penalty. A penalty must reduce attractiveness: subtract it in maximisation or add it to a minimised cost. Do not silently remove the original constraint because you used more bits.
A maximisation problem has objective f(x)=x² and constraint x≤10. The given penalised fitness is F(x)=x²−100 max(0,x−10). Compare x=9 and x=12. [3 marks · Practice]
Penalty coefficient 100; bound 10; candidates 9 and 12.
Penalised fitness of both candidates.
Only excess above the bound receives a penalty.
![\"\\max(0,9-10)=0\"](\"math/a12fb07643a5aaa1.svg\")
![\"F(9)=81-100(0)=81\"](\"math/deb272d3158b70b9.svg\")
![\"\\max(0,12-10)=2\"](\"math/c2860199dab61091.svg\")
![\"F(12)=144-100(2)=-56\"](\"math/decf48b28bc4a229.svg\")
For this maximisation fitness, 81 > −56, so x=9 is preferred. Raw objective alone would misleadingly favour 12.
| Advantages | Limitations |
|---|---|
| Derivative-free; supports many encodings | Many objective evaluations may be costly |
| Population explores several candidates | Premature convergence is possible |
| Crossover and mutation can create new candidates | Parameter and encoding choices affect results |
| Applicable to discontinuous or combinatorial searches | No guaranteed global optimum; variation can worsen fitness |
Yaad rakho: Do generation = do baar poora cycle. Sirf children banana aur decode karna do alag iterations nahi.
"},{"id":"two-generations","title":"Two generations with given selection, cut and mutation","html":"\r\n![\"This](\"fig/v2_ga_generation.fa7bf86c05.svg\")
Maximise ![\"f(x)=x^2\"](\"math/e23769019c76d37b.svg\")
over unsigned 4-bit integers using the supplied GA settings. Perform two complete generations and compare the population fitness. [3 marks · Practice; complete two-generation solution]
Initial ordered population: [3, 5, 10, 12]. Use 4 bits, crossover after bit 2, pair mating-pool slots 1–2 and 3–4. Flip bit 4 of children [1, 2, 3, 4] in every generation. Replace the whole population with the four children; no elitism.
| Generation | Selection instruction |
|---|---|
| 1 | Current individual row numbers [3, 4, 4, 3] |
| 2 | Current individual row numbers [2, 3, 2, 3] |
Both generations: evaluation, selection, crossover, mutation, decoding and comparison.
Encoding represents x; it does not change the objective. Crossover exchanges tails. Mutation flips specified bits. Fitness must be recomputed after mutation. Use exactly the supplied selection rule; replacing the weakest by the strongest is not a universal GA step.
| Individual | x | Binary | Objective calculation |
|---|---|---|---|
| 1 | 3 | 0011 | 3 × 3 = 9 |
| 2 | 5 | 0101 | 5 × 5 = 25 |
| 3 | 10 | 1010 | 10 × 10 = 100 |
| 4 | 12 | 1100 | 12 × 12 = 144 |
The question directly supplies individual numbers [3, 4, 4, 3]. These refer to the current generation’s row numbers, not their x values.
| Mating-pool slot | Chosen individual | x | Chromosome |
|---|---|---|---|
| 1 | 3 | 10 | 1010 |
| 2 | 4 | 12 | 1100 |
| 3 | 4 | 12 | 1100 |
| 4 | 3 | 10 | 1010 |
Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1.
| Pool slots paired | Parent A | Parent B | Child A | Child B |
|---|---|---|---|---|
| 1 & 2 | 10 | 10 | 11 | 00 | 1000 | 1110 |
| 3 & 4 | 11 | 00 | 10 | 10 | 1110 | 1000 |
| Child | Before mutation | Instruction | After mutation |
|---|---|---|---|
| 1 | 1000 | Flip bit 4 | 1001 |
| 2 | 1110 | Flip bit 4 | 1111 |
| 3 | 1110 | Flip bit 4 | 1111 |
| 4 | 1000 | Flip bit 4 | 1001 |
| Chromosome | Decode using place values | Objective |
|---|---|---|
| 1001 | 8 + 0 + 0 + 1 = 9 | 9² = 81 |
| 1111 | 8 + 4 + 2 + 1 = 15 | 15² = 225 |
| 1111 | 8 + 4 + 2 + 1 = 15 | 15² = 225 |
| 1001 | 8 + 0 + 0 + 1 = 9 | 9² = 81 |
Add the objective scores in each population, then divide by its number of individuals.
![\"\\overline](\"math/ac14e9c27d115d0e.svg\")
![\"\\overline](\"math/b4cd0bde2e1de484.svg\")
Best objective: 144 → 225. Higher is better. The next population is [9, 15, 15, 9]. No elitism is imposed, so the best previous individual is not automatically protected.
| Individual | x | Binary | Objective calculation |
|---|---|---|---|
| 1 | 9 | 1001 | 9 × 9 = 81 |
| 2 | 15 | 1111 | 15 × 15 = 225 |
| 3 | 15 | 1111 | 15 × 15 = 225 |
| 4 | 9 | 1001 | 9 × 9 = 81 |
The question directly supplies individual numbers [2, 3, 2, 3]. These refer to the current generation’s row numbers, not their x values.
| Mating-pool slot | Chosen individual | x | Chromosome |
|---|---|---|---|
| 1 | 2 | 15 | 1111 |
| 2 | 3 | 15 | 1111 |
| 3 | 2 | 15 | 1111 |
| 4 | 3 | 15 | 1111 |
Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1.
| Pool slots paired | Parent A | Parent B | Child A | Child B |
|---|---|---|---|---|
| 1 & 2 | 11 | 11 | 11 | 11 | 1111 | 1111 |
| 3 & 4 | 11 | 11 | 11 | 11 | 1111 | 1111 |
| Child | Before mutation | Instruction | After mutation |
|---|---|---|---|
| 1 | 1111 | Flip bit 4 | 1110 |
| 2 | 1111 | Flip bit 4 | 1110 |
| 3 | 1111 | Flip bit 4 | 1110 |
| 4 | 1111 | Flip bit 4 | 1110 |
| Chromosome | Decode using place values | Objective |
|---|---|---|
| 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 |
| 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 |
| 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 |
| 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 |
Add the objective scores in each population, then divide by its number of individuals.
![\"\\overline](\"math/1526400ff09c5372.svg\")
![\"\\overline](\"math/83698ff516912ccd.svg\")
Best objective: 225 → 196. Higher is better. The next population is [14, 14, 14, 14]. No elitism is imposed, so the best previous individual is not automatically protected.
Maximise ![\"f(x)=x(15-x)+1\"](\"math/163a21622d906b2b.svg\")
over unsigned 4-bit integers using the supplied GA settings. Perform two complete generations and compare the population fitness. [3 marks · Practice; complete two-generation solution]
Initial ordered population: [2, 5, 9, 12]. Use 4 bits, crossover after bit 2, pair mating-pool slots 1–2 and 3–4. Flip bit 4 of children [2, 4] in every generation. Replace the whole population with the four children; no elitism.
| Generation | Selection instruction |
|---|---|
| 1 | Random draws [0.1, 0.4, 0.65, 0.9] |
| 2 | Random draws [0.2, 0.4, 0.6, 0.95] |
Both generations: evaluation, selection, crossover, mutation, decoding and comparison.
Encoding represents x; it does not change the objective. Crossover exchanges tails. Mutation flips specified bits. Fitness must be recomputed after mutation. Use exactly the supplied selection rule; replacing the weakest by the strongest is not a universal GA step.
| Individual | x | Binary | Objective calculation |
|---|---|---|---|
| 1 | 2 | 0010 | 2 × (15 − 2) + 1 = 27 |
| 2 | 5 | 0101 | 5 × (15 − 5) + 1 = 51 |
| 3 | 9 | 1001 | 9 × (15 − 9) + 1 = 55 |
| 4 | 12 | 1100 | 12 × (15 − 12) + 1 = 37 |
Fitness weights are nonnegative. The nonnegative objective itself is the fitness in this maximisation example.
![\"\\sum](\"math/3c9eb5a0f63c3e40.svg\")
| Individual | Probability: weight ÷ total | Cumulative: old boundary + probability |
|---|---|---|
| 1 | 27 ÷ 170 ≈ 0.158823529 | 0 + 0.158823529 ≈ 0.158823529 |
| 2 | 51 ÷ 170 ≈ 0.3 | 0.158823529 + 0.3 ≈ 0.458823529 |
| 3 | 55 ÷ 170 ≈ 0.323529412 | 0.458823529 + 0.323529412 ≈ 0.782352941 |
| 4 | 37 ÷ 170 ≈ 0.217647059 | 0.782352941 + 0.217647059 ≈ 1 |
| Random draw | Interval containing it | Selected individual | Selected x |
|---|---|---|---|
| 0.1 | 0 ≤ 0.1 < 0.158823529 | 1 | 2 |
| 0.4 | 0.158823529 ≤ 0.4 < 0.458823529 | 2 | 5 |
| 0.65 | 0.458823529 ≤ 0.65 < 0.782352941 | 3 | 9 |
| 0.9 | 0.782352941 ≤ 0.9 < 1 | 4 | 12 |
Intervals are left-closed and right-open, except that a draw of exactly 1 selects the last interval. No draw here is on a boundary. Draws select parents with replacement: selecting a parent once does not remove it. Calculations use unrounded weights; the table displays rounded numbers.
Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1.
| Pool slots paired | Parent A | Parent B | Child A | Child B |
|---|---|---|---|---|
| 1 & 2 | 00 | 10 | 01 | 01 | 0001 | 0110 |
| 3 & 4 | 10 | 01 | 11 | 00 | 1000 | 1101 |
| Child | Before mutation | Instruction | After mutation |
|---|---|---|---|
| 1 | 0001 | No mutation | 0001 |
| 2 | 0110 | Flip bit 4 | 0111 |
| 3 | 1000 | No mutation | 1000 |
| 4 | 1101 | Flip bit 4 | 1100 |
| Chromosome | Decode using place values | Objective |
|---|---|---|
| 0001 | 0 + 0 + 0 + 1 = 1 | 1(15−1)+1 = 15 |
| 0111 | 0 + 4 + 2 + 1 = 7 | 7(15−7)+1 = 57 |
| 1000 | 8 + 0 + 0 + 0 = 8 | 8(15−8)+1 = 57 |
| 1100 | 8 + 4 + 0 + 0 = 12 | 12(15−12)+1 = 37 |
Add the objective scores in each population, then divide by its number of individuals.
![\"\\overline](\"math/22945ccb0391cac7.svg\")
![\"\\overline](\"math/4832c61f6dee4ab2.svg\")
Best objective: 55 → 57. Higher is better. The next population is [1, 7, 8, 12]. No elitism is imposed, so the best previous individual is not automatically protected.
| Individual | x | Binary | Objective calculation |
|---|---|---|---|
| 1 | 1 | 0001 | 1 × (15 − 1) + 1 = 15 |
| 2 | 7 | 0111 | 7 × (15 − 7) + 1 = 57 |
| 3 | 8 | 1000 | 8 × (15 − 8) + 1 = 57 |
| 4 | 12 | 1100 | 12 × (15 − 12) + 1 = 37 |
Fitness weights are nonnegative. The nonnegative objective itself is the fitness in this maximisation example.
![\"\\sum](\"math/74f6736e95b5ff63.svg\")
| Individual | Probability: weight ÷ total | Cumulative: old boundary + probability |
|---|---|---|
| 1 | 15 ÷ 166 ≈ 0.090361446 | 0 + 0.090361446 ≈ 0.090361446 |
| 2 | 57 ÷ 166 ≈ 0.343373494 | 0.090361446 + 0.343373494 ≈ 0.43373494 |
| 3 | 57 ÷ 166 ≈ 0.343373494 | 0.43373494 + 0.343373494 ≈ 0.777108434 |
| 4 | 37 ÷ 166 ≈ 0.222891566 | 0.777108434 + 0.222891566 ≈ 1 |
| Random draw | Interval containing it | Selected individual | Selected x |
|---|---|---|---|
| 0.2 | 0.090361446 ≤ 0.2 < 0.43373494 | 2 | 7 |
| 0.4 | 0.090361446 ≤ 0.4 < 0.43373494 | 2 | 7 |
| 0.6 | 0.43373494 ≤ 0.6 < 0.777108434 | 3 | 8 |
| 0.95 | 0.777108434 ≤ 0.95 < 1 | 4 | 12 |
Intervals are left-closed and right-open, except that a draw of exactly 1 selects the last interval. No draw here is on a boundary. Draws select parents with replacement: selecting a parent once does not remove it. Calculations use unrounded weights; the table displays rounded numbers.
Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1.
| Pool slots paired | Parent A | Parent B | Child A | Child B |
|---|---|---|---|---|
| 1 & 2 | 01 | 11 | 01 | 11 | 0111 | 0111 |
| 3 & 4 | 10 | 00 | 11 | 00 | 1000 | 1100 |
| Child | Before mutation | Instruction | After mutation |
|---|---|---|---|
| 1 | 0111 | No mutation | 0111 |
| 2 | 0111 | Flip bit 4 | 0110 |
| 3 | 1000 | No mutation | 1000 |
| 4 | 1100 | Flip bit 4 | 1101 |
| Chromosome | Decode using place values | Objective |
|---|---|---|
| 0111 | 0 + 4 + 2 + 1 = 7 | 7(15−7)+1 = 57 |
| 0110 | 0 + 4 + 2 + 0 = 6 | 6(15−6)+1 = 55 |
| 1000 | 8 + 0 + 0 + 0 = 8 | 8(15−8)+1 = 57 |
| 1101 | 8 + 4 + 0 + 1 = 13 | 13(15−13)+1 = 27 |
Add the objective scores in each population, then divide by its number of individuals.
![\"\\overline](\"math/9b9333d937330aaa.svg\")
![\"\\overline](\"math/5b337e1a852cf7fa.svg\")
Best objective: 57 → 57. Higher is better. The next population is [7, 6, 8, 13]. No elitism is imposed, so the best previous individual is not automatically protected.
Roulette favours a larger weight. A minimisation objective instead favours a smaller cost. The stated transformation F=1/(1+cost) reverses their order for nonnegative costs: cost 0 → weight 1/1=1; cost 3 → 1/4=.25; cost 24 → 1/25=.04. The +1 prevents division by zero. This is one specified transformation, not a universal rule for every minimisation problem.
Minimise ![\"f(x)=(x-7)^2\"](\"math/caf5baf4d59a93e5.svg\")
over unsigned 4-bit integers using the supplied GA settings. Perform two complete generations and compare the population fitness. [3 marks · Practice; complete two-generation solution]
Initial ordered population: [2, 4, 8, 12]. Use 4 bits, crossover after bit 2, pair mating-pool slots 1–2 and 3–4. Flip bit 4 of children [2, 4] in every generation. Replace the whole population with the four children; no elitism.
| Generation | Selection instruction |
|---|---|
| 1 | Random draws [0.1, 0.4, 0.65, 0.9] |
| 2 | Random draws [0.2, 0.4, 0.6, 0.95] |
Both generations: evaluation, selection, crossover, mutation, decoding and comparison.
Encoding represents x; it does not change the objective. Crossover exchanges tails. Mutation flips specified bits. Fitness must be recomputed after mutation. Roulette uses Fᵢ/ΣF. For minimisation here, use Fᵢ=1/(1+f(xᵢ)); do not directly favour large costs.
| Individual | x | Binary | Objective calculation |
|---|---|---|---|
| 1 | 2 | 0010 | (2 − 7)² = 25 |
| 2 | 4 | 0100 | (4 − 7)² = 9 |
| 3 | 8 | 1000 | (8 − 7)² = 1 |
| 4 | 12 | 1100 | (12 − 7)² = 25 |
Fitness weights are nonnegative. For this minimisation example use F=1/(1+objective), so a smaller cost receives a larger weight.
| Individual | Cost | Fitness calculation |
|---|---|---|
| 1 | 25 | 1 / (1 + 25) = 1 / 26 ≈ 0.038461538 |
| 2 | 9 | 1 / (1 + 9) = 1 / 10 ≈ 0.1 |
| 3 | 1 | 1 / (1 + 1) = 1 / 2 ≈ 0.5 |
| 4 | 25 | 1 / (1 + 25) = 1 / 26 ≈ 0.038461538 |
![\"\\sum](\"math/3a777f87cb72ab3e.svg\")
| Individual | Probability: weight ÷ total | Cumulative: old boundary + probability |
|---|---|---|
| 1 | 0.038461538 ÷ 0.676923077 ≈ 0.056818182 | 0 + 0.056818182 ≈ 0.056818182 |
| 2 | 0.1 ÷ 0.676923077 ≈ 0.147727273 | 0.056818182 + 0.147727273 ≈ 0.204545455 |
| 3 | 0.5 ÷ 0.676923077 ≈ 0.738636364 | 0.204545455 + 0.738636364 ≈ 0.943181818 |
| 4 | 0.038461538 ÷ 0.676923077 ≈ 0.056818182 | 0.943181818 + 0.056818182 ≈ 1 |
| Random draw | Interval containing it | Selected individual | Selected x |
|---|---|---|---|
| 0.1 | 0.056818182 ≤ 0.1 < 0.204545455 | 2 | 4 |
| 0.4 | 0.204545455 ≤ 0.4 < 0.943181818 | 3 | 8 |
| 0.65 | 0.204545455 ≤ 0.65 < 0.943181818 | 3 | 8 |
| 0.9 | 0.204545455 ≤ 0.9 < 0.943181818 | 3 | 8 |
Intervals are left-closed and right-open, except that a draw of exactly 1 selects the last interval. No draw here is on a boundary. Draws select parents with replacement: selecting a parent once does not remove it. Calculations use unrounded weights; the table displays rounded numbers.
Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1.
| Pool slots paired | Parent A | Parent B | Child A | Child B |
|---|---|---|---|---|
| 1 & 2 | 01 | 00 | 10 | 00 | 0100 | 1000 |
| 3 & 4 | 10 | 00 | 10 | 00 | 1000 | 1000 |
| Child | Before mutation | Instruction | After mutation |
|---|---|---|---|
| 1 | 0100 | No mutation | 0100 |
| 2 | 1000 | Flip bit 4 | 1001 |
| 3 | 1000 | No mutation | 1000 |
| 4 | 1000 | Flip bit 4 | 1001 |
| Chromosome | Decode using place values | Objective |
|---|---|---|
| 0100 | 0 + 4 + 0 + 0 = 4 | (4−7)² = 9 |
| 1001 | 8 + 0 + 0 + 1 = 9 | (9−7)² = 4 |
| 1000 | 8 + 0 + 0 + 0 = 8 | (8−7)² = 1 |
| 1001 | 8 + 0 + 0 + 1 = 9 | (9−7)² = 4 |
Add the objective scores in each population, then divide by its number of individuals.
![\"\\overline](\"math/a688edfd604798b3.svg\")
![\"\\overline](\"math/f966da034fd08628.svg\")
Best objective: 1 → 1. Lower is better. The next population is [4, 9, 8, 9]. No elitism is imposed, so the best previous individual is not automatically protected.
| Individual | x | Binary | Objective calculation |
|---|---|---|---|
| 1 | 4 | 0100 | (4 − 7)² = 9 |
| 2 | 9 | 1001 | (9 − 7)² = 4 |
| 3 | 8 | 1000 | (8 − 7)² = 1 |
| 4 | 9 | 1001 | (9 − 7)² = 4 |
Fitness weights are nonnegative. For this minimisation example use F=1/(1+objective), so a smaller cost receives a larger weight.
| Individual | Cost | Fitness calculation |
|---|---|---|
| 1 | 9 | 1 / (1 + 9) = 1 / 10 ≈ 0.1 |
| 2 | 4 | 1 / (1 + 4) = 1 / 5 ≈ 0.2 |
| 3 | 1 | 1 / (1 + 1) = 1 / 2 ≈ 0.5 |
| 4 | 4 | 1 / (1 + 4) = 1 / 5 ≈ 0.2 |
![\"\\sum](\"math/8709c94ae272ba4a.svg\")
| Individual | Probability: weight ÷ total | Cumulative: old boundary + probability |
|---|---|---|
| 1 | 0.1 ÷ 1 ≈ 0.1 | 0 + 0.1 ≈ 0.1 |
| 2 | 0.2 ÷ 1 ≈ 0.2 | 0.1 + 0.2 ≈ 0.3 |
| 3 | 0.5 ÷ 1 ≈ 0.5 | 0.3 + 0.5 ≈ 0.8 |
| 4 | 0.2 ÷ 1 ≈ 0.2 | 0.8 + 0.2 ≈ 1 |
| Random draw | Interval containing it | Selected individual | Selected x |
|---|---|---|---|
| 0.2 | 0.1 ≤ 0.2 < 0.3 | 2 | 9 |
| 0.4 | 0.3 ≤ 0.4 < 0.8 | 3 | 8 |
| 0.6 | 0.3 ≤ 0.6 < 0.8 | 3 | 8 |
| 0.95 | 0.8 ≤ 0.95 < 1 | 4 | 9 |
Intervals are left-closed and right-open, except that a draw of exactly 1 selects the last interval. No draw here is on a boundary. Draws select parents with replacement: selecting a parent once does not remove it. Calculations use unrounded weights; the table displays rounded numbers.
Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1.
| Pool slots paired | Parent A | Parent B | Child A | Child B |
|---|---|---|---|---|
| 1 & 2 | 10 | 01 | 10 | 00 | 1000 | 1001 |
| 3 & 4 | 10 | 00 | 10 | 01 | 1001 | 1000 |
| Child | Before mutation | Instruction | After mutation |
|---|---|---|---|
| 1 | 1000 | No mutation | 1000 |
| 2 | 1001 | Flip bit 4 | 1000 |
| 3 | 1001 | No mutation | 1001 |
| 4 | 1000 | Flip bit 4 | 1001 |
| Chromosome | Decode using place values | Objective |
|---|---|---|
| 1000 | 8 + 0 + 0 + 0 = 8 | (8−7)² = 1 |
| 1000 | 8 + 0 + 0 + 0 = 8 | (8−7)² = 1 |
| 1001 | 8 + 0 + 0 + 1 = 9 | (9−7)² = 4 |
| 1001 | 8 + 0 + 0 + 1 = 9 | (9−7)² = 4 |
Add the objective scores in each population, then divide by its number of individuals.
![\"\\overline](\"math/b6ef25b43ff486a0.svg\")
![\"\\overline](\"math/818ac9c6d2d2746f.svg\")
Best objective: 1 → 1. Lower is better. The next population is [8, 8, 9, 9]. No elitism is imposed, so the best previous individual is not automatically protected.
Swarm intelligence is collective problem-solving behaviour that emerges from interactions among many relatively simple agents and their environment. It is inspired by systems such as bird flocks, fish schools and ant colonies.
\r\nParticle swarm optimization (PSO) is a population-based search method inspired by social movement. Each particle represents a candidate solution. It moves through the search space using a velocity influenced by its own best previous position and the best position known to the swarm.
\r\n| Term | Meaning | Example |
|---|---|---|
| Particle | One candidate solution | Particle A |
| Position x | The candidate’s coordinates | xA = 3 |
| Objective f(x) | The problem’s score for that candidate | f(3) = 3² = 9 |
| Velocity v | Proposed change of position | vA = −1 means move left by 1 |
| pbest | Best position that this particle has visited | A once visited x=1 with score 1 |
| gbest | Best position found by any particle | Among all personal bests, x=0.5 scores 0.25 |
![\"A](\"fig/v2_pso_memory.fcc9783a95.svg\")
For minimising x², suppose a particle visits x=3, then x=1, then x=2. Its current position is 2, but its personal best stays 1 because 1² is less than 2². The particle remembers a better past result even after a bad move.
A particle visits x=3, x=1, x=2 in that order while minimising f(x)=x². Find pbest after each visit. [2 marks · Practice]
Positions in chronological order: 3 → 1 → 2.
The best-so-far position and score.
Keep the location with the lowest objective encountered so far.
![\"f(3)=9\"](\"math/7f1b6387a96fa5e5.svg\")
No history: pbest=3, best score=9.
![\"f(1)=1<9\"](\"math/c0e5424c98d7c06b.svg\")
Replace pbest by 1.
![\"f(2)=4>1\"](\"math/e8b8ef3456701e2b.svg\")
Keep pbest=1 even though the current position is 2.
Define swarm intelligence and briefly explain one nature-inspired algorithm. [2 marks · Practice]
Swarm intelligence is collective problem-solving that emerges from interactions among simple agents. Particle swarm optimization is inspired by flocking behaviour. Each particle is a candidate solution with a position and velocity. It remembers its own best position and uses the swarm’s best position to guide movement. Repeated evaluation and movement search for a low-cost or high-fitness solution.
Yaad rakho: Position abhi kahan ho; pbest apna best past; gbest sabka best past.
"},{"id":"equation","title":"Read the PSO equation and work three individual updates","html":"\r\nNew velocity = retained old velocity + pull toward personal best + pull toward the group best. After finding that velocity, add it to the old position. You are calculating a movement, then applying it.
![\"A](\"fig/depth_pso.88e9c2a672.svg\")
Example: current x=4 and personal best=1 gives the difference 1−4=−3. A coefficient of .5 gives personal pull .5×(−3)=−1.5. The negative sign says “move left”; it does not mean a negative probability. If current x=−2 and best=1, the difference is 1−(−2)=1+2=3, which pulls right.
In vi,jt, i chooses the particle, j chooses its coordinate, and t chooses the recorded iteration. For particle B, coordinate 2, iteration 0, look at row B and the second velocity entry in the initial table. The letters identify a stored number; they are not extra multiplication factors.
![\"v_{i,j}^{t+1}=w](\"math/dac83013fb52dd1c.svg\")
![\"x_{i,j}^{t+1}=x_{i,j}^{t}+v_{i,j}^{t+1}\"](\"math/60fe3c0d98c51b4c.svg\")
| Symbol | Meaning |
|---|---|
| i | Which particle |
| j | Which coordinate of that particle |
| t / t+1 | Before / after this iteration; these are labels, not powers |
| w | Inertia weight: how much old velocity is kept |
| c₁r₁ | Combined strength of the pull to personal best |
| c₂r₂ | Combined strength of the pull to global best |
| r₁,r₂ | Random factors, normally in [0,1]; use the values supplied in an exam |
A best position to the left gives a negative difference and pulls left; a best position to the right gives a positive difference and pulls right. The multiplication sets the strength of each pull.
For one coordinate, x=4, v=0, pbest=4, gbest=2. With w=0.5, c₁=c₂=1 and r₁=r₂=0.5, find the next velocity and position. [3 marks · Practice]
x=4, v=0, pbest=4, gbest=2; w=.5; c₁r₁=.5; c₂r₂=.5.
New velocity followed by new position.
Use the three-term velocity equation, then add the new velocity to x.
![\"v_{new}=0.5v+0.5(pbest-x)+0.5(gbest-x)\"](\"math/a86502ce620dafcc.svg\")
![\"0.5\\times0=0\"](\"math/e71e2d0a61e9d301.svg\")
![\"0.5(4-(4))=0\"](\"math/a6908257a641f5f9.svg\")
![\"0.5(2-(4))=-1\"](\"math/572c7c6589c1b745.svg\")
![\"v_{new}=0+0+(-1)=-1\"](\"math/96abcf8e044d8408.svg\")
![\"x_{new}=4+(-1)=3\"](\"math/7ce7c2323916220b.svg\")
For one coordinate, x=4, v=-2, pbest=3, gbest=1. With w=0.5, c₁=c₂=1 and r₁=r₂=0.5, find the next velocity and position. [3 marks · Practice]
x=4, v=-2, pbest=3, gbest=1; w=.5; c₁r₁=.5; c₂r₂=.5.
New velocity followed by new position.
Use the three-term velocity equation, then add the new velocity to x.
![\"0.5\\times(-2)=-1\"](\"math/493f21e1dcf9ae65.svg\")
![\"0.5(3-(4))=-0.5\"](\"math/4914c1155f2d3a69.svg\")
![\"0.5(1-(4))=-1.5\"](\"math/b774cba7be997b80.svg\")
![\"v_{new}=(-1)+(-0.5)+(-1.5)=-3\"](\"math/fb1a87666bd393ac.svg\")
![\"x_{new}=4+(-3)=1\"](\"math/87db5516b0eef4a5.svg\")
For one coordinate, x=-2, v=1, pbest=-1, gbest=1. With w=0.5, c₁=c₂=1 and r₁=r₂=0.5, find the next velocity and position. [3 marks · Practice]
x=-2, v=1, pbest=-1, gbest=1; w=.5; c₁r₁=.5; c₂r₂=.5.
New velocity followed by new position.
Use the three-term velocity equation, then add the new velocity to x.
![\"0.5\\times1=0.5\"](\"math/12ea92d29476784c.svg\")
![\"0.5(-1-(-2))=0.5\"](\"math/61d7bcc89fa9fd8a.svg\")
![\"0.5(1-(-2))=1.5\"](\"math/85cc697d8abbcf2c.svg\")
![\"v_{new}=0.5+0.5+1.5=2.5\"](\"math/bc8dd34ee10c4d4d.svg\")
![\"x_{new}=(-2)+2.5=0.5\"](\"math/264c82a55bf4620a.svg\")
First-iteration shortcut: if pbest was just initialised to x, the personal difference is zero. This follows from the initialisation; it is not a permanent property of PSO. A gbest particle can still move because its inertia term can be nonzero.
Yaad rakho: Old speed ka hissa + apne best ki pull + group best ki pull; phir new speed ko position mein add.
"},{"id":"two-particles","title":"Two particles, two iterations: start with the simplest minimisation","html":"\r\nA synchronous iteration is a class photograph of the old swarm: every particle uses that same snapshot of gbest. First calculate all new positions, then compare all new scores and update memories. If particle A discovers a better position halfway through the calculations, B still uses the previous snapshot during this iteration. B can use the improved gbest in the next iteration.
Work down the page: each particle is fully updated, then all scores and memories are compared. Keep at least six digits internally for longer decimal examples; the final table displays rounded values.
Minimise using 2 particles. Use the initial data below and perform exactly two PSO iterations, reporting position, velocity, pbest and gbest after each iteration. [3 marks · Practice; expanded teaching solution]
| Particle | Initial position x⁰ | Initial velocity v⁰ |
|---|---|---|
| A | (-2) | (0) |
| B | (4) | (0) |
w=0.5, c₁=1, c₂=1, r₁=0.5, r₂=0.5. Initialise each pbest at its initial position. Use these random coefficients unchanged in both iterations. No bounds or velocity clamp are imposed in this example. Updates are synchronous: every particle uses the same previous-iteration gbest.
The complete swarm state after iterations 1 and 2, and the best location found.
PSO combines inertia, personal memory and shared memory. Subtract positions coordinate by coordinate. Evaluate the objective only after moving. For minimisation, replace a best only when its score decreases.
![\"v^{t+1}=wv^t+c_1r_1(pbest^t-x^t)+c_2r_2(gbest^t-x^t)\"](\"math/8a9f3b6c402bae75.svg\")
![\"x^{t+1}=x^t+v^{t+1}\"](\"math/3a35d97963567317.svg\")
Particle A
![\"f(-2)=(-2)^2=4=4\"](\"math/73d4f3be5484ad52.svg\")
Particle B
![\"f(4)=4^2=16=16\"](\"math/312433d364d9cfab.svg\")
pbest is each initial position because there is no earlier history. The smallest objective decides gbest.
![\"gbest^0=(-2)\"](\"math/130e21267aeccc61.svg\")
Its score is 4. Store the position, not only the score.
All particles in this iteration use gbest = (-2). pbest values are A: (-2), B: (4).
Only coordinate
Inertia = keep a fraction of the old velocity.
![\"I=0.5\\times0=0\"](\"math/6f4daee36ea96185.svg\")
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=-2-(-2)=0\"](\"math/1d782a69e1d0fcb8.svg\")
Multiply the two personal coefficients, then multiply by the difference.
![\"c_1r_1=1\\times0.5=0.5\"](\"math/19e8c86525a0eadb.svg\")
![\"C=0.5\\times(0)=0\"](\"math/660e9363a712569b.svg\")
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=-2-(-2)=0\"](\"math/6f82367c11e33631.svg\")
Multiply the social coefficients, then multiply by that difference.
![\"c_2r_2=1\\times0.5=0.5\"](\"math/1cc595b44ce899fa.svg\")
![\"S=0.5\\times(0)=0\"](\"math/d00ab76bdfccf855.svg\")
Add the three velocity contributions.
![\"v^{(1)}=0+0+0=0\"](\"math/dec9907b2180e0af.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(1)}=(-2)+0=-2\"](\"math/4e58cfc2f623e47e.svg\")
Only coordinate
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=4-(4)=0\"](\"math/ad073051db57b851.svg\")
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=-2-(4)=-6\"](\"math/eb7d3c2c066600f2.svg\")
Multiply the social coefficients, then multiply by that difference.
![\"S=0.5\\times(-6)=-3\"](\"math/4dd7f304ffa4f233.svg\")
Add the three velocity contributions.
![\"v^{(1)}=0+0+(-3)=-3\"](\"math/c239eeeeeca7f890.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(1)}=4+(-3)=1\"](\"math/d71458dcf56b7a77.svg\")
Particle A
Particle B
![\"f(1)=1^2=1=1\"](\"math/872b6624767f4385.svg\")
| Particle | Old pbest score | New score | Decision | pbest position |
|---|---|---|---|---|
| A | 4 | 4 | Keep old: tie | (-2) |
| B | 16 | 1 | Replace: new score is lower | (1) |
Compare the retained pbest scores. gbest = (1), with objective 1.
| Particle | Position | Velocity | pbest |
|---|---|---|---|
| A | (-2) | (0) | (-2) |
| B | (1) | (-3) | (1) |
All particles in this iteration use gbest = (1). pbest values are A: (-2), B: (1).
Only coordinate
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=1-(-2)=3\"](\"math/9102a4aae4f9ab55.svg\")
Multiply the social coefficients, then multiply by that difference.
![\"S=0.5\\times(3)=1.5\"](\"math/9efd684a8627ac87.svg\")
Add the three velocity contributions.
![\"v^{(2)}=0+0+1.5=1.5\"](\"math/25e29c2b2380aded.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(2)}=(-2)+1.5=-0.5\"](\"math/0af67fb865c5f5c7.svg\")
Only coordinate
Inertia = keep a fraction of the old velocity.
![\"I=0.5\\times(-3)=-1.5\"](\"math/02f316319733ea90.svg\")
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=1-(1)=0\"](\"math/a352ebe71945e68e.svg\")
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=1-(1)=0\"](\"math/d4af3bfbae14d68f.svg\")
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
![\"v^{(2)}=(-1.5)+0+0=-1.5\"](\"math/4e2008e440fe556c.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(2)}=1+(-1.5)=-0.5\"](\"math/700c5bb73aa33b26.svg\")
Particle A
![\"f(-0.5)=(-0.5)^2=0.25=0.25\"](\"math/6c453bb065df6e4a.svg\")
Particle B
| Particle | Old pbest score | New score | Decision | pbest position |
|---|---|---|---|---|
| A | 4 | 0.25 | Replace: new score is lower | (-0.5) |
| B | 1 | 0.25 | Replace: new score is lower | (-0.5) |
Compare the retained pbest scores. gbest = (-0.5), with objective 0.25.
| Particle | Position | Velocity | pbest |
|---|---|---|---|
| A | (-0.5) | (1.5) | (-0.5) |
| B | (-0.5) | (-1.5) | (-0.5) |
Work down the page: each particle is fully updated, then all scores and memories are compared. Keep at least six digits internally for longer decimal examples; the final table displays rounded values.
Minimise using 2 particles. Use the initial data below and perform exactly two PSO iterations, reporting position, velocity, pbest and gbest after each iteration. [3 marks · Practice; expanded teaching solution]
| Particle | Initial position x⁰ | Initial velocity v⁰ |
|---|---|---|
| A | (1) | (4) |
| B | (3) | (-1) |
w=0.5, c₁=1, c₂=1, r₁=0.5, r₂=0.5. Initialise each pbest at its initial position. Use these random coefficients unchanged in both iterations. No bounds or velocity clamp are imposed in this example. Updates are synchronous: every particle uses the same previous-iteration gbest.
The complete swarm state after iterations 1 and 2, and the best location found.
PSO combines inertia, personal memory and shared memory. Subtract positions coordinate by coordinate. Evaluate the objective only after moving. For minimisation, replace a best only when its score decreases.
Particle A
Particle B
![\"f(3)=3^2=9=9\"](\"math/519522f49346438a.svg\")
pbest is each initial position because there is no earlier history. The smallest objective decides gbest.
![\"gbest^0=(1)\"](\"math/d6d8eee6e10c4922.svg\")
Its score is 1. Store the position, not only the score.
All particles in this iteration use gbest = (1). pbest values are A: (1), B: (3).
Only coordinate
Inertia = keep a fraction of the old velocity.
![\"I=0.5\\times4=2\"](\"math/e009c778bab94586.svg\")
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
![\"v^{(1)}=2+0+0=2\"](\"math/25c7782eb44a678a.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(1)}=1+2=3\"](\"math/8e886f0f9174fead.svg\")
Only coordinate
Inertia = keep a fraction of the old velocity.
![\"I=0.5\\times(-1)=-0.5\"](\"math/dc7ed342b0db88bc.svg\")
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=3-(3)=0\"](\"math/114e8d23bb26d062.svg\")
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=1-(3)=-2\"](\"math/57f4bea8a1c6b79f.svg\")
Multiply the social coefficients, then multiply by that difference.
![\"S=0.5\\times(-2)=-1\"](\"math/96dd8f64f9d9a341.svg\")
Add the three velocity contributions.
![\"v^{(1)}=(-0.5)+0+(-1)=-1.5\"](\"math/e15db2dfd9b255cf.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(1)}=3+(-1.5)=1.5\"](\"math/76c197431121b5e1.svg\")
Particle A
Particle B
![\"f(1.5)=1.5^2=2.25=2.25\"](\"math/b61e97d213b18164.svg\")
| Particle | Old pbest score | New score | Decision | pbest position |
|---|---|---|---|---|
| A | 1 | 9 | Keep old: new score is worse | (1) |
| B | 9 | 2.25 | Replace: new score is lower | (1.5) |
Compare the retained pbest scores. gbest = (1), with objective 1.
| Particle | Position | Velocity | pbest |
|---|---|---|---|
| A | (3) | (2) | (1) |
| B | (1.5) | (-1.5) | (1.5) |
All particles in this iteration use gbest = (1). pbest values are A: (1), B: (1.5).
Only coordinate
Inertia = keep a fraction of the old velocity.
![\"I=0.5\\times2=1\"](\"math/4d6486404a7509ba.svg\")
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=1-(3)=-2\"](\"math/e56ef95c25fb8bde.svg\")
Multiply the two personal coefficients, then multiply by the difference.
![\"C=0.5\\times(-2)=-1\"](\"math/a08a6d09414479ce.svg\")
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
![\"v^{(2)}=1+(-1)+(-1)=-1\"](\"math/a51eec1438d949c1.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(2)}=3+(-1)=2\"](\"math/60104aa04f06aad8.svg\")
Only coordinate
Inertia = keep a fraction of the old velocity.
![\"I=0.5\\times(-1.5)=-0.75\"](\"math/16e3f0369a50c03f.svg\")
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=1.5-(1.5)=0\"](\"math/dcb7c97c6295715d.svg\")
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=1-(1.5)=-0.5\"](\"math/2e6ff91ff778f21f.svg\")
Multiply the social coefficients, then multiply by that difference.
![\"S=0.5\\times(-0.5)=-0.25\"](\"math/418a15cd3fa3bbcc.svg\")
Add the three velocity contributions.
![\"v^{(2)}=(-0.75)+0+(-0.25)=-1\"](\"math/c528696fa4ba4081.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(2)}=1.5+(-1)=0.5\"](\"math/905c77866baf10ae.svg\")
Particle A
![\"f(2)=2^2=4=4\"](\"math/00c58d851ae8132e.svg\")
Particle B
![\"f(0.5)=0.5^2=0.25=0.25\"](\"math/4c640081b99e1be6.svg\")
| Particle | Old pbest score | New score | Decision | pbest position |
|---|---|---|---|---|
| A | 1 | 4 | Keep old: new score is worse | (1) |
| B | 2.25 | 0.25 | Replace: new score is lower | (0.5) |
Compare the retained pbest scores. gbest = (0.5), with objective 0.25.
| Particle | Position | Velocity | pbest |
|---|---|---|---|
| A | (2) | (-1) | (1) |
| B | (0.5) | (-1) | (0.5) |
Yaad rakho: Naya result kharab ho toh position toh badlegi, pbest nahi.
"},{"id":"two-coordinates","title":"Two-dimensional PSO: keep the coordinates separate","html":"\r\nA two-dimensional particle has two entries in its position and two in its velocity. Apply the scalar velocity calculation once to coordinate 1 and once to coordinate 2. Keep them in the same order to form the new pair. Only after both entries are ready do you insert the pair into f(x₁,x₂). Adding x₁ and x₂ into a single position would lose information.
Work down the page: each particle is fully updated, then all scores and memories are compared. Keep at least six digits internally for longer decimal examples; the final table displays rounded values.
Minimise ![\"f(x)=x_1^2+x_2^2\"](\"math/a02144dad2ff9c8a.svg\")
using 2 particles. Use the initial data below and perform exactly two PSO iterations, reporting position, velocity, pbest and gbest after each iteration. [3 marks · Practice; expanded teaching solution]
| Particle | Initial position x⁰ | Initial velocity v⁰ |
|---|---|---|
| A | (1, 1) | (0, 0) |
| B | (3, 2) | (0, 0) |
w=0.5, c₁=1, c₂=1, r₁=0.5, r₂=0.5. Initialise each pbest at its initial position. Use these random coefficients unchanged in both iterations. No bounds or velocity clamp are imposed in this example. Updates are synchronous: every particle uses the same previous-iteration gbest.
The complete swarm state after iterations 1 and 2, and the best location found.
PSO combines inertia, personal memory and shared memory. Subtract positions coordinate by coordinate. Evaluate the objective only after moving. For minimisation, replace a best only when its score decreases.
Particle A
![\"f(1,](\"math/e7fb8b8ea59ba406.svg\")
Particle B
![\"f(3,](\"math/952bb71a534d7b48.svg\")
pbest is each initial position because there is no earlier history. The smallest objective decides gbest.
![\"gbest^0=(1,](\"math/bae2559b2265b2a0.svg\")
Its score is 2. Store the position, not only the score.
All particles in this iteration use gbest = (1, 1). pbest values are A: (1, 1), B: (3, 2).
Coordinate 1
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(1)}=1+0=1\"](\"math/bc4c5112ff057d33.svg\")
Coordinate 2
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Coordinate 1
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
![\"v^{(1)}=0+0+(-1)=-1\"](\"math/000e26bf52e8a50e.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(1)}=3+(-1)=2\"](\"math/a890dd5de2855674.svg\")
Coordinate 2
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=2-(2)=0\"](\"math/e7f9deeb68816f5a.svg\")
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=1-(2)=-1\"](\"math/2fe1a995f4b4afef.svg\")
Multiply the social coefficients, then multiply by that difference.
![\"S=0.5\\times(-1)=-0.5\"](\"math/38f7adbc6f0de36b.svg\")
Add the three velocity contributions.
![\"v^{(1)}=0+0+(-0.5)=-0.5\"](\"math/a489ee8326d49a33.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(1)}=2+(-0.5)=1.5\"](\"math/51c2ec0b6a29b186.svg\")
Particle A
Particle B
![\"f(2,](\"math/d342f4df799385a1.svg\")
| Particle | Old pbest score | New score | Decision | pbest position |
|---|---|---|---|---|
| A | 2 | 2 | Keep old: tie | (1, 1) |
| B | 13 | 6.25 | Replace: new score is lower | (2, 1.5) |
Compare the retained pbest scores. gbest = (1, 1), with objective 2.
| Particle | Position | Velocity | pbest |
|---|---|---|---|
| A | (1, 1) | (0, 0) | (1, 1) |
| B | (2, 1.5) | (-1, -0.5) | (2, 1.5) |
All particles in this iteration use gbest = (1, 1). pbest values are A: (1, 1), B: (2, 1.5).
Coordinate 1
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
![\"v^{(2)}=0+0+0=0\"](\"math/8ba6a1d6c9203afb.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(2)}=1+0=1\"](\"math/a7841f9dac4f92c7.svg\")
Coordinate 2
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Coordinate 1
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
![\"v^{(2)}=(-0.5)+0+(-0.5)=-1\"](\"math/31b640bb301f8dce.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(2)}=2+(-1)=1\"](\"math/144d4f69ce96cfe2.svg\")
Coordinate 2
Inertia = keep a fraction of the old velocity.
![\"I=0.5\\times(-0.5)=-0.25\"](\"math/84dfd42953d2811a.svg\")
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
![\"v^{(2)}=(-0.25)+0+(-0.25)=-0.5\"](\"math/8789b37bca837a83.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(2)}=1.5+(-0.5)=1\"](\"math/b71d05464709fcf3.svg\")
Particle A
Particle B
| Particle | Old pbest score | New score | Decision | pbest position |
|---|---|---|---|---|
| A | 2 | 2 | Keep old: tie | (1, 1) |
| B | 6.25 | 2 | Replace: new score is lower | (1, 1) |
Compare the retained pbest scores. gbest = (1, 1), with objective 2.
| Particle | Position | Velocity | pbest |
|---|---|---|---|
| A | (1, 1) | (0, 0) | (1, 1) |
| B | (1, 1) | (-1, -0.5) | (1, 1) |
Work down the page: each particle is fully updated, then all scores and memories are compared. Keep at least six digits internally for longer decimal examples; the final table displays rounded values.
The positions, velocities and coefficients are transcribed from the class photo. We use f(x₁,x₂)=x₁²+x₂²−x₁x₂, minimisation, initial pbest=x, fixed supplied random factors and synchronous updates. The photo does not itself specify all these conventions; they are stated here to make the two-iteration extension unambiguous.
Minimise ![\"f(x)=x_1^2+x_2^2-x_1x_2\"](\"math/ef9d55cf7c043920.svg\")
using 3 particles. Use the initial data below and perform exactly two PSO iterations, reporting position, velocity, pbest and gbest after each iteration. [10 marks · Class data; two-iteration minimisation extension]
| Particle | Initial position x⁰ | Initial velocity v⁰ |
|---|---|---|
| A | (3.14, 7.21) | (0.1, 0.3) |
| B | (4.15, 7.31) | (0.5, 0.1) |
| C | (6.17, 8.27) | (0.2, 0) |
w=0.5, c₁=1.2, c₂=1.2, r₁=0.5, r₂=0.1. Initialise each pbest at its initial position. Use these random coefficients unchanged in both iterations. No bounds or velocity clamp are imposed in this example. Updates are synchronous: every particle uses the same previous-iteration gbest.
The complete swarm state after iterations 1 and 2, and the best location found.
PSO combines inertia, personal memory and shared memory. Subtract positions coordinate by coordinate. Evaluate the objective only after moving. For minimisation, replace a best only when its score decreases.
Particle A
![\"f(3.14,](\"math/bf321f61ed66837c.svg\")
![\"=9.8596+51.9841-22.6394=39.2043\"](\"math/d54138730c346bbf.svg\")
Particle B
![\"f(4.15,](\"math/1980ac03eb9e462c.svg\")
![\"=17.2225+53.4361-30.3365=40.3221\"](\"math/ba408f391fd916c4.svg\")
Particle C
![\"f(6.17,](\"math/ac0fc4a2897f8794.svg\")
![\"=38.0689+68.3929-51.0259=55.4359\"](\"math/558e7a3b9a40d34e.svg\")
pbest is each initial position because there is no earlier history. The smallest objective decides gbest.
![\"gbest^0=(3.14,](\"math/567a650862d29da5.svg\")
Its score is 39.2043. Store the position, not only the score.
All particles in this iteration use gbest = (3.14, 7.21). pbest values are A: (3.14, 7.21), B: (4.15, 7.31), C: (6.17, 8.27).
Coordinate 1
Inertia = keep a fraction of the old velocity.
![\"I=0.5\\times0.1=0.05\"](\"math/4f87680527f659f2.svg\")
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=3.14-(3.14)=0\"](\"math/d3ab4c94a5f284d2.svg\")
Multiply the two personal coefficients, then multiply by the difference.
![\"c_1r_1=1.2\\times0.5=0.6\"](\"math/16c273adcc74f25a.svg\")
![\"C=0.6\\times(0)=0\"](\"math/89df484c365b908d.svg\")
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=3.14-(3.14)=0\"](\"math/6212c336dbecc343.svg\")
Multiply the social coefficients, then multiply by that difference.
![\"c_2r_2=1.2\\times0.1=0.12\"](\"math/e19d636861b5bba6.svg\")
![\"S=0.12\\times(0)=0\"](\"math/16804d70bddc0ccc.svg\")
Add the three velocity contributions.
![\"v^{(1)}=0.05+0+0=0.05\"](\"math/6950d20a25859fe4.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(1)}=3.14+0.05=3.19\"](\"math/de8ab0140b454ff9.svg\")
Coordinate 2
Inertia = keep a fraction of the old velocity.
![\"I=0.5\\times0.3=0.15\"](\"math/eb88185c1e62a59c.svg\")
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=7.21-(7.21)=0\"](\"math/d7b5c8d3808f4c73.svg\")
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=7.21-(7.21)=0\"](\"math/ca02c52199da4f04.svg\")
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
![\"v^{(1)}=0.15+0+0=0.15\"](\"math/c7ca03a89be4beea.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(1)}=7.21+0.15=7.36\"](\"math/f693927b5eebad50.svg\")
Coordinate 1
Inertia = keep a fraction of the old velocity.
![\"I=0.5\\times0.5=0.25\"](\"math/db302571bbe707b4.svg\")
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=4.15-(4.15)=0\"](\"math/2ddd2b101f36cd26.svg\")
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=3.14-(4.15)=-1.01\"](\"math/1978fefc1be6f223.svg\")
Multiply the social coefficients, then multiply by that difference.
![\"S=0.12\\times(-1.01)=-0.1212\"](\"math/060e9302bdac4457.svg\")
Add the three velocity contributions.
![\"v^{(1)}=0.25+0+(-0.1212)=0.1288\"](\"math/08fa8ab93dce4fd3.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(1)}=4.15+0.1288=4.2788\"](\"math/bb72c3734ea8a809.svg\")
Coordinate 2
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=7.31-(7.31)=0\"](\"math/bceb9619d4320629.svg\")
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=7.21-(7.31)=-0.1\"](\"math/2fc0a8e88db078cd.svg\")
Multiply the social coefficients, then multiply by that difference.
![\"S=0.12\\times(-0.1)=-0.012\"](\"math/f8e69de5d9161da2.svg\")
Add the three velocity contributions.
![\"v^{(1)}=0.05+0+(-0.012)=0.038\"](\"math/3dca081cd95c2dd0.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(1)}=7.31+0.038=7.348\"](\"math/644604a12b5bcc31.svg\")
Coordinate 1
Inertia = keep a fraction of the old velocity.
![\"I=0.5\\times0.2=0.1\"](\"math/3c5f4d0f2cda8136.svg\")
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=6.17-(6.17)=0\"](\"math/13f27d08182e6e02.svg\")
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=3.14-(6.17)=-3.03\"](\"math/93416da8e366b396.svg\")
Multiply the social coefficients, then multiply by that difference.
![\"S=0.12\\times(-3.03)=-0.3636\"](\"math/a59d670bba6f5dac.svg\")
Add the three velocity contributions.
![\"v^{(1)}=0.1+0+(-0.3636)=-0.2636\"](\"math/3fdb1b4bffd3874d.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(1)}=6.17+(-0.2636)=5.9064\"](\"math/40f8a4a790d7094f.svg\")
Coordinate 2
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=8.27-(8.27)=0\"](\"math/d343c09e8978bf19.svg\")
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=7.21-(8.27)=-1.06\"](\"math/ea093f799c7411f8.svg\")
Multiply the social coefficients, then multiply by that difference.
![\"S=0.12\\times(-1.06)=-0.1272\"](\"math/b9fa5298c9c7cb16.svg\")
Add the three velocity contributions.
![\"v^{(1)}=0+0+(-0.1272)=-0.1272\"](\"math/28af9633e743719d.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(1)}=8.27+(-0.1272)=8.1428\"](\"math/dba6d056546c8535.svg\")
Particle A
![\"f(3.19,](\"math/3ae29bc527e20248.svg\")
![\"=10.1761+54.1696-23.4784=40.8673\"](\"math/b9b0c1a8c0f513a8.svg\")
Particle B
![\"f(4.2788,](\"math/c476266715700a96.svg\")
![\"=18.30812944+53.993104-31.4406224=40.86061104\"](\"math/1baa4a09739c8180.svg\")
Particle C
![\"f(5.9064,](\"math/3a9fac3b191a1ffa.svg\")
![\"=34.88556096+66.30519184-48.09463392=53.09611888\"](\"math/273655b1059b7aef.svg\")
| Particle | Old pbest score | New score | Decision | pbest position |
|---|---|---|---|---|
| A | 39.2043 | 40.8673 | Keep old: new score is worse | (3.14, 7.21) |
| B | 40.3221 | 40.86061104 | Keep old: new score is worse | (4.15, 7.31) |
| C | 55.4359 | 53.09611888 | Replace: new score is lower | (5.9064, 8.1428) |
Compare the retained pbest scores. gbest = (3.14, 7.21), with objective 39.2043.
| Particle | Position | Velocity | pbest |
|---|---|---|---|
| A | (3.19, 7.36) | (0.05, 0.15) | (3.14, 7.21) |
| B | (4.2788, 7.348) | (0.1288, 0.038) | (4.15, 7.31) |
| C | (5.9064, 8.1428) | (-0.2636, -0.1272) | (5.9064, 8.1428) |
All particles in this iteration use gbest = (3.14, 7.21). pbest values are A: (3.14, 7.21), B: (4.15, 7.31), C: (5.9064, 8.1428).
Coordinate 1
Inertia = keep a fraction of the old velocity.
![\"I=0.5\\times0.05=0.025\"](\"math/e4bd35a53c831c76.svg\")
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=3.14-(3.19)=-0.05\"](\"math/cc2af53bc1a4a60e.svg\")
Multiply the two personal coefficients, then multiply by the difference.
![\"C=0.6\\times(-0.05)=-0.03\"](\"math/db60d8d854aa5ee6.svg\")
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=3.14-(3.19)=-0.05\"](\"math/1dbfd156a9aafbb3.svg\")
Multiply the social coefficients, then multiply by that difference.
![\"S=0.12\\times(-0.05)=-0.006\"](\"math/e7739769871dabe8.svg\")
Add the three velocity contributions.
![\"v^{(2)}=0.025+(-0.03)+(-0.006)=-0.011\"](\"math/4e6287c93cb39658.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(2)}=3.19+(-0.011)=3.179\"](\"math/04d374313aaab72d.svg\")
Coordinate 2
Inertia = keep a fraction of the old velocity.
![\"I=0.5\\times0.15=0.075\"](\"math/e9c97812dc9ecc86.svg\")
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=7.21-(7.36)=-0.15\"](\"math/08869d3b9cefb65b.svg\")
Multiply the two personal coefficients, then multiply by the difference.
![\"C=0.6\\times(-0.15)=-0.09\"](\"math/b80cfa7b094e421e.svg\")
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=7.21-(7.36)=-0.15\"](\"math/8346d41c793ff815.svg\")
Multiply the social coefficients, then multiply by that difference.
![\"S=0.12\\times(-0.15)=-0.018\"](\"math/bb6864a422495c83.svg\")
Add the three velocity contributions.
![\"v^{(2)}=0.075+(-0.09)+(-0.018)=-0.033\"](\"math/df1003fde626354c.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(2)}=7.36+(-0.033)=7.327\"](\"math/04ced9552ec0a2fc.svg\")
Coordinate 1
Inertia = keep a fraction of the old velocity.
![\"I=0.5\\times0.1288=0.0644\"](\"math/c7a3d7f19f863ddf.svg\")
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=4.15-(4.2788)=-0.1288\"](\"math/a86b46d750156354.svg\")
Multiply the two personal coefficients, then multiply by the difference.
![\"C=0.6\\times(-0.1288)=-0.07728\"](\"math/9088271ed20d40af.svg\")
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=3.14-(4.2788)=-1.1388\"](\"math/ad7db7d0feab17e6.svg\")
Multiply the social coefficients, then multiply by that difference.
![\"S=0.12\\times(-1.1388)=-0.136656\"](\"math/115797e01c1867dd.svg\")
Add the three velocity contributions.
![\"v^{(2)}=0.0644+(-0.07728)+(-0.136656)=-0.149536\"](\"math/ebf74e4d56d1b7b9.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(2)}=4.2788+(-0.149536)=4.129264\"](\"math/eae7d898d5a551e1.svg\")
Coordinate 2
Inertia = keep a fraction of the old velocity.
![\"I=0.5\\times0.038=0.019\"](\"math/45baaa3ae928b8b4.svg\")
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=7.31-(7.348)=-0.038\"](\"math/c86909b46639b9bb.svg\")
Multiply the two personal coefficients, then multiply by the difference.
![\"C=0.6\\times(-0.038)=-0.0228\"](\"math/241c822838613f93.svg\")
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=7.21-(7.348)=-0.138\"](\"math/fe0c41420480a542.svg\")
Multiply the social coefficients, then multiply by that difference.
![\"S=0.12\\times(-0.138)=-0.01656\"](\"math/e9a578685c025bc9.svg\")
Add the three velocity contributions.
![\"v^{(2)}=0.019+(-0.0228)+(-0.01656)=-0.02036\"](\"math/9363048bb26f273e.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(2)}=7.348+(-0.02036)=7.32764\"](\"math/d30e07ac98dd2bba.svg\")
Coordinate 1
Inertia = keep a fraction of the old velocity.
![\"I=0.5\\times(-0.2636)=-0.1318\"](\"math/cfc400d9e06f7358.svg\")
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=5.9064-(5.9064)=0\"](\"math/786ba18b3b62c512.svg\")
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=3.14-(5.9064)=-2.7664\"](\"math/6b3316f35248ffdf.svg\")
Multiply the social coefficients, then multiply by that difference.
![\"S=0.12\\times(-2.7664)=-0.331968\"](\"math/b32a5de2e3b2cefc.svg\")
Add the three velocity contributions.
![\"v^{(2)}=(-0.1318)+0+(-0.331968)=-0.463768\"](\"math/de7212cc169aa2f9.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(2)}=5.9064+(-0.463768)=5.442632\"](\"math/ab68ef2fa1b527a9.svg\")
Coordinate 2
Inertia = keep a fraction of the old velocity.
![\"I=0.5\\times(-0.1272)=-0.0636\"](\"math/d56f94fb044781a6.svg\")
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=8.1428-(8.1428)=0\"](\"math/9923de46bc28d517.svg\")
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=7.21-(8.1428)=-0.9328\"](\"math/024d71da7fe2cc3c.svg\")
Multiply the social coefficients, then multiply by that difference.
![\"S=0.12\\times(-0.9328)=-0.111936\"](\"math/71d9611ef2b3e668.svg\")
Add the three velocity contributions.
![\"v^{(2)}=(-0.0636)+0+(-0.111936)=-0.175536\"](\"math/1a832b53cf0e106d.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(2)}=8.1428+(-0.175536)=7.967264\"](\"math/17e63ccf7082af52.svg\")
Particle A
![\"f(3.179,](\"math/76bb4ad3f6895999.svg\")
![\"=10.106041+53.684929-23.292533=40.498437\"](\"math/e1463199434aa355.svg\")
Particle B
![\"f(4.129264,](\"math/0ac428234bdb63ad.svg\")
![\"=17.050821182+53.69430797-30.257760057=40.487369094\"](\"math/a32da9cd0f25ac36.svg\")
Particle C
![\"f(5.442632,](\"math/515e8ddbaa301792.svg\")
![\"=29.622243087+63.477295646-43.362885999=49.736652734\"](\"math/a9051d35de42b6a1.svg\")
| Particle | Old pbest score | New score | Decision | pbest position |
|---|---|---|---|---|
| A | 39.2043 | 40.498437 | Keep old: new score is worse | (3.14, 7.21) |
| B | 40.3221 | 40.487369094 | Keep old: new score is worse | (4.15, 7.31) |
| C | 53.09611888 | 49.736652734 | Replace: new score is lower | (5.442632, 7.967264) |
Compare the retained pbest scores. gbest = (3.14, 7.21), with objective 39.2043.
| Particle | Position | Velocity | pbest |
|---|---|---|---|
| A | (3.179, 7.327) | (-0.011, -0.033) | (3.14, 7.21) |
| B | (4.129264, 7.32764) | (-0.149536, -0.02036) | (4.15, 7.31) |
| C | (5.442632, 7.967264) | (-0.463768, -0.175536) | (5.442632, 7.967264) |
![\"Motion](\"fig/v2_pso_class_paths.961831cda1.svg\")
| Advantages | Limitations |
|---|---|
| Simple position/velocity updates; does not require objective derivatives | Can converge prematurely to a local region |
| Uses both individual and shared experience | Results depend on coefficients, initialisation and random factors |
| Useful for many continuous optimisation problems | Constraints and discrete encodings need suitable handling |
| Keeps best-so-far memories | Does not guarantee the global optimum in a finite run |
Explain one application of PSO. [3 marks · Practice]
In controller tuning, a particle can represent the gain values of a controller. The objective is a performance cost, such as tracking error plus an overshoot penalty. Each proposed gain vector is evaluated in the model. PSO updates candidate gains using individual and swarm bests. The best feasible gains found are selected and then validated on the actual system. A lower training cost alone does not prove safe real-world behaviour.
| Exam situation | Action |
|---|---|
| Minimisation | Lower f is better |
| Maximisation | Higher f is better; do not reverse velocity differences |
| Tie in best score | Keep the previous best unless a tie rule is supplied |
| Different random factors each iteration | Read the relevant row; do not reuse old factors |
| Synchronous update | Freeze gbest for all particles within one iteration |
| Position/velocity outside a limit | Use the limit-handling rule explicitly supplied; state an assumption if absent |
Yaad rakho: Velocity → position → score → pbest → gbest. Ye order har iteration mein same.
"}]},{"id":"fuzzy-relations","title":"Fuzzy relations and matrices","unit":"Unit 2","scope":"Longer Unit 2 numerical practice, also useful for TEE.","sections":[{"id":"cartesian","title":"Cartesian product: every cell explained","html":"\r\nLet A₁=.6 and A₂=.2 label two rows. Let B₁=.3, B₂=.8 and B₃=.5 label three columns. Cell R₂₃ belongs to row 2 and column 3, so use A₂=.2 and B₃=.5. Their min is .2. You do not multiply 2×3; those subscripts only locate the cell.
![\"R_{23}=\\min(A_2,B_3)=\\min(0.2,0.5)=0.2\"](\"math/59fc3a309a7ab813.svg\")
There are two choices of row and three of column, so fill six cells in a 2×3 rectangle. Keep the row and column labels when reporting a relation.
A fuzzy set gives a membership to one element. A fuzzy relation gives a degree to an ordered pair. A matrix is a convenient table: rows name the first element and columns name the second.
For the standard min-based Cartesian product R=A×B, put the A memberships down the rows and the B memberships across the columns. Each cell is the smaller of its row and column memberships.
![\"R_{ij}=\\min(\\mu_A(a_i),\\mu_B(b_j))\"](\"math/1132359c819cab8a.svg\")
Find the fuzzy Cartesian product A×B for memberships A=[0.6, 0.2] and B=[0.3, 0.8, 0.5]. [3 marks · Practice]
Ordered A memberships [0.6, 0.2]; B memberships [0.3, 0.8, 0.5].
A 2×3 relation matrix.
Every row corresponds to an A element and every column to a B element; apply min in every cell.
![\"R_{ij}=\\min(A_i,B_j)\"](\"math/50b8f8fcf2c15000.svg\")
2 row elements × 3 column elements.
| A ↓ / B → | 0.3 | 0.8 | 0.5 |
|---|---|---|---|
| 0.6 | min(0.6,0.3) = 0.3 | min(0.6,0.8) = 0.6 | min(0.6,0.5) = 0.5 |
| 0.2 | min(0.2,0.3) = 0.2 | min(0.2,0.8) = 0.2 | min(0.2,0.5) = 0.2 |
![\"R=\\begin{bmatrix}0.3&0.6&0.5\\\\0.2&0.2&0.2\\end{bmatrix}\"](\"math/eb5e24dcbf5263d2.svg\")
Find the fuzzy Cartesian product A×B for memberships A=[1, 0.4] and B=[0.7, 0]. [3 marks · Practice]
Ordered A memberships [1, 0.4]; B memberships [0.7, 0].
A 2×2 relation matrix.
Every row corresponds to an A element and every column to a B element; apply min in every cell.
2 row elements × 2 column elements.
| A ↓ / B → | 0.7 | 0 |
|---|---|---|
| 1 | min(1,0.7) = 0.7 | min(1,0) = 0 |
| 0.4 | min(0.4,0.7) = 0.4 | min(0.4,0) = 0 |
![\"R=\\begin{bmatrix}0.7&0\\\\0.4&0\\end{bmatrix}\"](\"math/00a9ec4b6529a3a5.svg\")
Find the fuzzy Cartesian product A×B for memberships A=[0.2, 0.9, 0.5] and B=[0.6, 0.1]. [3 marks · Practice]
Ordered A memberships [0.2, 0.9, 0.5]; B memberships [0.6, 0.1].
A 3×2 relation matrix.
Every row corresponds to an A element and every column to a B element; apply min in every cell.
3 row elements × 2 column elements.
| A ↓ / B → | 0.6 | 0.1 |
|---|---|---|
| 0.2 | min(0.2,0.6) = 0.2 | min(0.2,0.1) = 0.1 |
| 0.9 | min(0.9,0.6) = 0.6 | min(0.9,0.1) = 0.1 |
| 0.5 | min(0.5,0.6) = 0.5 | min(0.5,0.1) = 0.1 |
![\"R=\\begin{bmatrix}0.2&0.1\\\\0.6&0.1\\\\0.5&0.1\\end{bmatrix}\"](\"math/587e0aaf2cb10cc7.svg\")
Yaad rakho: Row ka degree aur column ka degree—dono ka min us cell mein.
"},{"id":"composition","title":"Max–min composition: three complete matrix examples","html":"\r\nFor T₁₁, start at u₁ and end at w₁. If the middle domain has v₁ and v₂, there are two candidate paths. Their strengths come from matching row 1 of R with column 1 of S.
| Path | First link from R | Second link from S | Path strength |
|---|---|---|---|
| u₁ → v₁ → w₁ | .2 | .5 | min(.2,.5)=.2 |
| u₁ → v₂ → w₁ | .8 | .7 | min(.8,.7)=.7 |
Choose the stronger path: max(.2,.7)=.7, so T₁₁=.7. The middle label j runs through v₁ and v₂. The labels i and k stay fixed at 1 while this one cell is calculated. Then move to the next result cell.
Yaad rakho: Ek path mein dono links chahiye, isliye min. Alag paths mein best chalega, isliye max.
Suppose R relates U to V and S relates V to W. To find how strongly u connects to w through each middle v, take min of the two links. Then choose the strongest such path using max.
![\"T_{ik}=\\max_j\\min(R_{ij},S_{jk})\"](\"math/56cc213f59fd4c68.svg\")
We call this matrix operation R composed with S here. Some books write the symbolic order in reverse. State the formula and row/column labels, especially when a question says S∘R. Composition requires the intermediate domains and their ordering to match; having two 3×3 matrices alone does not prove their labels match.
Calculate the max–min composition using ![\"T_{ik}=\\max_j\\min(R_{ij},S_{jk})\"](\"math/7ad9b78ea1d9260b.svg\")
. [3 marks · Practice]
![\"R=\\begin{bmatrix}0.2&0.8\\\\0.6&0.3\\end{bmatrix}\"](\"math/6a3980f8eb0faad9.svg\")
![\"S=\\begin{bmatrix}0.5&0.9\\\\0.7&0.1\\end{bmatrix}\"](\"math/ae4133e873ae6a3f.svg\")
Every entry of T.
Pair each row entry with the corresponding column entry; min for a two-link path, max across the possible middle elements. This is not ordinary matrix multiplication.
![\"T_{11}=\\max\\{\\min(0.2,0.5),\\min(0.8,0.7)\\}\"](\"math/ce290d4a2b6414c6.svg\")
![\"=\\max\\{0.2,0.7\\}=0.7\"](\"math/5d7c26e3f57345f9.svg\")
![\"T_{12}=\\max\\{\\min(0.2,0.9),\\min(0.8,0.1)\\}\"](\"math/ce2b2b77de4033df.svg\")
![\"=\\max\\{0.2,0.1\\}=0.2\"](\"math/197ec23ba477e3ac.svg\")
![\"T_{21}=\\max\\{\\min(0.6,0.5),\\min(0.3,0.7)\\}\"](\"math/07f583df1f70440b.svg\")
![\"=\\max\\{0.5,0.3\\}=0.5\"](\"math/5d84612aa0c3dabf.svg\")
![\"T_{22}=\\max\\{\\min(0.6,0.9),\\min(0.3,0.1)\\}\"](\"math/e2fa4ca72e9eecdf.svg\")
![\"=\\max\\{0.6,0.1\\}=0.6\"](\"math/b6bb46f767031fd8.svg\")
![\"T=\\begin{bmatrix}0.7&0.2\\\\0.5&0.6\\end{bmatrix}\"](\"math/c434184a79340d55.svg\")
.Calculate the max–min composition using . [3 marks · Practice]
![\"R=\\begin{bmatrix}1&0\\\\0.4&0.7\\end{bmatrix}\"](\"math/5525680a26d9df18.svg\")
![\"S=\\begin{bmatrix}0.2&0.6\\\\0.8&0.3\\end{bmatrix}\"](\"math/377016d2bce333c2.svg\")
Every entry of T.
Pair each row entry with the corresponding column entry; min for a two-link path, max across the possible middle elements. This is not ordinary matrix multiplication.
![\"T_{11}=\\max\\{\\min(1,0.2),\\min(0,0.8)\\}\"](\"math/cf3e1892a786961d.svg\")
![\"=\\max\\{0.2,0\\}=0.2\"](\"math/9ba8671f889808c7.svg\")
![\"T_{12}=\\max\\{\\min(1,0.6),\\min(0,0.3)\\}\"](\"math/d6b9af12c569f647.svg\")
![\"=\\max\\{0.6,0\\}=0.6\"](\"math/234802688b1bc952.svg\")
![\"T_{21}=\\max\\{\\min(0.4,0.2),\\min(0.7,0.8)\\}\"](\"math/145badb953d03e7a.svg\")
![\"T_{22}=\\max\\{\\min(0.4,0.6),\\min(0.7,0.3)\\}\"](\"math/7580af7f56a5d58b.svg\")
![\"=\\max\\{0.4,0.3\\}=0.4\"](\"math/fed46295bb39d225.svg\")
![\"T=\\begin{bmatrix}0.2&0.6\\\\0.7&0.4\\end{bmatrix}\"](\"math/303c2bbbbf128653.svg\")
.Calculate the max–min composition using . [3 marks · Practice]
![\"R=\\begin{bmatrix}0.3&0.6&0.9\\end{bmatrix}\"](\"math/12f5f2edbde65b9d.svg\")
![\"S=\\begin{bmatrix}0.8&0.1\\\\0.2&0.7\\\\0.5&0.4\\end{bmatrix}\"](\"math/1823a7e672a05e76.svg\")
Every entry of T.
Pair each row entry with the corresponding column entry; min for a two-link path, max across the possible middle elements. This is not ordinary matrix multiplication.
![\"T_{11}=\\max\\{\\min(0.3,0.8),\\min(0.6,0.2),\\min(0.9,0.5)\\}\"](\"math/559127d73e85d555.svg\")
![\"=\\max\\{0.3,0.2,0.5\\}=0.5\"](\"math/ebeaf1e5ce8f9612.svg\")
![\"T_{12}=\\max\\{\\min(0.3,0.1),\\min(0.6,0.7),\\min(0.9,0.4)\\}\"](\"math/843b8913b1abe9b7.svg\")
![\"=\\max\\{0.1,0.6,0.4\\}=0.6\"](\"math/8ce46b5da48e6fa8.svg\")
![\"T=\\begin{bmatrix}0.5&0.6\\end{bmatrix}\"](\"math/fc69cd14a88a24b1.svg\")
.Yaad rakho: Har path mein min, phir saare paths mein max. Multiply-and-add nahi.
"}]}],"formulas":[{"id":"m2-formulas-v2","title":"All M2 formulas","unit":"M2 revision","scope":"One place for the formulas, meanings, conditions and links to full worked examples.","sections":[{"id":"fuzzy","title":"Fuzzy formulas, with meanings and worked-example links","html":"![\"\\mu(x)=\\max\\left(0,\\min\\left(\\frac{x-a}{b-a},\\frac{c-x}{c-b}\\right)\\right)\"](\"math/f0ca7a326bd064e8.svg\")
a=left zero, b=peak, c=right zero, with a<b<c. Locate x before choosing the rising/falling branch.
Read the explanation and worked sums →
![\"\\mu_{A\\cup](\"math/eaf940ee2fdf1f17.svg\")
Compare memberships at the same element. These formulas use standard max–min operations.
Read the explanation and worked sums →
![\"\\mu_{A^c}=1-\\mu_A\\quad\\mu_{A\\setminus](\"math/ac545c5022ffd6a1.svg\")
A difference means A AND NOT B; it is not ordinary subtraction.
Read the explanation and worked sums →
![\"\\alpha_r=\\min(a_r,b_r)\\quad\\mu_r(z)=\\min(\\alpha_r,\\mu_{C_r}(z))\\quad\\mu_{out}(z)=\\max_r\\mu_r(z)\"](\"math/043f511911bca932.svg\")
a,b are input degrees; α is the strength; z is a candidate output value. OR conditions use max instead.
Read the explanation and worked sums →
![\"z^*=\\frac{\\sum_{z_i\\in](\"math/9ed17a341041965a.svg\")
M is the set of discrete output locations attaining maximum membership. The plateau shortcut is for one continuous maximum interval.
Read the explanation and worked sums →
![\"z^*=\\frac{\\sum_i](\"math/79ddc9ad485491a0.svg\")
Use actual discrete memberships or the full continuous aggregate. Denominator must be positive.
Read the explanation and worked sums →
![\"z^*=\\frac{\\sum_r\\alpha_r](\"math/0fb4cda8fcd0e1d8.svg\")
c is a supplied representative such as a peak. Do not call this the exact area centroid of a clipped Mamdani aggregate without justification.
Read the explanation and worked sums →
![\"R_{ij}=\\min(A_i,B_j)\\quad](\"math/dfabdb4f1a2b77c3.svg\")
Relations must have a common middle domain in the same order. Min along a path, max across paths.
Read the explanation and worked sums →
\r\n\r\n\r\n![\"\\mu(x)=\\max\\left(0,\\min\\left(\\frac{x-a}{b-a},1,\\frac{d-x}{d-c}\\right)\\right)\"](\"math/61dfda51f019f3c9.svg\")
For a<b≤c<d: a is the left zero, b the start of the height-1 plateau, c its end and d the right zero. Vertical shoulder edges need the specified piecewise definition. Branch examples →
![\"\\mu(x)=e^{-(x-c)^2/(2\\sigma^2)}\"](\"math/3a8ce5565c9dd1ff.svg\")
c is the centre and σ>0 sets the width. At x=c the exponent is zero and membership is 1. Worked exponential calculations →
![\"\\mu(x)=\\frac{1}{1+|(x-c)/a|^{2b}}\"](\"math/37da3dd6f213b8e7.svg\")
a>0 controls width; b>0 controls steepness; c is the centre. First subtract c, divide by a, take absolute value, apply the power, add 1, then take its reciprocal.
![\"A_\\alpha=\\{x:\\mu_A(x)\\ge\\alpha\\}\"](\"math/34e8388304c6f4b2.svg\")
α is the chosen membership threshold. For 0<α≤1 keep inputs meeting or exceeding it, including threshold-equality endpoints. Discrete and triangular examples →
![\"x=\\sum_{i=1}^{n}b_i2^{n-i}\"](\"math/4fb94e1436769fba.svg\")
bᵢ is a bit; the maximum n-bit unsigned integer is 2ⁿ−1.
Full explanation and worked solutions →
![\"P_i=F_i/\\sum_jF_j\\quad](\"math/a3d3e670e29c240d.svg\")
Fitness weights must be nonnegative with a positive total. For minimisation use the specified transformation or selection method. C is cumulative probability.
Full explanation and worked solutions →
![\"\\bar](\"math/5a16aa84fc40bbd3.svg\")
N is the number of individuals. Compare both best and average; mutation need not improve either.
Full explanation and worked solutions →
w weights inertia; c₁r₁ weights personal pull; c₂r₂ weights social pull. Apply the same equation separately to each coordinate.
Full explanation and worked solutions →
Use the new velocity, not the old one. Then reevaluate the objective.
Full explanation and worked solutions →
Memory updates: minimise → replace a pbest only for a lower score; maximise → only for a higher score. gbest is a position selected from the personal bests. Store the corresponding objective score alongside it. Keep previous best on ties unless instructed otherwise.
Yaad rakho: Formula ke pehle objective direction aur old state likho. Iteration 2 mein fresh positions aur retained memories use karo.
\r\n\r\n\r\nFor the specified uniform real-range encoding: decode n bits into unsigned integer d first. L and U are the physical limits. Endpoint examples →
![\"F=\\frac{1}{1+f(x)}\"](\"math/361a654c14fe241e.svg\")
This is the stated roulette transformation in our nonnegative-cost minimisation example. It is not required for every GA. It converts a smaller cost into a larger selection weight. Complete minimisation example →
Question. Consider two fuzzy sets X = {(c1, 0.6), (c2, 0.8), (c3, 0.3), (c4, 0.5), (c5, 0.7)} and Y = {(c1, 0.4), (c2, 0.5), (c3, 0.9), (c4, 0.6), (c5, 0.2)}. Find 1) X ∪ Y 2) X ∩ Y 3) ¬X ∪ ¬Y 4) ¬X ∩ ¬Y. [5 marks]
The aligned X and Y memberships printed in this question.
The four requested membership lists in the original element order.
Use max for OR, min for AND, and 1−μ for NOT.
![\"\\mu_{X\\cup](\"math/e32ad02c5ba3dae5.svg\")
¬X = {0.4, 0.2, 0.7, 0.5, 0.3}; ¬Y = {0.6, 0.5, 0.1, 0.4, 0.8}.
| c1 | c2 | c3 | c4 | c5 | |
|---|---|---|---|---|---|
| X | 0.6 | 0.8 | 0.3 | 0.5 | 0.7 |
| Y | 0.4 | 0.5 | 0.9 | 0.6 | 0.2 |
| X ∪ Y = max | 0.6 | 0.8 | 0.9 | 0.6 | 0.7 |
| X ∩ Y = min | 0.4 | 0.5 | 0.3 | 0.5 | 0.2 |
| ¬X ∪ ¬Y | 0.6 | 0.5 | 0.7 | 0.5 | 0.8 |
| ¬X ∩ ¬Y | 0.4 | 0.2 | 0.1 | 0.4 | 0.3 |
¬X ∪ ¬Y must equal 1 − (X ∩ Y) = {0.6, 0.5, 0.7, 0.5, 0.8} ✓, and ¬X ∩ ¬Y must equal 1 − (X ∪ Y) = {0.4, 0.2, 0.1, 0.4, 0.3} ✓.
Official marking: 1 + 1 + 1.5 + 1.5.
\r\n![\"\\neg](\"math/ab521a108b232136.svg\")
![\"X\\cup](\"math/9fba815e535802c7.svg\")
![\"X\\cap](\"math/580180c90f774b30.svg\")
![\"\\neg](\"math/ac5d994d1c80e0dd.svg\")
![\"\\neg](\"math/f8ad2bf1c74a8449.svg\")
![\"\\neg](\"math/994b18481ef2566a.svg\")
![\"X\\cup](\"math/34ae0707b407acaa.svg\")
![\"X\\cap](\"math/276fcc7ea7945a48.svg\")
![\"\\neg](\"math/e450f31b46fa1e42.svg\")
![\"\\neg](\"math/503c7cc18cb10c65.svg\")
![\"\\neg](\"math/abbf5f7c84a2a614.svg\")
![\"X\\cup](\"math/b75d7a85a782c820.svg\")
![\"X\\cap](\"math/33c018c6e76860e7.svg\")
![\"\\neg](\"math/b158ec14569973da.svg\")
![\"\\neg](\"math/7b8935d4bd0523c1.svg\")
![\"\\neg](\"math/9b11f9a14b9e240f.svg\")
![\"X\\cup](\"math/30d486e82175d463.svg\")
![\"X\\cap](\"math/dd263e46a0f4e96c.svg\")
![\"\\neg](\"math/bc8abd6badb711b9.svg\")
![\"\\neg](\"math/2acca2b0b18cfada.svg\")
![\"\\neg](\"math/b115216f7121911a.svg\")
![\"X\\cup](\"math/ca6e5562fab61517.svg\")
![\"X\\cap](\"math/3de60cd86ef1a257.svg\")
![\"\\neg](\"math/0712a083103a814e.svg\")
![\"\\neg](\"math/6274e157a87eaf7b.svg\")
Question. Maximize ![\"F(x_1,x_2)=2x_1^2-3x_1x_2+x_2^2+5x_1+2x_2+7\"](\"math/51d7b4f8ba214898.svg\")
using Particle Swarm Optimization. ![\"X=\\begin{bmatrix}2.5&-3.0\\\\3.5&-1.5\\end{bmatrix},\\;V=\\begin{bmatrix}0.2&0.3\\\\0.1&0.25\\end{bmatrix}\"](\"math/cf25c61c163c59c7.svg\")
. Take ![\"W=0.9,\\;c_1=c_2=1.5,\\;r_1=0.5949,\\;r_2=0.0855\"](\"math/d99737a5fc79423b.svg\")
. Optimize and give the solution after 1 iteration showing the step-by-step process. [10 marks]
Particle 1: ![\"x=(2.5,\\,-3.0)\"](\"math/85ac77f48027edb2.svg\")
, ![\"v=(0.2,\\,0.3)\"](\"math/438166430b0200d4.svg\")
. Particle 2: ![\"x=(3.5,\\,-1.5)\"](\"math/a92c6e5b7c6f437f.svg\")
, ![\"v=(0.1,\\,0.25)\"](\"math/3bf1b4b0830b2535.svg\")
.
![\"c_1r_1=1.5\\times0.5949=0.89235,\\qquad](\"math/24c25a9332d3129e.svg\")
Fitness, pbest, gbest; new velocities and positions; updated pbest and gbest.
![\"v_{ij}\\leftarrow](\"math/15c9c9495dc886c9.svg\")
Each particle is pulled by its own memory and by the swarm's best; we maximise, so the larger ![\"F\"](\"math/277a47abb82a8255.svg\")
is better.
![\"F(2.5,-3)=2(6.25)-3(2.5)(-3)+9+5(2.5)+2(-3)+7=12.5+22.5+9+12.5-6+7=57.5\"](\"math/3938546a658fac5b.svg\")
![\"F(3.5,-1.5)=2(12.25)-3(3.5)(-1.5)+2.25+17.5-3+7=24.5+15.75+2.25+17.5-3+7=64\"](\"math/4d48c40dfb866c17.svg\")
| Particle | Position | F | pbest |
|---|---|---|---|
| 1 | (2.5, −3.0) | 57.5 | (2.5, −3.0) |
| 2 | (3.5, −1.5) | 64 | (3.5, −1.5) |
No previous iteration, so pbest = current position. Largest fitness is 64, so gbest = (3.5, −1.5).
![\"v_{11}=0.9(0.2)+0.89235(2.5-2.5)+0.12825(3.5-2.5)=0.18+0+0.12825=0.30825\"](\"math/c5345abf4a118373.svg\")
![\"x_{11}=2.5+0.30825=2.80825\"](\"math/aa0bacfeb906b753.svg\")
![\"v_{12}=0.9(0.3)+0.89235(0)+0.12825(-1.5-(-3.0))=0.27+0.192375=0.462375\"](\"math/a3af7b24f214c46d.svg\")
![\"x_{12}=-3.0+0.462375=-2.537625\"](\"math/c09ac9a314e4985c.svg\")
![\"v_{21}=0.9(0.1)+0+0=0.09,\\qquad](\"math/d5ef6ea1fa183f88.svg\")
![\"v_{22}=0.9(0.25)+0+0=0.225,\\qquad](\"math/cdc998e6ae01e141.svg\")
![\"F(2.80825,-2.537625)=15.772536+21.378856+6.439541+14.04125-5.07525+7=59.5569\"](\"math/d85fb6275e35f72d.svg\")
![\"F(3.59,-1.275)=25.7762+13.73175+1.625625+17.95-2.55+7=63.5336\"](\"math/df29fd84a6cdf039.svg\")
| Particle | New position | New F | Old pbest F | pbest after iteration |
|---|---|---|---|---|
| 1 | (2.80825, −2.537625) | 59.5569 | 57.5 | (2.80825, −2.537625) — improved |
| 2 | (3.59, −1.275) | 63.5336 | 64 | (3.5, −1.5) — kept |
The best pbest fitness is still 64, so gbest stays (3.5, −1.5).
![\"V=\\begin{bmatrix}0.30825&0.462375\\\\0.09&0.225\\end{bmatrix}\"](\"math/4cf56ad79b60f4ff.svg\")
, ![\"X=\\begin{bmatrix}2.80825&-2.537625\\\\3.59&-1.275\\end{bmatrix}\"](\"math/befb5808089844a6.svg\")
. After one iteration particle 1 improved to F = 59.5569; the best solution so far is gbest = (3.5, −1.5) with F = 64.Official key: fitness and gbest (5 marks), velocity and position updates (5 marks). The calculations above use the question data. The key's summary matrix writes 0.462 in place of 2.808 for ![\"x_{11}\"](\"math/e82facc124656ec6.svg\")
; that is a copying slip.
A bank uses fuzzy decision support with Credit Score and Income Stability, both on 1–100. Loan Approval is a score on 1–100 with labels Reject, Consider and Approve. Define triangular membership functions, construct nine rules, explain Mamdani inference and defuzzify for Credit Score=55 and Income Stability=35. [10 marks]
| Variable | Crisp value |
|---|---|
| Credit Score | 55 |
| Income Stability | 35 |
The question asks us to design the membership functions. This solution chooses Low=Tri(1,25,50), Medium=Tri(25,50,75), High=Tri(50,75,100) for both inputs. Output labels use the same triangles in their respective order. These are explicit design choices, not additional printed givens.
A complete stated design and its continuous Mamdani centroid.
Use min for AND and implication, max for aggregation and the true area centroid of the resulting shape. A peak-only weighted average is a different defuzzifier.
![\"z^*=\\frac{\\int](\"math/5aa4acd6aab4215a.svg\")
| Input / output position | Triangle | Peak |
|---|---|---|
| Low / Reject | (1,25,50) | 25 |
| Medium / Consider | (25,50,75) | 50 |
| High / Approve | (50,75,100) | 75 |
Use 0 outside each triangle, the rising fraction before the peak and the falling fraction after it. This design is evaluated at the requested input; a deployed controller would also need coverage/fallback at all boundary values.
| Credit Score / Income Stability | Low | Medium | High |
|---|---|---|---|
| Low | Reject | Reject | Consider |
| Medium | Reject | Consider | Approve |
| High | Consider | Approve | Approve |
Read each cell as an IF–AND–THEN rule. For example, the centre cell combines Medium with Medium.
![\"\\mu_{Low}(55)=0\"](\"math/b1adf4cfc2121094.svg\")
![\"\\mu_{Medium}(55)=\\frac{75-55}{75-50}=0.8\"](\"math/0625f25570fc580f.svg\")
![\"\\mu_{High}(55)=\\frac{55-50}{75-50}=0.2\"](\"math/6f64cb8c41f3eca0.svg\")
![\"\\mu_{Low}(35)=\\frac{50-35}{50-25}=0.6\"](\"math/c3eeae587085fbae.svg\")
![\"\\mu_{Medium}(35)=\\frac{35-25}{50-25}=0.4\"](\"math/d607a558d66f8791.svg\")
![\"\\mu_{High}(35)=0\"](\"math/1509e75effd66d75.svg\")
| Credit Score | Income Stability | Firing strength | Consequent |
|---|---|---|---|
| Low | Low | min(0,0.6) = 0 | Reject |
| Low | Medium | min(0,0.4) = 0 | Reject |
| Low | High | min(0,0) = 0 | Consider |
| Medium | Low | min(0.8,0.6) = 0.6 | Reject |
| Medium | Medium | min(0.8,0.4) = 0.4 | Consider |
| Medium | High | min(0.8,0) = 0 | Approve |
| High | Low | min(0.2,0.6) = 0.2 | Consider |
| High | Medium | min(0.2,0.4) = 0.2 | Approve |
| High | High | min(0.2,0) = 0 | Approve |
Aggregated label strengths: Reject = 0.6, Consider = 0.4, Approve = 0.2.
![\"\\mu_{out}(z)=\\max_r\\min(\\alpha_r,\\mu_{C_r}(z))\"](\"math/d1c2a08444481d0c.svg\")
![\"The](\"fig/v2_loan_aggregate.2c5ffa32a7.svg\")
For a strip from a to b with endpoint heights u and v, its trapezoid area is A. Its first moment M is the integral of z times the line height.
