![\"One](\"fig/journey.a5c52083fd.svg\")
Portion: Pure ALOHA onward in Unit 3, followed by complete Unit 4. You first learn only the vocabulary and mathematics needed for this portion. Each numerical is fully solved; the question stays above its answer.
| Question | Marks | Rule |
|---|---|---|
| Q1(a) | 2 | Compulsory |
| Q1(b) | 2 | Compulsory |
| Q2, Q3, Q4 | 3 each | Attempt any two |
Attempted total = 2 + 2 + 3 + 3 = 10 marks. The mock papers follow this pattern. No CN-specific time limit was supplied, so the guide does not invent one.
The full CN answer-book is a secondary reference, not the definition of your M2 portion. CRC, Hamming codes and ARQ occur before ALOHA; transport and application protocols occur after Unit 4. They are excluded here. IP packet-format material is included because it directly appears in the supplied network-layer PYQs. Fragmentation is labelled an extension.
The current course code is 702AI0C023. The two supplied older papers use 702AI0C007; only questions matching the current M2 topics are included. This is an M2 course, not a claim that the entire CN TEE syllabus is covered.
"},{"id":"map","title":"Exactly what this M2 guide covers","html":"| Current syllabus topic | Where to study |
|---|---|
| Pure ALOHA and Slotted ALOHA | ALOHA |
| CSMA/CD and collision avoidance | CSMA and its variants |
| Switching techniques | Switching |
| IPv4 addressing, subnet mask and CIDR | IPv4 and subnetting |
| IPv6 | IPv6 |
| ARP, RARP and DHCP | Address mapping |
| Shortest paths, Bellman–Ford, distance vector | Routing foundations |
| Link state, Dijkstra and OSPF | Routing protocols |
| RIP | RIP |
The syllabus prints “CDMA/CA” in the random-access list. The standard collision-avoidance protocol corresponding to CSMA/CD is CSMA/CA, taught here. Code-division multiple access (CDMA) is a different channelization method; do not expand CSMA/CA as CDMA.
Foundations are support lessons, not extra units. The PYQ tab contains 14 matching parts from two actual TEE papers, with original marks retained. M2 mocks use the confirmed 2/2/3/3/3 structure.
"}]},{"id":"basics","title":"Only the basics you need","scope":"Prerequisites for CN M2","sections":[{"id":"network","title":"Device, link, frame, packet and router","html":"Definition. A computer network is a collection of interconnected devices that exchange data using agreed communication rules called protocols.
Simple meaning. Device number bhejta hai; link usko carry karta hai; protocol batata hai kaise bhejna hai.
A frame is a data-link-layer unit for one-link delivery; an IP packet is a network-layer unit for delivery across interconnected networks. A router examines the destination IP address, selects a next hop and forwards the packet in a new link-layer frame.
Yaad rakho: Frame = ek link; packet = networks ke paar.
"},{"id":"bits","title":"Bits, bytes and rate: read the units first","html":"Definition. A bit is one binary digit. A byte contains eight bits. A data rate measures the number of bits transmitted per second.
Simple meaning. Capital B = byte; small b = bit. Size aur speed ki units same karo.
![\"T_{\\mathrm{frame}}=\\frac{L}{R}\"](\"math/045768b5f250c7fa.svg\")
![\"T_p=\\frac{d}{v}\"](\"math/b9c12990b79f75b6.svg\")
Transmission delay is L/R, where L is the number of bits and R is the bit rate. Propagation delay is d/v, where d is link length and v is propagation speed. Larger frames increase transmission delay; a longer link increases propagation delay.
Yaad rakho: Bits / bits-per-second = seconds; metres / metres-per-second = seconds.
"},{"id":"delay-1","title":"Delay example 1","html":"| Quantity | Value |
|---|---|
| Frame length | 1000 bytes |
| Rate | 2 Mbps |
| Link length | 2,000 m |
| Signal speed | 200,000,000 m/s |
The two separate delays, in seconds and microseconds.
![\"T_{\\mathrm{frame}}=L/R\"](\"math/4bcb5bb2e87de6c2.svg\")
![\"T_p=d/v\"](\"math/9c2209baf93e272e.svg\")
L must be in bits and R in bits/s. Link length and signal speed must use the same distance unit.
1000 × 8 = 8,000 bits.
2 × 1,000,000 = 2,000,000 bits/s.
8,000 ÷ 2,000,000 = 0.004000 s = 4000 μs.
