Intelligent Systems II · M2 · mocks

Q2 · GA through two generations [3]

M2 mock 1 · 10 marks

Q1(a)+Q1(b) compulsory; any two of Q2–Q4. Answers immediately below each question.

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 $f(x)=x^2$ 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]

Given

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]

To find

Both generations: evaluation, selection, crossover, mutation, decoding and comparison.

Formula and why it applies

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.

Solution

Step 1. Generation 1: evaluate and encode the current population

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

Step 2. Generation 1: apply the supplied selection order

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

Step 3. Generation 1: crossover after bit 2

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

Step 4. Generation 1: mutate only the specified bits

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

Step 5. Generation 1: decode and evaluate every child

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

Step 6. Generation 1: compare and form the next population

Add the objective scores in each population, then divide by its number of individuals.

\overline f_{old}=\frac{9+25+100+144}{4}=\frac{278}{4}=69.5

\overline f_{new}=\frac{81+225+225+81}{4}=\frac{612}{4}=153

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.

Step 7. Generation 2: evaluate and encode the current population

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

Step 8. Generation 2: apply the supplied selection order

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

Step 9. Generation 2: crossover after bit 2

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

Step 10. Generation 2: mutate only the specified bits

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

Step 11. Generation 2: decode and evaluate every child

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

Step 12. Generation 2: compare and form the next population

Add the objective scores in each population, then divide by its number of individuals.

\overline f_{old}=\frac{81+225+225+81}{4}=\frac{612}{4}=153

\overline f_{new}=\frac{196+196+196+196}{4}=\frac{784}{4}=196

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.

Final answer. After two generations the ordered population is [14, 14, 14, 14]. Best final objective = 196 at x = [14].