![\"A=(b-a)(u+v)/2\"](\"math/a1493aef29965a09.svg\")
![\"M=\\frac{b-a}{6}\\big[(2a+b)u+(a+2b)v\\big]\"](\"math/d68335c9ce0df62c.svg\")
| a | b | u | v | Area A | Moment M |
|---|---|---|---|---|---|
| 1 | 15.4 | 0 | 0.6 | 4.32 | 45.792 |
| 15.4 | 25 | 0.6 | 0.6 | 5.76 | 116.352 |
| 25 | 35 | 0.6 | 0.6 | 6 | 180 |
| 35 | 40 | 0.6 | 0.4 | 2.5 | 93.333333333 |
| 40 | 50 | 0.4 | 0.4 | 4 | 180 |
| 50 | 55 | 0.4 | 0.4 | 2 | 105 |
| 55 | 65 | 0.4 | 0.4 | 4 | 240 |
| 65 | 70 | 0.4 | 0.2 | 1.5 | 100.833333333 |
| 70 | 75 | 0.2 | 0.2 | 1 | 72.5 |
| 75 | 95 | 0.2 | 0.2 | 4 | 340 |
| 95 | 100 | 0.2 | 0 | 0.5 | 48.333333333 |
![\"\\sum](\"math/04fe6e4450abbfe1.svg\")
![\"\\sum](\"math/db33614c63a7fe19.svg\")
![\"z^*=\\frac{1522.144}{35.58}\\approx42.780888139\"](\"math/dd239737d0e6ec58.svg\")
Because the paper asks for a design, a different explicitly specified membership/rule system can yield a different numerical output. Explain the choices and use the chosen defuzzification method consistently; no particular examiner mark outcome is promised.
\r\n\r\n\r\nFor each output triangle, solve “line height = clip height”. Reject is clipped at .6: rising endpoint 1+.6(25−1)=15.4 and falling endpoint 50−.6(50−25)=35. Consider is clipped at .4: endpoints 25+.4(50−25)=35 and 75−.4(75−50)=65. Approve is clipped at .2: endpoints 50+.2(75−50)=55 and 100−.2(100−75)=95.
Now take the maximum of the three clipped curves. The declining Reject edge reaches the Consider height .4 at (50−z)/25=.4, hence z=40. The declining Consider edge reaches the Approve height .2 at (75−z)/25=.2, hence z=70. These crossover locations must be included when splitting the aggregate. Original peaks and other clip endpoints may add extra splits without changing the total area.
For the first aggregate strip from a=1 to b=15.4, endpoint heights are u=0 and v=0.6. Find its area and first moment. [3 marks · Worked example]
Strip endpoints a=1 and b=15.4; left height u=0 and right height v=0.6. This is a straight rising edge of the clipped Reject set.
A and M for one table row.
![\"M=\\frac{b-a}{6}[(2a+b)u+(a+2b)v]\"](\"math/ee3fb707739e8b69.svg\")
A is trapezoid area. M weights every small area by its horizontal position; this formula integrates a straight line.
![\"b-a=15.4-1=14.4\\qquad](\"math/49c4a9b06649aac9.svg\")
![\"A=14.4\\times0.6/2=8.64/2=4.32\"](\"math/344db3e57e5c18d1.svg\")
![\"(2a+b)u=(2+15.4)\\times0=0\"](\"math/f077b4473411d432.svg\")
![\"(a+2b)v=(1+30.8)\\times0.6=31.8\\times0.6=19.08\"](\"math/79c1733b7155a919.svg\")
![\"M=(14.4/6)(0+19.08)=2.4\\times19.08=45.792\"](\"math/26cef63984606712.svg\")
Choose the output label having the greatest membership at the centroid; this is an additional stated interpretation rule. At z=42.780888139:
![\"\\mu_{Reject}=(50-z)/25\\approx0.288764474,\\quad\\mu_{Consider}=(z-25)/25\\approx0.711235526,\\quad\\mu_{Approve}=0\"](\"math/388fb9433b0d08db.svg\")
The resulting recommendation is Consider / manual review; this design does not recommend automatic approval. If a binary approve/reject answer is required, the question must supply a threshold or you must state the threshold you adopt. A different decision threshold is separate from defuzzification.
"},{"id":"q4a","title":"Q4(a) Cartesian product and max–min composition [10]","html":"Question. Consider fuzzy sets A = {0.4/small + 0.3/medium + 0.1/large} and B = {0.2/slow + 0.7/moderate + 0.3/fast}. 1. Find R = A × B. 2. Given C = {0.3/small + 0.1/medium + 0.4/large}, find S = C × B. 3. Calculate the max–min composition S ∘ R. [10 marks]
The fuzzy sets or relation matrices and their listed row/column order.
The relation matrix/matrices and the composition specified in the question.
For Cartesian products take min of row and column membership. For the ordered matrix composition shown here:
![\"T_{ik}=\\max_j\\min(S_{ij},R_{jk})\"](\"math/1a39e95150df91b7.svg\")
![\"R=\\begin{bmatrix}\\min(.4,.2)&\\min(.4,.7)&\\min(.4,.3)\\\\\\min(.3,.2)&\\min(.3,.7)&\\min(.3,.3)\\\\\\min(.1,.2)&\\min(.1,.7)&\\min(.1,.3)\\end{bmatrix}=\\begin{bmatrix}0.2&0.4&0.3\\\\0.2&0.3&0.3\\\\0.1&0.1&0.1\\end{bmatrix}\"](\"math/9db67d15872888f6.svg\")
![\"S=\\begin{bmatrix}0.2&0.3&0.3\\\\0.1&0.1&0.1\\\\0.2&0.4&0.3\\end{bmatrix}\"](\"math/03c1b332a4b4facf.svg\")
The printed relation domains do not directly compose: both relate size to speed. For the numerical exercise, explicitly assume positional alignment and compute row of S with column of R. Equal dimensions alone are not a justification for semantic relation composition.
\r\n![\"(1,1)=\\max[\\min(.2,.2),\\min(.3,.2),\\min(.3,.1)]=\\max(.2,.2,.1)=0.2\"](\"math/2134a14de60dd459.svg\")
![\"(1,2)=\\max[\\min(.2,.4),\\min(.3,.3),\\min(.3,.1)]=\\max(.2,.3,.1)=0.3\"](\"math/b81568f26ae092d4.svg\")
![\"(1,3)=\\max[\\min(.2,.3),\\min(.3,.3),\\min(.3,.1)]=0.3\"](\"math/0857c1f332a31d0c.svg\")
Row 2 of S is all 0.1, so every entry of row 2 is 0.1.
\r\n![\"(3,1)=\\max[\\min(.2,.2),\\min(.4,.2),\\min(.3,.1)]=0.2,\\;(3,2)=\\max[.2,.3,.1]=0.3,\\;(3,3)=\\max[.2,.3,.1]=0.3\"](\"math/7cbca4576287a20a.svg\")
![\"S\\circ](\"math/564480fc64ab9d52.svg\")
.Both printed relations run from size to speed; their middle labels do not match for genuine relation composition. The numerical working below follows the question’s intended row-of-S with column-of-R operation using explicitly assumed positional alignment of the three middle entries.
| Cell | Row degree | Column degree | min result |
|---|---|---|---|
| R11 | 0.4 | 0.2 | 0.2 |
| R12 | 0.4 | 0.7 | 0.4 |
| R13 | 0.4 | 0.3 | 0.3 |
| R21 | 0.3 | 0.2 | 0.2 |
| R22 | 0.3 | 0.7 | 0.3 |
| R23 | 0.3 | 0.3 | 0.3 |
| R31 | 0.1 | 0.2 | 0.1 |
| R32 | 0.1 | 0.7 | 0.1 |
| R33 | 0.1 | 0.3 | 0.1 |
| Cell | Row degree | Column degree | min result |
|---|---|---|---|
| S11 | 0.3 | 0.2 | 0.2 |
| S12 | 0.3 | 0.7 | 0.3 |
| S13 | 0.3 | 0.3 | 0.3 |
| S21 | 0.1 | 0.2 | 0.1 |
| S22 | 0.1 | 0.7 | 0.1 |
| S23 | 0.1 | 0.3 | 0.1 |
| S31 | 0.4 | 0.2 | 0.2 |
| S32 | 0.4 | 0.7 | 0.4 |
| S33 | 0.4 | 0.3 | 0.3 |
Result cell (1,1)
![\"\\max\\{\\min(0.2,0.2),\\min(0.3,0.2),\\min(0.3,0.1)\\}\"](\"math/04a79caa9a98c1c4.svg\")
![\"=\\max\\{0.2,0.2,0.1\\}=0.2\"](\"math/e40652bfd4dc8015.svg\")
Result cell (1,2)
![\"\\max\\{\\min(0.2,0.4),\\min(0.3,0.3),\\min(0.3,0.1)\\}\"](\"math/fb42face9e153c77.svg\")
![\"=\\max\\{0.2,0.3,0.1\\}=0.3\"](\"math/6d394999310022b5.svg\")
Result cell (1,3)
![\"\\max\\{\\min(0.2,0.3),\\min(0.3,0.3),\\min(0.3,0.1)\\}\"](\"math/4535d46543709132.svg\")
Result cell (2,1)
![\"\\max\\{\\min(0.1,0.2),\\min(0.1,0.2),\\min(0.1,0.1)\\}\"](\"math/b8bd41528e9d1a36.svg\")
![\"=\\max\\{0.1,0.1,0.1\\}=0.1\"](\"math/2caac33b19e2f71f.svg\")
Result cell (2,2)
![\"\\max\\{\\min(0.1,0.4),\\min(0.1,0.3),\\min(0.1,0.1)\\}\"](\"math/9eda900daa88d0f0.svg\")
Result cell (2,3)
![\"\\max\\{\\min(0.1,0.3),\\min(0.1,0.3),\\min(0.1,0.1)\\}\"](\"math/a0554ac89675d583.svg\")
Result cell (3,1)
![\"\\max\\{\\min(0.2,0.2),\\min(0.4,0.2),\\min(0.3,0.1)\\}\"](\"math/7738d5099a9e3ab1.svg\")
Result cell (3,2)
![\"\\max\\{\\min(0.2,0.4),\\min(0.4,0.3),\\min(0.3,0.1)\\}\"](\"math/952b0167063f5e45.svg\")
Result cell (3,3)
![\"\\max\\{\\min(0.2,0.3),\\min(0.4,0.3),\\min(0.3,0.1)\\}\"](\"math/859d029599adbb59.svg\")
Implement a GA to maximise ![\"f(x)=2x^3-5x^2+7x\"](\"math/aacd674038ffd6be.svg\")
for x in [0,15]. Take 4-bit representations of 13, 22, 18, 23. Crossover after the second bit and mutate by flipping any one bit. Produce two offspring and comment on fitness. [10 marks]
The printed data are inconsistent. 22, 18 and 23 are outside 0–15 and cannot fit into four bits. Increasing the bit count would encode the numbers but would still violate the stated range. There is no single exact solution satisfying all printed conditions.
The complete worked variant below explicitly changes the invalid parents to their last four bits (values modulo 16), producing [13,6,2,7]. This is one declared repair, not a silent interpretation of the printed question. If the instructor clarifies different corrected data, follow those data.
Printed parents: [13, 22, 18, 23]. Worked variant: [13, 6, 2, 7] with 4 bits. For the unspecified selection step, select the two highest-fitness candidates. For the unspecified mutation location, flip the last bit of each child. Crossover is after bit 2.
Two valid children under the declared assumptions and their fitness.
Encoding must respect the candidate range. Crossover swaps tails; mutation flips the chosen bit. Fitness is evaluated after mutation.
![\"f(x)=2x^3-5x^2+7x\"](\"math/fb065d212fc56261.svg\")
| x | Binary | Fitness substitution |
|---|---|---|
| 13 | 1101 | 2(13³) − 5(13²) + 7(13) = 3640 |
| 6 | 0110 | 2(6³) − 5(6²) + 7(6) = 294 |
| 2 | 0010 | 2(2³) − 5(2²) + 7(2) = 10 |
| 7 | 0111 | 2(7³) − 5(7²) + 7(7) = 490 |
[13, 7] with fitness [3640, 490].
| Parent 1 | Parent 2 | Child 1 | Child 2 |
|---|---|---|---|
| 11 | 01 | 01 | 11 | 1111 | 0101 |
| Before | After | Decoded x |
|---|---|---|
| 1111 | 1110 | 14 |
| 0101 | 0100 | 4 |
| x | Fitness calculation |
|---|---|
| 14 | 2(2744) − 5(196) + 7(14) = 4606 |
| 4 | 2(64) − 5(16) + 7(4) = 76 |
Best parent score = 3640; best child score = 4606. Parent average = 2065; child average = 2341. Both the best score and the average score increased in this worked variant. An individual child can still be worse than a parent; GA does not guarantee improvement after every random operator.
The printed wording about “two offspring” is not an explicit requirement for two complete generations. The M2 examples separately work two full generations because your class update asks for iteration 2.
\r\n\r\n\r\nCalculate x²=x×x, then x³=x²×x. Multiply each by its coefficient and combine with the written signs. A bit-string repair in the worked answer is an explicitly stated response to inconsistent printed givens; it is not a new instruction in the original paper.
![\"x^2=13\\times13=169\\qquad](\"math/3b2d51c91ac2cf33.svg\")
![\"2x^3=4394\\quad](\"math/b223b59b645fb4b2.svg\")
![\"f(13)=4394-845+91=3640\"](\"math/729d7b573feaaf95.svg\")
![\"x^2=6\\times6=36\\qquad](\"math/4a05c63ed6e04319.svg\")
![\"2x^3=432\\quad](\"math/402fc08edced6189.svg\")
![\"f(6)=432-180+42=294\"](\"math/a5f219480ddd154a.svg\")
![\"x^2=2\\times2=4\\qquad](\"math/3a35f3beb07d2871.svg\")
![\"2x^3=16\\quad](\"math/42f55d86ed2302e7.svg\")
![\"f(2)=16-20+14=10\"](\"math/235889b9921ff9b5.svg\")
![\"x^2=7\\times7=49\\qquad](\"math/eac1f56ad09d694c.svg\")
![\"2x^3=686\\quad](\"math/e9a8667a88c96f77.svg\")
![\"f(7)=686-245+49=490\"](\"math/8c54966ea16cb628.svg\")
![\"x^2=15\\times15=225\\qquad](\"math/8c2aaf117e33ac77.svg\")
![\"2x^3=6750\\quad](\"math/9652ff9e90959bb5.svg\")
![\"f(15)=6750-1125+105=5730\"](\"math/1609c9dac811fd7a.svg\")
![\"x^2=5\\times5=25\\qquad](\"math/63703f64d73bd95d.svg\")
![\"2x^3=250\\quad](\"math/c7ab1b91c2eced73.svg\")
![\"f(5)=250-125+35=160\"](\"math/b80fc8e2e02f0c86.svg\")
![\"x^2=14\\times14=196\\qquad](\"math/13e67d57f1606f3f.svg\")
![\"2x^3=5488\\quad](\"math/7a11853924c667dd.svg\")
![\"f(14)=5488-980+98=4606\"](\"math/4be8ca3a4ac49622.svg\")
![\"x^2=4\\times4=16\\qquad](\"math/9c1546066fc40eda.svg\")
![\"2x^3=128\\quad](\"math/a4dd7bf5365d3c7e.svg\")
![\"f(4)=128-80+28=76\"](\"math/818af979753c0a98.svg\")
“Modulo 16” means keep the remainder after dividing by 16: 22=1×16+6, so its remainder is 6; 18=1×16+2; 23=1×16+7. This is only the declared repair used here, not an unspoken general encoding rule.
List the applications of fuzzy logic. Explain any one application in detail. [10 marks]
Fuzzy logic is a multi-valued logic, introduced by Lotfi Zadeh (1965), in which truth values range over [0, 1]; it lets machines reason with vague, human-like terms such as “slightly dirty” or “very hot”.
\r\n| Area | Examples |
|---|---|
| Home appliances | Washing machines (wash time from dirt and load), air conditioners, refrigerators, rice cookers, vacuum cleaners |
| Automotive | Automatic transmission, anti-lock braking (ABS), cruise control, engine control |
| Transportation | Traffic-signal control, Sendai subway train control (smooth braking and acceleration) |
| Consumer electronics | Camera autofocus and auto-exposure, image stabilisation |
| Industrial control | Cement kilns, pH, temperature and distillation control in chemical plants |
| Aerospace | Satellite and spacecraft attitude control |
| Medicine | Diagnosis support, anaesthesia dosage control |
| Finance and decisions | Loan approval, credit and fire-risk scoring, stock trading rules |
| AI and robotics | Robot navigation, natural-language interfaces, expert systems |
Advantages: handles imprecise or noisy data; mimics human reasoning; easy to design and modify; no exact model needed; low computational cost. Disadvantages: rules and membership functions depend on expert judgement; results are hard to verify formally; tuning can take time; not suited where very high precision is required.
\r\nList the applications of Particle Swarm Optimization and explain any one application in detail. [10 marks]
Particle Swarm Optimization is a population-based metaheuristic, inspired by bird flocking and fish schooling, in which particles move through the search space using their own best position (pbest) and the swarm's best position (gbest) to find optimal or near-optimal solutions.
\r\n\r\n| Advantages | Disadvantages |
|---|---|
| Simple to implement; few parameters (w, c₁, c₂) | Can converge prematurely to a local optimum |
| No derivatives needed | Sensitive to parameter choice |
| Fast convergence in early iterations; memory (pbest, gbest) | No guarantee of the global optimum |
| Objective evaluations can be parallel; particles share information through gbest | Less suited to discrete problems without modification |
Marking pattern used for ACO in the same paper: listing 3 marks + explanation 2 marks (5-mark version); listing 5 + explanation 5 (10-mark version).
\r\n\r\nList the applications of Particle Swarm Optimization and explain any one application in detail. [5 marks]
Particle Swarm Optimization is a population-based metaheuristic, inspired by bird flocking and fish schooling, in which particles move through the search space using their own best position (pbest) and the swarm's best position (gbest) to find optimal or near-optimal solutions.
\r\n\r\n| Advantages | Disadvantages |
|---|---|
| Simple to implement; few parameters (w, c₁, c₂) | Can converge prematurely to a local optimum |
| No derivatives needed | Sensitive to parameter choice |
| Fast convergence in early iterations; memory (pbest, gbest) | No guarantee of the global optimum |
| Objective evaluations can be parallel; particles share information through gbest | Less suited to discrete problems without modification |
Marking pattern used for ACO in the same paper: listing 3 marks + explanation 2 marks (5-mark version); listing 5 + explanation 5 (10-mark version).
\r\n\r\nQuestion. X = {(c1, 0.7), (c2, 0.9), (c3, 0.2), (c4, 0.4), (c5, 0.9)}, Y = {(c1, 0.6), (c2, 0.3), (c3, 0.8), (c4, 0.7), (c5, 0.1)}. Find X ∪ Y, X ∩ Y, ¬X ∪ ¬Y, ¬X ∩ ¬Y. [5 marks]
The aligned X and Y memberships printed in this question.
The four requested membership lists in the original element order.
Use max for OR, min for AND, and 1−μ for NOT.
| c1 | c2 | c3 | c4 | c5 | |
|---|---|---|---|---|---|
| ¬X | 0.3 | 0.1 | 0.8 | 0.6 | 0.1 |
| ¬Y | 0.4 | 0.7 | 0.2 | 0.3 | 0.9 |
| X ∪ Y | 0.7 | 0.9 | 0.8 | 0.7 | 0.9 |
| X ∩ Y | 0.6 | 0.3 | 0.2 | 0.4 | 0.1 |
| ¬X ∪ ¬Y | 0.4 | 0.7 | 0.8 | 0.6 | 0.9 |
| ¬X ∩ ¬Y | 0.3 | 0.1 | 0.2 | 0.3 | 0.1 |
![\"\\neg](\"math/c590062072bce4a3.svg\")
![\"X\\cup](\"math/a2eea0fe6bbb6b87.svg\")
![\"X\\cap](\"math/ecd4a7a60c39fa85.svg\")
![\"\\neg](\"math/1aca856a2f38fa09.svg\")
![\"\\neg](\"math/4ef33b6f553800c5.svg\")
![\"\\neg](\"math/ce24f9cbda57e56d.svg\")
![\"X\\cup](\"math/05dee517321dfca3.svg\")
![\"X\\cap](\"math/4a2bc31233f313e7.svg\")
![\"\\neg](\"math/b188e2e037566b1e.svg\")
![\"\\neg](\"math/82b3199ed4befd72.svg\")
![\"\\neg](\"math/24c8a43adb44c324.svg\")
![\"X\\cup](\"math/ddba21d8acd6b044.svg\")
![\"X\\cap](\"math/247907e255730431.svg\")
![\"\\neg](\"math/41e12e8e143ca16d.svg\")
![\"\\neg](\"math/ea5cfbc43203761d.svg\")
![\"\\neg](\"math/0b5f5f89785edd94.svg\")
![\"X\\cup](\"math/05b7177b2b3fbf81.svg\")
![\"X\\cap](\"math/1cfa1c76382cbe6b.svg\")
![\"\\neg](\"math/1568254a4bf6e796.svg\")
![\"\\neg](\"math/edebdcb00abb9544.svg\")
![\"\\neg](\"math/f594ae61dd359b85.svg\")
![\"X\\cup](\"math/da89b631e703aa27.svg\")
![\"X\\cap](\"math/5461384d5ad733c2.svg\")
![\"\\neg](\"math/2a8b87917a1340f9.svg\")
![\"\\neg](\"math/601ec64fa8926431.svg\")
A building safety department wants a fuzzy logic system to predict fire hazards. Inputs are Temperature (1–100°C) and Smoke Level (1–100). The output Fire Risk (1–100) has labels Low, Moderate and High. Define triangular membership functions, construct at least nine rules, explain Mamdani inference and defuzzify for Temperature=65°C and Smoke=80. [10 marks]
| Variable | Crisp value |
|---|---|
| Temperature | 65 |
| Smoke Level | 80 |
The question asks us to design the membership functions. This solution chooses Low=Tri(1,25,50), Medium=Tri(25,50,75), High=Tri(50,75,100) for both inputs. Output labels use the same triangles in their respective order. These are explicit design choices, not additional printed givens.
A complete stated design and its continuous Mamdani centroid.
Use min for AND and implication, max for aggregation and the true area centroid of the resulting shape. A peak-only weighted average is a different defuzzifier.
| Input / output position | Triangle | Peak |
|---|---|---|
| Low / Low | (1,25,50) | 25 |
| Medium / Moderate | (25,50,75) | 50 |
| High / High | (50,75,100) | 75 |
Use 0 outside each triangle, the rising fraction before the peak and the falling fraction after it. This design is evaluated at the requested input; a deployed controller would also need coverage/fallback at all boundary values.
| Temperature / Smoke Level | Low | Medium | High |
|---|---|---|---|
| Low | Low | Low | Moderate |
| Medium | Low | Moderate | High |
| High | Moderate | High | High |
Read each cell as an IF–AND–THEN rule. For example, the centre cell combines Medium with Medium.
![\"\\mu_{Low}(65)=0\"](\"math/9e3859fafc31a24f.svg\")
![\"\\mu_{Medium}(65)=\\frac{75-65}{75-50}=0.4\"](\"math/f35f654b9795c088.svg\")
![\"\\mu_{High}(65)=\\frac{65-50}{75-50}=0.6\"](\"math/de5666dfbbcb02fc.svg\")
![\"\\mu_{Low}(80)=0\"](\"math/0a1689e705282396.svg\")
![\"\\mu_{Medium}(80)=0\"](\"math/af4f190a786e2c05.svg\")
![\"\\mu_{High}(80)=\\frac{100-80}{100-75}=0.8\"](\"math/ec0cdd51e84c6204.svg\")
| Temperature | Smoke Level | Firing strength | Consequent |
|---|---|---|---|
| Low | Low | min(0,0) = 0 | Low |
| Low | Medium | min(0,0) = 0 | Low |
| Low | High | min(0,0.8) = 0 | Moderate |
| Medium | Low | min(0.4,0) = 0 | Low |
| Medium | Medium | min(0.4,0) = 0 | Moderate |
| Medium | High | min(0.4,0.8) = 0.4 | High |
| High | Low | min(0.6,0) = 0 | Moderate |
| High | Medium | min(0.6,0) = 0 | High |
| High | High | min(0.6,0.8) = 0.6 | High |
Aggregated label strengths: Low = 0, Moderate = 0, High = 0.6.
![\"The](\"fig/v2_fire_aggregate.e0cab4ad42.svg\")
For a strip from a to b with endpoint heights u and v, its trapezoid area is A. Its first moment M is the integral of z times the line height.
| a | b | u | v | Area A | Moment M |
|---|---|---|---|---|---|
| 50 | 65 | 0 | 0.6 | 4.5 | 270 |
| 65 | 85 | 0.6 | 0.6 | 12 | 900 |
| 85 | 100 | 0.6 | 0 | 4.5 | 405 |
![\"\\sum](\"math/2ffb3645f0d4cc76.svg\")
![\"\\sum](\"math/0a14da79e18490bb.svg\")
![\"z^*=\\frac{1575}{21}\\approx75\"](\"math/647a203082aabc5c.svg\")
Because the paper asks for a design, a different explicitly specified membership/rule system can yield a different numerical output. Explain the choices and use the chosen defuzzification method consistently; no particular examiner mark outcome is promised.
\r\n\r\n\r\nOnly the High output survives, at height .6. Its triangle is (50,75,100). Solve its rising edge: (z−50)/25=.6, so z=50+15=65. Solve its falling edge: (100−z)/25=.6, so z=100−15=85.
| Piece | Area calculation | Centre | Moment |
|---|---|---|---|
| Left triangle [50,65] | ½×15×.6=4.5 | (50+65+65)/3=60 | 4.5×60=270 |
| Rectangle [65,85] | 20×.6=12 | (65+85)/2=75 | 12×75=900 |
| Right triangle [85,100] | ½×15×.6=4.5 | (85+85+100)/3=90 | 4.5×90=405 |
The horizontal centroid of a triangle is the average of the three vertex x-coordinates. Add areas: 4.5+12+4.5=21. Add moments:270+900+405=1575. Divide 1575/21=75, agreeing with symmetry.
Question. A = {0.6/small + 0.4/medium + 0.1/large}, B = {0.4/slow + 0.8/moderate + 0.3/fast}, C = {0.5/small + 0.2/medium + 0.4/large}. Find R = A × B, S = C × B and the max–min composition S ∘ R. [10 marks]
The fuzzy sets or relation matrices and their listed row/column order.
The relation matrix/matrices and the composition specified in the question.
For Cartesian products take min of row and column membership. For the ordered matrix composition shown here:
![\"R=\\begin{bmatrix}0.4&0.6&0.3\\\\0.4&0.4&0.3\\\\0.1&0.1&0.1\\end{bmatrix},\\qquad](\"math/463247054a5e2b55.svg\")
![\"(1,1)=\\max[\\min(.4,.4),\\min(.5,.4),\\min(.3,.1)]=\\max(.4,.4,.1)=0.4\"](\"math/07a16abae3efd072.svg\")
![\"(1,2)=\\max[\\min(.4,.6),\\min(.5,.4),\\min(.3,.1)]=\\max(.4,.4,.1)=0.4\"](\"math/8e54167083aa61b1.svg\")
![\"(1,3)=\\max[\\min(.4,.3),\\min(.5,.3),\\min(.3,.1)]=0.3\"](\"math/6ce0ff88a64ad0ef.svg\")
![\"\\text{row](\"math/c87a743bfa211789.svg\")
![\"R=\\begin{bmatrix}.4&.6&.3\\\\.4&.4&.3\\\\.1&.1&.1\\end{bmatrix}\"](\"math/f54af3b992bbd86c.svg\")
, ![\"S=\\begin{bmatrix}.4&.5&.3\\\\.2&.2&.2\\\\.4&.4&.3\\end{bmatrix}\"](\"math/acc98fbd37745e57.svg\")
, ![\"S\\circ](\"math/291bb354e326dcfa.svg\")
. (The key gives R and S only.)Both printed relations run from size to speed; their middle labels do not match for genuine relation composition. The numerical working below follows the question’s intended row-of-S with column-of-R operation using explicitly assumed positional alignment of the three middle entries.
| Cell | Row degree | Column degree | min result |
|---|---|---|---|
| R11 | 0.6 | 0.4 | 0.4 |
| R12 | 0.6 | 0.8 | 0.6 |
| R13 | 0.6 | 0.3 | 0.3 |
| R21 | 0.4 | 0.4 | 0.4 |
| R22 | 0.4 | 0.8 | 0.4 |
| R23 | 0.4 | 0.3 | 0.3 |
| R31 | 0.1 | 0.4 | 0.1 |
| R32 | 0.1 | 0.8 | 0.1 |
| R33 | 0.1 | 0.3 | 0.1 |
| Cell | Row degree | Column degree | min result |
|---|---|---|---|
| S11 | 0.5 | 0.4 | 0.4 |
| S12 | 0.5 | 0.8 | 0.5 |
| S13 | 0.5 | 0.3 | 0.3 |
| S21 | 0.2 | 0.4 | 0.2 |
| S22 | 0.2 | 0.8 | 0.2 |
| S23 | 0.2 | 0.3 | 0.2 |
| S31 | 0.4 | 0.4 | 0.4 |
| S32 | 0.4 | 0.8 | 0.4 |
| S33 | 0.4 | 0.3 | 0.3 |
Result cell (1,1)
![\"\\max\\{\\min(0.4,0.4),\\min(0.5,0.4),\\min(0.3,0.1)\\}\"](\"math/21233f6c94b2b173.svg\")
![\"=\\max\\{0.4,0.4,0.1\\}=0.4\"](\"math/891d26c6d9632710.svg\")
Result cell (1,2)
![\"\\max\\{\\min(0.4,0.6),\\min(0.5,0.4),\\min(0.3,0.1)\\}\"](\"math/999b4f88f5256b56.svg\")
Result cell (1,3)
![\"\\max\\{\\min(0.4,0.3),\\min(0.5,0.3),\\min(0.3,0.1)\\}\"](\"math/d50a44d1852d902c.svg\")
![\"=\\max\\{0.3,0.3,0.1\\}=0.3\"](\"math/f11631b96248ce3d.svg\")
Result cell (2,1)
![\"\\max\\{\\min(0.2,0.4),\\min(0.2,0.4),\\min(0.2,0.1)\\}\"](\"math/c683e55ca4bccac0.svg\")
Result cell (2,2)
![\"\\max\\{\\min(0.2,0.6),\\min(0.2,0.4),\\min(0.2,0.1)\\}\"](\"math/5eeb6f5598722d4c.svg\")
Result cell (2,3)
![\"\\max\\{\\min(0.2,0.3),\\min(0.2,0.3),\\min(0.2,0.1)\\}\"](\"math/524a01bd31cd2a0e.svg\")
Result cell (3,1)
![\"\\max\\{\\min(0.4,0.4),\\min(0.4,0.4),\\min(0.3,0.1)\\}\"](\"math/4c61c54bcac1075e.svg\")
Result cell (3,2)
![\"\\max\\{\\min(0.4,0.6),\\min(0.4,0.4),\\min(0.3,0.1)\\}\"](\"math/cedfe000344abbe9.svg\")
Result cell (3,3)
![\"\\max\\{\\min(0.4,0.3),\\min(0.4,0.3),\\min(0.3,0.1)\\}\"](\"math/98186c591a86ee1c.svg\")
Question. Maximize ![\"F(x)=x_1^2-x_1x_2+x_2^2+3x_1+4x_2+5\"](\"math/523616224185a5ce.svg\")
using Particle Swarm Optimization. ![\"X=\\begin{bmatrix}3.1472&-4.0246\\\\4.0579&-2.2150\\end{bmatrix},\\;V=\\begin{bmatrix}0.2194&0.2449\\\\0.1908&0.2228\\end{bmatrix}\"](\"math/0ba0d9983a41cf83.svg\")
. Take . Optimize and give the solution after 1 iteration showing the step-by-step process. [10 marks]
P1: ![\"x=(3.1472,\\,-4.0246)\"](\"math/f58367ac748ee874.svg\")
, ![\"v=(0.2194,\\,0.2449)\"](\"math/582ec15d8b22479d.svg\")
. P2: ![\"x=(4.0579,\\,-2.2150)\"](\"math/1588eb472c257b22.svg\")
, ![\"v=(0.1908,\\,0.2228)\"](\"math/ea7da7e3a3880e01.svg\")
. ![\"c_1r_1=0.89235\"](\"math/37a512f090105dd3.svg\")
, ![\"c_2r_2=0.12825\"](\"math/c760fda5e7728aea.svg\")
.
Updated velocities, positions, pbest and gbest after one iteration.
![\"v^{new}=wv+c_1r_1(pbest-x)+c_2r_2(gbest-x),\\quad](\"math/d17ec7429fc5876d.svg\")
Initial personal bests equal the starting positions. Use one frozen global best for all particles. This question maximises fitness, so keep a memory only when its score is larger.
![\"x_1^2=9.904868,\\;-x_1x_2=+12.666221,\\;x_2^2=16.197405,\\;3x_1=9.4416,\\;4x_2=-16.0984\"](\"math/3c47a0a1f1465e31.svg\")
![\"F_1=9.904868+12.666221+16.197405+9.4416-16.0984+5=37.1117\"](\"math/10f900e5ecd1d222.svg\")
![\"F_2=16.466552+8.988249+4.906225+12.1737-8.86+5=38.6747\"](\"math/323572bf8ccf630b.svg\")
pbest = current positions. ![\"38.6747>37.1117\"](\"math/0858b2bce4269ef8.svg\")
, so gbest = P2 = (4.0579, −2.2150).
![\"v_{11}=0.9(0.2194)+0+0.12825(4.0579-3.1472)=0.19746+0.116797=0.314257\"](\"math/0c6669f7602148c0.svg\")
![\"x_{11}=3.1472+0.314257=3.461457\"](\"math/c789efe14e1349c9.svg\")
![\"v_{12}=0.9(0.2449)+0+0.12825(-2.2150+4.0246)=0.22041+0.232081=0.452491\"](\"math/82dc2ed77545eeb8.svg\")
![\"x_{12}=-4.0246+0.452491=-3.572109\"](\"math/daa1f4fff8808c55.svg\")
![\"v_{21}=0.9(0.1908)=0.17172,\\quad](\"math/42d42b7358982c8c.svg\")
Use the same objective at each new position. Keep full precision internally; displayed coordinates are rounded.
![\"F(3.461457275,-3.5721088)=(3.461457275)^2-(3.461457275)(-3.5721088)+(-3.5721088)^2+3(3.461457275)+4(-3.5721088)+5\"](\"math/b030a3c1723e90f0.svg\")
![\"=11.981686467+12.364701993+12.759961279+10.384371825+(-14.2884352)+5\\approx38.202286364\"](\"math/246cb1e4add2ab6e.svg\")
![\"F(4.22962,-2.01448)=(4.22962)^2-(4.22962)(-2.01448)+(-2.01448)^2+3(4.22962)+4(-2.01448)+5\"](\"math/3fe27d20d1a3f2c2.svg\")
![\"=17.889685344+8.520484898+4.05812967+12.68886+(-8.05792)+5\\approx40.099239912\"](\"math/a64c8baf9521e960.svg\")
| Particle | New position | New F | Old F | pbest |
|---|---|---|---|---|
| 1 | (3.461457, −3.572109) | 38.2023 | 37.1117 | updated |
| 2 | (4.22962, −2.01448) | 40.0992 | 38.6747 | updated |
The best pbest is now P2's with F = 40.0992, so gbest moves to (4.22962, −2.01448).
![\"V=\\begin{bmatrix}0.314257&0.452491\\\\0.17172&0.20052\\end{bmatrix}\"](\"math/181c38752c9bafe9.svg\")
, ![\"X=\\begin{bmatrix}3.461457&-3.572109\\\\4.22962&-2.01448\\end{bmatrix}\"](\"math/751b3a44f8f94dbc.svg\")
. Both particles improved; new gbest = (4.22962, −2.01448) with F = 40.0992.This X and V are particles 1 and 2 of your lecture's five-particle example; only the function differs (+3x₁ and +5 here instead of +2x₁ and +3).
\r\nImplement a GA to maximise ![\"f(x)=2x^3-3x^2+8x\"](\"math/d678d50d2d5c4ae4.svg\")
for x in [0,50]. Take 4-bit representations of 12, 20, 17, 13. Crossover after the second bit and mutate by flipping any one bit. Produce two offspring and comment on fitness. [10 marks]
The printed data are inconsistent. 20 and 17 do not fit into four bits, and covering 0–50 needs six bits. We use six-bit encoding to preserve the supplied parent values and the stated range.
Printed parents: [12, 20, 17, 13]. Worked variant: [12, 20, 17, 13] with 6 bits. For the unspecified selection step, select the two highest-fitness candidates. For the unspecified mutation location, flip the last bit of each child. Crossover is after bit 2.
Two valid children under the declared assumptions and their fitness.
Encoding must respect the candidate range. Crossover swaps tails; mutation flips the chosen bit. Fitness is evaluated after mutation.
![\"f(x)=2x^3-3x^2+8x\"](\"math/68385220447af94f.svg\")
| x | Binary | Fitness substitution |
|---|---|---|
| 12 | 001100 | 2(12³) − 3(12²) + 8(12) = 3120 |
| 20 | 010100 | 2(20³) − 3(20²) + 8(20) = 14960 |
| 17 | 010001 | 2(17³) − 3(17²) + 8(17) = 9095 |
| 13 | 001101 | 2(13³) − 3(13²) + 8(13) = 3991 |
[20, 17] with fitness [14960, 9095].
| Parent 1 | Parent 2 | Child 1 | Child 2 |
|---|---|---|---|
| 01 | 0100 | 01 | 0001 | 010001 | 010100 |
| Before | After | Decoded x |
|---|---|---|
| 010001 | 010000 | 16 |
| 010100 | 010101 | 21 |
| x | Fitness calculation |
|---|---|
| 16 | 2(4096) − 3(256) + 8(16) = 7552 |
| 21 | 2(9261) − 3(441) + 8(21) = 17367 |
Best parent score = 14960; best child score = 17367. Parent average = 12027.5; child average = 12459.5. Both the best score and the average score increased in this worked variant. An individual child can still be worse than a parent; GA does not guarantee improvement after every random operator.
The printed wording about “two offspring” is not an explicit requirement for two complete generations. The M2 examples separately work two full generations because your class update asks for iteration 2.
\r\n\r\n\r\nCalculate x²=x×x, then x³=x²×x. Multiply each by its coefficient and combine with the written signs. A bit-string repair in the worked answer is an explicitly stated response to inconsistent printed givens; it is not a new instruction in the original paper.
![\"x^2=12\\times12=144\\qquad](\"math/68dcfd678cb95894.svg\")
![\"2x^3=3456\\quad](\"math/5f6a8611548b73ce.svg\")
![\"f(12)=3456-432+96=3120\"](\"math/ece74f225f6dc343.svg\")
![\"x^2=20\\times20=400\\qquad](\"math/b143d22c2e5129d2.svg\")
![\"2x^3=16000\\quad](\"math/581d9ac708a9edf4.svg\")
![\"f(20)=16000-1200+160=14960\"](\"math/af62ea37a6cc0979.svg\")
![\"x^2=17\\times17=289\\qquad](\"math/8106fc4d3ea404bb.svg\")
![\"2x^3=9826\\quad](\"math/3f446d5c18f6525b.svg\")
![\"f(17)=9826-867+136=9095\"](\"math/33f7934e87e644ca.svg\")
![\"2x^3=4394\\quad](\"math/963b6763e9667406.svg\")
![\"f(13)=4394-507+104=3991\"](\"math/fb78bd652840ab88.svg\")
![\"x^2=16\\times16=256\\qquad](\"math/1a7d6636e629e211.svg\")
![\"2x^3=8192\\quad](\"math/116d99ec87e1c7e9.svg\")
![\"f(16)=8192-768+128=7552\"](\"math/f0bfce15122fad53.svg\")
![\"x^2=21\\times21=441\\qquad](\"math/db60547636676c53.svg\")
![\"2x^3=18522\\quad](\"math/7a74ad5911e6f0d8.svg\")
![\"f(21)=18522-1323+168=17367\"](\"math/da828d5642e53dc9.svg\")
List the applications of fuzzy logic. Explain any one application in detail. [10 marks]
Fuzzy logic is a multi-valued logic, introduced by Lotfi Zadeh (1965), in which truth values range over [0, 1]; it lets machines reason with vague, human-like terms such as “slightly dirty” or “very hot”.
\r\n| Area | Examples |
|---|---|
| Home appliances | Washing machines (wash time from dirt and load), air conditioners, refrigerators, rice cookers, vacuum cleaners |
| Automotive | Automatic transmission, anti-lock braking (ABS), cruise control, engine control |
| Transportation | Traffic-signal control, Sendai subway train control (smooth braking and acceleration) |
| Consumer electronics | Camera autofocus and auto-exposure, image stabilisation |
| Industrial control | Cement kilns, pH, temperature and distillation control in chemical plants |
| Aerospace | Satellite and spacecraft attitude control |
| Medicine | Diagnosis support, anaesthesia dosage control |
| Finance and decisions | Loan approval, credit and fire-risk scoring, stock trading rules |
| AI and robotics | Robot navigation, natural-language interfaces, expert systems |
Advantages: handles imprecise or noisy data; mimics human reasoning; easy to design and modify; no exact model needed; low computational cost. Disadvantages: rules and membership functions depend on expert judgement; results are hard to verify formally; tuning can take time; not suited where very high precision is required.
\r\nConfirmed M2 pattern: Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer.
Q1(a). On U = {u1, u2, u3}, A has memberships [0.2, 0.8, 0.6] and B has memberships [0.7, 0.3, 0.5]. Find A ∪ B, A ∩ B, Bᶜ and A ∖ B using standard max–min fuzzy operations. [2 marks · Practice]
| Element | u1 | u2 | u3 |
|---|---|---|---|
| A | 0.2 | 0.8 | 0.6 |
| B | 0.7 | 0.3 | 0.5 |
Four membership lists in exactly the same element order.
Union is OR → max. Intersection is AND → min. Complement is NOT → 1 minus membership. Difference A ∖ B means A AND NOT B, so first complement B, then intersect with A.
All operations compare the two memberships of the same u. Do not compare A at u₁ with B at u₂.
| Element | A | B | Union | Intersection | Bᶜ | A ∖ B |
|---|---|---|---|---|---|---|
| u1 | 0.2 | 0.7 | max(0.2, 0.7) = 0.7 | min(0.2, 0.7) = 0.2 | 1 − 0.7 = 0.3 | min(0.2, 0.3) = 0.2 |
| u2 | 0.8 | 0.3 | max(0.8, 0.3) = 0.8 | min(0.8, 0.3) = 0.3 | 1 − 0.3 = 0.7 | min(0.8, 0.7) = 0.7 |
| u3 | 0.6 | 0.5 | max(0.6, 0.5) = 0.6 | min(0.6, 0.5) = 0.5 | 1 − 0.5 = 0.5 | min(0.6, 0.5) = 0.5 |
Confirmed M2 pattern: Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer.
Q1(b). Define swarm intelligence. Briefly explain one nature-inspired algorithm. [2 marks · Practice]
Swarm intelligence is collective problem-solving arising from interactions among simple agents. PSO is inspired by social movement in flocks. Each particle represents a candidate solution and has a position and velocity. It moves using its own best visited position and the swarm’s best position. Repeated objective evaluation guides the search.
Confirmed M2 pattern: Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer.
Q2. Maximise over unsigned 4-bit integers using the supplied GA settings. Perform two complete generations and compare the population fitness. [3 marks · Practice; complete two-generation solution]
Initial ordered population: [3, 5, 10, 12]. Use 4 bits, crossover after bit 2, pair mating-pool slots 1–2 and 3–4. Flip bit 4 of children [1, 2, 3, 4] in every generation. Replace the whole population with the four children; no elitism.
| Generation | Selection instruction |
|---|---|
| 1 | Current individual row numbers [3, 4, 4, 3] |
| 2 | Current individual row numbers [2, 3, 2, 3] |
Both generations: evaluation, selection, crossover, mutation, decoding and comparison.
Encoding represents x; it does not change the objective. Crossover exchanges tails. Mutation flips specified bits. Fitness must be recomputed after mutation. Use exactly the supplied selection rule; replacing the weakest by the strongest is not a universal GA step.
| Individual | x | Binary | Objective calculation |
|---|---|---|---|
| 1 | 3 | 0011 | 3 × 3 = 9 |
| 2 | 5 | 0101 | 5 × 5 = 25 |
| 3 | 10 | 1010 | 10 × 10 = 100 |
| 4 | 12 | 1100 | 12 × 12 = 144 |
The question directly supplies individual numbers [3, 4, 4, 3]. These refer to the current generation’s row numbers, not their x values.
| Mating-pool slot | Chosen individual | x | Chromosome |
|---|---|---|---|
| 1 | 3 | 10 | 1010 |
| 2 | 4 | 12 | 1100 |
| 3 | 4 | 12 | 1100 |
| 4 | 3 | 10 | 1010 |
Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1.
| Pool slots paired | Parent A | Parent B | Child A | Child B |
|---|---|---|---|---|
| 1 & 2 | 10 | 10 | 11 | 00 | 1000 | 1110 |
| 3 & 4 | 11 | 00 | 10 | 10 | 1110 | 1000 |
| Child | Before mutation | Instruction | After mutation |
|---|---|---|---|
| 1 | 1000 | Flip bit 4 | 1001 |
| 2 | 1110 | Flip bit 4 | 1111 |
| 3 | 1110 | Flip bit 4 | 1111 |
| 4 | 1000 | Flip bit 4 | 1001 |
| Chromosome | Decode using place values | Objective |
|---|---|---|
| 1001 | 8 + 0 + 0 + 1 = 9 | 9² = 81 |
| 1111 | 8 + 4 + 2 + 1 = 15 | 15² = 225 |
| 1111 | 8 + 4 + 2 + 1 = 15 | 15² = 225 |
| 1001 | 8 + 0 + 0 + 1 = 9 | 9² = 81 |
Add the objective scores in each population, then divide by its number of individuals.
Best objective: 144 → 225. Higher is better. The next population is [9, 15, 15, 9]. No elitism is imposed, so the best previous individual is not automatically protected.
| Individual | x | Binary | Objective calculation |
|---|---|---|---|
| 1 | 9 | 1001 | 9 × 9 = 81 |
| 2 | 15 | 1111 | 15 × 15 = 225 |
| 3 | 15 | 1111 | 15 × 15 = 225 |
| 4 | 9 | 1001 | 9 × 9 = 81 |
The question directly supplies individual numbers [2, 3, 2, 3]. These refer to the current generation’s row numbers, not their x values.
| Mating-pool slot | Chosen individual | x | Chromosome |
|---|---|---|---|
| 1 | 2 | 15 | 1111 |
| 2 | 3 | 15 | 1111 |
| 3 | 2 | 15 | 1111 |
| 4 | 3 | 15 | 1111 |
Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1.
| Pool slots paired | Parent A | Parent B | Child A | Child B |
|---|---|---|---|---|
| 1 & 2 | 11 | 11 | 11 | 11 | 1111 | 1111 |
| 3 & 4 | 11 | 11 | 11 | 11 | 1111 | 1111 |
| Child | Before mutation | Instruction | After mutation |
|---|---|---|---|
| 1 | 1111 | Flip bit 4 | 1110 |
| 2 | 1111 | Flip bit 4 | 1110 |
| 3 | 1111 | Flip bit 4 | 1110 |
| 4 | 1111 | Flip bit 4 | 1110 |
| Chromosome | Decode using place values | Objective |
|---|---|---|
| 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 |
| 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 |
| 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 |
| 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 |
Add the objective scores in each population, then divide by its number of individuals.
Best objective: 225 → 196. Higher is better. The next population is [14, 14, 14, 14]. No elitism is imposed, so the best previous individual is not automatically protected.
Confirmed M2 pattern: Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer.
Q3. Minimise using 2 particles. Use the initial data below and perform exactly two PSO iterations, reporting position, velocity, pbest and gbest after each iteration. [3 marks · Practice; expanded teaching solution]
| Particle | Initial position x⁰ | Initial velocity v⁰ |
|---|---|---|
| A | (-1) | (0) |
| B | (3) | (0) |
w=0.5, c₁=1, c₂=1, r₁=0.5, r₂=0.5. Initialise each pbest at its initial position. Use these random coefficients unchanged in both iterations. No bounds or velocity clamp are imposed in this example. Updates are synchronous: every particle uses the same previous-iteration gbest.
The complete swarm state after iterations 1 and 2, and the best location found.
PSO combines inertia, personal memory and shared memory. Subtract positions coordinate by coordinate. Evaluate the objective only after moving. For minimisation, replace a best only when its score decreases.
Particle A
![\"f(-1)=(-1)^2=1=1\"](\"math/cb918b9a5ef7b82c.svg\")
Particle B
pbest is each initial position because there is no earlier history. The smallest objective decides gbest.
![\"gbest^0=(-1)\"](\"math/4bb3271e5427b8e5.svg\")
Its score is 1. Store the position, not only the score.
All particles in this iteration use gbest = (-1). pbest values are A: (-1), B: (3).
Only coordinate
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=-1-(-1)=0\"](\"math/726263f8c97a9f8e.svg\")
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=-1-(-1)=0\"](\"math/6223039cb154c72e.svg\")
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(1)}=(-1)+0=-1\"](\"math/92761bdfeb82fb87.svg\")
Only coordinate
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=-1-(3)=-4\"](\"math/4224cdc3d1fd7742.svg\")
Multiply the social coefficients, then multiply by that difference.
![\"S=0.5\\times(-4)=-2\"](\"math/6d9a3c8d94a7312e.svg\")
Add the three velocity contributions.
![\"v^{(1)}=0+0+(-2)=-2\"](\"math/146db9adf4c34c65.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(1)}=3+(-2)=1\"](\"math/39edc010c99575f0.svg\")
Particle A
Particle B
| Particle | Old pbest score | New score | Decision | pbest position |
|---|---|---|---|---|
| A | 1 | 1 | Keep old: tie | (-1) |
| B | 9 | 1 | Replace: new score is lower | (1) |
Compare the retained pbest scores. gbest = (-1), with objective 1.
| Particle | Position | Velocity | pbest |
|---|---|---|---|
| A | (-1) | (0) | (-1) |
| B | (1) | (-2) | (1) |
All particles in this iteration use gbest = (-1). pbest values are A: (-1), B: (1).
Only coordinate
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(2)}=(-1)+0=-1\"](\"math/a4857620c6480702.svg\")
Only coordinate
Inertia = keep a fraction of the old velocity.
![\"I=0.5\\times(-2)=-1\"](\"math/8dba29f0d505433a.svg\")
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=-1-(1)=-2\"](\"math/de044b0c7b70c868.svg\")
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
![\"v^{(2)}=(-1)+0+(-1)=-2\"](\"math/b437ff946e9d4080.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(2)}=1+(-2)=-1\"](\"math/6aabd1c885e75a7f.svg\")
Particle A
Particle B
| Particle | Old pbest score | New score | Decision | pbest position |
|---|---|---|---|---|
| A | 1 | 1 | Keep old: tie | (-1) |
| B | 1 | 1 | Keep old: tie | (1) |
Compare the retained pbest scores. gbest = (-1), with objective 1.
| Particle | Position | Velocity | pbest |
|---|---|---|---|
| A | (-1) | (0) | (-1) |
| B | (-1) | (-2) | (1) |
Confirmed M2 pattern: Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer.
Q4. Explain the architecture of a Mamdani fuzzy inference system with a fan-controller example. [3 marks · Practice]
The fuzzifier maps crisp temperature and occupancy inputs to membership degrees. The knowledge base contains membership functions and IF–THEN rules. The inference engine combines antecedent degrees, applies each rule’s consequent and aggregates the inferred outputs. The defuzzifier converts this aggregate to a numerical fan speed. For example, IF temperature is High AND occupancy is High THEN fan speed is Very High.
Confirmed M2 pattern: Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer.
Q1(a). On U = {u1, u2, u3}, A has memberships [0.1, 0.5, 1] and B has memberships [0.9, 0.4, 0]. Find A ∪ B, A ∩ B, Bᶜ and A ∖ B using standard max–min fuzzy operations. [2 marks · Practice]
| Element | u1 | u2 | u3 |
|---|---|---|---|
| A | 0.1 | 0.5 | 1 |
| B | 0.9 | 0.4 | 0 |
Four membership lists in exactly the same element order.
Union is OR → max. Intersection is AND → min. Complement is NOT → 1 minus membership. Difference A ∖ B means A AND NOT B, so first complement B, then intersect with A.
All operations compare the two memberships of the same u. Do not compare A at u₁ with B at u₂.
![\"\\mu_{B^c}(u_1)=1-0.9=0.1\"](\"math/3e895fda5c8f4e29.svg\")
![\"\\mu_{A\\setminus](\"math/4ec160ba0bee68fe.svg\")
| Element | A | B | Union | Intersection | Bᶜ | A ∖ B |
|---|---|---|---|---|---|---|
| u1 | 0.1 | 0.9 | max(0.1, 0.9) = 0.9 | min(0.1, 0.9) = 0.1 | 1 − 0.9 = 0.1 | min(0.1, 0.1) = 0.1 |
| u2 | 0.5 | 0.4 | max(0.5, 0.4) = 0.5 | min(0.5, 0.4) = 0.4 | 1 − 0.4 = 0.6 | min(0.5, 0.6) = 0.5 |
| u3 | 1 | 0 | max(1, 0) = 1 | min(1, 0) = 0 | 1 − 0 = 1 | min(1, 1) = 1 |
Confirmed M2 pattern: Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer.
Q1(b). Define swarm intelligence. Briefly explain one nature-inspired algorithm. [2 marks · Practice]
Swarm intelligence is collective problem-solving arising from interactions among simple agents. PSO is inspired by social movement in flocks. Each particle represents a candidate solution and has a position and velocity. It moves using its own best visited position and the swarm’s best position. Repeated objective evaluation guides the search.
Confirmed M2 pattern: Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer.
Q2. Maximise over unsigned 4-bit integers using the supplied GA settings. Perform two complete generations and compare the population fitness. [3 marks · Practice; complete two-generation solution]
Initial ordered population: [1, 6, 9, 14]. Use 4 bits, crossover after bit 2, pair mating-pool slots 1–2 and 3–4. Flip bit 3 of children [1, 2, 3, 4] in every generation. Replace the whole population with the four children; no elitism.
| Generation | Selection instruction |
|---|---|
| 1 | Current individual row numbers [3, 4, 4, 3] |
| 2 | Current individual row numbers [2, 3, 2, 3] |
Both generations: evaluation, selection, crossover, mutation, decoding and comparison.
Encoding represents x; it does not change the objective. Crossover exchanges tails. Mutation flips specified bits. Fitness must be recomputed after mutation. Use exactly the supplied selection rule; replacing the weakest by the strongest is not a universal GA step.
| Individual | x | Binary | Objective calculation |
|---|---|---|---|
| 1 | 1 | 0001 | 1 × 1 = 1 |
| 2 | 6 | 0110 | 6 × 6 = 36 |
| 3 | 9 | 1001 | 9 × 9 = 81 |
| 4 | 14 | 1110 | 14 × 14 = 196 |
The question directly supplies individual numbers [3, 4, 4, 3]. These refer to the current generation’s row numbers, not their x values.
| Mating-pool slot | Chosen individual | x | Chromosome |
|---|---|---|---|
| 1 | 3 | 9 | 1001 |
| 2 | 4 | 14 | 1110 |
| 3 | 4 | 14 | 1110 |
| 4 | 3 | 9 | 1001 |
Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1.
| Pool slots paired | Parent A | Parent B | Child A | Child B |
|---|---|---|---|---|
| 1 & 2 | 10 | 01 | 11 | 10 | 1010 | 1101 |
| 3 & 4 | 11 | 10 | 10 | 01 | 1101 | 1010 |
| Child | Before mutation | Instruction | After mutation |
|---|---|---|---|
| 1 | 1010 | Flip bit 3 | 1000 |
| 2 | 1101 | Flip bit 3 | 1111 |
| 3 | 1101 | Flip bit 3 | 1111 |
| 4 | 1010 | Flip bit 3 | 1000 |
| Chromosome | Decode using place values | Objective |
|---|---|---|
| 1000 | 8 + 0 + 0 + 0 = 8 | 8² = 64 |
| 1111 | 8 + 4 + 2 + 1 = 15 | 15² = 225 |
| 1111 | 8 + 4 + 2 + 1 = 15 | 15² = 225 |
| 1000 | 8 + 0 + 0 + 0 = 8 | 8² = 64 |
Add the objective scores in each population, then divide by its number of individuals.
![\"\\overline](\"math/7aeb7f13264a06b2.svg\")
![\"\\overline](\"math/534a2fd2aa176655.svg\")
Best objective: 196 → 225. Higher is better. The next population is [8, 15, 15, 8]. No elitism is imposed, so the best previous individual is not automatically protected.
| Individual | x | Binary | Objective calculation |
|---|---|---|---|
| 1 | 8 | 1000 | 8 × 8 = 64 |
| 2 | 15 | 1111 | 15 × 15 = 225 |
| 3 | 15 | 1111 | 15 × 15 = 225 |
| 4 | 8 | 1000 | 8 × 8 = 64 |
The question directly supplies individual numbers [2, 3, 2, 3]. These refer to the current generation’s row numbers, not their x values.
| Mating-pool slot | Chosen individual | x | Chromosome |
|---|---|---|---|
| 1 | 2 | 15 | 1111 |
| 2 | 3 | 15 | 1111 |
| 3 | 2 | 15 | 1111 |
| 4 | 3 | 15 | 1111 |
Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1.
| Pool slots paired | Parent A | Parent B | Child A | Child B |
|---|---|---|---|---|
| 1 & 2 | 11 | 11 | 11 | 11 | 1111 | 1111 |
| 3 & 4 | 11 | 11 | 11 | 11 | 1111 | 1111 |
| Child | Before mutation | Instruction | After mutation |
|---|---|---|---|
| 1 | 1111 | Flip bit 3 | 1101 |
| 2 | 1111 | Flip bit 3 | 1101 |
| 3 | 1111 | Flip bit 3 | 1101 |
| 4 | 1111 | Flip bit 3 | 1101 |
| Chromosome | Decode using place values | Objective |
|---|---|---|
| 1101 | 8 + 4 + 0 + 1 = 13 | 13² = 169 |
| 1101 | 8 + 4 + 0 + 1 = 13 | 13² = 169 |
| 1101 | 8 + 4 + 0 + 1 = 13 | 13² = 169 |
| 1101 | 8 + 4 + 0 + 1 = 13 | 13² = 169 |
Add the objective scores in each population, then divide by its number of individuals.
![\"\\overline](\"math/290d9d3cf2ac1342.svg\")
![\"\\overline](\"math/d460690a1684b5f8.svg\")
Best objective: 225 → 169. Higher is better. The next population is [13, 13, 13, 13]. No elitism is imposed, so the best previous individual is not automatically protected.
Confirmed M2 pattern: Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer.
Q3. Minimise using 2 particles. Use the initial data below and perform exactly two PSO iterations, reporting position, velocity, pbest and gbest after each iteration. [3 marks · Practice; expanded teaching solution]
| Particle | Initial position x⁰ | Initial velocity v⁰ |
|---|---|---|
| A | (2) | (0) |
| B | (5) | (0) |
w=0.5, c₁=1, c₂=1, r₁=0.5, r₂=0.5. Initialise each pbest at its initial position. Use these random coefficients unchanged in both iterations. No bounds or velocity clamp are imposed in this example. Updates are synchronous: every particle uses the same previous-iteration gbest.
The complete swarm state after iterations 1 and 2, and the best location found.
PSO combines inertia, personal memory and shared memory. Subtract positions coordinate by coordinate. Evaluate the objective only after moving. For minimisation, replace a best only when its score decreases.
Particle A
Particle B
![\"f(5)=5^2=25=25\"](\"math/12fb2bbb149af9d6.svg\")
pbest is each initial position because there is no earlier history. The smallest objective decides gbest.
![\"gbest^0=(2)\"](\"math/a00ef2dd53de0008.svg\")
Its score is 4. Store the position, not only the score.
All particles in this iteration use gbest = (2). pbest values are A: (2), B: (5).
Only coordinate
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=2-(2)=0\"](\"math/df4d0e49facbaae8.svg\")
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(1)}=2+0=2\"](\"math/bfaa1b2d391c5f4e.svg\")
Only coordinate
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=5-(5)=0\"](\"math/302db7bc64d6f9e9.svg\")
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=2-(5)=-3\"](\"math/c53a1760c2aca6d6.svg\")
Multiply the social coefficients, then multiply by that difference.
![\"S=0.5\\times(-3)=-1.5\"](\"math/13ae8508c1795085.svg\")
Add the three velocity contributions.
![\"v^{(1)}=0+0+(-1.5)=-1.5\"](\"math/4e26de872fe8bfca.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(1)}=5+(-1.5)=3.5\"](\"math/83a647473c6fa205.svg\")
Particle A
Particle B
![\"f(3.5)=3.5^2=12.25=12.25\"](\"math/4c8cdadcd9350fe6.svg\")
| Particle | Old pbest score | New score | Decision | pbest position |
|---|---|---|---|---|
| A | 4 | 4 | Keep old: tie | (2) |
| B | 25 | 12.25 | Replace: new score is lower | (3.5) |
Compare the retained pbest scores. gbest = (2), with objective 4.
| Particle | Position | Velocity | pbest |
|---|---|---|---|
| A | (2) | (0) | (2) |
| B | (3.5) | (-1.5) | (3.5) |
All particles in this iteration use gbest = (2). pbest values are A: (2), B: (3.5).
Only coordinate
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(2)}=2+0=2\"](\"math/2189c0d7cea22f3c.svg\")
Only coordinate
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=3.5-(3.5)=0\"](\"math/dfcb6b54b5bd04e2.svg\")
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=2-(3.5)=-1.5\"](\"math/01ad6e560abc36e6.svg\")
Multiply the social coefficients, then multiply by that difference.
![\"S=0.5\\times(-1.5)=-0.75\"](\"math/59d60f341dbf8e9b.svg\")
Add the three velocity contributions.
![\"v^{(2)}=(-0.75)+0+(-0.75)=-1.5\"](\"math/02342391a534b540.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(2)}=3.5+(-1.5)=2\"](\"math/24dbc384ce5eb239.svg\")
Particle A
Particle B
| Particle | Old pbest score | New score | Decision | pbest position |
|---|---|---|---|---|
| A | 4 | 4 | Keep old: tie | (2) |
| B | 12.25 | 4 | Replace: new score is lower | (2) |
Compare the retained pbest scores. gbest = (2), with objective 4.
| Particle | Position | Velocity | pbest |
|---|---|---|---|
| A | (2) | (0) | (2) |
| B | (2) | (-1.5) | (2) |
Confirmed M2 pattern: Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer.
Q4. A fuzzy set A has triangular membership Tri(15, 25, 35). Find membership at x = 32. [3 marks · Practice]
Left endpoint a = 15; peak b = 25; right endpoint c = 35; input x = 32.
The membership degree μA(x), a number from 0 to 1.
Locate x relative to a, b, c before choosing the branch.
25 < 32 < 35, so membership decreases as x moves away from the peak.
Falling side
![\"\\mu_A(32)=\\frac{35-32}{35-25}=\\frac{3}{10}=0.3\"](\"math/3ed9e45724bceb7e.svg\")
The degree of belonging to A is 0.3. It is a membership degree, not a probability.
Confirmed M2 pattern: Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer.
Q1(a). On U = {u1, u2, u3}, A has memberships [0.7, 0.2, 0.9] and B has memberships [0.3, 0.8, 0.4]. Find A ∪ B, A ∩ B, Bᶜ and A ∖ B using standard max–min fuzzy operations. [2 marks · Practice]
| Element | u1 | u2 | u3 |
|---|---|---|---|
| A | 0.7 | 0.2 | 0.9 |
| B | 0.3 | 0.8 | 0.4 |
Four membership lists in exactly the same element order.
Union is OR → max. Intersection is AND → min. Complement is NOT → 1 minus membership. Difference A ∖ B means A AND NOT B, so first complement B, then intersect with A.
All operations compare the two memberships of the same u. Do not compare A at u₁ with B at u₂.
![\"\\mu_{B^c}(u_1)=1-0.3=0.7\"](\"math/0d89f5d2d8ca13ed.svg\")
![\"\\mu_{A\\setminus](\"math/c98cc0f742bee1be.svg\")
| Element | A | B | Union | Intersection | Bᶜ | A ∖ B |
|---|---|---|---|---|---|---|
| u1 | 0.7 | 0.3 | max(0.7, 0.3) = 0.7 | min(0.7, 0.3) = 0.3 | 1 − 0.3 = 0.7 | min(0.7, 0.7) = 0.7 |
| u2 | 0.2 | 0.8 | max(0.2, 0.8) = 0.8 | min(0.2, 0.8) = 0.2 | 1 − 0.8 = 0.2 | min(0.2, 0.2) = 0.2 |
| u3 | 0.9 | 0.4 | max(0.9, 0.4) = 0.9 | min(0.9, 0.4) = 0.4 | 1 − 0.4 = 0.6 | min(0.9, 0.6) = 0.6 |
Confirmed M2 pattern: Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer.
Q1(b). Define swarm intelligence. Briefly explain one nature-inspired algorithm. [2 marks · Practice]
Swarm intelligence is collective problem-solving arising from interactions among simple agents. PSO is inspired by social movement in flocks. Each particle represents a candidate solution and has a position and velocity. It moves using its own best visited position and the swarm’s best position. Repeated objective evaluation guides the search.
Confirmed M2 pattern: Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer.
Q2. Maximise over unsigned 4-bit integers using the supplied GA settings. Perform two complete generations and compare the population fitness. [3 marks · Practice; complete two-generation solution]
Initial ordered population: [2, 7, 8, 13]. Use 4 bits, crossover after bit 2, pair mating-pool slots 1–2 and 3–4. Flip bit 4 of children [1, 2, 3, 4] in every generation. Replace the whole population with the four children; no elitism.
| Generation | Selection instruction |
|---|---|
| 1 | Current individual row numbers [3, 4, 4, 3] |
| 2 | Current individual row numbers [2, 3, 2, 3] |
Both generations: evaluation, selection, crossover, mutation, decoding and comparison.
Encoding represents x; it does not change the objective. Crossover exchanges tails. Mutation flips specified bits. Fitness must be recomputed after mutation. Use exactly the supplied selection rule; replacing the weakest by the strongest is not a universal GA step.
| Individual | x | Binary | Objective calculation |
|---|---|---|---|
| 1 | 2 | 0010 | 2 × 2 = 4 |
| 2 | 7 | 0111 | 7 × 7 = 49 |
| 3 | 8 | 1000 | 8 × 8 = 64 |
| 4 | 13 | 1101 | 13 × 13 = 169 |
The question directly supplies individual numbers [3, 4, 4, 3]. These refer to the current generation’s row numbers, not their x values.
| Mating-pool slot | Chosen individual | x | Chromosome |
|---|---|---|---|
| 1 | 3 | 8 | 1000 |
| 2 | 4 | 13 | 1101 |
| 3 | 4 | 13 | 1101 |
| 4 | 3 | 8 | 1000 |
Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1.
| Pool slots paired | Parent A | Parent B | Child A | Child B |
|---|---|---|---|---|
| 1 & 2 | 10 | 00 | 11 | 01 | 1001 | 1100 |
| 3 & 4 | 11 | 01 | 10 | 00 | 1100 | 1001 |
| Child | Before mutation | Instruction | After mutation |
|---|---|---|---|
| 1 | 1001 | Flip bit 4 | 1000 |
| 2 | 1100 | Flip bit 4 | 1101 |
| 3 | 1100 | Flip bit 4 | 1101 |
| 4 | 1001 | Flip bit 4 | 1000 |
| Chromosome | Decode using place values | Objective |
|---|---|---|
| 1000 | 8 + 0 + 0 + 0 = 8 | 8² = 64 |
| 1101 | 8 + 4 + 0 + 1 = 13 | 13² = 169 |
| 1101 | 8 + 4 + 0 + 1 = 13 | 13² = 169 |
| 1000 | 8 + 0 + 0 + 0 = 8 | 8² = 64 |
Add the objective scores in each population, then divide by its number of individuals.
![\"\\overline](\"math/6171658d65ed4b49.svg\")
![\"\\overline](\"math/866b6bb4aa32a9a4.svg\")
Best objective: 169 → 169. Higher is better. The next population is [8, 13, 13, 8]. No elitism is imposed, so the best previous individual is not automatically protected.
| Individual | x | Binary | Objective calculation |
|---|---|---|---|
| 1 | 8 | 1000 | 8 × 8 = 64 |
| 2 | 13 | 1101 | 13 × 13 = 169 |
| 3 | 13 | 1101 | 13 × 13 = 169 |
| 4 | 8 | 1000 | 8 × 8 = 64 |
The question directly supplies individual numbers [2, 3, 2, 3]. These refer to the current generation’s row numbers, not their x values.
| Mating-pool slot | Chosen individual | x | Chromosome |
|---|---|---|---|
| 1 | 2 | 13 | 1101 |
| 2 | 3 | 13 | 1101 |
| 3 | 2 | 13 | 1101 |
| 4 | 3 | 13 | 1101 |
Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1.
| Pool slots paired | Parent A | Parent B | Child A | Child B |
|---|---|---|---|---|
| 1 & 2 | 11 | 01 | 11 | 01 | 1101 | 1101 |
| 3 & 4 | 11 | 01 | 11 | 01 | 1101 | 1101 |
| Child | Before mutation | Instruction | After mutation |
|---|---|---|---|
| 1 | 1101 | Flip bit 4 | 1100 |
| 2 | 1101 | Flip bit 4 | 1100 |
| 3 | 1101 | Flip bit 4 | 1100 |
| 4 | 1101 | Flip bit 4 | 1100 |
| Chromosome | Decode using place values | Objective |
|---|---|---|
| 1100 | 8 + 4 + 0 + 0 = 12 | 12² = 144 |
| 1100 | 8 + 4 + 0 + 0 = 12 | 12² = 144 |
| 1100 | 8 + 4 + 0 + 0 = 12 | 12² = 144 |
| 1100 | 8 + 4 + 0 + 0 = 12 | 12² = 144 |
Add the objective scores in each population, then divide by its number of individuals.
![\"\\overline](\"math/fcafd07eb412a037.svg\")
![\"\\overline](\"math/cd00806a61b8ba0c.svg\")
Best objective: 169 → 144. Higher is better. The next population is [12, 12, 12, 12]. No elitism is imposed, so the best previous individual is not automatically protected.
Confirmed M2 pattern: Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer.
Q3. Minimise using 2 particles. Use the initial data below and perform exactly two PSO iterations, reporting position, velocity, pbest and gbest after each iteration. [3 marks · Practice; expanded teaching solution]
| Particle | Initial position x⁰ | Initial velocity v⁰ |
|---|---|---|
| A | (1) | (2) |
| B | (4) | (-1) |
w=0.5, c₁=1, c₂=1, r₁=0.5, r₂=0.5. Initialise each pbest at its initial position. Use these random coefficients unchanged in both iterations. No bounds or velocity clamp are imposed in this example. Updates are synchronous: every particle uses the same previous-iteration gbest.
The complete swarm state after iterations 1 and 2, and the best location found.
PSO combines inertia, personal memory and shared memory. Subtract positions coordinate by coordinate. Evaluate the objective only after moving. For minimisation, replace a best only when its score decreases.
Particle A
Particle B
pbest is each initial position because there is no earlier history. The smallest objective decides gbest.
Its score is 1. Store the position, not only the score.
All particles in this iteration use gbest = (1). pbest values are A: (1), B: (4).
Only coordinate
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
![\"v^{(1)}=1+0+0=1\"](\"math/9a2ea8db49561d57.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(1)}=1+1=2\"](\"math/f9d52a6143caaf7b.svg\")
Only coordinate
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
![\"d_g=1-(4)=-3\"](\"math/7cc359443926f7d3.svg\")
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
![\"v^{(1)}=(-0.5)+0+(-1.5)=-2\"](\"math/1329839a69ba9e13.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(1)}=4+(-2)=2\"](\"math/2b24c838f14c9776.svg\")
Particle A
Particle B
| Particle | Old pbest score | New score | Decision | pbest position |
|---|---|---|---|---|
| A | 1 | 4 | Keep old: new score is worse | (1) |
| B | 16 | 4 | Replace: new score is lower | (2) |
Compare the retained pbest scores. gbest = (1), with objective 1.
| Particle | Position | Velocity | pbest |
|---|---|---|---|
| A | (2) | (1) | (1) |
| B | (2) | (-2) | (2) |
All particles in this iteration use gbest = (1). pbest values are A: (1), B: (2).
Only coordinate
Inertia = keep a fraction of the old velocity.
![\"I=0.5\\times1=0.5\"](\"math/15c0369c89ebba10.svg\")
Personal pull: first subtract the current coordinate from the personal-best coordinate.
![\"d_p=1-(2)=-1\"](\"math/72aafd3f3c32f9a7.svg\")
Multiply the two personal coefficients, then multiply by the difference.
![\"C=0.5\\times(-1)=-0.5\"](\"math/5fb29aabbbcf9b90.svg\")
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
![\"v^{(2)}=0.5+(-0.5)+(-0.5)=-0.5\"](\"math/cd06fdfee86fe231.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(2)}=2+(-0.5)=1.5\"](\"math/7929aa22070f7f4e.svg\")
Only coordinate
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
![\"v^{(2)}=(-1)+0+(-0.5)=-1.5\"](\"math/88e7bce915d4dcc4.svg\")
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
![\"x^{(2)}=2+(-1.5)=0.5\"](\"math/5721079871bdab79.svg\")
Particle A
Particle B
| Particle | Old pbest score | New score | Decision | pbest position |
|---|---|---|---|---|
| A | 1 | 2.25 | Keep old: new score is worse | (1) |
| B | 4 | 0.25 | Replace: new score is lower | (0.5) |
Compare the retained pbest scores. gbest = (0.5), with objective 0.25.
| Particle | Position | Velocity | pbest |
|---|---|---|---|
| A | (1.5) | (-0.5) | (1) |
| B | (0.5) | (-1.5) | (0.5) |
Confirmed M2 pattern: Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer.
Q4. Using the weighted-average method, combine output representatives [20, 50, 80] % with respective strengths [0.2, 0.6, 0.2]. [3 marks · Practice]
| Output representative | Strength |
|---|---|
| 20 | 0.2 |
| 50 | 0.6 |
| 80 | 0.2 |
One crisp output by the specified weighted-average method.
Each representative is multiplied by its strength; add these products and divide by the total strength. Here α is a strength and z is a representative value. These are not automatically the full area of a clipped Mamdani set.
| Strength × value | Product |
|---|---|
| 0.2 × 20 | 4 |
| 0.6 × 50 | 30 |
| 0.2 × 80 | 16 |
![\"z^*=\\frac{50}{1}=50\"](\"math/9c1293c80ea00bab.svg\")
| Asked for | Write |
|---|---|
| Union / intersection | Same element, max / min respectively. |
| Complement / difference | 1−μA / min(μA,1−μB). |
| Triangle membership | Locate x; rising (x−a)/(b−a), falling (c−x)/(c−b), 0 outside, 1 at peak. |
| Swarm intelligence | Collective problem-solving emerging from interacting simple agents; explain PSO through position, velocity, pbest and gbest. |
Individual → current x → bit string → fitness → selected parent → crossover child → mutated child → decoded x → new fitness. Use this resulting population as the input to the second generation. The cut position and flipped bit are counted from the stated end; this guide counts from the left.
\r\nDo not change the mutation instruction midway because an answer looks worse. Without elitism, fitness can worsen. Do not confuse a mating-pool index with the decimal x value.
\r\nParticle → old position → old velocity → pbest → gbest → inertia/personal/social terms → new velocity → new position → new objective → retained pbest. After all particles, update gbest. For synchronous updates, freeze the old gbest throughout one iteration.
\r\nAt initialisation pbest=x, hence personal pull=0. In iteration 2 it need not be zero. For minimisation, smaller objective wins; a worse move must not overwrite a better memory.
\r\nMamdani: crisp inputs → fuzzifier → knowledge base and inference engine → aggregation → defuzzifier → crisp output. Rules are interpretable, but require careful membership/rule design and can grow in number.
GA: population search inspired by evolution, using evaluation, selection, crossover, mutation and replacement. Flexible and derivative-free, but may be expensive or converge prematurely.
PSO: particles move using inertia, personal best and swarm best. Simple and derivative-free, but parameter-dependent and not guaranteed to find a global optimum.
\r\nEnter a fraction as (whole numerator) ÷ (whole denominator). For a negative square use parentheses: (−2)²=4. Retain full calculator precision through iterations and round only the displayed result. Binary place values 8,4,2,1 can be checked manually without a special calculator mode.
\r\nYaad rakho: Question ki instruction follow karo; answer ko sundar banane ke liye data ya operator mat badlo.
"}]}],"about":[{"id":"about","title":"Portion, sources and corrections","unit":"0","scope":"What this guide covers, where it came from, and every error found in the slides and answer keys","sections":[{"id":"portion-sources","title":"Portion, source priority and what was checked","html":"The 8 October class update identifies Fuzzy Logic, GA and PSO, including rules, architecture, advantages/disadvantages, fuzzy operations and numerical work through iteration 2. The confirmed pattern is Q1(a) 2 marks and Q1(b) 2 marks compulsory, plus any two of three 3-mark questions. ACO and GA-for-TSP remain in TEE, not the M2 core.
\r\nThe course policy specifies a 100-mark, three-hour TEE scaled to 50, and 10-mark class tests. The available Final 24 November 2025 and Re-exam 31 January 2026 papers have Q1 compulsory plus any four of Q2–Q7. The next paper’s instructions take precedence.
\r\nPrimary course sources are the AY 2026–27 syllabus and course-policy files, faculty slides and solved-problem material in the 8 October OneDrive archive, the two actual question papers, and the class data supplied in the conversation. The old generated IS2 draft and the separate generated midterm PDF are not authorities for the new teaching sequence.
\r\nHandwritten photos can contain inconsistent arithmetic or incomplete conventions. The fan example states its chosen membership functions; the PSO example states minimisation and synchronous updates. Missing selection choices, mutation bits or inconsistent encodings in older papers are labelled explicitly rather than presented as unique answers.
\r\nNew worked numerical examples are calculated by reproducible scripts and checked separately. Build checks validate formula rendering and internal routes. This does not justify a promise that no typo can remain, that every future numerical is predictable, or that reading alone guarantees a score.
"},{"id":"corrections","title":"How inconsistent source questions are handled","html":"The final GA paper requests four-bit parents larger than 15 while also restricting x to 0–15. Those conditions cannot all hold. Its worked example declares a repaired parent set. The re-exam GA paper needs six bits to cover the stated 0–50 range. An encoding change must not silently change the legal range.
\r\nThe final joint table sums to 1.002 rather than 1. Its lesson distinguishes ratios from the printed weights, normalisation, and an assumed typo correction. None is presented as a guaranteed marking-scheme interpretation.
\r\nSome fuzzy classroom solutions calculate a peak-weighted output and call it a centroid. The new lessons name that operation and separately teach the true area centroid. The class fan photographs do not supply one consistent complete set of membership functions, so its worked design makes every breakpoint explicit.
\r\nFor relation composition, check the intermediate labels and the author’s convention for S∘R. A numerical matrix product alone does not resolve a mismatched relation domain.
"}]}]},"tee":{"read":[{"id":"tee-start","title":"Start here · TEE plan","unit":"0","scope":"Term-end examination · 100 marks · 3 hours · all four units","sections":[{"id":"tee-plan","title":"How the TEE is set and how to prepare","html":"| Topic | Unit | Nov 2025 | Jan 2026 |
|---|---|---|---|
| Joint table (5) | 1 | Q1a | Q1a |
| GMM: k-means, log-likelihood, parameters (10) | 1 | Q2b | Q2b |
| Bayesian network query (10) | 1 | Q3a | Q3a |
| Naive Bayes (10) | 1 | Q4b | Q4b |
| Fuzzy set operations (5) | 2 | Q1c | Q1c |
| Mamdani design + defuzzify (10) | 2 | Q3b | Q2a |
| Fuzzy relation composition (10) | 2 | Q4a | Q3b |
| Fuzzy applications (10) | 2 | Q6a | Q7a |
| PSO numerical (10) | 3 | Q2a | Q4a |
| GA numerical (10) | 3 | Q5b | Q5a |
| ACO numerical (10) | 3 | Q7b | Q6a |
| PSO / ACO applications (5–10) | 3 | Q1b, Q7a | Q1b |
| Markov 2-state (5) | 4 | Q1d | Q1d |
| Markov 3-state, 5 parts (10) | 4 | Q5a, Q6b | Q5b, Q6b |
| Bellman + RL applications (10) | 4 | — | Q7b |
These two available papers repeat many question families. They guide practice but do not establish an exhaustive list for the next examination. The reading route also covers the official syllabus topics not asked in these two papers.
\r\n\r\nThe TEE reading route covers Units 1–4; if continuing after M2, begin with Unit 1, then the TEE-only GA-for-TSP/ACO material and Unit 4, solve both papers in the “Solved papers” tab without looking, compare, and finish with the formula sheets and the last-day revision.
The table above describes the actual November 2025 final and January 2026 re-exam. The solved papers retain their original marks. Other worked examples labelled Practice are authored learning exercises; their marks are suggested practice sizes.
Both papers ask GMM comparison with k-means, log-likelihood derivation and a K=5 parameter count. The small EM-update examples explain the faculty topic but are not additional observed TEE questions. Q-learning is an optional extension, outside the core formula list. Retain the full syllabus theory even when these two papers did not ask a particular variant.
"},{"id":"question-map","title":"Every paper: question, unit, marks and solved answer","html":"Each row links directly to the full question and its answer. Offered marks count all printed choices; attempted marks are the marks you are allowed to answer. Unit totals below describe what appeared on that paper, not a prediction or a compulsory unit-wise distribution.
\r\nTypes: Numerical = calculations; Theory = explain/compare; Mixed = calculations with explanation; Case = apply or design a system. A case can contain substantial calculation.
\r\n100 marks to attempt; 140 marks offered. Q1 compulsory; choose any four complete questions from Q2–Q7. Each whole question carries 20 marks.
\r\nOffered marks by current unit: Unit 1: 35, Unit 2: 35, Unit 3: 35, Unit 4: 35. Unspecified internal task marks are not split or guessed.
\r\n| Question | Marks | Current unit and topic | Type | Answer |
|---|---|---|---|---|
| Q1(a) | 5 | Unit 1 · Joint probability: disease, fever and test | Numerical | Full solution |
| Q1(b) | 5 | Unit 3 · PSO applications | Theory | Full solution |
| Q1(c) | 5 | Unit 2 · Fuzzy set operations | Numerical | Full solution |
| Q1(d) | 5 | Unit 4 · Two-state Markov bank model | Numerical | Full solution |
| Q2(a) | 10 | Unit 2 · Mamdani fire-controller design and centroid | Case | Full solution |
| Q2(b) | 10 | Unit 1 · GMM, k-means, likelihood and parameters | Mixed | Full solution |
| Q3(a) | 10 | Unit 1 · Bayesian network: fever given no flu | Numerical | Full solution |
| Q3(b) | 10 | Unit 2 · Fuzzy Cartesian product and composition | Numerical | Full solution |
| Q4(a) | 10 | Unit 3 · PSO one iteration and best memories | Numerical | Full solution |
| Q4(b) | 10 | Unit 1 · Naive Bayes job classification | Numerical | Full solution |
| Q5(a) | 10 | Unit 3 · Genetic algorithm offspring and fitness | Numerical | Full solution |
| Q5(b) | 10 | Unit 4 · Markov booking channels and stationary distribution | Numerical | Full solution |
| Q6(a) | 10 | Unit 3 · ACO travelling salesman tour and pheromones | Numerical | Full solution |
| Q6(b) | 10 | Unit 4 · Markov aircraft movement and stationary distribution | Numerical | Full solution |
| Q7(a) | 10 | Unit 2 · Fuzzy logic applications | Theory | Full solution |
| Q7(b) | 10 | Unit 4 · Bellman equation and RL applications | Theory | Full solution |
100 marks to attempt; 140 marks offered. Q1 compulsory; choose any four complete questions from Q2–Q7. Each whole question carries 20 marks.
\r\nOffered marks by current unit: Unit 1: 35, Unit 2: 35, Unit 3: 45, Unit 4: 25. Unspecified internal task marks are not split or guessed.
\r\n| Question | Marks | Current unit and topic | Type | Answer |
|---|---|---|---|---|
| Q1(a) | 5 | Unit 1 · Joint probability and conditional inference | Numerical | Full solution |
| Q1(b) | 5 | Unit 3 · ACO applications | Theory | Full solution |
| Q1(c) | 5 | Unit 2 · Fuzzy set operations | Numerical | Full solution |
| Q1(d) | 5 | Unit 4 · Two-state Markov transition model | Numerical | Full solution |
| Q2(a) | 10 | Unit 3 · PSO one iteration and best memories | Numerical | Full solution |
| Q2(b) | 10 | Unit 1 · GMM, k-means, likelihood and parameters | Mixed | Full solution |
| Q3(a) | 10 | Unit 1 · Bayesian network: wet grass given no rain | Numerical | Full solution |
| Q3(b) | 10 | Unit 2 · Mamdani loan-controller design and centroid | Case | Full solution |
| Q4(a) | 10 | Unit 2 · Fuzzy Cartesian product and composition | Numerical | Full solution |
| Q4(b) | 10 | Unit 1 · Naive Bayes flu classification | Numerical | Full solution |
| Q5(a) | 10 | Unit 4 · Markov hotel movement and stationary distribution | Numerical | Full solution |
| Q5(b) | 10 | Unit 3 · Genetic algorithm offspring and fitness | Numerical | Full solution |
| Q6(a) | 10 | Unit 2 · Fuzzy logic applications | Theory | Full solution |
| Q6(b) | 10 | Unit 4 · Markov learner progression and stationary state | Numerical | Full solution |
| Q7(a) | 10 | Unit 3 · PSO applications | Theory | Full solution |
| Q7(b) | 10 | Unit 3 · ACO travelling salesman tour and pheromones | Numerical | Full solution |
100 marks to attempt; 140 marks offered. Q1 compulsory; choose any four complete questions from Q2–Q7. Each whole question carries 20 marks.
\r\nOffered marks by current unit: Unit 1: 35, Unit 2: 35, Unit 3: 35, Unit 4: 35. Unspecified internal task marks are not split or guessed.
\r\n| Question | Marks | Current unit and topic | Type | Answer |
|---|---|---|---|---|
| Q1(a) | 5 | Unit 1 · Joint masses, normalization and conditional probability | Numerical | Full solution |
| Q1(b) | 5 | Unit 3 · Genetic algorithm applications | Theory | Full solution |
| Q1(c) | 5 | Unit 2 · Fuzzy union, intersection and complements | Numerical | Full solution |
| Q1(d) | 5 | Unit 4 · Cable-company Markov matrix and diagram | Numerical | Full solution |
| Q2(a) | 10 | Unit 2 · Fuzzy investment-controller design and centroid | Numerical | Full solution |
| Q2(b) | 10 | Unit 1 · GMM versus k-means, log likelihood and parameter count | Theory | Full solution |
| Q3(a) | 10 | Unit 1 · Bayesian-network conditional cough probability | Numerical | Full solution |
| Q3(b) | 10 | Unit 2 · Fuzzy Cartesian relations and composition-domain ambiguity | Numerical | Full solution |
| Q4(a) | 10 | Unit 3 · PSO one iteration with missing initialization | Numerical | Full solution |
| Q4(b) | 10 | Unit 1 · Naive Bayes product-purchase classification | Numerical | Full solution |
| Q5(a) | 10 | Unit 3 · Binary GA crossover, one-bit mutation and fitness | Numerical | Full solution |
| Q5(b) | 10 | Unit 4 · Bike-sharing Markov transitions, retention and gains | Numerical | Full solution |
| Q6(a) | 10 | Unit 3 · ACO tour lengths, evaporation and pheromone deposition | Numerical | Full solution |
| Q6(b) | 10 | Unit 4 · Two-step walking/cycling Markov transition | Numerical | Full solution |
| Q7(a) | 10 | Unit 2 · Fuzzy logic applications and complete controller example | Theory | Full solution |
| Q7(b) | 10 | Unit 4 · Bellman equations and reinforcement-learning applications | Theory | Full solution |
Source qualification: Some questions omit parameter choices or contain ambiguous/inconsistent source data. The relevant answer distinguishes printed data from assumptions and calculates valid alternatives; all source questions remain represented.
\r\n100 marks to attempt; 140 marks offered. Q1 compulsory; choose any four complete questions from Q2–Q7. Each whole question carries 20 marks.
\r\nOffered marks by current unit: Unit 1: 30, Unit 2: 40, Unit 3: 40, Unit 4: 30. Unspecified internal task marks are not split or guessed.