2,000 ÷ 200,000,000 = 0.000010 s = 10 μs.
Final answer: Transmission = 4000 μs; propagation = 10 μs.
| Quantity | Value |
|---|---|
| Frame length | 500 bytes |
| Rate | 10 Mbps |
| Link length | 10,000 m |
| Signal speed | 200,000,000 m/s |
The two separate delays, in seconds and microseconds.
L must be in bits and R in bits/s. Link length and signal speed must use the same distance unit.
500 × 8 = 4,000 bits.
10 × 1,000,000 = 10,000,000 bits/s.
4,000 ÷ 10,000,000 = 0.000400 s = 400 μs.
10,000 ÷ 200,000,000 = 0.000050 s = 50 μs.
Final answer: Transmission = 400 μs; propagation = 50 μs.
| Quantity | Value |
|---|---|
| Frame length | 1500 bytes |
| Rate | 100 Mbps |
| Link length | 100,000 m |
| Signal speed | 250,000,000 m/s |
The two separate delays, in seconds and microseconds.
L must be in bits and R in bits/s. Link length and signal speed must use the same distance unit.
1500 × 8 = 12,000 bits.
100 × 1,000,000 = 100,000,000 bits/s.
12,000 ÷ 100,000,000 = 0.000120 s = 120 μs.
100,000 ÷ 250,000,000 = 0.000400 s = 400 μs.
Final answer: Transmission = 120 μs; propagation = 400 μs.
Definition. Binary is a base-two number system. Each position represents a power of 2; its digit says whether that value is present (1) or absent (0).
Simple meaning. Right se 1 shuru karo; left jaate hi har box double: 1, 2, 4, 8…
![\"Binary](\"fig/binary.f629a09a71.svg\")
Assign weights 128,64,32,16,8,4,2,1 from left to right. Multiply each weight by the bit in that position and add the eight products. For 11000000, the result is 128+64 = 192.
Yaad rakho: Position ka weight likho; jahan 1 hai bas woh add karo.
"},{"id":"binary-1","title":"Binary to decimal · example 1","html":"Binary octet = 00000101.
Its decimal value.
![\"\\text{Decimal](\"math/2d0b01052ad0f77f.svg\")
The symbol Σ means add all the terms. bₖ is the bit at position k; the rightmost position is k=0. The table expands the entire sum.
| Position from right | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |
|---|---|---|---|---|---|---|---|---|
| Place value | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
| Bit | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 |
| Place value | Bit | Product |
|---|---|---|
| 128 | 0 | 128 × 0 = 0 |
| 64 | 0 | 64 × 0 = 0 |
| 32 | 0 | 32 × 0 = 0 |
| 16 | 0 | 16 × 0 = 0 |
| 8 | 0 | 8 × 0 = 0 |
| 4 | 1 | 4 × 1 = 4 |
| 2 | 0 | 2 × 0 = 0 |
| 1 | 1 | 1 × 1 = 1 |
0 + 0 + 0 + 0 + 0 + 4 + 0 + 1 = 5.
Final answer: 5
Binary octet = 10100110.
Its decimal value.
The symbol Σ means add all the terms. bₖ is the bit at position k; the rightmost position is k=0. The table expands the entire sum.
| Position from right | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |
|---|---|---|---|---|---|---|---|---|
| Place value | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
| Bit | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 |
| Place value | Bit | Product |
|---|---|---|
| 128 | 1 | 128 × 1 = 128 |
| 64 | 0 | 64 × 0 = 0 |
| 32 | 1 | 32 × 1 = 32 |
| 16 | 0 | 16 × 0 = 0 |
| 8 | 0 | 8 × 0 = 0 |
| 4 | 1 | 4 × 1 = 4 |
| 2 | 1 | 2 × 1 = 2 |
| 1 | 0 | 1 × 0 = 0 |
128 + 0 + 32 + 0 + 0 + 4 + 2 + 0 = 166.
Final answer: 166
Binary octet = 11100000.
Its decimal value.
The symbol Σ means add all the terms. bₖ is the bit at position k; the rightmost position is k=0. The table expands the entire sum.
| Position from right | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |
|---|---|---|---|---|---|---|---|---|
| Place value | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
| Bit | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
| Place value | Bit | Product |
|---|---|---|
| 128 | 1 | 128 × 1 = 128 |
| 64 | 1 | 64 × 1 = 64 |
| 32 | 1 | 32 × 1 = 32 |
| 16 | 0 | 16 × 0 = 0 |
| 8 | 0 | 8 × 0 = 0 |
| 4 | 0 | 4 × 0 = 0 |
| 2 | 0 | 2 × 0 = 0 |
| 1 | 0 | 1 × 0 = 0 |
128 + 64 + 32 + 0 + 0 + 0 + 0 + 0 = 224.