\r\n| Question | Marks | Current unit and topic | Type | Answer |
|---|---|---|---|---|
| Q1(a) | 5 | Unit 3 · Traditional versus evolutionary optimization | Theory | Full solution |
| Q1(b) | 5 | Unit 1 · Bayes posterior for red ball | Numerical | Full solution |
| Q1(c) | 5 | Unit 2 · Fuzzy union, intersection and complements | Numerical | Full solution |
| Q1(d) | 5 | Unit 4 · Irreducibility and stationary distribution | Numerical | Full solution |
| Q2(a) | 10 | Unit 1 · GMM modified log likelihood and parameters | Mixed | Full solution |
| Q2(b) | 10 | Unit 2 · Fuzzy washing-machine controller | Design | Full solution |
| Q3(a) | 10 | Unit 2 · Fuzzy relation and max-product comparison | Numerical | Full solution |
| Q3(b) | 10 | Unit 3 · PSO maximization, supplied parameters | Numerical | Full solution |
| Q4(a) | 10 | Unit 1 · Naive Bayes and Laplace smoothing | Numerical | Full solution |
| Q4(b) | 10 | Unit 4 · Transition matrix properties and two-step probabilities | Numerical | Full solution |
| Q5(a) | 10 | Unit 3 · GA selection, crossover and survivor replacement | Numerical | Full solution |
| Q5(b) | 10 | Unit 2 · Mean of maxima and centroid of trapezoids | Numerical | Full solution |
| Q6(a) | 10 | Unit 3 · Genetic algorithm for four-city TSP | Theory | Full solution |
| Q6(b) | 10 | Unit 4 · Markov process and homework transition paths | Mixed | Full solution |
| Q7(a) | 5 | Unit 1 · Expectation Maximization algorithm | Theory | Full solution |
| Q7(b) | 5 | Unit 2 · Lambda-cut of a fuzzy relation | Numerical | Full solution |
| Q7(c) | 5 | Unit 3 · ACO path length and pheromone update | Numerical | Full solution |
| Q7(d) | 5 | Unit 4 · Bellman equation | Theory | Full solution |
Source qualification: Some questions omit parameter choices or contain ambiguous/inconsistent source data. The relevant answer distinguishes printed data from assumptions and calculates valid alternatives; all source questions remain represented.
\r\n100 marks to attempt; 140 marks offered. Q1 compulsory; choose any four complete questions from Q2–Q7. Each whole question carries 20 marks.
\r\nOffered marks by current unit: Unit 1: 30, Unit 2: 40, Unit 3: 40, Unit 4: 30. Unspecified internal task marks are not split or guessed.
\r\n| Question | Marks | Current unit and topic | Type | Answer |
|---|---|---|---|---|
| Q1(a) | 5 | Unit 1 · Conditional probability from fruit counts | Numerical | Full solution |
| Q1(b) | 5 | Unit 2 · Membership function and age sketch | Theory | Full solution |
| Q1(c) | 5 | Unit 3 · Mutation operators | Theory | Full solution |
| Q1(d) | 5 | Unit 4 · Markov chain diagram and path probability | Numerical | Full solution |
| Q2(a) | 10 | Unit 1 · Bayesian network joint probabilities | Numerical | Full solution |
| Q2(b) | 10 | Unit 2 · Fuzzy Cartesian product and composition | Numerical | Full solution |
| Q3(a) | 10 | Unit 1 · GMM versus K-means | Theory | Full solution |
| Q3(b) | 10 | Unit 2 · Center of sums defuzzification | Numerical | Full solution |
| Q4(a) | 10 | Unit 2 · Fuzzy quality-control design | Design | Full solution |
| Q4(b) | 10 | Unit 3 · ACO components and roles | Theory | Full solution |
| Q5(a) | 10 | Unit 3 · PSO inspiration, parameters and one iteration | Mixed | Full solution |
| Q5(b) | 10 | Unit 4 · Brand-switching transition matrix | Numerical | Full solution |
| Q6(a) | 10 | Unit 3 · GA ranking and three crossover methods | Numerical | Full solution |
| Q6(b) | 10 | Unit 4 · Markov property and weather path | Mixed | Full solution |
| Q7(a) | 5 | Unit 1 · Bayes bag posterior | Numerical | Full solution |
| Q7(b) | 5 | Unit 2 · Max–min relational composition | Numerical | Full solution |
| Q7(c) | 5 | Unit 3 · Point versus uniform crossover | Theory | Full solution |
| Q7(d) | 5 | Unit 4 · Applications of reinforcement learning | Theory | Full solution |
Source qualification: Some questions omit parameter choices or contain ambiguous/inconsistent source data. The relevant answer distinguishes printed data from assumptions and calculates valid alternatives; all source questions remain represented.
\r\nPrepare each complete question: choices apply to whole questions, so a strong numerical alone may not cover its paired theory. Repeated topics deserve practice, but the next paper can change the combination. Historical course codes and syllabus changes are kept visible; out-of-syllabus questions are not counted as missing current topics.
\r\nSame-date papers can be different programme or batch variants. They are listed separately, so their marks must not be added as if one examination offered all those questions. Duplicate scans are reconciled to one represented paper.
"}]},{"id":"core-maths","title":"1. The maths these three topics use","unit":"Start here","scope":"Short prerequisites, with worked examples. Read once before Fuzzy → GA → PSO.","sections":[{"id":"reading-formulas","title":"Symbols, brackets and iteration labels: how to read a formula","html":"A variable is a name for a number that can change. In f(x)=x²+1, x is the input, f is the name of the scoring rule, and f(x) is its result. An equation says that the expressions on both sides have the same value. A formula tells you how to calculate a quantity.
| Read this | Meaning | Concrete example |
|---|---|---|
| x² | x multiplied by itself | 3² = 3 × 3 = 9 |
| 2x or 2(x) | 2 multiplied by x | For x=3: 2x=6, while x²=9 |
| f(3) | Apply the rule named f to input 3 | If f(x)=x²+1, then f(3)=10 |
| x₁ and x₂ | First and second named entries; the small lower number is a label | If a position is (3,7), x₁=3 and x₂=7 |
| x at time t | A version of x at a particular iteration | Old x=3; after a move x=2. The time label is not squaring. |
| = and ≈ | Exactly equal; approximately equal after rounding | 1/3 ≈ 0.333333 |
| ∈ and { } | Belongs to; a collection of values | x ∈ {2,5} means x may be 2 or 5 |
Evaluate f(x)=2x²+1 at x=−3. [2 marks · Worked example]
The rule is f(x)=2x²+1. The supplied input is x=−3.
f(−3): the output of this rule for input −3.
The question supplies the rule. Replace x by −3, square that whole input, multiply the square by 2, and add 1. Parentheses keep the negative input together.
Evaluate g(x)=(x−2)/4 at x=10. [2 marks · Worked example]
The rule is g(x)=(x−2)/4. The supplied input is x=10.
g(10): the output of this rule for input 10.
The fraction bar groups the numerator x−2. Subtract 2 from the input first, then divide the entire difference by 4.
At iteration 0, x=4. The rule is new x = old x − 1. Find x after two iterations. [2 marks · Worked example]
Initial value x=4 at iteration 0. Every update subtracts 1 from the current value. Perform two updates.
The value after update 1 and after update 2.
t labels the current iteration, and t+1 labels the next. The parenthesised upper label records the step; it is not a power. Apply the supplied rule once, then use that new answer as the input to the second update.
x at iteration 2 is 2. Squaring the initial value would give 4²=16; that is a different operation.
Yaad rakho: Symbol sirf naam hai. Pehle bolo kaunsa number chahiye, phir uski jagah value likho.
"},{"id":"objective","title":"Position, score, minimisation and maximisation","html":"An intelligent system uses information and a decision procedure to choose an output or an action. In M2, you study three different procedures.
| Topic | Input | What the procedure does | Output |
|---|---|---|---|
| Fuzzy logic | Temperature and occupancy | Applies rules such as IF hot AND crowded THEN fast | A fan-speed setting |
| Genetic algorithm (GA) | A population of candidate answers | Selects, combines and changes candidate answers | A good solution and its fitness |
| Particle swarm optimization (PSO) | Positions and velocities of particles | Moves candidate answers using individual and shared experience | A best position and its objective value |
An objective function is the calculation used to score an answer. In minimisation, a smaller score is better; in maximisation, a larger score is better. The letter x names a candidate answer. The notation f(x), read “f of x”, means its score. Parentheses here do not mean f multiplied by x.
For the minimisation problem , find the score at . [2 marks · Practice]
,
The objective value f(x).
Replace x by the supplied number. Square means multiply the result by itself.
For the minimisation problem , find the score at . [2 marks · Practice]
,
The objective value f(x).
Replace x by the supplied number. Square means multiply the result by itself.
For the minimisation problem , find the score at . [2 marks · Practice]
,
The objective value f(x).
Replace x by the supplied number. Square means multiply the result by itself.
Comparing these candidates: x = 2 has score 1, x = 5 has score 4, and x = −1 has score 16. For minimisation, x = 2 is the best of these three. A negative position is allowed unless the question gives a restriction; a low score, not a low coordinate, decides the winner.
Yaad rakho: x = answer ki jagah; f(x) = us answer ka score. Min mein chhota score jeetega.
"},{"id":"arithmetic","title":"Fractions, brackets, minimum and maximum","html":"A fraction means division: the top number is the numerator, and the bottom is the denominator. Calculate each bracket first. min selects the smallest listed number; max selects the largest. Neither word means addition.
Evaluate . [2 marks · Practice]
Top: 32 − 25; bottom: 35 − 25.
The quotient.
Subtract separately in the numerator and denominator, then divide.
Evaluate . [2 marks · Practice]
Top: 35 − 32; bottom: 35 − 25.
The quotient.
Subtract separately in the numerator and denominator, then divide.
Evaluate . [2 marks · Practice]
Top: 7 − 6; bottom: 8 − 6.
The quotient.
Subtract separately in the numerator and denominator, then divide.
Find . [2 marks · Practice]
Two inner min operations followed by one outer max.
One number.
Work from the innermost brackets outward. This exact pattern occurs in fuzzy relation composition.
Yaad rakho: Pehle bracket, phir divide. AND ka min; OR ka max.
"},{"id":"probability","title":"Percentages, probabilities and cumulative intervals","html":"A probability is a number between 0 and 1. In these finite examples, 0 means an event has no probability mass and 1 means it occurs with certainty under the model. In a continuous distribution, a single exact value can have probability zero without being logically impossible; probabilities then describe intervals. 25% means 25 out of 100, so it equals 25÷100=0.25. For equally likely outcomes, probability is favourable outcomes divided by all possible outcomes. For GA selection, it is the candidate’s weight divided by the sum of weights.
A bag contains 3 red balls and 7 blue balls. One ball is drawn uniformly at random. Find the probability of red and of blue. [2 marks · Worked example]
Red=3; blue=7. Each ball is equally likely.
Both probabilities and their percentages.
All balls are equally likely, so counting them gives probability. The denominator counts all balls; the numerator counts the requested colour.
A weight need not itself be a probability. Weights 1, 2, 3, 4 add to 10. Dividing every weight by 10 makes their total 1: this is normalisation. Cumulative means “total accumulated so far”. It turns separate shares into boundaries along a line from 0 to 1.
Candidates A, B, C and D have weights 1, 2, 3 and 4. Find probabilities and cumulative probabilities. Which candidate is selected by random draw r=0.55? [2 marks · Worked example]
Weights [1,2,3,4], draw 0.55. Use intervals including their lower endpoint and excluding their upper endpoint.
Four probabilities, their running totals and the selected candidate.
F is weight. P is its share. Add P values in row order for the cumulative boundary.
| Candidate | Division | Probability |
|---|---|---|
| A | 1÷10 | 0.1 |
| B | 2÷10 | 0.2 |
| C | 3÷10 | 0.3 |
| D | 4÷10 | 0.4 |
| Candidate | Running addition | Interval |
|---|---|---|
| A | 0+0.1=0.1 | 0 ≤ r < 0.1 |
| B | 0.1+0.2=0.3 | 0.1 ≤ r < 0.3 |
| C | 0.3+0.3=0.6 | 0.3 ≤ r < 0.6 |
| D | 0.6+0.4=1 | 0.6 ≤ r ≤ 1 |
0.55 is at least 0.3 and less than 0.6. It lies in C’s interval. The draw is not compared with C’s weight 3.
Two unnormalised class scores are 0.12 and 0.08. Convert them to probabilities. [2 marks · Worked example]
Scores A=0.12 and B=0.08; they describe the only two alternatives.
Normalised probabilities.
Divide each score by the sum, preserving their ratio.
Yaad rakho: Probability = apna hissa ÷ sabka total. Cumulative = ab tak ke saare hisse jodna.
"},{"id":"binary","title":"Binary from the beginning: positions, place values and bits","html":"A position is simply a place occupied by one digit in a number. A digit is the symbol written there. The place value tells us how much one unit in that position is worth.
\r\nStart with an ordinary number you already know: 23. Its two digits are 2 and 3. They do not contribute the same kind of unit.
| LEFT: tens place | RIGHT: ones place | |
|---|---|---|
| Value of this place | 10 | 1 |
| Digit written here | 2 | 3 |
| Contribution | 2 × 10 = 20 | 3 × 1 = 3 |
The 2 means two tens because of where it sits. In 32, the same digit 2 sits on the right and means only two ones. Changing a digit’s position can change its contribution.
Yaad rakho: Position = digit ki jagah. Place value = us jagah par likhe ONE ki keemat. 23 mein 2 ka contribution 20 hai, kyunki woh tens ki jagah par hai.
Our ordinary number system is decimal: it has ten possible digits, 0 to 9. Its places are ones, tens, hundreds, and so on. Moving one place left multiplies the place value by 10.
\r\nBinary has only two possible digits: 0 and 1. One binary digit is called a bit. Therefore, a 4-bit number means four digit boxes, each holding either 0 or 1.
\r\nIn binary, grouping happens in twos. The rightmost box counts single units. After one single, a second single cannot be written as digit “2”, because binary has no digit 2. Instead, the two singles become one group of two in the next box on the left.
\r\nBinary 10 means one group of two and zero single units.
It represents the ordinary number 2. Read its digits as “one-zero” when discussing binary, so you do not accidentally read it as decimal ten.
Two groups of two make 4, so the next box to the left is the fours place. Two groups of four make 8, so the next box is the eights place. This is why the values double.
| Ordinary number | Binary digits | Meaning |
|---|---|---|
| 0 | 0 | No single units |
| 1 | 1 | One single unit |
| 2 | 10 | One two + zero ones = 2 |
| 3 | 11 | One two + one one = 3 |
| 4 | 100 | One four + zero twos + zero ones = 4 |
Yaad rakho: Decimal mein 10 ka group; binary mein 2 ka group. Do singles = ek pair; do pairs = ek group of four.
First draw four empty boxes. Start with the box furthest to the right. Give that place the value 1. Move left one box at a time, doubling the previous value.
\r\nPlace value: what ONE in this box is worth
← Double each time you move one box LEFT
We built the values by moving right to left: 1 → 2 → 4 → 8. Once the boxes are on paper, reading them in the usual left-to-right direction shows 8, 4, 2, 1. These are the same four places, viewed in the opposite direction.
Position number is not place value. The third box from the right has position number 3, but its place value is 4. “Third” tells you where the box is; “4” tells you what a 1 in it contributes.
If you add a fifth box on the left, its value is 8 × 2 = 16. The rightmost place stays 1; adding a box on the left does not change the existing places.
Yaad rakho: Sabse RIGHT se 1 likho. LEFT jaate jaate double karo: 1, 2, 4, 8, 16. Question ke digits baad mein in boxes mein rakho.
Keep the digit 1 the same and move it into a different place. Put 0 in every other box. Watch how its position changes the number it represents:
| Binary digits | Where the 1 sits | Decimal value |
|---|---|---|
| 0001 | 1st place from the right: ones | 1 × 1 = 1 |
| 0010 | 2nd place from the right: twos | 1 × 2 = 2 |
| 0100 | 3rd place from the right: fours | 1 × 4 = 4 |
| 1000 | 4th place from the right: eights | 1 × 8 = 8 |
In all four rows, the written digit is still 1. Its position decides whether that 1 contributes one, two, four or eight.
\r\nNow put 1 0 1 1 in the four boxes. The values 8, 4, 2 and 1 are labels for the boxes. The written digits tell us which values to include:
\r\nPlace value: what ONE in this box is worth
Digits written in those boxes
Each box contributes
The 0 above does not mean the place value changes from 4 to 0. The place is still worth 4 per unit; the digit says to take zero units of it. Its contribution is therefore 0 × 4 = 0.
The three 1s are the same digit, but they sit in different places. They contribute 8, 2 and 1. Therefore, binary 1011 represents decimal 11.
Yaad rakho: Box ki value fixed. Andar 1 hai toh value lo; 0 hai toh skip karo. Phir li hui values jodo.
Decode means convert the binary digits into an ordinary decimal number. In exam notation, a small 2 below a number means binary; a small 10 means decimal. These small labels identify the number system. They are not multipliers.
This line says: “one-zero-one-one in binary and eleven in decimal represent the same quantity.”
Convert the binary number to decimal. [2 marks · Practice]
Binary digits: 1011, in this exact order.
The ordinary decimal number represented by these digits.
Each digit tells us how many of that place value to include. Multiply each digit by its place value, then add.
| Place value | 8 | 4 | 2 | 1 |
|---|---|---|---|---|
| Binary digit | 1 | 0 | 1 | 1 |
The digit is 1, so include the value of this place.
The digit is 0, so this place contributes zero. Its place value still stays the same.
The digit is 1, so include the value of this place.
The digit is 1, so include the value of this place.
.Convert the binary number to decimal. [2 marks · Practice]
Binary digits: 0110, in this exact order.
The ordinary decimal number represented by these digits.
Each digit tells us how many of that place value to include. Multiply each digit by its place value, then add.
| Place value | 8 | 4 | 2 | 1 |
|---|---|---|---|---|
| Binary digit | 0 | 1 | 1 | 0 |
The digit is 0, so this place contributes zero. Its place value still stays the same.
The digit is 1, so include the value of this place.
The digit is 1, so include the value of this place.
The digit is 0, so this place contributes zero. Its place value still stays the same.
.Convert the binary number to decimal. [2 marks · Practice]
Binary digits: 1001, in this exact order.
The ordinary decimal number represented by these digits.
Each digit tells us how many of that place value to include. Multiply each digit by its place value, then add.
| Place value | 8 | 4 | 2 | 1 |
|---|---|---|---|---|
| Binary digit | 1 | 0 | 0 | 1 |
The digit is 1, so include the value of this place.
The digit is 0, so this place contributes zero. Its place value still stays the same.
The digit is 0, so this place contributes zero. Its place value still stays the same.
The digit is 1, so include the value of this place.
.Convert the binary number to decimal. [2 marks · Practice]
Binary digits: 10011, in this exact order.
The ordinary decimal number represented by these digits.
Each digit tells us how many of that place value to include. Multiply each digit by its place value, then add.
| Place value | 16 | 8 | 4 | 2 | 1 |
|---|---|---|---|---|---|
| Binary digit | 1 | 0 | 0 | 1 | 1 |
The digit is 1, so include the value of this place.
The digit is 0, so this place contributes zero. Its place value still stays the same.
The digit is 0, so this place contributes zero. Its place value still stays the same.
The digit is 1, so include the value of this place.
The digit is 1, so include the value of this place.
.Encode means write an ordinary number using binary digits. The direction of the task is now reversed: the number is known, and you have to choose the 0s and 1s.
\r\nFor four bits, write the places 8, 4, 2, 1. Start with the largest place on the left. Use it only if it does not exceed the amount still needed. Whenever you use a place, subtract it before moving to the next one.
Yaad rakho: Number banana hai. Badi value se start: jitna bacha hai usmein value fit ho toh 1 aur subtract; fit nahi ho toh 0.
Write the decimal number 6 as an unsigned 4-bit binary number. [2 marks · Practice]
Decimal number: 6. Required length: 4 bits, meaning exactly 4 digit boxes.
The binary digits, including any zero at the beginning needed to keep the required length.
At each place, compare its value with the amount still needed. If that amount is at least the place value, write 1 and subtract that value. Otherwise write 0 and keep the remaining amount unchanged.
This works because each place can be used either once (1) or not at all (0). The selected places must add up to the given number.
| Place value | 8 | 4 | 2 | 1 |
|---|---|---|---|---|
| Digit to choose | □ | □ | □ | □ |
Start with the whole number 6 still to make. Work from the largest place on the LEFT to the smallest on the RIGHT.
We still need 6. Taking 8 would be too much, so skip this place. Write 0 in this box.
We used nothing from this place. The remaining amount stays 6.
We still need 6. This is at least 4, so take one 4. Write 1 in this box.
We still need 2. This is at least 2, so take one 2. Write 1 in this box.
We still need 0. Taking 1 would be too much, so skip this place. Write 0 in this box.
We used nothing from this place. The remaining amount stays 0.
The boxes now contain 0 1 1 0. Join them in the same order: 0110.
The remaining amount is 0, so we have made the required number exactly.
. The answer has 4 bits.Write the decimal number 11 as an unsigned 4-bit binary number. [2 marks · Practice]
Decimal number: 11. Required length: 4 bits, meaning exactly 4 digit boxes.
The binary digits, including any zero at the beginning needed to keep the required length.
At each place, compare its value with the amount still needed. If that amount is at least the place value, write 1 and subtract that value. Otherwise write 0 and keep the remaining amount unchanged.
This works because each place can be used either once (1) or not at all (0). The selected places must add up to the given number.
| Place value | 8 | 4 | 2 | 1 |
|---|---|---|---|---|
| Digit to choose | □ | □ | □ | □ |
Start with the whole number 11 still to make. Work from the largest place on the LEFT to the smallest on the RIGHT.
We still need 11. This is at least 8, so take one 8. Write 1 in this box.
We still need 3. Taking 4 would be too much, so skip this place. Write 0 in this box.
We used nothing from this place. The remaining amount stays 3.
We still need 3. This is at least 2, so take one 2. Write 1 in this box.
We still need 1. This is at least 1, so take one 1. Write 1 in this box.
The boxes now contain 1 0 1 1. Join them in the same order: 1011.
The remaining amount is 0, so we have made the required number exactly.
. The answer has 4 bits.Write the decimal number 19 as an unsigned 5-bit binary number. [2 marks · Practice]
Decimal number: 19. Required length: 5 bits, meaning exactly 5 digit boxes.
The binary digits, including any zero at the beginning needed to keep the required length.
At each place, compare its value with the amount still needed. If that amount is at least the place value, write 1 and subtract that value. Otherwise write 0 and keep the remaining amount unchanged.
This works because each place can be used either once (1) or not at all (0). The selected places must add up to the given number.
| Place value | 16 | 8 | 4 | 2 | 1 |
|---|---|---|---|---|---|
| Digit to choose | □ | □ | □ | □ | □ |
Start with the whole number 19 still to make. Work from the largest place on the LEFT to the smallest on the RIGHT.
We still need 19. This is at least 16, so take one 16. Write 1 in this box.
We still need 3. Taking 8 would be too much, so skip this place. Write 0 in this box.
We used nothing from this place. The remaining amount stays 3.
We still need 3. Taking 4 would be too much, so skip this place. Write 0 in this box.
We used nothing from this place. The remaining amount stays 3.
We still need 3. This is at least 2, so take one 2. Write 1 in this box.
We still need 1. This is at least 1, so take one 1. Write 1 in this box.
The boxes now contain 1 0 0 1 1. Join them in the same order: 10011.
The remaining amount is 0, so we have made the required number exactly.
. The answer has 5 bits.Write the decimal number 0 as an unsigned 4-bit binary number. [2 marks · Practice]
Decimal number: 0. Required length: 4 bits, meaning exactly 4 digit boxes.
The binary digits, including any zero at the beginning needed to keep the required length.
At each place, compare its value with the amount still needed. If that amount is at least the place value, write 1 and subtract that value. Otherwise write 0 and keep the remaining amount unchanged.
This works because each place can be used either once (1) or not at all (0). The selected places must add up to the given number.
| Place value | 8 | 4 | 2 | 1 |
|---|---|---|---|---|
| Digit to choose | □ | □ | □ | □ |
Start with the whole number 0 still to make. Work from the largest place on the LEFT to the smallest on the RIGHT.
We still need 0. Taking 8 would be too much, so skip this place. Write 0 in this box.
We used nothing from this place. The remaining amount stays 0.
We still need 0. Taking 4 would be too much, so skip this place. Write 0 in this box.
We used nothing from this place. The remaining amount stays 0.
We still need 0. Taking 2 would be too much, so skip this place. Write 0 in this box.
We used nothing from this place. The remaining amount stays 0.
We still need 0. Taking 1 would be too much, so skip this place. Write 0 in this box.
We used nothing from this place. The remaining amount stays 0.
The boxes now contain 0 0 0 0. Join them in the same order: 0000.
The remaining amount is 0, so we have made the required number exactly.
. The answer has 4 bits.Leading zeros are zeros at the beginning, on the left. Binary 110 and 0110 both represent decimal 6:
The added leftmost place contributes zero. It changes the length from three bits to four bits, but not the value. If a GA question requires four bits, write 0110.
This rule is specifically about zeros on the left. Appending a zero on the right shifts the existing digits into places worth twice as much. For example, binary 1100 is 8 + 4 = 12, not 6.
For these GA conversions, unsigned means the bits represent zero or a positive whole number, with no negative sign. The smallest 4-bit value is 0000. The largest is 1111 because all four places are included.
Find the largest unsigned integer that can be represented using 4 bits. Can decimal 19 fit in 4 bits? [2 marks · Practice]
Four places with values 8, 4, 2 and 1. Each digit is either 0 or 1.
The largest possible value and whether 19 can be represented.
A digit contributes the most when it is 1. Set every digit to 1 and add all the place values.
The largest 4-bit string is 1111.
19 is greater than 15, so four bits are insufficient. Add the 16-place on the left. Then 19 = 16 + 2 + 1, giving 10011 in five bits.
The general rule is . Here n means the number of bits. The power means multiplying n copies of 2; for four bits, , so the maximum is 16 − 1 = 15. There are 16 different values, from 0 through 15, because zero is also one value.
| Bits | Place values from left to right | Unsigned values |
|---|---|---|
| 3 | 4, 2, 1 | 0 to 7 |
| 4 | 8, 4, 2, 1 | 0 to 15 |
| 5 | 16, 8, 4, 2, 1 | 0 to 31 |
If an exam gives a number outside the capacity of the stated bit length, state that inconsistency. Do not silently cut off a digit: removing the 16-place from 10011 would change 19 into 3. Also retain any separate allowed range stated in the question.
A chromosome is the representation of one candidate solution in a genetic algorithm. With the direct unsigned binary encoding used here, decimal 6 is represented by the four-bit chromosome 0110.
\r\nDuring crossover or mutation, the algorithm changes these bits. Afterwards, decode the resulting chromosome to find its new value of x. Then use that x in the given objective or fitness function. The binary string, its decimal value, and its fitness score are different things.
Yaad rakho: Encode: number se bits. Decode: bits se number. GA mein bits badle toh x dobara decode karo, phir uska score nikalo.
"},{"id":"coordinates","title":"Vectors, coordinates, matrices and transpose from the first entry","html":"A scalar is one number, such as 5. A vector is an ordered list, such as (3,7). Order matters: the first number may be horizontal position and the second vertical position. A matrix is a rectangular table of numbers. These words describe how numbers are arranged, not a new kind of arithmetic.
| Arrangement | Example | How to read it |
|---|---|---|
| Scalar | 5 | One value |
| Row vector | [3,7] | One row, two entries |
| Column vector | | Two rows, one entry |
| Matrix | | Two rows, two columns |
For a matrix A, A₂₁ means the entry in row 2, column 1. The first lower label is the row; the second is the column. Matrix size is written number of rows × number of columns. A 2×3 matrix has two rows containing three entries each.
Let A have first row [2,5,8] and second row [1,4,7]. State its size and find A₂₃ and A₁₂. [2 marks · Worked example]
Matrix dimensions and two entries.
Read the lower labels in row–column order.
There are 2 horizontal rows and 3 vertical columns. Size=2×3.
Row 2 is [1,4,7]. Its third entry is 7.
Row 1 is [2,5,8]. Its second entry is 5.
A particle has position x=(3,7), velocity v=(−1,2), and best position p=(1,8). Find p−x, half of that difference, and x+v. [2 marks · Worked example]
Position x=(3,7); velocity v=(−1,2); personal-best position p=(1,8). First and second coordinates stay in the same order.
The difference p−x, the scaled difference 0.5(p−x), and the updated position x+v.
Match first with first and second with second. A scalar multiplies every coordinate.
The superscript T in Aᵀ means transpose. It does not mean multiply or raise A to a power. Row 1 of A becomes column 1 of Aᵀ; row 2 becomes column 2. An m×n arrangement becomes n×m.
Find the transpose of A = [[2,5,8],[1,4,7]]. [2 marks · Worked example]
Aᵀ.
Turn each old row into a new column, keeping its order.
[2,5,8] becomes the first column from top to bottom.
[1,4,7] becomes the second column from top to bottom.
Calculate the dot product of [2,3] and [4,5]. [2 marks · Worked example]
First vector a=[2,3]; second vector b=[4,5]. Each has two entries.
The dot product a·b, a single number obtained by adding the two matching products.
Multiply corresponding entries, then add the products. Markov matrix multiplication later repeats this for each destination column.
Yaad rakho: Vector = order wali list. Matrix = rows-columns ka table. Entry padhte waqt pehle row, phir column.
"},{"id":"graphs","title":"Graph axes, intervals and the fractions inside membership functions","html":"The horizontal axis runs left to right and shows input x. The vertical axis runs bottom to top and shows the output. A point (3,0.4) means: input 3 gives output 0.4. In a membership graph, the vertical output is a degree between 0 and 1.
To read the graph at x=3, start at 3 on the bottom axis, move vertically until you reach the line, then move horizontally to the vertical axis. The height you read is the membership.
| Read this | Meaning | Concrete example |
|---|---|---|
| x < 5 | Strictly less than 5 | 4.9 qualifies; 5 does not |
| x ≤ 5 | Less than or equal to 5 | Both 4.9 and 5 qualify |
| 2 ≤ x ≤ 8 | x is between 2 and 8 including both ends | The interval is [2,8] |
| [2,8) | Include 2; exclude 8 | Useful for roulette interval boundaries |
| Piecewise formula | Choose one calculation according to where x lies | Rising branch, falling branch or outside |
A straight membership line rises from (2,0) to (6,1). Find its height at x=3. [2 marks · Worked example]
Left input=2; peak input=6; required input=3.
Membership height.
A straight line changes by the same proportion of its height as the proportion travelled along its width.
A straight membership line rises from (2,0) to (6,1). Find its height at x=4. [2 marks · Worked example]
Left input=2; peak input=6; required input=4.
Membership height.
A straight line changes by the same proportion of its height as the proportion travelled along its width.
A straight membership line rises from (10,0) to (30,1). Find its height at x=25. [2 marks · Worked example]
Left input=10; peak input=30; required input=25.
Membership height.
A straight line changes by the same proportion of its height as the proportion travelled along its width.
A membership line falls from (5,1) to (8,0). Find its height at x=7. [2 marks · Worked example]
Peak=5; right zero=8; x=7.
Membership.
Yaad rakho: Chadhti line: left se kitna aage? Utarti line: right tak kitna baaki? Dono mein poori width se divide.
"},{"id":"sums-averages","title":"Sum notation, ordinary averages and weighted averages","html":"The capital Greek letter Σ is read “sum”. The lower label says where to start and the upper label where to stop. The small index i is just a counter: first item, second item, third item.
For a₁=2, a₂=5 and a₃=8, calculate Σaᵢ and their mean. [2 marks · Worked example]
Three numbers: 2,5,8.
Sum and ordinary average.
The bar means average. Each number has equal weight.
If values 20 and 80 have weights 3 and 1, this is equivalent to averaging [20,20,20,80]. Therefore you divide by the sum of weights, not by the number of distinct values. Weights may also be decimals. Their total need not initially be 1.
Find the weighted average of [20, 80] with corresponding weights [3, 1]. [2 marks · Worked example]
| Value | Its weight |
|---|---|
| 20 | 3 |
| 80 | 1 |
The weighted average of the listed values.
zᵢ is a value and wᵢ is its corresponding weight. Multiply each value by its own weight, add, then divide by total weight. Here all weights are nonnegative and their total is positive.
Find the weighted average of [20, 50, 80] with corresponding weights [0.2, 0.6, 0.2]. [2 marks · Worked example]
| Value | Its weight |
|---|---|
| 20 | 0.2 |
| 50 | 0.6 |
| 80 | 0.2 |
The weighted average of the listed values.
zᵢ is a value and wᵢ is its corresponding weight. Multiply each value by its own weight, add, then divide by total weight. Here all weights are nonnegative and their total is positive.
Find the weighted average of [65, 85] with corresponding weights [0.3, 0.5]. [2 marks · Worked example]
| Value | Its weight |
|---|---|
| 65 | 0.3 |
| 85 | 0.5 |
The weighted average of the listed values.
zᵢ is a value and wᵢ is its corresponding weight. Multiply each value by its own weight, add, then divide by total weight. Here all weights are nonnegative and their total is positive.
For [0.5,0.2,0.4], first 0.5×0.2=0.1, then 0.1×0.4=0.04. Σ would instead give 1.1. Naive Bayes multiplies likelihood factors; it does not add them.
Yaad rakho: Sigma = sab jodo. Product Pi = sab multiply. Weighted mean = value×weight ke total ko weights ke total se divide.
"},{"id":"powers-exponentials","title":"Powers, roots, absolute values and exp on the calculator","html":"A positive integer power repeats multiplication. 2³=2×2×2=8; it does not mean 2×3. A nonzero number raised to power 0 equals 1. A negative power means a reciprocal: 2⁻²=1/(2²)=1/4. A square root reverses a square for a nonnegative result: √9=3.
| Read this | Meaning | Concrete example |
|---|---|---|
| 2⁰, 2¹, 2², 2³ | Binary place values | 1,2,4,8 |
| |x| | Absolute value: distance from zero | |−3|=3 and |3|=3 |
| e ≈ 2.71828 | A fixed mathematical constant | e⁰=1 |
| exp(u) | Another way of writing eᵘ | exp(−2)=e⁻²≈0.135335 |
| ln(x) | The natural logarithm; the exponent giving x from e | ln(e²)=2; ln(1)=0 |
Calculate exp(−x²/2) at x=-2. [2 marks · Worked example]
x=-2; exp means e raised to the bracketed exponent.
The exponential value.
Calculate the complete exponent first. Then use the calculator’s e-to-the-power function.
Calculate exp(−x²/2) at x=0. [2 marks · Worked example]
x=0; exp means e raised to the bracketed exponent.
The exponential value.
Calculate the complete exponent first. Then use the calculator’s e-to-the-power function.
Calculate exp(−x²/2) at x=2. [2 marks · Worked example]
x=2; exp means e raised to the bracketed exponent.
The exponential value.
Calculate the complete exponent first. Then use the calculator’s e-to-the-power function.
On an fx-991ES-style calculator, eˣ is normally the shifted function above ln. Enter the complete exponent in brackets. Use the negative-number key for −2 where required. Confirm against e⁰=1 before a longer calculation. The exact key labels depend on the model.
Yaad rakho: exp ka matlab e ki power. Pehle exponent ka poora hisaab, phir eˣ.
"}]},{"id":"u1-zero","title":"Unit 1 · Probability from zero (AND, OR, given, Bayes)","unit":"1","scope":"Start here if probability is new: counting, AND/OR, conditional probability and Bayes with pictures","sections":[{"id":"and-or","title":"AND, OR and NOT with a picture","html":"\r\n| Read this | Meaning | Concrete example |
|---|---|---|
| Experiment | An observation with possible outcomes | One coin toss |
| Outcome | One complete result | Heads |
| Event | A set of outcomes satisfying a statement | Die result is even: {2,4,6} |
| A AND B | Both statements hold | Likes cricket and football |
| A OR B | At least one statement holds, including both | Likes either sport or both |
| NOT A | All outcomes outside A | Does not like cricket |
Before this: read Probability from zero. Here we combine two events.
\r\n\r\nIn a class of 100, 40 play cricket, 30 play football, and 10 of them play both. Put the 10 “both” in the overlap first, then fill the rest: only cricket = 40 − 10 = 30, only football = 30 − 10 = 20, neither = 100 − 30 − 10 − 20 = 40.
\r\n![\"AND](\"fig/prob_venn.84491b9acc.svg\")
![\"P(C\\text{](\"math/a166ec0150b9d8e6.svg\")
Count: 30 + 10 + 20 = 60, so P(C or F) = 0.6. With a formula: add both circles, then subtract the overlap because it was counted twice:
\r\n![\"P(C\\cup](\"math/e851c0d30cd17e85.svg\")
![\"P(\\text{not](\"math/fd0572e2636f2a68.svg\")
Two events are independent if one does not change the chance of the other (two separate coin tosses). Then
\r\n![\"P(A\\text{](\"math/9d043d505b5382fc.svg\")
Cricket and football are not independent here: 0.4 × 0.3 = 0.12 ≠ 0.1.
\r\n| Word in the question | Symbol | What to do |
|---|---|---|
| and / both | ∩, ∧, comma | overlap; multiply only if independent |
| or / at least one | ∪, ∨ | add, subtract the overlap |
| not | ¬, ′, c | 1 minus |
| given / if we know | | | next lesson: divide |
In one line: “P(A given B)” means: forget everyone who is not B, and ask what fraction of the B group is also A.
\r\n\r\nQuestion: if a student plays cricket, what is the probability that they also play football? “Given cricket” means we look only at the 40 cricket players. Of those, 10 play football.
\r\n![\"P(F\\mid](\"math/e504a0ad42ea83d7.svg\")
The same with probabilities instead of counts:
\r\n![\"\\boxed{P(A\\mid](\"math/06d91b8a011bacbf.svg\")
The thing after the bar “|” always goes in the denominator. Notice ![\"P(C\\mid](\"math/a402991a380b1d2f.svg\")
is different: the world you shrink to changes.
A joint table lists the probability of every combination. Example: 200 patients.
\r\n| Counts | Flu | No flu | Row total |
|---|---|---|---|
| Fever | 30 | 20 | 50 |
| No fever | 10 | 140 | 150 |
| Column total | 40 | 160 | 200 |
Divide every cell by 200 to get the probability table:
\r\n| P | Flu | No flu | Total |
|---|---|---|---|
| Fever | 0.15 | 0.10 | 0.25 |
| No fever | 0.05 | 0.70 | 0.75 |
| Total | 0.20 | 0.80 | 1 |
Exam tables have three variables (e.g. disease, fever, test), with two possible values each, so there are 8 cells, but the method is identical: add the cells you need, divide by the cells of the “given” part. Solved PYQs: full joint distribution.
\r\n\r\n![\"P(A\\text{](\"math/aec0a3a12cca75ad.svg\")
Bayesian networks use this again and again: multiply a “given” probability by the probability of the thing given.
\r\nA table of 200 patients has 160 patients without flu; 140 of those also have no fever. Find P(no fever | no flu). [2 marks · Practice]
No-flu group=160; both no fever and no flu=140.
The conditional probability.
Keep only the evidence group. The denominator measures that group; the numerator measures the part satisfying both events.
![\"P(A\\mid](\"math/af10b31123342cc5.svg\")
Numerator=140; denominator=160. The quantity after the bar identifies the denominator.
![\"\\frac{140}{160}=0.875\"](\"math/ed912694e6123813.svg\")
Given P(A and B)=0.12 and P(B)=0.30, find P(A | B). [2 marks · Practice]
Joint event probability=0.12; evidence probability=0.30.
The conditional probability.
Keep only the evidence group. The denominator measures that group; the numerator measures the part satisfying both events.
Numerator=0.12; denominator=0.3. The quantity after the bar identifies the denominator.
![\"\\frac{0.12}{0.3}=0.4\"](\"math/2ef396ae4f7c0a1c.svg\")
A probability tree branches by successive conditions. Multiply along one complete path because all its events must occur together. Add different complete paths when either path qualifies. Counts such as 0.99 out of 1,000 are expected counts in the illustration, not fractions of an actual person. Scaling the same model to 100,000 gives 100 diseased people, 99 positive among them, and 1,998 false positives among 99,900 others. The posterior is 99/(99+1998)=99/2097≈0.0472103.
In one line: to reverse a “given”, imagine a crowd, split it by the cause, then by the evidence, and count.
\r\n\r\nA disease affects 1 person in 1000. A test is positive for 99% of sick people, and wrongly positive for 2% of healthy people. Your test is positive. How likely is it that you are sick?
\r\n![\"Imagine](\"fig/prob_tree.201d46f630.svg\")
Only about 4.7%, because the healthy group is so much bigger that its small false-alarm rate still produces most positives.
\r\n\r\n![\"P(\\text{sick}\\mid+)=\\frac{P(+\\mid\\text{sick})\\,P(\\text{sick})}{P(+\\mid\\text{sick})P(\\text{sick})+P(+\\mid\\text{healthy})P(\\text{healthy})}=\\frac{0.99\\times0.001}{0.99\\times0.001+0.02\\times0.999}=0.047\"](\"math/7999e36c046cbb22.svg\")
Each part of the formula is one branch of the tree. Names: P(sick) is the prior (before the test), P(+ | sick) the likelihood, the denominator the evidence (all positives), and the answer the posterior (after the test).
\r\nNext: the formal version and more examples in Bayes' rule.
"}]},{"id":"u1-prob","title":"Unit 1 · Uncertainty, joint distributions and Bayes rule","unit":"1","scope":"Unit 1 · Quantifying uncertainty, inference using the full joint distribution, Bayes rule","sections":[{"id":"uncertainty","title":"Why agents need probability (and the basic rules)","html":"\r\nAction A pays 10 points with probability 0.6 and −2 points with probability 0.4. Action B always pays 4 points. Which has greater expected reward? [2 marks · Worked example]
Action A: reward 10 with probability 0.6, reward −2 with probability 0.4. Action B: reward 4 with probability 1. The two probabilities for A add to 1.
Expected score for each action.
![\"\\mathbb](\"math/81d37eff7c373c7f.svg\")
Expectation is a probability-weighted average. It need not equal any one actual outcome.
![\"0.6\\times10=6\\qquad0.4\\times(-2)=-0.8\"](\"math/6978caf34fd95e0c.svg\")
![\"\\mathbb](\"math/95811b7af645e1fd.svg\")
![\"\\mathbb](\"math/bbe9161115867c07.svg\")
5.2>4, so A has the greater expected reward. A single attempt can still return −2.
Ω names the sample space (all outcomes). ω names one outcome. The symbol E or 𝔼 means expectation here; it does not mean an observed reward. A rational probabilistic decision uses both how likely outcomes are and how desirable they are.
In one line: when an agent cannot be certain, it should hold a degree of belief between 0 and 1 for each possibility and update it with evidence; probability theory is the tool for that.
\r\n\r\nAn agent is uncertain because of partial observability (it cannot see everything) and non-determinism (actions do not always have the same result). Pure logic fails in such domains. Example: “Toothache ⇒ Cavity” is false (gum disease also causes toothache), and listing every possible cause is impossible. Logic fails for three reasons:
\r\nSo the agent assigns a degree of belief (probability), e.g. “80% of toothache patients have cavities”. Decision theory = probability theory + utility theory: a rational agent chooses the action with the maximum expected utility (e.g. a taxi choosing when to leave for the airport balances the chance of missing the flight against waiting time).
\r\nSources of uncertainty in AI: incomplete data, noise, ambiguity, stochastic environments, model and algorithm limits.
\r\n\r\n| Term | Meaning / rule |
|---|---|
| Sample space Ω | All possible worlds; ![\"0\\le](\"math/7e932392a2d209b9.svg\") , ![\"\\sum_\\omega](\"math/2e04a098f6f00102.svg\") |
| Prior (unconditional) | ![\"P(a)\"](\"math/3ff7ba191ece3ffd.svg\") : belief before evidence, e.g. ![\"P(cavity)=0.2\"](\"math/422e41a24d087e55.svg\") |
| Posterior (conditional) | ![\"P(a\\mid](\"math/b7aba0396dc1aeb0.svg\") , ![\"P(b)>0\"](\"math/5a47a3fc657b0d7c.svg\") |
| Product rule | ![\"P(a\\land](\"math/7b86112b191ef76c.svg\") |
| Complement | ![\"P(\\lnot](\"math/1afe967401ee19dd.svg\") |
| Inclusion–exclusion | ![\"P(a\\lor](\"math/f67fe6f37e8347cc.svg\") |
| Marginalisation | ![\"P(Y)=\\sum_z](\"math/e6ca224bc18b80c0.svg\") |
| Total probability | ![\"P(b)=P(b\\mid](\"math/2fd503c9ba9e8f6f.svg\") |
| Independence | ![\"P(a\\land](\"math/ed94fa4d4b6553f6.svg\") , equivalently ![\"P(a\\mid](\"math/1b8c2e9773a67bd2.svg\") |
| Distribution | Bold ![\"\\mathbf](\"math/ffe772be82ede0e4.svg\") lists all values |
A joint-probability table is a list of complete combinations. Three Yes/No variables have 2×2×2=8 combinations. “Marginalise” means ignore a variable by adding its possible cases. “Condition” means first restrict attention to the evidence cases, then divide by their total. For unnormalised weights .12 and .08, the total is .20; α=1/.20=5, so α×.12=.6 and α×.08=.4. α is only a rescaling constant.
In one line: a full joint table lists the probability of every combination; to answer any question, add the cells where it is true, and for “given” questions divide by the cells of the evidence.
\r\n\r\n: add every cell in which a is true.![\"P(a\\mid](\"math/e6e697dab9ccb462.svg\")
: add cells where both hold, divide by the sum of cells where e holds.![\"\\mathbf](\"math/10d12eaae749bfb3.svg\")
with ![\"\\alpha=1/P(e)\"](\"math/c3f28a2b8d2fb918.svg\")
: compute the unnormalised values for every value of A, then scale them to sum to 1.Question. For the dentist table below find P(cavity ∨ toothache), P(cavity | toothache) and P(¬cavity | toothache). [class example · 3 marks]
The complete joint table or cell values printed in this question. The rows and columns identify the event combinations.
All requested marginal and conditional probabilities.
Marginal: add cells satisfying the event. Conditional: retain the evidence cells and divide the joint-event sum by the evidence sum.
![\"P(A\\mid](\"math/3b42af132ac8057a.svg\")
| toothache, catch | toothache, ¬catch | ¬toothache, catch | ¬toothache, ¬catch | |
|---|---|---|---|---|
| cavity | 0.108 | 0.012 | 0.072 | 0.008 |
| ¬cavity | 0.016 | 0.064 | 0.144 | 0.576 |
![\"P(cavity\\lor](\"math/93d535e63ac5c87c.svg\")
![\"P(toothache)=0.108+0.012+0.016+0.064=0.2,\\qquad](\"math/9b888308d7a5d7e4.svg\")
With α: ![\"\\mathbf](\"math/473e1cb48309ce75.svg\")
, ![\"\\alpha=1/0.2=5\"](\"math/7ce6fe6ef36e3167.svg\")
.
Question. A supermarket recorded the joint distribution of Product type (Y: premium = 1, regular = 0), Payment mode (X: credit = 1, debit = 0) and Gender (Z: female = 1, male = 0). (a) If a customer bought a premium item, find the probability that (1) they are female, (2) they paid by credit card. (b) Find (1) P(premium) and (2) P(credit card). (c) Given a male customer paid by debit card, find the probability he bought a regular item. [5 marks · Final Nov 2025 Q1a]
The printed eight entries are reproduced below. Their total is 1.002, not 1. Therefore the table is inconsistent as a probability distribution.
| Credit, female | Credit, male | Debit, female | Debit, male | |
|---|---|---|---|---|
| Premium (Y=1) | 0.08 | 0.10 | 0.072 | 0.05 |
| Regular (Y=0) | 0.10 | 0.15 | 0.25 | 0.20 |
The five requested probabilities, stating how the inconsistent data are interpreted.
For this worked solution, treat all printed entries as nonnegative weights and normalise by their total 1.002. This preserves the printed ratios. It is a declared interpretation, not a claim about the examiner’s intended correction.
![\"P(E)=\\frac{\\text{sum](\"math/e29a33f81d36c3d3.svg\")
![\"P(A\\mid](\"math/656f97e9f0aa20fc.svg\")
![\"0.08+0.10+0.072+0.05+0.10+0.15+0.25+0.20=1.002\"](\"math/96cf20835aaf1e81.svg\")
Use total weight 1.002 consistently for unconditional probabilities.
![\"w(premium)=0.08+0.10+0.072+0.05=0.302\"](\"math/455884f740d33473.svg\")
![\"P(female\\mid](\"math/61d38ae71dd8a06e.svg\")
![\"P(credit\\mid](\"math/f1511f5b6ec07db0.svg\")
![\"P(premium)=\\frac{0.302}{1.002}\\approx0.301397\"](\"math/3465c1e9a29a79c1.svg\")
![\"P(credit)=\\frac{0.08+0.10+0.10+0.15}{1.002}=\\frac{0.43}{1.002}\\approx0.429142\"](\"math/594c62fd5c571144.svg\")
![\"P(regular\\mid](\"math/7f500d88c2aab86f.svg\")
Alternative correction: if 0.072 is corrected to 0.070, the total becomes 1 and the corresponding answers are (a) 0.50 and 0.60; (b) 0.30 and 0.43; (c) 0.8. This agrees with the interpretation used in the supplied key. State the correction; do not call raw totals 0.302 and 0.43 valid marginals of the original inconsistent table.
Question. For a disease Y with fever X as a symptom, a test Z is conducted. Joint probabilities: Disease (Y=1): fever & test+ 0.576, fever & test− 0.144, no fever & test+ 0.072, no fever & test− 0.016; No disease (Y=0): 0.008, 0.064, 0.012, 0.108. (1) If fever occurs and the test is positive, what is the probability of disease? (2) What is the probability that a random person tests positive? (3) Given no fever, what is the probability of still having the disease? [5 marks · Re-exam Jan 2026 Q1a]
The complete joint table or cell values printed in this question. The rows and columns identify the event combinations.
All requested marginal and conditional probabilities.
Marginal: add cells satisfying the event. Conditional: retain the evidence cells and divide the joint-event sum by the evidence sum.
![\"P(Y{=}1\\mid](\"math/e2428bb7484a2222.svg\")
![\"P(Z{=}1)=0.576+0.072+0.008+0.012=0.668\"](\"math/4ae537c83ce53b71.svg\")
![\"P(X{=}0)=0.072+0.016+0.012+0.108=0.208,\\qquad](\"math/ab59aa2401b72e20.svg\")
A prior is the probability of a cause before the new evidence. A likelihood is the probability of that evidence if the cause is true. A posterior is the probability of the cause after observing the evidence. These are different conditional directions: P(positive | disease) is not P(disease | positive).
\r\n![\"P(A\\cap](\"math/174a2d1dd45a6fd5.svg\")
Divide both sides by P(B), provided that P(B)>0.
![\"P(A\\mid](\"math/f27b652409444deb.svg\")
If A and its complement are the only two possibilities, evidence B can arise by either path. Add both paths in the denominator.
![\"P(A\\mid](\"math/715ffe0029a7c804.svg\")
| Symbol | Read it as |
|---|---|
| P(A) | Prior: how common is A? |
| P(B | A) | Likelihood: if A is true, how likely is B? |
| P(B) | Evidence: how common is B through all possible causes? |
| P(A | B) | Posterior: among B cases, what fraction come from A? |
A hypothetical population has a disease prevalence of 1 in 1000. A test is positive in 99% of diseased cases and 2% of healthy cases. Given a positive result, calculate the probability of disease. [4 marks · Worked probability example]
A = disease; B = positive test. P(A)=0.001; P(B|A)=0.99; P(B|not A)=0.02.
The reversed conditional probability P(A | B).
The question gives evidence given each cause, but asks for cause given evidence. Bayes’ rule combines the prior with that likelihood and divides by all ways of obtaining the evidence.
![\"P(\\neg](\"math/eba03740537f2993.svg\")
Multiply the prior for disease by the likelihood within that group.
![\"P(A\\cap](\"math/44ef59701ff8e69a.svg\")
Now include the path through healthy.
![\"P(\\neg](\"math/f8f7a41ed738725f.svg\")
![\"P(B)=0.00099+0.01998=0.02097\"](\"math/588fc60be824b4ec.svg\")
![\"P(A\\mid](\"math/b015ca384b1a70e6.svg\")
In a hypothetical population, 49% are female and 51% male. 8% of females and 12% of males are classified high-risk in the supplied data. Given that a person is high-risk, find the probability that the person is male. [3 marks · Worked probability example]
A = male; B = high-risk category. P(A)=0.51; P(B|A)=0.12; P(B|not A)=0.08.
The reversed conditional probability P(A | B).
The question gives evidence given each cause, but asks for cause given evidence. Bayes’ rule combines the prior with that likelihood and divides by all ways of obtaining the evidence.
![\"P(\\neg](\"math/8cc383cc77727275.svg\")
Multiply the prior for male by the likelihood within that group.
![\"P(A\\cap](\"math/b7d733145715c7d9.svg\")
Now include the path through female.
![\"P(\\neg](\"math/916824e2b1c49a5f.svg\")
![\"P(B)=0.0612+0.0392=0.1004\"](\"math/eafd0d70d15c9138.svg\")
![\"P(A\\mid](\"math/35fed556321f4dd0.svg\")
Machine A makes 60% of items and Machine B makes 40%. Defect rates are 4% for A and 1% for B. A randomly chosen item is defective. Find the probability that Machine A made it. [3 marks · Worked probability example]
Cause A = Machine A; evidence B = defective item. P(A)=0.60; P(B|A)=0.04; P(B|not A)=0.01.
The reversed conditional probability P(A | B).
The question gives evidence given each cause, but asks for cause given evidence. Bayes’ rule combines the prior with that likelihood and divides by all ways of obtaining the evidence.
![\"P(\\neg](\"math/2d95ba220d546868.svg\")
Multiply the prior for Machine A by the likelihood within that group.
![\"P(A\\cap](\"math/69ed4b855796ade5.svg\")
Now include the path through Machine B.
![\"P(\\neg](\"math/43a91fadefff9097.svg\")
![\"P(B)=0.024+0.004=0.028\"](\"math/7297bb21152d3d31.svg\")
![\"P(A\\mid](\"math/c6539c9bc8ea7676.svg\")
A rare prior can make a positive result much less conclusive than its sensitivity suggests. The posterior still depends on the actual prior and both test rates; rarity alone does not determine the answer.
Yaad rakho: Pehle requested cause ka path multiply. Neeche evidence tak pahunchne wale saare paths add.
"}]},{"id":"u1-nb","title":"Unit 1 · Naive Bayes classifier","unit":"1","scope":"Unit 1 · Naive Bayes model · 10-mark PYQ in both papers","sections":[{"id":"model","title":"The naive Bayes model (worked on PlayTennis)","html":"\r\nA training row is one past case whose correct class is known. Its features are the information used to predict that class. In the flu exercise, Fever, Cough and Temperature are features; Flu Yes/No is the class. A query is the new case whose features are known but whose class must be predicted.
A prior is the class share before using the query features. A likelihood asks how common one query feature is inside that class. A posterior is the class probability after accounting for the query. The naive assumption makes the feature likelihoods multiply once the class is fixed; it does not assert that all features are independent in the whole population.
| Read this | Meaning | Concrete example |
|---|---|---|
| Π likelihoods | Multiply every feature factor | With factors .5,.2,.4: .5×.2=.1; .1×.4=.04 |
| score(C) | An unnormalised class weight | Prior .3 times .04 gives .012 |
| max | The largest numerical score | max(.012,.008)=.012 |
| argmax | The label attaining the largest score | The winning label is the class with score .012 |
| Laplace denominator N꜀+k | Class count plus number of possible values for this feature | Temperature has three values, so k=3; a Yes/No feature has k=2 |
In one line: assume the features are independent given the class; then score each class as prior × product of feature likelihoods and pick the largest.
\r\n\r\nNaive Bayes is a supervised probabilistic classifier based on Bayes' rule with the naive assumption that all features are conditionally independent given the class:
\r\n![\"P(C\\mid](\"math/b770ed6380567e45.svg\")
“Naive” because real features are often correlated; it still works very well for spam filtering, sentiment analysis and medical screening.
\r\n![\"P(c)=\"](\"math/73533725519e6dc2.svg\")
(rows of class c) / (total rows).![\"P(a_i\\mid](\"math/51ff85a257c034ba.svg\")
(rows of class c with that value) / (rows of class c).![\"P(a\\mid](\"math/55066f62fb8b94f2.svg\")
, k = number of values of the attribute. Mention it in one line; it can change the answer.Question. Using the 14-day PlayTennis data (9 Yes, 5 No), classify ⟨Outlook = Sunny, Temperature = Cool, Humidity = High, Wind = Strong⟩. [class example · 6 marks]
Class counts: Yes=9, No=5. Within Yes, counts for Sunny/Cool/High/Strong are 2/3/3/3; within No they are 3/1/4/3. Classify Sunny, Cool, High, Strong.
Both class scores and the predicted class.
Conditional independence given the class allows the product of feature likelihoods.
![\"score(C)=P(C)\\prod_jP(a_j\\mid](\"math/e0195b1e4ddb0473.svg\")
![\"P(yes)=\\tfrac9{14}\\approx0.643,\\qquad](\"math/be7de04ac2f36060.svg\")
| Sunny | Cool | High | Strong | |
|---|---|---|---|---|
| Yes | 2/9 | 3/9 | 3/9 | 3/9 |
| No | 3/5 | 1/5 | 4/5 | 3/5 |
![\"v(yes)=\\tfrac9{14}\\cdot\\tfrac29\\cdot\\tfrac39\\cdot\\tfrac39\\cdot\\tfrac39\\approx0.005291005,\\qquad](\"math/21600d7754c6d954.svg\")
![\"P(yes\\mid](\"math/183b5657f9957d15.svg\")
Question. A hospital wants to predict flu from Fever (Yes/No), Cough (Yes/No) and Temperature Level (Low/Normal/High). Training data: (Yes, Yes, High, Yes), (No, No, Normal, No), (Yes, No, Low, No), (Yes, Yes, High, Yes), (No, Yes, Normal, No), (Yes, Yes, High, Yes), (No, No, Low, No), (Yes, No, Normal, No). Using naive Bayes, predict whether a patient with Fever = Yes, Cough = Yes, Temperature = Normal has flu. [10 marks · Final Nov 2025 Q4b]
The full training data are in Step 1. Query: Fever=Yes, Cough=Yes, Temperature=Normal. The main calculation uses unsmoothed counts.
Both class scores and the predicted class.
C is a class label. Each aⱼ is one query feature value. First count inside a class, then multiply the prior and all three likelihoods.
| Row | Fever | Cough | Temperature | Class |
|---|---|---|---|---|
| 1 | Yes | Yes | High | Yes |
| 2 | No | No | Normal | No |
| 3 | Yes | No | Low | No |
| 4 | Yes | Yes | High | Yes |
| 5 | No | Yes | Normal | No |
| 6 | Yes | Yes | High | Yes |
| 7 | No | No | Low | No |
| 8 | Yes | No | Normal | No |
| Class | Rows in this group | Class count | Prior |
|---|---|---|---|
| Yes | 1, 4, 6 | 3 | 3/8 = 0.375 |
| No | 2, 3, 5, 7, 8 | 5 | 5/8 = 0.625 |
Look only at class Yes rows 1, 4, 6. The matching rows are 1, 4, 6.
![\"P(\\text{Fever}=\\text{Yes}\\mid](\"math/bf71c2b5401bc336.svg\")
Look only at class Yes rows 1, 4, 6. The matching rows are 1, 4, 6.
![\"P(\\text{Cough}=\\text{Yes}\\mid](\"math/a6ffcdb19b7e5a36.svg\")
Look only at class Yes rows 1, 4, 6. No row in this class has that feature value.
![\"P(\\text{Temperature}=\\text{Normal}\\mid](\"math/2bcd33037561157b.svg\")
![\"P(Yes)=3/8=0.375\"](\"math/c3f053c53b40a9b7.svg\")
![\"0.375\\times1=0.375\"](\"math/5f8249f6f69de64a.svg\")
![\"0.375\\times0=0\"](\"math/e5245fc04fe5be62.svg\")
Look only at class No rows 2, 3, 5, 7, 8. The matching rows are 3, 8.
![\"P(\\text{Fever}=\\text{Yes}\\mid](\"math/2feb71df5b426bfc.svg\")
Look only at class No rows 2, 3, 5, 7, 8. The matching rows are 5.
![\"P(\\text{Cough}=\\text{Yes}\\mid](\"math/d406a69469adaf9b.svg\")
Look only at class No rows 2, 3, 5, 7, 8. The matching rows are 2, 5, 8.
![\"P(\\text{Temperature}=\\text{Normal}\\mid](\"math/738acde5b7f3afea.svg\")
![\"P(No)=5/8=0.625\"](\"math/b83bf0fcbd8f3bee.svg\")
![\"0.625\\times0.4=0.25\"](\"math/ccabec7597d2610a.svg\")
![\"0.25\\times0.2=0.05\"](\"math/4b7931095eb1884c.svg\")
![\"0.05\\times0.6=0.03\"](\"math/fcec3b47d13c61a4.svg\")
![\"score(Yes)=0\\qquad](\"math/9f7a6514218ef651.svg\")
The larger score belongs to No. Normalise only if a posterior probability is requested: divide each score by their total 0.03.
The zero Yes score is caused by an absent training combination. If Laplace smoothing is explicitly required, apply it to all feature likelihoods in both classes. That is a different calculation and may change the prediction; do not mix smoothed and unsmoothed sides.
Yaad rakho: Class ke rows alag karo. Har feature ka count usi class ke total se divide. Saare factors multiply karo.
"},{"id":"job-offer","title":"Solved PYQ: naive Bayes for a job offer (Re-exam 2026, 10 marks)","html":"Question. Predict whether a candidate accepts a job offer from Experience (Fresher/Mid/Senior), Expected Salary (Low/Medium/High) and Company Reputation (Good/Average). Data: (Fresher, Low, Good, Yes), (Mid, High, Good, Yes), (Senior, High, Average, No), (Fresher, Medium, Average, Yes), (Senior, Low, Good, No), (Mid, Medium, Good, Yes), (Fresher, High, Average, No), (Senior, Medium, Average, No). New candidate: Mid, Low, Good. [10 marks · Re-exam Jan 2026 Q4b]
The full training data are in Step 1. Query: Experience=Mid, Salary=Low, Reputation=Good. The main calculation uses unsmoothed counts.
Both class scores and the predicted class.
C is a class label. Each aⱼ is one query feature value. First count inside a class, then multiply the prior and all three likelihoods.
| Row | Experience | Salary | Reputation | Class |
|---|---|---|---|---|
| 1 | Fresher | Low | Good | Yes |
| 2 | Mid | High | Good | Yes |
| 3 | Senior | High | Average | No |
| 4 | Fresher | Medium | Average | Yes |
| 5 | Senior | Low | Good | No |
| 6 | Mid | Medium | Good | Yes |
| 7 | Fresher | High | Average | No |
| 8 | Senior | Medium | Average | No |
| Class | Rows in this group | Class count | Prior |
|---|---|---|---|
| Yes | 1, 2, 4, 6 | 4 | 4/8 = 0.5 |
| No | 3, 5, 7, 8 | 4 | 4/8 = 0.5 |
Look only at class Yes rows 1, 2, 4, 6. The matching rows are 2, 6.
![\"P(\\text{Experience}=\\text{Mid}\\mid](\"math/d14d2e65e85aa98c.svg\")
Look only at class Yes rows 1, 2, 4, 6. The matching rows are 1.
![\"P(\\text{Salary}=\\text{Low}\\mid](\"math/b6e871091f36d8f9.svg\")
Look only at class Yes rows 1, 2, 4, 6. The matching rows are 1, 2, 6.
![\"P(\\text{Reputation}=\\text{Good}\\mid](\"math/00e0bac1e035a9f7.svg\")
![\"P(Yes)=4/8=0.5\"](\"math/fa6d94e12c8c8a44.svg\")
![\"0.5\\times0.5=0.25\"](\"math/cd8320ff28f21681.svg\")
![\"0.25\\times0.25=0.0625\"](\"math/646a455657cc6928.svg\")
![\"0.0625\\times0.75=0.046875\"](\"math/9df28e0200ccc17b.svg\")
Look only at class No rows 3, 5, 7, 8. No row in this class has that feature value.
![\"P(\\text{Experience}=\\text{Mid}\\mid](\"math/210f557f91613842.svg\")
Look only at class No rows 3, 5, 7, 8. The matching rows are 5.
![\"P(\\text{Salary}=\\text{Low}\\mid](\"math/e95fc4843fd44476.svg\")
Look only at class No rows 3, 5, 7, 8. The matching rows are 5.
![\"P(\\text{Reputation}=\\text{Good}\\mid](\"math/6f502d0eec66aa0e.svg\")
![\"P(No)=4/8=0.5\"](\"math/7ee037eb525f597a.svg\")
![\"0\\times0.25=0\"](\"math/cf25ab85b70dd28a.svg\")
![\"score(Yes)=0.046875\\qquad](\"math/14926697646660d6.svg\")
The larger score belongs to Yes. Normalise only if a posterior probability is requested: divide each score by their total 0.046875.
Yaad rakho: Class ke rows alag karo. Har feature ka count usi class ke total se divide. Saare factors multiply karo.
"}]},{"id":"u1-bn","title":"Unit 1 · Bayesian networks","unit":"1","scope":"Unit 1 · Semantics of Bayesian networks and exact inference · 10-mark PYQ in both papers","sections":[{"id":"semantics","title":"Bayesian networks: structure, CPTs and the chain rule","html":"\r\nA node is a variable, such as Rain. An edge is an arrow representing a direct dependency in the model. A parent has an arrow into a node. A root has no parents. Directed acyclic graph (DAG) means arrows have directions and you cannot follow them around a loop back to the starting node. The graph alone does not establish real-world causation.
A CPT is a conditional probability table: for each parent configuration, it states the probabilities of the child’s values. A Boolean variable has two values, true and false. If a CPT gives P(Wet=true|Rain=true)=.9, its complementary entry is P(Wet=false|Rain=true)=1−.9=.1.
In Rain → Wet, P(Rain)=0.2, P(Wet|Rain)=0.9 and P(Wet|no Rain)=0.1. Find P(Rain AND Wet), P(Wet), and P(Rain|Wet). [2 marks · Worked example]
P(Rain)=0.2; P(Wet|Rain)=0.9; P(Wet|no Rain)=0.1. Each variable has two possible values: true or false. Rain is the root; Wet depends on Rain.
A joint probability, a marginal probability and a posterior.
Multiply down one complete case, add the alternative ways to get Wet, then divide by the evidence probability.
![\"P(R,W)=0.2\\times0.9=0.18\"](\"math/7395138baece4728.svg\")
![\"P(\\neg](\"math/1b0b61dc6cadc7a7.svg\")
![\"P(W)=0.18+0.08=0.26\"](\"math/6c276fca73f350e2.svg\")
![\"P(R\\mid](\"math/000872ced145bc03.svg\")
In one line: a Bayesian network is a DAG of random variables in which each node stores P(node | its parents); multiplying those tables gives the full joint distribution.
\r\n\r\nA Bayesian network (belief network) is a directed acyclic graph in which:
\r\n![\"\\mathbf](\"math/e2e32da4799205fe.svg\")
. A root node stores its prior.![\"Burglary](\"fig/bn_burglary.9802d6d53d.svg\")
![\"P(x_1,\\dots,x_n)=\\prod_{i=1}^{n}P\\big(x_i\\mid](\"math/228b443a51a8c577.svg\")
So the network is a compact description of the full joint distribution. For the burglary network:
\r\n![\"P(J,M,A,B,E)=P(J\\mid](\"math/d065035a53d0c75c.svg\")
Conditional independence: each node is conditionally independent of its non-descendants given its parents (e.g. JohnCalls is independent of Burglary once Alarm is known).
\r\nCompactness: a Boolean node with k Boolean parents needs ![\"2^k\"](\"math/1e088370ad90412f.svg\")
numbers. The burglary network needs 1 + 1 + 4 + 2 + 2 = 10 numbers instead of ![\"2^5-1=31\"](\"math/2bd250ce49e6fced.svg\")
for the full joint.
For burglary given both calls, B is the query, J=true and M=true are evidence, and E and A are hidden. “Hidden” only means their values are not supplied in the query. A sum over a Boolean variable adds its true and false cases.
| Calculation | Burglary true | Burglary false |
|---|---|---|
| Average alarm probability over E | .95×.002 + .94×.998 = .94002 | .29×.002 + .001×.998 = .001578 |
| No-alarm probability | 1−.94002=.05998 | 1−.001578=.998422 |
| Both calls if alarm | .9×.7=.63 | .9×.7=.63 |
| Both calls if no alarm | .05×.01=.0005 | .05×.01=.0005 |
| Calls and this B value | .001×(.63×.94002 + .0005×.05998)=.00059224259 | .999×(.63×.001578 + .0005×.998422)=.001491857649 |
Add the final two weights: .00059224259+.001491857649=.002084100239. Divide the burglary weight by that total: .00059224259/.002084100239≈.284171835. No step equates P(B|calls) with P(calls|B); the prior P(B) is included explicitly.
In one line: to answer P(query | evidence), sum the chain-rule product over every hidden variable, do it for each value of the query, and normalise with α.
\r\n\r\n| Question type | Method |
|---|---|
| Every variable's value given | Multiply the CPT entries (chain rule) |
| One marginal, e.g. P(J) | Sum over hidden variables; no α |
| Query given evidence, e.g. P(B | j, m) | Enumerate for each value of B, then ![\"\\alpha=1/\\text{(sum)}\"](\"math/0666d2fea1f0f562.svg\") |
Question. In the burglary network find (a) P(j, m, a, ¬b, ¬e), (b) P(JohnCalls), (c) P(Burglary | John and Mary call). [class examples · 10 marks]
For the burglary example: B and E are independent roots with P(B)=0.001 and P(E)=0.002. Alarm A has parents B,E. John J and Mary M each have parent A. P(A|B,E)=0.95, P(A|B,¬E)=0.94, P(A|¬B,E)=0.29, P(A|¬B,¬E)=0.001. P(J|A)=0.9, P(J|¬A)=0.05; P(M|A)=0.7, P(M|¬A)=0.01. False probabilities are the complements of the listed true probabilities.
The network, root probabilities and conditional probability tables stated in the question.
The requested joint, marginal or evidence-conditioned probability.
A Bayesian network factorises the joint using each node conditioned on its parents. Sum over hidden values; normalise for evidence.
![\"P(x_1,\\ldots,x_n)=\\prod_iP(x_i\\mid](\"math/fbf52d09e73db054.svg\")
![\"P=P(j\\mid](\"math/ac4abea66468645f.svg\")
![\"P(a)=0.95(0.001)(0.002)+0.94(0.001)(0.998)+0.29(0.999)(0.002)+0.001(0.999)(0.998)=0.002516\"](\"math/c16479d9b9398167.svg\")
![\"P(j)=P(j\\mid](\"math/6aa5c702fa516ce7.svg\")
![\"P(b\\mid](\"math/5648415bb994a47e.svg\")
![\"=\\alpha\\,(0.001)\\big[0.9(0.7)(0.95\\times0.002+0.94\\times0.998)+0.05(0.01)(0.05\\times0.002+0.06\\times0.998)\\big]=\\alpha\\times0.000592\"](\"math/c977798f6840354f.svg\")
![\"P(\\lnot](\"math/57244f2b9ccfa07b.svg\")
![\"\\alpha=\\frac1{0.000592+0.001492}=479.8\\;\\Rightarrow\\;P(b\\mid](\"math/57f47697add29181.svg\")
![\"P(\\lnot](\"math/f2ae2e9bf6ce6704.svg\")
; it is 1 − 0.94 = 0.06 (corrected by hand on the slide).Enumeration repeats the same products many times (exponential, ![\"O(2^n)\"](\"math/4c7e79dfd318e20b.svg\")
). Variable elimination pushes each sum as far inside as possible, computes it once and stores the result as a factor; hidden variables are “summed out” one at a time:
![\"P(J)=\\sum_AP(J\\mid](\"math/5ea704a1fb57ef3a.svg\")
Any sum of a CPT over its own variable is 1 (e.g. ![\"\\sum_MP(M\\mid](\"math/6230fbd1828345e4.svg\")
), so irrelevant (barren) nodes drop out.
Question. Heart-disease network: Exercise → HD ← Diet; HD → BP. P(E) = 0.7, P(D = healthy) = 0.25; P(HD | E, D) = 0.25 (E, healthy), 0.45 (E, unhealthy), 0.55 (¬E, healthy), 0.75 (¬E, unhealthy); P(BP high | HD) = 0.85, P(BP high | ¬HD) = 0.2. A patient has high BP. Find P(HD | BP). [class example · 6 marks]
The network, root probabilities and conditional probability tables stated in the question.
The requested joint, marginal or evidence-conditioned probability.
A Bayesian network factorises the joint using each node conditioned on its parents. Sum over hidden values; normalise for evidence.
![\"P(hd)=0.7(0.25\\times0.25+0.45\\times0.75)+0.3(0.55\\times0.25+0.75\\times0.75)=0.7(0.4)+0.3(0.7)=0.49\"](\"math/3485875632aa8ada.svg\")
![\"\\alpha\\,P(bp\\mid](\"math/9429a00abcd3cccf.svg\")
![\"\\alpha=\\frac1{0.5185},\\quad](\"math/13277c3f7f4e8b01.svg\")
The following table enumerates the values of Cloudy and Sprinkler that were not fixed by the question. Two Boolean choices give four rows: true/true, true/false, false/true, false/false. Each row is one mutually exclusive complete case. Read the CPT entry using that row’s parent values, multiply its factors, and add the resulting row weights. The denominator must include all cases matching the evidence, including both possible values of the queried outcome.
Question. Cloudy → Sprinkler, Cloudy → Rain, Sprinkler and Rain → Wet Grass. P(Cloudy) = 0.5; P(Sprinkler | C) = 0.10, P(Sprinkler | ¬C) = 0.50; P(Rain | C) = 0.80, P(Rain | ¬C) = 0.20; P(Wet | S, R) = 0.99, P(Wet | S, ¬R) = 0.90, P(Wet | ¬S, R) = 0.90, P(Wet | ¬S, ¬R) = 0.01. Calculate the probability of the grass being wet given it doesn't rain. [10 marks · Final Nov 2025 Q3a]
C→S, C→R, (S,R)→W. Root and conditional probabilities are listed in the question. Evidence R=false.
P(W=true | R=false). Hidden variables C and S must be summed out.
Use the network joint and divide by the evidence probability.
![\"P(W\\mid\\neg](\"math/ec8db3b28ef749de.svg\")
![\"Final](\"fig/bn_sprinkler.a251583272.svg\")
![\"P(C,S,R,W)=P(C)\\,P(S\\mid](\"math/62a8884ca57ac2e6.svg\")
| C | S | P(C) | P(S|C) | P(¬R|C) | P(W|S,¬R) | Product |
|---|---|---|---|---|---|---|
| T | T | 0.5 | 0.1 | 0.2 | 0.90 | 0.0090 |
| T | F | 0.5 | 0.9 | 0.2 | 0.01 | 0.0009 |
| F | T | 0.5 | 0.5 | 0.8 | 0.90 | 0.1800 |
| F | F | 0.5 | 0.5 | 0.8 | 0.01 | 0.0020 |
| P(W, ¬R) | 0.1919 |
![\"P(\\lnot](\"math/97de213766352c7b.svg\")
![\"P(W\\mid\\lnot](\"math/c18c31810569b62d.svg\")
The four no-rain cases already total 0.1919. Now add the four rain cases, keeping the same Cloudy/Sprinkler order.
\r\n![\"P(W,R)=0.5(0.1)(0.8)(0.99)+0.5(0.9)(0.8)(0.90)+0.5(0.5)(0.2)(0.99)+0.5(0.5)(0.2)(0.90)\"](\"math/fa867c388246e93b.svg\")
![\"P(W,R)=0.0396+0.324+0.0495+0.045=0.4581\"](\"math/43750abdd00f3218.svg\")
![\"P(W)=P(W,R)+P(W,\\neg](\"math/9357fadf4faa85ed.svg\")
![\"P(R\\mid](\"math/a03cc788f3ded23d.svg\")
Meaning: the unconditional chance of Wet is 0.65, the chance of Rain among wet cases is about 0.704769, and the requested chance of Wet among no-rain cases is 0.3838. These are three different questions.
\r\nThe following table enumerates the values of Weather and Cough that were not fixed by the question. Two Boolean choices give four rows: true/true, true/false, false/true, false/false. Each row is one mutually exclusive complete case. Read the CPT entry using that row’s parent values, multiply its factors, and add the resulting row weights. The denominator must include all cases matching the evidence, including both possible values of the queried outcome.
Question. Weather → Flu, Weather → Cough, Flu and Cough → Fever. P(Weather) = 0.75; P(Flu | W) = 0.2, P(Flu | ¬W) = 0.5; P(Cough | W) = 0.70, P(Cough | ¬W) = 0.25; P(Fever | Flu, Cough) = 0.60, P(Fever | Flu, ¬Cough) = 0.02, P(Fever | ¬Flu, Cough) = 0.60, P(Fever | ¬Flu, ¬Cough) = 0.01. Calculate the probability of a person having fever given he doesn't have flu. [10 marks · Re-exam Jan 2026 Q3a]
W→F, W→C, (F,C)→V. The complete probabilities are in the question. Evidence F=false.
P(V=true | F=false), summing hidden W and C.
Use the network joint and normalise by the no-Flu evidence.
![\"P(V\\mid\\neg](\"math/e2104f63677bf585.svg\")
![\"Re-exam](\"fig/bn_flu.1d732f0fb6.svg\")
![\"P(W,F,C,V)=P(W)\\,P(F\\mid](\"math/73135e72ac19cca4.svg\")
| W | C | P(W) | P(¬F|W) | P(C|W) | P(V|¬F,C) | Product |
|---|---|---|---|---|---|---|
| T | T | 0.75 | 0.8 | 0.70 | 0.60 | 0.252 |
| T | F | 0.75 | 0.8 | 0.30 | 0.01 | 0.0018 |
| F | T | 0.25 | 0.5 | 0.25 | 0.60 | 0.01875 |
| F | F | 0.25 | 0.5 | 0.75 | 0.01 | 0.0009375 |
| P(V, ¬F) | 0.2734875 |
![\"P(\\lnot](\"math/a396371c860e59b0.svg\")
![\"P(V\\mid\\lnot](\"math/a27699b35a63a308.svg\")
A Gaussian component describes a cluster using its mean (centre) and variance (spread). The standard deviation is the square root of variance. The Greek letters μ and σ name these two quantities; σ² is the variance.
Calculate the mean, population variance and standard deviation of [1, 3, 5]. [2 marks · Worked example]
Data values: [1, 3, 5]. Number of observations n=3. Use population variance, dividing by n.
The mean μ, population variance σ², and standard deviation σ.
![\"\\mu=\\frac{\\sum_i](\"math/13c92c5c73ee1c2f.svg\")
This descriptive/maximum-likelihood convention divides by n. The n−1 sample-variance estimator is a different convention.
![\"\\mu=\\frac{1+3+5}{3}=3\"](\"math/c0bd139e7dbf3b1f.svg\")
| x | x−mean | Squared deviation |
|---|---|---|
| 1 | -2 | 4 |
| 3 | 0 | 0 |
| 5 | 2 | 4 |
![\"\\sigma^2=\\frac{4+0+4}{3}=2.666666667\"](\"math/f8055b1e2e609fe7.svg\")
![\"\\sigma=\\sqrt{2.666666667}\\approx1.632993162\"](\"math/f193c4f2114bfb50.svg\")
Calculate the mean, population variance and standard deviation of [2, 4]. [2 marks · Worked example]
Data values: [2, 4]. Number of observations n=2. Use population variance, dividing by n.
The mean μ, population variance σ², and standard deviation σ.
This descriptive/maximum-likelihood convention divides by n. The n−1 sample-variance estimator is a different convention.
![\"\\mu=\\frac{2+4}{2}=3\"](\"math/0c66b5b185711660.svg\")
| x | x−mean | Squared deviation |
|---|---|---|
| 2 | -1 | 1 |
| 4 | 1 | 1 |
![\"\\sigma^2=\\frac{1+1}{2}=1\"](\"math/7740370cf600b4a5.svg\")
![\"\\sigma=\\sqrt{1}\\approx1\"](\"math/dc84a3fdebee073d.svg\")
Calculate the mean, population variance and standard deviation of [4, 4, 4]. [2 marks · Worked example]
Data values: [4, 4, 4]. Number of observations n=3. Use population variance, dividing by n.
The mean μ, population variance σ², and standard deviation σ.
This descriptive/maximum-likelihood convention divides by n. The n−1 sample-variance estimator is a different convention.
![\"\\mu=\\frac{4+4+4}{3}=4\"](\"math/133ab42db9ac4f4c.svg\")
| x | x−mean | Squared deviation |
|---|---|---|
| 4 | 0 | 0 |
| 4 | 0 | 0 |
| 4 | 0 | 0 |
![\"\\sigma^2=\\frac{0+0+0}{3}=0\"](\"math/8148ccf99111a269.svg\")
![\"\\sigma=\\sqrt{0}\\approx0\"](\"math/56fce38f6a89699b.svg\")
The constant data [4,4,4] have variance zero. A standard Gaussian density formula requires positive variance; do not divide by zero when fitting a component.
A density is the height of a continuous distribution. Probability over an interval is area, so density height can exceed 1 without violating probability rules. A likelihood treats the observed data as fixed and scores candidate model parameters. For independent observations, their densities multiply.
A model assigns density 0.2 to one observation and 0.5 to a second independent observation. Find likelihood and natural log-likelihood. [2 marks · Worked example]
Two density values .2 and .5.
Product and its natural logarithm.
![\"L=p(x_1)p(x_2)\\qquad\\ln](\"math/8e3c44857ba18435.svg\")
The logarithm turns a product into a sum.
![\"L=0.2\\times0.5=0.1\"](\"math/b651f1b57b4503e2.svg\")
![\"\\ln](\"math/47a26d7f62e313c6.svg\")
![\"\\ln(0.2)+\\ln(0.5)\\approx-1.609437912-0.693147181=-2.302585093\"](\"math/5a43c9ee348bb3a2.svg\")
Do not turn a sum inside a log into a sum of logs: ln(.2+.5)=ln(.7)≈−.356675, whereas ln(.2)+ln(.5)≈−2.302585. A GMM has an inner sum over components, which must stay inside each observation’s logarithm.
Yaad rakho: Mean = centre. Variance = squared distance ka average. Likelihood = fixed data ko model kitna support karta hai.
"},{"id":"mixture-models","title":"GMM prerequisites: mean, variance, density and mixture","html":"\r\n| Read this | Meaning | Concrete example |
|---|---|---|
| Component k | One Gaussian group in the mixture | Component 1 may have mean 2; component 2 mean 4 |
| Mixing weight πₖ | The prior share assigned to component k | π₁=.4 and π₂=.6 sum to 1 |
| N(x|μ,σ²) | The Gaussian density evaluated at x | The vertical bar supplies parameters; it is not division |
| Σ over k | Add contributions of every component | p(x)=.4N₁(x)+.6N₂(x) |
A mean is the centre, found by adding observations and dividing by their count. A variance measures squared spread around the mean. In one dimension, a Gaussian component is described by mean μ and positive variance σ². Its standard deviation σ is the square root of that variance.
A density is a height, not the probability of one exact real-valued observation. Areas under a density give probabilities. Density heights can exceed 1; their total area must be 1.
![\"\\mathcal](\"math/a9b2cdf84258b1c9.svg\")
Here π inside 2π is the constant 3.14159…; the separately written πₖ in a mixture denotes a component weight. exp(a)=eᵃ, and the calculator eˣ key evaluates it. The minus sign in the exponent reduces density away from the mean.
For a standard Gaussian with μ=0 and σ²=1, find the density at x=0. [3 marks · Practice]
x=0; μ=0; σ=1.
Density N(x|0,1).
Use the Gaussian density with the supplied mean and variance.
![\"(0-0)^2=0\"](\"math/faf8b496eb6d2931.svg\")
![\"-0/(2\\times1)=0\"](\"math/52577cebaa3ae73f.svg\")
![\"e^{0}\\approx1\"](\"math/86787607e49b39df.svg\")
![\"\\frac{1}{\\sqrt{2\\pi}}\\approx0.39894228\"](\"math/187fc723b4884bed.svg\")
For a standard Gaussian with μ=0 and σ²=1, find the density at x=1. [3 marks · Practice]
x=1; μ=0; σ=1.
Density N(x|0,1).
Use the Gaussian density with the supplied mean and variance.
![\"(1-0)^2=1\"](\"math/39a0d42c254cae5b.svg\")
![\"-1/(2\\times1)=-0.5\"](\"math/a4066df8f72b8058.svg\")
![\"e^{-0.5}\\approx0.60653066\"](\"math/9fa1a1619e482083.svg\")
![\"\\frac{0.60653066}{\\sqrt{2\\pi}}\\approx0.241970725\"](\"math/7893b72b8b5aa648.svg\")
For a standard Gaussian with μ=0 and σ²=1, find the density at x=2. [3 marks · Practice]
x=2; μ=0; σ=1.
Density N(x|0,1).
Use the Gaussian density with the supplied mean and variance.
![\"(2-0)^2=4\"](\"math/9da874308fbfbb30.svg\")
![\"-4/(2\\times1)=-2\"](\"math/aa2983139384fc61.svg\")
![\"e^{-2}\\approx0.135335283\"](\"math/37f9ca8ef53fee4b.svg\")
![\"\\frac{0.135335283}{\\sqrt{2\\pi}}\\approx0.053990967\"](\"math/70e444d7db1b7149.svg\")
A mixture model combines component densities. A Gaussian mixture model (GMM) uses K Gaussian components. A hidden component label says which component generated a point; it is not observed in ordinary clustering. A weight πₖ is the prior probability of choosing component k. All weights are nonnegative and sum to 1.
![\"p(x)=\\sum_{k=1}^K\\pi_k\\mathcal](\"math/8c4c98f40f54268b.svg\")
For two components this means π₁×density₁ + π₂×density₂. A GMM can represent several concentrations of data rather than forcing all points around one mean.
![\"The](\"fig/gmm_mix.84e959698a.svg\")
At x, two component densities are 0.2 and 0.05, with weights 0.6 and 0.4. Find the mixture density. [2 marks · Practice]
π₁=.6, π₂=.4; densities=.2,.05.
p(x).
Weight each component density, then add.
![\"p(x)=\\pi_1N_1+\\pi_2N_2\"](\"math/91ef77408ecb18fc.svg\")
![\"0.6(0.2)=0.12\"](\"math/b90648f46ba2060f.svg\")
![\"0.4(0.05)=0.02\"](\"math/1cee8e50d6767b0f.svg\")
![\"p(x)=0.12+0.02=0.14\"](\"math/70eb8d3c0588a981.svg\")
Yaad rakho: Har Gaussian ek group ka model; mixture = weights ke saath sab densities ka sum.
"},{"id":"em","title":"EM for GMM: every responsibility and parameter update","html":"\r\nAt one observation, the two weighted component densities are q₁=0.12 and q₂=0.03. Find the responsibilities. Then use responsibilities [0.8,0.2] for data [1,5] to find a component’s new mean. [2 marks · Worked example]
Part A: weighted component densities q₁=0.12 and q₂=0.03 for one observation. Part B: data x₁=1 and x₂=5, with component responsibilities 0.8 and 0.2 respectively. Part A compares components for one point; Part B combines points for one component.
Two responsibilities and a weighted mean.
![\"\\gamma_{ik}=q_{ik}/\\sum_jq_{ij}\\qquad\\mu_k^{new}=\\frac{\\sum_i\\gamma_{ik}x_i}{\\sum_i\\gamma_{ik}}\"](\"math/5add87e9c47f7469.svg\")
i identifies an observation; k identifies a component. Notice which index is being added in each formula.
![\"q_1+q_2=0.12+0.03=0.15\"](\"math/b318c8654f7774af.svg\")
![\"\\gamma_1=0.12/0.15=0.8\\quad\\gamma_2=0.03/0.15=0.2\"](\"math/867693405ef3ab51.svg\")
![\"N_k=0.8+0.2=1\"](\"math/860f0958878c117d.svg\")
![\"0.8\\times1=0.8\\qquad0.2\\times5=1\"](\"math/bf3d6ca02caff691.svg\")
![\"\\mu_k^{new}=(0.8+1)/1=1.8\"](\"math/5699fad0064a323f.svg\")
Expectation–maximisation (EM) alternates two calculations when component assignments are hidden. The E-step estimates how responsible each component is for each observation. The M-step uses those responsibilities as weights to update the model parameters.