Final answer: 224
Binary octet = 11000000.
Its decimal value.
The symbol Σ means add all the terms. bₖ is the bit at position k; the rightmost position is k=0. The table expands the entire sum.
| Position from right | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |
|---|---|---|---|---|---|---|---|---|
| Place value | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
| Bit | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
| Place value | Bit | Product |
|---|---|---|
| 128 | 1 | 128 × 1 = 128 |
| 64 | 1 | 64 × 1 = 64 |
| 32 | 0 | 32 × 0 = 0 |
| 16 | 0 | 16 × 0 = 0 |
| 8 | 0 | 8 × 0 = 0 |
| 4 | 0 | 4 × 0 = 0 |
| 2 | 0 | 2 × 0 = 0 |
| 1 | 0 | 1 × 0 = 0 |
128 + 64 + 0 + 0 + 0 + 0 + 0 + 0 = 192.
Final answer: 192
Decimal value = 13.
Eight bits, including leading zeroes.
Use the weights 128,64,32,16,8,4,2,1.
At each weight, write 1 and subtract if the remaining value is at least that weight; otherwise write 0.
| Weight | Remaining before | Bit decision | Remaining after |
|---|---|---|---|
| 128 | 13 | 0: too large | 13 |
| 64 | 13 | 0: too large | 13 |
| 32 | 13 | 0: too large | 13 |
| 16 | 13 | 0: too large | 13 |
| 8 | 13 | 1: subtract | 5 |
| 4 | 5 | 1: subtract | 1 |
| 2 | 1 | 0: too large | 1 |
| 1 | 1 | 1: subtract | 0 |
00001101.
8 + 4 + 1 = 13.
Final answer: 00001101
Decimal value = 192.
Eight bits, including leading zeroes.
Use the weights 128,64,32,16,8,4,2,1.
At each weight, write 1 and subtract if the remaining value is at least that weight; otherwise write 0.
| Weight | Remaining before | Bit decision | Remaining after |
|---|---|---|---|
| 128 | 192 | 1: subtract | 64 |
| 64 | 64 | 1: subtract | 0 |
| 32 | 0 | 0: too large | 0 |
| 16 | 0 | 0: too large | 0 |
| 8 | 0 | 0: too large | 0 |
| 4 | 0 | 0: too large | 0 |
| 2 | 0 | 0: too large | 0 |
| 1 | 0 | 0: too large | 0 |
11000000.
128 + 64 = 192.
Final answer: 11000000
Decimal value = 224.
Eight bits, including leading zeroes.
Use the weights 128,64,32,16,8,4,2,1.
At each weight, write 1 and subtract if the remaining value is at least that weight; otherwise write 0.
| Weight | Remaining before | Bit decision | Remaining after |
|---|---|---|---|
| 128 | 224 | 1: subtract | 96 |
| 64 | 96 | 1: subtract | 32 |
| 32 | 32 | 1: subtract | 0 |
| 16 | 0 | 0: too large | 0 |
| 8 | 0 | 0: too large | 0 |
| 4 | 0 | 0: too large | 0 |
| 2 | 0 | 0: too large | 0 |
| 1 | 0 | 0: too large | 0 |
11100000.
128 + 64 + 32 = 224.
Final answer: 11100000
Definition. A power describes repeated multiplication. The number e is approximately 2.718281828; e raised to a negative power is a positive number below 1.
Simple meaning. 2 ki power bits ke combinations batati hai; e ki negative power ALOHA ki collision probability mein aati hai.
| Power | Value |
|---|---|
| 20 | 1 |
| 21 | 2 |
| 22 | 4 |
| 23 | 8 |
| 24 | 16 |
| 25 | 32 |
| 26 | 64 |
| 27 | 128 |
| 28 | 256 |
| 29 | 512 |
| 210 | 1024 |
| 211 | 2048 |
| 212 | 4096 |
| 213 | 8192 |
| 214 | 16384 |
| 215 | 32768 |
| 216 | 65536 |
Each bit has two choices; six independent bit positions provide 2×2×2×2×2×2 = 64 combinations. A fraction is converted to percent by multiplying by 100, so 0.18394 corresponds to 18.394%.