A responsibility γᵢₖ is a posterior component probability: for observation i, how likely is component k under the current model? For each observation, the responsibilities sum to 1. Unlike fuzzy memberships, these probabilities are explicitly normalised across mixture components.
| Symbol | Meaning |
|---|---|
| i / k | Observation number / component number |
| γᵢₖ | Responsibility of component k for observation i |
| Nₖ | Sum of responsibilities for component k: an effective count |
| μₖ | Component mean |
| σ²ₖ | Component variance |
| πₖ | Component mixing weight |
![\"\\gamma_{ik}=\\frac{\\pi_k\\mathcal](\"math/1d66a1a364cc52ce.svg\")
![\"N_k=\\sum_i\\gamma_{ik},\\quad\\mu_k^{new}=\\frac{\\sum_i\\gamma_{ik}x_i}{N_k}\"](\"math/3aff5f591828cac6.svg\")
![\"(\\sigma_k^2)^{new}=\\frac{\\sum_i\\gamma_{ik}(x_i-\\mu_k^{new})^2}{N_k},\\quad\\pi_k^{new}=N_k/n\"](\"math/10983e175fb2b145.svg\")
The E-step uses all old parameters. The M-step mean uses those fixed responsibilities; the variance uses the newly calculated mean. One exact EM update cannot lower the likelihood when its optimisation steps are performed correctly; numerical approximations and regularisation need their own interpretation. EM can still reach a local solution.
For observations {1,3,5}, initialise a two-component GMM with means 2 and 4, both variances 1, and both weights 0.5. Perform one complete EM iteration. [6 marks · Practice]
n=3, K=2, μ₁=2, μ₂=4; σ₁²=σ₂²=1; π₁=π₂=.5.
Six responsibilities and the new weights, means and variances.
Apply the E-step and M-step equations above. Calculate with full precision; decimals below are displayed to nine places.
![\"q_1=0.5\\frac{e^{-(1-2)^2/2}}{\\sqrt{2\\pi}}\\approx0.120985362\"](\"math/8c570fbfdd0a7367.svg\")
![\"q_2=0.5\\frac{e^{-(1-4)^2/2}}{\\sqrt{2\\pi}}\\approx0.002215924\"](\"math/00ad8a0accc6492e.svg\")
![\"q_1=0.5\\frac{e^{-(3-2)^2/2}}{\\sqrt{2\\pi}}\\approx0.120985362\"](\"math/abfe7f04db4f042d.svg\")
![\"q_2=0.5\\frac{e^{-(3-4)^2/2}}{\\sqrt{2\\pi}}\\approx0.120985362\"](\"math/5c2d2f7bb6375554.svg\")
![\"q_1=0.5\\frac{e^{-(5-2)^2/2}}{\\sqrt{2\\pi}}\\approx0.002215924\"](\"math/1433e6810e632ac7.svg\")
![\"q_2=0.5\\frac{e^{-(5-4)^2/2}}{\\sqrt{2\\pi}}\\approx0.120985362\"](\"math/ad6c931c7f9ef8f8.svg\")
![\"\\gamma_1=\\frac{0.120985362}{0.120985362+0.002215924}\\approx0.98201379\"](\"math/aded404e73fc7586.svg\")
![\"\\gamma_2=1-\\gamma_1\\approx0.01798621\"](\"math/e951eddaa67ec02a.svg\")
![\"\\gamma_1=\\frac{0.120985362}{0.120985362+0.120985362}\\approx0.5\"](\"math/4db037db96f3bfa7.svg\")
![\"\\gamma_2=1-\\gamma_1\\approx0.5\"](\"math/4ed7d74583102629.svg\")
![\"\\gamma_1=\\frac{0.002215924}{0.002215924+0.120985362}\\approx0.01798621\"](\"math/f0ca8e430dd11e63.svg\")
![\"\\gamma_2=1-\\gamma_1\\approx0.98201379\"](\"math/50f16fe4d69df396.svg\")
![\"N_1=0.98201379+0.5+0.01798621=1.5\"](\"math/eb730836d82ad0e1.svg\")
![\"\\mu_1^{new}=\\frac{0.98201379(1)+0.5(3)+0.01798621(5)}{1.5}\\approx1.714629893\"](\"math/eac77d4f47719aec.svg\")
| x | Responsibility | Squared deviation from new mean | Weighted squared deviation |
|---|---|---|---|
| 1 | 0.98201379 | 0.510695884 | 0.501510401 |
| 3 | 0.5 | 1.652176311 | 0.826088156 |
| 5 | 0.01798621 | 10.793656738 | 0.194136976 |
![\"(\\sigma_1^2)^{new}=\\frac{1.521735533}{1.5}\\approx1.014490355\"](\"math/23888cceffe7f895.svg\")
![\"\\pi_1^{new}=1.5/3=0.5\"](\"math/221d1a6298a7ea65.svg\")
![\"N_2=0.01798621+0.5+0.98201379=1.5\"](\"math/04b9a1a1ff3b0c25.svg\")
![\"\\mu_2^{new}=\\frac{0.01798621(1)+0.5(3)+0.98201379(5)}{1.5}\\approx4.285370107\"](\"math/c4600c4965dd3721.svg\")
| x | Responsibility | Squared deviation from new mean | Weighted squared deviation |
|---|---|---|---|
| 1 | 0.01798621 | 10.793656738 | 0.194136976 |
| 3 | 0.5 | 1.652176311 | 0.826088156 |
| 5 | 0.98201379 | 0.510695884 | 0.501510401 |
![\"(\\sigma_2^2)^{new}=\\frac{1.521735533}{1.5}\\approx1.014490355\"](\"math/a7522d8383a399aa.svg\")
![\"\\pi_2^{new}=1.5/3=0.5\"](\"math/a4b477c2cff182c7.svg\")
Yaad rakho: E: har point ki responsibility. M: responsibilities se weighted mean, variance, weight.
"},{"id":"ten-mark-answer","title":"10-mark answer: GMM vs k-means, log-likelihood derivation, parameters for K = 5","html":"\r\n“i.i.d.” means the observations are modelled as independent draws from the same distribution. The product Π runs over observations; the inner sum Σ runs over components. λki in some notes is the same responsibility that other sections call γik. A superscript i here labels the observation, not an exponent.
Jensen’s inequality gives a lower bound on the log of the component sum. EM chooses responsibilities making that bound touch the current likelihood, then updates the parameters to maximise the bound with those responsibilities fixed. This explains the two alternating steps; λ values are not arbitrary new model parameters.
| Parameter | Number in a 1D K-component GMM | Why |
|---|---|---|
| Means | K | One centre for each component |
| Variances | K | One spread for each component |
| Independent mixing weights | K−1 | The last weight is 1 minus the others |
| Total | 3K−1 | K+K+(K−1) |
For K=2, choose μ₁,μ₂,σ₁²,σ₂² and π₁: five free numbers. Then π₂=1−π₁. For K=5, 5+5+4=14. In 2D with a full covariance matrix, each component instead has two means and three independent covariance entries (two variances and one shared off-diagonal covariance). Five such components have 5×(2+3)+4=29 free parameters.
Asked twice as: “With reference to GMM: compare GMM with the k-means algorithm; derive the equation for the log-likelihood function; what will be the number of parameters to be computed if K = 5?” [10 marks · Final 2025 Q2b and Re-exam 2026 Q2b; key: up to the maximum-likelihood equation 5, estimation 5]
\r\n\r\n| Aspect | k-means | GMM |
|---|---|---|
| Assignment | Hard: each point in exactly one cluster | Soft: probability ![\"\\lambda_k^i\"](\"math/d61decbc739ada16.svg\") of belonging to each cluster |
| Model | Geometric: nearest centroid | Probabilistic generative model ![\"\\sum_k\\pi_k\\mathcal](\"math/04ce4e1e89065414.svg\") |
| Cluster shape | Spherical, similar size | Elliptical, any size/orientation; overlapping clusters |
| Parameters | K centroids | Weights, means, covariances (1-D: 3K − 1) |
| Algorithm / objective | Assign–update; minimise squared distances | EM; maximise log-likelihood |
| Output | Hard labels and fitted centroids | Labels, membership probabilities and a density (useful for anomaly detection) |
| Relation | The hard-assignment, equal spherical variance, small-variance limit of a Gaussian-mixture objective leads to the k-means objective. Ordinary finite-variance EM is not the same algorithm as k-means. Both need K in advance and depend on initialisation. |
Step 1. Model. Each point is generated by choosing component k with probability ![\"\\pi_k\"](\"math/e60970d7047375ff.svg\")
and drawing ![\"x_i\\sim\\mathcal](\"math/f66cfb5e517110e8.svg\")
, so its density is ![\"f_{mix}(x_i)=\\sum_k\\pi_k\\mathcal](\"math/6550a4aaf9e87286.svg\")
.
Step 2. Likelihood. Points are i.i.d., so the likelihood is a product (θ = all parameters):
\r\n![\"L(\\theta)=\\prod_{i=1}^{n}\\Big[\\sum_{k=1}^{K}\\pi_k\\frac{1}{\\sqrt{2\\pi}\\,\\sigma_k}e^{-\\frac{(x_i-\\mu_k)^2}{2\\sigma_k^2}}\\Big]\"](\"math/41dc390b2652cd78.svg\")
Step 3. Log-likelihood. Logs turn the product into a sum:
\r\n![\"\\log](\"math/9a0bb26e3ef13adb.svg\")
Step 4. Why it cannot be solved directly. The sum sits inside the log, so setting derivatives to zero gives no closed-form solution.
\r\nStep 5. Jensen's inequality. log is concave, so for weights ![\"\\lambda_k\\ge0,\\;\\sum\\lambda_k=1\"](\"math/5006fcb64f6c3cf6.svg\")
: ![\"\\log\\big(\\sum_k\\lambda_ka_k\\big)\\ge\\sum_k\\lambda_k\\log](\"math/742b7c09a1ce32a4.svg\")
. Introduce for every point (![\"\\sum_k\\lambda_k^i=1\"](\"math/fbe0441b97e0b9c1.svg\")
), multiply and divide inside the log:
![\"\\log](\"math/06dff5bae3d2cfe1.svg\")
Step 6. Estimation (EM). Maximise the lower bound alternately. Fixing θ gives the E-step ![\"\\lambda_k^i=\\pi_k\\mathcal](\"math/662cdc674fe7b2c5.svg\")
. Fixing λ and differentiating (with a Lagrange multiplier for ![\"\\sum\\pi_k=1\"](\"math/131aa1c87311a5f9.svg\")
) gives the M-step:
![\"\\hat\\mu_k=\\frac{\\sum_i\\lambda_k^ix_i}{\\sum_i\\lambda_k^i},\\qquad\\hat\\sigma_k^2=\\frac{\\sum_i\\lambda_k^i(x_i-\\hat\\mu_k)^2}{\\sum_i\\lambda_k^i},\\qquad\\hat\\pi_k=\\frac{\\sum_i\\lambda_k^i}{n}\"](\"math/ffbe3f9b626d2a86.svg\")
The dimension and covariance structure must also be specified. For the one-dimensional, independent-variance model used in the derivation:
\r\n![\"2K+(K-1)=3K-1=3(5)-1=\\mathbf{14}\"](\"math/5d57bdfd8f83239c.svg\")
(5 means + 5 variances + 4 free weights, since the 5 weights must add to 1.) In two dimensions with full covariance it would be 29.
\r\n![\"f(\\sum\\lambda](\"math/2ddf537dc44d74a6.svg\")
(≥, not >).; the largest gives its cluster, and the values show how confident the assignment is.![\"p(x)\"](\"math/8f52f259a1ffabd6.svg\")
under every component are anomalies.Example (class). Marks 45, 48, 50, 52, 80, 82, 85, 88, 90 form two groups, “average” near 50 and “high performers” near 85. A two-component GMM gives each student a probability of belonging to each group; students in between (none here) would get split probabilities instead of a forced label.
\r\n| Advantages | Limitations |
|---|---|
| Soft assignments with probabilities | K must be chosen in advance |
| Flexible elliptical, overlapping clusters | Sensitive to initialisation; can stop at a local optimum |
| Interpretable parameters (weights, means, covariances) | Assumes each cluster is Gaussian |
| Gives a full density model | Costly for many dimensions (covariance matrices) |
Probability vs likelihood (lab question): probability fixes the parameters and asks how likely the data is; likelihood fixes the observed data and asks how plausible each parameter value is. Maximum-likelihood estimation chooses the parameters that make the observed data most probable.
\r\nA Gaussian centred on one training observation can make its variance approach zero, causing its density there and the mixture likelihood to grow without a finite bound. This is a singularity, not evidence of a useful perfect cluster. Practical fitting therefore uses a positive variance/covariance floor or a prior, sensible initialisation, and checks for nearly empty components. Different starts can reach different local solutions. These qualifications do not change the stated one-iteration classroom calculation.
Primary reference: Bishop, Pattern Recognition and Machine Learning, chapter 9.
"}]},{"id":"fuzzy-core","title":"2. Fuzzy logic, from meaning to full solutions","unit":"Unit 2","scope":"Definitions → membership graphs → operations → rules → defuzzification → complete controller.","sections":[{"id":"membership-meaning","title":"Fuzzy sets: what membership actually means","html":"\r\nSuppose the input is temperature 32°C and the fuzzy label is Hot. A membership function is the rule that converts that temperature into a degree of matching the label. If μHot(32)=0.7, read it aloud as “the membership of 32 degrees in Hot is 0.7”. The lower word Hot names the set; 32 is the input; 0.7 is the output.
The same temperature can have membership 0.3 in Medium and 0.7 in Hot. These are two descriptions of one input. Fuzzy memberships do not generally have to sum to 1. By contrast, probabilities of mutually exclusive, exhaustive outcomes do.
| Read this | Meaning | Concrete example |
|---|---|---|
| Universe U | All input values being considered | Temperatures from 0 to 40°C |
| Crisp value | One definite measured number | 32°C |
| Linguistic term | A word naming a fuzzy category | Low, Medium, High |
| μA(x) | Degree to which input x fits category A | μHot(32)=0.7 |
Yaad rakho: 32 temperature hai; 0.7 us temperature ka Hot label mein degree hai. Dono ko mix mat karo.
Suppose a controller says a room is hot only when its temperature is at least 30°C. At 29.9°C it says “not hot”; at 30°C it says “hot”. That sharp boundary is a crisp rule. Fuzzy logic allows a gradual degree of belonging to the category “Hot”.
\r\nDefinition. A fuzzy set A on a universe U assigns every element x a membership degree μA(x) between 0 and 1. The symbol μ is the Greek letter “mu”; read μA(x) as “the degree to which x belongs to set A”.
| Term | Meaning | Temperature example |
|---|---|---|
| Universe U | All allowed input values | Temperatures from 0°C to 40°C |
| Element x | The particular input being considered | 32°C |
| Fuzzy set A | A category described by a membership function | Hot |
| Membership μA(x) | How strongly x fits that category | μHot(32) = 0.7 |
Membership 0 means no belonging; 1 means full belonging; 0.7 means partial belonging under the chosen definition. It does not mean a 70% chance that the measured temperature is 32°C. Nor is it a 70% chance that the sensor is correct. Fuzziness concerns the gradual meaning of “Hot”; probability concerns uncertainty about which event occurs.
| Input | Warm membership | Hot membership | Interpretation |
|---|---|---|---|
| 28 | 0.8 | 0.1 | Fits Warm strongly and Hot weakly |
| 32 | 0.3 | 0.7 | Fits Hot more than Warm |
| 35 | 0 | 1 | Fully Hot in this chosen model |
These memberships need not add to 1. Overlapping categories are allowed. A crisp set is the special case in which membership can only be 0 or 1. A crisp set can also contain infinitely many elements; “crisp” does not mean “finite”.
On U = {20, 25, 30, 35}, the Hot memberships are [0, 0.2, 0.6, 1]. Write the fuzzy set Hot and identify the fully and partly belonging inputs. [2 marks · Practice]
Ordered inputs [20, 25, 30, 35]; memberships [0, 0.2, 0.6, 1].
The set notation and interpretation.
In the common fuzzy-set notation μ/x, the slash separates membership from element; it is not division.
| x | μHot(x) |
|---|---|
| 20 | 0 |
| 25 | 0.2 |
| 30 | 0.6 |
| 35 | 1 |
35 fully belongs; 25 and 30 partly belong; 20 does not belong.
Distinguish crisp and fuzzy sets. [2 marks · Practice]
| Crisp set | Fuzzy set |
|---|---|
| Membership is 0 or 1 | Membership is any value in [0,1] |
| A sharp boundary separates members | A membership function permits gradual boundaries |
| Example: temperature ≥ 30°C | Example: degree of being Hot increases gradually |
Yaad rakho: μ ka matlab: is category mein kitna fit hota hai. Chance nahi.
"},{"id":"triangle","title":"Triangular membership: choose the side, then calculate","html":"\r\nTri(2,5,8) means membership starts at zero at x=2, rises to 1 at x=5, then falls to zero at x=8. The first three numbers are input positions, not three membership degrees.
| Where is x? | Meaning | Calculation |
|---|---|---|
| x≤2 or x≥8 | Outside, or at a zero-height endpoint | 0 |
| 2<x<5 | Part-way up the rising edge | (x−2)/(5−2) |
| x=5 | At the peak | 1 |
| 5<x<8 | Part-way down the falling edge | (8−x)/(8−5) |
For x=3: distance travelled is 3−2=1; total rising width is 5−2=3; the height is 1÷3. For x=7: distance still remaining is 8−7=1; total falling width is 8−5=3; again the height is 1÷3. Use the branch containing the input before doing any division.
Yaad rakho: a = start, b = peak, c = end. x kahan pada hai, wahi fraction decide karega.
The horizontal axis gives the input x. The vertical axis gives membership μ. To read μ at an input, go vertically up from that x to the line, then horizontally to the membership axis.
A triangle has three parameters: a is the left zero, b is the peak with membership 1, and c is the right zero. We assume a < b < c.
On the rising side, x − a is how far you have climbed across the horizontal base; b − a is the full rising width. Their ratio goes from 0 to 1. On the falling side, c − x is the remaining horizontal distance to the right zero; c − b is the full falling width. That ratio goes from 1 to 0.
A fuzzy set A has triangular membership Tri(2, 5, 8). Find membership at x = 3. [2 marks · Practice]
Left endpoint a = 2; peak b = 5; right endpoint c = 8; input x = 3.
The membership degree μA(x), a number from 0 to 1.
Locate x relative to a, b, c before choosing the branch.
2 < 3 ≤ 5, so membership increases as x moves toward the peak.
Rising side
The degree of belonging to A is 0.333333333. It is a membership degree, not a probability.
A fuzzy set A has triangular membership Tri(2, 5, 8). Find membership at x = 7. [2 marks · Practice]
Left endpoint a = 2; peak b = 5; right endpoint c = 8; input x = 7.
The membership degree μA(x), a number from 0 to 1.
Locate x relative to a, b, c before choosing the branch.
5 < 7 < 8, so membership decreases as x moves away from the peak.
Falling side
The degree of belonging to A is 0.333333333. It is a membership degree, not a probability.
A fuzzy set A has triangular membership Tri(2, 5, 8). Find membership at x = 5. [2 marks · Practice]
Left endpoint a = 2; peak b = 5; right endpoint c = 8; input x = 5.
The membership degree μA(x), a number from 0 to 1.
Locate x relative to a, b, c before choosing the branch.
2 < 5 ≤ 5, so membership increases as x moves toward the peak.
Rising side
The degree of belonging to A is 1. This is full membership at the peak.
A fuzzy set A has triangular membership Tri(2, 5, 8). Find membership at x = 9. [2 marks · Practice]
Left endpoint a = 2; peak b = 5; right endpoint c = 8; input x = 9.
The membership degree μA(x), a number from 0 to 1.
Locate x relative to a, b, c before choosing the branch.
9 is outside the open support (2, 8) or at an endpoint.
Outside the triangle
The degree of belonging to A is 0. It is a membership degree, not a probability.
Yaad rakho: Peak se left: x − a. Peak se right: c − x. Neeche usi side ki poori width.
"},{"id":"trapezoid-intuition","title":"Trapezoids, shoulders and intuition-based fuzzification","html":"\r\nTrap(a,b,c,d) has four input positions: start rising at a, reach height 1 at b, stay at height 1 until c, and fall to zero at d. For Trap(0,2,6,8), any input between 2 and 6 has full membership. The rising width is 2−0=2; the falling width is 8−6=2.
| Input | Locate it | Numerator | Denominator | Membership |
|---|---|---|---|---|
| 1 | Rising: 0<1<2 | 1−0=1 | 2−0=2 | 1/2=0.5 |
| 4 | Flat top: 2≤4≤6 | No division | No division | 1 |
| 7 | Falling: 6<7<8 | 8−7=1 | 8−6=2 | 1/2=0.5 |
| 9 | Beyond right endpoint | No division | No division | 0 |
A left shoulder can stay at height 1 for all sufficiently small inputs, then fall. A right shoulder does the reverse. If a diagram uses a vertical edge, read its stated endpoint convention; do not divide by a zero-width edge.
A trapezoid has four parameters a < b ≤ c < d: rise from a to b, stay at 1 between b and c, then fall from c to d. The flat top means a whole interval fully belongs.
For Trap(10,20,30,40), calculate μ(15). [2 marks · Practice]
a=10, b=20, c=30, d=40; x=15.
Membership at the input.
Choose the piece of the trapezoid containing x. A plateau needs no division.
x = 15 lies on the rising side.
For Trap(10,20,30,40), calculate μ(25). [2 marks · Practice]
a=10, b=20, c=30, d=40; x=25.
Membership at the input.
Choose the piece of the trapezoid containing x. A plateau needs no division.
x = 25 lies on the flat top.
For Trap(10,20,30,40), calculate μ(35). [2 marks · Practice]
a=10, b=20, c=30, d=40; x=35.
Membership at the input.
Choose the piece of the trapezoid containing x. A plateau needs no division.
x = 35 lies on the falling side.
For Trap(10,20,30,40), calculate μ(45). [2 marks · Practice]
a=10, b=20, c=30, d=40; x=45.
Membership at the input.
Choose the piece of the trapezoid containing x. A plateau needs no division.
x = 45 lies on the outside support.
A left shoulder remains fully true at the low end; a right shoulder remains fully true at the high end. A Gaussian has a smooth rounded peak. In a numerical, use the supplied graph or expression: do not invent breakpoints from a label such as “High”.
\r\nFuzzification converts a measured crisp input into its membership degrees. In the intuition method, a domain expert defines useful linguistic categories and membership functions from experience. For example, the designer chooses the Warm triangle (15,25,35); after that choice, μWarm(32) = (35−32)/(35−25) = 0.3 is a calculation, not another guess.
\r\nYaad rakho: Shape banana design choice; shape milne ke baad membership calculate karni hai.
"},{"id":"set-operations","title":"Union, intersection, complement and difference","html":"\r\nIf A assigns element u₁ a degree 0.7 and B assigns the same element degree 0.4, every operation compares these two degrees. Do not compare u₁ from A with u₂ from B. ∪ means union (“A OR B”); ∩ means intersection (“A AND B”); superscript c means complement (“NOT”).
At one element u, μA(u)=0.7 and μB(u)=0.4. Find union, intersection, complement of B and A\\B. [2 marks · Worked example]
At the same element u, A assigns membership 0.7 and B assigns membership 0.4.
The membership at u in A∪B, A∩B, Bᶜ and A minus B.
Use max for OR, min for AND, and 1 minus membership for NOT. Difference is A AND NOT B.
Use the standard max–min convention unless the question specifies another t-norm or t-conorm. “T-norm” simply names a fuzzy AND operator; here it is min. “T-conorm” names a fuzzy OR operator; here it is max.
| Operation | Meaning | Rule |
|---|---|---|
| Union A ∪ B | A OR B | Take the larger membership |
| Intersection A ∩ B | A AND B | Take the smaller membership |
| Complement Aᶜ | NOT A | 1 − membership of A |
| Difference A ∖ B | A AND NOT B | min(μA, 1 − μB) |
On U = {u1, u2, u3}, A has memberships [0.2, 0.8, 0.6] and B has memberships [0.7, 0.3, 0.5]. Find A ∪ B, A ∩ B, Bᶜ and A ∖ B using standard max–min fuzzy operations. [3 marks · Practice]
| Element | u1 | u2 | u3 |
|---|---|---|---|
| A | 0.2 | 0.8 | 0.6 |
| B | 0.7 | 0.3 | 0.5 |
Four membership lists in exactly the same element order.
Union is OR → max. Intersection is AND → min. Complement is NOT → 1 minus membership. Difference A ∖ B means A AND NOT B, so first complement B, then intersect with A.
All operations compare the two memberships of the same u. Do not compare A at u₁ with B at u₂.
| Element | A | B | Union | Intersection | Bᶜ | A ∖ B |
|---|---|---|---|---|---|---|
| u1 | 0.2 | 0.7 | max(0.2, 0.7) = 0.7 | min(0.2, 0.7) = 0.2 | 1 − 0.7 = 0.3 | min(0.2, 0.3) = 0.2 |
| u2 | 0.8 | 0.3 | max(0.8, 0.3) = 0.8 | min(0.8, 0.3) = 0.3 | 1 − 0.3 = 0.7 | min(0.8, 0.7) = 0.7 |
| u3 | 0.6 | 0.5 | max(0.6, 0.5) = 0.6 | min(0.6, 0.5) = 0.5 | 1 − 0.5 = 0.5 | min(0.6, 0.5) = 0.5 |
On U = {u1, u2, u3}, A has memberships [0, 1, 0.4] and B has memberships [1, 0, 0.4]. Find A ∪ B, A ∩ B, Bᶜ and A ∖ B using standard max–min fuzzy operations. [3 marks · Practice]
| Element | u1 | u2 | u3 |
|---|---|---|---|
| A | 0 | 1 | 0.4 |
| B | 1 | 0 | 0.4 |
Four membership lists in exactly the same element order.
Union is OR → max. Intersection is AND → min. Complement is NOT → 1 minus membership. Difference A ∖ B means A AND NOT B, so first complement B, then intersect with A.
All operations compare the two memberships of the same u. Do not compare A at u₁ with B at u₂.
| Element | A | B | Union | Intersection | Bᶜ | A ∖ B |
|---|---|---|---|---|---|---|
| u1 | 0 | 1 | max(0, 1) = 1 | min(0, 1) = 0 | 1 − 1 = 0 | min(0, 0) = 0 |
| u2 | 1 | 0 | max(1, 0) = 1 | min(1, 0) = 0 | 1 − 0 = 1 | min(1, 1) = 1 |
| u3 | 0.4 | 0.4 | max(0.4, 0.4) = 0.4 | min(0.4, 0.4) = 0.4 | 1 − 0.4 = 0.6 | min(0.4, 0.6) = 0.4 |
On U = {u1, u2, u3, u4}, A has memberships [0.9, 0.2, 0.5, 0.7] and B has memberships [0.6, 0.8, 0.1, 0.7]. Find A ∪ B, A ∩ B, Bᶜ and A ∖ B using standard max–min fuzzy operations. [3 marks · Practice]
| Element | u1 | u2 | u3 | u4 |
|---|---|---|---|---|
| A | 0.9 | 0.2 | 0.5 | 0.7 |
| B | 0.6 | 0.8 | 0.1 | 0.7 |
Four membership lists in exactly the same element order.
Union is OR → max. Intersection is AND → min. Complement is NOT → 1 minus membership. Difference A ∖ B means A AND NOT B, so first complement B, then intersect with A.
All operations compare the two memberships of the same u. Do not compare A at u₁ with B at u₂.
| Element | A | B | Union | Intersection | Bᶜ | A ∖ B |
|---|---|---|---|---|---|---|
| u1 | 0.9 | 0.6 | max(0.9, 0.6) = 0.9 | min(0.9, 0.6) = 0.6 | 1 − 0.6 = 0.4 | min(0.9, 0.4) = 0.4 |
| u2 | 0.2 | 0.8 | max(0.2, 0.8) = 0.8 | min(0.2, 0.8) = 0.2 | 1 − 0.8 = 0.2 | min(0.2, 0.2) = 0.2 |
| u3 | 0.5 | 0.1 | max(0.5, 0.1) = 0.5 | min(0.5, 0.1) = 0.1 | 1 − 0.1 = 0.9 | min(0.5, 0.9) = 0.5 |
| u4 | 0.7 | 0.7 | max(0.7, 0.7) = 0.7 | min(0.7, 0.7) = 0.7 | 1 − 0.7 = 0.3 | min(0.7, 0.3) = 0.3 |
Why not subtract A − B? With A=0.2 and B=0.7, raw subtraction gives −0.5, which is not a membership degree. Under this convention, A ∖ B = min(0.2, 0.3) = 0.2. Some alternative fuzzy algebras define other difference operators; always identify the specified convention.
Yaad rakho: Union bada; intersection chhota; complement 1 se ghata; difference = pehle B ka complement, phir min.
"},{"id":"properties","title":"Fuzzy properties, De Morgan laws and α-cuts","html":"\r\n| Read this | Meaning | Concrete example |
|---|---|---|
| Commutative | Changing order changes nothing | max(0.3,0.8)=max(0.8,0.3)=0.8 |
| Associative | Regroup the same operation | max(max(0.2,0.7),0.4)=max(0.2,max(0.7,0.4))=0.7 |
| Idempotent | Combining a membership with itself keeps it | min(0.6,0.6)=0.6 |
| Involution | Complement twice returns the original | 1−(1−0.4)=1−0.6=0.4 |
| De Morgan | NOT swaps AND and OR while complementing each operand | 1−max(a,b)=min(1−a,1−b) |
| Property | Statement for max–min operations |
|---|---|
| Commutative | A ∪ B = B ∪ A; A ∩ B = B ∩ A |
| Associative | (A ∪ B) ∪ C = A ∪ (B ∪ C); likewise for intersection |
| Idempotent | A ∪ A = A; A ∩ A = A |
| Distributive | A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C); and its dual |
| Involution | (Aᶜ)ᶜ = A |
| De Morgan | (A ∪ B)ᶜ = Aᶜ ∩ Bᶜ; (A ∩ B)ᶜ = Aᶜ ∪ Bᶜ |
For fuzzy sets, A ∪ Aᶜ need not equal the entire universe at membership 1; A ∩ Aᶜ need not be the empty set at membership 0.
At an element x, μA(x)=0.4. Calculate μ(A ∪ Aᶜ) and μ(A ∩ Aᶜ). [2 marks · Practice]
μA(x)=0.4; standard complement/max/min.
Whether excluded middle and non-contradiction hold at this element.
Complement first, then use max or min.
Verify the first De Morgan law for μA=0.3 and μB=0.8. [2 marks · Practice]
A=0.3, B=0.8 at the same element.
Both sides of (A ∪ B)ᶜ = Aᶜ ∩ Bᶜ.
Evaluate the operations independently on each side.
For U={a,b,c,d} and memberships A=[0,0.3,1,0.7], find support, core, boundary, height and the 0.7-cut. [2 marks · Worked example]
Universe U={a,b,c,d}. Memberships: A(a)=0, A(b)=0.3, A(c)=1, A(d)=0.7. Cut level α=0.7.
The support, core, boundary and 0.7-cut as sets of elements, and the height as one membership value.
Support: degree>0; core: degree=1; boundary: 0<degree<1; height: largest degree; α-cut: degree≥α.
| Element | Degree | Positive? | Exactly 1? | At least 0.7? |
|---|---|---|---|---|
| a | 0 | No | No | No |
| b | 0.3 | Yes | No | No |
| c | 1 | Yes | Yes | Yes |
| d | 0.7 | Yes | No | Yes |
Support={b,c,d}. Core={c}. Boundary={b,d}. The 0.7-cut={c,d}.
The maximum degree is 1, so the height is 1 and the set is normal.
Find the 0.5-cut of the triangular fuzzy set Tri(2,5,8). [2 marks · Worked example]
Triangle start a=2, peak b=5, end c=8; cut level α=0.5. Retain all inputs whose membership is at least 0.5.
Left and right endpoints of the retained interval.
The triangle rises past 0.5, remains above it, then falls through 0.5. Solve height=0.5 on both slopes.
The graph is at or above 0.5 for 3.5≤x≤6.5. Include the endpoints because the condition is ≥.
Support: inputs with μ > 0. Core: inputs with μ = 1. Boundary: inputs with 0 < μ < 1. Height: largest membership (more generally the supremum). A set is normal when its height is 1; it is subnormal when its height is below 1. A crossover point has membership 0.5. An α-cut contains inputs with μ ≥ α.
For Tri(2,5,8), support is (2,8), core is {5}, height is 1, and the 0.5-cut is [3.5,6.5]. Solve (x−2)/3=0.5 → x=3.5 and (8−x)/3=0.5 → x=6.5.
A one-dimensional fuzzy set is convex if every α-cut is an interval (or empty). The membership between two inputs must not dip below the smaller endpoint membership. A single triangle is convex; two separated peaks with a lower valley may not be.
Yaad rakho: Fuzzy A aur NOT A dono partial ho sakte hain. Isliye crisp waali 1 aur 0 identities blindly nahi lagani.
"},{"id":"mamdani-architecture","title":"Mamdani architecture: what every block does","html":"\r\nA controller turns measurements into an action setting. Consider temperature 32°C and occupancy 7 people. The input numbers first become membership degrees. The rules then combine those degrees. Only the final defuzzification returns one fan-speed number.
| Stage | What goes in | Operation | What comes out |
|---|---|---|---|
| Fuzzifier | 32°C and 7 people | Use the supplied membership graphs | Temperature Medium=.3, High=.7; occupancy High=.5 |
| Knowledge base | Chosen labels and their shapes | Store graphs and IF–THEN rules | A reusable definition of how the controller works |
| Inference | Medium=.3 AND High=.5 | Rule strength=min(.3,.5)=.3 | “Fan High” fires at .3 |
| Aggregation | All clipped output sets | Take the largest output degree at each speed | One combined fuzzy output shape |
| Defuzzifier | Combined shape | Apply the specified MoM, centroid or representative method | One numerical speed |
The antecedent is the IF part (“temperature is Medium AND occupancy is High”). The consequent is the THEN part (“fan speed is High”). A rule strength of 0.3 is a membership height; it does not mean the fan runs at 30%.
A fuzzy inference system (FIS) maps inputs to an output by evaluating fuzzy IF–THEN rules. A Mamdani FIS uses fuzzy sets in the rule consequents, combines their inferred output sets, and defuzzifies the result.
\r\n| Block | What it stores or calculates | Concrete meaning |
|---|---|---|
| Fuzzifier | Membership values of the measured inputs | 32°C may be Medium to degree 0.3 and High to degree 0.7 |
| Knowledge base: database | Ranges and membership functions | Medium = Tri(15,25,35) |
| Knowledge base: rule base | IF–THEN rules | IF temperature High AND occupancy High THEN fan Very High |
| Inference engine | Combines conditions and applies consequents | min for AND; clips the output triangle at the rule strength |
| Aggregator | Combines all rule outputs | At every output location, take the maximum clipped membership |
| Defuzzifier | Converts the aggregate to one crisp value | Use centroid or mean of maxima, as requested |
In “IF temperature is High AND occupancy is High THEN fan speed is Very High”, the IF part is the antecedent (condition), and the THEN part is the consequent (output category). The input variable temperature, the label High, and the numerical membership μHigh(32) are different things.
\r\nA strength such as 0.3 says how strongly a rule fires. It is on the membership scale 0–1. Fan speed might be on a 0–100% scale. Defuzzification is the step that produces a number on that output scale.
Explain the architecture of a Mamdani fuzzy inference system. [3 marks · Practice]
A Mamdani FIS contains a fuzzifier, knowledge base, inference engine and defuzzifier. The database specifies membership functions and the rule base stores IF–THEN rules. Crisp inputs are fuzzified. The inference engine calculates rule firing strengths and constructs output fuzzy sets. These sets are aggregated, then defuzzified into a crisp action. For a fan controller, temperature and occupancy are inputs and fan speed is the output.
Yaad rakho: Input ka degree → rule ka strength → output ka shape → final number.
"},{"id":"rules-clipping","title":"Rule firing, clipping and aggregation: four solved cases","html":"\r\nA rule firing at strength 0.3 cannot give its consequent a membership above 0.3. At every candidate speed z, keep the smaller of 0.3 and the original output-set membership. The speed axis is unchanged.
At speeds 40, 60 and 80, output set A has memberships [0.1,0.8,0.2] and set B has [0,0.4,1]. Rule A fires at 0.3 and rule B at 0.5. Clip and max-aggregate the sets. [2 marks · Worked example]
| Speed | Original A membership | Original B membership |
|---|---|---|
| 40 | 0.1 | 0 |
| 60 | 0.8 | 0.4 |
| 80 | 0.2 | 1 |
Rule A strength=0.3; rule B strength=0.5. Compare memberships at the same speed.
The combined membership at each speed.
Inner min limits a rule’s height; outer max keeps the stronger support at that speed.
At 40: min(.3,.1)=.1. At 60: min(.3,.8)=.3. At 80: min(.3,.2)=.2.
At 40: min(.5,0)=0. At 60: min(.5,.4)=.4. At 80: min(.5,1)=.5.
| Speed | Clipped A | Clipped B | max(A,B) |
|---|---|---|---|
| 40 | 0.1 | 0 | 0.1 |
| 60 | 0.3 | 0.4 | 0.4 |
| 80 | 0.2 | 0.5 | 0.5 |
Yaad rakho: Clip = height ki ceiling. Aggregate = har same x par sabse upar wala height.
AND/OR combine conditions within a rule. Implication uses the resulting strength to modify that rule’s output membership function. Aggregation merges different rule outputs. Here we use min-AND, max-OR, min-implication and max-aggregation.
α (“alpha”) is the rule firing strength. z names a possible output value. The subscript r labels a rule, not a multiplication.
A rule has two condition degrees 0.7 and 0.5, joined by AND. Find its firing strength. [2 marks · Practice]
a=0.7; b=0.5; connective=AND.
The strength α.
Use the smaller degree because both conditions must hold.
A rule has two condition degrees 0.3 and 0, joined by AND. Find its firing strength. [2 marks · Practice]
a=0.3; b=0; connective=AND.
The strength α.
Use the smaller degree because both conditions must hold.
A rule has two condition degrees 0.2 and 0.8, joined by OR. Find its firing strength. [2 marks · Practice]
a=0.2; b=0.8; connective=OR.
The strength α.
Use the larger degree because either condition is sufficient.
Rule 1 fires at 0.3 and its original output membership at z=60 is 0.8. Rule 2 fires at 0.6 and its original membership at z=60 is 0.2. Find the aggregate membership at z=60 using min-implication and max-aggregation. [3 marks · Practice]
Rule 1: α₁=0.3, μ₁(60)=0.8. Rule 2: α₂=0.6, μ₂(60)=0.2.
Membership at the single output location z=60 after implication and aggregation.
Clip each original degree at its rule strength; then take the larger clipped degree.
Yaad rakho: AND ke liye min; output clip karne ke liye bhi min; rules ko jodne ke liye max. Har min ka role alag hai.
"},{"id":"mom","title":"Mean of maxima: discrete peaks and clipped plateaus","html":"\r\nMaximum membership is a height. Mean of maxima averages the horizontal locations attaining that height. If heights at speeds [20,50,80] are [.2,.6,.6], the maximum height is .6 and the qualifying speeds are 50 and 80; MoM=(50+80)/2=65. Do not average [.6,.6], which would return a degree rather than a speed. A continuous flat maximum from 50 to 80 also has midpoint 65.
Defuzzification converts the aggregate membership function into one crisp output. For MoM, first find the largest membership height, then average the output locations attaining that height. Do not average the membership values themselves.
For a continuous flat maximum from a to b, the mean is its midpoint:
For the following discrete aggregate, find the mean-of-maxima output. [2 marks · Practice]
| Output z | Membership μ |
|---|---|
| 10 | 0.2 |
| 20 | 0.8 |
| 30 | 0.8 |
| 40 | 0.3 |
MoM crisp output.
Find all locations at the largest height, then average only those locations.
20, 30
For the following discrete aggregate, find the mean-of-maxima output. [2 marks · Practice]
| Output z | Membership μ |
|---|---|
| 1 | 0.1 |
| 2 | 0.4 |
| 3 | 0.9 |
| 4 | 0.2 |
MoM crisp output.
Find all locations at the largest height, then average only those locations.
3
A single triangular output Tri(50,65,80) is clipped at α=0.3. Find its crisp value by mean of maxima. [3 marks · Practice]
a=50, b=65, c=80; clip height α=0.3.
Both plateau endpoints and their mean.
Set the rising and falling membership fractions equal to 0.3. Every point between those intersections has the maximum height.
For several disjoint continuous maximum plateaus, a continuous MoM uses a length-weighted mean over all maximum intervals; simply averaging two outside endpoints is not generally valid. In tabulated questions, average the listed locations at the maximum.
Yaad rakho: Sabse upar ka level dekho; us level par x-values ka mean lo.
"},{"id":"centroid","title":"Centroid: three table sums and one area calculation","html":"\r\nImagine the area under the output graph as a thin sheet of uniform material. The centroid is its horizontal balance point. More area on one side pulls the answer toward that side. In a discrete problem, each membership is a weight. In a continuous problem, the entire area supplies the weights.
The integral sign ∫ means adding very thin strips across a continuous interval. The denominator ∫μ(z)dz is the total area. The numerator ∫zμ(z)dz is the first moment: position multiplied by area, accumulated across the shape. Their ratio has units of the output, such as fan-speed percent.
A non-overlapping aggregate consists of a rectangle of width 4 and height 0.5 centred at z=2, plus a triangle of base 4 and height 0.5 centred horizontally at z=6. Find the area centroid. [2 marks · Worked example]
Rectangle: width 4, height 0.5, horizontal centre 2. Triangle: base 4, height 0.5, horizontal centroid 6. Their areas do not overlap.
The combined horizontal centroid.
Each piece contributes area times its own centre. Use this only for a valid decomposition of the aggregate.
For overlapping clipped sets, first form their max-aggregate. Adding the areas of the original overlapping sets would count the overlap twice.
The centroid uses the whole aggregate membership distribution. For a discrete output table, multiply each output location by its membership, add the products, then divide by total membership.
The index i simply numbers table entries. Σ (“sigma”) means add all listed terms. The denominator must be nonzero.
Find the discrete centroid of the aggregate membership table. [3 marks · Practice]
| z | μ(z) |
|---|---|
| 10 | 0.2 |
| 20 | 0.6 |
| 30 | 0.4 |
One centroid value on the output scale.
This question supplies memberships at discrete output locations; use the discrete centroid, not a peak-only approximation.
| z | μ | z × μ |
|---|---|---|
| 10 | 0.2 | 2 |
| 20 | 0.6 | 12 |
| 30 | 0.4 | 12 |
Find the discrete centroid of the aggregate membership table. [3 marks · Practice]
| z | μ(z) |
|---|---|
| 0 | 0.5 |
| 5 | 1 |
| 10 | 0.5 |
One centroid value on the output scale.
This question supplies memberships at discrete output locations; use the discrete centroid, not a peak-only approximation.
| z | μ | z × μ |
|---|---|---|
| 0 | 0.5 | 0 |
| 5 | 1 | 5 |
| 10 | 0.5 | 5 |
Find the discrete centroid of the aggregate membership table. [3 marks · Practice]
| z | μ(z) |
|---|---|
| 20 | 0 |
| 40 | 0.3 |
| 60 | 0.6 |
| 80 | 0.1 |
One centroid value on the output scale.
This question supplies memberships at discrete output locations; use the discrete centroid, not a peak-only approximation.
| z | μ | z × μ |
|---|---|---|
| 20 | 0 | 0 |
| 40 | 0.3 | 12 |
| 60 | 0.6 | 36 |
| 80 | 0.1 | 8 |
The denominator is the area under the aggregate. The numerator is its first moment about zero: area weighted by horizontal location. For a shape made of non-overlapping rectangles and triangles, compute each part’s area A and its own centroid c, then use . Overlapping regions of a max-aggregate must not be counted twice.
An aggregate is a rectangle of height 0.5 from z=0 to z=4, followed by a right triangle decreasing from height 0.5 at z=4 to 0 at z=6. Find its area centroid. [3 marks · Practice]
Rectangle [0,4], height 0.5; triangle with vertices (4,0),(4,0.5),(6,0).
Centroid of the combined area.
The parts do not overlap except at their boundary. Weight each part’s horizontal centroid by its area.
Yaad rakho: Centroid = har location × uska weight, total karke total weight se divide.
"},{"id":"weighted-average","title":"Weighted average and centroid are not automatically the same","html":"\r\nThis is the same weighted-average calculation taught in Essential maths. Here a representative output value c is weighted by a rule strength α. Keep the pair together: strength .3 belonging to speed 65 contributes .3×65. The denominator is the total strength. It is not automatically the number of rules, and the representative is not automatically the area-centre of a clipped set.
Some classroom solutions use one representative (often the peak) per output label and weight it by the firing strength. That is a weighted-average or height-based calculation. It is exact for a specified weighted combination of singleton outputs, but is not generally the centroid of a max-aggregated, clipped Mamdani shape.
Always name the method. A table of memberships at actual output locations supports the discrete centroid. A list of category peaks and strengths supports the specified representative-weighted calculation. Full continuous output shapes support area centroid.
Using the weighted-average method, combine output representatives [25, 75] mph with respective strengths [0.1, 0.7]. [3 marks · Practice]
| Output representative | Strength |
|---|---|
| 25 | 0.1 |
| 75 | 0.7 |
One crisp output by the specified weighted-average method.
Each representative is multiplied by its strength; add these products and divide by the total strength. Here α is a strength and z is a representative value. These are not automatically the full area of a clipped Mamdani set.
| Strength × value | Product |
|---|---|
| 0.1 × 25 | 2.5 |
| 0.7 × 75 | 52.5 |
Using the weighted-average method, combine output representatives [40, 65, 85] % fan speed with respective strengths [0, 0.3, 0.5]. [3 marks · Practice]
| Output representative | Strength |
|---|---|
| 40 | 0 |
| 65 | 0.3 |
| 85 | 0.5 |
One crisp output by the specified weighted-average method.
Each representative is multiplied by its strength; add these products and divide by the total strength. Here α is a strength and z is a representative value. These are not automatically the full area of a clipped Mamdani set.
| Strength × value | Product |
|---|---|
| 0 × 40 | 0 |
| 0.3 × 65 | 19.5 |
| 0.5 × 85 | 42.5 |
Using the weighted-average method, combine output representatives [25, 40, 60] minutes with respective strengths [0.6, 0.4, 0.2]. [3 marks · Practice]
| Output representative | Strength |
|---|---|
| 25 | 0.6 |
| 40 | 0.4 |
| 60 | 0.2 |
One crisp output by the specified weighted-average method.
Each representative is multiplied by its strength; add these products and divide by the total strength. Here α is a strength and z is a representative value. These are not automatically the full area of a clipped Mamdani set.
| Strength × value | Product |
|---|---|
| 0.6 × 25 | 15 |
| 0.4 × 40 | 16 |
| 0.2 × 60 | 12 |
| Method | What it uses | Limitation |
|---|---|---|
| MoM | Locations at maximum membership | Ignores lower portions of the shape; can change abruptly |
| Area centroid | Entire aggregated output area | Needs an area/moment calculation or numerical integration |
| Representative weighted average | Supplied representative values and strengths | Not interchangeable with a clipped Mamdani area centroid |
No method is universally “most accurate”. The correct exam calculation is the method specified by the question. If all memberships or all weights are zero, division by zero is undefined; a controller needs an explicit fallback rule.
Yaad rakho: Formula similar dikhega, par weight kya hai? Membership, area, ya rule strength—pehle woh identify karo.
\r\nThis is a different example from the two-input fan design. With Warm=Tri(20,30,40) and Hot=Tri(30,40,50), temperature 32 gives Warm=(40−32)/10=0.8 and Hot=(32−30)/10=0.2. The slide uses output representatives 50 and 80.
Using the weighted-average method, combine output representatives [50, 80] % fan speed with respective strengths [0.8, 0.2]. [3 marks · Practice]
| Output representative | Strength |
|---|---|
| 50 | 0.8 |
| 80 | 0.2 |
One crisp output by the specified weighted-average method.
Each representative is multiplied by its strength; add these products and divide by the total strength. Here α is a strength and z is a representative value. These are not automatically the full area of a clipped Mamdani set.
| Strength × value | Product |
|---|---|
| 0.8 × 50 | 40 |
| 0.2 × 80 | 16 |
There are 3 temperature labels and 3 occupancy labels. Pair each temperature label with each occupancy label: 3×3=9 possible AND rules. At the specified input, any rule containing a zero input membership has strength min(0,other)=0 and contributes no output. The rules with nonzero strength remain; their output labels come from the supplied rule table, not from the magnitude of the strength.
This is the fan-controller design question shared in class: temperature 32°C and occupancy 7. The photos contain inconsistent graph breakpoints and arithmetic. A design question permits a stated design; below, every membership function is specified explicitly. These choices make the solution reproducible and should not be read as a uniquely recoverable answer from the photos.
Design a Mamdani fan controller for temperature 0–40°C and occupancy 0–10, with three input labels each and five fan-speed labels on 0–100%. For T=32°C and occupancy=7, use the membership functions and nine rules below. Calculate all memberships, firing strengths, and active output sets. Find fan speed by (a) mean of maxima and (b) the specified representative-weighted average. [10 marks · Class-question adaptation; full design]
| Variable | Low / Very Low | Medium / Low | High / Medium | High output | Very High output |
|---|---|---|---|---|---|
| Temperature | Tri(0,10,20) | Tri(15,25,35) | Tri(25,35,40) | — | — |
| Occupancy | Tri(0,2,4) | Tri(3,5,7) | Tri(6,8,10) | — | — |
| Fan speed | Tri(0,10,20) | Tri(10,25,40) | Tri(20,40,60) | Tri(50,65,80) | Tri(70,85,100) |
Input T=32°C; occupancy=7. Use min-AND, min-clipping and max-aggregation.
| Temperature | Occupancy | Fan output |
|---|---|---|
| Low | Low | Very Low |
| Low | Medium | Low |
| Low | High | Medium |
| Medium | Low | Low |
| Medium | Medium | Medium |
| Medium | High | High |
| High | Low | Medium |
| High | Medium | High |
| High | High | Very High |
Input memberships → nine rule strengths → clipped output → two named crisp outputs.
Use the appropriate triangle branch for membership. AND takes the smaller degree. Each output is clipped at that strength; rules with the same consequent combine by max. MoM uses only the highest plateau. The requested representative weighted average uses category peaks and strengths.
| T label / degree | O label / degree | min = strength | Output |
|---|---|---|---|
| Low / 0 | Low / 0 | 0 | Very Low |
| Low / 0 | Medium / 0 | 0 | Low |
| Low / 0 | High / 0.5 | 0 | Medium |
| Medium / 0.3 | Low / 0 | 0 | Low |
| Medium / 0.3 | Medium / 0 | 0 | Medium |
| Medium / 0.3 | High / 0.5 | 0.3 | High |
| High / 0.7 | Low / 0 | 0 | Medium |
| High / 0.7 | Medium / 0 | 0 | High |
| High / 0.7 | High / 0.5 | 0.5 | Very High |
The maximum is 0.5, belonging to Very High. Its triangle is (70,85,100).
How to adapt it in the exam: if breakpoints or output representatives differ, recompute the memberships from those data. The sequence of operations stays unchanged. If the question explicitly asks for the area centroid, integrate the max-aggregated shape; do not silently substitute the weighted average above.
Yaad rakho: Temperature aur people ka degree nikalo → 9 min → active outputs clip → jo method poocha hai woh lagao.
\r\n\r\n\r\nFor the stated Very High triangle (70,85,100) clipped at height 0.5, solve the rising edge: (z−70)/(85−70)=0.5, so z−70=0.5×15=7.5 and z=77.5. On the falling edge, (100−z)/(100−85)=0.5, so 100−z=7.5 and z=92.5. Thus the highest flat interval is [77.5,92.5]. Its midpoint is (77.5+92.5)/2=170/2=85.
Continue with the complete area-centroid calculation for this same fan design: 77.086956522%, with all overlap breakpoints and area/moment arithmetic shown.
"},{"id":"centroid-comparison","title":"The same fan controller: true area centroid versus MoM and weighted average","html":"The preceding class-fan lesson makes the input shapes and nine rules explicit because the photographs do not specify one internally consistent design. Keep that exact stated design here: temperature 32°C, occupancy 7, High output Tri(50,65,80) firing at 0.3 and Very High output Tri(70,85,100) firing at 0.5. This continuation calculates the actual area centroid of their max-aggregate. It does not reinterpret 77.5 as a centroid.
For the stated two-input class fan controller at T=32°C and occupancy=7, calculate the true continuous Mamdani area centroid of High=Tri(50,65,80) clipped at 0.3 and Very High=Tri(70,85,100) clipped at 0.5. Compare it with mean of maxima and the specified representative-weighted average. [6 marks]
Worked teaching continuation of the class-design adaptation; the mark allocation here is for practice.
All other output rules have strength zero. Use min for implication, max for aggregation and speed z on 0–100%. The two active rule strengths were calculated from input memberships: min(0.3,0.5)=0.3 and min(0.7,0.5)=0.5.
Every visible segment of the aggregate, its area and first moment, the centroid, and the differences between the three crisp answers.
Centroid balances all the area under the upper envelope. A straight strip with endpoints (a,u) and (b,v) has the following area A and first moment M; adding only disjoint strips counts overlap once.
High has a 0.3 plateau from 54.5 to 75.5. Very High has a 0.5 plateau from 77.5 to 92.5.
Very High rises through the High plateau at height 0.3. Solve its rising equation:
Before 74.5, clipped High supplies the upper envelope. After 74.5, Very High is above it. The descending High edge from 75.5 to 80 lies underneath and contributes no extra aggregate area.
Read left to right: (50,0) → (54.5,0.3) → (74.5,0.3) → (77.5,0.5) → (92.5,0.5) → (100,0). Membership is zero outside this support. Each adjacent pair is one straight strip in the table.
The first moment has units of speed multiplied by area. Dividing total moment by total area returns speed.
| a | b | u | v | Area A | First moment M |
|---|---|---|---|---|---|
| 50 | 54.5 | 0 | 0.3 | 0.675 | 35.775 |
| 54.5 | 74.5 | 0.3 | 0.3 | 6 | 387 |
| 74.5 | 77.5 | 0.3 | 0.5 | 1.2 | 91.35 |
| 77.5 | 92.5 | 0.5 | 0.5 | 7.5 | 637.5 |
| 92.5 | 100 | 0.5 | 0 | 1.875 | 178.125 |
| Method | Calculation | Answer |
|---|---|---|
| Area centroid | 1329.75 ÷ 17.25 | 77.086956522% |
| Mean of maxima | (77.5+92.5) ÷ 2 | 85% |
| Specified representative weighted average | (0.3×65+0.5×85) ÷ (0.3+0.5) | 77.5% |
MoM uses only the highest plateau. The representative method uses two chosen peaks and two rule strengths. The area centroid uses the full max-aggregate, so these are three legitimate but different calculations.
Boundary coverage of this teaching design. With the ordinary triangles stated in the class adaptation, every temperature degree is zero at T=0 or T=40, and every occupancy degree is zero at occupancy=0 or 10. Then no rule fires and an area centroid is undefined. To make a controller cover the entire declared input range, explicitly redesign outer sets as shoulders or specify an out-of-coverage rule; do not silently divide by zero. The worked T=32, occupancy=7 case has positive area and is unaffected.
Yaad rakho: Centroid mein poore shaded area ka balance chahiye. Peak × strength wala shortcut same calculation nahi hai.
"},{"id":"exam-theory","title":"Fuzzy theory: definitions, architecture, advantages and applications","html":"Explain fuzzy logic with one application. [2 marks · Practice]
Fuzzy logic represents gradual truth or category membership using values between 0 and 1. A temperature controller can use Low, Medium and High temperature sets. It evaluates rules such as IF temperature is High THEN fan speed is High, combines the rule outputs and defuzzifies them into a numerical fan speed. This permits gradual control around category boundaries.
Give the advantages and limitations of a Mamdani fuzzy controller. [3 marks · Practice]
| Advantages | Limitations |
|---|---|
| Human-readable rules express expert knowledge | Membership functions and rules require careful design |
| Can model gradual boundaries without an exact analytic plant model | Rule count can grow rapidly with more inputs and categories |
| Handles overlapping linguistic descriptions | A poorly chosen rule base can produce poor or uncovered outputs |
| Suitable for many control applications | Does not automatically learn or guarantee optimal control |
Describe the design of a fuzzy fan controller. [3 marks · Practice]
Choose temperature and occupancy as inputs and fan speed as output. Define their ranges and Low/Medium/High membership functions. Construct a rule table linking input labels to output labels. Fuzzify measured inputs, use min to obtain AND-rule strengths, clip the consequent sets and max-aggregate them. Apply the specified defuzzification method to obtain a fan speed. Test the design at normal inputs and range boundaries.
| Application | Inputs | Example rule | Output |
|---|---|---|---|
| Washing machine | Dirtiness, load | IF dirt High AND load High THEN wash time Long | Wash duration |
| Loan decision support | Income, credit score | IF income High AND credit Good THEN approval suitability High | Suitability score (not an automatic legal lending decision) |
| Fire warning | Temperature, smoke | IF temperature High OR smoke High THEN risk High | Alarm-risk score |
| Vehicle control | Distance, relative speed | IF distance Small AND closing speed High THEN braking Strong | Braking command |
Yaad rakho: Theory answer mein definition → blocks/steps → ek concrete example. Sirf headings se marks nahi milte.
"},{"id":"smooth-memberships","title":"Gaussian, singleton and bell memberships","html":"\r\nFor a Gaussian membership, c is the input at the centre and σ controls the width. Compute x−c, square it, divide by 2σ², put a minus sign in front, and apply eˣ. The maximum membership is 1 at x=c because the exponent becomes 0. Increasing σ makes the same distance from the centre count as a smaller relative distance, so the curve becomes wider. A Gaussian membership is not the same formula as a normal probability density: there is no 1/(√(2π)σ) multiplier here.
A Gaussian membership function reaches 1 at its centre c and decreases smoothly on both sides. Its width σ must be positive. This is a fuzzy membership curve, not a normalised probability density: there is no 1/(√(2π)σ) factor.
exp(u) means eᵘ. The calculator evaluates it using eˣ; calculate the entire negative exponent first. A squared distance treats equal left and right offsets equally.
For Gaussian membership with c=10 and σ=2, find μ(10). [2 marks · Practice]
x=10; centre c=10; width σ=2.
Membership degree.
Use squared distance from the centre in the Gaussian membership formula.
For Gaussian membership with c=10 and σ=2, find μ(12). [2 marks · Practice]
x=12; centre c=10; width σ=2.
Membership degree.
Use squared distance from the centre in the Gaussian membership formula.
For Gaussian membership with c=10 and σ=2, find μ(14). [2 marks · Practice]
x=14; centre c=10; width σ=2.
Membership degree.
Use squared distance from the centre in the Gaussian membership formula.
A singleton fuzzy set has membership 1 at exactly one specified point and 0 elsewhere. For a singleton at c=50: μ(50)=1, μ(49)=0 and μ(51)=0. A singleton consequent is represented by one numerical output; this is why a specified weighted average of singleton outputs is a legitimate calculation.
\r\nc is the centre, a>0 controls width and b>0 controls the slope. Absolute value |u| is the nonnegative magnitude: |−2|=2. At x=c membership is 1; at x=c±a membership is 1/2.
For a=2, b=1 and c=10, find generalised-bell membership at x=14. [2 marks · Practice]
a=2, b=1, c=10; x=14.
μ(14).
Subtract the centre, divide by a, take the absolute value, raise to 2b, then use the reciprocal.
Yaad rakho: Gaussian MF ka centre 1 hota hai. GMM density ke normalising factor ko yahan mat lagao.
"}]},{"id":"fuzzy-relations","title":"Fuzzy relations and matrices","unit":"Unit 2","scope":"Longer Unit 2 numerical practice, also useful for TEE.","sections":[{"id":"cartesian","title":"Cartesian product: every cell explained","html":"\r\nLet A₁=.6 and A₂=.2 label two rows. Let B₁=.3, B₂=.8 and B₃=.5 label three columns. Cell R₂₃ belongs to row 2 and column 3, so use A₂=.2 and B₃=.5. Their min is .2. You do not multiply 2×3; those subscripts only locate the cell.
There are two choices of row and three of column, so fill six cells in a 2×3 rectangle. Keep the row and column labels when reporting a relation.
A fuzzy set gives a membership to one element. A fuzzy relation gives a degree to an ordered pair. A matrix is a convenient table: rows name the first element and columns name the second.
For the standard min-based Cartesian product R=A×B, put the A memberships down the rows and the B memberships across the columns. Each cell is the smaller of its row and column memberships.
Find the fuzzy Cartesian product A×B for memberships A=[0.6, 0.2] and B=[0.3, 0.8, 0.5]. [3 marks · Practice]
Ordered A memberships [0.6, 0.2]; B memberships [0.3, 0.8, 0.5].
A 2×3 relation matrix.
Every row corresponds to an A element and every column to a B element; apply min in every cell.
2 row elements × 3 column elements.
| A ↓ / B → | 0.3 | 0.8 | 0.5 |
|---|---|---|---|
| 0.6 | min(0.6,0.3) = 0.3 | min(0.6,0.8) = 0.6 | min(0.6,0.5) = 0.5 |
| 0.2 | min(0.2,0.3) = 0.2 | min(0.2,0.8) = 0.2 | min(0.2,0.5) = 0.2 |
Find the fuzzy Cartesian product A×B for memberships A=[1, 0.4] and B=[0.7, 0]. [3 marks · Practice]
Ordered A memberships [1, 0.4]; B memberships [0.7, 0].
A 2×2 relation matrix.
Every row corresponds to an A element and every column to a B element; apply min in every cell.
2 row elements × 2 column elements.
| A ↓ / B → | 0.7 | 0 |
|---|---|---|
| 1 | min(1,0.7) = 0.7 | min(1,0) = 0 |
| 0.4 | min(0.4,0.7) = 0.4 | min(0.4,0) = 0 |
Find the fuzzy Cartesian product A×B for memberships A=[0.2, 0.9, 0.5] and B=[0.6, 0.1]. [3 marks · Practice]
Ordered A memberships [0.2, 0.9, 0.5]; B memberships [0.6, 0.1].
A 3×2 relation matrix.
Every row corresponds to an A element and every column to a B element; apply min in every cell.
3 row elements × 2 column elements.
| A ↓ / B → | 0.6 | 0.1 |
|---|---|---|
| 0.2 | min(0.2,0.6) = 0.2 | min(0.2,0.1) = 0.1 |
| 0.9 | min(0.9,0.6) = 0.6 | min(0.9,0.1) = 0.1 |
| 0.5 | min(0.5,0.6) = 0.5 | min(0.5,0.1) = 0.1 |
Yaad rakho: Row ka degree aur column ka degree—dono ka min us cell mein.
"},{"id":"composition","title":"Max–min composition: three complete matrix examples","html":"\r\nFor T₁₁, start at u₁ and end at w₁. If the middle domain has v₁ and v₂, there are two candidate paths. Their strengths come from matching row 1 of R with column 1 of S.
| Path | First link from R | Second link from S | Path strength |
|---|---|---|---|
| u₁ → v₁ → w₁ | .2 | .5 | min(.2,.5)=.2 |
| u₁ → v₂ → w₁ | .8 | .7 | min(.8,.7)=.7 |
Choose the stronger path: max(.2,.7)=.7, so T₁₁=.7. The middle label j runs through v₁ and v₂. The labels i and k stay fixed at 1 while this one cell is calculated. Then move to the next result cell.
Yaad rakho: Ek path mein dono links chahiye, isliye min. Alag paths mein best chalega, isliye max.
Suppose R relates U to V and S relates V to W. To find how strongly u connects to w through each middle v, take min of the two links. Then choose the strongest such path using max.
We call this matrix operation R composed with S here. Some books write the symbolic order in reverse. State the formula and row/column labels, especially when a question says S∘R. Composition requires the intermediate domains and their ordering to match; having two 3×3 matrices alone does not prove their labels match.
Calculate the max–min composition using . [3 marks · Practice]
Every entry of T.
Pair each row entry with the corresponding column entry; min for a two-link path, max across the possible middle elements. This is not ordinary matrix multiplication.
.Calculate the max–min composition using . [3 marks · Practice]
Every entry of T.
Pair each row entry with the corresponding column entry; min for a two-link path, max across the possible middle elements. This is not ordinary matrix multiplication.
.Calculate the max–min composition using . [3 marks · Practice]
Every entry of T.
Pair each row entry with the corresponding column entry; min for a two-link path, max across the possible middle elements. This is not ordinary matrix multiplication.
.Yaad rakho: Har path mein min, phir saare paths mein max. Multiply-and-add nahi.
"}]},{"id":"ga-core","title":"3. Genetic algorithms","unit":"Unit 3 · GA","scope":"Candidate → fitness → encoding → selection → crossover → mutation → two complete generations.","sections":[{"id":"meaning","title":"GA meaning, biological idea and every basic term","html":"\r\n| Object | Concrete value | Do not confuse it with |
|---|---|---|
| Individual number | Row 2 | The candidate’s value x |
| Candidate value x | 11 | Its chromosome 1011 |
| Chromosome | 1011 | Its fitness |
| Fitness when maximising x² | 11×11=121 | Its row number or bit string |
In this guide, “gene 3” counts the third bit from the left starting at 1. Binary place values were built from the right starting with weight 1. These are two labelling conventions, not contradictory place-value rules. Read the end specified in an exam.
A genetic algorithm (GA) is a population-based optimisation method inspired by biological evolution. It maintains candidate solutions, evaluates their fitness, selects parents, generates children through crossover and mutation, and repeats the process.
In biology, genes carry inherited information. In a GA, an encoding plays the role of inherited information. The analogy motivates the operators; it does not mean that a computer chromosome is literally DNA.
| Term | Meaning | Four-bit example |
|---|---|---|
| Individual | One candidate solution | x=11 |
| Chromosome | The complete encoded candidate | 1011 |
| Gene | One component of an encoding | The third bit |
| Allele | The value at a gene | 0 or 1 for binary encoding |
| Population | The collection of current individuals | [3,5,10,12] |
| Fitness | A score used to favour useful candidates | For max x², fitness of 11 is 121 |
| Generation | One cycle producing a new population | Select → cross → mutate → replace |
| Search space | All candidate solutions permitted | 4-bit integers: 0–15 |
| Elitism | Preserving best individuals unchanged | Copy the best into the next population |
An objective describes what the problem wants. Fitness describes what the selection mechanism favours. For maximising a nonnegative objective, they may be the same. For minimising cost, directly selecting in proportion to cost would favour bad candidates; use a specified transformation or a comparison-based selection method.
For a maximisation GA with objective f(x)=x², find the fitness of x=3. [2 marks · Practice]
x=3; fitness equals the nonnegative objective x².
Fitness, not the encoded chromosome.
Square x to evaluate its objective; binary encoding is a separate representation step.
For a maximisation GA with objective f(x)=x², find the fitness of x=7. [2 marks · Practice]
x=7; fitness equals the nonnegative objective x².
Fitness, not the encoded chromosome.
Square x to evaluate its objective; binary encoding is a separate representation step.
For a maximisation GA with objective f(x)=x², find the fitness of x=12. [2 marks · Practice]
x=12; fitness equals the nonnegative objective x².
Fitness, not the encoded chromosome.
Square x to evaluate its objective; binary encoding is a separate representation step.
Yaad rakho: Chromosome answer ka code; fitness us code wale answer ka score.
"},{"id":"encoding","title":"Encoding: binary, real values and permutations","html":"\r\nFor four bits, b₁ is the leftmost bit and receives weight 8; b₄ is the rightmost and receives weight 1. In general n−i gives the power belonging to bit i. For 1011, the four terms are 1×8, 0×4, 1×2, 1×1. Add 8+0+2+1=11.
For real-range mapping, d/(2ⁿ−1) is the fraction of the way through the possible codes. Multiplying by U−L travels that fraction of the physical range. Adding L shifts the result to start at the lower limit.
| 3-bit code | Decoded d | Fraction d/7 | Mapped to [−2,5] |
|---|---|---|---|
| 000 | 0 | 0/7=0 | −2+0×7=−2 |
| 011 | 3 | 3/7 | −2+(3/7)×7=1 |
| 111 | 7 | 7/7=1 | −2+1×7=5 |
Binary encoding uses bit strings; real-valued encoding stores real numbers directly; permutation encoding stores an order, such as a TSP city tour. An operator valid for bits may create invalid city tours, so encoding and operators must fit the problem.
| Representation | Candidate | Appropriate use |
|---|---|---|
| Binary | 1011 | An unsigned integer or a coded real value |
| Real-valued | [0.2, 1.7] | Continuous parameters |
| Permutation | [A,C,B,D] | Visit order in TSP |
b₁ is the leftmost bit, not a coefficient you have to learn. In four bits the weights are 8,4,2,1. The maximum is .
Decode the 4-bit chromosome 0011. [2 marks · Practice]
Bits 0011; place values 8,4,2,1.
Integer x.
Multiply each bit by its matching place value, then add.
| Bit | Place value | Contribution |
|---|---|---|
| 0 | 8 | 0 |
| 0 | 4 | 0 |
| 1 | 2 | 2 |
| 1 | 1 | 1 |
Decode the 4-bit chromosome 1010. [2 marks · Practice]
Bits 1010; place values 8,4,2,1.
Integer x.
Multiply each bit by its matching place value, then add.
| Bit | Place value | Contribution |
|---|---|---|
| 1 | 8 | 8 |
| 0 | 4 | 0 |
| 1 | 2 | 2 |
| 0 | 1 | 0 |
Decode the 4-bit chromosome 1111. [2 marks · Practice]
Bits 1111; place values 8,4,2,1.
Integer x.
Multiply each bit by its matching place value, then add.
| Bit | Place value | Contribution |
|---|---|---|
| 1 | 8 | 8 |
| 1 | 4 | 4 |
| 1 | 2 | 2 |
| 1 | 1 | 1 |
If the question maps an n-bit integer d to a real interval [L,U], first decode d, then use:
Map the 3-bit string 101 to the real interval [−2,5] using linear binary decoding. [3 marks · Practice]
n=3, L=−2, U=5; string 101.
Real x.
The smallest bit code maps to L and the largest to U; the remaining codes are equally spaced.
Yaad rakho: Pehle bits se integer. Real range mapping explicitly diya ho tabhi next formula lagao.
"},{"id":"selection","title":"Roulette selection, cumulative intervals and alternatives","html":"\r\nBefore the three complete sums below, read the running-total example in Percentages, probabilities and cumulative intervals. For weights [1,2,3,4], candidate C’s own probability is 0.3, but its interval is [0.3,0.6). A draw of 0.55 selects C because it lies inside that interval; it is not required to be less than C’s individual probability.
Roulette-wheel selection gives each candidate a probability proportional to a nonnegative fitness weight. A high weight occupies a larger interval on [0,1]. A supplied random number selects the interval containing it.
Fᵢ is the fitness weight of individual i. The sum must be positive. Probabilities add to 1. Cumulative probability is the running sum and supplies the interval boundaries.
Fitness weights of successive individuals are [1, 2, 3, 4]. Select parents using random numbers [0.05, 0.25, 0.55, 0.9]. [3 marks · Practice]
Fitness weights [1, 2, 3, 4]; draws [0.05, 0.25, 0.55, 0.9].
Parent row number for each draw.
Divide by the fitness sum, form cumulative intervals, then locate each random draw. Use half-open intervals [lower,upper); the final upper endpoint 1 is included.
| Individual | Fitness | Probability | Cumulative |
|---|---|---|---|
| 1 | 1 | 0.1 | 0.1 |
| 2 | 2 | 0.2 | 0.3 |
| 3 | 3 | 0.3 | 0.6 |
| 4 | 4 | 0.4 | 1 |
| Draw | Selected individual |
|---|---|
| 0.05 | 1 |
| 0.25 | 2 |
| 0.55 | 3 |
| 0.9 | 4 |
Fitness weights of successive individuals are [4, 4, 2]. Select parents using random numbers [0.2, 0.6, 0.85]. [3 marks · Practice]
Fitness weights [4, 4, 2]; draws [0.2, 0.6, 0.85].
Parent row number for each draw.
Divide by the fitness sum, form cumulative intervals, then locate each random draw. Use half-open intervals [lower,upper); the final upper endpoint 1 is included.
| Individual | Fitness | Probability | Cumulative |
|---|---|---|---|
| 1 | 4 | 0.4 | 0.4 |
| 2 | 4 | 0.4 | 0.8 |
| 3 | 2 | 0.2 | 1 |
| Draw | Selected individual |
|---|---|
| 0.2 | 1 |
| 0.6 | 2 |
| 0.85 | 3 |
Fitness weights of successive individuals are [0, 3, 1]. Select parents using random numbers [0.2, 0.8]. [3 marks · Practice]
Fitness weights [0, 3, 1]; draws [0.2, 0.8].
Parent row number for each draw.
Divide by the fitness sum, form cumulative intervals, then locate each random draw. Use half-open intervals [lower,upper); the final upper endpoint 1 is included.
| Individual | Fitness | Probability | Cumulative |
|---|---|---|---|
| 1 | 0 | 0 | 0 |
| 2 | 3 | 0.75 | 0.75 |
| 3 | 1 | 0.25 | 1 |
| Draw | Selected individual |
|---|---|
| 0.2 | 2 |
| 0.8 | 3 |
| Method | Procedure | Why it is used |
|---|---|---|
| Tournament | Sample a group and take its best candidate | Only objective comparisons are needed |
| Rank | Sort candidates; assign weights based on rank | Reduces sensitivity to a single huge raw fitness |
| Elitist survival | Preserve top individuals unchanged into the next population | Protects good solutions from destructive variation |
Selection is not automatically “replace the weakest with the fittest”. That is one possible classroom reproduction rule. If a problem gives selected parents, a rank rule, tournament contestants or roulette draws, follow that exact instruction.
Yaad rakho: Probability nikalo → cumulative banao → random number ka interval dekho.
"},{"id":"operators","title":"Crossover and mutation: four complete worked variants","html":"\r\nThe prefix is the part before the cut; the suffix is the part after it. “10 joined with 00 gives 1000” means writing strings next to each other. It is not the arithmetic addition 10+00.
| Operation | Parents / instruction | Construction | Children |
|---|---|---|---|
| One-point crossover | 10|11 and 01|00 | First prefix + second tail; second prefix + first tail | 1000 and 0111 |
| Two-point crossover | 1|01|10 and 0|10|01 | Exchange only the middle blocks | 11010 and 00101 |
| Uniform crossover | A=1010, B=0101, mask=1100 | Child 1 takes A where mask=1 and B where mask=0; child 2 reverses this | 1001 and 0110 |
| Mutation | 1000; flip position 3 from left | The third bit is 0; replace it by 1; keep every other bit | 1010 |
These crossover operators are alternatives. Apply the operator stated in the question; do not apply all three in one generation. Mutation is a separate step after crossover in the worked exam workflow.
In single-point crossover, cut both parents after the specified bit. Child 1 keeps parent 1’s prefix and takes parent 2’s suffix; child 2 does the reverse. A cut after bit 2 in a four-bit string looks like 10|11.
Cross parents 1011 and 0100 after bit 2. Then flip bit 4 of both children, counting from the left. Decode the final children. [3 marks · Practice]
Parents 1011, 0100; cut after bit 2; mutation bit 4.
Children after crossover, after mutation and their decimal values.
Exchange only suffixes. Mutation means 0→1 or 1→0; it does not mean set the bit to 1.
10 | 11 and 01 | 00
10 + 00 → 1000
01 + 11 → 0111
| Crossover child | Bit before | Bit after | Mutated child |
|---|---|---|---|
| 1000 | 0 | 1 | 1001 |
| 0111 | 1 | 0 | 0110 |
| String | Place-value sum | x |
|---|---|---|
| 1001 | 8 + 0 + 0 + 1 | 9 |
| 0110 | 0 + 4 + 2 + 0 | 6 |
Cross parents 0011 and 1100 after bit 1. Then flip bit 2 of both children, counting from the left. Decode the final children. [3 marks · Practice]
Parents 0011, 1100; cut after bit 1; mutation bit 2.
Children after crossover, after mutation and their decimal values.
Exchange only suffixes. Mutation means 0→1 or 1→0; it does not mean set the bit to 1.
0 | 011 and 1 | 100
0 + 100 → 0100
1 + 011 → 1011
| Crossover child | Bit before | Bit after | Mutated child |
|---|---|---|---|
| 0100 | 1 | 0 | 0000 |
| 1011 | 0 | 1 | 1111 |
| String | Place-value sum | x |
|---|---|---|
| 0000 | 0 + 0 + 0 + 0 | 0 |
| 1111 | 8 + 4 + 2 + 1 | 15 |
Cross parents 1110 and 0001 after bit 3. Then flip bit 1 of both children, counting from the left. Decode the final children. [3 marks · Practice]
Parents 1110, 0001; cut after bit 3; mutation bit 1.
Children after crossover, after mutation and their decimal values.
Exchange only suffixes. Mutation means 0→1 or 1→0; it does not mean set the bit to 1.
111 | 0 and 000 | 1
111 + 1 → 1111
000 + 0 → 0000
| Crossover child | Bit before | Bit after | Mutated child |
|---|---|---|---|
| 1111 | 1 | 0 | 0111 |
| 0000 | 0 | 1 | 1000 |
| String | Place-value sum | x |
|---|---|---|
| 0111 | 0 + 4 + 2 + 1 | 7 |
| 1000 | 8 + 0 + 0 + 0 | 8 |
Cross parents 1010 and 1010 after bit 2. Then flip bit 4 of both children, counting from the left. Decode the final children. [3 marks · Practice]
Parents 1010, 1010; cut after bit 2; mutation bit 4.
Children after crossover, after mutation and their decimal values.
Exchange only suffixes. Mutation means 0→1 or 1→0; it does not mean set the bit to 1.
10 | 10 and 10 | 10
10 + 10 → 1010
10 + 10 → 1010
| Crossover child | Bit before | Bit after | Mutated child |
|---|---|---|---|
| 1010 | 0 | 1 | 1011 |
| 1010 | 0 | 1 | 1011 |
| String | Place-value sum | x |
|---|---|---|
| 1011 | 8 + 0 + 2 + 1 | 11 |
| 1011 | 8 + 0 + 2 + 1 | 11 |
Two-point crossover exchanges the segment between two cuts. Uniform crossover uses a given mask to choose, at each position, which parent contributes a gene. Mutation may flip a bit, perturb a real value, or swap two positions in a permutation. These are different operators; apply the one named in the question.
Yaad rakho: Crossover = tail exchange. Mutation = specified bit flip. Flip ka matlab 1 bhi 0 banega.
"},{"id":"algorithm","title":"GA architecture, two-generation flow, constraints and theory","html":"\r\n| Read this | Meaning | Concrete example |
|---|---|---|
| Feasible | Obeys every stated constraint | x≤10: x=9 qualifies; x=12 does not |
| Penalty | A deduction for violating a constraint | For maximisation, subtract 100×excess |
| Elitism | Explicitly preserve a best old individual | Only apply if the question specifies it |
| Replacement | Choose what population the next generation uses | Our full-replacement examples use the new children |
| Stopping condition | A rule ending the repeated process | Stop after two complete generations |
A simple generational GA replaces the population in generations using a chosen encoding, selection rule and variation operators. A more general GA may use steady-state replacement, different encodings, adaptive operators or explicit elitism. Always state the choices used in your worked answer.
\r\nInitial population → first generation’s selection/crossover/mutation → new population → second generation’s selection/crossover/mutation → final population. Stopping after the first children and labelling them iteration 2 is not two full generations.
\r\nStop after a fixed generation budget, after reaching a specified acceptable objective, or after little improvement for a stated number of generations. A finite run does not prove the global optimum.
\r\nA constraint restricts allowed solutions, such as 0≤x≤15 or “visit every city once”. Common approaches are valid-by-design encoding, repairing an invalid child, rejecting it, or using a penalty. A penalty must reduce attractiveness: subtract it in maximisation or add it to a minimised cost. Do not silently remove the original constraint because you used more bits.
A maximisation problem has objective f(x)=x² and constraint x≤10. The given penalised fitness is F(x)=x²−100 max(0,x−10). Compare x=9 and x=12. [3 marks · Practice]
Penalty coefficient 100; bound 10; candidates 9 and 12.
Penalised fitness of both candidates.
Only excess above the bound receives a penalty.
For this maximisation fitness, 81 > −56, so x=9 is preferred. Raw objective alone would misleadingly favour 12.
| Advantages | Limitations |
|---|---|
| Derivative-free; supports many encodings | Many objective evaluations may be costly |
| Population explores several candidates | Premature convergence is possible |
| Crossover and mutation can create new candidates | Parameter and encoding choices affect results |
| Applicable to discontinuous or combinatorial searches | No guaranteed global optimum; variation can worsen fitness |
Yaad rakho: Do generation = do baar poora cycle. Sirf children banana aur decode karna do alag iterations nahi.
"},{"id":"two-generations","title":"Two generations with given selection, cut and mutation","html":"\r\nMaximise over unsigned 4-bit integers using the supplied GA settings. Perform two complete generations and compare the population fitness. [3 marks · Practice; complete two-generation solution]
Initial ordered population: [3, 5, 10, 12]. Use 4 bits, crossover after bit 2, pair mating-pool slots 1–2 and 3–4. Flip bit 4 of children [1, 2, 3, 4] in every generation. Replace the whole population with the four children; no elitism.
| Generation | Selection instruction |
|---|---|
| 1 | Current individual row numbers [3, 4, 4, 3] |
| 2 | Current individual row numbers [2, 3, 2, 3] |
Both generations: evaluation, selection, crossover, mutation, decoding and comparison.
Encoding represents x; it does not change the objective. Crossover exchanges tails. Mutation flips specified bits. Fitness must be recomputed after mutation. Use exactly the supplied selection rule; replacing the weakest by the strongest is not a universal GA step.
| Individual | x | Binary | Objective calculation |
|---|---|---|---|
| 1 | 3 | 0011 | 3 × 3 = 9 |
| 2 | 5 | 0101 | 5 × 5 = 25 |
| 3 | 10 | 1010 | 10 × 10 = 100 |
| 4 | 12 | 1100 | 12 × 12 = 144 |
The question directly supplies individual numbers [3, 4, 4, 3]. These refer to the current generation’s row numbers, not their x values.
| Mating-pool slot | Chosen individual | x | Chromosome |
|---|---|---|---|
| 1 | 3 | 10 | 1010 |
| 2 | 4 | 12 | 1100 |
| 3 | 4 | 12 | 1100 |
| 4 | 3 | 10 | 1010 |
Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1.
| Pool slots paired | Parent A | Parent B | Child A | Child B |
|---|---|---|---|---|
| 1 & 2 | 10 | 10 | 11 | 00 | 1000 | 1110 |
| 3 & 4 | 11 | 00 | 10 | 10 | 1110 | 1000 |
| Child | Before mutation | Instruction | After mutation |
|---|---|---|---|
| 1 | 1000 | Flip bit 4 | 1001 |
| 2 | 1110 | Flip bit 4 | 1111 |
| 3 | 1110 | Flip bit 4 | 1111 |
| 4 | 1000 | Flip bit 4 | 1001 |
| Chromosome | Decode using place values | Objective |
|---|---|---|
| 1001 | 8 + 0 + 0 + 1 = 9 | 9² = 81 |
| 1111 | 8 + 4 + 2 + 1 = 15 | 15² = 225 |
| 1111 | 8 + 4 + 2 + 1 = 15 | 15² = 225 |
| 1001 | 8 + 0 + 0 + 1 = 9 | 9² = 81 |
Add the objective scores in each population, then divide by its number of individuals.
Best objective: 144 → 225. Higher is better. The next population is [9, 15, 15, 9]. No elitism is imposed, so the best previous individual is not automatically protected.
| Individual | x | Binary | Objective calculation |
|---|---|---|---|
| 1 | 9 | 1001 | 9 × 9 = 81 |
| 2 | 15 | 1111 | 15 × 15 = 225 |
| 3 | 15 | 1111 | 15 × 15 = 225 |
| 4 | 9 | 1001 | 9 × 9 = 81 |
The question directly supplies individual numbers [2, 3, 2, 3]. These refer to the current generation’s row numbers, not their x values.
| Mating-pool slot | Chosen individual | x | Chromosome |
|---|---|---|---|
| 1 | 2 | 15 | 1111 |
| 2 | 3 | 15 | 1111 |
| 3 | 2 | 15 | 1111 |
| 4 | 3 | 15 | 1111 |
Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1.
| Pool slots paired | Parent A | Parent B | Child A | Child B |
|---|---|---|---|---|
| 1 & 2 | 11 | 11 | 11 | 11 | 1111 | 1111 |
| 3 & 4 | 11 | 11 | 11 | 11 | 1111 | 1111 |
| Child | Before mutation | Instruction | After mutation |
|---|---|---|---|
| 1 | 1111 | Flip bit 4 | 1110 |
| 2 | 1111 | Flip bit 4 | 1110 |
| 3 | 1111 | Flip bit 4 | 1110 |
| 4 | 1111 | Flip bit 4 | 1110 |
| Chromosome | Decode using place values | Objective |
|---|---|---|
| 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 |
| 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 |
| 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 |
| 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 |
Add the objective scores in each population, then divide by its number of individuals.
Best objective: 225 → 196. Higher is better. The next population is [14, 14, 14, 14]. No elitism is imposed, so the best previous individual is not automatically protected.
Maximise over unsigned 4-bit integers using the supplied GA settings. Perform two complete generations and compare the population fitness. [3 marks · Practice; complete two-generation solution]
Initial ordered population: [2, 5, 9, 12]. Use 4 bits, crossover after bit 2, pair mating-pool slots 1–2 and 3–4. Flip bit 4 of children [2, 4] in every generation. Replace the whole population with the four children; no elitism.
| Generation | Selection instruction |
|---|---|
| 1 | Random draws [0.1, 0.4, 0.65, 0.9] |
| 2 | Random draws [0.2, 0.4, 0.6, 0.95] |
Both generations: evaluation, selection, crossover, mutation, decoding and comparison.
Encoding represents x; it does not change the objective. Crossover exchanges tails. Mutation flips specified bits. Fitness must be recomputed after mutation. Use exactly the supplied selection rule; replacing the weakest by the strongest is not a universal GA step.
| Individual | x | Binary | Objective calculation |
|---|---|---|---|
| 1 | 2 | 0010 | 2 × (15 − 2) + 1 = 27 |
| 2 | 5 | 0101 | 5 × (15 − 5) + 1 = 51 |
| 3 | 9 | 1001 | 9 × (15 − 9) + 1 = 55 |
| 4 | 12 | 1100 | 12 × (15 − 12) + 1 = 37 |
Fitness weights are nonnegative. The nonnegative objective itself is the fitness in this maximisation example.
| Individual | Probability: weight ÷ total | Cumulative: old boundary + probability |
|---|---|---|
| 1 | 27 ÷ 170 ≈ 0.158823529 | 0 + 0.158823529 ≈ 0.158823529 |
| 2 | 51 ÷ 170 ≈ 0.3 | 0.158823529 + 0.3 ≈ 0.458823529 |
| 3 | 55 ÷ 170 ≈ 0.323529412 | 0.458823529 + 0.323529412 ≈ 0.782352941 |
| 4 | 37 ÷ 170 ≈ 0.217647059 | 0.782352941 + 0.217647059 ≈ 1 |
| Random draw | Interval containing it | Selected individual | Selected x |
|---|---|---|---|
| 0.1 | 0 ≤ 0.1 < 0.158823529 | 1 | 2 |
| 0.4 | 0.158823529 ≤ 0.4 < 0.458823529 | 2 | 5 |
| 0.65 | 0.458823529 ≤ 0.65 < 0.782352941 | 3 | 9 |
| 0.9 | 0.782352941 ≤ 0.9 < 1 | 4 | 12 |
Intervals are left-closed and right-open, except that a draw of exactly 1 selects the last interval. No draw here is on a boundary. Draws select parents with replacement: selecting a parent once does not remove it. Calculations use unrounded weights; the table displays rounded numbers.
Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1.
| Pool slots paired | Parent A | Parent B | Child A | Child B |
|---|---|---|---|---|
| 1 & 2 | 00 | 10 | 01 | 01 | 0001 | 0110 |
| 3 & 4 | 10 | 01 | 11 | 00 | 1000 | 1101 |
| Child | Before mutation | Instruction | After mutation |
|---|---|---|---|
| 1 | 0001 | No mutation | 0001 |
| 2 | 0110 | Flip bit 4 | 0111 |
| 3 | 1000 | No mutation | 1000 |
| 4 | 1101 | Flip bit 4 | 1100 |
| Chromosome | Decode using place values | Objective |
|---|---|---|
| 0001 | 0 + 0 + 0 + 1 = 1 | 1(15−1)+1 = 15 |
| 0111 | 0 + 4 + 2 + 1 = 7 | 7(15−7)+1 = 57 |
| 1000 | 8 + 0 + 0 + 0 = 8 | 8(15−8)+1 = 57 |
| 1100 | 8 + 4 + 0 + 0 = 12 | 12(15−12)+1 = 37 |
Add the objective scores in each population, then divide by its number of individuals.