Yaad rakho: Minus exponent andar hai; final exponential positive rehta hai.
"},{"id":"calculator","title":"Calculator steps for ALOHA and subnetting","html":"Use the normal calculation mode. Exact menu names vary across fx-991ES-family models; the mathematics below is the same. The eˣ function is usually the shifted function of ln. The EXP/×10ˣ key enters scientific notation and is not eˣ.
Yaad rakho: eˣ aur ×10ˣ alag keys/functions hain. Brackets se denominator safe rakho.
"},{"id":"graphs","title":"How to read a routing graph","html":"Definition. A weighted graph represents routers as vertices and links as edges labelled with costs. A directed edge may be used only in the arrow direction.
Simple meaning. Circle router hai, line link hai, number us link ka cost. Arrow ulta mat chalna.
![\"Directed](\"fig/graph-basics.ebd9ed3bc9.svg\")
Hop count counts links or router transitions according to the protocol. A weighted path cost adds the numerical metric assigned to each edge. The fewest-hop path need not have the lowest weighted cost; an indirect route costing 7 is cheaper than a direct link costing 10.
Yaad rakho: Shortest = smallest cost, zaroori nahi fewest arrows.
"}]},{"id":"aloha","title":"ALOHA: access, collisions and throughput","scope":"Unit 3 · confirmed M2 starting point","sections":[{"id":"pure","title":"Pure ALOHA: transmit first, retry if necessary","html":"Definition. Pure ALOHA is a random-access protocol in which a station transmits whenever a frame is ready and retransmits after a random delay if the transmission is unsuccessful.
Simple meaning. Frame ready → bhej do. Collision hua → random wait karke retry.
![\"Pure](\"fig/pure-aloha.8aa7732ee2.svg\")
In Pure ALOHA, a station transmits a ready frame without carrier sensing. If a successful acknowledgement is not received, it waits for a random interval and retransmits. Overlapping transmissions cause collisions in the basic model. Its maximum normalized throughput is 1/(2e) ≈ 18.4%, so it is inefficient at high offered load.
Yaad rakho: Pure = kisi bhi time start; retry ka wait random.
"},{"id":"vulnerable","title":"Why Pure ALOHA has a 2T vulnerable period","html":"Definition. The vulnerable period is the interval of start times of other frames that would cause a collision with a chosen frame.
Simple meaning. Apne frame se ek frame-time pehle start hua frame bhi takra sakta hai, aur baad mein bhi.
![\"Pure](\"fig/vulnerable.e2a01a5ab7.svg\")
A frame lasting T can collide with an equal-length frame that began less than T earlier, or one that begins less than T later. The two intervals together have length T+T=2T.
Yaad rakho: Pure: before T + after T = 2T.
"},{"id":"slotted","title":"Slotted ALOHA: starts only at slot boundaries","html":"Definition. Slotted ALOHA is a random-access protocol that divides time into slots of one frame duration and permits transmissions only at the beginning of a slot.
Simple meaning. Frame ready hai tab bhi next slot ki start-line ka wait karo.
![\"Slotted](\"fig/slotted-aloha.fbfbef9647.svg\")
| Feature | Pure | Slotted |
|---|---|---|
| Start time | Any time | Slot boundary only |
| Synchronization | No common slots required | Required |
| Vulnerable period | 2T | T |
| Maximum S | 1/(2e) ≈ 0.184 | 1/e ≈ 0.368 |
| G at maximum | 0.5 | 1 |
Yaad rakho: Slotted = aligned starts; collision risk half-window, not zero.
"},{"id":"formulas","title":"G, S and throughput: which number are you finding?","html":"Definition. Offered load G is the average number of transmission attempts per frame time. Normalized throughput S is the average number of successful frames per frame time.
Simple meaning. G = kitne attempts; S = unmein successful output per frame-time. Dono same nahi.
![\"ALOHA:](\"fig/aloha-curve.58889b65f2.svg\")
![\"T=L/R,\\qquad](\"math/85fb8d2da8040e20.svg\")
![\"S_{\\mathrm{pure}}=G](\"math/28ca9624d26f78f7.svg\")
![\"S_{\\mathrm{slotted}}=G](\"math/6d136a965636bb20.svg\")
![\"\\text{successful](\"math/751670fd3de42086.svg\")
![\"\\text{successful](\"math/58e5c110cc978ab9.svg\")
G denotes aggregate attempts per frame transmission time and S denotes successful frames per frame transmission time. Pure ALOHA: S=Ge⁻²ᴳ, with maximum 1/(2e) at G=0.5. Slotted ALOHA: S=Ge⁻ᴳ, with maximum 1/e at G=1. These results use the equal-length Poisson collision model.
Yaad rakho: P mein pair: −2G; S mein single: −G.
\r\nFor 200-bit frames on a 200000-bit/s channel, T = 200/200000 = 0.001 s. If 500 frames/s attempt transmission, G = 0.5 and pure-ALOHA S = 0.5e⁻¹ ≈ 0.1839397. Convert to frames/s by dividing by T: 0.1839397/0.001 ≈ 183.94 frames/s.
Do not multiply S by 500: S is successful frames per frame-time, not success probability per attempt. Some slide totals use that incorrect conversion. At 250 attempts/s the correct pure-ALOHA output is 151.63 frames/s. Use the definitions and units above.
"},{"id":"pure-1","title":"Pure ALOHA · load example 1","html":"Protocol = Pure ALOHA; G = 0.25 attempts/frame time.
S, 100S and the maximum possible S.
![\"S=G](\"math/66b7a6b6082e223e.svg\")
Select the formula from the protocol name. The exponent is −2G because Pure ALOHA has a 2T vulnerable period.
−2 × 0.25 = -0.5.
![\"e^{-0.5}](\"math/871b71246f6a730f.svg\")
S = 0.25 × 0.606530660 ≈ 0.151632665 successful frames/frame time.
0.151632665 × 100 ≈ 15.1633%.
The maximum is 1/(2e) ≈ 0.183940 = 18.3940%, at G=0.5.
Final answer: S ≈ 0.151633; percentage ≈ 15.1633%; Smax ≈ 0.183940 at G=0.5.
Protocol = Pure ALOHA; G = 0.5 attempts/frame time.
S, 100S and the maximum possible S.
Select the formula from the protocol name. The exponent is −2G because Pure ALOHA has a 2T vulnerable period.
−2 × 0.5 = -1.
![\"e^{-1}](\"math/f3e24ac70fcf4336.svg\")
S = 0.5 × 0.367879441 ≈ 0.183939721 successful frames/frame time.
0.183939721 × 100 ≈ 18.3940%.
The maximum is 1/(2e) ≈ 0.183940 = 18.3940%, at G=0.5.
Final answer: S ≈ 0.183940; percentage ≈ 18.3940%; Smax ≈ 0.183940 at G=0.5.
Protocol = Pure ALOHA; G = 1 attempts/frame time.
S, 100S and the maximum possible S.
Select the formula from the protocol name. The exponent is −2G because Pure ALOHA has a 2T vulnerable period.
−2 × 1 = -2.
![\"e^{-2}](\"math/6d4c1915157a6bc5.svg\")
S = 1 × 0.135335283 ≈ 0.135335283 successful frames/frame time.
0.135335283 × 100 ≈ 13.5335%.
The maximum is 1/(2e) ≈ 0.183940 = 18.3940%, at G=0.5.
Final answer: S ≈ 0.135335; percentage ≈ 13.5335%; Smax ≈ 0.183940 at G=0.5.
Protocol = Slotted ALOHA; G = 0.25 attempts/frame time.
S, 100S and the maximum possible S.
![\"S=G](\"math/01e46e63e1b256b9.svg\")
Select the formula from the protocol name. The exponent is −G because Slotted ALOHA has a T vulnerable period.
−1 × 0.25 = -0.25.
![\"e^{-0.25}](\"math/8ed0d2b7d5ee0b1f.svg\")
S = 0.25 × 0.778800783 ≈ 0.194700196 successful frames/frame time.
0.194700196 × 100 ≈ 19.4700%.
The maximum is 1/(1e) ≈ 0.367879 = 36.7879%, at G=1.
Final answer: S ≈ 0.194700; percentage ≈ 19.4700%; Smax ≈ 0.367879 at G=1.
Protocol = Slotted ALOHA; G = 1 attempts/frame time.
S, 100S and the maximum possible S.
Select the formula from the protocol name. The exponent is −G because Slotted ALOHA has a T vulnerable period.
−1 × 1 = -1.
S = 1 × 0.367879441 ≈ 0.367879441 successful frames/frame time.
0.367879441 × 100 ≈ 36.7879%.