Best objective: 55 → 57. Higher is better. The next population is [1, 7, 8, 12]. No elitism is imposed, so the best previous individual is not automatically protected.
| Individual | x | Binary | Objective calculation |
|---|---|---|---|
| 1 | 1 | 0001 | 1 × (15 − 1) + 1 = 15 |
| 2 | 7 | 0111 | 7 × (15 − 7) + 1 = 57 |
| 3 | 8 | 1000 | 8 × (15 − 8) + 1 = 57 |
| 4 | 12 | 1100 | 12 × (15 − 12) + 1 = 37 |
Fitness weights are nonnegative. The nonnegative objective itself is the fitness in this maximisation example.
| Individual | Probability: weight ÷ total | Cumulative: old boundary + probability |
|---|---|---|
| 1 | 15 ÷ 166 ≈ 0.090361446 | 0 + 0.090361446 ≈ 0.090361446 |
| 2 | 57 ÷ 166 ≈ 0.343373494 | 0.090361446 + 0.343373494 ≈ 0.43373494 |
| 3 | 57 ÷ 166 ≈ 0.343373494 | 0.43373494 + 0.343373494 ≈ 0.777108434 |
| 4 | 37 ÷ 166 ≈ 0.222891566 | 0.777108434 + 0.222891566 ≈ 1 |
| Random draw | Interval containing it | Selected individual | Selected x |
|---|---|---|---|
| 0.2 | 0.090361446 ≤ 0.2 < 0.43373494 | 2 | 7 |
| 0.4 | 0.090361446 ≤ 0.4 < 0.43373494 | 2 | 7 |
| 0.6 | 0.43373494 ≤ 0.6 < 0.777108434 | 3 | 8 |
| 0.95 | 0.777108434 ≤ 0.95 < 1 | 4 | 12 |
Intervals are left-closed and right-open, except that a draw of exactly 1 selects the last interval. No draw here is on a boundary. Draws select parents with replacement: selecting a parent once does not remove it. Calculations use unrounded weights; the table displays rounded numbers.
Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1.
| Pool slots paired | Parent A | Parent B | Child A | Child B |
|---|---|---|---|---|
| 1 & 2 | 01 | 11 | 01 | 11 | 0111 | 0111 |
| 3 & 4 | 10 | 00 | 11 | 00 | 1000 | 1100 |
| Child | Before mutation | Instruction | After mutation |
|---|---|---|---|
| 1 | 0111 | No mutation | 0111 |
| 2 | 0111 | Flip bit 4 | 0110 |
| 3 | 1000 | No mutation | 1000 |
| 4 | 1100 | Flip bit 4 | 1101 |
| Chromosome | Decode using place values | Objective |
|---|---|---|
| 0111 | 0 + 4 + 2 + 1 = 7 | 7(15−7)+1 = 57 |
| 0110 | 0 + 4 + 2 + 0 = 6 | 6(15−6)+1 = 55 |
| 1000 | 8 + 0 + 0 + 0 = 8 | 8(15−8)+1 = 57 |
| 1101 | 8 + 4 + 0 + 1 = 13 | 13(15−13)+1 = 27 |
Add the objective scores in each population, then divide by its number of individuals.
Best objective: 57 → 57. Higher is better. The next population is [7, 6, 8, 13]. No elitism is imposed, so the best previous individual is not automatically protected.
Roulette favours a larger weight. A minimisation objective instead favours a smaller cost. The stated transformation F=1/(1+cost) reverses their order for nonnegative costs: cost 0 → weight 1/1=1; cost 3 → 1/4=.25; cost 24 → 1/25=.04. The +1 prevents division by zero. This is one specified transformation, not a universal rule for every minimisation problem.
Minimise over unsigned 4-bit integers using the supplied GA settings. Perform two complete generations and compare the population fitness. [3 marks · Practice; complete two-generation solution]
Initial ordered population: [2, 4, 8, 12]. Use 4 bits, crossover after bit 2, pair mating-pool slots 1–2 and 3–4. Flip bit 4 of children [2, 4] in every generation. Replace the whole population with the four children; no elitism.
| Generation | Selection instruction |
|---|---|
| 1 | Random draws [0.1, 0.4, 0.65, 0.9] |
| 2 | Random draws [0.2, 0.4, 0.6, 0.95] |
Both generations: evaluation, selection, crossover, mutation, decoding and comparison.
Encoding represents x; it does not change the objective. Crossover exchanges tails. Mutation flips specified bits. Fitness must be recomputed after mutation. Roulette uses Fᵢ/ΣF. For minimisation here, use Fᵢ=1/(1+f(xᵢ)); do not directly favour large costs.
| Individual | x | Binary | Objective calculation |
|---|---|---|---|
| 1 | 2 | 0010 | (2 − 7)² = 25 |
| 2 | 4 | 0100 | (4 − 7)² = 9 |
| 3 | 8 | 1000 | (8 − 7)² = 1 |
| 4 | 12 | 1100 | (12 − 7)² = 25 |
Fitness weights are nonnegative. For this minimisation example use F=1/(1+objective), so a smaller cost receives a larger weight.
| Individual | Cost | Fitness calculation |
|---|---|---|
| 1 | 25 | 1 / (1 + 25) = 1 / 26 ≈ 0.038461538 |
| 2 | 9 | 1 / (1 + 9) = 1 / 10 ≈ 0.1 |
| 3 | 1 | 1 / (1 + 1) = 1 / 2 ≈ 0.5 |
| 4 | 25 | 1 / (1 + 25) = 1 / 26 ≈ 0.038461538 |
| Individual | Probability: weight ÷ total | Cumulative: old boundary + probability |
|---|---|---|
| 1 | 0.038461538 ÷ 0.676923077 ≈ 0.056818182 | 0 + 0.056818182 ≈ 0.056818182 |
| 2 | 0.1 ÷ 0.676923077 ≈ 0.147727273 | 0.056818182 + 0.147727273 ≈ 0.204545455 |
| 3 | 0.5 ÷ 0.676923077 ≈ 0.738636364 | 0.204545455 + 0.738636364 ≈ 0.943181818 |
| 4 | 0.038461538 ÷ 0.676923077 ≈ 0.056818182 | 0.943181818 + 0.056818182 ≈ 1 |
| Random draw | Interval containing it | Selected individual | Selected x |
|---|---|---|---|
| 0.1 | 0.056818182 ≤ 0.1 < 0.204545455 | 2 | 4 |
| 0.4 | 0.204545455 ≤ 0.4 < 0.943181818 | 3 | 8 |
| 0.65 | 0.204545455 ≤ 0.65 < 0.943181818 | 3 | 8 |
| 0.9 | 0.204545455 ≤ 0.9 < 0.943181818 | 3 | 8 |
Intervals are left-closed and right-open, except that a draw of exactly 1 selects the last interval. No draw here is on a boundary. Draws select parents with replacement: selecting a parent once does not remove it. Calculations use unrounded weights; the table displays rounded numbers.
Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1.
| Pool slots paired | Parent A | Parent B | Child A | Child B |
|---|---|---|---|---|
| 1 & 2 | 01 | 00 | 10 | 00 | 0100 | 1000 |
| 3 & 4 | 10 | 00 | 10 | 00 | 1000 | 1000 |
| Child | Before mutation | Instruction | After mutation |
|---|---|---|---|
| 1 | 0100 | No mutation | 0100 |
| 2 | 1000 | Flip bit 4 | 1001 |
| 3 | 1000 | No mutation | 1000 |
| 4 | 1000 | Flip bit 4 | 1001 |
| Chromosome | Decode using place values | Objective |
|---|---|---|
| 0100 | 0 + 4 + 0 + 0 = 4 | (4−7)² = 9 |
| 1001 | 8 + 0 + 0 + 1 = 9 | (9−7)² = 4 |
| 1000 | 8 + 0 + 0 + 0 = 8 | (8−7)² = 1 |
| 1001 | 8 + 0 + 0 + 1 = 9 | (9−7)² = 4 |
Add the objective scores in each population, then divide by its number of individuals.
Best objective: 1 → 1. Lower is better. The next population is [4, 9, 8, 9]. No elitism is imposed, so the best previous individual is not automatically protected.
| Individual | x | Binary | Objective calculation |
|---|---|---|---|
| 1 | 4 | 0100 | (4 − 7)² = 9 |
| 2 | 9 | 1001 | (9 − 7)² = 4 |
| 3 | 8 | 1000 | (8 − 7)² = 1 |
| 4 | 9 | 1001 | (9 − 7)² = 4 |
Fitness weights are nonnegative. For this minimisation example use F=1/(1+objective), so a smaller cost receives a larger weight.
| Individual | Cost | Fitness calculation |
|---|---|---|
| 1 | 9 | 1 / (1 + 9) = 1 / 10 ≈ 0.1 |
| 2 | 4 | 1 / (1 + 4) = 1 / 5 ≈ 0.2 |
| 3 | 1 | 1 / (1 + 1) = 1 / 2 ≈ 0.5 |
| 4 | 4 | 1 / (1 + 4) = 1 / 5 ≈ 0.2 |
| Individual | Probability: weight ÷ total | Cumulative: old boundary + probability |
|---|---|---|
| 1 | 0.1 ÷ 1 ≈ 0.1 | 0 + 0.1 ≈ 0.1 |
| 2 | 0.2 ÷ 1 ≈ 0.2 | 0.1 + 0.2 ≈ 0.3 |
| 3 | 0.5 ÷ 1 ≈ 0.5 | 0.3 + 0.5 ≈ 0.8 |
| 4 | 0.2 ÷ 1 ≈ 0.2 | 0.8 + 0.2 ≈ 1 |
| Random draw | Interval containing it | Selected individual | Selected x |
|---|---|---|---|
| 0.2 | 0.1 ≤ 0.2 < 0.3 | 2 | 9 |
| 0.4 | 0.3 ≤ 0.4 < 0.8 | 3 | 8 |
| 0.6 | 0.3 ≤ 0.6 < 0.8 | 3 | 8 |
| 0.95 | 0.8 ≤ 0.95 < 1 | 4 | 9 |
Intervals are left-closed and right-open, except that a draw of exactly 1 selects the last interval. No draw here is on a boundary. Draws select parents with replacement: selecting a parent once does not remove it. Calculations use unrounded weights; the table displays rounded numbers.
Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1.
| Pool slots paired | Parent A | Parent B | Child A | Child B |
|---|---|---|---|---|
| 1 & 2 | 10 | 01 | 10 | 00 | 1000 | 1001 |
| 3 & 4 | 10 | 00 | 10 | 01 | 1001 | 1000 |
| Child | Before mutation | Instruction | After mutation |
|---|---|---|---|
| 1 | 1000 | No mutation | 1000 |
| 2 | 1001 | Flip bit 4 | 1000 |
| 3 | 1001 | No mutation | 1001 |
| 4 | 1000 | Flip bit 4 | 1001 |
| Chromosome | Decode using place values | Objective |
|---|---|---|
| 1000 | 8 + 0 + 0 + 0 = 8 | (8−7)² = 1 |
| 1000 | 8 + 0 + 0 + 0 = 8 | (8−7)² = 1 |
| 1001 | 8 + 0 + 0 + 1 = 9 | (9−7)² = 4 |
| 1001 | 8 + 0 + 0 + 1 = 9 | (9−7)² = 4 |
Add the objective scores in each population, then divide by its number of individuals.
Best objective: 1 → 1. Lower is better. The next population is [8, 8, 9, 9]. No elitism is imposed, so the best previous individual is not automatically protected.
A city is a location. An edge connects two cities and has a travel distance. A tour visits each city exactly once and returns to its start. A permutation is an ordered arrangement without omissions or repetitions: [A,B,D,C] is valid for four cities; [A,B,B,C] is not.
A distance matrix uses row=departure city and column=arrival city. Entry dAB=5 means the A→B leg costs 5. In a symmetric problem dBA=5 too. A diagonal 0 means no distance from a city to itself, not a free trip to another city. For A→B→D→C→A, add dAB+dBD+dDC+dCA.
Factorial counts arrangements: 3!=3×2×1=6. With one of four cities fixed as the start, the other three can be ordered in six ways. In a symmetric TSP reverse-direction tours have equal length, although they are different written orders.
In one line: encode a tour as a permutation of cities, use fitness = 1/(tour length), and use crossover and mutation operators that always produce valid tours.
\r\n\r\nTravelling salesman problem (TSP): given n cities and the distance between every pair, find the shortest route that visits each city exactly once and returns to the start. For a symmetric TSP with a fixed starting city, reversal-equivalent tours counted once, and n ≥ 3, the number of distinct tours is ![\"(n-1)!/2\"](\"math/eab927d7255210e1.svg\")
(12 for 5 cities, about ![\"6\\times10^{16}\"](\"math/1746ec49ffd70b58.svg\")
for 20 cities), so exhaustive search quickly becomes impossible.
| GA part | For TSP |
|---|---|
| Encoding | Permutation: C A E B D means C → A → E → B → D → back to C. |
| Fitness | ![\"F=1/L\"](\"math/7887e5a1e9ffa26d.svg\") , where ![\"L=\\sum](\"math/9c269c2e6ae2c088.svg\") includes the return edge. Shorter tour = higher fitness. |
| Selection | Roulette wheel on F, or tournament. |
| Crossover | Order crossover (OX), partially mapped crossover (PMX) or cycle crossover (CX). |
| Mutation | Swap two cities, or invert a segment (inversion), with small probability. |
This variant fills empty positions from left to right; standard cyclic OX instead starts scanning and filling after the second cut. Do not mix the two conventions. Example: P1 = 2 4 [1 8 3] 5 7 6, P2 = 7 5 6 8 3 4 2 1. Copy 1 8 3; P2 order without them is 7 5 6 4 2 → child = 7 5 1 8 3 6 4 2. (One GA slide prints 4 2 1 8 3 7 5 6, which uses a different fill order; state your rule in the exam.)
\r\n\r\nQuestion. Five cities with distances AB = 2, AC = 9, AD = 10, AE = 7, BC = 6, BD = 4, BE = 3, CD = 8, CE = 5, DE = 1. Parents are ABECD and ADBEC. (a) Find their lengths and fitness. (b) Apply the left-to-right order-preserving crossover variant described above with the segment at positions 2–3. (c) Apply swap mutation to positions 3 and 4 of the child. Comment. [practice · 10 marks]
All ten symmetric distances in the question; parents ABECD and ADBEC. Use the explicitly stated left-to-right order-preserving crossover at positions 2–3, then swap positions 3 and 4.
Parent lengths/fitness, the crossover child, and the mutated child.
Add all edges including the return to the start. A shorter positive length gives a larger reciprocal fitness.
![\"L=\\sum_{i=1}^{n-1}d(c_i,c_{i+1})+d(c_n,c_1),\\quad](\"math/7283c4b77917085c.svg\")
![\"L(ABECD)=2+3+5+8+10=28,\\quad](\"math/5403cdb93ef730d5.svg\")
![\"L(ADBEC)=10+4+3+5+9=31,\\quad](\"math/aede143ef375acd9.svg\")
Copy positions 2–3 from P1: _ B E _ _. P2's order without B, E: A, D, C → fill positions 1, 4, 5.
\r\n![\"Order](\"fig/ga_ox.8fd21cbc51.svg\")
![\"\\text{Child}=ABEDC,\\quad](\"math/ffc7b04e07bd4328.svg\")
![\"ABEDC\\to](\"math/421251db4f0614de.svg\")
PMX (partially mapped crossover): swap a middle segment and use the position mapping to repair duplicates. With the slide parents and segment 3–5 the mapping is 1↔6, giving children 7 5 1 8 3 4 2 6 and 2 4 6 8 3 5 7 1.
\r\nCycle crossover (CX): find cycles of positions between the parents and take alternate cycles from each parent; every city keeps a position it had in one of the parents.
\r\n\r\nIn one line: ants leave pheromone as they walk; shorter paths get refreshed more often, so their scent stays stronger, so more ants choose them: positive feedback can favour shorter paths.
\r\n\r\nAnt colony optimization (ACO) is a population-based, probabilistic metaheuristic, introduced by Marco Dorigo in 1992, that finds good paths through a graph by imitating how ants communicate with pheromone trails. It is especially suited to routing problems such as the travelling salesman problem (TSP).
\r\n\r\n![\"Why](\"fig/aco_bridge.3fc4d22f6d.svg\")
This indirect communication through changes in the environment is called stigmergy: no ant tells another where to go; the trail does.
\r\n\r\n| Real ants | Artificial ants in ACO |
|---|---|
| Walk from nest to food | Build a solution by moving from node to node in a graph |
| Drop pheromone on the path used | Increase the pheromone value ![\"\\tau_{ij}\"](\"math/bdc505e26b5c671c.svg\") on every edge used |
| Pheromone evaporates | All shrink by the evaporation rate ![\"\\rho\"](\"math/210f47ee7d9c0bf6.svg\") each iteration |
| Prefer stronger scent | Choose the next node with a probability based on pheromone and distance |
| Colony reinforces frequently used paths | The algorithm searches for low-cost tours; it may stagnate on a suboptimal tour |
| TSP | ACO |
|---|---|
| City | Node |
| Distance between two cities | Edge weight ![\"d_{ij}\"](\"math/bb784b3ff229be34.svg\") |
| Round trip visiting every city once | One ant's path |
| Tour length | Sum of the edges used, including the edge back to the start |
From city A an ant may choose B or C. Distances are AB=2 and AC=4. Pheromones are both 1 and α=β=1. Find the next-city probabilities. [2 marks · Worked example]
Current city A; allowed next cities B and C. Distances dAB=2 and dAC=4. Pheromones τAB=τAC=1, and exponents α=β=1. Visited cities are excluded.
P(A→B) and P(A→C).
![\"\\eta_{ij}=1/d_{ij}\\qquad](\"math/d98041105cc02612.svg\")
Shorter distance gives a larger reciprocal. Normalise the desirability weights to probabilities.
![\"\\eta_{AB}=1/2=0.5\\qquad\\eta_{AC}=1/4=0.25\"](\"math/d9c5253b54fcf92e.svg\")
![\"w_B=1\\times0.5=0.5\\qquad](\"math/7b0589337b7254cb.svg\")
![\"0.5+0.25=0.75\"](\"math/16a735913edad702.svg\")
![\"P_{AB}=0.5/0.75=2/3\"](\"math/371fe53440644f73.svg\")
![\"P_{AC}=0.25/0.75=1/3\"](\"math/b90f87b145809204.svg\")
Evaporation ρ=.5 means retain 1−.5=.5 of the old pheromone. A completed tour of length 20 deposits Q/20 on each edge it used; with Q=1 this is .05. An unused edge receives no deposit from that ant. Use complete tour length, not the length of the individual edge, in this denominator.
In one line: an ant picks the next city with probability ∝ pheromone × (1/distance); after all ants finish, pheromone evaporates and each ant adds Q/(its tour length) to every edge it used.
\r\n\r\nAn ant ![\"k\"](\"math/1dc6f9b1a0213681.svg\")
at city ![\"i\"](\"math/3c3a2b02c199adb2.svg\")
moves to an unvisited city ![\"j\"](\"math/7a5268d94266eaaf.svg\")
with probability
![\"P^{k}_{ij}=\\frac{\\tau_{ij}^{\\,\\alpha}\\,\\eta_{ij}^{\\,\\beta}}{\\displaystyle\\sum_{l\\in\\text{allowed}}\\tau_{il}^{\\,\\alpha}\\,\\eta_{il}^{\\,\\beta}},\\qquad](\"math/b2a12d0ebbd2c2db.svg\")
: pheromone on edge –; ![\"\\eta_{ij}\"](\"math/54ca51b8bfad2b71.svg\")
: visibility (heuristic), the inverse of distance.![\"\\alpha\"](\"math/ca2ff9ceb2ebf0ae.svg\")
: weight of pheromone; ![\"\\beta\"](\"math/fc31c14dd22861fd.svg\")
: weight of distance. In every class example ![\"\\alpha=\\beta=1\"](\"math/8fe45a81f0de6694.svg\")
.With all ![\"\\tau=1\"](\"math/5c811732c72735df.svg\")
and , the probabilities are simply proportional to ![\"1/d\"](\"math/cb175dc9c9dea9bf.svg\")
.
![\"r\"](\"math/b758343af751f858.svg\")
.![\"\\tau_{ij}\\leftarrow(1-\\rho)\\,\\tau_{ij}+\\sum_{k=1}^{m}\\Delta\\tau^{k}_{ij},\\qquad](\"math/7cdb88481f227e3c.svg\")
: evaporation rate (fraction lost). Class examples use ![\"\\rho=0.5\"](\"math/d7320dc626d3d883.svg\")
. ![\"Q\"](\"math/88c438d7bc523a5f.svg\")
: deposit constant, usually 1.![\"L_k\"](\"math/836949dff185b3ba.svg\")
: length of ant 's complete tour including the return edge. Shorter tours deposit more. and ![\"\\tau_{ji}\"](\"math/62cdaada6c1142ff.svg\")
together (the matrix stays symmetric).![\"\\tau_0\"](\"math/10a4957527d9bc35.svg\")
, maximum iterations. and record the best tour.![\"Q/L_k\"](\"math/0fe8d82420c3adda.svg\")
.| Parameter | Meaning | Typical value |
|---|---|---|
| m | Number of ants | number of cities (or 20) |
| α | Pheromone importance | 1 |
| β | Distance (visibility) importance | 1 (often 2–5 in practice) |
| ρ | Evaporation rate | 0.5 in class numericals (0.05–0.5 in practice) |
| Q | Deposit constant | 1 (100 in one slide example) |
| Initial pheromone | 1 in exam questions |
Question. For the four-city travelling salesman problem, the distance and pheromone matrices (A, B, C, D) are given. Apply Ant Colony Optimization. Calculate the path travelled by 2 ants to cover all cities starting from city A. Random numbers for Ant 1 are 0.3, 0.8 and for Ant 2 are 0.5, 0.7. Also update the pheromone matrix after 1 iteration. [10 marks · Final Nov 2025 Q7b]
\r\n![\"D=\\begin{array}{c|cccc}&A&B&C&D\\\\\\hline](\"math/ff22cd24b937efcc.svg\")
Assume , , ![\"Q=1\"](\"math/bd5a6aec418f8f8f.svg\")
(class convention; not stated in the question).
Tour of each ant, tour lengths, updated pheromone matrix.
![\"Final](\"fig/aco_final_graph.ac3cd4aeac.svg\")
![\"P_{ij}=\\frac{\\tau_{ij}\\,\\eta_{ij}}{\\sum_{l\\in\\text{allowed}}\\tau_{il}\\,\\eta_{il}},\\qquad\\eta_{ij}=\\frac1{d_{ij}}\"](\"math/4a0efb39e84f4c4e.svg\")
![\"\\tau_{ij}^{new}=(1-\\rho)\\tau_{ij}+\\sum_{k:\\,ij\\in\\mathrm{tour}_k}\\frac{Q}{L_k}\"](\"math/b8a3f5425a6d79b2.svg\")
The sum includes only ants whose completed tour contains edge ij. An unused edge receives no deposit but still evaporates. Here distances are symmetric, so a traversed edge updates both symmetric matrix cells. Both ants construct their tours using the old pheromones; update after both finish, including their return to A.
\r\n![\"\\tau\\eta:\\;A\\to](\"math/b5d4297d164fecda.svg\")
| Move | P | Cumulative | Range |
|---|---|---|---|
| A→B | 0.25 | 0.25 | 0–0.25 |
| A→C | 0.50 | 0.75 | 0.25–0.75 |
| A→D | 0.25 | 1.00 | 0.75–1 |
Ant 1: r = 0.3 → C. Ant 2: r = 0.5 → C.
![\"C\\to](\"math/52a7763a8065747e.svg\")
Cumulative in city order: B 0–0.4, D 0.4–1. Ant 1: r = 0.8 → D. Ant 2: r = 0.7 → D.
Only B is left, then return to A.
\r\n![\"\\text{Ant](\"math/d48c911f3133b5dc.svg\")
![\"Both](\"fig/aco_final_tour.1dbfdb8b98.svg\")
Each ant deposits ![\"Q/L=1/14=0.0714\"](\"math/799732b628eacf45.svg\")
on each edge of its tour; two ants used the same four edges (AC, CD, DB, BA), so each gets ![\"2/14=0.1429\"](\"math/2b49c5f9ee4dd7a9.svg\")
.
![\"\\text{used](\"math/beecdfda1143f6eb.svg\")
![\"\\tau_{\\text{new}}=\\begin{array}{c|cccc}&A&B&C&D\\\\\\hline](\"math/25cf4d02bee20e2a.svg\")
Official marking: path of Ant 1 (3) + path of Ant 2 (3) + pheromone matrix (4).
\r\nQuestion. For the four-city travelling salesman problem the distance matrix and pheromone matrix are given (all pheromones 1). Apply Ant Colony Optimization. Calculate the path travelled by 2 ants starting from city A; random numbers for Ant 1 are 0.3, 0.8 and for Ant 2 are 0.5, 0.7. Also update the pheromone matrix after 1 iteration. [10 marks · Re-exam Jan 2026 Q6a]
\r\n![\"D=\\begin{array}{c|cccc}&A&B&C&D\\\\\\hline](\"math/21fbc23ee8a38afa.svg\")
Assume ![\"\\alpha=\\beta=1,\\;\\rho=0.5,\\;Q=1\"](\"math/215be7d8bc049557.svg\")
.
Both tours, their lengths and the pheromone matrix after one iteration.
![\"P_{ij}=\\frac{\\tau_{ij}^{\\alpha}(1/d_{ij})^\\beta}{\\sum_{k\\in](\"math/5b888e9e5a3c73eb.svg\")
![\"\\tau_{ij}^{new}=(1-\\rho)\\tau_{ij}+\\sum_{k:\\,ij\\in](\"math/d5297204b4481c8e.svg\")
Use unvisited cities in alphabetical order for cumulative selection. Every ant completes its tour using old pheromones; then evaporate and deposit on each undirected edge, including the return edge.
![\"\\eta:\\;B=1,\\;C=0.5,\\;D=0.3333;\\;\\text{sum}=1.8333\\Rightarrow](\"math/d7a22769c0bad254.svg\")
Cumulative: B 0–0.5455, C 0.5455–0.8182, D 0.8182–1. Ant 1 (0.3) → B; Ant 2 (0.5) → B.
![\"\\eta:\\;C=0.25,\\;D=0.2;\\;\\text{sum}=0.45\\Rightarrow](\"math/9d1964928f02cbff.svg\")
Cumulative: C 0–0.5556, D 0.5556–1. Ant 1 (0.8) → D; Ant 2 (0.7) → D.
![\"A\\to](\"math/1409076465c0fee4.svg\")
![\"Re-exam](\"fig/aco_reexam_graph.02195c6ee9.svg\")
Used edges AB, BD, DC, CA get ![\"0.5(1)+2\\times\\frac1{14}=0.6429\"](\"math/a677db1cbc02a07c.svg\")
; unused edges AD and BC become ![\"0.5\"](\"math/97dede1883211cd7.svg\")
.
![\"1+0.1428=1.1428\"](\"math/7856892795a9fcf9.svg\")
) and its matrix is inconsistent: it increases the unused edge A–D, leaves the used return edge C–A at 1, and gives B–D and D–B different values. If your examiner follows the no-evaporation idea, the consistent version is 1.1429 on AB, BD, CD, AC and 1 on AD, BC. The paper does not specify ρ. State the convention you use; the resulting matrix depends on it. Grading depends on the examiner’s intended convention.Why do this: in the papers both ants happened to take the same tour. In the class example they take different tours, which shows how the deposit adds up when two lengths differ.
\r\nQuestion. Distances: AB = 5, AC = 6, AD = 7, BC = 2, BD = 9, CD = 3; all pheromones 1; . Two ants start at A. Ant 1 uses random numbers 0.45 then 0.5; Ant 2 uses 0.83 then 0.5. Find both tours and the updated pheromone matrix. [lecture example · 10 marks]
Symmetric distances AB=5, AC=6, AD=7, BC=2, BD=9, CD=3. Initial off-diagonal pheromone=1. α=β=1, ρ=0.5, Q=1. Ant draws: [0.45,0.5] and [0.83,0.5].
Two complete tours and every updated edge pheromone.
Prefer high pheromone and small distance, normalise over unvisited cities, then evaporate and deposit once per iteration.
![\"P_{ij}=\\frac{\\tau_{ij}^{\\alpha}(1/d_{ij})^{\\beta}}{\\sum_{k\\in](\"math/c9ac4c197173e498.svg\")
![\"\\tau_{ij}^{new}=(1-\\rho)\\tau_{ij}^{old}+\\sum_{ants\\](\"math/97bee3443d137b73.svg\")
![\"Lecture](\"fig/aco_lecture_graph.629672bf56.svg\")
![\"\\eta=\\tfrac15,\\tfrac16,\\tfrac17=0.2,\\;0.1667,\\;0.1429;\\;\\text{sum}=0.5095\\Rightarrow](\"math/e47560808b5cfa0e.svg\")
Cumulative: B 0–0.3925, C 0.3925–0.7196, D 0.7196–1. Ant 1 (0.45) → C. Ant 2 (0.83) → D.
![\"\\eta=\\tfrac12,\\tfrac13\\Rightarrow](\"math/1b6803f7bb0eede8.svg\")
r = 0.5 ≤ 0.6 → B; then D; then back to A.
\r\n![\"\\text{Ant](\"math/f2d71691d61a5d31.svg\")
![\"\\eta=\\tfrac19,\\tfrac13\\Rightarrow](\"math/b66f5122ccaaac24.svg\")
Cumulative B 0–0.25, C 0.25–1; r = 0.5 → C; then B; then A.
\r\n![\"\\text{Ant](\"math/161908e94cfc42c7.svg\")
| Edge | Used by | New τ |
|---|---|---|
| AD, BC | both ants | 0.5 + 1/24 + 1/17 = 0.6005 |
| AC, BD | Ant 1 only | 0.5 + 1/24 = 0.5417 |
| AB, CD | Ant 2 only | 0.5 + 1/17 = 0.5588 |
Ant 1 deposits 1/24≈.041666667; ant 2 deposits 1/17≈.058823529. Evaporation leaves .5 on every edge. Their shared edges AD and BC become .600490196. Ant-1-only edges AC and BD become .541666667; ant-2-only edges AB and CD become .558823529.
![\"\\tau^{new}\\approx\\begin{bmatrix}0&.558823529&.541666667&.600490196\\\\.558823529&0&.600490196&.541666667\\\\.541666667&.600490196&0&.558823529\\\\.600490196&.541666667&.558823529&0\\end{bmatrix}\"](\"math/9cb40cf52889b49c.svg\")
Row and column order is A,B,C,D. The diagonal 0 marks excluded self-edges. Off-diagonal entries mirror each other because this problem is undirected and symmetric.
Asked as: “List the applications of Ant Colony Optimization and explain any one application in detail.” [5 marks · Final 2025 Q1b; key: listing 3 + explanation 2]
\r\nAnt Colony Optimization is a probabilistic, population-based metaheuristic (Dorigo, 1992) in which artificial ants build solutions on a graph and communicate through pheromone trails. In the update rule used here, evaporation reduces all stored trails; deposits then reinforce edges of selected tours according to solution quality. Evaporation does not occur only on poor paths.
\r\n![\"\\propto\\tau_{ij}\\,(1/d_{ij})\"](\"math/7d200da3be6a78d4.svg\")
, skipping customers already served or that would exceed the vehicle capacity; when the vehicle is full it returns to the depot.![\"Q/L\"](\"math/afe42919eb73e68a.svg\")
on the routes of good solutions, so short connections are reinforced.| Advantages | Limitations |
|---|---|
| Naturally suited to graph and routing problems | Several parameters to tune (α, β, ρ, number of ants) |
| Can share useful route information and adapt search to changing graphs | Convergence time is uncertain; may stagnate on a sub-optimal path |
| Positive feedback + parallel ants | Theoretical analysis is difficult |
Swarm intelligence is collective problem-solving behaviour that emerges from interactions among many relatively simple agents and their environment. It is inspired by systems such as bird flocks, fish schools and ant colonies.
\r\nParticle swarm optimization (PSO) is a population-based search method inspired by social movement. Each particle represents a candidate solution. It moves through the search space using a velocity influenced by its own best previous position and the best position known to the swarm.
\r\n| Term | Meaning | Example |
|---|---|---|
| Particle | One candidate solution | Particle A |
| Position x | The candidate’s coordinates | xA = 3 |
| Objective f(x) | The problem’s score for that candidate | f(3) = 3² = 9 |
| Velocity v | Proposed change of position | vA = −1 means move left by 1 |
| pbest | Best position that this particle has visited | A once visited x=1 with score 1 |
| gbest | Best position found by any particle | Among all personal bests, x=0.5 scores 0.25 |
For minimising x², suppose a particle visits x=3, then x=1, then x=2. Its current position is 2, but its personal best stays 1 because 1² is less than 2². The particle remembers a better past result even after a bad move.
A particle visits x=3, x=1, x=2 in that order while minimising f(x)=x². Find pbest after each visit. [2 marks · Practice]
Positions in chronological order: 3 → 1 → 2.
The best-so-far position and score.
Keep the location with the lowest objective encountered so far.
No history: pbest=3, best score=9.
Replace pbest by 1.
Keep pbest=1 even though the current position is 2.
Define swarm intelligence and briefly explain one nature-inspired algorithm. [2 marks · Practice]
Swarm intelligence is collective problem-solving that emerges from interactions among simple agents. Particle swarm optimization is inspired by flocking behaviour. Each particle is a candidate solution with a position and velocity. It remembers its own best position and uses the swarm’s best position to guide movement. Repeated evaluation and movement search for a low-cost or high-fitness solution.
Yaad rakho: Position abhi kahan ho; pbest apna best past; gbest sabka best past.
"},{"id":"equation","title":"Read the PSO equation and work three individual updates","html":"\r\nNew velocity = retained old velocity + pull toward personal best + pull toward the group best. After finding that velocity, add it to the old position. You are calculating a movement, then applying it.
Example: current x=4 and personal best=1 gives the difference 1−4=−3. A coefficient of .5 gives personal pull .5×(−3)=−1.5. The negative sign says “move left”; it does not mean a negative probability. If current x=−2 and best=1, the difference is 1−(−2)=1+2=3, which pulls right.
In vi,jt, i chooses the particle, j chooses its coordinate, and t chooses the recorded iteration. For particle B, coordinate 2, iteration 0, look at row B and the second velocity entry in the initial table. The letters identify a stored number; they are not extra multiplication factors.
| Symbol | Meaning |
|---|---|
| i | Which particle |
| j | Which coordinate of that particle |
| t / t+1 | Before / after this iteration; these are labels, not powers |
| w | Inertia weight: how much old velocity is kept |
| c₁r₁ | Combined strength of the pull to personal best |
| c₂r₂ | Combined strength of the pull to global best |
| r₁,r₂ | Random factors, normally in [0,1]; use the values supplied in an exam |
A best position to the left gives a negative difference and pulls left; a best position to the right gives a positive difference and pulls right. The multiplication sets the strength of each pull.
For one coordinate, x=4, v=0, pbest=4, gbest=2. With w=0.5, c₁=c₂=1 and r₁=r₂=0.5, find the next velocity and position. [3 marks · Practice]
x=4, v=0, pbest=4, gbest=2; w=.5; c₁r₁=.5; c₂r₂=.5.
New velocity followed by new position.
Use the three-term velocity equation, then add the new velocity to x.
For one coordinate, x=4, v=-2, pbest=3, gbest=1. With w=0.5, c₁=c₂=1 and r₁=r₂=0.5, find the next velocity and position. [3 marks · Practice]
x=4, v=-2, pbest=3, gbest=1; w=.5; c₁r₁=.5; c₂r₂=.5.
New velocity followed by new position.
Use the three-term velocity equation, then add the new velocity to x.
For one coordinate, x=-2, v=1, pbest=-1, gbest=1. With w=0.5, c₁=c₂=1 and r₁=r₂=0.5, find the next velocity and position. [3 marks · Practice]
x=-2, v=1, pbest=-1, gbest=1; w=.5; c₁r₁=.5; c₂r₂=.5.
New velocity followed by new position.
Use the three-term velocity equation, then add the new velocity to x.
First-iteration shortcut: if pbest was just initialised to x, the personal difference is zero. This follows from the initialisation; it is not a permanent property of PSO. A gbest particle can still move because its inertia term can be nonzero.
Yaad rakho: Old speed ka hissa + apne best ki pull + group best ki pull; phir new speed ko position mein add.
"},{"id":"two-particles","title":"Two particles, two iterations: start with the simplest minimisation","html":"\r\nA synchronous iteration is a class photograph of the old swarm: every particle uses that same snapshot of gbest. First calculate all new positions, then compare all new scores and update memories. If particle A discovers a better position halfway through the calculations, B still uses the previous snapshot during this iteration. B can use the improved gbest in the next iteration.
Work down the page: each particle is fully updated, then all scores and memories are compared. Keep at least six digits internally for longer decimal examples; the final table displays rounded values.
Minimise using 2 particles. Use the initial data below and perform exactly two PSO iterations, reporting position, velocity, pbest and gbest after each iteration. [3 marks · Practice; expanded teaching solution]
| Particle | Initial position x⁰ | Initial velocity v⁰ |
|---|---|---|
| A | (-2) | (0) |
| B | (4) | (0) |
w=0.5, c₁=1, c₂=1, r₁=0.5, r₂=0.5. Initialise each pbest at its initial position. Use these random coefficients unchanged in both iterations. No bounds or velocity clamp are imposed in this example. Updates are synchronous: every particle uses the same previous-iteration gbest.
The complete swarm state after iterations 1 and 2, and the best location found.
PSO combines inertia, personal memory and shared memory. Subtract positions coordinate by coordinate. Evaluate the objective only after moving. For minimisation, replace a best only when its score decreases.
Particle A
Particle B
pbest is each initial position because there is no earlier history. The smallest objective decides gbest.
Its score is 4. Store the position, not only the score.
All particles in this iteration use gbest = (-2). pbest values are A: (-2), B: (4).
Only coordinate
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Only coordinate
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Particle A
Particle B
| Particle | Old pbest score | New score | Decision | pbest position |
|---|---|---|---|---|
| A | 4 | 4 | Keep old: tie | (-2) |
| B | 16 | 1 | Replace: new score is lower | (1) |
Compare the retained pbest scores. gbest = (1), with objective 1.
| Particle | Position | Velocity | pbest |
|---|---|---|---|
| A | (-2) | (0) | (-2) |
| B | (1) | (-3) | (1) |
All particles in this iteration use gbest = (1). pbest values are A: (-2), B: (1).
Only coordinate
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Only coordinate
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Particle A
Particle B
| Particle | Old pbest score | New score | Decision | pbest position |
|---|---|---|---|---|
| A | 4 | 0.25 | Replace: new score is lower | (-0.5) |
| B | 1 | 0.25 | Replace: new score is lower | (-0.5) |
Compare the retained pbest scores. gbest = (-0.5), with objective 0.25.
| Particle | Position | Velocity | pbest |
|---|---|---|---|
| A | (-0.5) | (1.5) | (-0.5) |
| B | (-0.5) | (-1.5) | (-0.5) |
Work down the page: each particle is fully updated, then all scores and memories are compared. Keep at least six digits internally for longer decimal examples; the final table displays rounded values.
Minimise using 2 particles. Use the initial data below and perform exactly two PSO iterations, reporting position, velocity, pbest and gbest after each iteration. [3 marks · Practice; expanded teaching solution]
| Particle | Initial position x⁰ | Initial velocity v⁰ |
|---|---|---|
| A | (1) | (4) |
| B | (3) | (-1) |
w=0.5, c₁=1, c₂=1, r₁=0.5, r₂=0.5. Initialise each pbest at its initial position. Use these random coefficients unchanged in both iterations. No bounds or velocity clamp are imposed in this example. Updates are synchronous: every particle uses the same previous-iteration gbest.
The complete swarm state after iterations 1 and 2, and the best location found.
PSO combines inertia, personal memory and shared memory. Subtract positions coordinate by coordinate. Evaluate the objective only after moving. For minimisation, replace a best only when its score decreases.
Particle A
Particle B
pbest is each initial position because there is no earlier history. The smallest objective decides gbest.
Its score is 1. Store the position, not only the score.
All particles in this iteration use gbest = (1). pbest values are A: (1), B: (3).
Only coordinate
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Only coordinate
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Particle A
Particle B
| Particle | Old pbest score | New score | Decision | pbest position |
|---|---|---|---|---|
| A | 1 | 9 | Keep old: new score is worse | (1) |
| B | 9 | 2.25 | Replace: new score is lower | (1.5) |
Compare the retained pbest scores. gbest = (1), with objective 1.
| Particle | Position | Velocity | pbest |
|---|---|---|---|
| A | (3) | (2) | (1) |
| B | (1.5) | (-1.5) | (1.5) |
All particles in this iteration use gbest = (1). pbest values are A: (1), B: (1.5).
Only coordinate
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Only coordinate
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Particle A
Particle B
| Particle | Old pbest score | New score | Decision | pbest position |
|---|---|---|---|---|
| A | 1 | 4 | Keep old: new score is worse | (1) |
| B | 2.25 | 0.25 | Replace: new score is lower | (0.5) |
Compare the retained pbest scores. gbest = (0.5), with objective 0.25.
| Particle | Position | Velocity | pbest |
|---|---|---|---|
| A | (2) | (-1) | (1) |
| B | (0.5) | (-1) | (0.5) |
Yaad rakho: Naya result kharab ho toh position toh badlegi, pbest nahi.
"},{"id":"two-coordinates","title":"Two-dimensional PSO: keep the coordinates separate","html":"\r\nA two-dimensional particle has two entries in its position and two in its velocity. Apply the scalar velocity calculation once to coordinate 1 and once to coordinate 2. Keep them in the same order to form the new pair. Only after both entries are ready do you insert the pair into f(x₁,x₂). Adding x₁ and x₂ into a single position would lose information.
Work down the page: each particle is fully updated, then all scores and memories are compared. Keep at least six digits internally for longer decimal examples; the final table displays rounded values.
Minimise using 2 particles. Use the initial data below and perform exactly two PSO iterations, reporting position, velocity, pbest and gbest after each iteration. [3 marks · Practice; expanded teaching solution]
| Particle | Initial position x⁰ | Initial velocity v⁰ |
|---|---|---|
| A | (1, 1) | (0, 0) |
| B | (3, 2) | (0, 0) |
w=0.5, c₁=1, c₂=1, r₁=0.5, r₂=0.5. Initialise each pbest at its initial position. Use these random coefficients unchanged in both iterations. No bounds or velocity clamp are imposed in this example. Updates are synchronous: every particle uses the same previous-iteration gbest.
The complete swarm state after iterations 1 and 2, and the best location found.
PSO combines inertia, personal memory and shared memory. Subtract positions coordinate by coordinate. Evaluate the objective only after moving. For minimisation, replace a best only when its score decreases.
Particle A
Particle B
pbest is each initial position because there is no earlier history. The smallest objective decides gbest.
Its score is 2. Store the position, not only the score.
All particles in this iteration use gbest = (1, 1). pbest values are A: (1, 1), B: (3, 2).
Coordinate 1
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Coordinate 2
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Coordinate 1
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Coordinate 2
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Particle A
Particle B
| Particle | Old pbest score | New score | Decision | pbest position |
|---|---|---|---|---|
| A | 2 | 2 | Keep old: tie | (1, 1) |
| B | 13 | 6.25 | Replace: new score is lower | (2, 1.5) |
Compare the retained pbest scores. gbest = (1, 1), with objective 2.
| Particle | Position | Velocity | pbest |
|---|---|---|---|
| A | (1, 1) | (0, 0) | (1, 1) |
| B | (2, 1.5) | (-1, -0.5) | (2, 1.5) |
All particles in this iteration use gbest = (1, 1). pbest values are A: (1, 1), B: (2, 1.5).
Coordinate 1
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Coordinate 2
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Coordinate 1
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Coordinate 2
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Particle A
Particle B
| Particle | Old pbest score | New score | Decision | pbest position |
|---|---|---|---|---|
| A | 2 | 2 | Keep old: tie | (1, 1) |
| B | 6.25 | 2 | Replace: new score is lower | (1, 1) |
Compare the retained pbest scores. gbest = (1, 1), with objective 2.
| Particle | Position | Velocity | pbest |
|---|---|---|---|
| A | (1, 1) | (0, 0) | (1, 1) |
| B | (1, 1) | (-1, -0.5) | (1, 1) |
Work down the page: each particle is fully updated, then all scores and memories are compared. Keep at least six digits internally for longer decimal examples; the final table displays rounded values.
The positions, velocities and coefficients are transcribed from the class photo. We use f(x₁,x₂)=x₁²+x₂²−x₁x₂, minimisation, initial pbest=x, fixed supplied random factors and synchronous updates. The photo does not itself specify all these conventions; they are stated here to make the two-iteration extension unambiguous.
Minimise using 3 particles. Use the initial data below and perform exactly two PSO iterations, reporting position, velocity, pbest and gbest after each iteration. [10 marks · Class data; two-iteration minimisation extension]
| Particle | Initial position x⁰ | Initial velocity v⁰ |
|---|---|---|
| A | (3.14, 7.21) | (0.1, 0.3) |
| B | (4.15, 7.31) | (0.5, 0.1) |
| C | (6.17, 8.27) | (0.2, 0) |
w=0.5, c₁=1.2, c₂=1.2, r₁=0.5, r₂=0.1. Initialise each pbest at its initial position. Use these random coefficients unchanged in both iterations. No bounds or velocity clamp are imposed in this example. Updates are synchronous: every particle uses the same previous-iteration gbest.
The complete swarm state after iterations 1 and 2, and the best location found.
PSO combines inertia, personal memory and shared memory. Subtract positions coordinate by coordinate. Evaluate the objective only after moving. For minimisation, replace a best only when its score decreases.
Particle A
Particle B
Particle C
pbest is each initial position because there is no earlier history. The smallest objective decides gbest.
Its score is 39.2043. Store the position, not only the score.
All particles in this iteration use gbest = (3.14, 7.21). pbest values are A: (3.14, 7.21), B: (4.15, 7.31), C: (6.17, 8.27).
Coordinate 1
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Coordinate 2
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Coordinate 1
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Coordinate 2
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Coordinate 1
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Coordinate 2
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Particle A
Particle B
Particle C
| Particle | Old pbest score | New score | Decision | pbest position |
|---|---|---|---|---|
| A | 39.2043 | 40.8673 | Keep old: new score is worse | (3.14, 7.21) |
| B | 40.3221 | 40.86061104 | Keep old: new score is worse | (4.15, 7.31) |
| C | 55.4359 | 53.09611888 | Replace: new score is lower | (5.9064, 8.1428) |
Compare the retained pbest scores. gbest = (3.14, 7.21), with objective 39.2043.
| Particle | Position | Velocity | pbest |
|---|---|---|---|
| A | (3.19, 7.36) | (0.05, 0.15) | (3.14, 7.21) |
| B | (4.2788, 7.348) | (0.1288, 0.038) | (4.15, 7.31) |
| C | (5.9064, 8.1428) | (-0.2636, -0.1272) | (5.9064, 8.1428) |
All particles in this iteration use gbest = (3.14, 7.21). pbest values are A: (3.14, 7.21), B: (4.15, 7.31), C: (5.9064, 8.1428).
Coordinate 1
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Coordinate 2
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Coordinate 1
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Coordinate 2
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Coordinate 1
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Coordinate 2
Inertia = keep a fraction of the old velocity.
Personal pull: first subtract the current coordinate from the personal-best coordinate.
Multiply the two personal coefficients, then multiply by the difference.
Social pull: subtract the current coordinate from the frozen global-best coordinate.
Multiply the social coefficients, then multiply by that difference.
Add the three velocity contributions.
Add the new velocity to the old position. The parenthesised upper label is the iteration number.
Particle A
Particle B
Particle C
| Particle | Old pbest score | New score | Decision | pbest position |
|---|---|---|---|---|
| A | 39.2043 | 40.498437 | Keep old: new score is worse | (3.14, 7.21) |
| B | 40.3221 | 40.487369094 | Keep old: new score is worse | (4.15, 7.31) |
| C | 53.09611888 | 49.736652734 | Replace: new score is lower | (5.442632, 7.967264) |
Compare the retained pbest scores. gbest = (3.14, 7.21), with objective 39.2043.
| Particle | Position | Velocity | pbest |
|---|---|---|---|
| A | (3.179, 7.327) | (-0.011, -0.033) | (3.14, 7.21) |
| B | (4.129264, 7.32764) | (-0.149536, -0.02036) | (4.15, 7.31) |
| C | (5.442632, 7.967264) | (-0.463768, -0.175536) | (5.442632, 7.967264) |
| Advantages | Limitations |
|---|---|
| Simple position/velocity updates; does not require objective derivatives | Can converge prematurely to a local region |
| Uses both individual and shared experience | Results depend on coefficients, initialisation and random factors |
| Useful for many continuous optimisation problems | Constraints and discrete encodings need suitable handling |
| Keeps best-so-far memories | Does not guarantee the global optimum in a finite run |
Explain one application of PSO. [3 marks · Practice]
In controller tuning, a particle can represent the gain values of a controller. The objective is a performance cost, such as tracking error plus an overshoot penalty. Each proposed gain vector is evaluated in the model. PSO updates candidate gains using individual and swarm bests. The best feasible gains found are selected and then validated on the actual system. A lower training cost alone does not prove safe real-world behaviour.
| Exam situation | Action |
|---|---|
| Minimisation | Lower f is better |
| Maximisation | Higher f is better; do not reverse velocity differences |
| Tie in best score | Keep the previous best unless a tie rule is supplied |
| Different random factors each iteration | Read the relevant row; do not reuse old factors |
| Synchronous update | Freeze gbest for all particles within one iteration |
| Position/velocity outside a limit | Use the limit-handling rule explicitly supplied; state an assumption if absent |
Yaad rakho: Velocity → position → score → pbest → gbest. Ye order har iteration mein same.
"}]},{"id":"u4-rl","title":"Unit 4 · What reinforcement learning is","unit":"4","scope":"Unit 4 · Introduction to reinforcement learning · TEE","sections":[{"id":"what-is-rl","title":"Reinforcement learning: learning from rewards","html":"\r\n| Time | Current state | Chosen action | Observed reward | Next state |
|---|---|---|---|---|
| 0 | Robot in room A | Move right | −1 for the move | Room B |
| 1 | Room B | Pick up parcel | +5 for pickup | Carrying parcel |
| 2 | Carrying parcel | Finish | +2 for delivery | Terminal |
The agent is the robot; the environment supplies the next state and reward after an action. A policy tells the robot which action to take in each state. An episode is one run up to termination. The immediate reward at time 0 is −1; the undiscounted return for the whole displayed run is −1+5+2=6. A value estimates future return, not just the next reward.
For ε-greedy with two actions, ε=.1 and a unique greedy action: exploit that action with probability .9; during the remaining .1, select uniformly from both actions. Thus the greedy action has probability .9+.1/2=.95 and the other .05. If a question uses a different exploration convention, follow that definition.
In one line: an agent learns what to do by trying actions, receiving rewards, and adjusting its behaviour so that the total reward over time is as large as possible.
\r\n\r\nReinforcement learning (RL) is a type of machine learning in which an agent learns a policy, a rule that maps situations (states) to actions, by interacting with an environment. After every action the environment returns a numerical reward and a new state. The agent's goal is to maximise the expected cumulative (discounted) reward, called the return. No labelled “correct action” is given; the agent discovers good actions by trial and error.
\r\n\r\n![\"At](\"fig/rl_loop_full.a9937cb961.svg\")
| Term | Meaning | Example: a robot vacuum |
|---|---|---|
| Agent | The learner and decision maker. | The vacuum's controller. |
| Environment | Everything outside the agent that responds to its actions. | The room, furniture, dust. |
State ![\"s\"](\"math/92329599456283d6.svg\") | A description of the current situation. | Position + battery level. |
Action ![\"a\"](\"math/2d4a1257492be872.svg\") | A choice the agent can make in a state. | Forward, turn left, return to dock. |
| Reward | Immediate scalar feedback after an action. | +1 per dust patch, −10 for hitting a wall. |
Policy ![\"\\pi\"](\"math/7e7e8c545536f6a2.svg\") | The agent's behaviour: ![\"\\pi(a\\mid](\"math/f155e46af3f71ce4.svg\") = probability of action in state . | “If battery < 15%, go to dock.” |
Return ![\"G_t\"](\"math/418bfe298764c6bc.svg\") | Total discounted reward from time ![\"t\"](\"math/b84a36f811c3dd68.svg\") onward. | All future dust collected, minus penalties. |
Value ![\"v(s)\"](\"math/214f0238bdf479cc.svg\") | Expected return starting from state . | How good it is to be at the centre with a full battery. |
| Model | The agent's knowledge of ![\"P(s'\\mid](\"math/8818d06a66002156.svg\") and ![\"R\"](\"math/486a3031e7ac622b.svg\") (optional). | A map of the room. |
| Supervised | Unsupervised | Reinforcement | |
|---|---|---|---|
| Feedback | Correct label for every input | None | A reward, often delayed |
| Goal | Predict labels | Find structure (clusters) | Maximise long-term return |
| Data | Fixed dataset (i.i.d.) | Fixed dataset | Generated by the agent's own actions (sequential) |
| Example | Spam classifier | Customer segments (GMM) | Game-playing agent |
Delayed reward. A move in chess earns nothing now but may win the game 30 moves later. RL must assign credit to earlier actions.
\r\nExploration versus exploitation. To earn reward the agent should exploit the best action it knows, but to find better actions it must explore untried ones. A common rule is ε-greedy: with probability ![\"1-\\varepsilon\"](\"math/bad648db5f9b2803.svg\")
take the best-known action; with probability ![\"\\varepsilon\"](\"math/e34d6bfa22904706.svg\")
take a random one.
A state is the current category, such as Savings or Current. A transition is one step from one category to another. The Markov assumption says that, once the current state is known, older states provide no additional information about the next-state probabilities in this model. It does not claim that the past never mattered; the current state must capture the relevant information.
In one line: in a Markov process the next state depends only on the present state, so the whole system is described by one table of “from → to” probabilities.
\r\n\r\nMarkov property. A sequence of states ![\"S_1,](\"math/a4c47d7eb261ab94.svg\")
has the Markov property if the future is independent of the past given the present:
![\"P(S_{t+1}\\mid](\"math/8b44c46b24e14f80.svg\")
Markov process (Markov chain). A memoryless random process given by a pair ![\"\\langle](\"math/bd68aa54ee45e3af.svg\")
: a finite set of states ![\"\\mathcal](\"math/54f9d81745f0a6da.svg\")
and a transition probability matrix ![\"P\"](\"math/43eba13e48b4ac63.svg\")
with entries
![\"P_{ij}=P(S_{t+1}=j\\mid](\"math/8f271383e8f91295.svg\")
Row is the state you are in now (“from”); column is the state you go to next. Two rules always hold:
![\"0\\le](\"math/e864eb4c19313d10.svg\")
.![\"\\sum_j](\"math/4864b97aaf5ebb33.svg\")
(from any state you must go somewhere, possibly staying).A state transition diagram draws each state as a circle and each non-zero ![\"P_{ij}\"](\"math/fe1d46e7593684d1.svg\")
as an arrow from to labelled with the probability. Staying in the same state is a self-loop.
Question. A survey shows that users switch between Brand X and Brand Y each year: 25% of Brand X users switch to Brand Y and 75% remain with X; 40% of Brand Y users switch to X and 60% remain with Y. (a) Write the transition probability matrix. (b) Draw the state transition diagram. [5 marks · Final Nov 2025 Q1d]
\r\nTwo states: X, Y. From X: stay 0.75, to Y 0.25. From Y: to X 0.40, stay 0.60.
Matrix and its diagram.
![\"P_{ij}=P(\\text{next}=j\\mid\\text{now}=i)\"](\"math/7f4c1d9519ac27a9.svg\")
. Each yearly switch depends only on the brand used this year, so the process is Markov.
![\"P=\\begin{array}{c|cc}](\"math/550d8027e65ec3eb.svg\")
Row X: 0.75 + 0.25 = 1. Row Y: 0.40 + 0.60 = 1. Valid.
![\"Self-loops](\"fig/mk_brand.c2c4be3e6a.svg\")
![\"P=\\begin{bmatrix}0.75&0.25\\\\0.40&0.60\\end{bmatrix}\"](\"math/1028784b72801945.svg\")
with the diagram above.Marking in the official key: matrix 3 marks, diagram 2 marks.
\r\nQuestion. A bank offers Savings and Current accounts. Each year 20% of Savings holders switch to Current and 80% stay; 10% of Current holders switch to Savings and 90% stay. (a) Construct the transition probability matrix. (b) Draw the state diagram. [5 marks · Re-exam Jan 2026 Q1d]
For Savings: stay 80%, switch 20%. For Current: switch to Savings 10%, stay 90%. State order: Savings, Current.
The transition matrix and labelled state diagram.
Convert percentages to fractions of 100. A row fixes the present state, while columns list possible destinations.
![\"P=\\begin{array}{c|cc}](\"math/ad78149edee2ba2b.svg\")
![\"Savings](\"fig/mk_bank.879499f6eb.svg\")
![\"P=\\begin{bmatrix}0.8&0.2\\\\0.1&0.9\\end{bmatrix}\"](\"math/e333fb4ce294957a.svg\")
; diagram as shown. Each part carries 2.5 marks in the key.A bank has 60 Savings customers and 40 Current customers. In one period, 80% of Savings stay and 20% switch to Current; 10% of Current switch to Savings and 90% stay. Find expected counts after one and two periods. [3 marks · Worked example]
![\"P=\\begin{bmatrix}0.8&0.2\\\\0.1&0.9\\end{bmatrix}\"](\"math/a352d238bb02bc14.svg\")
Row/column order is Savings, Current; initial counts [60,40].
Two successive count vectors.
Each destination receives incoming contributions from every origin. Multiply origin count by the corresponding transition probability, then add.
![\"60\\times0.8=48\\qquad40\\times0.1=4\"](\"math/fb4be71da02b8086.svg\")
![\"S_1=48+4=52\"](\"math/637f4b18747624bc.svg\")
![\"60\\times0.2=12\\qquad40\\times0.9=36\"](\"math/365204994bb40e62.svg\")
![\"C_1=12+36=48\"](\"math/ea2f674cf5833607.svg\")
![\"S_2=52\\times0.8+48\\times0.1=41.6+4.8=46.4\"](\"math/413733c9cae5c40e.svg\")
![\"C_2=52\\times0.2+48\\times0.9=10.4+43.2=53.6\"](\"math/61e8c23afa57bd0c.svg\")
Matrix notation [60,40]P is shorthand for these destination-column calculations. P² means P multiplied by P using row-by-column multiplication; it does not mean squaring every individual entry.
In one line: write today's numbers as a row, multiply by the matrix, and you get tomorrow's numbers.
\r\n\r\nLet ![\"x_t\"](\"math/c5e02241cdaf66e6.svg\")
be a row vector of today's shares (or counts) in each state, in the same order as the matrix rows. Then
![\"x_{t+1}=x_t\\,P,\\qquad](\"math/efd1264e0038146b.svg\")
Entry of ![\"x_{t+1}\"](\"math/de9a8c96e0ee7db9.svg\")
is the column calculation: everyone who arrives in , from every state.
![\"x_{t+1}(j)=\\sum_i](\"math/030f0c9b46caf90b.svg\")
![\"P^{\\,n}\"](\"math/9ef45de9aca6f359.svg\")
is the n-step transition matrix: ![\"(P^{\\,n})_{ij}\"](\"math/270ba67bd7a265ef.svg\")
is the probability of being in after ![\"n\"](\"math/0a03c1424e475329.svg\")
steps when starting in (the Chapman–Kolmogorov result). Counts stay counts (the total is preserved); shares stay shares (they still add to 1).
, not its rows. A quick check: the answer's total must equal the starting total (100 guests stay 100 guests).Question. A hotel chain has properties in Paris (P), London (L) and Rome (R). Daily movement: from Paris 60% stay, 25% to London, 15% to Rome; from London 30% to Paris, 50% stay, 20% to Rome; from Rome 20% to Paris, 30% to London, 50% stay. Today's guest counts are [50, 30, 20] in P, L, R. Find the expected counts tomorrow. [part of 10 marks · Final Nov 2025 Q5a]
\r\n![\"P=\\begin{bmatrix}0.60&0.25&0.15\\\\0.30&0.50&0.20\\\\0.20&0.30&0.50\\end{bmatrix},\\qquad](\"math/a2936167c6c14fec.svg\")
![\"x_1=x_0P\"](\"math/fe018f0ddfd05790.svg\")
.
![\"x_{t+1}=x_tP,\\qquad](\"math/8b9618ddde204a22.svg\")
For each destination, multiply each present-state share by its probability of arriving there, then add all arrivals.
![\"50(0.60)+30(0.30)+20(0.20)=30+9+4=43\"](\"math/396ed931211b6fc9.svg\")
![\"50(0.25)+30(0.50)+20(0.30)=12.5+15+6=33.5\"](\"math/66d48e5543eb7f89.svg\")
![\"50(0.15)+30(0.20)+20(0.50)=7.5+6+10=23.5\"](\"math/1005a6b946cdda94.svg\")
43 + 33.5 + 23.5 = 100 = 50 + 30 + 20. Correct.
Question. Using the brand-switching matrix , a customer uses Brand X this year. Find the probability that the customer uses each brand after two years. [practice · 3 marks]
Transition matrix P is printed in the question. For the two-year brand example the initial state is certainly X, so x₀=[1,0].
The requested next-step or two-step distribution.
For each destination, multiply each present-state share by its probability of arriving there, then add all arrivals.
![\"x_0=[1,\\;0]\"](\"math/b6383902c21eeee0.svg\")
(certainly X now).
![\"x_1=[1,0]P=[0.75,\\;0.25]\"](\"math/05a31cec6643ce3f.svg\")
![\"x_2=x_1P=[0.75(0.75)+0.25(0.40),\\;0.75(0.25)+0.25(0.60)]=[0.6625,\\;0.3375]\"](\"math/c991023e560f0430.svg\")
![\"P^2=\\begin{bmatrix}0.6625&0.3375\\\\0.54&0.46\\end{bmatrix}\"](\"math/6708ca668a7d9dcd.svg\")
.)| Paper question | Start | Next step |
|---|---|---|
| Learning levels B, I, A (Final Q6b) | [50, 30, 20] | [36, 32, 32] |
| Booking channels W, A, C (Re-exam Q5b) | [0.40, 0.35, 0.25] | [0.39, 0.335, 0.275] |
| Aircraft at hubs A, B, C (Re-exam Q6b) | [30, 40, 30] | [38.5, 32.5, 29] |
The complete working for each is in Solved papers.
\r\n\r\nFor transitions Savings→Current=0.2 and Current→Savings=0.1, derive the stationary shares. [3 marks · Worked example]
Savings→Current=0.2, so Savings→Savings=0.8. Current→Savings=0.1, so Current→Current=0.9. Let s and c be the Savings and Current shares; s+c=1.
s and c that remain unchanged after a step.
At stationarity total incoming share equals current share.
![\"s=0.8s+0.1c\"](\"math/193b341ebcb587b5.svg\")
![\"s-0.8s=0.1c\\quad\\Rightarrow\\quad0.2s=0.1c\"](\"math/3dfa7c84f62b2e86.svg\")
![\"2s=c\"](\"math/63aa27a487626a55.svg\")
![\"s+c=1\\quad\\Rightarrow\\quad](\"math/191d2d48b18264ce.svg\")
![\"s=1/3\\qquad](\"math/2788e62b0b2007f8.svg\")
In one line: a stationary distribution is unchanged by one transition; that fixed row satisfies ![\"\\pi=\\pi](\"math/e595caf895fa199c.svg\")
, and comparing it with today's shares tells you who gains and who loses in the long run.
A steady-state (stationary) distribution is a row vector ![\"\\pi=[\\pi_1,\\dots,\\pi_n]\"](\"math/118db8b67dc77bc6.svg\")
such that
![\"\\pi=\\pi](\"math/e3293de0e0ea04b1.svg\")
Once the system has these shares, one more step leaves them unchanged. For a chain in which every state can reach every other state and that does not cycle with a fixed period (true for every paper question so far), ![\"x_t\\to\\pi\"](\"math/45f2d2d1b7d524a8.svg\")
from any starting distribution.
![\"\\pi_j=\\sum_i\\pi_iP_{ij}\"](\"math/ea9b496718656056.svg\")
.![\"\\sum\\pi_i=1\"](\"math/b39286a8c89942c9.svg\")
. in terms of one of them, then use the sum to find it.Two-state shortcut. When a+b>0, if ![\"P=\\begin{bmatrix}1-a&a\\\\b&1-b\\end{bmatrix}\"](\"math/3eacaa69615e9607.svg\")
then
![\"\\pi=\\left[\\frac{b}{a+b},\\;\\frac{a}{a+b}\\right]\"](\"math/2511ae4b7c84ed79.svg\")
Brand switching: ![\"a=0.25,\\;b=0.40\\Rightarrow\\pi=[0.40/0.65,\\;0.25/0.65]=[0.6154,\\;0.3846]\"](\"math/6003dba221d13feb.svg\")
. Bank: ![\"a=0.2,\\;b=0.1\\Rightarrow\\pi=[1/3,\\;2/3]\"](\"math/119a700d02c295a5.svg\")
.
![\"P_{ii}\"](\"math/d21fe82fdd25b21f.svg\")
= retention: the fraction that stays. = transfer/loss from to ; looking down column shows inflow (gain) into .![\"\\pi_j\"](\"math/e0c070505a488343.svg\")
with the current share. Larger → gains; smaller → loses.![\"P_{ii}=1\"](\"math/698cf8e871a3b571.svg\")
: once entered it is never left.Question. Learners move daily between Beginner (B), Intermediate (I) and Advanced (A): from B 60% stay, 30% to I, 10% to A; from I 20% to B, 50% stay, 30% to A; from A 10% drop to I, 80% stay and 10% stop learning (treated as remaining in A). Find the steady-state proportions and the level with the largest long-run share. [part of 10 marks · Final Nov 2025 Q6b]
\r\n![\"P=\\begin{bmatrix}0.6&0.3&0.1\\\\0.2&0.5&0.3\\\\0&0.1&0.9\\end{bmatrix}\"](\"math/ad531c204822e0b1.svg\")
Row A is 0 + 0.1 + (0.8 + 0.1) = 1.
![\"\\pi=[\\pi_B,\\pi_I,\\pi_A]\"](\"math/031664350e5c144f.svg\")
.
![\"\\pi=\\pi](\"math/690921434c640510.svg\")
At stationarity each destination receives exactly its current share. Solve the balance equations together with the total-probability equation.
![\"\\pi_B=0.6\\pi_B+0.2\\pi_I+0\\cdot\\pi_A\\;\\Rightarrow\\;0.4\\pi_B=0.2\\pi_I\\;\\Rightarrow\\;\\pi_I=2\\pi_B\"](\"math/445f3c8d74cadf82.svg\")
![\"\\pi_A=0.1\\pi_B+0.3\\pi_I+0.9\\pi_A\\;\\Rightarrow\\;0.1\\pi_A=0.1\\pi_B+0.3(2\\pi_B)=0.7\\pi_B\\;\\Rightarrow\\;\\pi_A=7\\pi_B\"](\"math/15df96f89f27c195.svg\")
![\"\\pi_B+2\\pi_B+7\\pi_B=1\\;\\Rightarrow\\;\\pi_B=0.1,\\quad\\pi_I=0.2,\\quad\\pi_A=0.7\"](\"math/0d4ea6d007ce74ae.svg\")
![\"0.3(0.1)+0.5(0.2)+0.1(0.7)=0.03+0.10+0.07=0.20=\\pi_I\\;\\checkmark\"](\"math/6fea0316b61a8fc1.svg\")
(column B gives ![\"0.6(0.163)+0.2(0.255)=0.149\\ne0.163\"](\"math/c4d46c0025cc2225.svg\")
). The correct steady state is 0.1, 0.2, 0.7.Question. For the hotel chain (Paris, London, Rome) with today's counts [50, 30, 20], find the long-run distribution and say which city gains or loses guests. [part of 10 marks · Final Nov 2025 Q5a]
\r\n![\"P=\\begin{bmatrix}0.60&0.25&0.15\\\\0.30&0.50&0.20\\\\0.20&0.30&0.50\\end{bmatrix}\"](\"math/a6093f9acfe22fe1.svg\")
Stationary shares and the requested comparison with the initial distribution.
At stationarity each destination receives exactly its current share. Solve the balance equations together with the total-probability equation.
![\"\\pi_P=0.6\\pi_P+0.3\\pi_L+0.2\\pi_R\\;\\Rightarrow\\;\\pi_P=0.75\\pi_L+0.5\\pi_R\"](\"math/1708781ef111507a.svg\")
![\"\\pi_L=0.25\\pi_P+0.5\\pi_L+0.3\\pi_R\\;\\Rightarrow\\;0.5\\pi_L=0.25\\pi_P+0.3\\pi_R\"](\"math/5f029acce8c76d6c.svg\")
Substitute Step 1: ![\"0.5\\pi_L=0.1875\\pi_L+0.125\\pi_R+0.3\\pi_R\\Rightarrow0.3125\\pi_L=0.425\\pi_R\\Rightarrow\\pi_L=1.36\\,\\pi_R\"](\"math/bb6389f4842051c2.svg\")
.
Then ![\"\\pi_P=0.75(1.36\\pi_R)+0.5\\pi_R=1.52\\,\\pi_R\"](\"math/ea975cda05879034.svg\")
.
![\"(1.52+1.36+1)\\pi_R=3.88\\,\\pi_R=1\\Rightarrow\\pi_R=0.2577,\\;\\pi_L=0.3505,\\;\\pi_P=0.3918\"](\"math/40a070a0bf109760.svg\")
Exactly: ![\"\\pi=[38/97,\\;34/97,\\;25/97]\"](\"math/3b7300b6febd286b.svg\")
.
| City | Today | Long run | Verdict |
|---|---|---|---|
| Paris | 0.50 | 0.392 | Loses guests |
| London | 0.30 | 0.351 | Gains |
| Rome | 0.20 | 0.258 | Gains |
![\"\\pi\\approx[0.392,\\;0.351,\\;0.258]\"](\"math/a2ee97608735415b.svg\")
: Paris loses share; London and Rome gain. (The key's 0.3928 for Paris is a rounding slip; the three values must add to 1.)| Question | Steady state | Compared with now |
|---|---|---|
| Brand X/Y (Final Q1d) | [8/13, 5/13] = [0.615, 0.385] | — |
| Savings/Current (Re-exam Q1d) | [1/3, 2/3] | Current dominates |
| Booking W, A, C (Re-exam Q5b) | [8/21, 1/3, 2/7] = [0.381, 0.333, 0.286] | Now [0.40, 0.35, 0.25]: Website and App shrink slightly; Counter grows. |
| Aircraft A, B, C (Re-exam Q6b) | [25/63, 20/63, 2/7] = [0.397, 0.317, 0.286] | Now [0.3, 0.4, 0.3]: A gains; B loses; C loses slightly. |
![\"\\pi](\"math/3ac2346fee4256a7.svg\")
aur total 1. Gain/loss = steady share vs aaj ka share.γ (gamma) is a discount factor. Immediate reward has weight 1; one-step-later reward has weight γ; two-step-later reward has weight γ². This is repeated multiplication, not subtracting γ from a reward. An expectation averages possible future trajectories. A return G is the discounted total along one realised trajectory.
In one line: a Markov reward process is a Markov chain that also pays a reward in each state; the return adds those rewards, shrinking later ones by γ.
\r\n\r\nA Markov reward process (MRP) is a tuple ![\"\\langle\\mathcal](\"math/1cf28b140fd79e47.svg\")
:
: finite set of states; : transition matrix, ![\"P_{ss'}=P(S_{t+1}=s'\\mid](\"math/0ebf21d8b931d2d0.svg\")
;: reward function, ![\"R(s)=\\mathbb](\"math/bcb8b1797c06bfd8.svg\")
(the expected reward for leaving );![\"\\gamma\\in[0,1]\"](\"math/389866bfd285136c.svg\")
: the discount factor.There are no actions yet; the process simply drifts and collects rewards. (Add actions and you get an MDP.)
\r\n\r\nThe return is the total discounted reward from time :
![\"G_t=R_{t+1}+\\gamma](\"math/61d2488b0a270706.svg\")
Why discount? (1) With bounded rewards, it keeps the infinite discounted sum finite when ![\"\\gamma<1\"](\"math/4ec13e93147870ec.svg\")
. (2) Future rewards are uncertain. (3) Immediate rewards are usually preferred (like interest on money). ![\"\\gamma=0\"](\"math/d06cb90e005f54d4.svg\")
makes the agent short-sighted (only the next reward counts); ![\"\\gamma\\to1\"](\"math/cd9ca536b7f46490.svg\")
makes it far-sighted.
Question. An episode gives rewards ![\"R_1=-2,\\;R_2=-1,\\;R_3=-2,\\;R_4=0\"](\"math/49e74a4d062c7fe7.svg\")
and then ends. Find ![\"G_0\"](\"math/5322d8ff87ec1002.svg\")
with ![\"\\gamma=0.9\"](\"math/c491a3a52610e919.svg\")
. [practice · 2 marks]
Rewards [−2,−1,−2,0]; discount γ=0.9; episode ends after the fourth reward.
Return G₀.
The first reward is undiscounted. Each later reward receives one extra factor of γ.
![\"G_0=R_1+\\gamma](\"math/92c2d569c2ff0264.svg\")
![\"G_0=-2+0.9(-1)+0.81(-2)+0.729(0)=-2-0.9-1.62=-4.52\"](\"math/9a77a967b32f363f.svg\")
![\"G_0=-4.52\"](\"math/54ecfef1a5e374f5.svg\")
.The value of a state is the expected return when starting there:
\r\n![\"v(s)=\\mathbb](\"math/3a1a21a071450128.svg\")
The return of one episode is random; the value is its average over all possible futures.
\r\n\r\nAn episode gives rewards [2, 4, 8] and then ends. Find the return from the start with γ=0.5. [2 marks · Worked example]
Rewards in order: R₁=2, R₂=4, R₃=8. Discount factor γ=0.5. The episode then ends, so there are no later rewards.
G₀: the discounted sum of all three rewards, measured from the start.
![\"G_0=R_1+\\gamma](\"math/7e7a8fe5c95dbdc8.svg\")
Keep the first reward undiscounted; multiply each later reward by the appropriate power.
![\"1\\times(2)=2\"](\"math/c1fb0fa1e275d956.svg\")
![\"0.5\\times(4)=2\"](\"math/8f4e70094d023847.svg\")
![\"0.5^2\\times(8)=2\"](\"math/d314ab19535cf0b6.svg\")
![\"G_0=(2)+(2)+(2)=6\"](\"math/f58d97b5beb6cfcc.svg\")
An episode gives rewards [-1, 5, 2] and then ends. Find the return from the start with γ=0.9. [2 marks · Worked example]
Rewards in order: R₁=-1, R₂=5, R₃=2. Discount factor γ=0.9. The episode then ends, so there are no later rewards.
G₀: the discounted sum of all three rewards, measured from the start.
Keep the first reward undiscounted; multiply each later reward by the appropriate power.
![\"1\\times(-1)=-1\"](\"math/01d91edc60cbc82c.svg\")
![\"0.9\\times(5)=4.5\"](\"math/307a4fc23f4ba8e0.svg\")
![\"0.9^2\\times(2)=1.62\"](\"math/b276421b6d8cf042.svg\")
![\"G_0=(-1)+(4.5)+(1.62)=5.12\"](\"math/dc498059f2b74dca.svg\")
An episode gives rewards [2, 4, 8] and then ends. Find the return from the start with γ=0. [2 marks · Worked example]
Rewards in order: R₁=2, R₂=4, R₃=8. Discount factor γ=0. The episode then ends, so there are no later rewards.
G₀: the discounted sum of all three rewards, measured from the start.
Keep the first reward undiscounted; multiply each later reward by the appropriate power.
![\"0\\times(4)=0\"](\"math/2e86b7550d7cec9c.svg\")
![\"0^2\\times(8)=0\"](\"math/3f0045412e0c9af2.svg\")
![\"G_0=(2)+(0)+(0)=2\"](\"math/6433acabcea5b54f.svg\")
A state pays immediate reward 2. Next it reaches a state of value 10 with probability 0.6 or a state of value 0 with probability 0.4. With γ=0.5, calculate its Bellman value. [2 marks · Worked example]
Immediate reward R=2; discount γ=0.5. Next-state possibilities: value 10 with probability 0.6, or value 0 with probability 0.4. These supplied continuation values already include rewards after reaching those states.
Immediate reward plus discounted expected next value.
![\"v(s)=R(s)+\\gamma\\sum_{s'}P(s'\\mid](\"math/811e452127929f3e.svg\")
s′ is read “next state”; the prime is not a derivative. The sum is a weighted average of next-state values.
![\"0.6\\times10=6\"](\"math/6862aff895141018.svg\")
![\"0.4\\times0=0\"](\"math/2711ae2397112a4c.svg\")
![\"6+0=6\\qquad0.5\\times6=3\"](\"math/386c4e1d4519186c.svg\")
![\"v(s)=2+3=5\"](\"math/aed5ac659044d3f1.svg\")
In the matrix expression, I is the identity matrix (ones on the diagonal and zero elsewhere); multiplying by I leaves a vector unchanged. The inverse solves simultaneous equations when it exists. For the small exam examples below, ordinary substitution is sufficient.
In one line: the value of a state = the reward you get now + γ × the average value of where you land next.
\r\n\r\n![\"\\begin{aligned}v(s)&=\\mathbb](\"math/b890be2892d436ac.svg\")
The key step is ![\"G_t=R_{t+1}+\\gamma](\"math/837aecc6afff8d86.svg\")
: the return splits into the immediate reward plus the discounted return from the next state. That recursive split is the whole idea of every Bellman equation.
Matrix form. Stack the values in a column ![\"v\"](\"math/fe332ce60af0f3c3.svg\")
: ![\"v=R+\\gamma](\"math/8248280819c17d38.svg\")
. For small problems just solve the linear equations.
![\"Done](\"fig/mrp_student.be69eb4eb6.svg\")
Question. In the MRP shown, Class pays −2 and Lab pays −1 when left; Done pays 0 and is terminal. From Class: 0.5 to Lab, 0.5 to Done. From Lab: 0.4 to Class, 0.6 to Done. With , find v(Class) and v(Lab). [practice · 5 marks]
![\"R(C)=-2,\\;R(L)=-1,\\;v(D)=0,\\;\\gamma=0.9\"](\"math/b06d57b864effc2d.svg\")
.
![\"v(C),\\;v(L)\"](\"math/3d1e99a2714e55b1.svg\")
.
![\"v(s)=R(s)+\\gamma\\sum_{s'}P_{ss'}v(s')\"](\"math/eb908cf2d66f9fa5.svg\")
It holds for every state, so we get one equation per non-terminal state.
\r\n![\"v(C)=-2+0.9\\,[\\,0.5\\,v(L)+0.5\\,(0)\\,]=-2+0.45\\,v(L)\"](\"math/c27d02b92c11fd77.svg\")
![\"v(L)=-1+0.9\\,[\\,0.4\\,v(C)+0.6\\,(0)\\,]=-1+0.36\\,v(C)\"](\"math/7b9d5ae6e490a014.svg\")
![\"v(C)=-2+0.45(-1+0.36\\,v(C))=-2.45+0.162\\,v(C)\\Rightarrow0.838\\,v(C)=-2.45\"](\"math/ebb39216a5765bf5.svg\")
![\"v(C)=\\frac{-2.45}{0.838}=-2.9236,\\qquad](\"math/2b295ea148a86b91.svg\")
![\"-2+0.45(-2.0525)=-2-0.9236=-2.9236\\;\\checkmark\"](\"math/6e971d442b2340c9.svg\")
![\"v(\\text{Class})\\approx-2.924,\\;v(\\text{Lab})\\approx-2.053,\\;v(\\text{Done})=0\"](\"math/c86b63e5ffd55eb7.svg\")
. Lab is “less bad” because it is more likely to finish quickly.![\"v_C=-2+0.45(-1+0.36v_C)\"](\"math/637ce97b6476a63f.svg\")
Multiply .45 into both bracket terms: .45×(−1)=−.45 and .45×.36=.162.
![\"v_C=-2-0.45+0.162v_C=-2.45+0.162v_C\"](\"math/bd1b2a47575bc95d.svg\")
Subtract .162vC from both sides; the coefficient on the left is 1−.162=.838.
![\"0.838v_C=-2.45\\quad\\Rightarrow\\quad](\"math/591d6eadcfd21cd8.svg\")
Now use the Lab equation: .36×(−2.923627685)≈−1.052505967; adding −1 gives vL≈−2.052505967.
| Read this | Meaning | Concrete example |
|---|---|---|
| π(a|s) | Probability the policy chooses action a in state s | In room A: left .3, right .7 |
| P(s′|s,a) | Probability of the environment’s next state after that action | Right action reaches room B with probability .8 |
| vπ(s) | Expected return starting in state s and following policy π | A number scoring the state |
| qπ(s,a) | Expected return if action a is taken first, then policy π is followed | One score for each possible first action |
| π in this chapter | A policy | Unlike π in a Markov-chain stationary distribution; context distinguishes them |
The policy chooses an action; the transition model describes what happens after the action. They are two separate probability distributions. The worked examples use stationary Markov policies, depending on the current state and unchanged over time. More general policies may depend on time or history.
In one line: an MDP is an MRP in which the agent chooses actions; a policy says which action to choose, and value functions score how good a state or a state–action pair is under that policy.
\r\n\r\nA Markov decision process (MDP) is a tuple ![\"\\langle\\mathcal](\"math/2a8c7332f3030a47.svg\")
:
: states; ![\"\\mathcal](\"math/ce88a93ac568728b.svg\")
: actions;![\"P^{a}_{ss'}=P(S_{t+1}=s'\\mid](\"math/9a04b612a1d5ca7f.svg\")
: transition probabilities, now depending on the action;![\"R^{a}_{s}=\\mathbb](\"math/5d475467732358d0.svg\")
: expected reward for taking in ;![\"\\gamma\"](\"math/8a226d1a655f7384.svg\")
: discount factor.The Markov property still holds: the next state and reward depend only on the current state and action. An MDP is a common model for RL when the chosen state captures the relevant information; partially observed problems require additional treatment.
\r\n\r\n![\"In](\"fig/mdp_two.331bc03880.svg\")
A policy is a distribution over actions given states:
![\"\\pi(a\\mid](\"math/3bcff52e9ca32562.svg\")
![\"a=\\pi(s)\"](\"math/c6437d9ecddc37a5.svg\")
.![\"\\pi(\\text{stay}\\mid](\"math/6b25a58130fefd91.svg\")
.The policies used in these examples are stationary Markov policies. The term policy more generally also permits time-dependent or history-dependent decision rules.
\r\n\r\nState-value function: expected return from state when following :
![\"v_\\pi(s)=\\mathbb](\"math/ca7b3acce010f57a.svg\")
Action-value function (Q-function): expected return when taking action in and following afterwards:
![\"q_\\pi(s,a)=\\mathbb](\"math/830f0f5268ecd933.svg\")
They are linked by averaging over the policy's choices:
\r\n![\"v_\\pi(s)=\\sum_{a\\in\\mathcal](\"math/9ede9bf563f2c6c9.svg\")
Optimal value functions take the best policy: ![\"v_*(s)=\\max_\\pi](\"math/ed035ab67f88529a.svg\")
and ![\"q_*(s,a)=\\max_\\pi](\"math/c4640b6ec47433ce.svg\")
. An optimal policy acts greedily on ![\"q_*\"](\"math/3ccaa9f377c85da8.svg\")
: ![\"\\pi_*(s)=\\arg\\max_a](\"math/f3cf791d9745c384.svg\")
.
| Markov process | MRP | MDP | |
|---|---|---|---|
| Tuple | ![\"\\langle\\mathcal](\"math/f25474192d3bf3f0.svg\") | | |
| Rewards | No | Yes | Yes |
| Decisions | No | No | Yes (actions) |
| Value | — | | ![\"v_\\pi(s),\\;q_\\pi(s,a)\"](\"math/16476f59173bd3a8.svg\") |
At a state, action Left has value 3 and action Right has value 7. A policy chooses Left with probability 0.25 and Right with probability 0.75. Find the value under this policy, the maximum action value and the greedy action. [2 marks · Worked example]
Action values: Left=3 and Right=7. Policy probabilities: Left=0.25 and Right=0.75. The supplied action values already include subsequent returns.
A weighted average, a maximum and the label attaining that maximum.
![\"v_\\pi(s)=\\sum_a\\pi(a\\mid](\"math/46eb24952a8c1ab7.svg\")
The expectation averages according to policy; maximisation selects a largest action score.
![\"0.25\\times3=0.75\"](\"math/653f73dba2be3445.svg\")
![\"0.75\\times7=5.25\"](\"math/5c142a10f628b6eb.svg\")
![\"v_\\pi=0.75+5.25=6\"](\"math/c56792e7802f455e.svg\")
max(3,7)=7 is a number. argmax is Right, the action with that number.
In one line: the expectation equations score a given policy; the optimality equations replace the policy's average by a max and describe the best possible behaviour.
\r\n\r\nSplit the return into “reward now + discounted value later”, and average over the policy and the environment:
\r\n![\"v_\\pi(s)=\\sum_{a}\\pi(a\\mid](\"math/7b8a3aea9af35f5f.svg\")
![\"q_\\pi(s,a)=\\sum_{s'}P(s'\\mid](\"math/4789f0f7def0f243.svg\")
Read the first one aloud: for each action the policy might take, for each next state the environment might give, add the reward and γ × the next state's value, weighted by both probabilities.
\r\n\r\n![\"v_*(s)=\\max_{a}\\sum_{s'}P(s'\\mid](\"math/9a59f92e2e358c6a.svg\")
![\"q_*(s,a)=\\sum_{s'}P(s'\\mid](\"math/6f856204425ff6bb.svg\")
The only change is ![\"\\sum_a\\pi(a\\mid](\"math/935b09896577bd3b.svg\")
: the optimal agent picks the best action instead of averaging. These equations are non-linear (because of max), so they are solved iteratively (value iteration, policy iteration, Q-learning).
Question. In the two-state MDP (figure in the previous lesson), s₂ is terminal with ![\"v(s_2)=0\"](\"math/14eba4db728727d6.svg\")
and . In s₁, “stay” returns to s₁ with reward +1; “go” reaches s₂ with probability 0.8 and reward +5, or returns to s₁ with probability 0.2 and reward 0. (a) Find ![\"v_\\pi(s_1)\"](\"math/05a49a1a1019f0f7.svg\")
for the random policy ![\"\\pi(\\text{stay})=\\pi(\\text{go})=0.5\"](\"math/8c95286b27a764a3.svg\")
. (b) Find ![\"v_*(s_1)\"](\"math/ba05a03037bd4feb.svg\")
and the optimal action. [practice · 6 marks]
![\"q(s_1,\\text{stay})=1+0.9\\,v(s_1)\"](\"math/bebd3888f239dad7.svg\")
; ![\"q(s_1,\\text{go})=0.8(5+0.9\\cdot0)+0.2(0+0.9\\,v(s_1))=4+0.18\\,v(s_1)\"](\"math/3b7159339f0c3987.svg\")
.
The value under the random policy, the optimal value and the optimal action.
![\"v_\\pi(s)=\\sum_a\\pi(a\\mid](\"math/a5e70ba1f7aaa949.svg\")
The expectation equation averages actions using the given policy. The optimality equation chooses the largest action value. Each action value already includes its expected immediate reward and discounted continuation.
![\"v=0.5(1+0.9v)+0.5(4+0.18v)=2.5+0.54v\\Rightarrow0.46v=2.5\\Rightarrow](\"math/0b711572c2407738.svg\")
![\"v=4+0.18v\\Rightarrow](\"math/c5eca38ec463b77a.svg\")
Check whether switching helps: ![\"q(\\text{stay})=1+0.9(4.878)=5.390>4.878\"](\"math/0eff67e823e28a45.svg\")
. So “go” is not optimal.
![\"v=1+0.9v\\Rightarrow](\"math/1f1b525d33de57d2.svg\")
Check: ![\"q(\\text{go})=4+0.18(10)=5.8<10=q(\\text{stay})\"](\"math/15cfa2584493bce2.svg\")
. No improvement possible.
![\"v_\\pi(s_1)\\approx5.43\"](\"math/80d83715ae3f22a7.svg\")
. (b) ![\"v_*(s_1)=10\"](\"math/379f03e15709665d.svg\")
with ![\"q_*(s_1,\\text{stay})=10\"](\"math/26abf1418e7a596a.svg\")
, ![\"q_*(s_1,\\text{go})=5.8\"](\"math/defe0522e073671d.svg\")
: the optimal action is stay. A steady +1 forever is worth ![\"1/(1-0.9)=10\"](\"math/ad897de8c397fc5d.svg\")
, more than a one-time +5.Start with ![\"v_0(s_1)=0\"](\"math/392369c0ba50f924.svg\")
and apply the optimality equation repeatedly: ![\"v_{k+1}=\\max(1+0.9v_k,\\;4+0.18v_k)\"](\"math/db13b37224954848.svg\")
.
| k | stay = 1 + 0.9vₖ | go = 4 + 0.18vₖ | vₖ₊₁ | best |
|---|---|---|---|---|
| 0 | 1.000 | 4.000 | 4.000 | go |
| 1 | 4.600 | 4.720 | 4.720 | go |
| 2 | 5.248 | 4.850 | 5.248 | stay |
| 3 | 5.723 | 4.945 | 5.723 | stay |
| … | converges | 10 | stay |
Early on the agent prefers the quick +5; as it “looks further ahead”, the steady reward wins. This is why γ and long-term value matter.
\r\n\r\nFor the full MDP question above, “stay” gives 1+.9v. “Go” has two possible transitions: .8×(5+.9×0)=4 and .2×(0+.9v)=.18v, so its total is 4+.18v. The random policy then averages these two action values.
![\"v=0.5(1+0.9v)+0.5(4+0.18v)\"](\"math/886e804c14fdc761.svg\")
![\"v=0.5+0.45v+2+0.09v=2.5+0.54v\"](\"math/6e4eca135d9eae35.svg\")
Subtract .54v from both sides, then divide by .46.
![\"(1-0.54)v=2.5\\quad0.46v=2.5\\quad](\"math/294b5e0b255011dd.svg\")
A backup means applying the Bellman equation once to improve a value estimate. In value iteration, each sweep uses the previous sweep’s values for all states. In policy iteration, first solve values for the current policy; then choose the action with the largest one-step action value using those solved continuation values. Repeat if the policy changes. The “try go, then improve to stay” calculation in the previous lesson is one complete policy-improvement example.
In one line: policy/value iteration use Bellman equations with a model; Q-learning uses sampled experience. These are examples of methods, not a claim about every RL algorithm.
\r\n\r\nGiven , compute ![\"v_\\pi\"](\"math/59ce33a75ce0983e.svg\")
by repeatedly applying the expectation equation until values stop changing:
![\"v_{k+1}(s)\\leftarrow\\sum_a\\pi(a\\mid](\"math/3bbeb8acbcb4f4f5.svg\")
![\"\\pi'(s)=\\arg\\max_a](\"math/d82f4ea4f553e1e7.svg\")
For a finite discounted MDP with exact policy evaluation and consistent tie handling, policy improvement does not reduce value and policy iteration can terminate at an optimal policy.
\r\n\r\nSkip full evaluation; apply the optimality equation directly:
\r\n![\"v_{k+1}(s)\\leftarrow\\max_a\\sum_{s'}P(s'\\mid](\"math/b41a10ce7769c4dd.svg\")
Stop when the change is tiny, then act greedily. The table in the previous lesson is value iteration on one state.
\r\n\r\nThe two supplied TEE papers do not ask a Q-learning update. The January 2026 paper asks an explanation of the Bellman equation and RL applications. This extra worked example can wait until those requirements are covered.
Q-learning is included to illustrate a Bellman target; it is not separately named in the supplied introductory RL syllabus.
\r\nWhen and are unknown, the agent learns from its own experience ![\"(s,a,r,s')\"](\"math/fc6fae0a339bc5e8.svg\")
:
![\"Q(s,a)\\leftarrow](\"math/d6e7feae89b08cf3.svg\")
is the learning rate. The bracket is the temporal-difference error: (Bellman target) − (current guess). The agent usually explores with ε-greedy actions while learning.
Question. A Q-table has ![\"Q(s,a)=2\"](\"math/efb25431f54f8edc.svg\")
. The agent takes in , receives ![\"r=3\"](\"math/e61c4165fe299ce3.svg\")
and lands in ![\"s'\"](\"math/45840389386c62f8.svg\")
where ![\"\\max_{a'}Q(s',a')=5\"](\"math/3f677ab805df68d2.svg\")
. With ![\"\\alpha=0.5,\\;\\gamma=0.9\"](\"math/32cf597702a3bb21.svg\")
, find the updated ![\"Q(s,a)\"](\"math/0caba73ef360f772.svg\")
. [Optional authored example]
Old Q=2, reward=3, best next-state Q=5, α=0.5, γ=0.9; the next state is nonterminal.
The updated Q estimate.
Move partway from the old estimate toward the one-step Bellman target.
![\"Q_{new}=Q_{old}+\\alpha[r+\\gamma\\max](\"math/bf626aee4ce371f4.svg\")
![\"r+\\gamma\\max_{a'}Q(s',a')=3+0.9(5)=7.5\"](\"math/9f91a54deb2a51d6.svg\")
![\"Q\\leftarrow2+0.5(7.5-2)=2+2.75=4.75\"](\"math/fb6236e8c0963234.svg\")
![\"Q(s,a)=4.75\"](\"math/12b09999f026b089.svg\")
.Asked as: “Explain Bellman equation in reinforcement learning. Also give applications of reinforcement learning.” [10 marks · Re-exam Jan 2026 Q7b; marking: Bellman explanation 5, applications 5]
\n\nWrite the answer below in this order. It fits in about two pages.
\n\n1. Setting. Reinforcement learning problems are modelled as a Markov decision process . The agent follows a policy and wants to maximise the return ![\"G_t=\\sum_{k\\ge0}\\gamma^kR_{t+k+1}\"](\"math/3f482d8ea694678d.svg\")
.
2. Value functions. ![\"v_\\pi(s)=\\mathbb](\"math/a27ddca95f461e82.svg\")
and ![\"q_\\pi(s,a)=\\mathbb](\"math/0885e6b22d6c67c2.svg\")
.
3. Key idea. The return is recursive: . Taking expectations gives the Bellman expectation equation:
![\"v_\\pi(s)=\\sum_a\\pi(a\\mid](\"math/cd7f793810f9ae1e.svg\")
![\"q_\\pi(s,a)=\\sum_{s'}P(s'\\mid](\"math/0aaca96cab3cf147.svg\")
4. Optimality. For the optimal policy the average over actions becomes a maximum (Bellman optimality equation):
\n![\"v_*(s)=\\max_a\\sum_{s'}P(s'\\mid](\"math/cc767eadfe69fc8c.svg\")
5. Meaning and use. The value of a state equals the immediate reward plus the discounted value of the successor state. This breaks a long-horizon problem into one-step sub-problems (dynamic programming). It is the basis of policy evaluation, policy iteration, value iteration and Q-learning; the optimal policy is .
6. Small illustration (adds clarity): if staying in a state gives +1 each step with , then ![\"v=1+0.9v\\Rightarrow](\"math/64606eb39a2dfd24.svg\")
.
| Agent | Controller at an intersection. |
|---|---|
| State | Queue length and waiting time on each approach, current signal phase. |
| Actions | Keep the current phase or switch to the next green phase. |
| Reward | Negative of total waiting time (or vehicles cleared per cycle). |
| Learning | Q-learning updates from observed traffic; ε-greedy exploration tries alternative timings. |
| Benefit | Adapts to rush hours and incidents; fixed timers cannot. |
| Challenge | Safety during exploration, so agents are trained in simulation first. |
a=left zero, b=peak, c=right zero, with a<b<c. Locate x before choosing the rising/falling branch.
Read the explanation and worked sums →
Compare memberships at the same element. These formulas use standard max–min operations.
Read the explanation and worked sums →
A difference means A AND NOT B; it is not ordinary subtraction.
Read the explanation and worked sums →
a,b are input degrees; α is the strength; z is a candidate output value. OR conditions use max instead.
Read the explanation and worked sums →
M is the set of discrete output locations attaining maximum membership. The plateau shortcut is for one continuous maximum interval.
Read the explanation and worked sums →
Use actual discrete memberships or the full continuous aggregate. Denominator must be positive.
Read the explanation and worked sums →
c is a supplied representative such as a peak. Do not call this the exact area centroid of a clipped Mamdani aggregate without justification.
Read the explanation and worked sums →
Relations must have a common middle domain in the same order. Min along a path, max across paths.
Read the explanation and worked sums →
\r\n\r\n\r\nFor a<b≤c<d: a is the left zero, b the start of the height-1 plateau, c its end and d the right zero. Vertical shoulder edges need the specified piecewise definition. Branch examples →
c is the centre and σ>0 sets the width. At x=c the exponent is zero and membership is 1. Worked exponential calculations →
a>0 controls width; b>0 controls steepness; c is the centre. First subtract c, divide by a, take absolute value, apply the power, add 1, then take its reciprocal.
α is the chosen membership threshold. For 0<α≤1 keep inputs meeting or exceeding it, including threshold-equality endpoints. Discrete and triangular examples →
bᵢ is a bit; the maximum n-bit unsigned integer is 2ⁿ−1.
Full explanation and worked solutions →
Fitness weights must be nonnegative with a positive total. For minimisation use the specified transformation or selection method. C is cumulative probability.
Full explanation and worked solutions →
N is the number of individuals. Compare both best and average; mutation need not improve either.
Full explanation and worked solutions →
w weights inertia; c₁r₁ weights personal pull; c₂r₂ weights social pull. Apply the same equation separately to each coordinate.
Full explanation and worked solutions →
Use the new velocity, not the old one. Then reevaluate the objective.
Full explanation and worked solutions →
Memory updates: minimise → replace a pbest only for a lower score; maximise → only for a higher score. gbest is a position selected from the personal bests. Store the corresponding objective score alongside it. Keep previous best on ties unless instructed otherwise.
Yaad rakho: Formula ke pehle objective direction aur old state likho. Iteration 2 mein fresh positions aur retained memories use karo.
\r\n\r\n\r\nFor the specified uniform real-range encoding: decode n bits into unsigned integer d first. L and U are the physical limits. Endpoint examples →
This is the stated roulette transformation in our nonnegative-cost minimisation example. It is not required for every GA. It converts a smaller cost into a larger selection weight. Complete minimisation example →
Conditional probabilities require nonzero evidence probability. “Prior” means the class probability before the features; “likelihood” is the feature probability given a class; “posterior” is the class probability after evidence. Σ adds alternatives; Π multiplies factors; argmax returns the winning label. For Laplace smoothing, Nc is the class count and k is the number of possible values of the particular feature.
For the Jensen expression below, λ weights are nonnegative, sum to 1, and the logged arguments are positive. For a Gaussian, σ>0. GMM weights πk are nonnegative and sum to 1; γik or λki are responsibilities, not the mixing weights themselves.
Worked explanations: Naive Bayes · CPTs and Bayesian networks · Mean, variance and likelihood · EM update.
| Conditional | ![\"P(a\\mid](\"math/b69a34e61294d724.svg\") |
|---|---|
| Product rule | ![\"P(a\\land](\"math/68a6596a80d9767b.svg\") |
| Marginalisation | ![\"P(Y)=\\sum_zP(Y,z)\"](\"math/b6dbba8c2c4c1da9.svg\") |
| Normalisation | ![\"\\mathbf](\"math/6a7c981fbbb8b9db.svg\") |
| Bayes' rule | ![\"P(A\\mid](\"math/21c4330d83c41c29.svg\") |
| Naive Bayes | ![\"v_{NB}=\\arg\\max_cP(c)\\prod_iP(a_i\\mid](\"math/2c5a9cfc42c5e274.svg\") ; Laplace: ![\"\\frac{n_{ic}+1}{N_c+k}\"](\"math/c521050ae7174687.svg\") |
| BN chain rule | ![\"P(x_1..x_n)=\\prod_iP(x_i\\mid](\"math/44d1eaaf40faab74.svg\") |
| BN query | ![\"P(X\\mid](\"math/49b2c852ff5f290f.svg\") |
| CPT size | Boolean node with k Boolean parents: independent probability parameters = numbers |
| GMM density | ![\"p(x)=\\sum_k\\pi_k\\mathcal](\"math/68bcea7fd026bedf.svg\") |
| Gaussian | ![\"\\mathcal](\"math/0a27dae1a77ea038.svg\") |
| Log-likelihood | ![\"\\log](\"math/7852bd4ef8cbe433.svg\") |
| Jensen (log concave) | ![\"\\log\\sum_k\\lambda_ka_k\\ge\\sum_k\\lambda_k\\log](\"math/e188562c73c0cd46.svg\") |
| E-step | |
| M-step | ![\"\\mu_k=\\frac{\\sum\\lambda_k^ix_i}{\\sum\\lambda_k^i},\\;\\sigma_k^2=\\frac{\\sum\\lambda_k^i(x_i-\\mu_k)^2}{\\sum\\lambda_k^i},\\;\\pi_k=\\frac{\\sum_i\\lambda_k^i}{n}\"](\"math/94b0dd17291e3c1a.svg\") |
| GMM parameters | 1-D: ![\"3K-1\"](\"math/eec5afc3e1c07c8c.svg\") (K = 5 → 14) |
P is a transition matrix with origin rows and destination columns. π denotes a stationary distribution in the first equations, but a policy in the MDP equations. s′ is a next state; a′ a next action. R is the specified immediate or expected reward; γ discounts future rewards. E means expectation (probability-weighted average). I is the identity matrix.
The two-state stationary shortcut needs a+b>0. The inverse (I−γP)⁻¹ exists for a finite stochastic P with 0≤γ<1; at γ=1, check the problem’s terminal structure instead of assuming the full matrix is invertible. The continuing discounted-return examples use γ<1. A finite episode can also use γ=1.
Q-learning is supplementary here; the explicitly listed course focus is the Markov/MDP and Bellman material. α in Q-learning is a learning rate; α in Bayes normalisation was a different constant.
Worked explanations: Matrix multiplication · Stationary algebra · Discounted return · Bellman equations.
| Markov property | ![\"P(S_{t+1}\\mid](\"math/e99a6e85c6b029e3.svg\") |
|---|---|
| Transition matrix | ; rows sum to 1 |
| Next step | ![\"x_{t+1}=x_tP\"](\"math/8b6e8d7e9dedda16.svg\") (row vector); ![\"x_{t+n}=x_tP^n\"](\"math/3229af8e13e5a4dd.svg\") |
| Steady state | ![\"\\pi=\\pi](\"math/fa1b266acbaaacb0.svg\") ; two states ![\"\\begin{bmatrix}1-a&a\\\\b&1-b\\end{bmatrix}\\Rightarrow\\pi=\\left[\\frac](\"math/b54df1e7f5634ea5.svg\") |
| Return | |
| MRP Bellman | ![\"v(s)=R(s)+\\gamma\\sum_{s'}P_{ss'}v(s')\"](\"math/7c2858b3388b1fca.svg\") ; ![\"v=(I-\\gamma](\"math/d7e46a6497b0a4a6.svg\") |
| Policy | ![\"\\pi(a\\mid](\"math/556c05bd826349e2.svg\") |
| Value functions | ![\"v_\\pi(s)=\\mathbb](\"math/b08e042bdd23bc61.svg\") ; ![\"q_\\pi(s,a)=\\mathbb](\"math/99133acdbfaa7afa.svg\") ; ![\"v_\\pi=\\sum_a\\pi(a\\mid](\"math/3ce6dda6fd8fe296.svg\") |
| Bellman expectation | ![\"v_\\pi(s)=\\sum_a\\pi(a\\mid](\"math/e0207f466cf8054f.svg\") |
| Bellman optimality | ![\"v_*(s)=\\max_a\\sum_{s'}P(s'\\mid](\"math/7e6827fd1d3ef854.svg\") ; ![\"q_*(s,a)=\\sum_{s'}P(s'\\mid](\"math/d84f2e508180aed8.svg\") |
Optional further reading: the Q-learning extension is kept with the Bellman lesson, outside this core formula list.
"},{"id":"aco-tsp","title":"TEE-only formulas: ACO and GA for TSP","html":"| Transition probability | ![\"P_{ij}=\\dfrac{\\tau_{ij}^\\alpha\\eta_{ij}^\\beta}{\\sum_{l\\in\\text{allowed}}\\tau_{il}^\\alpha\\eta_{il}^\\beta},\\;\\eta_{ij}=1/d_{ij}\"](\"math/8df8f5c198d8cd6b.svg\") |
|---|---|
| Choice rule | cumulative probabilities in city order; first city with cumulative ≥ r |
| Pheromone update | ![\"\\tau_{ij}\\leftarrow(1-\\rho)\\tau_{ij}+\\sum_{k:\\,ij\\in](\"math/0cc34ae2992158a9.svg\") on every edge ant k used (return edge included); symmetric only for an undirected symmetric TSP |
| Values assumed in the worked PYQs | ![\"\\alpha=\\beta=1,\\;\\rho=0.5,\\;Q=1,\\;\\tau_0=1\"](\"math/0a0a3aed7677e436.svg\") |
α weights pheromone influence; β weights inverse-distance influence; ρ is the evaporation rate; Q is the deposit constant; L is the complete tour length. These must be supplied or explicitly assumed. Do not treat the example values as universal defaults.
![\"L=\\sum_{i=1}^{n-1}d(c_i,c_{i+1})+d(c_n,c_1)\"](\"math/316d88916534af37.svg\")
![\"F_{TSP}=1/L\"](\"math/85040bacae1471d9.svg\")
The return edge is part of the tour. Permutation crossover/mutation must keep each city exactly once. For positive lengths, shorter length gives larger reciprocal fitness.
"}]}],"papers":[{"id":"paper-final-2025","title":"Final examination · 24 Nov 2025","unit":"paper","scope":"Actual TEE paper · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged)","sections":[{"id":"q1a","title":"Q1(a) Joint probability table [5]","html":"Question. A supermarket recorded the joint distribution of Product type (Y: premium = 1, regular = 0), Payment mode (X: credit = 1, debit = 0) and Gender (Z: female = 1, male = 0). (a) If a customer bought a premium item, find the probability that (1) they are female, (2) they paid by credit card. (b) Find (1) P(premium) and (2) P(credit card). (c) Given a male customer paid by debit card, find the probability he bought a regular item. [5 marks]
The printed eight entries are reproduced below. Their total is 1.002, not 1. Therefore the table is inconsistent as a probability distribution.
| Credit, female | Credit, male | Debit, female | Debit, male | |
|---|---|---|---|---|
| Premium (Y=1) | 0.08 | 0.10 | 0.072 | 0.05 |
| Regular (Y=0) | 0.10 | 0.15 | 0.25 | 0.20 |
The five requested probabilities, stating how the inconsistent data are interpreted.
For this worked solution, treat all printed entries as nonnegative weights and normalise by their total 1.002. This preserves the printed ratios. It is a declared interpretation, not a claim about the examiner’s intended correction.
Use total weight 1.002 consistently for unconditional probabilities.
Alternative correction: if 0.072 is corrected to 0.070, the total becomes 1 and the corresponding answers are (a) 0.50 and 0.60; (b) 0.30 and 0.43; (c) 0.8. This agrees with the interpretation used in the supplied key. State the correction; do not call raw totals 0.302 and 0.43 valid marginals of the original inconsistent table.
List the applications of Ant Colony Optimization and explain any one application in detail. [5 marks]
Ant Colony Optimization is a probabilistic, population-based metaheuristic (Dorigo, 1992) in which artificial ants build solutions on a graph and communicate through pheromone trails. In the update rule used here, evaporation reduces all stored trails; deposits then reinforce edges of selected tours according to solution quality. Evaporation does not occur only on poor paths.
\r\n, skipping customers already served or that would exceed the vehicle capacity; when the vehicle is full it returns to the depot. on the routes of good solutions, so short connections are reinforced.| Advantages | Limitations |
|---|---|
| Naturally suited to graph and routing problems | Several parameters to tune (α, β, ρ, number of ants) |
| Can share useful route information and adapt search to changing graphs | Convergence time is uncertain; may stagnate on a sub-optimal path |
| Positive feedback + parallel ants | Theoretical analysis is difficult |
Question. Consider two fuzzy sets X = {(c1, 0.6), (c2, 0.8), (c3, 0.3), (c4, 0.5), (c5, 0.7)} and Y = {(c1, 0.4), (c2, 0.5), (c3, 0.9), (c4, 0.6), (c5, 0.2)}. Find 1) X ∪ Y 2) X ∩ Y 3) ¬X ∪ ¬Y 4) ¬X ∩ ¬Y. [5 marks]
The aligned X and Y memberships printed in this question.
The four requested membership lists in the original element order.
Use max for OR, min for AND, and 1−μ for NOT.
¬X = {0.4, 0.2, 0.7, 0.5, 0.3}; ¬Y = {0.6, 0.5, 0.1, 0.4, 0.8}.
| c1 | c2 | c3 | c4 | c5 | |
|---|---|---|---|---|---|
| X | 0.6 | 0.8 | 0.3 | 0.5 | 0.7 |
| Y | 0.4 | 0.5 | 0.9 | 0.6 | 0.2 |
| X ∪ Y = max | 0.6 | 0.8 | 0.9 | 0.6 | 0.7 |
| X ∩ Y = min | 0.4 | 0.5 | 0.3 | 0.5 | 0.2 |
| ¬X ∪ ¬Y | 0.6 | 0.5 | 0.7 | 0.5 | 0.8 |
| ¬X ∩ ¬Y | 0.4 | 0.2 | 0.1 | 0.4 | 0.3 |
¬X ∪ ¬Y must equal 1 − (X ∩ Y) = {0.6, 0.5, 0.7, 0.5, 0.8} ✓, and ¬X ∩ ¬Y must equal 1 − (X ∪ Y) = {0.4, 0.2, 0.1, 0.4, 0.3} ✓.
Official marking: 1 + 1 + 1.5 + 1.5.
\r\nQuestion. Users switch between Brand X and Brand Y each year: 25% of X users switch to Y and 75% stay; 40% of Y users switch to X and 60% stay. (a) Write the transition probability matrix. (b) Draw the state transition diagram. [5 marks]
All transition values are given in the full question. State order: X, Y.
The transition matrix and state diagram.
Rows mean current state; columns mean next state. For each destination, multiply each origin amount by its transition probability, then add.
| From / To | X | Y |
|---|---|---|
| X | 0.75 | 0.25 |
| Y | 0.4 | 0.6 |
Each row sums to 1 because it lists all possible destinations for that origin. A percentage is divided by 100: 25%=.25.
Diagonal entries are the fractions staying at their current state; off-diagonal entries are fractions moving to a different state. Each arrow uses its origin row and destination column. These fractions have different origin populations, so adding raw column probabilities does not give the number arriving.
Question. Maximize using Particle Swarm Optimization. . Take . Optimize and give the solution after 1 iteration showing the step-by-step process. [10 marks]
Particle 1: , . Particle 2: , .
Fitness, pbest, gbest; new velocities and positions; updated pbest and gbest.
Each particle is pulled by its own memory and by the swarm's best; we maximise, so the larger is better.
| Particle | Position | F | pbest |
|---|---|---|---|
| 1 | (2.5, −3.0) | 57.5 | (2.5, −3.0) |
| 2 | (3.5, −1.5) | 64 | (3.5, −1.5) |
No previous iteration, so pbest = current position. Largest fitness is 64, so gbest = (3.5, −1.5).
| Particle | New position | New F | Old pbest F | pbest after iteration |
|---|---|---|---|---|
| 1 | (2.80825, −2.537625) | 59.5569 | 57.5 | (2.80825, −2.537625) — improved |
| 2 | (3.59, −1.275) | 63.5336 | 64 | (3.5, −1.5) — kept |
The best pbest fitness is still 64, so gbest stays (3.5, −1.5).
, . After one iteration particle 1 improved to F = 59.5569; the best solution so far is gbest = (3.5, −1.5) with F = 64.Official key: fitness and gbest (5 marks), velocity and position updates (5 marks). The calculations above use the question data. The key's summary matrix writes 0.462 in place of 2.808 for ; that is a copying slip.
With reference to the Gaussian Mixture Model (GMM): compare GMM with the k-means algorithm; derive the equation for the log-likelihood function; state the number of parameters to be computed if K = 5. [10 marks]
“i.i.d.” means the observations are modelled as independent draws from the same distribution. The product Π runs over observations; the inner sum Σ runs over components. λki in some notes is the same responsibility that other sections call γik. A superscript i here labels the observation, not an exponent.
Jensen’s inequality gives a lower bound on the log of the component sum. EM chooses responsibilities making that bound touch the current likelihood, then updates the parameters to maximise the bound with those responsibilities fixed. This explains the two alternating steps; λ values are not arbitrary new model parameters.
| Parameter | Number in a 1D K-component GMM | Why |
|---|---|---|
| Means | K | One centre for each component |
| Variances | K | One spread for each component |
| Independent mixing weights | K−1 | The last weight is 1 minus the others |
| Total | 3K−1 | K+K+(K−1) |
For K=2, choose μ₁,μ₂,σ₁²,σ₂² and π₁: five free numbers. Then π₂=1−π₁. For K=5, 5+5+4=14. In 2D with a full covariance matrix, each component instead has two means and three independent covariance entries (two variances and one shared off-diagonal covariance). Five such components have 5×(2+3)+4=29 free parameters.
| Aspect | k-means | GMM |
|---|---|---|
| Assignment | Hard: each point in exactly one cluster | Soft: probability of belonging to each cluster |
| Model | Geometric: nearest centroid | Probabilistic generative model |
| Cluster shape | Spherical, similar size | Elliptical, any size/orientation; overlapping clusters |
| Parameters | K centroids | Weights, means, covariances (1-D: 3K − 1) |
| Algorithm / objective | Assign–update; minimise squared distances | EM; maximise log-likelihood |
| Output | Hard labels and fitted centroids | Labels, membership probabilities and a density (useful for anomaly detection) |
| Relation | The hard-assignment, equal spherical variance, small-variance limit of a Gaussian-mixture objective leads to the k-means objective. Ordinary finite-variance EM is not the same algorithm as k-means. Both need K in advance and depend on initialisation. |
Step 1. Model. Each point is generated by choosing component k with probability and drawing , so its density is .
Step 2. Likelihood. Points are i.i.d., so the likelihood is a product (θ = all parameters):
\r\nStep 3. Log-likelihood. Logs turn the product into a sum:
\r\nStep 4. Why it cannot be solved directly. The sum sits inside the log, so setting derivatives to zero gives no closed-form solution.
\r\nStep 5. Jensen's inequality. log is concave, so for weights : . Introduce for every point (), multiply and divide inside the log:
Step 6. Estimation (EM). Maximise the lower bound alternately. Fixing θ gives the E-step . Fixing λ and differentiating (with a Lagrange multiplier for ) gives the M-step:
The dimension and covariance structure must also be specified. For the one-dimensional, independent-variance model used in the derivation:
\r\n(5 means + 5 variances + 4 free weights, since the 5 weights must add to 1.) In two dimensions with full covariance it would be 29.
\r\n (≥, not >).Question. Cloudy → Sprinkler, Cloudy → Rain, Sprinkler and Rain → Wet Grass. P(Cloudy) = 0.5; P(Sprinkler | C) = 0.10, P(Sprinkler | ¬C) = 0.50; P(Rain | C) = 0.80, P(Rain | ¬C) = 0.20; P(Wet | S, R) = 0.99, P(Wet | S, ¬R) = 0.90, P(Wet | ¬S, R) = 0.90, P(Wet | ¬S, ¬R) = 0.01. Calculate the probability of the grass being wet given it doesn't rain. [10 marks]
The following table enumerates the values of Cloudy and Sprinkler that were not fixed by the question. Two Boolean choices give four rows: true/true, true/false, false/true, false/false. Each row is one mutually exclusive complete case. Read the CPT entry using that row’s parent values, multiply its factors, and add the resulting row weights. The denominator must include all cases matching the evidence, including both possible values of the queried outcome.
C→S, C→R, (S,R)→W. Root and conditional probabilities are listed in the question. Evidence R=false.
P(W=true | R=false). Hidden variables C and S must be summed out.
Use the network joint and divide by the evidence probability.
| C | S | P(C) | P(S|C) | P(¬R|C) | P(W|S,¬R) | Product |
|---|---|---|---|---|---|---|
| T | T | 0.5 | 0.1 | 0.2 | 0.90 | 0.0090 |
| T | F | 0.5 | 0.9 | 0.2 | 0.01 | 0.0009 |
| F | T | 0.5 | 0.5 | 0.8 | 0.90 | 0.1800 |
| F | F | 0.5 | 0.5 | 0.8 | 0.01 | 0.0020 |
| P(W, ¬R) | 0.1919 |
The four no-rain cases already total 0.1919. Now add the four rain cases, keeping the same Cloudy/Sprinkler order.
\r\nMeaning: the unconditional chance of Wet is 0.65, the chance of Rain among wet cases is about 0.704769, and the requested chance of Wet among no-rain cases is 0.3838. These are three different questions.
\r\nA bank uses fuzzy decision support with Credit Score and Income Stability, both on 1–100. Loan Approval is a score on 1–100 with labels Reject, Consider and Approve. Define triangular membership functions, construct nine rules, explain Mamdani inference and defuzzify for Credit Score=55 and Income Stability=35. [10 marks]
| Variable | Crisp value |
|---|---|
| Credit Score | 55 |
| Income Stability | 35 |
The question asks us to design the membership functions. This solution chooses Low=Tri(1,25,50), Medium=Tri(25,50,75), High=Tri(50,75,100) for both inputs. Output labels use the same triangles in their respective order. These are explicit design choices, not additional printed givens.
A complete stated design and its continuous Mamdani centroid.
Use min for AND and implication, max for aggregation and the true area centroid of the resulting shape. A peak-only weighted average is a different defuzzifier.
| Input / output position | Triangle | Peak |
|---|---|---|
| Low / Reject | (1,25,50) | 25 |
| Medium / Consider | (25,50,75) | 50 |
| High / Approve | (50,75,100) | 75 |
Use 0 outside each triangle, the rising fraction before the peak and the falling fraction after it. This design is evaluated at the requested input; a deployed controller would also need coverage/fallback at all boundary values.
| Credit Score / Income Stability | Low | Medium | High |
|---|---|---|---|
| Low | Reject | Reject | Consider |
| Medium | Reject | Consider | Approve |
| High | Consider | Approve | Approve |
Read each cell as an IF–AND–THEN rule. For example, the centre cell combines Medium with Medium.
| Credit Score | Income Stability | Firing strength | Consequent |
|---|---|---|---|
| Low | Low | min(0,0.6) = 0 | Reject |
| Low | Medium | min(0,0.4) = 0 | Reject |
| Low | High | min(0,0) = 0 | Consider |
| Medium | Low | min(0.8,0.6) = 0.6 | Reject |
| Medium | Medium | min(0.8,0.4) = 0.4 | Consider |
| Medium | High | min(0.8,0) = 0 | Approve |
| High | Low | min(0.2,0.6) = 0.2 | Consider |
| High | Medium | min(0.2,0.4) = 0.2 | Approve |
| High | High | min(0.2,0) = 0 | Approve |
Aggregated label strengths: Reject = 0.6, Consider = 0.4, Approve = 0.2.
For a strip from a to b with endpoint heights u and v, its trapezoid area is A. Its first moment M is the integral of z times the line height.
| a | b | u | v | Area A | Moment M |
|---|---|---|---|---|---|
| 1 | 15.4 | 0 | 0.6 | 4.32 | 45.792 |
| 15.4 | 25 | 0.6 | 0.6 | 5.76 | 116.352 |
| 25 | 35 | 0.6 | 0.6 | 6 | 180 |
| 35 | 40 | 0.6 | 0.4 | 2.5 | 93.333333333 |
| 40 | 50 | 0.4 | 0.4 | 4 | 180 |
| 50 | 55 | 0.4 | 0.4 | 2 | 105 |
| 55 | 65 | 0.4 | 0.4 | 4 | 240 |
| 65 | 70 | 0.4 | 0.2 | 1.5 | 100.833333333 |
| 70 | 75 | 0.2 | 0.2 | 1 | 72.5 |
| 75 | 95 | 0.2 | 0.2 | 4 | 340 |
| 95 | 100 | 0.2 | 0 | 0.5 | 48.333333333 |
Because the paper asks for a design, a different explicitly specified membership/rule system can yield a different numerical output. Explain the choices and use the chosen defuzzification method consistently; no particular examiner mark outcome is promised.
\r\n\r\n\r\nFor each output triangle, solve “line height = clip height”. Reject is clipped at .6: rising endpoint 1+.6(25−1)=15.4 and falling endpoint 50−.6(50−25)=35. Consider is clipped at .4: endpoints 25+.4(50−25)=35 and 75−.4(75−50)=65. Approve is clipped at .2: endpoints 50+.2(75−50)=55 and 100−.2(100−75)=95.
Now take the maximum of the three clipped curves. The declining Reject edge reaches the Consider height .4 at (50−z)/25=.4, hence z=40. The declining Consider edge reaches the Approve height .2 at (75−z)/25=.2, hence z=70. These crossover locations must be included when splitting the aggregate. Original peaks and other clip endpoints may add extra splits without changing the total area.
For the first aggregate strip from a=1 to b=15.4, endpoint heights are u=0 and v=0.6. Find its area and first moment. [3 marks · Worked example]
Strip endpoints a=1 and b=15.4; left height u=0 and right height v=0.6. This is a straight rising edge of the clipped Reject set.
A and M for one table row.
A is trapezoid area. M weights every small area by its horizontal position; this formula integrates a straight line.
Choose the output label having the greatest membership at the centroid; this is an additional stated interpretation rule. At z=42.780888139:
The resulting recommendation is Consider / manual review; this design does not recommend automatic approval. If a binary approve/reject answer is required, the question must supply a threshold or you must state the threshold you adopt. A different decision threshold is separate from defuzzification.
"},{"id":"q4a","title":"Q4(a) Cartesian product and max–min composition [10]","html":"Question. Consider fuzzy sets A = {0.4/small + 0.3/medium + 0.1/large} and B = {0.2/slow + 0.7/moderate + 0.3/fast}. 1. Find R = A × B. 2. Given C = {0.3/small + 0.1/medium + 0.4/large}, find S = C × B. 3. Calculate the max–min composition S ∘ R. [10 marks]
The fuzzy sets or relation matrices and their listed row/column order.
The relation matrix/matrices and the composition specified in the question.
For Cartesian products take min of row and column membership. For the ordered matrix composition shown here:
The printed relation domains do not directly compose: both relate size to speed. For the numerical exercise, explicitly assume positional alignment and compute row of S with column of R. Equal dimensions alone are not a justification for semantic relation composition.
\r\nRow 2 of S is all 0.1, so every entry of row 2 is 0.1.
\r\n.Both printed relations run from size to speed; their middle labels do not match for genuine relation composition. The numerical working below follows the question’s intended row-of-S with column-of-R operation using explicitly assumed positional alignment of the three middle entries.
| Cell | Row degree | Column degree | min result |
|---|---|---|---|
| R11 | 0.4 | 0.2 | 0.2 |
| R12 | 0.4 | 0.7 | 0.4 |
| R13 | 0.4 | 0.3 | 0.3 |
| R21 | 0.3 | 0.2 | 0.2 |
| R22 | 0.3 | 0.7 | 0.3 |
| R23 | 0.3 | 0.3 | 0.3 |
| R31 | 0.1 | 0.2 | 0.1 |
| R32 | 0.1 | 0.7 | 0.1 |
| R33 | 0.1 | 0.3 | 0.1 |
| Cell | Row degree | Column degree | min result |
|---|---|---|---|
| S11 | 0.3 | 0.2 | 0.2 |
| S12 | 0.3 | 0.7 | 0.3 |
| S13 | 0.3 | 0.3 | 0.3 |
| S21 | 0.1 | 0.2 | 0.1 |
| S22 | 0.1 | 0.7 | 0.1 |
| S23 | 0.1 | 0.3 | 0.1 |
| S31 | 0.4 | 0.2 | 0.2 |
| S32 | 0.4 | 0.7 | 0.4 |
| S33 | 0.4 | 0.3 | 0.3 |
Result cell (1,1)
Result cell (1,2)
Result cell (1,3)
Result cell (2,1)
Result cell (2,2)
Result cell (2,3)
Result cell (3,1)
Result cell (3,2)
Result cell (3,3)
Question. A hospital wants to predict flu from Fever (Yes/No), Cough (Yes/No) and Temperature Level (Low/Normal/High). Training data: (Yes, Yes, High, Yes), (No, No, Normal, No), (Yes, No, Low, No), (Yes, Yes, High, Yes), (No, Yes, Normal, No), (Yes, Yes, High, Yes), (No, No, Low, No), (Yes, No, Normal, No). Using naive Bayes, predict whether a patient with Fever = Yes, Cough = Yes, Temperature = Normal has flu. [10 marks]
The full training data are in Step 1. Query: Fever=Yes, Cough=Yes, Temperature=Normal. The main calculation uses unsmoothed counts.
Both class scores and the predicted class.
C is a class label. Each aⱼ is one query feature value. First count inside a class, then multiply the prior and all three likelihoods.
| Row | Fever | Cough | Temperature | Class |
|---|---|---|---|---|
| 1 | Yes | Yes | High | Yes |
| 2 | No | No | Normal | No |
| 3 | Yes | No | Low | No |
| 4 | Yes | Yes | High | Yes |
| 5 | No | Yes | Normal | No |
| 6 | Yes | Yes | High | Yes |
| 7 | No | No | Low | No |
| 8 | Yes | No | Normal | No |
| Class | Rows in this group | Class count | Prior |
|---|---|---|---|
| Yes | 1, 4, 6 | 3 | 3/8 = 0.375 |
| No | 2, 3, 5, 7, 8 | 5 | 5/8 = 0.625 |
Look only at class Yes rows 1, 4, 6. The matching rows are 1, 4, 6.
Look only at class Yes rows 1, 4, 6. The matching rows are 1, 4, 6.
Look only at class Yes rows 1, 4, 6. No row in this class has that feature value.
Look only at class No rows 2, 3, 5, 7, 8. The matching rows are 3, 8.
Look only at class No rows 2, 3, 5, 7, 8. The matching rows are 5.
Look only at class No rows 2, 3, 5, 7, 8. The matching rows are 2, 5, 8.
The larger score belongs to No. Normalise only if a posterior probability is requested: divide each score by their total 0.03.
The zero Yes score is caused by an absent training combination. If Laplace smoothing is explicitly required, apply it to all feature likelihoods in both classes. That is a different calculation and may change the prediction; do not mix smoothed and unsmoothed sides.
Yaad rakho: Class ke rows alag karo. Har feature ka count usi class ke total se divide. Saare factors multiply karo.
"},{"id":"q5a","title":"Q5(a) Hotel guest movement (Markov chain) [10]","html":"Question. A hotel chain has properties in Paris (P), London (L) and Rome (R). Of guests in Paris, 60% stay, 25% go to London, 15% to Rome; in London 30% go to Paris, 50% stay, 20% go to Rome; in Rome 20% go to Paris, 30% to London, 50% stay. 1. Construct the transition matrix. 2. Explain the diagonal and off-diagonal entries. 3. Draw the state diagram. 4. Today's counts are [50, 30, 20]; find tomorrow's expected counts. 5. Which city gains or loses guests in the long run? [10 marks]
All transition values are given in the full question. State order: P, L, R. Initial vector [50, 30, 20].
The matrix, diagram, next-step vector and long-run comparison.
Rows mean current state; columns mean next state. For each destination, multiply each origin amount by its transition probability, then add.
![\"x_{next}=x_{now}P\\qquad\\pi=\\pi](\"math/b179f364b70ac72c.svg\")
| From / To | P | L | R |
|---|---|---|---|
| P | 0.6 | 0.25 | 0.15 |
| L | 0.3 | 0.5 | 0.2 |
| R | 0.2 | 0.3 | 0.5 |
Each row sums to 1 because it lists all possible destinations for that origin. A percentage is divided by 100: 25%=.25.
Diagonal entries are the fractions staying at their current state; off-diagonal entries are fractions moving to a different state. Each arrow uses its origin row and destination column. These fractions have different origin populations, so adding raw column probabilities does not give the number arriving.
![\"Each](\"fig/mk_hotel.55f5204329.svg\")
| Origin | Current count/share | Transition to P | Contribution |
|---|---|---|---|
| P | 50 | 0.6 | 50 × 0.6 = 30 |
| L | 30 | 0.3 | 30 × 0.3 = 9 |
| R | 20 | 0.2 | 20 × 0.2 = 4 |
![\"P_{new}=30+9+4=43\"](\"math/26b7dba609bb77e3.svg\")
| Origin | Current count/share | Transition to L | Contribution |
|---|---|---|---|
| P | 50 | 0.25 | 50 × 0.25 = 12.5 |
| L | 30 | 0.5 | 30 × 0.5 = 15 |
| R | 20 | 0.3 | 20 × 0.3 = 6 |
![\"L_{new}=12.5+15+6=33.5\"](\"math/141ce72082fe6ed8.svg\")
| Origin | Current count/share | Transition to R | Contribution |
|---|---|---|---|
| P | 50 | 0.15 | 50 × 0.15 = 7.5 |
| L | 30 | 0.2 | 30 × 0.2 = 6 |
| R | 20 | 0.5 | 20 × 0.5 = 10 |
![\"R_{new}=7.5+6+10=23.5\"](\"math/1cfa32a9ab8dc27f.svg\")
![\"p=0.6p+0.3l+0.2r\"](\"math/a8c5e214edbb94e4.svg\")
![\"0.4p=0.3l+0.2r\"](\"math/bf5256a21829e5b4.svg\")
Divide every term by .4.
![\"p=0.75l+0.5r\"](\"math/e7f829189002f918.svg\")
![\"l=0.25p+0.5l+0.3r\"](\"math/809047184e208e3d.svg\")
![\"0.5l=0.25(0.75l+0.5r)+0.3r\"](\"math/f791f1205fe97449.svg\")
![\"0.5l=0.1875l+0.125r+0.3r\"](\"math/bc67c26ec7fc1e90.svg\")
![\"(0.5-0.1875)l=(0.125+0.3)r\"](\"math/ac05d4785e692e11.svg\")
![\"0.3125l=0.425r\\quad\\Rightarrow\\quad](\"math/d21c86c6a141bc54.svg\")
![\"p=0.75(1.36r)+0.5r=1.52r\"](\"math/af26ae261b198416.svg\")
![\"p+l+r=1\\quad\\Rightarrow\\quad1.52r+1.36r+r=1\"](\"math/936703629ebeff23.svg\")
![\"3.88r=1\\quad](\"math/76317a80ed92e6f4.svg\")
![\"l=1.36(25/97)=34/97\\quad](\"math/442f9f9b9a48c0ce.svg\")
| State | Initial share | Stationary share | Change in share | Conclusion |
|---|---|---|---|---|
| P | 0.5 | 0.391752577 | -0.108247423 | Loses |
| L | 0.3 | 0.350515464 | 0.050515464 | Gains |
| R | 0.2 | 0.257731959 | 0.057731959 | Gains |
Compare like with like: divide counts by their total before comparing with stationary probabilities. “Stationary” means one more step leaves the shares unchanged.
Implement a GA to maximise for x in [0,15]. Take 4-bit representations of 13, 22, 18, 23. Crossover after the second bit and mutate by flipping any one bit. Produce two offspring and comment on fitness. [10 marks]
The printed data are inconsistent. 22, 18 and 23 are outside 0–15 and cannot fit into four bits. Increasing the bit count would encode the numbers but would still violate the stated range. There is no single exact solution satisfying all printed conditions.
The complete worked variant below explicitly changes the invalid parents to their last four bits (values modulo 16), producing [13,6,2,7]. This is one declared repair, not a silent interpretation of the printed question. If the instructor clarifies different corrected data, follow those data.
Printed parents: [13, 22, 18, 23]. Worked variant: [13, 6, 2, 7] with 4 bits. For the unspecified selection step, select the two highest-fitness candidates. For the unspecified mutation location, flip the last bit of each child. Crossover is after bit 2.
Two valid children under the declared assumptions and their fitness.
Encoding must respect the candidate range. Crossover swaps tails; mutation flips the chosen bit. Fitness is evaluated after mutation.
| x | Binary | Fitness substitution |
|---|---|---|
| 13 | 1101 | 2(13³) − 5(13²) + 7(13) = 3640 |
| 6 | 0110 | 2(6³) − 5(6²) + 7(6) = 294 |
| 2 | 0010 | 2(2³) − 5(2²) + 7(2) = 10 |
| 7 | 0111 | 2(7³) − 5(7²) + 7(7) = 490 |
[13, 7] with fitness [3640, 490].
| Parent 1 | Parent 2 | Child 1 | Child 2 |
|---|---|---|---|
| 11 | 01 | 01 | 11 | 1111 | 0101 |
| Before | After | Decoded x |
|---|---|---|
| 1111 | 1110 | 14 |
| 0101 | 0100 | 4 |
| x | Fitness calculation |
|---|---|
| 14 | 2(2744) − 5(196) + 7(14) = 4606 |
| 4 | 2(64) − 5(16) + 7(4) = 76 |
Best parent score = 3640; best child score = 4606. Parent average = 2065; child average = 2341. Both the best score and the average score increased in this worked variant. An individual child can still be worse than a parent; GA does not guarantee improvement after every random operator.
The printed wording about “two offspring” is not an explicit requirement for two complete generations. The M2 examples separately work two full generations because your class update asks for iteration 2.
\r\n\r\n\r\nCalculate x²=x×x, then x³=x²×x. Multiply each by its coefficient and combine with the written signs. A bit-string repair in the worked answer is an explicitly stated response to inconsistent printed givens; it is not a new instruction in the original paper.
“Modulo 16” means keep the remainder after dividing by 16: 22=1×16+6, so its remainder is 6; 18=1×16+2; 23=1×16+7. This is only the declared repair used here, not an unspoken general encoding rule.
List the applications of fuzzy logic. Explain any one application in detail. [10 marks]
Fuzzy logic is a multi-valued logic, introduced by Lotfi Zadeh (1965), in which truth values range over [0, 1]; it lets machines reason with vague, human-like terms such as “slightly dirty” or “very hot”.
\r\n| Area | Examples |
|---|---|
| Home appliances | Washing machines (wash time from dirt and load), air conditioners, refrigerators, rice cookers, vacuum cleaners |
| Automotive | Automatic transmission, anti-lock braking (ABS), cruise control, engine control |
| Transportation | Traffic-signal control, Sendai subway train control (smooth braking and acceleration) |
| Consumer electronics | Camera autofocus and auto-exposure, image stabilisation |
| Industrial control | Cement kilns, pH, temperature and distillation control in chemical plants |
| Aerospace | Satellite and spacecraft attitude control |
| Medicine | Diagnosis support, anaesthesia dosage control |
| Finance and decisions | Loan approval, credit and fire-risk scoring, stock trading rules |
| AI and robotics | Robot navigation, natural-language interfaces, expert systems |
Advantages: handles imprecise or noisy data; mimics human reasoning; easy to design and modify; no exact model needed; low computational cost. Disadvantages: rules and membership functions depend on expert judgement; results are hard to verify formally; tuning can take time; not suited where very high precision is required.
\r\nQuestion. Students are Beginner (B), Intermediate (I) or Advanced (A). Of Beginners, 60% remain, 30% become Intermediate, 10% go to Advanced. Of Intermediates, 20% fall back, 50% remain, 30% progress. Of Advanced, 10% drop to Intermediate, 80% remain and 10% stop (treated as remaining in A). 1. Construct the matrix. 2. Interpret retention, progression and regression. 3. Draw the diagram. 4. For [50, 30, 20], find the distribution after one day. 5. Which level has the largest steady-state share? [10 marks]
All transition values are given in the full question. State order: B, I, A. Initial vector [50, 30, 20].
The matrix, diagram, next-step vector and long-run comparison.
Rows mean current state; columns mean next state. For each destination, multiply each origin amount by its transition probability, then add.
| From / To | B | I | A |
|---|---|---|---|
| B | 0.6 | 0.3 | 0.1 |
| I | 0.2 | 0.5 | 0.3 |
| A | 0 | 0.1 | 0.9 |
Each row sums to 1 because it lists all possible destinations for that origin. A percentage is divided by 100: 25%=.25.
Diagonal entries are the fractions staying at their current state; off-diagonal entries are fractions moving to a different state. Each arrow uses its origin row and destination column. These fractions have different origin populations, so adding raw column probabilities does not give the number arriving.
![\"Each](\"fig/mk_learning.61a789edc7.svg\")
| Origin | Current count/share | Transition to B | Contribution |
|---|---|---|---|
| B | 50 | 0.6 | 50 × 0.6 = 30 |
| I | 30 | 0.2 | 30 × 0.2 = 6 |
| A | 20 | 0 | 20 × 0 = 0 |
![\"B_{new}=30+6+0=36\"](\"math/60af91b5a47e34d2.svg\")
| Origin | Current count/share | Transition to I | Contribution |
|---|---|---|---|
| B | 50 | 0.3 | 50 × 0.3 = 15 |
| I | 30 | 0.5 | 30 × 0.5 = 15 |
| A | 20 | 0.1 | 20 × 0.1 = 2 |
![\"I_{new}=15+15+2=32\"](\"math/d10fe140b0e969ab.svg\")
| Origin | Current count/share | Transition to A | Contribution |
|---|---|---|---|
| B | 50 | 0.1 | 50 × 0.1 = 5 |
| I | 30 | 0.3 | 30 × 0.3 = 9 |
| A | 20 | 0.9 | 20 × 0.9 = 18 |
![\"A_{new}=5+9+18=32\"](\"math/e07af325efaf974e.svg\")
![\"b=0.6b+0.2i\"](\"math/e6252d668732cace.svg\")
![\"0.4b=0.2i\\quad\\Rightarrow\\quad](\"math/3a1b2778c1d42930.svg\")
![\"a=0.1b+0.3i+0.9a\"](\"math/6e1d2349f082467b.svg\")
![\"0.1a=0.1b+0.3(2b)=0.7b\"](\"math/e730f86d7c6fb366.svg\")
![\"a=0.7b/0.1=7b\"](\"math/4e3c76ca7c5b3101.svg\")
![\"b+i+a=1\\quad](\"math/8e00c1c8e9302c7a.svg\")
![\"10b=1\\quad](\"math/401c3cb4b3606dfc.svg\")
Advanced is not an absorbing state: its self-transition is .9, not 1.
| State | Initial share | Stationary share | Change in share | Conclusion |
|---|---|---|---|---|
| B | 0.5 | 0.1 | -0.4 | Loses |
| I | 0.3 | 0.2 | -0.1 | Loses |
| A | 0.2 | 0.7 | 0.5 | Gains |
Compare like with like: divide counts by their total before comparing with stationary probabilities. “Stationary” means one more step leaves the shares unchanged.
List the applications of Particle Swarm Optimization and explain any one application in detail. [10 marks]
Particle Swarm Optimization is a population-based metaheuristic, inspired by bird flocking and fish schooling, in which particles move through the search space using their own best position (pbest) and the swarm's best position (gbest) to find optimal or near-optimal solutions.
\r\n\r\n| Advantages | Disadvantages |
|---|---|
| Simple to implement; few parameters (w, c₁, c₂) | Can converge prematurely to a local optimum |
| No derivatives needed | Sensitive to parameter choice |
| Fast convergence in early iterations; memory (pbest, gbest) | No guarantee of the global optimum |
| Objective evaluations can be parallel; particles share information through gbest | Less suited to discrete problems without modification |
Marking pattern used for ACO in the same paper: listing 3 marks + explanation 2 marks (5-mark version); listing 5 + explanation 5 (10-mark version).
\r\n\r\nQuestion. For the four-city travelling salesman problem, the distance and pheromone matrices (A, B, C, D) are given. Apply Ant Colony Optimization. Calculate the path travelled by 2 ants to cover all cities starting from city A. Random numbers for Ant 1 are 0.3, 0.8 and for Ant 2 are 0.5, 0.7. Also update the pheromone matrix after 1 iteration. [10 marks]
Assume , , (class convention; not stated in the question).
Tour of each ant, tour lengths, updated pheromone matrix.
The sum includes only ants whose completed tour contains edge ij. An unused edge receives no deposit but still evaporates. Here distances are symmetric, so a traversed edge updates both symmetric matrix cells. Both ants construct their tours using the old pheromones; update after both finish, including their return to A.
\r\n| Move | P | Cumulative | Range |
|---|---|---|---|
| A→B | 0.25 | 0.25 | 0–0.25 |
| A→C | 0.50 | 0.75 | 0.25–0.75 |
| A→D | 0.25 | 1.00 | 0.75–1 |
Ant 1: r = 0.3 → C. Ant 2: r = 0.5 → C.
Cumulative in city order: B 0–0.4, D 0.4–1. Ant 1: r = 0.8 → D. Ant 2: r = 0.7 → D.
Only B is left, then return to A.
\r\nEach ant deposits on each edge of its tour; two ants used the same four edges (AC, CD, DB, BA), so each gets .
Official marking: path of Ant 1 (3) + path of Ant 2 (3) + pheromone matrix (4).
\r\nQuestion. For a disease Y with fever X as a symptom, a test Z is conducted. Joint probabilities: Disease (Y=1): fever & test+ 0.576, fever & test− 0.144, no fever & test+ 0.072, no fever & test− 0.016; No disease (Y=0): 0.008, 0.064, 0.012, 0.108. (1) If fever occurs and the test is positive, what is the probability of disease? (2) What is the probability that a random person tests positive? (3) Given no fever, what is the probability of still having the disease? [5 marks]
The complete joint table or cell values printed in this question. The rows and columns identify the event combinations.
All requested marginal and conditional probabilities.
Marginal: add cells satisfying the event. Conditional: retain the evidence cells and divide the joint-event sum by the evidence sum.
List the applications of Particle Swarm Optimization and explain any one application in detail. [5 marks]
Particle Swarm Optimization is a population-based metaheuristic, inspired by bird flocking and fish schooling, in which particles move through the search space using their own best position (pbest) and the swarm's best position (gbest) to find optimal or near-optimal solutions.
\r\n\r\n| Advantages | Disadvantages |
|---|---|
| Simple to implement; few parameters (w, c₁, c₂) | Can converge prematurely to a local optimum |
| No derivatives needed | Sensitive to parameter choice |
| Fast convergence in early iterations; memory (pbest, gbest) | No guarantee of the global optimum |
| Objective evaluations can be parallel; particles share information through gbest | Less suited to discrete problems without modification |
Marking pattern used for ACO in the same paper: listing 3 marks + explanation 2 marks (5-mark version); listing 5 + explanation 5 (10-mark version).
\r\n\r\nQuestion. X = {(c1, 0.7), (c2, 0.9), (c3, 0.2), (c4, 0.4), (c5, 0.9)}, Y = {(c1, 0.6), (c2, 0.3), (c3, 0.8), (c4, 0.7), (c5, 0.1)}. Find X ∪ Y, X ∩ Y, ¬X ∪ ¬Y, ¬X ∩ ¬Y. [5 marks]
The aligned X and Y memberships printed in this question.
The four requested membership lists in the original element order.
Use max for OR, min for AND, and 1−μ for NOT.
| c1 | c2 | c3 | c4 | c5 | |
|---|---|---|---|---|---|
| ¬X | 0.3 | 0.1 | 0.8 | 0.6 | 0.1 |
| ¬Y | 0.4 | 0.7 | 0.2 | 0.3 | 0.9 |
| X ∪ Y | 0.7 | 0.9 | 0.8 | 0.7 | 0.9 |
| X ∩ Y | 0.6 | 0.3 | 0.2 | 0.4 | 0.1 |
| ¬X ∪ ¬Y | 0.4 | 0.7 | 0.8 | 0.6 | 0.9 |
| ¬X ∩ ¬Y | 0.3 | 0.1 | 0.2 | 0.3 | 0.1 |
Question. Each year 20% of Savings holders switch to Current and 80% stay; 10% of Current holders switch to Savings and 90% stay. (a) Construct the transition matrix. (b) Draw the state diagram. [5 marks]
All transition values are given in the full question. State order: S, C.
The transition matrix and state diagram.
Rows mean current state; columns mean next state. For each destination, multiply each origin amount by its transition probability, then add.
| From / To | S | C |
|---|---|---|
| S | 0.8 | 0.2 |
| C | 0.1 | 0.9 |
Each row sums to 1 because it lists all possible destinations for that origin. A percentage is divided by 100: 25%=.25.
Diagonal entries are the fractions staying at their current state; off-diagonal entries are fractions moving to a different state. Each arrow uses its origin row and destination column. These fractions have different origin populations, so adding raw column probabilities does not give the number arriving.
A building safety department wants a fuzzy logic system to predict fire hazards. Inputs are Temperature (1–100°C) and Smoke Level (1–100). The output Fire Risk (1–100) has labels Low, Moderate and High. Define triangular membership functions, construct at least nine rules, explain Mamdani inference and defuzzify for Temperature=65°C and Smoke=80. [10 marks]
| Variable | Crisp value |
|---|---|
| Temperature | 65 |
| Smoke Level | 80 |
The question asks us to design the membership functions. This solution chooses Low=Tri(1,25,50), Medium=Tri(25,50,75), High=Tri(50,75,100) for both inputs. Output labels use the same triangles in their respective order. These are explicit design choices, not additional printed givens.
A complete stated design and its continuous Mamdani centroid.
Use min for AND and implication, max for aggregation and the true area centroid of the resulting shape. A peak-only weighted average is a different defuzzifier.
| Input / output position | Triangle | Peak |
|---|---|---|
| Low / Low | (1,25,50) | 25 |
| Medium / Moderate | (25,50,75) | 50 |
| High / High | (50,75,100) | 75 |
Use 0 outside each triangle, the rising fraction before the peak and the falling fraction after it. This design is evaluated at the requested input; a deployed controller would also need coverage/fallback at all boundary values.
| Temperature / Smoke Level | Low | Medium | High |
|---|---|---|---|
| Low | Low | Low | Moderate |
| Medium | Low | Moderate | High |
| High | Moderate | High | High |
Read each cell as an IF–AND–THEN rule. For example, the centre cell combines Medium with Medium.
| Temperature | Smoke Level | Firing strength | Consequent |
|---|---|---|---|
| Low | Low | min(0,0) = 0 | Low |
| Low | Medium | min(0,0) = 0 | Low |
| Low | High | min(0,0.8) = 0 | Moderate |
| Medium | Low | min(0.4,0) = 0 | Low |
| Medium | Medium | min(0.4,0) = 0 | Moderate |
| Medium | High | min(0.4,0.8) = 0.4 | High |
| High | Low | min(0.6,0) = 0 | Moderate |
| High | Medium | min(0.6,0) = 0 | High |
| High | High | min(0.6,0.8) = 0.6 | High |
Aggregated label strengths: Low = 0, Moderate = 0, High = 0.6.
For a strip from a to b with endpoint heights u and v, its trapezoid area is A. Its first moment M is the integral of z times the line height.
| a | b | u | v | Area A | Moment M |
|---|---|---|---|---|---|
| 50 | 65 | 0 | 0.6 | 4.5 | 270 |
| 65 | 85 | 0.6 | 0.6 | 12 | 900 |
| 85 | 100 | 0.6 | 0 | 4.5 | 405 |
Because the paper asks for a design, a different explicitly specified membership/rule system can yield a different numerical output. Explain the choices and use the chosen defuzzification method consistently; no particular examiner mark outcome is promised.
\r\n\r\n\r\nOnly the High output survives, at height .6. Its triangle is (50,75,100). Solve its rising edge: (z−50)/25=.6, so z=50+15=65. Solve its falling edge: (100−z)/25=.6, so z=100−15=85.
| Piece | Area calculation | Centre | Moment |
|---|---|---|---|
| Left triangle [50,65] | ½×15×.6=4.5 | (50+65+65)/3=60 | 4.5×60=270 |
| Rectangle [65,85] | 20×.6=12 | (65+85)/2=75 | 12×75=900 |
| Right triangle [85,100] | ½×15×.6=4.5 | (85+85+100)/3=90 | 4.5×90=405 |
The horizontal centroid of a triangle is the average of the three vertex x-coordinates. Add areas: 4.5+12+4.5=21. Add moments:270+900+405=1575. Divide 1575/21=75, agreeing with symmetry.
With reference to the Gaussian Mixture Model (GMM): compare GMM with the k-means algorithm; derive the equation for the log-likelihood function; state the number of parameters to be computed if K = 5. [10 marks]
“i.i.d.” means the observations are modelled as independent draws from the same distribution. The product Π runs over observations; the inner sum Σ runs over components. λki in some notes is the same responsibility that other sections call γik. A superscript i here labels the observation, not an exponent.
Jensen’s inequality gives a lower bound on the log of the component sum. EM chooses responsibilities making that bound touch the current likelihood, then updates the parameters to maximise the bound with those responsibilities fixed. This explains the two alternating steps; λ values are not arbitrary new model parameters.
| Parameter | Number in a 1D K-component GMM | Why |
|---|---|---|
| Means | K | One centre for each component |
| Variances | K | One spread for each component |
| Independent mixing weights | K−1 | The last weight is 1 minus the others |
| Total | 3K−1 | K+K+(K−1) |
For K=2, choose μ₁,μ₂,σ₁²,σ₂² and π₁: five free numbers. Then π₂=1−π₁. For K=5, 5+5+4=14. In 2D with a full covariance matrix, each component instead has two means and three independent covariance entries (two variances and one shared off-diagonal covariance). Five such components have 5×(2+3)+4=29 free parameters.
| Aspect | k-means | GMM |
|---|---|---|
| Assignment | Hard: each point in exactly one cluster | Soft: probability of belonging to each cluster |
| Model | Geometric: nearest centroid | Probabilistic generative model |
| Cluster shape | Spherical, similar size | Elliptical, any size/orientation; overlapping clusters |
| Parameters | K centroids | Weights, means, covariances (1-D: 3K − 1) |
| Algorithm / objective | Assign–update; minimise squared distances | EM; maximise log-likelihood |
| Output | Hard labels and fitted centroids | Labels, membership probabilities and a density (useful for anomaly detection) |
| Relation | The hard-assignment, equal spherical variance, small-variance limit of a Gaussian-mixture objective leads to the k-means objective. Ordinary finite-variance EM is not the same algorithm as k-means. Both need K in advance and depend on initialisation. |
Step 1. Model. Each point is generated by choosing component k with probability and drawing , so its density is .
Step 2. Likelihood. Points are i.i.d., so the likelihood is a product (θ = all parameters):
\r\nStep 3. Log-likelihood. Logs turn the product into a sum:
\r\nStep 4. Why it cannot be solved directly. The sum sits inside the log, so setting derivatives to zero gives no closed-form solution.
\r\nStep 5. Jensen's inequality. log is concave, so for weights : . Introduce for every point (), multiply and divide inside the log:
Step 6. Estimation (EM). Maximise the lower bound alternately. Fixing θ gives the E-step . Fixing λ and differentiating (with a Lagrange multiplier for ) gives the M-step:
The dimension and covariance structure must also be specified. For the one-dimensional, independent-variance model used in the derivation:
\r\n(5 means + 5 variances + 4 free weights, since the 5 weights must add to 1.) In two dimensions with full covariance it would be 29.
\r\n (≥, not >).Question. Weather → Flu, Weather → Cough, Flu and Cough → Fever. P(Weather) = 0.75; P(Flu | W) = 0.2, P(Flu | ¬W) = 0.5; P(Cough | W) = 0.70, P(Cough | ¬W) = 0.25; P(Fever | Flu, Cough) = 0.60, P(Fever | Flu, ¬Cough) = 0.02, P(Fever | ¬Flu, Cough) = 0.60, P(Fever | ¬Flu, ¬Cough) = 0.01. Calculate the probability of a person having fever given he doesn't have flu. [10 marks]
The following table enumerates the values of Weather and Cough that were not fixed by the question. Two Boolean choices give four rows: true/true, true/false, false/true, false/false. Each row is one mutually exclusive complete case. Read the CPT entry using that row’s parent values, multiply its factors, and add the resulting row weights. The denominator must include all cases matching the evidence, including both possible values of the queried outcome.
W→F, W→C, (F,C)→V. The complete probabilities are in the question. Evidence F=false.
P(V=true | F=false), summing hidden W and C.
Use the network joint and normalise by the no-Flu evidence.
| W | C | P(W) | P(¬F|W) | P(C|W) | P(V|¬F,C) | Product |
|---|---|---|---|---|---|---|
| T | T | 0.75 | 0.8 | 0.70 | 0.60 | 0.252 |
| T | F | 0.75 | 0.8 | 0.30 | 0.01 | 0.0018 |
| F | T | 0.25 | 0.5 | 0.25 | 0.60 | 0.01875 |
| F | F | 0.25 | 0.5 | 0.75 | 0.01 | 0.0009375 |
| P(V, ¬F) | 0.2734875 |
Question. A = {0.6/small + 0.4/medium + 0.1/large}, B = {0.4/slow + 0.8/moderate + 0.3/fast}, C = {0.5/small + 0.2/medium + 0.4/large}. Find R = A × B, S = C × B and the max–min composition S ∘ R. [10 marks]
The fuzzy sets or relation matrices and their listed row/column order.
The relation matrix/matrices and the composition specified in the question.
For Cartesian products take min of row and column membership. For the ordered matrix composition shown here:
, , . (The key gives R and S only.)Both printed relations run from size to speed; their middle labels do not match for genuine relation composition. The numerical working below follows the question’s intended row-of-S with column-of-R operation using explicitly assumed positional alignment of the three middle entries.
| Cell | Row degree | Column degree | min result |
|---|---|---|---|
| R11 | 0.6 | 0.4 | 0.4 |
| R12 | 0.6 | 0.8 | 0.6 |
| R13 | 0.6 | 0.3 | 0.3 |
| R21 | 0.4 | 0.4 | 0.4 |
| R22 | 0.4 | 0.8 | 0.4 |
| R23 | 0.4 | 0.3 | 0.3 |
| R31 | 0.1 | 0.4 | 0.1 |
| R32 | 0.1 | 0.8 | 0.1 |
| R33 | 0.1 | 0.3 | 0.1 |
| Cell | Row degree | Column degree | min result |
|---|---|---|---|
| S11 | 0.5 | 0.4 | 0.4 |
| S12 | 0.5 | 0.8 | 0.5 |
| S13 | 0.5 | 0.3 | 0.3 |
| S21 | 0.2 | 0.4 | 0.2 |
| S22 | 0.2 | 0.8 | 0.2 |
| S23 | 0.2 | 0.3 | 0.2 |
| S31 | 0.4 | 0.4 | 0.4 |
| S32 | 0.4 | 0.8 | 0.4 |
| S33 | 0.4 | 0.3 | 0.3 |
Result cell (1,1)
Result cell (1,2)
Result cell (1,3)
Result cell (2,1)
Result cell (2,2)
Result cell (2,3)
Result cell (3,1)
Result cell (3,2)
Result cell (3,3)
Question. Maximize using Particle Swarm Optimization. . Take . Optimize and give the solution after 1 iteration showing the step-by-step process. [10 marks]
P1: , . P2: , . , .
Updated velocities, positions, pbest and gbest after one iteration.
Initial personal bests equal the starting positions. Use one frozen global best for all particles. This question maximises fitness, so keep a memory only when its score is larger.
pbest = current positions. , so gbest = P2 = (4.0579, −2.2150).
Use the same objective at each new position. Keep full precision internally; displayed coordinates are rounded.
| Particle | New position | New F | Old F | pbest |
|---|---|---|---|---|
| 1 | (3.461457, −3.572109) | 38.2023 | 37.1117 | updated |
| 2 | (4.22962, −2.01448) | 40.0992 | 38.6747 | updated |
The best pbest is now P2's with F = 40.0992, so gbest moves to (4.22962, −2.01448).
, . Both particles improved; new gbest = (4.22962, −2.01448) with F = 40.0992.This X and V are particles 1 and 2 of your lecture's five-particle example; only the function differs (+3x₁ and +5 here instead of +2x₁ and +3).
\r\nQuestion. Predict whether a candidate accepts a job offer from Experience (Fresher/Mid/Senior), Expected Salary (Low/Medium/High) and Company Reputation (Good/Average). Data: (Fresher, Low, Good, Yes), (Mid, High, Good, Yes), (Senior, High, Average, No), (Fresher, Medium, Average, Yes), (Senior, Low, Good, No), (Mid, Medium, Good, Yes), (Fresher, High, Average, No), (Senior, Medium, Average, No). New candidate: Mid, Low, Good. [10 marks]
The full training data are in Step 1. Query: Experience=Mid, Salary=Low, Reputation=Good. The main calculation uses unsmoothed counts.
Both class scores and the predicted class.
C is a class label. Each aⱼ is one query feature value. First count inside a class, then multiply the prior and all three likelihoods.
| Row | Experience | Salary | Reputation | Class |
|---|---|---|---|---|
| 1 | Fresher | Low | Good | Yes |
| 2 | Mid | High | Good | Yes |
| 3 | Senior | High | Average | No |
| 4 | Fresher | Medium | Average | Yes |
| 5 | Senior | Low | Good | No |
| 6 | Mid | Medium | Good | Yes |
| 7 | Fresher | High | Average | No |
| 8 | Senior | Medium | Average | No |
| Class | Rows in this group | Class count | Prior |
|---|---|---|---|
| Yes | 1, 2, 4, 6 | 4 | 4/8 = 0.5 |
| No | 3, 5, 7, 8 | 4 | 4/8 = 0.5 |
Look only at class Yes rows 1, 2, 4, 6. The matching rows are 2, 6.
Look only at class Yes rows 1, 2, 4, 6. The matching rows are 1.
Look only at class Yes rows 1, 2, 4, 6. The matching rows are 1, 2, 6.
Look only at class No rows 3, 5, 7, 8. No row in this class has that feature value.
Look only at class No rows 3, 5, 7, 8. The matching rows are 5.
Look only at class No rows 3, 5, 7, 8. The matching rows are 5.
The larger score belongs to Yes. Normalise only if a posterior probability is requested: divide each score by their total 0.046875.
Yaad rakho: Class ke rows alag karo. Har feature ka count usi class ke total se divide. Saare factors multiply karo.
"},{"id":"q5a","title":"Q5(a) Genetic algorithm [10]","html":"Implement a GA to maximise for x in [0,50]. Take 4-bit representations of 12, 20, 17, 13. Crossover after the second bit and mutate by flipping any one bit. Produce two offspring and comment on fitness. [10 marks]
The printed data are inconsistent. 20 and 17 do not fit into four bits, and covering 0–50 needs six bits. We use six-bit encoding to preserve the supplied parent values and the stated range.
Printed parents: [12, 20, 17, 13]. Worked variant: [12, 20, 17, 13] with 6 bits. For the unspecified selection step, select the two highest-fitness candidates. For the unspecified mutation location, flip the last bit of each child. Crossover is after bit 2.
Two valid children under the declared assumptions and their fitness.
Encoding must respect the candidate range. Crossover swaps tails; mutation flips the chosen bit. Fitness is evaluated after mutation.
| x | Binary | Fitness substitution |
|---|---|---|
| 12 | 001100 | 2(12³) − 3(12²) + 8(12) = 3120 |
| 20 | 010100 | 2(20³) − 3(20²) + 8(20) = 14960 |
| 17 | 010001 | 2(17³) − 3(17²) + 8(17) = 9095 |
| 13 | 001101 | 2(13³) − 3(13²) + 8(13) = 3991 |
[20, 17] with fitness [14960, 9095].
| Parent 1 | Parent 2 | Child 1 | Child 2 |
|---|---|---|---|
| 01 | 0100 | 01 | 0001 | 010001 | 010100 |
| Before | After | Decoded x |
|---|---|---|
| 010001 | 010000 | 16 |
| 010100 | 010101 | 21 |
| x | Fitness calculation |
|---|---|
| 16 | 2(4096) − 3(256) + 8(16) = 7552 |
| 21 | 2(9261) − 3(441) + 8(21) = 17367 |
Best parent score = 14960; best child score = 17367. Parent average = 12027.5; child average = 12459.5. Both the best score and the average score increased in this worked variant. An individual child can still be worse than a parent; GA does not guarantee improvement after every random operator.
The printed wording about “two offspring” is not an explicit requirement for two complete generations. The M2 examples separately work two full generations because your class update asks for iteration 2.
\r\n\r\n\r\nCalculate x²=x×x, then x³=x²×x. Multiply each by its coefficient and combine with the written signs. A bit-string repair in the worked answer is an explicitly stated response to inconsistent printed givens; it is not a new instruction in the original paper.
Question. Bookings come via Website (W), App (A) or Counter (C). From W: 50% W, 30% A, 20% C. From A: 40% A, 40% W, 20% C. From C: 20% W, 30% A, 50% C. 1. Develop the transition matrix. 2. Interpret retention and gain/loss. 3. Draw the diagram. 4. If now 40% W, 35% A, 25% C, estimate next week. 5. Which channel grows or shrinks in the long term? [10 marks]
All transition values are given in the full question. State order: W, A, C. Initial vector [0.4, 0.35, 0.25].
The matrix, diagram, next-step vector and long-run comparison.
Rows mean current state; columns mean next state. For each destination, multiply each origin amount by its transition probability, then add.
| From / To | W | A | C |
|---|---|---|---|
| W | 0.5 | 0.3 | 0.2 |
| A | 0.4 | 0.4 | 0.2 |
| C | 0.2 | 0.3 | 0.5 |
Each row sums to 1 because it lists all possible destinations for that origin. A percentage is divided by 100: 25%=.25.
Diagonal entries are the fractions staying at their current state; off-diagonal entries are fractions moving to a different state. Each arrow uses its origin row and destination column. These fractions have different origin populations, so adding raw column probabilities does not give the number arriving.
![\"Each](\"fig/mk_booking.a7176dff51.svg\")
| Origin | Current count/share | Transition to W | Contribution |
|---|---|---|---|
| W | 0.4 | 0.5 | 0.4 × 0.5 = 0.2 |
| A | 0.35 | 0.4 | 0.35 × 0.4 = 0.14 |
| C | 0.25 | 0.2 | 0.25 × 0.2 = 0.05 |
![\"W_{new}=0.2+0.14+0.05=0.39\"](\"math/0367611df103826d.svg\")
| Origin | Current count/share | Transition to A | Contribution |
|---|---|---|---|
| W | 0.4 | 0.3 | 0.4 × 0.3 = 0.12 |
| A | 0.35 | 0.4 | 0.35 × 0.4 = 0.14 |
| C | 0.25 | 0.3 | 0.25 × 0.3 = 0.075 |
![\"A_{new}=0.12+0.14+0.075=0.335\"](\"math/5f59f5d8bf197b8d.svg\")
| Origin | Current count/share | Transition to C | Contribution |
|---|---|---|---|
| W | 0.4 | 0.2 | 0.4 × 0.2 = 0.08 |
| A | 0.35 | 0.2 | 0.35 × 0.2 = 0.07 |
| C | 0.25 | 0.5 | 0.25 × 0.5 = 0.125 |
![\"C_{new}=0.08+0.07+0.125=0.275\"](\"math/7892e8414e63bbd0.svg\")
![\"c=0.2w+0.2a+0.5c\"](\"math/39cb7355944fbe4a.svg\")
Since w+a+c=1, replace w+a with 1−c.
![\"c=0.2(1-c)+0.5c=0.2+0.3c\"](\"math/49f2613d4b58dc75.svg\")
![\"0.7c=0.2\\quad](\"math/eaaea4df9a4e16b6.svg\")
![\"a=0.3w+0.4a+0.3c\"](\"math/ac7bb8ac13980636.svg\")
![\"0.6a=0.3w+0.3c\\quad](\"math/5e60a8a8d562c8fe.svg\")
![\"w+(0.5w+0.5c)+c=1\"](\"math/748747a4be5b6069.svg\")
![\"1.5w=1-1.5c=1-3/7=4/7\"](\"math/9499078f217c4146.svg\")
![\"w=(4/7)/1.5=8/21\"](\"math/38376c6530974a73.svg\")
![\"a=1-w-c=1-8/21-6/21=7/21=1/3\"](\"math/afad15a4af697ec7.svg\")
| State | Initial share | Stationary share | Change in share | Conclusion |
|---|---|---|---|---|
| W | 0.4 | 0.380952381 | -0.019047619 | Loses |
| A | 0.35 | 0.333333333 | -0.016666667 | Loses |
| C | 0.25 | 0.285714286 | 0.035714286 | Gains |
Compare like with like: divide counts by their total before comparing with stationary probabilities. “Stationary” means one more step leaves the shares unchanged.
Question. For the four-city travelling salesman problem the distance matrix and pheromone matrix are given (all pheromones 1). Apply Ant Colony Optimization. Calculate the path travelled by 2 ants starting from city A; random numbers for Ant 1 are 0.3, 0.8 and for Ant 2 are 0.5, 0.7. Also update the pheromone matrix after 1 iteration. [10 marks]
Assume .
Both tours, their lengths and the pheromone matrix after one iteration.
Use unvisited cities in alphabetical order for cumulative selection. Every ant completes its tour using old pheromones; then evaporate and deposit on each undirected edge, including the return edge.
Cumulative: B 0–0.5455, C 0.5455–0.8182, D 0.8182–1. Ant 1 (0.3) → B; Ant 2 (0.5) → B.
Cumulative: C 0–0.5556, D 0.5556–1. Ant 1 (0.8) → D; Ant 2 (0.7) → D.
Used edges AB, BD, DC, CA get ; unused edges AD and BC become .
) and its matrix is inconsistent: it increases the unused edge A–D, leaves the used return edge C–A at 1, and gives B–D and D–B different values. If your examiner follows the no-evaporation idea, the consistent version is 1.1429 on AB, BD, CD, AC and 1 on AD, BC. The paper does not specify ρ. State the convention you use; the resulting matrix depends on it. Grading depends on the examiner’s intended convention.Question. Planes at A: 50% stay, 30% to B, 20% to C. At B: 40% to A, 40% stay, 20% to C. At C: 25% to A, 25% to B, 50% stay. 1. Construct the matrix. 2. Interpret retention and movement. 3. Draw the diagram. 4. For [30, 40, 30] find the next night's distribution. 5. Which airport is a long-term net gainer or loser? [10 marks]
All transition values are given in the full question. State order: A, B, C. Initial vector [30, 40, 30].
The matrix, diagram, next-step vector and long-run comparison.
Rows mean current state; columns mean next state. For each destination, multiply each origin amount by its transition probability, then add.
| From / To | A | B | C |
|---|---|---|---|
| A | 0.5 | 0.3 | 0.2 |
| B | 0.4 | 0.4 | 0.2 |
| C | 0.25 | 0.25 | 0.5 |
Each row sums to 1 because it lists all possible destinations for that origin. A percentage is divided by 100: 25%=.25.
Diagonal entries are the fractions staying at their current state; off-diagonal entries are fractions moving to a different state. Each arrow uses its origin row and destination column. These fractions have different origin populations, so adding raw column probabilities does not give the number arriving.
![\"Each](\"fig/mk_airline.779a8c5fd6.svg\")
| Origin | Current count/share | Transition to A | Contribution |
|---|---|---|---|
| A | 30 | 0.5 | 30 × 0.5 = 15 |
| B | 40 | 0.4 | 40 × 0.4 = 16 |
| C | 30 | 0.25 | 30 × 0.25 = 7.5 |
![\"A_{new}=15+16+7.5=38.5\"](\"math/67abb9b77cf134f6.svg\")
| Origin | Current count/share | Transition to B | Contribution |
|---|---|---|---|
| A | 30 | 0.3 | 30 × 0.3 = 9 |
| B | 40 | 0.4 | 40 × 0.4 = 16 |
| C | 30 | 0.25 | 30 × 0.25 = 7.5 |
![\"B_{new}=9+16+7.5=32.5\"](\"math/c43c2e3add5e9f7a.svg\")
| Origin | Current count/share | Transition to C | Contribution |
|---|---|---|---|
| A | 30 | 0.2 | 30 × 0.2 = 6 |
| B | 40 | 0.2 | 40 × 0.2 = 8 |
| C | 30 | 0.5 | 30 × 0.5 = 15 |
![\"C_{new}=6+8+15=29\"](\"math/e1d10b6213a4f6c2.svg\")
![\"c=0.2(a+b)+0.5c\"](\"math/063ad466582b38bf.svg\")
![\"b=0.3a+0.4b+0.25c\"](\"math/8b2ff5dead1627ab.svg\")
![\"0.6b=0.3a+0.25c\"](\"math/463b4822655ff527.svg\")
![\"b=0.5a+\\frac5{12}c\"](\"math/c6b1f0d1f25d8a7c.svg\")
![\"a+0.5a+\\frac5{12}\\frac27=1-\\frac27=\\frac57\"](\"math/2e4dfec7416e59d8.svg\")
![\"1.5a=\\frac57-\\frac5{42}=\\frac{25}{42}\"](\"math/dbf72a6e6a9bedfd.svg\")
![\"a=\\frac{25}{42}\\div\\frac32=\\frac{25}{63}\"](\"math/8653487298ec14a0.svg\")
![\"b=1-\\frac{25}{63}-\\frac{18}{63}=\\frac{20}{63}\"](\"math/369b239ecbe26cbf.svg\")
| State | Initial share | Stationary share | Change in share | Conclusion |
|---|---|---|---|---|
| A | 0.3 | 0.396825397 | 0.096825397 | Gains |
| B | 0.4 | 0.317460317 | -0.082539683 | Loses |
| C | 0.3 | 0.285714286 | -0.014285714 | Loses |
Compare like with like: divide counts by their total before comparing with stationary probabilities. “Stationary” means one more step leaves the shares unchanged.
List the applications of fuzzy logic. Explain any one application in detail. [10 marks]
Fuzzy logic is a multi-valued logic, introduced by Lotfi Zadeh (1965), in which truth values range over [0, 1]; it lets machines reason with vague, human-like terms such as “slightly dirty” or “very hot”.
\r\n| Area | Examples |
|---|---|
| Home appliances | Washing machines (wash time from dirt and load), air conditioners, refrigerators, rice cookers, vacuum cleaners |
| Automotive | Automatic transmission, anti-lock braking (ABS), cruise control, engine control |
| Transportation | Traffic-signal control, Sendai subway train control (smooth braking and acceleration) |
| Consumer electronics | Camera autofocus and auto-exposure, image stabilisation |
| Industrial control | Cement kilns, pH, temperature and distillation control in chemical plants |
| Aerospace | Satellite and spacecraft attitude control |
| Medicine | Diagnosis support, anaesthesia dosage control |
| Finance and decisions | Loan approval, credit and fire-risk scoring, stock trading rules |
| AI and robotics | Robot navigation, natural-language interfaces, expert systems |
Advantages: handles imprecise or noisy data; mimics human reasoning; easy to design and modify; no exact model needed; low computational cost. Disadvantages: rules and membership functions depend on expert judgement; results are hard to verify formally; tuning can take time; not suited where very high precision is required.
\r\nExplain the Bellman equation in reinforcement learning. Also give applications of reinforcement learning. [10 marks]
Write the answer below in this order. It fits in about two pages.
\n\n1. Setting. Reinforcement learning problems are modelled as a Markov decision process . The agent follows a policy and wants to maximise the return .
2. Value functions. and .
3. Key idea. The return is recursive: . Taking expectations gives the Bellman expectation equation:
4. Optimality. For the optimal policy the average over actions becomes a maximum (Bellman optimality equation):
\n5. Meaning and use. The value of a state equals the immediate reward plus the discounted value of the successor state. This breaks a long-horizon problem into one-step sub-problems (dynamic programming). It is the basis of policy evaluation, policy iteration, value iteration and Q-learning; the optimal policy is .
6. Small illustration (adds clarity): if staying in a state gives +1 each step with , then .
| Agent | Controller at an intersection. |
|---|---|
| State | Queue length and waiting time on each approach, current signal phase. |
| Actions | Keep the current phase or switch to the next green phase. |
| Reward | Negative of total waiting time (or vehicles cleared per cycle). |
| Learning | Q-learning updates from observed traffic; ε-greedy exploration tries alternative timings. |
| Benefit | Adapts to rush hours and incidents; fixed timers cannot. |
| Challenge | Safety during exploration, so agents are trained in simulation first. |
| Machine | No Fault (X = 0) | Fault (X = 1) |
|---|---|---|
| Machine A | 0.40 | 0.05 |
| Machine B | 0.25 | 0.10 |
| Machine C | 0.20 | 0.08 |
| Machine D | 0.05 | 0.07 |
A factory uses four machines (Machine A, Machine B, Machine C, and Machine D) for production. The above table shows the joint probability distribution of a fault occurring (X) and the machine type (Y).
The eight printed entries are used exactly as shown. X=1 means fault; Y identifies the machine.
P(C | fault), P(fault), and P(fault | B).
![\"P(A\\mid](\"math/b482646983d755cb.svg\")
Conditioning restricts the denominator to the given group. Before treating a whole table as a joint probability distribution, its entries must sum to 1.
![\"\\text{No-fault](\"math/6388bf0e029e7de3.svg\")
![\"\\text{Fault](\"math/3c42285ce038d0e3.svg\")
![\"\\text{Grand](\"math/017ee19616d63679.svg\")
Source inconsistency: as printed, this is not a valid normalized probability table. Treating the entries as proportional nonnegative masses is one explicit repair; dividing every entry by 1.20 preserves conditional ratios.
![\"P(C\\mid](\"math/ec2f2af190c229ec.svg\")
Among all fault mass, 0.08 belongs to C. The common normalization factor cancels.
![\"P(X=1)=\\frac{0.30}{1.20}=0.25\"](\"math/2d7f62f0b50223d0.svg\")
Answer under mass normalization: 25%. The raw column sum 0.30 is only the unnormalized fault mass. If the examiner intended another corrected entry, the unconditional probability depends on that correction; 0.30 cannot be asserted as a valid probability while retaining all eight printed numbers.
![\"P(X=1\\mid](\"math/a88bec71b29e2cf3.svg\")
Machine B is the denominator because the product is known to come from B. Final conditional answers: 26.6667% and 28.5714%.
List the applications of genetic algorithm and explain any one application in detail.
Genetic algorithm (GA) is a population-based search method that evolves encoded candidate solutions using fitness evaluation, selection, crossover and mutation.
| Stage | How it is used |
|---|---|
| Representation | A chromosome assigns a time slot and room to every examination. A gene identifies one exam's assignment. |
| Fitness | Penalize student clashes, room-capacity violations and unavailable slots; among feasible schedules, reward spread-out exams and efficient room use. |
| Initial population | Create several different candidate timetables, using random assignments or constructive rules. |
| Selection | Prefer feasible and lower-penalty timetables while retaining diversity. |
| Crossover | Combine assignments from two parents; repair conflicts if the operator can violate feasibility. |
| Mutation | Move an exam to another permitted slot or swap compatible assignments. Recheck constraints. |
| Replacement / stopping | Keep useful candidates, often protecting the best feasible schedule; stop at a generation/time limit or sufficient improvement. |
Illustrative example: two exams sharing students cannot occupy the same slot. A candidate with that clash receives a strong penalty or is repaired before acceptance. This shows why fitness must reflect the actual constraints, not merely how many rooms are occupied.
Benefit: explores many combinations without exhaustively listing all timetables. Limitation: stochastic search can stagnate and does not guarantee a global optimum; poorly designed penalties can reward unusable schedules.
Consider two fuzzy sets
X={(c1,0.6),(c2,0.8),(c3,0.3),(c4,0.5),(c5,0.7)}
Y={(c1,0.4),(c2,0.5),(c3,0.9),(c4,0.6),(c5,0.2)}
Find 1) X∪Y 2) X∩Y 3) ¬X∪¬Y 4) ¬X∩¬Y
The same universe {c1, c2, c3, c4, c5} is used for both sets.
All four membership vectors, keeping the element order fixed.
![\"\\mu_{X\\cup](\"math/f1157faae69b5acb.svg\")
![\"\\mu_{\\neg](\"math/094fb09a37beb92f.svg\")
Union takes the larger membership, intersection the smaller, and standard complement subtracts from 1. Apply each operation separately at the same element.
| Element | X | Y | 1−X | 1−Y |
|---|---|---|---|---|
| c1 | 0.6 | 0.4 | 1−0.6=0.4 | 1−0.4=0.6 |
| c2 | 0.8 | 0.5 | 1−0.8=0.2 | 1−0.5=0.5 |
| c3 | 0.3 | 0.9 | 1−0.3=0.7 | 1−0.9=0.1 |
| c4 | 0.5 | 0.6 | 1−0.5=0.5 | 1−0.6=0.4 |
| c5 | 0.7 | 0.2 | 1−0.7=0.3 | 1−0.2=0.8 |
| Element | X∪Y = max (X, Y) | X∩Y = min (X, Y) | ¬X∪¬Y | ¬X∩¬Y |
|---|---|---|---|---|
| c1 | max(0.6,0.4)=0.6 | min(0.6,0.4)=0.4 | max(0.4,0.6)=0.6 | min(0.4,0.6)=0.4 |
| c2 | max(0.8,0.5)=0.8 | min(0.8,0.5)=0.5 | max(0.2,0.5)=0.5 | min(0.2,0.5)=0.2 |
| c3 | max(0.3,0.9)=0.9 | min(0.3,0.9)=0.3 | max(0.7,0.1)=0.7 | min(0.7,0.1)=0.1 |
| c4 | max(0.5,0.6)=0.6 | min(0.5,0.6)=0.5 | max(0.5,0.4)=0.5 | min(0.5,0.4)=0.4 |
| c5 | max(0.7,0.2)=0.7 | min(0.7,0.2)=0.2 | max(0.3,0.8)=0.8 | min(0.3,0.8)=0.3 |
X∪Y={(c1,0.6),(c2,0.8),(c3,0.9),(c4,0.6),(c5,0.7)}.
X∩Y={(c1,0.4),(c2,0.5),(c3,0.3),(c4,0.5),(c5,0.2)}.
¬X∪¬Y={(c1,0.6),(c2,0.5),(c3,0.7),(c4,0.5),(c5,0.8)}.
¬X∩¬Y={(c1,0.4),(c2,0.2),(c3,0.1),(c4,0.4),(c5,0.3)}.
Yaad rakho: union bada, intersection chhota, complement one minus.
A city is served by two cable TV companies, BestTV and CableCast. Due to their aggressive sales tactics, each year 40% of BestTV customers switch to CableCast; the other 60% of BestTV customers stay with BestTV. On the other hand, 30% of the CableCast customers switch to Best TV. The two states in this example are BestTV and CableCast.
Order the states as B=BestTV and C=CableCast.
The one-year transition matrix and all directed state transitions.
![\"P_{ij}=P(\\text{next](\"math/ac80fe221bb10516.svg\")
Rows mean current state; columns mean next state. Each row includes every possible destination.
BestTV stays =1−0.40=0.60. CableCast stays =1−0.30=0.70.
![\"P=\\begin{bmatrix}0.60&0.40\\\\0.30&0.70\\end{bmatrix}_{(B,C)}\"](\"math/f73b0238f5d9fefe.svg\")
| Current → next | BestTV | CableCast | Row total |
|---|---|---|---|
| BestTV | 0.6 | 0.4 | 1 |
| CableCast | 0.3 | 0.7 | 1 |
![\"BestTV](\"fig/legacy25-cable.c15b41f741.svg\")
Create a fuzzy logic system to assist in investment decision-making. The system should consider variables such as risk tolerance, and historical performance to provide recommendations on asset allocation and investment strategies.
Input variables: Risk Tolerance (range 1-100) and historical performance (range 1-100). Output variable is Asset allocation (range 1-100) (Conservative, Balanced and Aggressive). Use 3 descriptors for input as well as output. Set rules to accurately predict the output. Show defuzzification for risk 40 and historical performance 8.
Printed inputs: risk=40, historical performance=8, both on 1–100; output allocation score on 1–100. Membership shapes, breakpoints, rules and inference method are design choices, not printed data.
One fully specified fuzzy system and the crisp output for these inputs.
Use a Mamdani system: fuzzify inputs → apply min for AND → clip each output membership at its rule strength → aggregate by max → centroid. The result is a defined model output, not a guaranteed real investment return.
![\"z^*=\\frac{\\int_1^{100}z\\mu_{agg}(z)\\,dz}{\\int_1^{100}\\mu_{agg}(z)\\,dz}\"](\"math/3619dbb6f7443588.svg\")
For either input t∈[1,100], choose Low as a left shoulder, Medium as a triangle and High as a right shoulder:
![\"\\mu_L(t)=\\begin{cases}(50-t)/49&1\\le](\"math/6b2ac0fe5ac34f60.svg\")
![\"\\mu_M(t)=\\begin{cases}(t-1)/49&1\\le](\"math/6ad9728f32af11c4.svg\")
![\"\\mu_H(t)=\\begin{cases}0&1\\le](\"math/e4429c897c410965.svg\")
Use the same three shapes on output z, named Conservative, Balanced and Aggressive. The score is an increasing aggressiveness index; it is not automatically a percentage in one specified asset. Higher scores recommend a more aggressive strategy within this illustrative model.
![\"Declared](\"fig/legacy25-memberships.904ed2cacf.svg\")
| Risk tolerance | Historical performance | Output |
|---|---|---|
| Low | Low | Conservative |
| Low | Medium | Conservative |
| Low | High | Conservative |
| Medium | Low | Conservative |
| Medium | Medium | Balanced |
| Medium | High | Balanced |
| High | Low | Balanced |
| High | Medium | Aggressive |
| High | High | Aggressive |
The rule base respects low risk tolerance even if past performance is high. Medium risk plus poor past performance moves towards Conservative. High risk with medium/high performance allows Aggressive. These are declared illustrative design judgments; other valid designs can yield other outputs.
| Input | Low | Medium | High |
|---|---|---|---|
| Risk 40 | (50−40)/49=10/49=0.204082 | (40−1)/49=39/49=0.795918 | 0 |
| Performance 8 | (50−8)/49=42/49=0.857143 | (8−1)/49=7/49=1/7=0.142857 | 0 |
8 is retained exactly; it is not replaced with 80.
| Rule | AND calculation | Firing strength | Consequent |
|---|---|---|---|
| Low & Low | min(10/49,42/49) | 10/49 | Conservative |
| Low & Medium | min(10/49,7/49) | 7/49 | Conservative |
| Medium & Low | min(39/49,42/49) | 39/49 | Conservative |
| Medium & Medium | min(39/49,7/49) | 7/49 | Balanced |
The other five rules have a zero antecedent and do not fire. Maximum aggregation gives Conservative height 39/49, Balanced height 1/7 and Aggressive height 0.
![\"\\mu_{agg}(z)=\\max\\{\\min(39/49,\\mu_L(z)),\\min(1/7,\\mu_M(z))\\}\"](\"math/b4a4496253a59f9d.svg\")
The Conservative clip meets its descending slope at z=11. It falls to 1/7 at z=43. The Balanced descending side reaches 1/7 at z=650/7≈92.857143.
![\"\\mu_{agg}(z)=\\begin{cases}39/49&1\\le](\"math/b2489917bec1091e.svg\")
![\"Actual](\"fig/legacy25-centroid.6d2be43cdb.svg\")
For a segment µ(z)=mz+c between a and b, integrate its area A and first moment M as follows:
![\"A=\\frac](\"math/1c36be6d1316a809.svg\")
| Interval | Membership | Area A | First moment M |
|---|---|---|---|
| 1 to 11 | 39/49 | 7.959183673 | 47.755102041 |
| 11 to 43 | (50−z)/49 | 15.020408163 | 349.823129252 |
| 43 to 650/7 | 1/7 | 7.122448980 | 483.817784257 |
| 650/7 to 100 | (100−z)/50 | 0.510204082 | 48.590864917 |
For example, the first rectangle has area (11−1)×39/49=390/49 and moment (39/49)×(11²−1²)/2=2340/49. The other areas and moments follow explicitly: interval 11–43 has A=[−z²/98+50z/49] from 11 to 43=15.020408163 and M=[−z³/147+25z²/49] from 11 to 43=349.823129252. Interval 43–650/7 has A=(650/7−43)/7=7.122448980 and M=((650/7)²−43²)/14=483.817784257. Interval 650/7–100 has A=[2z−z²/100] from 650/7 to 100=0.510204082 and M=[z²−z³/150] from 650/7 to 100=48.590864917.
![\"\\sum](\"math/302b0a0b55fb1874.svg\")
![\"\\sum](\"math/5cd3d9be4d829cea.svg\")
![\"z^*=\\frac{637971/686}{1500/49}=\\frac{212657}{7000}\\approx\\boxed{30.379571}\"](\"math/19f8c5be600656c4.svg\")
The resulting allocation score leans Conservative for this declared rule base. A centroid is obtained from the aggregated area; do not replace this Mamdani calculation with an unexplained weighted average of label names.
With reference to Gaussian Mixture model (GMM), answer the following
Gaussian mixture model: a probability density formed by a weighted sum of K Gaussian components. Each observation has a latent component label; its membership is generally represented by responsibilities rather than a single compulsory hard assignment.
| Aspect | K-means | GMM |
|---|---|---|
| Assignment | Hard: choose one closest centroid | Soft: a posterior responsibility for each component |
| Parameters | K centroid vectors | Mixture weights, mean vectors and covariance matrices |
| Objective | Minimize within-cluster squared Euclidean distances | Maximize data likelihood or log likelihood |
| Shape | Nearest-centroid partitions suit roughly spherical/equal-scale groups | Full covariance allows tilted ellipsoids; restricted covariance gives simpler shapes |
| Updates | Alternate assignment and centroid averaging | EM: responsibility calculation and weighted parameter updates |
| Uncertainty | No probability from basic assignment alone | Responsibilities quantify ambiguity under the fitted model |
![\"p(x_i\\mid\\theta)=\\sum_{k=1}^K\\pi_k\\,\\mathcal](\"math/b313ea1944e92313.svg\")
![\"\\mathcal](\"math/e8919c0c3aee3455.svg\")
πk is the probability of component k; µk is its centre; Σk describes spread and correlation. Assume positive-definite full covariances for this density.
![\"\\mathcal](\"math/e3640e9995db4c3c.svg\")
The product follows from the usual independent, identically distributed training-sample assumption. The sum inside handles the unobserved component choice.
![\"\\ell(\\theta)=\\log\\mathcal](\"math/ac7ba81fe410766e.svg\")
Log turns the product across samples into a sum. It does not turn the mixture sum inside each logarithm into a sum of logarithms. The coupled terms motivate iterative optimization such as EM.
The question gives K=5 but does not give dimension d or covariance structure, so one universal integer cannot be determined.
| Component | Independent parameters |
|---|---|
| Means | Kd |
| Full symmetric covariances | K d (d+1)/2 |
| Mixture weights | K−1, because their sum is 1 |
![\"n_{free}=Kd+K\\frac{d(d+1)}2+(K-1)\"](\"math/24096c582fbc12b9.svg\")
![\"K=5:\\quad](\"math/d72da4a5707e5b6a.svg\")
If one-dimensional data are intended: five means+five variances+four independent weights=14 free parameters. One implementation stores 15 numbers (including all five weights), but the sum-to-one constraint removes one independent degree of freedom. For d-dimensional diagonal covariances the count is 10d+4; state the selected covariance assumption.
![\"Original](\"fig/legacy25-bayes.7f566dbf34.svg\")
| Smoking S | P(S) |
|---|---|
| T | 0.3 |
| F | 0.7 |
| Smoking S | P(C=T|S) | P(C=F|S) |
|---|---|---|
| T | 0.1 | 0.9 |
| F | 0.01 | 0.99 |
| Pneumonia P | P(P) |
|---|---|
| T | 0.3 |
| F | 0.7 |
| C | P | P(Co=T|C, P) | P(Co=F|C, P) |
|---|---|---|---|
| T | T | 0.95 | 0.05 |
| T | F | 0.8 | 0.2 |
| F | T | 0.7 | 0.3 |
| F | F | 0.1 | 0.9 |
| P | P(Fever=T|P) | P(Fever=F|P) |
|---|---|---|
| T | 0.9 | 0.1 |
| F | 0.01 | 0.99 |
Given the above tables for the Bayesian network, calculate Probability of a person having cough given he doesn’t have lung cancer but Smokes.
Evidence: C=F (no lung cancer), S=T (smokes). Query: Co=T. Pneumonia is not observed; sum over its two states.
P(Co=T | C=F, S=T).
![\"P(Co\\mid\\neg](\"math/09e203ebfdf53e2b.svg\")
The network says cough depends on its parents C and P. Once they are known, smoking provides no additional direct information about cough. The pneumonia root is independent of S and C in this network when no descendant evidence such as cough/fever is conditioned on.
![\"P(S,C,P,Co,F)=P(S)P(C\\mid](\"math/dfc06603bb26a3be.svg\")
This factorization expresses the model, not a general medical truth. Fever is unobserved and its two conditional probabilities sum to 1.
| Hidden pneumonia state | Cough probability with C=F | Pneumonia weight |
|---|---|---|
| P=T | P(Co=T|C=F, P=T)=0.70 | 0.30 |
| P=F | P(Co=T|C=F, P=F)=0.10 | 0.70 |
Do not use 0.95 or 0.80: those rows assume lung cancer is present, which contradicts the given evidence.
![\"0.70\\times0.30=0.21,\\qquad0.10\\times0.70=0.07\"](\"math/e8bf2d5761946110.svg\")
![\"P(Co=T\\mid](\"math/b36e6cc824d0d47d.svg\")
Final answer:28%. We do not multiply the final answer byP(smoking)=0.3: smoking is already a condition, not an additional joint event.
![\"P(\\neg](\"math/27b16cf13dc13276.svg\")
![\"P(Co,\\neg](\"math/51760d268143a4b5.svg\")
![\"P(Co\\mid\\neg](\"math/ad2987ff1355d124.svg\")
Consider 2 fuzzy sets
A={0.8/small+0.5/medium+0.2/large}
B={0.6/slow+0.9/moderate+0.4/fast}
A and C use size universe X={small, medium, large}; B uses speed universe Y={slow, moderate, fast}. Use the listed orders for rows and columns.
R and S; then explain and resolve the composition convention explicitly.
![\"\\mu_{A\\times](\"math/e9a9f703d571640c.svg\")
Fuzzy Cartesian product uses minimum here, not multiplication. A relation composition additionally needs the first relation's output universe to match the second's input universe.
| Size / speed | slow 0.6 | moderate 0.9 | fast 0.4 |
|---|---|---|---|
| small 0.8 | min(0.8,0.6)=0.6 | min(0.8,0.9)=0.8 | min(0.8,0.4)=0.4 |
| medium 0.5 | min(0.5,0.6)=0.5 | min(0.5,0.9)=0.5 | min(0.5,0.4)=0.4 |
| large 0.2 | min(0.2,0.6)=0.2 | min(0.2,0.9)=0.2 | min(0.2,0.4)=0.2 |
![\"R=\\begin{bmatrix}0.6&0.8&0.4\\\\0.5&0.5&0.4\\\\0.2&0.2&0.2\\end{bmatrix}\"](\"math/52316a1420d445ac.svg\")
| Size / speed | slow 0.6 | moderate 0.9 | fast 0.4 |
|---|---|---|---|
| small 0.7 | min(0.7,0.6)=0.6 | min(0.7,0.9)=0.7 | min(0.7,0.4)=0.4 |
| medium 0.3 | min(0.3,0.6)=0.3 | min(0.3,0.9)=0.3 | min(0.3,0.4)=0.3 |
| large 0.5 | min(0.5,0.6)=0.5 | min(0.5,0.9)=0.5 | min(0.5,0.4)=0.4 |
![\"S=\\begin{bmatrix}0.6&0.7&0.4\\\\0.3&0.3&0.3\\\\0.5&0.5&0.4\\end{bmatrix}\"](\"math/0a4c516a6ffc1e2a.svg\")
As typed relations, both R and S map X→Y. S cannot directly follow R because S expects a size while R ends in a speed. Equal matrix dimensions do not make the universes equal. Thus the requested composition has no unique meaningful typed interpretation as written.
Conditional classroom calculation: temporarily identify intermediate labels by position 1,2,3 and define the requested notation here as T=S⊙R, where⊙ is max–min matrix multiplication. This is an index convention, not a claim that “slow” equals “small”.
| Entry | Three minimum terms | Maximum |
|---|---|---|
| T11 | min(0.6,0.6)=0.6, min(0.7,0.5)=0.5, min(0.4,0.2)=0.2 | 0.6 |
| T12 | min(0.6,0.8)=0.6, min(0.7,0.5)=0.5, min(0.4,0.2)=0.2 | 0.6 |
| T13 | min(0.6,0.4)=0.4, min(0.7,0.4)=0.4, min(0.4,0.2)=0.2 | 0.4 |
| T21 | min(0.3,0.6)=0.3, min(0.3,0.5)=0.3, min(0.3,0.2)=0.2 | 0.3 |
| T22 | min(0.3,0.8)=0.3, min(0.3,0.5)=0.3, min(0.3,0.2)=0.2 | 0.3 |
| T23 | min(0.3,0.4)=0.3, min(0.3,0.4)=0.3, min(0.3,0.2)=0.2 | 0.3 |
| T31 | min(0.5,0.6)=0.5, min(0.5,0.5)=0.5, min(0.4,0.2)=0.2 | 0.5 |
| T32 | min(0.5,0.8)=0.5, min(0.5,0.5)=0.5, min(0.4,0.2)=0.2 | 0.5 |
| T33 | min(0.5,0.4)=0.4, min(0.5,0.4)=0.4, min(0.4,0.2)=0.2 | 0.4 |
![\"T=S\\odot](\"math/fef1c19f50df11d5.svg\")
If the course instead writes S∘R to mean first R then S, its untyped matrix computation is R⊙S, which yields rows(0.6,0.6,0.4), (0.5,0.5,0.4), (0.2,0.2,0.2) for these values. The convention must be declared; silently reversing the factors changes the answer.
One possible corrected task is to compose R:X→Y with the inverse relation Sᵀ:Y→X. Then compare size elements through a common speed:
![\"U=R\\odot](\"math/3a687cf430317d18.svg\")
![\"U_{11}=\\max\\{\\min(0.6,0.6),\\min(0.8,0.7),\\min(0.4,0.4)\\}=0.7\"](\"math/62612d777f597842.svg\")
This last matrix answers the explicitly repaired task, not an unannounced correction to the printed question.
Maximize the F(x) = x1² − x1.x2 + x2² + 3x1 + 4x2 + 5 using Particle Swarm Optimization.
Take values W=0.9, c1=c2=1.5, r1=0.5949, r2=0.0855. Optimize and give solution after 1 iteration showing step by step process.
F(x1, x2)=x1²−x1x2+x2²+3x1+4x2+5; inertia 0.9; c1=c2=1.5; r1=0.5949; r2=0.0855.
One PSO update, while stating why the printed information does not determine a unique swarm result.
![\"v_{ij}^{t+1}=wv_{ij}^t+c_1r_1(pbest_{ij}-x_{ij}^t)+c_2r_2(gbest_j-x_{ij}^t)\"](\"math/7d9ccf789fe769f1.svg\")
![\"x_{ij}^{t+1}=x_{ij}^t+v_{ij}^{t+1}\"](\"math/566cb0a85349654b.svg\")
Velocity combines inertia, attraction to a particle's own best and attraction to the swarm's best. For maximization, a larger objective updates the best memories.
The question gives no population size, initial positions, velocities, pbest/gbest or coordinate bounds. Different initializations therefore give different one-iteration results. Also, along x1=x2=t:
![\"F(t,t)=t^2+7t+5\\longrightarrow+\\infty\\quad(t\\to+\\infty)\"](\"math/b88e59abd3293f46.svg\")
There is no finite global maximum on an unrestricted plane. A bounded domain would be required for a finite maximization target.
Using the paper's instruction to assume suitable data where necessary, assume two particles: A=(0,0), B=(1,1); both velocities=(0,0); initial pbest equals the starting position. Use the printed random values in both coordinates and freeze the old gbest during this synchronous update.
| Particle | Position | Velocity | Objective | Initial pbest |
|---|---|---|---|---|
| A | (0,0) | (0,0) | 5 | (0,0) |
| B | (1,1) | (0,0) | 1−1+1+3+4+5=13 | (1,1) |
The old gbest is B=(1,1), with value 13. These particle coordinates are assumptions, not data from the scan.
![\"c_1r_1=1.5(0.5949)=0.89235,\\quad](\"math/df63d53634007eba.svg\")
The cognitive term is zero for both particles because each starts at its own pbest.
![\"v_{A1}^{1}=0.9(0)+0.89235(0-0)+0.12825(1-0)=0.12825\"](\"math/9c531ea472a5b2fe.svg\")
![\"v_{A2}^{1}=0.9(0)+0.89235(0-0)+0.12825(1-0)=0.12825\"](\"math/f5e85d68eed20ee1.svg\")
![\"x_A^1=(0+0.12825,0+0.12825)=(0.12825,0.12825)\"](\"math/3d974cd8e38814b0.svg\")
![\"F(x_A^1)=(0.12825)^2+7(0.12825)+5=5.9141980625\"](\"math/7930f3e2597545dd.svg\")
This exceeds 5, so A's pbest becomes its new position.
![\"v_{B1}^{1}=v_{B2}^{1}=0.9(0)+0.89235(1-1)+0.12825(1-1)=0\"](\"math/86465ec4ca89266b.svg\")
B remains at(1,1), with objective 13. Compare 13 against 5.9141980625: the best found after this assumed iteration is still (1,1), F=13. This is only the best of the two evaluated particles, not the nonexistent unrestricted global maximum.
| Age | Income Level | Previous Purchase | Buy Product |
|---|---|---|---|
| Young | Low | No | No |
| Middle-aged | Medium | Yes | Yes |
| Senior | High | No | No |
| Young | Medium | Yes | Yes |
| Senior | Low | No | No |
| Middle-aged | High | Yes | Yes |
| Young | Low | Yes | Yes |
| Senior | Medium | No | No |
A company wants to predict whether a customer will buy a product, based on the following features: Age (Young, Middle-aged, Senior), Income Level (Low, Medium, High), Previous Purchase (Yes, No). The output is Buy Product (Yes, No). Given a new customer with the following attributes: Age=Young, Income Level=High, Previous Purchase=No. Use Naive Bayes to predict whether this customer will buy the product.
Eight training rows; query=(Young, High, No). No smoothing instruction is printed.
The two class scores and the larger-posterior class.
![\"s_c=P(c)P(Young\\mid](\"math/0629d7338a1ecd30.svg\")
Naive Bayes assumes conditional independence of these features given the class. The evidence denominator is common to both classes, so compare the unnormalized scores first.
Yes occurs in rows 2,4,6,7:4 rows. No occurs in rows 1,3,5,8:4 rows.
![\"P(Yes)=4/8=1/2,\\quad](\"math/c54c9411a1740830.svg\")
| Query feature | Count among Yes /4 | Count among No /4 |
|---|---|---|
| Young | 2/4 | 1/4 |
| High income | 1/4 | 1/4 |
| Previous purchase No | 0/4 | 4/4 |
The denominator is the class count 4, not the full dataset 8.
![\"s_{Yes}=\\frac12\\times\\frac24\\times\\frac14\\times\\frac04=0\"](\"math/7af34167018bca92.svg\")
The zero comes from no training Yes row having Previous Purchase=No.
![\"s_{No}=\\frac12\\times\\frac14\\times\\frac14\\times\\frac44=\\frac1{32}=0.03125\"](\"math/e376b189fb5efc12.svg\")
![\"P(Yes\\mid](\"math/6e2299fa87aa4056.svg\")
Predicted class: No, under the unsmoothed model. A posterior of 1 here is a consequence of tiny-data maximum-likelihood estimates and a zero conditional count; it is not certain knowledge about a real customer.
With add-one smoothing, Age and Income each have 3 categories and Previous Purchase has 2. Keep the class priors 1/2. Then:
![\"s_{Yes}^{(+1)}=\\frac12\\frac37\\frac27\\frac16=\\frac1{98},\\quad](\"math/df5992f5cc97daef.svg\")
![\"P_{+1}(Yes\\mid](\"math/c94cdf7b26810ac7.svg\")
The decision is still No. This is a labeled alternative, not a silent change to the printed task.
Implement genetic algorithm to maximize the function f (x)=x³−4x²+6x with x in the range [0,15]. Take 4 bit representation of 10,12,7 and 3 as the parent chromosomes. Crossover is after 2 nd bit and mutation is by flipping any one bit. Repeat the process for 2 offspring also comment on the fitness of the offspring.
Parents 10,12,7,3; unsigned 4-bit encoding; crossover after the second bit; one bit may be flipped. Pairing and mutation positions are not specified.
Crossover children, explicit mutated chromosomes and objective comparison.
![\"x=8b_1+4b_2+2b_3+b_4,\\qquad](\"math/7000daead38032a9.svg\")
Split after the second bit, swap the right halves, then flip exactly the declared mutation bit. Fitness is the stated objective; maximize it.
| Parent x | 4-bit code | Fitness calculation | f (x) |
|---|---|---|---|
| 10 | 1010 | 10³−4(10²)+6(10) | 660 |
| 12 | 1100 | 12³−4(12²)+6(12) | 1224 |
| 7 | 0111 | 7³−4(7²)+6(7) | 189 |
| 3 | 0011 | 3³−4(3²)+6(3) | 9 |
Pair the parents in their listed order: (10,12) and(7,3). Each pair produces two reciprocal children. This uses all four supplied parents; “2 offspring” is interpreted as the two children from a crossover pair, not as two complete generations.
| Pair | Split parents | Child 1 | Child 2 |
|---|---|---|---|
| 10 with 12 | 10|10 and 11|00 | 10|00=1000₂=8 | 11|10=1110₂=14 |
| 7 with 3 | 01|11 and 00|11 | 01|11=0111₂=7 | 00|11=0011₂=3 |
Choose the rightmost bit (bit 4 from the left, weight 1) for every child. This is allowed by “any one bit”; it is not the only possible mutation outcome.
| Before mutation | Flip bit 4 | After mutation | Decimal |
|---|---|---|---|
| 1000 | 0→1 | 1001 | 9 |
| 1110 | 0→1 | 1111 | 15 |
| 0111 | 1→0 | 0110 | 6 |
| 0011 | 1→0 | 0010 | 2 |
| Child x before mutation | f (x) before | Child x after mutation | f (x) after |
|---|---|---|---|
| 8 | 304 | 9 | 459 |
| 14 | 2044 | 15 | 2565 |
| 7 | 189 | 6 | 108 |
| 3 | 9 | 2 | 4 |
![\"f(15)=15^3-4(15^2)+6(15)=3375-900+90=2565\"](\"math/5600d463abeea9e2.svg\")
The best supplied parent is 12 with 1224. The best mutated child is 15 with 2565, an improvement. Mutation did not help every child:7→6 lowers 189 to 108, and 3→2 lowers 9 to 4. Crossover/mutation generate candidates; fitness evaluation decides which are retained.
For these declared pairings and mutations, offspring are 9,15,6,2 and the best is 15. Different allowed mutation positions can give different offspring. If only two children are requested in a short answer, the fully worked first pair supplies 9 and 15; the second pair shows how the remaining supplied parents are handled. On the finite integer domain 0…15, direct evaluation confirms 15 gives the largest objective, but one generic GA generation is not a proof of global optimality for arbitrary problems.
Around 600 cities globally have bike-share programs where users borrow and return bikes at various stations. The distribution of bikes changes daily as they move between stations. Considering three stations (A, B, C), we can track bike movement from midnight to the next midnight.
States ordered A, B, C; one step is one day. The “around 600 cities” sentence is preserved as historical question context, not asserted as a current statistic.
Transition matrix, retention/outflow/inflow interpretation and state diagram.
![\"\\mathbf](\"math/f39ae5af858943b8.svg\")
Use row vectors: rows of P are departure stations, columns arrival stations. Multiplying the initial counts by P gives expected next-day counts. Without starting counts, incoming percentages from different source groups cannot simply be added as a station's probability of gain.
![\"P=\\begin{bmatrix}0.30&0.50&0.20\\\\0.10&0.60&0.30\\\\0.10&0.10&0.80\\end{bmatrix}\"](\"math/18c74fdbd43cdfe0.svg\")
Each row sums to 1:0.3+0.5+0.2=1;0.1+0.6+0.3=1;0.1+0.1+0.8=1.
| Origin | Retained at origin | Leaves origin | Where the losses go |
|---|---|---|---|
| A | 30% | 70% | 50% toB;20% toC |
| B | 60% | 40% | 10% toA;30% toC |
| C | 80% | 20% | 10% toA;10% toB |
Let a, b, c be the starting numbers of bikes. Then:
![\"a_{next}=0.30a+0.10b+0.10c\"](\"math/aee58f980f95f13c.svg\")
![\"b_{next}=0.50a+0.60b+0.10c\"](\"math/24b806632ba7642d.svg\")
![\"c_{next}=0.20a+0.30b+0.80c\"](\"math/8fa41e7634228ffe.svg\")
| Station | Retained component | Incoming gain from other stations | Net change |
|---|---|---|---|
| A | 0.30a | 0.10b+0.10c | −0.70a+0.10b+0.10c |
| B | 0.60b | 0.50a+0.10c | 0.50a−0.40b+0.10c |
| C | 0.80c | 0.20a+0.30b | 0.20a+0.30b−0.20c |
Whether a station gains bikes overall depends on a, b, c. A column need not sum to 1 because it combines conditional probabilities from different origins.
For illustration, assume 100 bikes initially at each station. Then next counts areA=30+10+10=50;B=50+60+10=120;C=20+30+80=130. Total remains 300. These initial counts are not printed data.
![\"Three-state](\"fig/legacy25-bikes.7db03f5dcb.svg\")
For the four city Travelling salesman problem, the pheromone and cost matrix (A, B, C, D in that order given by row no, and column no) are given by
| Pheromone | A | B | C | D |
|---|---|---|---|---|
| A | 0 | 15 | 5 | 4 |
| B | 15 | 0 | 4 | 8 |
| C | 5 | 4 | 0 | 1 |
| D | 4 | 8 | 1 | 0 |
| Cost | A | B | C | D |
|---|---|---|---|---|
| A | 0 | 10 | 4 | 3 |
| B | 10 | 0 | 1 | 2 |
| C | 4 | 1 | 0 | 1 |
| D | 3 | 2 | 1 | 0 |
Ant L1 travels from A-B-C-D-A and Ant L1 travels from A-C-B-D-A and Taking vaporization co-efficient as 0.5, calculate the following:1.The length of each path 2.Pheromone level for each path
Initial pheromones and symmetric costs are printed above; evaporationρ=0.5. The two listed tours are treated as ant 1 and ant 2, although the scan repeats the nameL 1. Deposit constantQ is not supplied.
Tour lengths and updated edge pheromones after evaporation and both deposits.
![\"L_k=\\sum_{(i,j)\\in](\"math/d62616a1401fbf2a.svg\")
![\"\\tau_{ij}^{new}=(1-\\rho)\\tau_{ij}^{old}+\\sum_k\\Delta\\tau_{ij}^{k}\"](\"math/657b62b2f2dbf71a.svg\")
Declare the standard Ant System update: evaporate once, then add both ants' deposits. Because the matrices are symmetric, treat each city pair as an undirected edge. Give the answer inQ, then illustrateQ=1; noQ value is silently attributed to the paper.
![\"L_1=d_{AB}+d_{BC}+d_{CD}+d_{DA}=10+1+1+3=15\"](\"math/ac9a78a66f280841.svg\")
![\"L_2=d_{AC}+d_{CB}+d_{BD}+d_{DA}=4+1+2+3=10\"](\"math/cd7de0bca5f15da5.svg\")
Tour 2 is shorter and deposits more per used edge:Q/10 instead ofQ/15. Use the cost matrix for lengths, not the pheromone matrix.
| Edge | Old τ | After evaporation 0.5τ | Deposits | New τ |
|---|---|---|---|---|
| AB | 15 | 7.5 | Q/15 | 7.5+Q/15 |
| AC | 5 | 2.5 | Q/10 | 2.5+Q/10 |
| AD | 4 | 2 | Q/15+Q/10=Q/6 | 2+Q/6 |
| BC | 4 | 2 | Q/15+Q/10=Q/6 | 2+Q/6 |
| BD | 8 | 4 | Q/10 | 4+Q/10 |
| CD | 1 | 0.5 | Q/15 | 0.5+Q/15 |
BC andCB are the same undirected edge here; both tours also useDA. Both deposits must be included on those shared edges.
![\"\\tau^{new}\\approx\\begin{bmatrix}0&7.566667&2.600000&2.166667\\\\7.566667&0&2.166667&4.100000\\\\2.600000&2.166667&0&0.566667\\\\2.166667&4.100000&0.566667&0\\end{bmatrix}\"](\"math/b935c574afe72660.svg\")
For example,τAD=0.5×4+1/15+1/10=2+1/6=2.166667. Pheromone is maintained on edges; a single scalar “pheromone of the whole tour” is not defined by this rule. Each path's edge levels can be read from the matrix. The diagonal stays unused at 0.
A student walks to school, or he rides his bicycle. If he walks to school one day, then the next day he will walk or cycle with equal probability. But if he bicycles one day, then the probability that he will walk next day is 1/4.
State orderW=walk, C=cycle. One transition is one day.
P for one day andP² for Monday→Wednesday.
![\"(P^2)_{ij}=\\sum_kP_{ik}P_{kj}\"](\"math/1fa028a29cde1ac6.svg\")
There are two transitions, Monday→Tuesday and Tuesday→Wednesday. Multiply along each possible intermediate Tuesday state and add the alternatives.
From walking:W→W=1/2 andW→C=1/2. From cycling:C→W=1/4, soC→C=1−1/4=3/4.
![\"P=\\begin{bmatrix}1/2&1/2\\\\1/4&3/4\\end{bmatrix}\"](\"math/ed5dd48ea3dd8f9e.svg\")
| Monday→Wednesday | Sum over Tuesday W or C | Result |
|---|---|---|
| W→W | (1/2)(1/2)+(1/2)(1/4)=1/4+1/8 | 3/8 |
| W→C | (1/2)(1/2)+(1/2)(3/4)=1/4+3/8 | 5/8 |
| C→W | (1/4)(1/2)+(3/4)(1/4)=1/8+3/16 | 5/16 |
| C→C | (1/4)(1/2)+(3/4)(3/4)=1/8+9/16 | 11/16 |
![\"P^2=\\begin{bmatrix}3/8&5/8\\\\5/16&11/16\\end{bmatrix}=\\begin{bmatrix}0.375&0.625\\\\0.3125&0.6875\\end{bmatrix}\"](\"math/a532335af3e981b5.svg\")
Rows still specify the Monday state; columns specify Wednesday. Each row sums to 1. If Monday's actual state is not given, this conditional matrix is the complete answer; an unconditional Wednesday distribution additionally requires Monday's starting distribution.
List the applications of Fuzzy logic. Explain one in detail.
Fuzzy logic represents gradual membership in linguistic categories and reasons with graded rules. A temperature can partly belong to both “warm” and “hot”; this is degree of membership, not a probability that a hidden label is true.
![\"Crisp](\"fig/legacy25-controller.80f7bae29e.svg\")
| Block | Role in this example |
|---|---|
| Sensors / crisp input | Measure room temperature and occupancy |
| Fuzzifier | Convert each measurement to memberships such asCool, Warm, Hot andLow, High occupancy |
| Knowledge base | Store membership-function definitions and IF–THEN rules |
| Inference engine | Evaluate each rule's antecedent strength using the stated AND/OR operations |
| Defuzzifier / output computation | Convert the activated outputs to one fan command |
| Actuator | Apply the bounded fan-speed command; sensors provide fresh input for the next cycle |
For a simple single-input numerical illustration, hold occupancy fixed. On 20≤T≤40°C chooseµWarm=(40−T)/20 andµHot=(T−20)/20. AtT=30°C both memberships are 0.5.
Use a zero-order Sugeno design with these constant consequents, so its weighted output is:
![\"fan=\\frac{0.5(50)+0.5(90)}{0.5+0.5}=\\frac{25+45}{1}=70\\%\"](\"math/fb20d690460ddd1a.svg\")
Here 50% and 90% are explicit illustrative output constants, not discovered from a source dataset. A Mamdani controller instead defines fuzzy output shapes and may use their aggregated centroid; the two procedures must not be mixed without saying which design is used.
Yaad rakho: measured number→kitnaWarm/Hot→rules→one usable fan command.
Explain Bellman equation in reinforcement learning. Also give applications of reinforcement learning.
Reinforcement learning (RL) studies an agent that chooses actions, receives rewards and learns a policy to maximize expected cumulative reward through interaction. A policyπ(a|s) describes which action is chosen in a state; the value function measures expected future return, not only the immediate reward.
| Symbol | Meaning |
|---|---|
| s, a | Current state and selected action |
| s′, r | Next state and one-step reward |
| γ | Discount factor; usually 0≤γ<1 for discounted continuing tasks |
| π(a|s) | Probability of actiona under the policy |
| Vπ(s) | Expected discounted return starting in state s and followingπ |
| Qπ(s, a) | Expected discounted return after first choosinga in s, then followingπ |
![\"G_t=R_{t+1}+\\gamma](\"math/97caee2822bc6475.svg\")
This one-step decomposition is the basis of the Bellman equation: value now equals expected immediate reward plus discounted value of what follows.
![\"V^\\pi(s)=\\sum_a\\pi(a\\mid](\"math/80ef2d8261fa9e17.svg\")
The inner sum averages over uncertain transitions and rewards. The outer sum averages over the policy's actions. For a deterministic action and transition, it reduces toV(s)=r+γV(s′).
![\"V^*(s)=\\max_a\\sum_{s',r}p(s',r\\mid](\"math/863310fc89ae0de3.svg\")
![\"Q^*(s,a)=\\sum_{s',r}p(s',r\\mid](\"math/1f89cb0381d5b714.svg\")
Expectation evaluates a given policy; the maximum chooses the best available action. Value iteration applies this backup repeatedly. Q-learning estimates a corresponding action-value backup from sampled transitions instead of requiring the complete transition model.
Assumeγ=0.9 and deterministic transitions. ActionA gives reward 2 and a next-state value 10. ActionB gives reward 5 and a next-state value 4.
| Action | Bellman backup | Value |
|---|---|---|
| A | 2+0.9×10 | 11 |
| B | 5+0.9×4 | 8.6 |
The optimality backup choosesA because 11>8.6, even thoughB gives the larger immediate reward. Under a policy selecting both equally, the expectation backup is 0.5×11+0.5×8.6=9.8. These are one-step backups using the given continuation values.
| Application | State / action / reward example |
|---|---|
| Robot navigation | State:location/sensors; action:move; reward:goal reached with penalties for collision and delay |
| Game playing | State:board configuration; action:legal move; reward:win/loss or a carefully designed game score |
| Traffic-signal control | State:queue lengths; action:signal phase; reward:reduced delay with safety constraints |
| Resource scheduling | State:workload/resources; action:allocate jobs; reward:throughput or energy objective |
| Adaptive recommendation | State:user/context history; action:recommend an item; reward:measured engagement or another validated long-term outcome |
Rewards and exploration policies matter. An agent can optimize the wrong behavior if the reward ignores safety, fairness or the true task objective. A Bellman equation defines the value relation; it does not guarantee that a learned policy is already optimal.
Compare traditional optimization and evolutionary optimization algorithms. Illustrate the basic framework of evolutionary algorithms.
Optimization searches for a feasible candidate that maximizes or minimizes an objective. Evolutionary algorithms use a population, fitness-based selection and variation inspired by evolution.
| Aspect | Traditional methods (typical examples) | Evolutionary methods |
|---|---|---|
| Search state | Often one current candidate, e.g. gradient descent | Usually a population of candidates |
| Information | Gradient/Hessian in derivative methods; not required by all traditional methods | Primarily objective/fitness evaluations |
| Variation | A mathematically derived step or deterministic search rule | Crossover and mutation with stochastic choices |
| Landscape | Strong efficiency/guarantees possible under assumptions such as convexity | Useful for discrete, non-smooth, mixed or multimodal search |
| Guarantee | Depends on the algorithm and assumptions | No general finite-run global-optimum guarantee |
| Cost | May require fewer evaluations on suitable smooth problems | Often needs many evaluations and parameter tuning |
Framework: encode candidates → initialize feasible population → evaluate fitness → select parents → crossover → mutate → evaluate offspring → choose survivors → repeat until stopping criterion → return best observed candidate.
For example, in a delivery-route GA a chromosome is a permutation of cities; fitness can be inverse route length. Shorter valid tours are favored. A swap mutation preserves city uniqueness. Selection exploits good candidates, while variation introduces exploration. Do not claim every traditional method requires derivatives or every evolutionary method escapes all local optima.
Bag A contains 3 red and 4 green balls. Bag B contains 4 red and 5 green balls. One ball is drawn at random from one of the bags and found red. Find the probability it came from Bag A.
A has 7 balls, B has 9; red likelihoods 3/7 and 4/9.
P(A|red).
Reverse the direction of the evidence using Bayes. The probability of choosing each bag is a prior; the probability of red inside a chosen bag is a likelihood.
![\"P(A\\mid](\"math/ba4aa6be7e570107.svg\")
![\"P(R,A)=\\frac37\\frac12=\\frac3{14},\\quad](\"math/a518125ef5d2bed8.svg\")
![\"P(R)=\\frac3{14}+\\frac29=\\frac{27+28}{126}=\\frac{55}{126}\"](\"math/8383f96e6caef877.svg\")
![\"P(A\\mid](\"math/7b275c539000ed1c.svg\")
A={(x₁,.7),(x₂,.3),(x₃,.9),(x₄,.1)} and B={(x₁,.2),(x₂,.5),(x₃,.7),(x₄,.4)}. Find (1) A∪B, (2) A∩B, (3) Ā∩B̄, (4) Ā∪B̄.
Memberships A and B as printed; use the listed element order.
Union, intersection and the requested operations on the complements.
Membership is a degree between 0 and 1. Union means choose the larger membership; intersection means choose the smaller. Complement means subtract the membership from 1.
![\"\\mu_{A\\cup](\"math/8ef43c082b1d8a4a.svg\")
At x₁: 1−0.7=0.3, and 1−0.2=0.8. Never subtract the element label; subtract its membership.
| Element | A | B | 1−A | 1−B | max (A, B) | min (A, B) | max(1−A,1−B) | min(1−A,1−B) |
|---|---|---|---|---|---|---|---|---|
| x₁ | 0.7 | 0.2 | 0.3 | 0.8 | 0.7 | 0.2 | 0.8 | 0.3 |
| x₂ | 0.3 | 0.5 | 0.7 | 0.5 | 0.5 | 0.3 | 0.7 | 0.5 |
| x₃ | 0.9 | 0.7 | 0.1 | 0.3 | 0.9 | 0.7 | 0.3 | 0.1 |
| x₄ | 0.1 | 0.4 | 0.9 | 0.6 | 0.4 | 0.1 | 0.9 | 0.6 |
Keep the labels unchanged. Only the membership next to each label changes. De Morgan checks: complement of the union equals intersection of complements; complement of the intersection equals union of complements.
For a two-state Markov chain with P=[[0.8,0.2],[0.3,0.7]], (a) is the chain irreducible? (b) find its stationary distribution.
Two states, row-stochastic matrix [[.8,.2],[.3,.7]].
Mutual reachability and a probability row vector π with πP=π.
Irreducible means every state can reach every other with positive probability in some number of steps. Stationary means one transition leaves the distribution unchanged.
![\"\\pi=\\pi](\"math/bf18c709d7b4ad37.svg\")
![\"Redrawn](\"fig/legacy24r-stationary.41b587e21f.svg\")
1→2 has probability .2>0 and 2→1 has probability .3>0. Each state reaches the other in one step; the chain is irreducible. Self-loops also make it aperiodic.
![\"\\pi_1=0.8\\pi_1+0.3\\pi_2\\Rightarrow0.2\\pi_1=0.3\\pi_2\\Rightarrow\\pi_1=1.5\\pi_2\"](\"math/bd5f35e7318e17ca.svg\")
![\"1.5\\pi_2+\\pi_2=1\\Rightarrow\\pi_2=0.4,\\quad\\pi_1=0.6\"](\"math/d630dbfb75f1cbad.svg\")
![\"[0.6,0.4]P=[0.48+0.12,\\;0.12+0.28]=[0.6,0.4]\"](\"math/ff5d9937f5d667c5.svg\")
With reference to the Gaussian mixture model: (1) explain the need for the modified log-likelihood function; (2) how is it solved? (3) for K=6 Gaussian distributions estimate the number of parameters to compute.
A Gaussian mixture models a population using K Gaussian components. A hidden component label says which component generated a sample. The labels are not observed, so the likelihood sums over all possible components.
![\"p(x_i\\mid\\theta)=\\sum_{k=1}^{K}\\pi_k\\mathcal](\"math/34e74a21f3e93c71.svg\")
For independent observations multiply their densities; taking the logarithm turns the product over observations into a sum. The sum over mixture components remains inside the log. This log-of-sum is why a direct closed-form joint maximization is difficult.
![\"\\ell(\\theta)=\\log\\prod_{i=1}^{N}\\sum_k\\pi_k\\mathcal](\"math/68bab3da61241a5e.svg\")
Initialize valid weights, means and covariances. Using the old parameters, compute responsibility γᵢₖ: the posterior fraction of sample i assigned to component k. Each row of responsibilities adds to 1.
![\"\\gamma_{ik}=\\frac{\\pi_k\\mathcal](\"math/3463c7734eab2353.svg\")
The “modified” objective in EM is the expectation of the complete-data log likelihood. During this M-step the responsibilities are fixed, so the difficult log of a component sum is replaced by a weighted sum of component log densities.
![\"Q(\\theta\\mid\\theta^{old})=\\sum_i\\sum_k\\gamma_{ik}\\left[\\log\\pi_k+\\log\\mathcal](\"math/da8bce8c31e000fa.svg\")
![\"N_k=\\sum_i\\gamma_{ik},\\quad](\"math/c63266009fda4bdb.svg\")
![\"\\Sigma_k^{new}=\\frac{\\sum_i\\gamma_{ik}(x_i-\\mu_k^{new})(x_i-\\mu_k^{new})^T}{N_k}\"](\"math/433755f43e0968ba.svg\")
Repeat E and M until the observed log likelihood changes by less than a chosen tolerance or an iteration limit is reached. EM improves or preserves likelihood under exact steps, but can converge to a local optimum. Try several initializations and avoid singular covariance matrices.
The question gives K=6, but not the data dimension d or covariance type. For full covariance, each component has d mean entries and d (d+1)/2 independent covariance entries. There are K−1 independent mixture weights.
For a stated one-dimensional assumption: 6 means + 6 variances + 5 independent weights = 17 free parameters. A table may store all 6 weights, giving 18 stored values, but their sum-to-one constraint removes one degree of freedom. With diagonal covariance the general count is 2 Kd+K−1.
Design a domestic washing-machine controller with Dirt and Grease inputs, both 0–100, and Wash time output 0–100. Use low/medium/high input terms and very short/short/medium/high/very high output terms. Derive rules, support the design with graphs and show that larger dirt gives longer wash time. Defuzzify for Dirt=60% and Grease=70%.
Dirt=60, Grease=70; ranges 0–100; three input terms and five output terms.
A complete controller: membership graphs, rules, inference and crisp output.
Fuzzification translates measured values into degrees. Min implements AND, max combines rules, and centroid converts the fuzzy output to one numerical command.
![\"\\alpha_r=\\min(\\mu_{A_r}(x),\\mu_{B_r}(y)),\\quad](\"math/07c525af02eb813b.svg\")
Dirt and Grease each span 0–100. Use low=left shoulder (0,0,50), medium=triangle (0,50,100), high=right shoulder (50,100,100). Output is wash-time index on 0–100; larger means longer washing. The question gives no physical time unit, so do not invent minutes.
![\"Explicit](\"fig/legacy24-wash-inputs.c512a5765f.svg\")
![\"Five](\"fig/legacy24-wash-outputs.bb86080be7.svg\")
| Dirt / Grease | Low | Medium | High |
|---|---|---|---|
| Low | Very short | Short | Medium |
| Medium | Short | Medium | High |
| High | Medium | High | Very high |
Greater dirt or grease never reduces the rule consequent. This implements the requested monotonic relationship.
| Input | Low membership | Medium membership | High membership |
|---|---|---|---|
| Dirt=60 | 0 | 4/5 | 1/5 |
| Grease=70 | 0 | 3/5 | 2/5 |
For Dirt=60: low=max(0,(50−60)/50)=0; medium=(100−60)/50=0.8; high=max(0,(60−50)/50)=0.2. For Grease=70: low=0, medium=(100−70)/50=0.6, high=(70−50)/50=0.4.
| Dirt | Grease | AND strength | Output descriptor |
|---|---|---|---|
| Medium | Medium | min(0.8, 0.6) = 0.6 | Medium |
| Medium | High | min(0.8, 0.4) = 0.4 | High |
| High | Medium | min(0.2, 0.6) = 0.2 | High |
| High | High | min(0.2, 0.4) = 0.2 | Very high |
The other rules have a zero antecedent membership and therefore fire at zero.
| Descriptor | Maximum firing strength |
|---|---|
| Very short | 0.0 |
| Short | 0.0 |
| Medium | 0.6 |
| High | 0.4 |
| Very high | 0.2 |
![\"\\mu_{out}(z)=\\max_r\\min(\\alpha_r,\\mu_{B_r}(z))\"](\"math/fa8dd4982ac39cbc.svg\")
![\"Clipped](\"fig/legacy24-wash-aggregate.b750073d41.svg\")
Split the upper outline at every corner and intersection. On a segment [a, b] with endpoint heights u and v, integrate the straight line. This counts overlap once.
![\"A=(b-a)\\frac{u+v}{2},\\qquad](\"math/ed18c6126c4569a0.svg\")
| Segment [a, b] | Heights [u, v] | Area A | Moment M |
|---|---|---|---|
| [25, 40] | [0, 0.6] | 4.500000 | 157.500000 |
| [40, 50] | [0.6, 0.6] | 6.000000 | 270.000000 |
| [50, 60] | [0.6, 0.6] | 6.000000 | 330.000000 |
| [60, 65] | [0.6, 0.4] | 2.500000 | 155.833333 |
| [65, 75] | [0.4, 0.4] | 4.000000 | 280.000000 |
| [75, 80] | [0.4, 0.4] | 2.000000 | 155.000000 |
| [80, 90] | [0.4, 0.4] | 4.000000 | 340.000000 |
| [90, 95] | [0.4, 0.2] | 1.500000 | 138.333333 |
| [95, 100] | [0.2, 0.2] | 1.000000 | 97.500000 |
![\"z^*=\\frac{1924.166667}{31.500000}\\approx](\"math/8c383994ebbe02fc.svg\")
Hold grease fixed at 70 and repeat the same minimum, maximum and centroid steps for three dirt readings. This compares outputs without changing the rules or membership graphs.
| Dirt | Grease | Exact centroid | Wash-time index |
|---|---|---|---|
| 20 | 70 | 1925/41 | 46.951220 |
| 60 | 70 | 11545/189 | 61.084656 |
| 80 | 70 | 6580/99 | 66.464646 |
The calculated wash-time index increases from 46.951220 to 61.084656 to 66.464646 as dirt rises from 20 to 60 to 80. These comparisons demonstrate the requested behavior for the stated test inputs. At the target reading (60,70), only medium, high and very-high consequents fire.
D={0.3/0 + 0.7/1 + 1/2}, I={0.5/20 + 0.6/30 + 0.8/40}, V={0.6/20 + 0.5/30 + 0.4/40}. Determine (1) R=D×I, (2) max–min composition of V∘R, (3) max–product composition of V∘R.
D labels 0,1,2; I and V labels 20,30,40.
The 3×3 relation and the two resulting memberships on D’s universe.
R connects a D label to an I label. V supplies a membership on the I side. For each D row, combine its three relation memberships with V at the same I labels, then take the maximum. Max–product changes only the inner operation.
![\"R_{ij}=\\min(D_i,I_j),\\quad](\"math/322f826c4ddc94dd.svg\")
| D label | D membership | Min with.5 | Min with.6 | Min with.8 |
|---|---|---|---|---|
| 0 | 0.3 | 0.3 | 0.3 | 0.3 |
| 1 | 0.7 | 0.5 | 0.6 | 0.7 |
| 2 | 1 | 0.5 | 0.6 | 0.8 |
![\"R=\\begin{bmatrix}0.3&0.3&0.3\\\\0.5&0.6&0.7\\\\0.5&0.6&0.8\\end{bmatrix}\"](\"math/a6d0138a2e4b14ec.svg\")
| Output cell | Compare every matching middle index | Take the largest result |
|---|---|---|
| (1, 1) | min(0.3, 0.6) = 0.3, min(0.3, 0.5) = 0.3, min(0.3, 0.4) = 0.3 | max(0.3, 0.3, 0.3) = 0.3 |
| (2, 1) | min(0.5, 0.6) = 0.5, min(0.6, 0.5) = 0.5, min(0.7, 0.4) = 0.4 | max(0.5, 0.5, 0.4) = 0.5 |
| (3, 1) | min(0.5, 0.6) = 0.5, min(0.6, 0.5) = 0.5, min(0.8, 0.4) = 0.4 | max(0.5, 0.5, 0.4) = 0.5 |
| Output cell | Compare every matching middle index | Take the largest result |
|---|---|---|
| (1, 1) | 0.3 × 0.6 = 0.18, 0.3 × 0.5 = 0.15, 0.3 × 0.4 = 0.12 | max(0.18, 0.15, 0.12) = 0.18 |
| (2, 1) | 0.5 × 0.6 = 0.3, 0.6 × 0.5 = 0.3, 0.7 × 0.4 = 0.28 | max(0.3, 0.3, 0.28) = 0.3 |
| (3, 1) | 0.5 × 0.6 = 0.3, 0.6 × 0.5 = 0.3, 0.8 × 0.4 = 0.32 | max(0.3, 0.3, 0.32) = 0.32 |
The max–min output is {0.3/0 + 0.5/1 + 0.5/2}. The max–product output is {0.18/0 + 0.30/1 + 0.32/2}. These are fuzzy sets, so the slash notation means membership/label, not division.
Find the maximum of f (x)=−x²+5x+20 on −6≤x≤4 using PSO. Use four initial positions −5.6,−5.2,−1.1,2.3. Show detailed computations for iteration 1, with c₁=c₂=1, r₁=r₂=.5, w=1 and all initial velocities zero.
The four x values; zero velocities; w=1, c₁=c₂=1, r₁=r₂=.5; bounded domain[−6,4].
One iteration and the best value actually reached.
![\"v_i^{new}=wv_i+c_1r_1(pbest_i-x_i)+c_2r_2(gbest-x_i),\\quad](\"math/99893bd63a6f8a01.svg\")
This is maximization: larger f is better. Initialize each pbest to its own initial position, and freeze the old gbest for the whole simultaneous update.
| Particle | Position | −x² | 5x | f (x) |
|---|---|---|---|---|
| 1 | -5.6 | -31.36 | -28.0 | -39.36 |
| 2 | -5.2 | -27.04 | -26.0 | -33.04 |
| 3 | -1.1 | -1.21 | -5.5 | 13.29 |
| 4 | 2.3 | -5.29 | 11.5 | 26.21 |
Initial values [−39.36,−33.04,13.29,26.21]. Maximum 26.21 occurs at x=2.3.
The first two terms are zero: old velocity=0 and pbest−x=0. Thus vnew=.5(2.3−x).
| Particle | Social difference | Velocity | New position |
|---|---|---|---|
| 1 | 2.3−(-5.6)=7.9 | .5×(7.9)=3.95 | -5.6 + (3.95)=-1.65 |
| 2 | 2.3−(-5.2)=7.5 | .5×(7.5)=3.75 | -5.2 + (3.75)=-1.45 |
| 3 | 2.3−(-1.1)=3.4 | .5×(3.4)=1.7 | -1.1 + (1.7)=0.6 |
| 4 | 2.3−(2.3)=0 | .5×(0)=0 | 2.3 + (0)=2.3 |
| New x | −x²+5x+20 | New fitness |
|---|---|---|
| -1.65 | −(-1.65)²+5(-1.65)+20 | 9.0275 |
| -1.4500000000000002 | −(-1.45)²+5(-1.45)+20 | 10.6475 |
| 0.5999999999999999 | −(0.6)²+5(0.6)+20 | 22.64 |
| 2.3 | −(2.3)²+5(2.3)+20 | 26.21 |
All new positions are in [−6,4], so no boundary repair is required.
New positions [−1.65,−1.45,.6,2.3] give fitness [9.0275,10.6475,22.64,26.21]. The first three replace their worse personal records; the fourth ties its record. Global best remains 2.3 with 26.21.
As a separate analytical check, f (x)=26.25−(x−2.5)² has true maximum 26.25 at 2.5. PSO has not reached that point after this one iteration.
(1) Use Naive Bayes on the dataset to predict whether a person buys a computer for Age=Old, Income=Medium, Freelancer=Yes. (2) Explain Laplacian smoothing, whether it is required here and its impact.
| ID | Age | Income | Freelancer | Buys computer |
|---|---|---|---|---|
| 1 | Young | High | No | No |
| 2 | Young | High | No | No |
| 3 | Middle Age | High | No | Yes |
| 4 | Old | Medium | No | Yes |
| 5 | Old | Low | Yes | Yes |
| 6 | Old | Low | Yes | No |
| 7 | Middle Age | Low | Yes | Yes |
| 8 | Young | Medium | No | No |
| 9 | Young | Low | Yes | Yes |
| 10 | Young | Medium | Yes | Yes |
Ten records; query (Old, Medium, Yes). Class counts Yes 6, No 4.
Class scores, normalized comparison and smoothing impact.
Naive Bayes multiplies the class prior and feature likelihoods, assuming features are conditionally independent given the class. The same evidence denominator cancels when comparing classes.
![\"s(c)=P(c)\\prod_j](\"math/839a41948edad8ea.svg\")
| Class | N | Old | Medium income | FreelancerYes |
|---|---|---|---|---|
| Yes | 6 | 2 | 2 | 4 |
| No | 4 | 1 | 1 | 1 |
Yes records are 3,4,5,7,9,10; No records 1,2,6,8. Count within the chosen class, not among all ten records.
![\"s(Y)=\\frac6{10}\\frac26\\frac26\\frac46=\\frac2{45}\\approx0.0444444\"](\"math/ff3411bced348831.svg\")
![\"s(N)=\\frac4{10}\\frac14\\frac14\\frac14=\\frac1{160}=0.00625\"](\"math/237fd941cadfe143.svg\")
![\"P(Y\\mid](\"math/d8ec61024e51e1a4.svg\")
![\"P(N\\mid](\"math/dd660d78450f1519.svg\")
Predict Yes because its score is larger.
A zero likelihood would make the whole class product zero, even if other features favor the class. Add-one smoothing prevents that by adding one to each category count and adding the number of possible categories to the denominator. No query likelihood is zero here, so smoothing is not needed to rescue a zero score; it still changes the estimates.
![\"P(x_j=v\\mid](\"math/08b794e505b1bd84.svg\")
There are 3 age categories,3 income categories and 2 freelancer categories. Retaining empirical priors:
![\"s_{+1}(Y)=\\frac6{10}\\frac{2+1}{6+3}\\frac{2+1}{6+3}\\frac{4+1}{6+2}=\\frac1{24}\\approx0.0416667\"](\"math/4ae4ba086320b944.svg\")
![\"s_{+1}(N)=\\frac4{10}\\frac{1+1}{4+3}\\frac{1+1}{4+3}\\frac{1+1}{4+2}=\\frac8{735}\\approx0.0108844\"](\"math/2e5b140a14fedb05.svg\")
![\"P_{+1}(Y\\mid](\"math/1a4ef59913a7687a.svg\")
The predicted class remains Yes, while certainty decreases. Stating whether priors are smoothed matters; here only feature likelihoods were add-one smoothed.
(1) Discuss how state transitions are represented by a transition matrix. (2) Cite and illustrate any 3 properties. (3) Given P=[[1/2,1/2],[3/4,1/4]] and initial row distribution p⁽⁰⁾=[1/4,3/4], define and find (a) P₂₁⁽²⁾, (b) P₁₂⁽²⁾, (c) P⁽²⁾, (d) p₁⁽²⁾.
P and p⁽⁰⁾ as printed; row-vector convention.
Two-step conditional transition probabilities, the full two-step matrix and unconditional state 1 probability at time 2.
Pᵢⱼ is next-state j probability given current-state i. Entries lie in[0,1], rows sum to 1, and the n-step matrix is Pⁿ for a time-homogeneous chain. Row sums need not equal column sums; a transition matrix need not be symmetric.
![\"(P^2)_{ij}=\\sum_kP_{ik}P_{kj},\\qquad](\"math/aef8325cefd5071a.svg\")
| Entry | Via state 1 | Via state 2 | Sum |
|---|---|---|---|
| (P²)11 | (1/2)(1/2)=1/4 | (1/2)(3/4)=3/8 | 5/8 |
| (P²)12 | (1/2)(1/2)=1/4 | (1/2)(1/4)=1/8 | 3/8 |
| (P²)21 | (3/4)(1/2)=3/8 | (1/4)(3/4)=3/16 | 9/16 |
| (P²)22 | (3/4)(1/2)=3/8 | (1/4)(1/4)=1/16 | 7/16 |
![\"P^2=\\begin{bmatrix}5/8&3/8\\\\9/16&7/16\\end{bmatrix}\"](\"math/063d8bbaa02b11f7.svg\")
Therefore (a) P₂₁⁽²⁾=9/16 and (b) P₁₂⁽²⁾=3/8. Each entry conditions on a specific starting state.
![\"p_1^{(2)}=\\frac14\\frac58+\\frac34\\frac9{16}=\\frac{10+27}{64}=\\frac{37}{64}=0.578125\"](\"math/223d07fbdc1aa402.svg\")
![\"p_2^{(2)}=1-37/64=27/64=0.421875\"](\"math/c5238d8cc5680149.svg\")
P² is 2×2 and gives two-step probabilities for either starting state. p⁽²⁾ is one row of two probabilities after averaging those starting states using p⁽⁰⁾. If part (c) was intended as the distribution rather than capital-P matrix, the supplied distribution is [37/64,27/64]. Both objects are shown.
Use the population table. (1) Rank-select the best two, then one-point crossover in the middle. (2) A three-way tournament on individuals 1,3,5 selects parent 1; a two-way tournament on 2,4 selects parent 2. Again crossover in the middle. (3) Replace using fitness-based survivor selection and compute the new average. (4) Repeat replacement from the original population using age-based survivor selection and compute its average.
| No. | Chromosome | Fitness | Age |
|---|---|---|---|
| 1 | 01000000 | 64 | 8 |
| 2 | 01011111 | 95 | 2 |
| 3 | 01010000 | 80 | 13 |
| 4 | 01011001 | 89 | 9 |
| 5 | 01011110 | 94 | 4 |
| 6 | 01001111 | 79 | 10 |
| 7 | 01010001 | 81 | 12 |
| 8 | 01000001 | 65 | 15 |
| 9 | 01011000 | 88 | 10 |
| 10 | 01000000 | 64 | 20 |
Ten 8-bit chromosomes with supplied fitness and ages; larger fitness is better.
Four offspring and two independent survivor-selection outcomes.
Rank or tournament selection chooses parents. Crossover exchanges suffixes after bit 4. Fitness-based replacement removes the weakest; age-based replacement removes the oldest. Treat the two requested replacements as separate experiments starting from the same initial population.
| Fitness rank | Individual | Fitness |
|---|---|---|
| 1 | 2 | 95 |
| 2 | 5 | 94 |
| 3 | 4 | 89 |
| 4 | 9 | 88 |
| 5 | 7 | 81 |
| 6 | 3 | 80 |
| 7 | 6 | 79 |
| 8 | 8 | 65 |
| 9 | 1 | 64 |
| 10 | 10 | 64 |
Select 2(95) and 5(94). Split 0101|1111 and 0101|1110. Swapping suffixes gives 01011110(94) and 01011111(95).
Tournament{1,3,5}: fitness 64,80,94 → choose 5. Tournament{2,4}:95,89 → choose 2. Split 0101|1110 and 0101|1111. Children are 01011111(95) and 01011110(94). No mutation is specified.
The four child fitnesses are 94,95,95,94, sum 378. Their strings already exist in the supplied table, so their supplied fitness values can be used directly; no new fitness function is invented.
Remove the four lowest-fitness originals:1(64),10(64),8(65),6(79). Keep 2,3,4,5,7,9 with fitness 95,80,89,94,81,88. Keep duplicates because the question does not impose uniqueness.
![\"\\text{retained](\"math/73527afb9489efc7.svg\")
![\"\\bar](\"math/df9902dbae821985.svg\")
Remove the oldest four:10(age 20),8(age 15),3(age 13),7(age 12). Keep 1,2,4,5,6,9 with fitness 64,95,89,94,79,88. Add the same four offspring.
![\"\\text{retained](\"math/b892f89037dda209.svg\")
![\"\\bar](\"math/5fdd0bc72d39766b.svg\")
![\"\\bar](\"math/66d0b787f8c6800c.svg\")
Fitness-based replacement improves the average more in this example because it directly removes weaker individuals. Age-based replacement promotes turnover, but age is not the same as quality. “Rank-based” selection is usually probabilistic; here the question explicitly says choose the best two, so that deterministic instruction is followed.
For the given membership functions determine the crisp output using (1) Mean of Maxima and (2) Centroid / Center of Area.
![\"First](\"fig/legacy24r-defuzz-a.4cd4ea6810.svg\")
![\"Second](\"fig/legacy24r-defuzz-b.b1f504dbf4.svg\")
Two separate printed graphs; graph 1 corners(0,0),(2,.75),(3,.75),(4,0); graph 2 corners(0,0),(3,1),(4,1),(7,0).
MOM and centroid for each graph.
MOM averages the x-values where membership reaches its greatest height. For a continuous flat maximum[a, b], MOM=(a+b)/2. Centroid uses all area: divide the sum of area moments by total area.
![\"z_{MOM}=\\frac{a+b}{2},\\qquad](\"math/083457b6b2635a4c.svg\")
Maximum height.75 occurs throughout[2,3].
![\"z_{MOM,1}=(2+3)/2=2.5\"](\"math/54f0d91e056c6e0b.svg\")
| Piece | Area | x centroid | Moment |
|---|---|---|---|
| Triangle 0–2 | ½×2×.75=.75 | (0+2+2)/3=4/3 | 1 |
| Rectangle 2–3 | 1×.75=.75 | 2.5 | 1.875 |
| Triangle 3–4 | ½×1×.75=.375 | (3+3+4)/3=10/3 | 1.25 |
![\"z_{COA,1}=\\frac{1+1.875+1.25}{.75+.75+.375}=\\frac{4.125}{1.875}=2.2\"](\"math/e3b301b5ea6a74cd.svg\")
![\"z_{MOM,2}=(3+4)/2=3.5\"](\"math/4ea913e911135351.svg\")
| Piece | Area | x centroid | Moment |
|---|---|---|---|
| Triangle 0–3 | 1.5 | 2 | 3 |
| Rectangle 3–4 | 1 | 3.5 | 3.5 |
| Triangle 4–7 | 1.5 | 5 | 7.5 |
![\"z_{COA,2}=\\frac{3+3.5+7.5}{1.5+1+1.5}=\\frac{14}{4}=3.5\"](\"math/656c9873a2d32c03.svg\")
The second graph is symmetric about 3.5, so its area centroid and plateau midpoint coincide. The first graph is asymmetric;2.2 and 2.5 need not agree.
For a genetic algorithm applied to a4-city TSP: (1) cite how to initialize a population; (2) explain fitness calculation; (3) describe common crossover and mutation operators.
Representation. A chromosome is a permutation of the cities. [A, B, C, D] means A→B→C→D→A; the return edge is implied. Each city appears exactly once before returning.
Example population:[A, B, C, D], [A, C, B, D], [A, B, D, C], [A, D, B, C]. Fix A at the start to avoid counting rotations as different encodings. Randomly shuffle the other cities or include a heuristic tour; check feasibility.
![\"L=d_{AB}+d_{BC}+d_{CD}+d_{DA},\\qquad](\"math/5d153218d012483f.svg\")
For an illustrative symmetric distance table AB 10, AC 4, AD 3, BC 1, BD 2, CD 1: tour A–B–C–D–A has length 15 and fitness 1/15; A–C–B–D–A has length 10 and fitness 1/10. The shorter tour has larger inverse-distance fitness. These sample costs illustrate the method; the question itself supplies no cost table.
Order crossover (OX) copies a segment from one parent, then fills empty positions with cities from the other parent in cyclic order, skipping copied cities. Example P1=A B C D, P2=C D A B. Copy P1 positions 2–3(B C), then scan P2 from position 4:B(skip), C(skip), D, A. Fill child positions 4 then 1 to obtain A B C D. A different cut or parent can yield a different valid tour. Partially matched crossover (PMX) uses a mapping to resolve duplicates. Ordinary bit-string crossover can give repeated/missing cities and is not automatically valid.
Swap: A B C D→A D C B by swapping B, D. Inversion: A | B C D→A | D C B. Insertion: remove D from A B C D and insert after A→A D B C. These preserve city uniqueness. Evaluate every child again because a valid route may still be longer.
Choose parents favoring shorter tours, apply crossover/mutation, evaluate offspring, retain a fixed-size population and optionally preserve the best tour (elitism). Stop after a generation limit or lack of improvement. Return the best observed tour, without claiming a guaranteed optimum unless all feasible tours have been checked.
(I) Define the Markov process and list applications. (II) A professor assigns full-credit F, half-credit H and no-credit N homework according to the transition diagram. (a) Construct the TPM; (b) find P(X₃=F|X₂=N); (c) P(X₂=N|X₁=H); (d) P(X₄=H|X₂=F).
![\"Redrawn](\"fig/legacy24r-homework.4001dfd62e.svg\")
The labeled arrows in the original diagram; state orderF, H, N.
The transition matrix and one-step/two-step probabilities.
A Markov process predicts the next state using the current state, conditionally independent of earlier history. Applications include weather regimes, brand switching, reliability, queueing and movement among locations. Here a state records the homework-credit category.
![\"P(X_{t+n}=j\\mid](\"math/23d2321daf991110.svg\")
![\"P=\\begin{bmatrix}0.3&0.25&0.45\\\\0.45&0.4&0.15\\\\0.65&0.25&0.1\\end{bmatrix}\"](\"math/17d3fd5cb7fb6fac.svg\")
| Row | Sum |
|---|---|
| F | .30+.25+.45=1 |
| H | .45+.40+.15=1 |
| N | .65+.25+.10=1 |
![\"P(X_3=F\\mid](\"math/b86a579af37c69bb.svg\")
![\"P(X_2=N\\mid](\"math/061a644323cfaa64.svg\")
The time difference is 4−2=2. Sum three mutually exclusive intermediate-state routes.
| Intermediate state | Path product |
|---|---|
| F | .30×.25=.075 |
| H | .25×.40=.100 |
| N | .45×.25=.1125 |
![\"(P^2)_{FH}=.075+.100+.1125=.2875\"](\"math/ff6357103803b6cd.svg\")
Discuss the Expectation Maximization (EM) algorithm.
A Gaussian mixture models a population using K Gaussian components. A hidden component label says which component generated a sample. The labels are not observed, so the likelihood sums over all possible components.
For independent observations multiply their densities; taking the logarithm turns the product over observations into a sum. The sum over mixture components remains inside the log. This log-of-sum is why a direct closed-form joint maximization is difficult.
Initialize valid weights, means and covariances. Using the old parameters, compute responsibility γᵢₖ: the posterior fraction of sample i assigned to component k. Each row of responsibilities adds to 1.
The “modified” objective in EM is the expectation of the complete-data log likelihood. During this M-step the responsibilities are fixed, so the difficult log of a component sum is replaced by a weighted sum of component log densities.
Repeat E and M until the observed log likelihood changes by less than a chosen tolerance or an iteration limit is reached. EM improves or preserves likelihood under exact steps, but can converge to a local optimum. Try several initializations and avoid singular covariance matrices.
Small responsibility example. If one sample has unnormalized component contributions 0.12 and 0.08, total 0.20 gives responsibilities 0.6 and 0.4. Its mean-update contribution is 0.6x for component 1 and 0.4x for component 2; it is not forced entirely into one cluster. This is why EM is called soft assignment in a GMM.
For the fuzzy relation R below, find the λ-cut relations for λ=.003 and λ=.5.
![\"R=\\begin{bmatrix}1&0.1&0&0.4\\\\0.001&0.5&0.5&0.6\\\\0.3&1&0.5&0\\\\0&0.4&0.002&0.1\\end{bmatrix}\"](\"math/fe401cd453e2d902.svg\")
A4×4 fuzzy relation with entries between 0 and 1.
Two crisp 0/1 relations retaining entries at or above the threshold.
A weak λ-cut includes an ordered pair when its membership is at least λ. Replace included entries by 1 and excluded entries by 0. A strong cut would use a strict inequality, but that is not requested.
![\"(R_\\lambda)_{ij}=\\begin{cases}1&R_{ij}\\geq\\lambda\\\\0&R_{ij}<\\lambda\\end{cases}\"](\"math/fcd0258f1fdf1909.svg\")
Zero,.001 and.002 are below.003 and become 0. All other positive entries are at least.1, so become 1.
![\"R_{.003}=\\begin{bmatrix}1&1&0&1\\\\0&1&1&1\\\\1&1&1&0\\\\0&1&0&1\\end{bmatrix}\"](\"math/69c4d4d1862cfea7.svg\")
Entries equal to.5 are retained. Values 1,.5,.6 become 1; values.4,.3,.1,.002,.001,0 become 0.
![\"R_{.5}=\\begin{bmatrix}1&0&0&0\\\\0&1&1&1\\\\0&1&1&0\\\\0&0&0&0\\end{bmatrix}\"](\"math/3083e2884959ed29.svg\")
R₂₃=.5≥.5, so output(2,3)=1. R₄₃=.002<.003, so output(4,3)=0. The row/column labels and 4×4 shape do not change.
For the four-city travelling salesman problem, row and column order is A, B, C, D. The pheromone and cost matrices are:
![\"\\tau=\\begin{bmatrix}0&15&5&4\\\\15&0&4&8\\\\5&4&0&1\\\\4&8&1&0\\end{bmatrix}\"](\"math/772dd747e10284ad.svg\")
![\"D=\\begin{bmatrix}0&10&4&3\\\\10&0&1&2\\\\4&1&0&1\\\\3&2&1&0\\end{bmatrix}\"](\"math/bddf9ec4c2c0baa6.svg\")
Ant L₁ travels A–B–C–D–A and ant L₂ travels A–C–B–D–A. Taking the vaporization coefficient as 0.5, calculate (1) the length of each path and (2) pheromone level for each path.
Two tours: A–B–C–D–A and A–C–B–D–A; evaporation ρ=0.5; printed matrices above.
Both tour lengths and updated pheromone on every edge.
Distance is read from the COST matrix, not the pheromone matrix. Sum all edges including the return home. Under the standard Ant System update, evaporate the old trail once, then add deposits from every ant that traversed the edge.
![\"L_k=\\sum_{(i,j)\\in](\"math/0e8184d1b87733c1.svg\")
![\"\\Delta\\tau_1=1/15\\approx0.0666667,\\qquad\\Delta\\tau_2=1/10=0.1\"](\"math/31ae3d58c7494e4d.svg\")
The shorter tour receives the larger deposit. Shared edges BC and AD receive both deposits; unused-by-an-ant edges receive zero from that ant.
| Edge | Old pheromone | After evaporation | Deposit from tour 1 | Deposit from tour 2 | New pheromone |
|---|---|---|---|---|---|
| AB | 15 | 7.5 | 1/15 | 0 | 7.566667 |
| AC | 5 | 2.5 | 0 | 1/10 | 2.600000 |
| AD | 4 | 2 | 1/15 | 1/10 | 2.166667 |
| BC | 4 | 2 | 1/15 | 1/10 | 2.166667 |
| BD | 8 | 4 | 0 | 1/10 | 4.100000 |
| CD | 1 | 0.5 | 1/15 | 0 | 0.566667 |
![\"\\begin{bmatrix}0&7.566667&2.6&2.166667\\\\7.566667&0&2.166667&4.1\\\\2.6&2.166667&0&0.566667\\\\2.166667&4.1&0.566667&0\\end{bmatrix}\"](\"math/4905d8df69aaed31.svg\")
The diagonal remains zero because an ant does not travel from a city to itself.
Explain the Bellman equation in reinforcement learning.
The Bellman equation expresses a state’s long-term value as immediate expected reward plus discounted value of the next state. It follows by splitting the return into its first reward and the remaining return.
![\"G_t=R_{t+1}+\\gamma](\"math/e076ba9a7dcf48af.svg\")
![\"V^\\pi(s)=\\sum_a\\pi(a\\mid](\"math/c42f67b85b9dd072.svg\")
| Symbol | Meaning |
|---|---|
| s, a, s′ | Current state, chosen action, next state |
| π(a|s) | Policy probability of choosing actiona |
| P(s′|s, a) | Environment transition probability |
| R(s, a, s′) | Expected immediate reward for that transition |
| γ | Discount factor, usually 0≤γ<1 for continuing tasks |
| Vπ(s) | Expected discounted return following policyπ |
For a fixed policy, average over its action probabilities. For the optimal value, replace the policy-weighted average by the best action:
![\"V^*(s)=\\max_a\\sum_{s\\prime}P(s\\prime\\mid](\"math/0d0ba7664438689d.svg\")
Numerical illustration. An action gives reward 2, then reaches a value 10 state with probability.6 or a value 4 state with probability.4. With γ=.9, its value is.6(2+.9×10)+.4(2+.9×4)=.6×11+.4×5.6=6.6+2.24=8.84. If another action has value 8, the optimality update chooses 8.84. An action value is not just the immediate reward. Bellman recursion underlies policy evaluation, value iteration and many temporal-difference methods.
Consider the given dataset, apply Bayes rule and predict if a fruit has the following properties. Find (a) P(Yellow | Mango), (b) P(Sweet | Banana).
| Fruit | Yellow | Sweet | Long | Total |
|---|---|---|---|---|
| Mango | 350 | 450 | 0 | 650 |
| Banana | 400 | 300 | 350 | 400 |
| Others | 50 | 100 | 50 | 150 |
| Total | 800 | 850 | 400 | 1200 |
650 mangoes, of which 350 are yellow; 400 bananas, of which 300 are sweet.
The two probabilities in the printed order.
The condition after the vertical bar restricts the denominator. For Yellow given Mango, only mangoes are eligible. This is direct conditional probability; reversing the condition would answer a different question.
![\"P(A\\mid](\"math/f893878b79f9bfdd.svg\")
![\"P(Yellow\\mid](\"math/730d82f2aaccc80b.svg\")
![\"P(Sweet\\mid](\"math/c65d5aa64de25ddd.svg\")
Using 800 in the first denominator would instead calculate P(Mango | Yellow)=350/800. Using 850 in the second would calculate P(Banana | Sweet)=300/850. Neither is the requested probability. The total 1200 cancels if you first convert every count to a joint/marginal probability.
Define a membership function in the context of fuzzy logic. Using your intuition, sketch fuzzy membership functions for the “age of people”.
A membership function assigns each element x of a universe X a degree μA(x) between 0 and 1, indicating compatibility with a linguistic concept A.
![\"\\mu_A:X\\rightarrow[0,1]\"](\"math/ba33a453b276adb3.svg\")
A value of 0 means no membership, 1 means full membership, and intermediate values represent partial membership. This is a degree of a vague property, not automatically a probability of a random event.
![\"Illustrative](\"fig/legacy24-age.140e92bbb3.svg\")
| Term | One explicit assumed shape |
|---|---|
| Young | 1 until age 20; then decreases linearly to 0 at age 40. |
| Middle-aged | 0 up to 25; rises to 1 at 45; falls to 0 at 65. |
| Old | 0 up to 50; rises to 1 at 70; then remains 1. |
For age 35, μYoung=(40−35)/(40−20)=0.25 and μMiddle=(35−25)/(45−25)=0.5. The person is partly young and partly middle-aged. Overlap permits smooth decisions; a crisp boundary would abruptly change the label at one exact birthday.
Remember: membership = “kitna fit hota hai is label mein”. State the axis labels and the assumed age breakpoints when sketching.
Cite and illustrate various mutation operators in the genetic algorithm.
Mutation randomly changes one or more genes to introduce diversity and explore solutions not obtained by crossover alone. Apply a small specified mutation probability; excessive mutation can destroy good structures.
| Operator | Before → after | Suitable representation |
|---|---|---|
| Bit flip | 10110 → 10010 (third bit 1→0) | Binary chromosome |
| Random resetting | [2,7,4] → [2,3,4] (replace gene 2 by a valid value) | Integer/discrete |
| Swap | A B C D E → A D C B E (swap positions 2 and 4) | Permutation / TSP |
| Inversion | A | B C D | E → A | D C B | E | Permutation / TSP |
| Insertion | A B C D E → A D B C E (remove D and insert after A) | Permutation / TSP |
| Gaussian perturbation | [1.2,2.0] → [1.2,2.0+0.1]=[1.2,2.1] | Real-valued; respect bounds |
For a TSP route, arbitrary bit or value replacement can duplicate a city and omit another. Swap/inversion preserve the permutation, so the resulting route remains valid. Mutation changes a chromosome; crossover combines material from two parents.
Consider the Markov chain with S={1,2,3} and transition matrix below. (a) Draw the state transition diagram. (b) The scan prints “P(X₁=1)=(X₂=2)=1/4”; calculate P(X₁=3, X₂=2, X₃=1).
![\"P=\\begin{bmatrix}1/2&1/4&1/4\\\\1/3&0&2/3\\\\1/2&1/2&0\\end{bmatrix}\"](\"math/1f388f49cad4444e.svg\")
The printed matrix; ambiguous condition reproduced above.
All nonzero transitions and the joint probability of the path 3→2→1.
Rows are current states and columns are next states. A specified path probability is initial-state probability multiplied by each transition along that path.
![\"P(X_1=3,X_2=2,X_3=1)=P(X_1=3)P_{32}P_{21}\"](\"math/250a6ac8bef84a28.svg\")
![\"Redrawn](\"fig/legacy24-three-state.d3ab43c06c.svg\")
| From | Outgoing destinations |
|---|---|
| 1 | 1 with 1/2; 2 with 1/4; 3 with 1/4 |
| 2 | 1 with 1/3; 3 with 2/3 |
| 3 | 1 with 1/2; 2 with 1/2 |
Let p₁=P(X₁=1)=1/4 and p₃=P(X₁=3). To be in state 2 at time 2, arrive from state 1 or state 3; transition 2→2 is zero.
![\"\\frac14=p_1P_{12}+p_2P_{22}+p_3P_{32}=\\frac14\\frac14+0+\\frac12p_3\"](\"math/052bbb970339e754.svg\")
![\"\\frac12p_3=\\frac14-\\frac1{16}=\\frac3{16}\\quad\\Rightarrow\\quad](\"math/4c1768a409855602.svg\")
![\"P(3,2,1)=\\frac38\\cdot\\frac12\\cdot\\frac13=\\frac1{16}=0.0625\"](\"math/c24f3d84f730db29.svg\")
If the intended statement was P(X₁=1)=P(X₁=2)=1/4, then P(X₁=3)=1−1/4−1/4=1/2.
![\"P(3,2,1)=\\frac12\\cdot\\frac12\\cdot\\frac13=\\frac1{12}\\approx0.0833333\"](\"math/6a536bdcde0eb5da.svg\")
Given the task of modelling a student’s marks for an exam, use the Bayesian network and conditional probability tables below. Exam E: 0 easy, 1 difficult. IQ I: 0 low, 1 high. Aptitude S: 0 low, 1 high. (a) Calculate the probability of a difficult exam, low IQ, low aptitude, passing and securing admission. (b) Calculate the probability of high IQ and aptitude, an easy exam, failing and no admission.
![\"Redrawn](\"fig/legacy24-exam-bn.9070a35002.svg\")
| Root | P(0) | P(1) |
|---|---|---|
| E | 0.7 | 0.3 |
| I | 0.8 | 0.2 |
| I | E | P(M=0) | P(M=1) |
|---|---|---|---|
| 0 | 0 | 0.6 | 0.4 |
| 0 | 1 | 0.9 | 0.1 |
| 1 | 0 | 0.5 | 0.5 |
| 1 | 1 | 0.8 | 0.2 |
| I | P(S=0) | P(S=1) |
|---|---|---|
| 0 | 0.75 | 0.25 |
| 1 | 0.4 | 0.6 |
| M | P(A=0) | P(A=1) |
|---|---|---|
| 0 | 0.6 | 0.4 |
| 1 | 0.9 | 0.1 |
All five variables and their CPTs are reproduced above.
The two complete joint-event probabilities.
For a Bayesian network, multiply each node’s probability conditional on its parents. Roots have no parents. Do not multiply an extra marginal for a node whose conditional factor is already present.
![\"P(E,I,M,S,A)=P(E)P(I)P(M\\mid](\"math/2567b59d7d2c83ad.svg\")
![\"(E,I,M,S,A)=(1,0,1,0,1)\"](\"math/8adff69e2c0b00ef.svg\")
| Factor | Why this entry | Value |
|---|---|---|
| P(E=1) | Difficult | 0.3 |
| P(I=0) | Low IQ | 0.8 |
| P(M=1|E=1, I=0) | Pass, difficult, low IQ | 0.1 |
| P(S=0|I=0) | Low aptitude given low IQ | 0.75 |
| P(A=1|M=1) | Admission given pass | 0.1 |
![\"0.3\\times0.8=0.24\\quad;\\quad0.24\\times0.1=0.024\"](\"math/79f6ccced168cd06.svg\")
![\"0.024\\times0.75=0.018\\quad;\\quad0.018\\times0.1=0.0018\"](\"math/e0cfdfc6521b1e91.svg\")
![\"(E,I,M,S,A)=(0,1,0,1,0)\"](\"math/fa60b6f67d870af3.svg\")
![\"0.7\\times0.2\\times0.5\\times0.6\\times0.6=0.0252\"](\"math/e1a6504a28d510dd.svg\")
Intermediate products: 0.7×0.2=0.14; ×0.5=0.07; ×0.6=0.042; ×0.6=0.0252. These are joint probabilities of the whole described event. If E/I/S were explicitly conditioned on, their factors would cancel; the conditional answers would be 0.1×0.1=0.01 and 0.5×0.6=0.30.
A={1/low + 0.2/medium + 0.6/high}; B={0.9/positive + 0.3/zero + 0.8/negative}. (1) Find R=A×B. (2) With C={0.3/low + 0.4/medium + 0.6/high}, find S=C×B. (3) Find S∘R using max–min composition.
A, B, C memberships as printed. Rows of R/S: low, medium, high; columns: positive, zero, negative.
Both Cartesian-product matrices and a carefully stated interpretation of the requested composition.
The usual fuzzy Cartesian product uses minimum, not ordinary multiplication. Max–min composition combines a row with a column: take each matching minimum, then the maximum.
![\"R_{ij}=\\min(A_i,B_j),\\quad](\"math/7ef66022cb4ba451.svg\")
| A row | Min with 0.9 | Min with 0.3 | Min with 0.8 |
|---|---|---|---|
| 1 | min(1,0.9)=0.9 | min(1,0.3)=0.3 | min(1,0.8)=0.8 |
| 0.2 | min(0.2,0.9)=0.2 | min(0.2,0.3)=0.2 | min(0.2,0.8)=0.2 |
| 0.6 | min(0.6,0.9)=0.6 | min(0.6,0.3)=0.3 | min(0.6,0.8)=0.6 |
![\"R=\\begin{bmatrix}0.9&0.3&0.8\\\\0.2&0.2&0.2\\\\0.6&0.3&0.6\\end{bmatrix}\"](\"math/ca9c5503d687e49c.svg\")
| C row | Min with 0.9 | Min with 0.3 | Min with 0.8 |
|---|---|---|---|
| 0.3 | min(0.3,0.9)=0.3 | min(0.3,0.3)=0.3 | min(0.3,0.8)=0.3 |
| 0.4 | min(0.4,0.9)=0.4 | min(0.4,0.3)=0.3 | min(0.4,0.8)=0.4 |
| 0.6 | min(0.6,0.9)=0.6 | min(0.6,0.3)=0.3 | min(0.6,0.8)=0.6 |
![\"S=\\begin{bmatrix}0.3&0.3&0.3\\\\0.4&0.3&0.4\\\\0.6&0.3&0.6\\end{bmatrix}\"](\"math/2c582f7ae6099f7e.svg\")
Here S∘R means apply R first, then S, so use row R and column S.
| Output cell | Compare every matching middle index | Take the largest result |
|---|---|---|
| (1, 1) | min(0.9, 0.3) = 0.3, min(0.3, 0.4) = 0.3, min(0.8, 0.6) = 0.6 | max(0.3, 0.3, 0.6) = 0.6 |
| (1, 2) | min(0.9, 0.3) = 0.3, min(0.3, 0.3) = 0.3, min(0.8, 0.3) = 0.3 | max(0.3, 0.3, 0.3) = 0.3 |
| (1, 3) | min(0.9, 0.3) = 0.3, min(0.3, 0.4) = 0.3, min(0.8, 0.6) = 0.6 | max(0.3, 0.3, 0.6) = 0.6 |
| (2, 1) | min(0.2, 0.3) = 0.2, min(0.2, 0.4) = 0.2, min(0.2, 0.6) = 0.2 | max(0.2, 0.2, 0.2) = 0.2 |
| (2, 2) | min(0.2, 0.3) = 0.2, min(0.2, 0.3) = 0.2, min(0.2, 0.3) = 0.2 | max(0.2, 0.2, 0.2) = 0.2 |
| (2, 3) | min(0.2, 0.3) = 0.2, min(0.2, 0.4) = 0.2, min(0.2, 0.6) = 0.2 | max(0.2, 0.2, 0.2) = 0.2 |
| (3, 1) | min(0.6, 0.3) = 0.3, min(0.3, 0.4) = 0.3, min(0.6, 0.6) = 0.6 | max(0.3, 0.3, 0.6) = 0.6 |
| (3, 2) | min(0.6, 0.3) = 0.3, min(0.3, 0.3) = 0.3, min(0.6, 0.3) = 0.3 | max(0.3, 0.3, 0.3) = 0.3 |
| (3, 3) | min(0.6, 0.3) = 0.3, min(0.3, 0.4) = 0.3, min(0.6, 0.6) = 0.6 | max(0.3, 0.3, 0.6) = 0.6 |
![\"T=\\begin{bmatrix}0.6&0.3&0.6\\\\0.2&0.2&0.2\\\\0.6&0.3&0.6\\end{bmatrix}\"](\"math/4a88f859d09d91ef.svg\")
If your course writes S∘R as the matrix product with S on the left, the index result is instead:
![\"\\begin{bmatrix}0.3&0.3&0.3\\\\0.4&0.3&0.4\\\\0.6&0.3&0.6\\end{bmatrix}\"](\"math/035832180398151c.svg\")
For a correctly typed relation between the input universes, compose R with the inverse relation Sᵀ; the row/column labels then match:
![\"R\\circ](\"math/c371ee91de2bed68.svg\")
Discuss the significance of using Gaussian Mixture Models (GMMs) over K-means clustering in capturing complex data distributions.
GMM models the density as a weighted sum of Gaussian components. K-means minimizes within-cluster squared Euclidean distance and assigns each point to one nearest centroid.
![\"p(x)=\\sum_{k=1}^K\\pi_k\\mathcal](\"math/3599d1d0f37ccb77.svg\")
| Aspect | K-means | Gaussian mixture |
|---|---|---|
| Assignment | Hard: one cluster label | Soft: posterior responsibility for each component |
| Cluster shape | Voronoi regions from Euclidean distance; suited to compact roughly spherical groups | Full covariance allows elongated and differently oriented ellipses |
| Spread | No explicit cluster-specific covariance in the standard objective | Separate covariance estimates model unequal spreads |
| Overlap | Boundary points assigned completely to one side | Uncertainty is represented by partial memberships |
| Output | Centroids and assignments | Density, mixing weights, means, covariances and responsibilities |
| Training | Alternate assignment and centroid mean | EM responsibility and parameter updates |
| Cost | Usually simpler and faster | More parameters and cost; can overfit or develop singular covariance |
| Initialization | Can reach local minima | Can reach local likelihood maxima; multiple restarts help |
Example. Two customer groups overlap in annual income/spending: one is a narrow diagonal band and one a broad ellipse. A spherical distance-only partition may split the diagonal group badly. Separate covariance matrices can align GMM ellipses with each group. A boundary customer might have responsibilities [0.55,0.45], showing uncertainty instead of an unjustifiably certain label.
Why this matters. The mixture can approximate multimodal distributions; unequal mixing weights represent different group sizes. Its density supports likelihood-based anomaly detection and probabilistic downstream decisions. However a Gaussian component is not automatically a real-world class, and arbitrary non-Gaussian shapes may require several components.
Selection and limitations. Choose K and covariance restrictions using validation/BIC/AIC as appropriate, scale features, regularize covariance and inspect convergence. K-means remains attractive for large, roughly spherical, similarly scaled clusters when hard assignments are enough. GMM is more flexible, not universally superior.
For the given membership functions, determine the defuzzified crisp output using the Center of Sums method.
![\"Source](\"fig/legacy24-centre-sums.6ec254a338.svg\")
A1 has height 0.5 and support [1,8], plateau [3,7]; A2 has height 0.3 and support [3,9], plateau [4,8].
Center of sums of the two clipped output sets.
Center of sums adds each set’s area and moment separately; overlapping portions are counted once per set. This differs from the centroid of the max-aggregated union. Split each trapezoid into triangles and a rectangle.
![\"z_{COS}=\\frac{A_1c_1+A_2c_2}{A_1+A_2}=\\frac{M_1+M_2}{A_1+A_2}\"](\"math/b24773ebdb90677f.svg\")
| Piece | Area | Centroid x | Area × centroid |
|---|---|---|---|
| Left triangle [1,3] | ½×2×0.5=0.5 | (1+3+3)/3=7/3 | 7/6 |
| Rectangle [3,7] | 4×0.5=2 | (3+7)/2=5 | 10 |
| Right triangle [7,8] | ½×1×0.5=0.25 | (7+7+8)/3=22/3 | 11/6 |
![\"A_1=0.5+2+0.25=2.75,\\quad](\"math/dcc4edef182cedf3.svg\")
| Piece | Area | Centroid x | Moment |
|---|---|---|---|
| Left triangle [3,4] | 0.15 | 11/3 | 0.55 |
| Rectangle [4,8] | 1.2 | 6 | 7.2 |
| Right triangle [8,9] | 0.15 | 25/3 | 1.25 |
![\"A_2=0.15+1.2+0.15=1.5,\\quad](\"math/6325190477396ab2.svg\")
![\"z_{COS}=\\frac{13+9}{2.75+1.5}=\\frac{22}{4.25}=\\frac{88}{17}\\approx5.176471\"](\"math/b790fc2742ec49ac.svg\")
Do not simply average 52/11 and 6: A1 has more area and must receive more weight. Do not subtract overlap for Center of Sums.
Design a fuzzy logic-based quality-control system for manufacturing that detects defects and decides whether to reject or accept a product. Inputs: Pressure [0,100] and Quality of material [0,100]. Output: Defect detection [0,100]. Use three descriptors per input and five per output; draw membership functions and derive rules. If pressure and material quality are high, accept; vice versa reject. Show defuzzification for Pressure 40 and Quality 80.
Pressure=40, Material quality=80; ranges 0–100; three input terms and five output terms.
A complete controller: membership graphs, rules, inference and crisp output.
Fuzzification translates measured values into degrees. Min implements AND, max combines rules, and centroid converts the fuzzy output to one numerical command.
Pressure and Material quality each span 0–100. Use low=left shoulder (0,0,50), medium=triangle (0,50,100), high=right shoulder (50,100,100). Output is defect severity on 0–100; larger means worse quality.
![\"Explicit](\"fig/legacy24-quality-inputs.c512a5765f.svg\")
![\"Five](\"fig/legacy24-quality-outputs.e8de6dab72.svg\")
| Pressure / Material quality | Low | Medium | High |
|---|---|---|---|
| Low | Very high defect | High defect | Medium defect |
| Medium | High defect | Medium defect | Low defect |
| High | Medium defect | Low defect | Very low defect |
Higher pressure and better material reduce defect severity, as stipulated in this question. This is an exam design assumption, not a universal manufacturing rule.
| Input | Low membership | Medium membership | High membership |
|---|---|---|---|
| Pressure=40 | 1/5 | 4/5 | 0 |
| Material quality=80 | 0 | 2/5 | 3/5 |
For Pressure=40: low=max(0,(50−40)/50)=0.2; medium=40/50=0.8; high=max(0,(40−50)/50)=0. For Material quality=80: low=0, medium=(100−80)/50=0.4, high=(80−50)/50=0.6.
| Pressure | Material quality | AND strength | Output descriptor |
|---|---|---|---|
| Low | Medium | min(0.2, 0.4) = 0.2 | High defect |
| Low | High | min(0.2, 0.6) = 0.2 | Medium defect |
| Medium | Medium | min(0.8, 0.4) = 0.4 | Medium defect |
| Medium | High | min(0.8, 0.6) = 0.6 | Low defect |
The other rules have a zero antecedent membership and therefore fire at zero.
| Descriptor | Maximum firing strength |
|---|---|
| Very low defect | 0.0 |
| Low defect | 0.6 |
| Medium defect | 0.4 |
| High defect | 0.2 |
| Very high defect | 0.0 |
![\"Clipped](\"fig/legacy24-quality-aggregate.dae0d30118.svg\")
Split the upper outline at every corner and intersection. On a segment [a, b] with endpoint heights u and v, integrate the straight line. This counts overlap once.
| Segment [a, b] | Heights [u, v] | Area A | Moment M |
|---|---|---|---|
| [0, 15] | [0, 0.6] | 4.500000 | 45.000000 |
| [15, 25] | [0.6, 0.6] | 6.000000 | 120.000000 |
| [25, 35] | [0.6, 0.6] | 6.000000 | 180.000000 |
| [35, 40] | [0.6, 0.4] | 2.500000 | 93.333333 |
| [40, 50] | [0.4, 0.4] | 4.000000 | 180.000000 |
| [50, 55] | [0.4, 0.4] | 2.000000 | 105.000000 |
| [55, 65] | [0.4, 0.4] | 4.000000 | 240.000000 |
| [65, 70] | [0.4, 0.2] | 1.500000 | 100.833333 |
| [70, 75] | [0.2, 0.2] | 1.000000 | 72.500000 |
| [75, 95] | [0.2, 0.2] | 4.000000 | 340.000000 |
| [95, 100] | [0.2, 0] | 0.500000 | 48.333333 |
![\"z^*=\\frac{1525.000000}{36.000000}\\approx](\"math/2cfff2284cb1a5a3.svg\")
Choose and state a threshold: accept when defect severity is below 50, reject at or above 50. The calculated output is below 50, so accept. High/high gives very low defects and low/low gives very high defects, satisfying the question.
Describe the key components of an Ant Colony Optimization algorithm and their respective roles in finding optimal solutions.
Ant Colony Optimization (ACO) is a population-based stochastic optimization method inspired by ants reinforcing useful paths with pheromone. Artificial ants construct feasible solutions, evaluate them and update a shared pheromone model.
| Component | Role | TSP interpretation |
|---|---|---|
| Problem graph / feasible moves | Defines allowable solution components and constraints | Cities are nodes, travel edges have costs |
| Artificial ants | Construct candidate solutions incrementally | Build tours visiting each city once then return |
| Pheromone τᵢⱼ | Learned desirability from prior successful tours | Strong trails favor often-rewarded edges |
| Heuristic ηᵢⱼ | Problem-specific local guidance | Often 1/dᵢⱼ: shorter edges look attractive |
| Tabu/visited set | Prevents invalid repeat decisions | Do not revisit a city before the closing edge |
| Probabilistic transition | Balances learned and local guidance | Normalize desirabilities over unvisited cities |
| Evaporation ρ | Forgets old trails; helps avoid stagnation | Multiply old pheromone by 1−ρ |
| Deposit / reinforcement | Rewards useful solutions | Ant on tour of length L deposits Q/L |
| Best-tour memory | Retains best solution discovered | Shorter tour replaces incumbent |
| Termination | Bounds computation | Iteration/time/no-improvement limit |
![\"p_{ij}^{(k)}=\\frac{\\tau_{ij}^{\\alpha}\\eta_{ij}^{\\beta}}{\\sum_{h\\in](\"math/d58589c662f1dcac.svg\")
α controls pheromone influence; β controls heuristic influence. If two allowed edges have desirabilities 2 and 6, their probabilities are 2/8=0.25 and 6/8=0.75. The less preferred edge can still be explored.
![\"\\tau_{ij}^{new}=(1-\\rho)\\tau_{ij}^{old}+\\sum_k](\"math/5e92452bfde69641.svg\")
Algorithm flow: initialize trails → construct all tours → compute lengths → evaporate → deposit → retain best tour → repeat. A population and random choice promote exploration; reinforcement promotes exploitation. High evaporation may forget useful trails too quickly, while too little evaporation can lock the colony into an inferior route. ACO is a heuristic and does not guarantee the global optimum after a fixed iteration count.
With reference to PSO: (1) justify its inspiration; (2) list and briefly explain its parameters; (3) minimize f (x)=3x²+4x+10 with five particles initially at −9, −6, −2, 8, 10. Assume suitable parameter values and estimate positions after one iteration.
PSO is inspired by coordinated movement in bird flocks/fish schools. Each particle remembers its own best position and responds to the swarm’s best position. A particle is one candidate solution, not one coordinate of a single shared solution.
| Symbol | Meaning |
|---|---|
| xᵢ | Current position / candidate solution |
| vᵢ | Velocity / proposed movement |
| pbestᵢ | Best position visited by this particle |
| gbest | Best position visited by the swarm |
| w | Inertia, retains part of previous velocity |
| c₁, c₂ | Cognitive and social acceleration weights |
| r₁, r₂ | Random multipliers in [0,1] |
| N / limit | Number of particles / stopping condition |
Five positions and objective as printed. The question explicitly permits parameter assumptions.
Updated velocities, positions, fitnesses and best records after one synchronous iteration.
For minimization, a smaller objective is better. Freeze gbest while moving the entire swarm; update best records only after evaluating the new positions.
| Particle | x | 3x²+4x+10 |
|---|---|---|
| 1 | -9 | 243 + (-36) + 10 = 217 |
| 2 | -6 | 108 + (-24) + 10 = 94 |
| 3 | -2 | 12 + (-8) + 10 = 14 |
| 4 | 8 | 192 + (32) + 10 = 234 |
| 5 | 10 | 300 + (40) + 10 = 350 |
Smallest value is 14 at x=−2, so freeze gbest=−2.
Initial velocity is zero and pbest equals the current position; therefore inertia and cognitive terms are both zero.
![\"v_i^{new}=0.5(-2-x_i)\"](\"math/4a6d06df6529bc16.svg\")
| Particle | Velocity calculation | New position | New fitness |
|---|---|---|---|
| 1 | 0.5×(−2−(-9))=3.5 | -9+(3.5)=-5.5 | 78.75 |
| 2 | 0.5×(−2−(-6))=2 | -6+(2)=-4 | 42 |
| 3 | 0.5×(−2−(-2))=0 | -2+(0)=-2 | 14 |
| 4 | 0.5×(−2−(8))=-5 | 8+(-5)=3 | 49 |
| 5 | 0.5×(−2−(10))=-6 | 10+(-6)=4 | 74 |
New positions are [−5.5,−4,−2,3,4], with objective values [78.75,42,14,49,74]. Every nonbest particle improves its own value, so its pbest becomes its new position; particle 3 remains −2. gbest is still −2 with value 14.
The analytical minimum is at x=−2/3 with f=26/3, but one PSO iteration has not reached it. Do not report the analytical optimum as the swarm’s achieved result.
Two cold-drink brands A and B are sold. An A buyer has 80% chance of buying A next. A B buyer has 90% chance of buying B next. (I) Develop the transition matrix. (II) Interpret retention and loss, and retention and gain. (III) Draw the transition diagram.
A→A=.80; B→B=.90. Two exhaustive states A and B.
Transition matrix, diagram, and correct flow interpretation.
Every row lists all possible destinations from one origin; its probabilities sum to one. Missing switch probability is 1 minus retention.
![\"P_{ij}=P(\\text{next}=j\\mid\\text{current}=i)\"](\"math/e570af7d131502f2.svg\")
![\"P_{AB}=1-0.8=0.2,\\quad](\"math/8f12974205c3ad53.svg\")
![\"Redrawn](\"fig/legacy24-brands.f41eb4de5f.svg\")
A retains 80% of its own current buyers and loses 20% to B. B retains 90% of its own buyers and loses 10% to A. A gains 10% of B’s current buyers; B gains 20% of A’s current buyers. These percentages have different denominators.
![\"A_{next}=0.8A+0.1B,\\quad](\"math/ddeb66c690da3b9e.svg\")
For an illustrative 100 buyers of each brand, next counts are A=80+10=90 and B=20+90=110. Without initial counts, “A loses 10% overall” is not justified.
A chromosome x=abcdefghij has ten genes, each a digit 0–9. Fitness f (x)=(a+b)−(c+d)+(e+f)+(g+h)−(i+j). Initial chromosomes: x₁=6541353253; x₂=8712660124; x₃=2392128546; x₄=4185209426. (1) Evaluate and rank fitness. (2) Cross the fittest two at the midpoint. (3) Cross second and third fittest using two points b and f. (4) Cross first and third using arithmetic crossover α=.5. (5) Evaluate all six offspring. (6) Has overall fitness improved?
Four ten-gene parents and a signed linear fitness function.
Ranking, two children per crossover, child fitness and fair population comparison.
One-point crossover swaps suffixes. Two-point crossover swaps the middle block. Arithmetic crossover takes weighted gene values.
![\"child_1=\\alpha](\"math/686b6d11dfecb89a.svg\")
| Parent | (a+b)−(c+d)+(e+f)+(g+h)−(i+j) | Fitness |
|---|---|---|
| x1 | 11−5+8+5−8 | 11 |
| x2 | 15−3+12+1−6 | 19 |
| x3 | 5−11+3+13−10 | 0 |
| x4 | 5−13+2+13−8 | -1 |
Descending order: x₂ (19), x₁ (11), x₃ (0), x₄ (−1).
x₂=87126 | 60124; x₁=65413 | 53253. Exchange the right blocks.
| Child | Genes | Fitness |
|---|---|---|
| C1 | 8712653253 | 20 |
| C2 | 6541360124 | 10 |
State cuts explicitly: positions 2 and 6. The parents are x₁=65 | 4135 | 3253 and x₃=23 | 9212 | 8546. Swap only the four-gene middle blocks.
| Child | Genes | Fitness |
|---|---|---|
| C3 | 6592123253 | 0 |
| C4 | 2341358546 | 11 |
| Position | x₂ | x₃ | 0.5x₂+0.5x₃ |
|---|---|---|---|
| 1 | 8 | 2 | 5.0 |
| 2 | 7 | 3 | 5.0 |
| 3 | 1 | 9 | 5.0 |
| 4 | 2 | 2 | 2.0 |
| 5 | 6 | 1 | 3.5 |
| 6 | 6 | 2 | 4.0 |
| 7 | 0 | 8 | 4.0 |
| 8 | 1 | 5 | 3.0 |
| 9 | 2 | 4 | 3.0 |
| 10 | 4 | 6 | 5.0 |
At α=0.5 both children are identical. C5=C6=[5.0, 5.0, 5.0, 2.0, 3.5, 4.0, 4.0, 3.0, 3.0, 5.0]. Their fitness is 9.5.
| Child | Pair sums with signs | Fitness |
|---|---|---|
| C1 | 15−3+11+5−8 | 20 |
| C2 | 11−5+9+1−6 | 10 |
| C3 | 11−11+3+5−8 | 0 |
| C4 | 5−5+8+13−10 | 11 |
| C5 | 10−7+7.5+7−8 | 9.5 |
| C6 | 10−7+7.5+7−8 | 9.5 |
![\"\\bar](\"math/07b0b7549d04f350.svg\")
![\"\\bar](\"math/539cf61121bb4c13.svg\")
Best old fitness is 19; best new fitness is 20. The mean improves, but the best individual may not. A larger total alone is not proof of improvement because six children replace four parents.
Describe the concept of a Markov process and its use in probabilistic systems. For the given state diagram S, C, R: (a) today is sunny; find probability tomorrow is sunny and the next day rainy. (b) today is cloudy and yesterday rainy; find probability tomorrow is sunny.
![\"Redrawn](\"fig/legacy24-weather.575ab21d0e.svg\")
State order Sunny, Cloudy, Rainy. The diagram gives P=[[.8,.05,.15],[.75,.2,.05],[.38,.02,.6]].
Definition and both requested conditional probabilities.
A Markov process has the memoryless property: after the current state is known, earlier states do not change the next-state probabilities. It models time evolution through a state distribution and transition law.
![\"P(X_{t+1}\\mid](\"math/acdbeb13aace7aac.svg\")
![\"P=\\begin{bmatrix}0.8&0.05&0.15\\\\0.75&0.2&0.05\\\\0.38&0.02&0.6\\end{bmatrix}\"](\"math/b7e30adc50b915ab.svg\")
For example P(Sunny→Rainy)=0.15 while P(Rainy→Sunny)=0.38; arrow directions cannot be interchanged. Each row sums to 1.
![\"P(X_{t+1}=S,X_{t+2}=R\\mid](\"math/f206e4e596cad948.svg\")
Do not sum all two-day routes ending in rain: tomorrow is specifically required to be sunny.
![\"P(X_{t+1}=S\\mid](\"math/72bc7a0f35b98e71.svg\")
Yesterday’s rain adds no further information under the stated Markov model. Such models are used for weather regimes, brand switching, queue states and reliability; the chosen state must capture the relevant predictive information.
Bag B₁ contains 3 white and 7 black balls. Bag B₂ contains 5 white and 3 black balls. A ball drawn from one of the bags is white. Find the probability it came from B₁.
P(W|B₁)=3/10; P(W|B₂)=5/8.
P(B₁|W).
Bayes reverses the condition. Weight the likelihood for each bag by its prior chance of being chosen, then normalize over all ways to obtain a white ball.
![\"P(B_1\\mid](\"math/e49e2acd2af82aac.svg\")
![\"P(W,B_1)=\\frac3{10}\\frac12=\\frac3{20}=0.15\"](\"math/797903c04a87ae2d.svg\")
![\"P(W,B_2)=\\frac58\\frac12=\\frac5{16}=0.3125\"](\"math/44aff1e67025be81.svg\")
![\"P(W)=\\frac3{20}+\\frac5{16}=\\frac{12+25}{80}=\\frac{37}{80}=0.4625\"](\"math/0310d0eb2cb65a99.svg\")
![\"P(B_1\\mid](\"math/6bf86459a0c31fc3.svg\")
For the following fuzzy relations R and S, obtain T using max–min composition.
![\"R=\\begin{bmatrix}0.6&0.3\\\\0.2&0.5\\end{bmatrix},\\quad](\"math/ab04f42254fa9175.svg\")
R is 2×2 and S is 2×3.
A 2×3 relation T connecting the first and last universes.
Use a row of R and a column of S. The common middle dimension is 2. Take two minima, then the larger of them.
![\"T_{ij}=\\max_{k=1,2}\\min(R_{ik},S_{kj})\"](\"math/e1639920f7163459.svg\")
(2×2) composed with (2×3) gives 2×3. Ordinary dot-product addition is not used.
| Output cell | Compare every matching middle index | Take the largest result |
|---|---|---|
| (1, 1) | min(0.6, 1) = 0.6, min(0.3, 0.8) = 0.3 | max(0.6, 0.3) = 0.6 |
| (1, 2) | min(0.6, 0.4) = 0.4, min(0.3, 0.3) = 0.3 | max(0.4, 0.3) = 0.4 |
| (1, 3) | min(0.6, 0.5) = 0.5, min(0.3, 0.8) = 0.3 | max(0.5, 0.3) = 0.5 |
| (2, 1) | min(0.2, 1) = 0.2, min(0.5, 0.8) = 0.5 | max(0.2, 0.5) = 0.5 |
| (2, 2) | min(0.2, 0.4) = 0.2, min(0.5, 0.3) = 0.3 | max(0.2, 0.3) = 0.3 |
| (2, 3) | min(0.2, 0.5) = 0.2, min(0.5, 0.8) = 0.5 | max(0.2, 0.5) = 0.5 |
![\"T=\\begin{bmatrix}0.6&0.4&0.5\\\\0.5&0.3&0.5\\end{bmatrix}\"](\"math/d993cd17e860e89f.svg\")
Differentiate between point crossover and uniform crossover.
Crossover recombines genes from two parents to form children. Its purpose is to combine useful building blocks; mutation introduces an independent local change.
| Aspect | Point crossover | Uniform crossover |
|---|---|---|
| Selection mechanism | Choose one or more cut locations; exchange contiguous segments | Choose parent at each position using a binary mask |
| Linkage | Tends to preserve neighboring gene blocks | Mixes genes more independently; may break useful adjacent blocks |
| Control | Number and position of cut points | Mask / per-gene mixing probability |
| Example parents | P1=111100, P2=000011 | Same parents |
| Rule | Cut after position 3: 111|100 and 000|011 | Mask 101010: child 1 takes P1 where mask=1, otherwise P2 |
| Children | 111011 and 000100 | 101001 and 010110 |
For uniform child 1, positions 1–6 are 1(fromP 1),0(P2),1(P1),0(P2),0(P1),1(P2). Child 2 uses the complementary choices. Both methods preserve chromosome length. For permutation encodings such as TSP, naive point/uniform crossover can repeat a city, so use a permutation-preserving method or a declared repair.
Discuss the applications of reinforcement learning.
Reinforcement learning learns a policy for sequential decisions by interacting with an environment and maximizing expected cumulative reward. Unlike supervised learning, each state need not have a supplied correct action label.
| Application | State and action | Reward / reason RL fits |
|---|---|---|
| Robot navigation | State: pose/obstacles; action: movement | Reward for reaching goal, penalties for collision; actions affect future states |
| Game playing | State: board; action: legal move | Win/loss and long-term strategy; a weak immediate move may yield a later win |
| Traffic signal control | State: queue lengths; action: phase/timing | Reduced delay and congestion over time |
| Resource scheduling | State: jobs/load; action: allocation | Throughput and latency rewards under changing demand |
| Recommendation sequences | State: interaction history; action: next item | Long-term engagement rather than only one click |
One worked interpretation: a delivery robot sees state “near junction”, chooses left/right, receives −1 per travel step and +20 on successful delivery. It learns a route balancing path length and successful arrival, using a discount factor for later rewards. Exploration must be controlled when real actions are costly or unsafe; a simulator can provide initial experience. RL is appropriate when actions influence subsequent opportunities, not merely for every static prediction task.
Question. A supermarket recorded the joint distribution of Product type (Y: premium = 1, regular = 0), Payment mode (X: credit = 1, debit = 0) and Gender (Z: female = 1, male = 0). (a) If a customer bought a premium item, find the probability that (1) they are female, (2) they paid by credit card. (b) Find (1) P(premium) and (2) P(credit card). (c) Given a male customer paid by debit card, find the probability he bought a regular item. [5 marks]
The printed eight entries are reproduced below. Their total is 1.002, not 1. Therefore the table is inconsistent as a probability distribution.
| Credit, female | Credit, male | Debit, female | Debit, male | |
|---|---|---|---|---|
| Premium (Y=1) | 0.08 | 0.10 | 0.072 | 0.05 |
| Regular (Y=0) | 0.10 | 0.15 | 0.25 | 0.20 |
The five requested probabilities, stating how the inconsistent data are interpreted.
For this worked solution, treat all printed entries as nonnegative weights and normalise by their total 1.002. This preserves the printed ratios. It is a declared interpretation, not a claim about the examiner’s intended correction.
Use total weight 1.002 consistently for unconditional probabilities.
Alternative correction: if 0.072 is corrected to 0.070, the total becomes 1 and the corresponding answers are (a) 0.50 and 0.60; (b) 0.30 and 0.43; (c) 0.8. This agrees with the interpretation used in the supplied key. State the correction; do not call raw totals 0.302 and 0.43 valid marginals of the original inconsistent table.
List the applications of Ant Colony Optimization and explain any one application in detail. [5 marks]
Ant Colony Optimization is a probabilistic, population-based metaheuristic (Dorigo, 1992) in which artificial ants build solutions on a graph and communicate through pheromone trails. In the update rule used here, evaporation reduces all stored trails; deposits then reinforce edges of selected tours according to solution quality. Evaporation does not occur only on poor paths.
\r\n, skipping customers already served or that would exceed the vehicle capacity; when the vehicle is full it returns to the depot. on the routes of good solutions, so short connections are reinforced.| Advantages | Limitations |
|---|---|
| Naturally suited to graph and routing problems | Several parameters to tune (α, β, ρ, number of ants) |
| Can share useful route information and adapt search to changing graphs | Convergence time is uncertain; may stagnate on a sub-optimal path |
| Positive feedback + parallel ants | Theoretical analysis is difficult |
Question. Consider two fuzzy sets X = {(c1, 0.6), (c2, 0.8), (c3, 0.3), (c4, 0.5), (c5, 0.7)} and Y = {(c1, 0.4), (c2, 0.5), (c3, 0.9), (c4, 0.6), (c5, 0.2)}. Find 1) X ∪ Y 2) X ∩ Y 3) ¬X ∪ ¬Y 4) ¬X ∩ ¬Y. [5 marks]
The aligned X and Y memberships printed in this question.
The four requested membership lists in the original element order.
Use max for OR, min for AND, and 1−μ for NOT.
¬X = {0.4, 0.2, 0.7, 0.5, 0.3}; ¬Y = {0.6, 0.5, 0.1, 0.4, 0.8}.
| c1 | c2 | c3 | c4 | c5 | |
|---|---|---|---|---|---|
| X | 0.6 | 0.8 | 0.3 | 0.5 | 0.7 |
| Y | 0.4 | 0.5 | 0.9 | 0.6 | 0.2 |
| X ∪ Y = max | 0.6 | 0.8 | 0.9 | 0.6 | 0.7 |
| X ∩ Y = min | 0.4 | 0.5 | 0.3 | 0.5 | 0.2 |
| ¬X ∪ ¬Y | 0.6 | 0.5 | 0.7 | 0.5 | 0.8 |
| ¬X ∩ ¬Y | 0.4 | 0.2 | 0.1 | 0.4 | 0.3 |
¬X ∪ ¬Y must equal 1 − (X ∩ Y) = {0.6, 0.5, 0.7, 0.5, 0.8} ✓, and ¬X ∩ ¬Y must equal 1 − (X ∪ Y) = {0.4, 0.2, 0.1, 0.4, 0.3} ✓.
Official marking: 1 + 1 + 1.5 + 1.5.
\r\nQuestion. Users switch between Brand X and Brand Y each year: 25% of X users switch to Y and 75% stay; 40% of Y users switch to X and 60% stay. (a) Write the transition probability matrix. (b) Draw the state transition diagram. [5 marks]
All transition values are given in the full question. State order: X, Y.
The transition matrix and state diagram.
Rows mean current state; columns mean next state. For each destination, multiply each origin amount by its transition probability, then add.
| From / To | X | Y |
|---|---|---|
| X | 0.75 | 0.25 |
| Y | 0.4 | 0.6 |
Each row sums to 1 because it lists all possible destinations for that origin. A percentage is divided by 100: 25%=.25.
Diagonal entries are the fractions staying at their current state; off-diagonal entries are fractions moving to a different state. Each arrow uses its origin row and destination column. These fractions have different origin populations, so adding raw column probabilities does not give the number arriving.
Question. Maximize using Particle Swarm Optimization. . Take . Optimize and give the solution after 1 iteration showing the step-by-step process. [10 marks]
Particle 1: , . Particle 2: , .
Fitness, pbest, gbest; new velocities and positions; updated pbest and gbest.
Each particle is pulled by its own memory and by the swarm's best; we maximise, so the larger is better.
| Particle | Position | F | pbest |
|---|---|---|---|
| 1 | (2.5, −3.0) | 57.5 | (2.5, −3.0) |
| 2 | (3.5, −1.5) | 64 | (3.5, −1.5) |
No previous iteration, so pbest = current position. Largest fitness is 64, so gbest = (3.5, −1.5).
| Particle | New position | New F | Old pbest F | pbest after iteration |
|---|---|---|---|---|
| 1 | (2.80825, −2.537625) | 59.5569 | 57.5 | (2.80825, −2.537625) — improved |
| 2 | (3.59, −1.275) | 63.5336 | 64 | (3.5, −1.5) — kept |
The best pbest fitness is still 64, so gbest stays (3.5, −1.5).
, . After one iteration particle 1 improved to F = 59.5569; the best solution so far is gbest = (3.5, −1.5) with F = 64.Official key: fitness and gbest (5 marks), velocity and position updates (5 marks). The calculations above use the question data. The key's summary matrix writes 0.462 in place of 2.808 for ; that is a copying slip.
With reference to the Gaussian Mixture Model (GMM): compare GMM with the k-means algorithm; derive the equation for the log-likelihood function; state the number of parameters to be computed if K = 5. [10 marks]
“i.i.d.” means the observations are modelled as independent draws from the same distribution. The product Π runs over observations; the inner sum Σ runs over components. λki in some notes is the same responsibility that other sections call γik. A superscript i here labels the observation, not an exponent.
Jensen’s inequality gives a lower bound on the log of the component sum. EM chooses responsibilities making that bound touch the current likelihood, then updates the parameters to maximise the bound with those responsibilities fixed. This explains the two alternating steps; λ values are not arbitrary new model parameters.
| Parameter | Number in a 1D K-component GMM | Why |
|---|---|---|
| Means | K | One centre for each component |
| Variances | K | One spread for each component |
| Independent mixing weights | K−1 | The last weight is 1 minus the others |
| Total | 3K−1 | K+K+(K−1) |
For K=2, choose μ₁,μ₂,σ₁²,σ₂² and π₁: five free numbers. Then π₂=1−π₁. For K=5, 5+5+4=14. In 2D with a full covariance matrix, each component instead has two means and three independent covariance entries (two variances and one shared off-diagonal covariance). Five such components have 5×(2+3)+4=29 free parameters.
| Aspect | k-means | GMM |
|---|---|---|
| Assignment | Hard: each point in exactly one cluster | Soft: probability of belonging to each cluster |
| Model | Geometric: nearest centroid | Probabilistic generative model |
| Cluster shape | Spherical, similar size | Elliptical, any size/orientation; overlapping clusters |
| Parameters | K centroids | Weights, means, covariances (1-D: 3K − 1) |
| Algorithm / objective | Assign–update; minimise squared distances | EM; maximise log-likelihood |
| Output | Hard labels and fitted centroids | Labels, membership probabilities and a density (useful for anomaly detection) |
| Relation | The hard-assignment, equal spherical variance, small-variance limit of a Gaussian-mixture objective leads to the k-means objective. Ordinary finite-variance EM is not the same algorithm as k-means. Both need K in advance and depend on initialisation. |
Step 1. Model. Each point is generated by choosing component k with probability and drawing , so its density is .
Step 2. Likelihood. Points are i.i.d., so the likelihood is a product (θ = all parameters):
\r\nStep 3. Log-likelihood. Logs turn the product into a sum:
\r\nStep 4. Why it cannot be solved directly. The sum sits inside the log, so setting derivatives to zero gives no closed-form solution.
\r\nStep 5. Jensen's inequality. log is concave, so for weights : . Introduce for every point (), multiply and divide inside the log:
Step 6. Estimation (EM). Maximise the lower bound alternately. Fixing θ gives the E-step . Fixing λ and differentiating (with a Lagrange multiplier for ) gives the M-step:
The dimension and covariance structure must also be specified. For the one-dimensional, independent-variance model used in the derivation:
\r\n(5 means + 5 variances + 4 free weights, since the 5 weights must add to 1.) In two dimensions with full covariance it would be 29.
\r\n (≥, not >).Question. Cloudy → Sprinkler, Cloudy → Rain, Sprinkler and Rain → Wet Grass. P(Cloudy) = 0.5; P(Sprinkler | C) = 0.10, P(Sprinkler | ¬C) = 0.50; P(Rain | C) = 0.80, P(Rain | ¬C) = 0.20; P(Wet | S, R) = 0.99, P(Wet | S, ¬R) = 0.90, P(Wet | ¬S, R) = 0.90, P(Wet | ¬S, ¬R) = 0.01. Calculate the probability of the grass being wet given it doesn't rain. [10 marks]
The following table enumerates the values of Cloudy and Sprinkler that were not fixed by the question. Two Boolean choices give four rows: true/true, true/false, false/true, false/false. Each row is one mutually exclusive complete case. Read the CPT entry using that row’s parent values, multiply its factors, and add the resulting row weights. The denominator must include all cases matching the evidence, including both possible values of the queried outcome.
C→S, C→R, (S,R)→W. Root and conditional probabilities are listed in the question. Evidence R=false.
P(W=true | R=false). Hidden variables C and S must be summed out.
Use the network joint and divide by the evidence probability.
| C | S | P(C) | P(S|C) | P(¬R|C) | P(W|S,¬R) | Product |
|---|---|---|---|---|---|---|
| T | T | 0.5 | 0.1 | 0.2 | 0.90 | 0.0090 |
| T | F | 0.5 | 0.9 | 0.2 | 0.01 | 0.0009 |
| F | T | 0.5 | 0.5 | 0.8 | 0.90 | 0.1800 |
| F | F | 0.5 | 0.5 | 0.8 | 0.01 | 0.0020 |
| P(W, ¬R) | 0.1919 |
The four no-rain cases already total 0.1919. Now add the four rain cases, keeping the same Cloudy/Sprinkler order.
\r\nMeaning: the unconditional chance of Wet is 0.65, the chance of Rain among wet cases is about 0.704769, and the requested chance of Wet among no-rain cases is 0.3838. These are three different questions.
\r\nA bank uses fuzzy decision support with Credit Score and Income Stability, both on 1–100. Loan Approval is a score on 1–100 with labels Reject, Consider and Approve. Define triangular membership functions, construct nine rules, explain Mamdani inference and defuzzify for Credit Score=55 and Income Stability=35. [10 marks]
| Variable | Crisp value |
|---|---|
| Credit Score | 55 |
| Income Stability | 35 |
The question asks us to design the membership functions. This solution chooses Low=Tri(1,25,50), Medium=Tri(25,50,75), High=Tri(50,75,100) for both inputs. Output labels use the same triangles in their respective order. These are explicit design choices, not additional printed givens.
A complete stated design and its continuous Mamdani centroid.
Use min for AND and implication, max for aggregation and the true area centroid of the resulting shape. A peak-only weighted average is a different defuzzifier.
| Input / output position | Triangle | Peak |
|---|---|---|
| Low / Reject | (1,25,50) | 25 |
| Medium / Consider | (25,50,75) | 50 |
| High / Approve | (50,75,100) | 75 |
Use 0 outside each triangle, the rising fraction before the peak and the falling fraction after it. This design is evaluated at the requested input; a deployed controller would also need coverage/fallback at all boundary values.
| Credit Score / Income Stability | Low | Medium | High |
|---|---|---|---|
| Low | Reject | Reject | Consider |
| Medium | Reject | Consider | Approve |
| High | Consider | Approve | Approve |
Read each cell as an IF–AND–THEN rule. For example, the centre cell combines Medium with Medium.
| Credit Score | Income Stability | Firing strength | Consequent |
|---|---|---|---|
| Low | Low | min(0,0.6) = 0 | Reject |
| Low | Medium | min(0,0.4) = 0 | Reject |
| Low | High | min(0,0) = 0 | Consider |
| Medium | Low | min(0.8,0.6) = 0.6 | Reject |
| Medium | Medium | min(0.8,0.4) = 0.4 | Consider |
| Medium | High | min(0.8,0) = 0 | Approve |
| High | Low | min(0.2,0.6) = 0.2 | Consider |
| High | Medium | min(0.2,0.4) = 0.2 | Approve |
| High | High | min(0.2,0) = 0 | Approve |
Aggregated label strengths: Reject = 0.6, Consider = 0.4, Approve = 0.2.
For a strip from a to b with endpoint heights u and v, its trapezoid area is A. Its first moment M is the integral of z times the line height.
| a | b | u | v | Area A | Moment M |
|---|---|---|---|---|---|
| 1 | 15.4 | 0 | 0.6 | 4.32 | 45.792 |
| 15.4 | 25 | 0.6 | 0.6 | 5.76 | 116.352 |
| 25 | 35 | 0.6 | 0.6 | 6 | 180 |
| 35 | 40 | 0.6 | 0.4 | 2.5 | 93.333333333 |
| 40 | 50 | 0.4 | 0.4 | 4 | 180 |
| 50 | 55 | 0.4 | 0.4 | 2 | 105 |
| 55 | 65 | 0.4 | 0.4 | 4 | 240 |
| 65 | 70 | 0.4 | 0.2 | 1.5 | 100.833333333 |
| 70 | 75 | 0.2 | 0.2 | 1 | 72.5 |
| 75 | 95 | 0.2 | 0.2 | 4 | 340 |
| 95 | 100 | 0.2 | 0 | 0.5 | 48.333333333 |
Because the paper asks for a design, a different explicitly specified membership/rule system can yield a different numerical output. Explain the choices and use the chosen defuzzification method consistently; no particular examiner mark outcome is promised.
\r\n\r\n\r\nFor each output triangle, solve “line height = clip height”. Reject is clipped at .6: rising endpoint 1+.6(25−1)=15.4 and falling endpoint 50−.6(50−25)=35. Consider is clipped at .4: endpoints 25+.4(50−25)=35 and 75−.4(75−50)=65. Approve is clipped at .2: endpoints 50+.2(75−50)=55 and 100−.2(100−75)=95.
Now take the maximum of the three clipped curves. The declining Reject edge reaches the Consider height .4 at (50−z)/25=.4, hence z=40. The declining Consider edge reaches the Approve height .2 at (75−z)/25=.2, hence z=70. These crossover locations must be included when splitting the aggregate. Original peaks and other clip endpoints may add extra splits without changing the total area.
For the first aggregate strip from a=1 to b=15.4, endpoint heights are u=0 and v=0.6. Find its area and first moment. [3 marks · Worked example]
Strip endpoints a=1 and b=15.4; left height u=0 and right height v=0.6. This is a straight rising edge of the clipped Reject set.
A and M for one table row.
A is trapezoid area. M weights every small area by its horizontal position; this formula integrates a straight line.
Choose the output label having the greatest membership at the centroid; this is an additional stated interpretation rule. At z=42.780888139:
The resulting recommendation is Consider / manual review; this design does not recommend automatic approval. If a binary approve/reject answer is required, the question must supply a threshold or you must state the threshold you adopt. A different decision threshold is separate from defuzzification.
"},{"id":"q4a","title":"Q4(a) Cartesian product and max–min composition [10]","html":"Question. Consider fuzzy sets A = {0.4/small + 0.3/medium + 0.1/large} and B = {0.2/slow + 0.7/moderate + 0.3/fast}. 1. Find R = A × B. 2. Given C = {0.3/small + 0.1/medium + 0.4/large}, find S = C × B. 3. Calculate the max–min composition S ∘ R. [10 marks]
The fuzzy sets or relation matrices and their listed row/column order.
The relation matrix/matrices and the composition specified in the question.
For Cartesian products take min of row and column membership. For the ordered matrix composition shown here:
The printed relation domains do not directly compose: both relate size to speed. For the numerical exercise, explicitly assume positional alignment and compute row of S with column of R. Equal dimensions alone are not a justification for semantic relation composition.
\r\nRow 2 of S is all 0.1, so every entry of row 2 is 0.1.
\r\n.Both printed relations run from size to speed; their middle labels do not match for genuine relation composition. The numerical working below follows the question’s intended row-of-S with column-of-R operation using explicitly assumed positional alignment of the three middle entries.
| Cell | Row degree | Column degree | min result |
|---|---|---|---|
| R11 | 0.4 | 0.2 | 0.2 |
| R12 | 0.4 | 0.7 | 0.4 |
| R13 | 0.4 | 0.3 | 0.3 |
| R21 | 0.3 | 0.2 | 0.2 |
| R22 | 0.3 | 0.7 | 0.3 |
| R23 | 0.3 | 0.3 | 0.3 |
| R31 | 0.1 | 0.2 | 0.1 |
| R32 | 0.1 | 0.7 | 0.1 |
| R33 | 0.1 | 0.3 | 0.1 |
| Cell | Row degree | Column degree | min result |
|---|---|---|---|
| S11 | 0.3 | 0.2 | 0.2 |
| S12 | 0.3 | 0.7 | 0.3 |
| S13 | 0.3 | 0.3 | 0.3 |
| S21 | 0.1 | 0.2 | 0.1 |
| S22 | 0.1 | 0.7 | 0.1 |
| S23 | 0.1 | 0.3 | 0.1 |
| S31 | 0.4 | 0.2 | 0.2 |
| S32 | 0.4 | 0.7 | 0.4 |
| S33 | 0.4 | 0.3 | 0.3 |
Result cell (1,1)
Result cell (1,2)
Result cell (1,3)
Result cell (2,1)
Result cell (2,2)
Result cell (2,3)
Result cell (3,1)
Result cell (3,2)
Result cell (3,3)
Question. A hospital wants to predict flu from Fever (Yes/No), Cough (Yes/No) and Temperature Level (Low/Normal/High). Training data: (Yes, Yes, High, Yes), (No, No, Normal, No), (Yes, No, Low, No), (Yes, Yes, High, Yes), (No, Yes, Normal, No), (Yes, Yes, High, Yes), (No, No, Low, No), (Yes, No, Normal, No). Using naive Bayes, predict whether a patient with Fever = Yes, Cough = Yes, Temperature = Normal has flu. [10 marks]
The full training data are in Step 1. Query: Fever=Yes, Cough=Yes, Temperature=Normal. The main calculation uses unsmoothed counts.
Both class scores and the predicted class.
C is a class label. Each aⱼ is one query feature value. First count inside a class, then multiply the prior and all three likelihoods.
| Row | Fever | Cough | Temperature | Class |
|---|---|---|---|---|
| 1 | Yes | Yes | High | Yes |
| 2 | No | No | Normal | No |
| 3 | Yes | No | Low | No |
| 4 | Yes | Yes | High | Yes |
| 5 | No | Yes | Normal | No |
| 6 | Yes | Yes | High | Yes |
| 7 | No | No | Low | No |
| 8 | Yes | No | Normal | No |
| Class | Rows in this group | Class count | Prior |
|---|---|---|---|
| Yes | 1, 4, 6 | 3 | 3/8 = 0.375 |
| No | 2, 3, 5, 7, 8 | 5 | 5/8 = 0.625 |
Look only at class Yes rows 1, 4, 6. The matching rows are 1, 4, 6.
Look only at class Yes rows 1, 4, 6. The matching rows are 1, 4, 6.
Look only at class Yes rows 1, 4, 6. No row in this class has that feature value.
Look only at class No rows 2, 3, 5, 7, 8. The matching rows are 3, 8.
Look only at class No rows 2, 3, 5, 7, 8. The matching rows are 5.
Look only at class No rows 2, 3, 5, 7, 8. The matching rows are 2, 5, 8.
The larger score belongs to No. Normalise only if a posterior probability is requested: divide each score by their total 0.03.
The zero Yes score is caused by an absent training combination. If Laplace smoothing is explicitly required, apply it to all feature likelihoods in both classes. That is a different calculation and may change the prediction; do not mix smoothed and unsmoothed sides.
Yaad rakho: Class ke rows alag karo. Har feature ka count usi class ke total se divide. Saare factors multiply karo.
"},{"id":"q5a","title":"Q5(a) Hotel guest movement (Markov chain) [10]","html":"Question. A hotel chain has properties in Paris (P), London (L) and Rome (R). Of guests in Paris, 60% stay, 25% go to London, 15% to Rome; in London 30% go to Paris, 50% stay, 20% go to Rome; in Rome 20% go to Paris, 30% to London, 50% stay. 1. Construct the transition matrix. 2. Explain the diagonal and off-diagonal entries. 3. Draw the state diagram. 4. Today's counts are [50, 30, 20]; find tomorrow's expected counts. 5. Which city gains or loses guests in the long run? [10 marks]
All transition values are given in the full question. State order: P, L, R. Initial vector [50, 30, 20].
The matrix, diagram, next-step vector and long-run comparison.
Rows mean current state; columns mean next state. For each destination, multiply each origin amount by its transition probability, then add.
| From / To | P | L | R |
|---|---|---|---|
| P | 0.6 | 0.25 | 0.15 |
| L | 0.3 | 0.5 | 0.2 |
| R | 0.2 | 0.3 | 0.5 |
Each row sums to 1 because it lists all possible destinations for that origin. A percentage is divided by 100: 25%=.25.
Diagonal entries are the fractions staying at their current state; off-diagonal entries are fractions moving to a different state. Each arrow uses its origin row and destination column. These fractions have different origin populations, so adding raw column probabilities does not give the number arriving.
| Origin | Current count/share | Transition to P | Contribution |
|---|---|---|---|
| P | 50 | 0.6 | 50 × 0.6 = 30 |
| L | 30 | 0.3 | 30 × 0.3 = 9 |
| R | 20 | 0.2 | 20 × 0.2 = 4 |
| Origin | Current count/share | Transition to L | Contribution |
|---|---|---|---|
| P | 50 | 0.25 | 50 × 0.25 = 12.5 |
| L | 30 | 0.5 | 30 × 0.5 = 15 |
| R | 20 | 0.3 | 20 × 0.3 = 6 |
| Origin | Current count/share | Transition to R | Contribution |
|---|---|---|---|
| P | 50 | 0.15 | 50 × 0.15 = 7.5 |
| L | 30 | 0.2 | 30 × 0.2 = 6 |
| R | 20 | 0.5 | 20 × 0.5 = 10 |
Divide every term by .4.
| State | Initial share | Stationary share | Change in share | Conclusion |
|---|---|---|---|---|
| P | 0.5 | 0.391752577 | -0.108247423 | Loses |
| L | 0.3 | 0.350515464 | 0.050515464 | Gains |
| R | 0.2 | 0.257731959 | 0.057731959 | Gains |
Compare like with like: divide counts by their total before comparing with stationary probabilities. “Stationary” means one more step leaves the shares unchanged.
Implement a GA to maximise for x in [0,15]. Take 4-bit representations of 13, 22, 18, 23. Crossover after the second bit and mutate by flipping any one bit. Produce two offspring and comment on fitness. [10 marks]
The printed data are inconsistent. 22, 18 and 23 are outside 0–15 and cannot fit into four bits. Increasing the bit count would encode the numbers but would still violate the stated range. There is no single exact solution satisfying all printed conditions.
The complete worked variant below explicitly changes the invalid parents to their last four bits (values modulo 16), producing [13,6,2,7]. This is one declared repair, not a silent interpretation of the printed question. If the instructor clarifies different corrected data, follow those data.
Printed parents: [13, 22, 18, 23]. Worked variant: [13, 6, 2, 7] with 4 bits. For the unspecified selection step, select the two highest-fitness candidates. For the unspecified mutation location, flip the last bit of each child. Crossover is after bit 2.
Two valid children under the declared assumptions and their fitness.
Encoding must respect the candidate range. Crossover swaps tails; mutation flips the chosen bit. Fitness is evaluated after mutation.
| x | Binary | Fitness substitution |
|---|---|---|
| 13 | 1101 | 2(13³) − 5(13²) + 7(13) = 3640 |
| 6 | 0110 | 2(6³) − 5(6²) + 7(6) = 294 |
| 2 | 0010 | 2(2³) − 5(2²) + 7(2) = 10 |
| 7 | 0111 | 2(7³) − 5(7²) + 7(7) = 490 |
[13, 7] with fitness [3640, 490].
| Parent 1 | Parent 2 | Child 1 | Child 2 |
|---|---|---|---|
| 11 | 01 | 01 | 11 | 1111 | 0101 |
| Before | After | Decoded x |
|---|---|---|
| 1111 | 1110 | 14 |
| 0101 | 0100 | 4 |
| x | Fitness calculation |
|---|---|
| 14 | 2(2744) − 5(196) + 7(14) = 4606 |
| 4 | 2(64) − 5(16) + 7(4) = 76 |
Best parent score = 3640; best child score = 4606. Parent average = 2065; child average = 2341. Both the best score and the average score increased in this worked variant. An individual child can still be worse than a parent; GA does not guarantee improvement after every random operator.
The printed wording about “two offspring” is not an explicit requirement for two complete generations. The M2 examples separately work two full generations because your class update asks for iteration 2.
\r\n\r\n\r\nCalculate x²=x×x, then x³=x²×x. Multiply each by its coefficient and combine with the written signs. A bit-string repair in the worked answer is an explicitly stated response to inconsistent printed givens; it is not a new instruction in the original paper.
“Modulo 16” means keep the remainder after dividing by 16: 22=1×16+6, so its remainder is 6; 18=1×16+2; 23=1×16+7. This is only the declared repair used here, not an unspoken general encoding rule.
List the applications of fuzzy logic. Explain any one application in detail. [10 marks]
Fuzzy logic is a multi-valued logic, introduced by Lotfi Zadeh (1965), in which truth values range over [0, 1]; it lets machines reason with vague, human-like terms such as “slightly dirty” or “very hot”.
\r\n| Area | Examples |
|---|---|
| Home appliances | Washing machines (wash time from dirt and load), air conditioners, refrigerators, rice cookers, vacuum cleaners |
| Automotive | Automatic transmission, anti-lock braking (ABS), cruise control, engine control |
| Transportation | Traffic-signal control, Sendai subway train control (smooth braking and acceleration) |
| Consumer electronics | Camera autofocus and auto-exposure, image stabilisation |
| Industrial control | Cement kilns, pH, temperature and distillation control in chemical plants |
| Aerospace | Satellite and spacecraft attitude control |
| Medicine | Diagnosis support, anaesthesia dosage control |
| Finance and decisions | Loan approval, credit and fire-risk scoring, stock trading rules |
| AI and robotics | Robot navigation, natural-language interfaces, expert systems |
Advantages: handles imprecise or noisy data; mimics human reasoning; easy to design and modify; no exact model needed; low computational cost. Disadvantages: rules and membership functions depend on expert judgement; results are hard to verify formally; tuning can take time; not suited where very high precision is required.
\r\nQuestion. Students are Beginner (B), Intermediate (I) or Advanced (A). Of Beginners, 60% remain, 30% become Intermediate, 10% go to Advanced. Of Intermediates, 20% fall back, 50% remain, 30% progress. Of Advanced, 10% drop to Intermediate, 80% remain and 10% stop (treated as remaining in A). 1. Construct the matrix. 2. Interpret retention, progression and regression. 3. Draw the diagram. 4. For [50, 30, 20], find the distribution after one day. 5. Which level has the largest steady-state share? [10 marks]
All transition values are given in the full question. State order: B, I, A. Initial vector [50, 30, 20].
The matrix, diagram, next-step vector and long-run comparison.
Rows mean current state; columns mean next state. For each destination, multiply each origin amount by its transition probability, then add.
| From / To | B | I | A |
|---|---|---|---|
| B | 0.6 | 0.3 | 0.1 |
| I | 0.2 | 0.5 | 0.3 |
| A | 0 | 0.1 | 0.9 |
Each row sums to 1 because it lists all possible destinations for that origin. A percentage is divided by 100: 25%=.25.
Diagonal entries are the fractions staying at their current state; off-diagonal entries are fractions moving to a different state. Each arrow uses its origin row and destination column. These fractions have different origin populations, so adding raw column probabilities does not give the number arriving.
| Origin | Current count/share | Transition to B | Contribution |
|---|---|---|---|
| B | 50 | 0.6 | 50 × 0.6 = 30 |
| I | 30 | 0.2 | 30 × 0.2 = 6 |
| A | 20 | 0 | 20 × 0 = 0 |
| Origin | Current count/share | Transition to I | Contribution |
|---|---|---|---|
| B | 50 | 0.3 | 50 × 0.3 = 15 |
| I | 30 | 0.5 | 30 × 0.5 = 15 |
| A | 20 | 0.1 | 20 × 0.1 = 2 |
| Origin | Current count/share | Transition to A | Contribution |
|---|---|---|---|
| B | 50 | 0.1 | 50 × 0.1 = 5 |
| I | 30 | 0.3 | 30 × 0.3 = 9 |
| A | 20 | 0.9 | 20 × 0.9 = 18 |
Advanced is not an absorbing state: its self-transition is .9, not 1.
| State | Initial share | Stationary share | Change in share | Conclusion |
|---|---|---|---|---|
| B | 0.5 | 0.1 | -0.4 | Loses |
| I | 0.3 | 0.2 | -0.1 | Loses |
| A | 0.2 | 0.7 | 0.5 | Gains |
Compare like with like: divide counts by their total before comparing with stationary probabilities. “Stationary” means one more step leaves the shares unchanged.
List the applications of Particle Swarm Optimization and explain any one application in detail. [10 marks]
Particle Swarm Optimization is a population-based metaheuristic, inspired by bird flocking and fish schooling, in which particles move through the search space using their own best position (pbest) and the swarm's best position (gbest) to find optimal or near-optimal solutions.
\r\n\r\n| Advantages | Disadvantages |
|---|---|
| Simple to implement; few parameters (w, c₁, c₂) | Can converge prematurely to a local optimum |
| No derivatives needed | Sensitive to parameter choice |
| Fast convergence in early iterations; memory (pbest, gbest) | No guarantee of the global optimum |
| Objective evaluations can be parallel; particles share information through gbest | Less suited to discrete problems without modification |
Marking pattern used for ACO in the same paper: listing 3 marks + explanation 2 marks (5-mark version); listing 5 + explanation 5 (10-mark version).
\r\n\r\nQuestion. For the four-city travelling salesman problem, the distance and pheromone matrices (A, B, C, D) are given. Apply Ant Colony Optimization. Calculate the path travelled by 2 ants to cover all cities starting from city A. Random numbers for Ant 1 are 0.3, 0.8 and for Ant 2 are 0.5, 0.7. Also update the pheromone matrix after 1 iteration. [10 marks]
Assume , , (class convention; not stated in the question).
Tour of each ant, tour lengths, updated pheromone matrix.
The sum includes only ants whose completed tour contains edge ij. An unused edge receives no deposit but still evaporates. Here distances are symmetric, so a traversed edge updates both symmetric matrix cells. Both ants construct their tours using the old pheromones; update after both finish, including their return to A.
\r\n| Move | P | Cumulative | Range |
|---|---|---|---|
| A→B | 0.25 | 0.25 | 0–0.25 |
| A→C | 0.50 | 0.75 | 0.25–0.75 |
| A→D | 0.25 | 1.00 | 0.75–1 |
Ant 1: r = 0.3 → C. Ant 2: r = 0.5 → C.
Cumulative in city order: B 0–0.4, D 0.4–1. Ant 1: r = 0.8 → D. Ant 2: r = 0.7 → D.
Only B is left, then return to A.
\r\nEach ant deposits on each edge of its tour; two ants used the same four edges (AC, CD, DB, BA), so each gets .
Official marking: path of Ant 1 (3) + path of Ant 2 (3) + pheromone matrix (4).
\r\nQuestion. For a disease Y with fever X as a symptom, a test Z is conducted. Joint probabilities: Disease (Y=1): fever & test+ 0.576, fever & test− 0.144, no fever & test+ 0.072, no fever & test− 0.016; No disease (Y=0): 0.008, 0.064, 0.012, 0.108. (1) If fever occurs and the test is positive, what is the probability of disease? (2) What is the probability that a random person tests positive? (3) Given no fever, what is the probability of still having the disease? [5 marks]
The complete joint table or cell values printed in this question. The rows and columns identify the event combinations.
All requested marginal and conditional probabilities.
Marginal: add cells satisfying the event. Conditional: retain the evidence cells and divide the joint-event sum by the evidence sum.
List the applications of Particle Swarm Optimization and explain any one application in detail. [5 marks]
Particle Swarm Optimization is a population-based metaheuristic, inspired by bird flocking and fish schooling, in which particles move through the search space using their own best position (pbest) and the swarm's best position (gbest) to find optimal or near-optimal solutions.
\r\n\r\n| Advantages | Disadvantages |
|---|---|
| Simple to implement; few parameters (w, c₁, c₂) | Can converge prematurely to a local optimum |
| No derivatives needed | Sensitive to parameter choice |
| Fast convergence in early iterations; memory (pbest, gbest) | No guarantee of the global optimum |
| Objective evaluations can be parallel; particles share information through gbest | Less suited to discrete problems without modification |
Marking pattern used for ACO in the same paper: listing 3 marks + explanation 2 marks (5-mark version); listing 5 + explanation 5 (10-mark version).
\r\n\r\nQuestion. X = {(c1, 0.7), (c2, 0.9), (c3, 0.2), (c4, 0.4), (c5, 0.9)}, Y = {(c1, 0.6), (c2, 0.3), (c3, 0.8), (c4, 0.7), (c5, 0.1)}. Find X ∪ Y, X ∩ Y, ¬X ∪ ¬Y, ¬X ∩ ¬Y. [5 marks]
The aligned X and Y memberships printed in this question.
The four requested membership lists in the original element order.
Use max for OR, min for AND, and 1−μ for NOT.
| c1 | c2 | c3 | c4 | c5 | |
|---|---|---|---|---|---|
| ¬X | 0.3 | 0.1 | 0.8 | 0.6 | 0.1 |
| ¬Y | 0.4 | 0.7 | 0.2 | 0.3 | 0.9 |
| X ∪ Y | 0.7 | 0.9 | 0.8 | 0.7 | 0.9 |
| X ∩ Y | 0.6 | 0.3 | 0.2 | 0.4 | 0.1 |
| ¬X ∪ ¬Y | 0.4 | 0.7 | 0.8 | 0.6 | 0.9 |
| ¬X ∩ ¬Y | 0.3 | 0.1 | 0.2 | 0.3 | 0.1 |
Question. Each year 20% of Savings holders switch to Current and 80% stay; 10% of Current holders switch to Savings and 90% stay. (a) Construct the transition matrix. (b) Draw the state diagram. [5 marks]
All transition values are given in the full question. State order: S, C.
The transition matrix and state diagram.
Rows mean current state; columns mean next state. For each destination, multiply each origin amount by its transition probability, then add.
| From / To | S | C |
|---|---|---|
| S | 0.8 | 0.2 |
| C | 0.1 | 0.9 |
Each row sums to 1 because it lists all possible destinations for that origin. A percentage is divided by 100: 25%=.25.
Diagonal entries are the fractions staying at their current state; off-diagonal entries are fractions moving to a different state. Each arrow uses its origin row and destination column. These fractions have different origin populations, so adding raw column probabilities does not give the number arriving.
A building safety department wants a fuzzy logic system to predict fire hazards. Inputs are Temperature (1–100°C) and Smoke Level (1–100). The output Fire Risk (1–100) has labels Low, Moderate and High. Define triangular membership functions, construct at least nine rules, explain Mamdani inference and defuzzify for Temperature=65°C and Smoke=80. [10 marks]
| Variable | Crisp value |
|---|---|
| Temperature | 65 |
| Smoke Level | 80 |
The question asks us to design the membership functions. This solution chooses Low=Tri(1,25,50), Medium=Tri(25,50,75), High=Tri(50,75,100) for both inputs. Output labels use the same triangles in their respective order. These are explicit design choices, not additional printed givens.
A complete stated design and its continuous Mamdani centroid.
Use min for AND and implication, max for aggregation and the true area centroid of the resulting shape. A peak-only weighted average is a different defuzzifier.
| Input / output position | Triangle | Peak |
|---|---|---|
| Low / Low | (1,25,50) | 25 |
| Medium / Moderate | (25,50,75) | 50 |
| High / High | (50,75,100) | 75 |
Use 0 outside each triangle, the rising fraction before the peak and the falling fraction after it. This design is evaluated at the requested input; a deployed controller would also need coverage/fallback at all boundary values.
| Temperature / Smoke Level | Low | Medium | High |
|---|---|---|---|
| Low | Low | Low | Moderate |
| Medium | Low | Moderate | High |
| High | Moderate | High | High |
Read each cell as an IF–AND–THEN rule. For example, the centre cell combines Medium with Medium.
| Temperature | Smoke Level | Firing strength | Consequent |
|---|---|---|---|
| Low | Low | min(0,0) = 0 | Low |
| Low | Medium | min(0,0) = 0 | Low |
| Low | High | min(0,0.8) = 0 | Moderate |
| Medium | Low | min(0.4,0) = 0 | Low |
| Medium | Medium | min(0.4,0) = 0 | Moderate |
| Medium | High | min(0.4,0.8) = 0.4 | High |
| High | Low | min(0.6,0) = 0 | Moderate |
| High | Medium | min(0.6,0) = 0 | High |
| High | High | min(0.6,0.8) = 0.6 | High |
Aggregated label strengths: Low = 0, Moderate = 0, High = 0.6.
For a strip from a to b with endpoint heights u and v, its trapezoid area is A. Its first moment M is the integral of z times the line height.
| a | b | u | v | Area A | Moment M |
|---|---|---|---|---|---|
| 50 | 65 | 0 | 0.6 | 4.5 | 270 |
| 65 | 85 | 0.6 | 0.6 | 12 | 900 |
| 85 | 100 | 0.6 | 0 | 4.5 | 405 |
Because the paper asks for a design, a different explicitly specified membership/rule system can yield a different numerical output. Explain the choices and use the chosen defuzzification method consistently; no particular examiner mark outcome is promised.
\r\n\r\n\r\nOnly the High output survives, at height .6. Its triangle is (50,75,100). Solve its rising edge: (z−50)/25=.6, so z=50+15=65. Solve its falling edge: (100−z)/25=.6, so z=100−15=85.
| Piece | Area calculation | Centre | Moment |
|---|---|---|---|
| Left triangle [50,65] | ½×15×.6=4.5 | (50+65+65)/3=60 | 4.5×60=270 |
| Rectangle [65,85] | 20×.6=12 | (65+85)/2=75 | 12×75=900 |
| Right triangle [85,100] | ½×15×.6=4.5 | (85+85+100)/3=90 | 4.5×90=405 |
The horizontal centroid of a triangle is the average of the three vertex x-coordinates. Add areas: 4.5+12+4.5=21. Add moments:270+900+405=1575. Divide 1575/21=75, agreeing with symmetry.
With reference to the Gaussian Mixture Model (GMM): compare GMM with the k-means algorithm; derive the equation for the log-likelihood function; state the number of parameters to be computed if K = 5. [10 marks]
“i.i.d.” means the observations are modelled as independent draws from the same distribution. The product Π runs over observations; the inner sum Σ runs over components. λki in some notes is the same responsibility that other sections call γik. A superscript i here labels the observation, not an exponent.
Jensen’s inequality gives a lower bound on the log of the component sum. EM chooses responsibilities making that bound touch the current likelihood, then updates the parameters to maximise the bound with those responsibilities fixed. This explains the two alternating steps; λ values are not arbitrary new model parameters.
| Parameter | Number in a 1D K-component GMM | Why |
|---|---|---|
| Means | K | One centre for each component |
| Variances | K | One spread for each component |
| Independent mixing weights | K−1 | The last weight is 1 minus the others |
| Total | 3K−1 | K+K+(K−1) |
For K=2, choose μ₁,μ₂,σ₁²,σ₂² and π₁: five free numbers. Then π₂=1−π₁. For K=5, 5+5+4=14. In 2D with a full covariance matrix, each component instead has two means and three independent covariance entries (two variances and one shared off-diagonal covariance). Five such components have 5×(2+3)+4=29 free parameters.
| Aspect | k-means | GMM |
|---|---|---|
| Assignment | Hard: each point in exactly one cluster | Soft: probability of belonging to each cluster |
| Model | Geometric: nearest centroid | Probabilistic generative model |
| Cluster shape | Spherical, similar size | Elliptical, any size/orientation; overlapping clusters |
| Parameters | K centroids | Weights, means, covariances (1-D: 3K − 1) |
| Algorithm / objective | Assign–update; minimise squared distances | EM; maximise log-likelihood |
| Output | Hard labels and fitted centroids | Labels, membership probabilities and a density (useful for anomaly detection) |
| Relation | The hard-assignment, equal spherical variance, small-variance limit of a Gaussian-mixture objective leads to the k-means objective. Ordinary finite-variance EM is not the same algorithm as k-means. Both need K in advance and depend on initialisation. |
Step 1. Model. Each point is generated by choosing component k with probability and drawing , so its density is .
Step 2. Likelihood. Points are i.i.d., so the likelihood is a product (θ = all parameters):
\r\nStep 3. Log-likelihood. Logs turn the product into a sum:
\r\nStep 4. Why it cannot be solved directly. The sum sits inside the log, so setting derivatives to zero gives no closed-form solution.
\r\nStep 5. Jensen's inequality. log is concave, so for weights : . Introduce for every point (), multiply and divide inside the log:
Step 6. Estimation (EM). Maximise the lower bound alternately. Fixing θ gives the E-step . Fixing λ and differentiating (with a Lagrange multiplier for ) gives the M-step:
The dimension and covariance structure must also be specified. For the one-dimensional, independent-variance model used in the derivation:
\r\n(5 means + 5 variances + 4 free weights, since the 5 weights must add to 1.) In two dimensions with full covariance it would be 29.
\r\n (≥, not >).Question. Weather → Flu, Weather → Cough, Flu and Cough → Fever. P(Weather) = 0.75; P(Flu | W) = 0.2, P(Flu | ¬W) = 0.5; P(Cough | W) = 0.70, P(Cough | ¬W) = 0.25; P(Fever | Flu, Cough) = 0.60, P(Fever | Flu, ¬Cough) = 0.02, P(Fever | ¬Flu, Cough) = 0.60, P(Fever | ¬Flu, ¬Cough) = 0.01. Calculate the probability of a person having fever given he doesn't have flu. [10 marks]
The following table enumerates the values of Weather and Cough that were not fixed by the question. Two Boolean choices give four rows: true/true, true/false, false/true, false/false. Each row is one mutually exclusive complete case. Read the CPT entry using that row’s parent values, multiply its factors, and add the resulting row weights. The denominator must include all cases matching the evidence, including both possible values of the queried outcome.
W→F, W→C, (F,C)→V. The complete probabilities are in the question. Evidence F=false.
P(V=true | F=false), summing hidden W and C.
Use the network joint and normalise by the no-Flu evidence.
| W | C | P(W) | P(¬F|W) | P(C|W) | P(V|¬F,C) | Product |
|---|---|---|---|---|---|---|
| T | T | 0.75 | 0.8 | 0.70 | 0.60 | 0.252 |
| T | F | 0.75 | 0.8 | 0.30 | 0.01 | 0.0018 |
| F | T | 0.25 | 0.5 | 0.25 | 0.60 | 0.01875 |
| F | F | 0.25 | 0.5 | 0.75 | 0.01 | 0.0009375 |
| P(V, ¬F) | 0.2734875 |
Question. A = {0.6/small + 0.4/medium + 0.1/large}, B = {0.4/slow + 0.8/moderate + 0.3/fast}, C = {0.5/small + 0.2/medium + 0.4/large}. Find R = A × B, S = C × B and the max–min composition S ∘ R. [10 marks]
The fuzzy sets or relation matrices and their listed row/column order.
The relation matrix/matrices and the composition specified in the question.
For Cartesian products take min of row and column membership. For the ordered matrix composition shown here:
, , . (The key gives R and S only.)Both printed relations run from size to speed; their middle labels do not match for genuine relation composition. The numerical working below follows the question’s intended row-of-S with column-of-R operation using explicitly assumed positional alignment of the three middle entries.
| Cell | Row degree | Column degree | min result |
|---|---|---|---|
| R11 | 0.6 | 0.4 | 0.4 |
| R12 | 0.6 | 0.8 | 0.6 |
| R13 | 0.6 | 0.3 | 0.3 |
| R21 | 0.4 | 0.4 | 0.4 |
| R22 | 0.4 | 0.8 | 0.4 |
| R23 | 0.4 | 0.3 | 0.3 |
| R31 | 0.1 | 0.4 | 0.1 |
| R32 | 0.1 | 0.8 | 0.1 |
| R33 | 0.1 | 0.3 | 0.1 |
| Cell | Row degree | Column degree | min result |
|---|---|---|---|
| S11 | 0.5 | 0.4 | 0.4 |
| S12 | 0.5 | 0.8 | 0.5 |
| S13 | 0.5 | 0.3 | 0.3 |
| S21 | 0.2 | 0.4 | 0.2 |
| S22 | 0.2 | 0.8 | 0.2 |
| S23 | 0.2 | 0.3 | 0.2 |
| S31 | 0.4 | 0.4 | 0.4 |
| S32 | 0.4 | 0.8 | 0.4 |
| S33 | 0.4 | 0.3 | 0.3 |
Result cell (1,1)
Result cell (1,2)
Result cell (1,3)
Result cell (2,1)
Result cell (2,2)
Result cell (2,3)
Result cell (3,1)
Result cell (3,2)
Result cell (3,3)
Question. Maximize using Particle Swarm Optimization. . Take . Optimize and give the solution after 1 iteration showing the step-by-step process. [10 marks]
P1: , . P2: , . , .
Updated velocities, positions, pbest and gbest after one iteration.
Initial personal bests equal the starting positions. Use one frozen global best for all particles. This question maximises fitness, so keep a memory only when its score is larger.
pbest = current positions. , so gbest = P2 = (4.0579, −2.2150).
Use the same objective at each new position. Keep full precision internally; displayed coordinates are rounded.
| Particle | New position | New F | Old F | pbest |
|---|---|---|---|---|
| 1 | (3.461457, −3.572109) | 38.2023 | 37.1117 | updated |
| 2 | (4.22962, −2.01448) | 40.0992 | 38.6747 | updated |
The best pbest is now P2's with F = 40.0992, so gbest moves to (4.22962, −2.01448).
, . Both particles improved; new gbest = (4.22962, −2.01448) with F = 40.0992.This X and V are particles 1 and 2 of your lecture's five-particle example; only the function differs (+3x₁ and +5 here instead of +2x₁ and +3).
\r\nQuestion. Predict whether a candidate accepts a job offer from Experience (Fresher/Mid/Senior), Expected Salary (Low/Medium/High) and Company Reputation (Good/Average). Data: (Fresher, Low, Good, Yes), (Mid, High, Good, Yes), (Senior, High, Average, No), (Fresher, Medium, Average, Yes), (Senior, Low, Good, No), (Mid, Medium, Good, Yes), (Fresher, High, Average, No), (Senior, Medium, Average, No). New candidate: Mid, Low, Good. [10 marks]
The full training data are in Step 1. Query: Experience=Mid, Salary=Low, Reputation=Good. The main calculation uses unsmoothed counts.
Both class scores and the predicted class.
C is a class label. Each aⱼ is one query feature value. First count inside a class, then multiply the prior and all three likelihoods.
| Row | Experience | Salary | Reputation | Class |
|---|---|---|---|---|
| 1 | Fresher | Low | Good | Yes |
| 2 | Mid | High | Good | Yes |
| 3 | Senior | High | Average | No |
| 4 | Fresher | Medium | Average | Yes |
| 5 | Senior | Low | Good | No |
| 6 | Mid | Medium | Good | Yes |
| 7 | Fresher | High | Average | No |
| 8 | Senior | Medium | Average | No |
| Class | Rows in this group | Class count | Prior |
|---|---|---|---|
| Yes | 1, 2, 4, 6 | 4 | 4/8 = 0.5 |
| No | 3, 5, 7, 8 | 4 | 4/8 = 0.5 |
Look only at class Yes rows 1, 2, 4, 6. The matching rows are 2, 6.
Look only at class Yes rows 1, 2, 4, 6. The matching rows are 1.
Look only at class Yes rows 1, 2, 4, 6. The matching rows are 1, 2, 6.
Look only at class No rows 3, 5, 7, 8. No row in this class has that feature value.
Look only at class No rows 3, 5, 7, 8. The matching rows are 5.
Look only at class No rows 3, 5, 7, 8. The matching rows are 5.
The larger score belongs to Yes. Normalise only if a posterior probability is requested: divide each score by their total 0.046875.
Yaad rakho: Class ke rows alag karo. Har feature ka count usi class ke total se divide. Saare factors multiply karo.
"},{"id":"q5a","title":"Q5(a) Genetic algorithm [10]","html":"Implement a GA to maximise for x in [0,50]. Take 4-bit representations of 12, 20, 17, 13. Crossover after the second bit and mutate by flipping any one bit. Produce two offspring and comment on fitness. [10 marks]
The printed data are inconsistent. 20 and 17 do not fit into four bits, and covering 0–50 needs six bits. We use six-bit encoding to preserve the supplied parent values and the stated range.
Printed parents: [12, 20, 17, 13]. Worked variant: [12, 20, 17, 13] with 6 bits. For the unspecified selection step, select the two highest-fitness candidates. For the unspecified mutation location, flip the last bit of each child. Crossover is after bit 2.
Two valid children under the declared assumptions and their fitness.
Encoding must respect the candidate range. Crossover swaps tails; mutation flips the chosen bit. Fitness is evaluated after mutation.
| x | Binary | Fitness substitution |
|---|---|---|
| 12 | 001100 | 2(12³) − 3(12²) + 8(12) = 3120 |
| 20 | 010100 | 2(20³) − 3(20²) + 8(20) = 14960 |
| 17 | 010001 | 2(17³) − 3(17²) + 8(17) = 9095 |
| 13 | 001101 | 2(13³) − 3(13²) + 8(13) = 3991 |
[20, 17] with fitness [14960, 9095].
| Parent 1 | Parent 2 | Child 1 | Child 2 |
|---|---|---|---|
| 01 | 0100 | 01 | 0001 | 010001 | 010100 |
| Before | After | Decoded x |
|---|---|---|
| 010001 | 010000 | 16 |
| 010100 | 010101 | 21 |
| x | Fitness calculation |
|---|---|
| 16 | 2(4096) − 3(256) + 8(16) = 7552 |
| 21 | 2(9261) − 3(441) + 8(21) = 17367 |
Best parent score = 14960; best child score = 17367. Parent average = 12027.5; child average = 12459.5. Both the best score and the average score increased in this worked variant. An individual child can still be worse than a parent; GA does not guarantee improvement after every random operator.
The printed wording about “two offspring” is not an explicit requirement for two complete generations. The M2 examples separately work two full generations because your class update asks for iteration 2.
\r\n\r\n\r\nCalculate x²=x×x, then x³=x²×x. Multiply each by its coefficient and combine with the written signs. A bit-string repair in the worked answer is an explicitly stated response to inconsistent printed givens; it is not a new instruction in the original paper.
Question. Bookings come via Website (W), App (A) or Counter (C). From W: 50% W, 30% A, 20% C. From A: 40% A, 40% W, 20% C. From C: 20% W, 30% A, 50% C. 1. Develop the transition matrix. 2. Interpret retention and gain/loss. 3. Draw the diagram. 4. If now 40% W, 35% A, 25% C, estimate next week. 5. Which channel grows or shrinks in the long term? [10 marks]
All transition values are given in the full question. State order: W, A, C. Initial vector [0.4, 0.35, 0.25].
The matrix, diagram, next-step vector and long-run comparison.
Rows mean current state; columns mean next state. For each destination, multiply each origin amount by its transition probability, then add.
| From / To | W | A | C |
|---|---|---|---|
| W | 0.5 | 0.3 | 0.2 |
| A | 0.4 | 0.4 | 0.2 |
| C | 0.2 | 0.3 | 0.5 |
Each row sums to 1 because it lists all possible destinations for that origin. A percentage is divided by 100: 25%=.25.
Diagonal entries are the fractions staying at their current state; off-diagonal entries are fractions moving to a different state. Each arrow uses its origin row and destination column. These fractions have different origin populations, so adding raw column probabilities does not give the number arriving.
| Origin | Current count/share | Transition to W | Contribution |
|---|---|---|---|
| W | 0.4 | 0.5 | 0.4 × 0.5 = 0.2 |
| A | 0.35 | 0.4 | 0.35 × 0.4 = 0.14 |
| C | 0.25 | 0.2 | 0.25 × 0.2 = 0.05 |
| Origin | Current count/share | Transition to A | Contribution |
|---|---|---|---|
| W | 0.4 | 0.3 | 0.4 × 0.3 = 0.12 |
| A | 0.35 | 0.4 | 0.35 × 0.4 = 0.14 |
| C | 0.25 | 0.3 | 0.25 × 0.3 = 0.075 |
| Origin | Current count/share | Transition to C | Contribution |
|---|---|---|---|
| W | 0.4 | 0.2 | 0.4 × 0.2 = 0.08 |
| A | 0.35 | 0.2 | 0.35 × 0.2 = 0.07 |
| C | 0.25 | 0.5 | 0.25 × 0.5 = 0.125 |
Since w+a+c=1, replace w+a with 1−c.
| State | Initial share | Stationary share | Change in share | Conclusion |
|---|---|---|---|---|
| W | 0.4 | 0.380952381 | -0.019047619 | Loses |
| A | 0.35 | 0.333333333 | -0.016666667 | Loses |
| C | 0.25 | 0.285714286 | 0.035714286 | Gains |
Compare like with like: divide counts by their total before comparing with stationary probabilities. “Stationary” means one more step leaves the shares unchanged.
Question. For the four-city travelling salesman problem the distance matrix and pheromone matrix are given (all pheromones 1). Apply Ant Colony Optimization. Calculate the path travelled by 2 ants starting from city A; random numbers for Ant 1 are 0.3, 0.8 and for Ant 2 are 0.5, 0.7. Also update the pheromone matrix after 1 iteration. [10 marks]
Assume .
Both tours, their lengths and the pheromone matrix after one iteration.
Use unvisited cities in alphabetical order for cumulative selection. Every ant completes its tour using old pheromones; then evaporate and deposit on each undirected edge, including the return edge.
Cumulative: B 0–0.5455, C 0.5455–0.8182, D 0.8182–1. Ant 1 (0.3) → B; Ant 2 (0.5) → B.
Cumulative: C 0–0.5556, D 0.5556–1. Ant 1 (0.8) → D; Ant 2 (0.7) → D.
Used edges AB, BD, DC, CA get ; unused edges AD and BC become .
) and its matrix is inconsistent: it increases the unused edge A–D, leaves the used return edge C–A at 1, and gives B–D and D–B different values. If your examiner follows the no-evaporation idea, the consistent version is 1.1429 on AB, BD, CD, AC and 1 on AD, BC. The paper does not specify ρ. State the convention you use; the resulting matrix depends on it. Grading depends on the examiner’s intended convention.Question. Planes at A: 50% stay, 30% to B, 20% to C. At B: 40% to A, 40% stay, 20% to C. At C: 25% to A, 25% to B, 50% stay. 1. Construct the matrix. 2. Interpret retention and movement. 3. Draw the diagram. 4. For [30, 40, 30] find the next night's distribution. 5. Which airport is a long-term net gainer or loser? [10 marks]
All transition values are given in the full question. State order: A, B, C. Initial vector [30, 40, 30].
The matrix, diagram, next-step vector and long-run comparison.
Rows mean current state; columns mean next state. For each destination, multiply each origin amount by its transition probability, then add.
| From / To | A | B | C |
|---|---|---|---|
| A | 0.5 | 0.3 | 0.2 |
| B | 0.4 | 0.4 | 0.2 |
| C | 0.25 | 0.25 | 0.5 |
Each row sums to 1 because it lists all possible destinations for that origin. A percentage is divided by 100: 25%=.25.
Diagonal entries are the fractions staying at their current state; off-diagonal entries are fractions moving to a different state. Each arrow uses its origin row and destination column. These fractions have different origin populations, so adding raw column probabilities does not give the number arriving.
| Origin | Current count/share | Transition to A | Contribution |
|---|---|---|---|
| A | 30 | 0.5 | 30 × 0.5 = 15 |
| B | 40 | 0.4 | 40 × 0.4 = 16 |
| C | 30 | 0.25 | 30 × 0.25 = 7.5 |
| Origin | Current count/share | Transition to B | Contribution |
|---|---|---|---|
| A | 30 | 0.3 | 30 × 0.3 = 9 |
| B | 40 | 0.4 | 40 × 0.4 = 16 |
| C | 30 | 0.25 | 30 × 0.25 = 7.5 |
| Origin | Current count/share | Transition to C | Contribution |
|---|---|---|---|
| A | 30 | 0.2 | 30 × 0.2 = 6 |
| B | 40 | 0.2 | 40 × 0.2 = 8 |
| C | 30 | 0.5 | 30 × 0.5 = 15 |
| State | Initial share | Stationary share | Change in share | Conclusion |
|---|---|---|---|---|
| A | 0.3 | 0.396825397 | 0.096825397 | Gains |
| B | 0.4 | 0.317460317 | -0.082539683 | Loses |
| C | 0.3 | 0.285714286 | -0.014285714 | Loses |
Compare like with like: divide counts by their total before comparing with stationary probabilities. “Stationary” means one more step leaves the shares unchanged.
List the applications of fuzzy logic. Explain any one application in detail. [10 marks]
Fuzzy logic is a multi-valued logic, introduced by Lotfi Zadeh (1965), in which truth values range over [0, 1]; it lets machines reason with vague, human-like terms such as “slightly dirty” or “very hot”.
\r\n| Area | Examples |
|---|---|
| Home appliances | Washing machines (wash time from dirt and load), air conditioners, refrigerators, rice cookers, vacuum cleaners |
| Automotive | Automatic transmission, anti-lock braking (ABS), cruise control, engine control |
| Transportation | Traffic-signal control, Sendai subway train control (smooth braking and acceleration) |
| Consumer electronics | Camera autofocus and auto-exposure, image stabilisation |
| Industrial control | Cement kilns, pH, temperature and distillation control in chemical plants |
| Aerospace | Satellite and spacecraft attitude control |
| Medicine | Diagnosis support, anaesthesia dosage control |
| Finance and decisions | Loan approval, credit and fire-risk scoring, stock trading rules |
| AI and robotics | Robot navigation, natural-language interfaces, expert systems |
Advantages: handles imprecise or noisy data; mimics human reasoning; easy to design and modify; no exact model needed; low computational cost. Disadvantages: rules and membership functions depend on expert judgement; results are hard to verify formally; tuning can take time; not suited where very high precision is required.
\r\nExplain the Bellman equation in reinforcement learning. Also give applications of reinforcement learning. [10 marks]
Write the answer below in this order. It fits in about two pages.
\n\n1. Setting. Reinforcement learning problems are modelled as a Markov decision process . The agent follows a policy and wants to maximise the return .
2. Value functions. and .
3. Key idea. The return is recursive: . Taking expectations gives the Bellman expectation equation:
4. Optimality. For the optimal policy the average over actions becomes a maximum (Bellman optimality equation):
\n5. Meaning and use. The value of a state equals the immediate reward plus the discounted value of the successor state. This breaks a long-horizon problem into one-step sub-problems (dynamic programming). It is the basis of policy evaluation, policy iteration, value iteration and Q-learning; the optimal policy is .
6. Small illustration (adds clarity): if staying in a state gives +1 each step with , then .
| Agent | Controller at an intersection. |
|---|---|
| State | Queue length and waiting time on each approach, current signal phase. |
| Actions | Keep the current phase or switch to the next green phase. |
| Reward | Negative of total waiting time (or vehicles cleared per cycle). |
| Learning | Q-learning updates from observed traffic; ε-greedy exploration tries alternative timings. |
| Benefit | Adapts to rush hours and incidents; fixed timers cannot. |
| Challenge | Safety during exploration, so agents are trained in simulation first. |
| Asked for | Write |
|---|---|
| Union / intersection | Same element, max / min respectively. |
| Complement / difference | 1−μA / min(μA,1−μB). |
| Triangle membership | Locate x; rising (x−a)/(b−a), falling (c−x)/(c−b), 0 outside, 1 at peak. |
| Swarm intelligence | Collective problem-solving emerging from interacting simple agents; explain PSO through position, velocity, pbest and gbest. |
Individual → current x → bit string → fitness → selected parent → crossover child → mutated child → decoded x → new fitness. Use this resulting population as the input to the second generation. The cut position and flipped bit are counted from the stated end; this guide counts from the left.
\r\nDo not change the mutation instruction midway because an answer looks worse. Without elitism, fitness can worsen. Do not confuse a mating-pool index with the decimal x value.
\r\nParticle → old position → old velocity → pbest → gbest → inertia/personal/social terms → new velocity → new position → new objective → retained pbest. After all particles, update gbest. For synchronous updates, freeze the old gbest throughout one iteration.
\r\nAt initialisation pbest=x, hence personal pull=0. In iteration 2 it need not be zero. For minimisation, smaller objective wins; a worse move must not overwrite a better memory.
\r\nMamdani: crisp inputs → fuzzifier → knowledge base and inference engine → aggregation → defuzzifier → crisp output. Rules are interpretable, but require careful membership/rule design and can grow in number.
GA: population search inspired by evolution, using evaluation, selection, crossover, mutation and replacement. Flexible and derivative-free, but may be expensive or converge prematurely.
PSO: particles move using inertia, personal best and swarm best. Simple and derivative-free, but parameter-dependent and not guaranteed to find a global optimum.
\r\nEnter a fraction as (whole numerator) ÷ (whole denominator). For a negative square use parentheses: (−2)²=4. Retain full calculator precision through iterations and round only the displayed result. Binary place values 8,4,2,1 can be checked manually without a special calculator mode.
\r\nYaad rakho: Question ki instruction follow karo; answer ko sundar banane ke liye data ya operator mat badlo.
"}]},{"id":"tee-revision","title":"Last-day revision · TEE","unit":"0","scope":"Read the night before the TEE","sections":[{"id":"units-1-and-4","title":"Units 1 and 4 in one page","html":" numbers per Boolean node.. With K separate means, K variances and K weights summing to 1, there are 3K − 1 free parameters (K=5 → 14). Higher-dimensional covariance models have different counts. EM alternates responsibilities and weighted parameter updates.The 8 October class update identifies Fuzzy Logic, GA and PSO, including rules, architecture, advantages/disadvantages, fuzzy operations and numerical work through iteration 2. The confirmed pattern is Q1(a) 2 marks and Q1(b) 2 marks compulsory, plus any two of three 3-mark questions. ACO and GA-for-TSP remain in TEE, not the M2 core.
\r\nThe course policy specifies a 100-mark, three-hour TEE scaled to 50, and 10-mark class tests. The available Final 24 November 2025 and Re-exam 31 January 2026 papers have Q1 compulsory plus any four of Q2–Q7. The next paper’s instructions take precedence.
\r\nPrimary course sources are the AY 2026–27 syllabus and course-policy files, faculty slides and solved-problem material in the 8 October OneDrive archive, the two actual question papers, and the class data supplied in the conversation. The old generated IS2 draft and the separate generated midterm PDF are not authorities for the new teaching sequence.
\r\nHandwritten photos can contain inconsistent arithmetic or incomplete conventions. The fan example states its chosen membership functions; the PSO example states minimisation and synchronous updates. Missing selection choices, mutation bits or inconsistent encodings in older papers are labelled explicitly rather than presented as unique answers.
\r\nNew worked numerical examples are calculated by reproducible scripts and checked separately. Build checks validate formula rendering and internal routes. This does not justify a promise that no typo can remain, that every future numerical is predictable, or that reading alone guarantees a score.
"},{"id":"corrections","title":"How inconsistent source questions are handled","html":"The final GA paper requests four-bit parents larger than 15 while also restricting x to 0–15. Those conditions cannot all hold. Its worked example declares a repaired parent set. The re-exam GA paper needs six bits to cover the stated 0–50 range. An encoding change must not silently change the legal range.
\r\nThe final joint table sums to 1.002 rather than 1. Its lesson distinguishes ratios from the printed weights, normalisation, and an assumed typo correction. None is presented as a guaranteed marking-scheme interpretation.
\r\nSome fuzzy classroom solutions calculate a peak-weighted output and call it a centroid. The new lessons name that operation and separately teach the true area centroid. The class fan photographs do not supply one consistent complete set of membership functions, so its worked design makes every breakpoint explicit.
\r\nFor relation composition, check the intermediate labels and the author’s convention for S∘R. A numerical matrix product alone does not resolve a mismatched relation domain.
"}]}]}}}