The maximum is 1/(1e) ≈ 0.367879 = 36.7879%, at G=1.
Final answer: S ≈ 0.367879; percentage ≈ 36.7879%; Smax ≈ 0.367879 at G=1.
Protocol = Slotted ALOHA; G = 2 attempts/frame time.
S, 100S and the maximum possible S.
Select the formula from the protocol name. The exponent is −G because Slotted ALOHA has a T vulnerable period.
−1 × 2 = -2.
S = 2 × 0.135335283 ≈ 0.270670566 successful frames/frame time.
0.270670566 × 100 ≈ 27.0671%.
The maximum is 1/(1e) ≈ 0.367879 = 36.7879%, at G=1.
Final answer: S ≈ 0.270671; percentage ≈ 27.0671%; Smax ≈ 0.367879 at G=1.
| Symbol | Given |
|---|---|
| L | 200 bits |
| R | 200 kbps = 200,000 bits/s |
| λ | 1000 aggregate attempts/s |
T, G, S, successful frames/s and bit/s.
![\"T=L/R\"](\"math/4ac7eb7c34287111.svg\")
![\"G=\\lambda](\"math/f8d7da692e922bd4.svg\")
![\"F_{\\mathrm{success}}=S/T,\\qquad](\"math/afb39a0ed04909f3.svg\")
First normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
T = 200/200,000 = 0.001 s = 1 ms.
G = 1000 × 0.001 = 1 attempts/frame time.
−2G = -2; e to that power ≈ 0.135335283.
S = 1 × 0.135335283 ≈ 0.135335283.
S/T = 0.135335283/0.001 ≈ 135.335283 frames/s.
SR = 0.135335283 × 200,000 ≈ 27067.056647 bits/s ≈ 27.067057 kbps.
Final answer: G=1, S≈0.135335, successful rate≈135.335 frames/s = 27.067 kbps.
| Symbol | Given |
|---|---|
| L | 200 bits |
| R | 200 kbps = 200,000 bits/s |
| λ | 500 aggregate attempts/s |
T, G, S, successful frames/s and bit/s.
First normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
T = 200/200,000 = 0.001 s = 1 ms.
G = 500 × 0.001 = 0.5 attempts/frame time.
−2G = -1; e to that power ≈ 0.367879441.
S = 0.5 × 0.367879441 ≈ 0.183939721.
S/T = 0.183939721/0.001 ≈ 183.939721 frames/s.
SR = 0.183939721 × 200,000 ≈ 36787.944117 bits/s ≈ 36.787944 kbps.
Final answer: G=0.5, S≈0.183940, successful rate≈183.940 frames/s = 36.788 kbps.
| Symbol | Given |
|---|---|
| L | 200 bits |
| R | 200 kbps = 200,000 bits/s |
| λ | 250 aggregate attempts/s |
T, G, S, successful frames/s and bit/s.
First normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
T = 200/200,000 = 0.001 s = 1 ms.
G = 250 × 0.001 = 0.25 attempts/frame time.
−2G = -0.5; e to that power ≈ 0.606530660.
S = 0.25 × 0.606530660 ≈ 0.151632665.
S/T = 0.151632665/0.001 ≈ 151.632665 frames/s.
SR = 0.151632665 × 200,000 ≈ 30326.532986 bits/s ≈ 30.326533 kbps.
Final answer: G=0.25, S≈0.151633, successful rate≈151.633 frames/s = 30.327 kbps.
| Symbol | Given |
|---|---|
| L | 400 bits |
| R | 200 kbps = 200,000 bits/s |
| λ | 500 aggregate attempts/s |
T, G, S, successful frames/s and bit/s.
First normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
T = 400/200,000 = 0.002 s = 2 ms.
G = 500 × 0.002 = 1 attempts/frame time.
−1G = -1; e to that power ≈ 0.367879441.
S = 1 × 0.367879441 ≈ 0.367879441.
S/T = 0.367879441/0.002 ≈ 183.939721 frames/s.
SR = 0.367879441 × 200,000 ≈ 73575.888234 bits/s ≈ 73.575888 kbps.
Final answer: G=1, S≈0.367879, successful rate≈183.940 frames/s = 73.576 kbps.
| Symbol | Given |
|---|---|
| L | 1600 bits |
| R | 800 kbps = 800,000 bits/s |
| λ | 250 aggregate attempts/s |
T, G, S, successful frames/s and bit/s.
First normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
T = 1600/800,000 = 0.002 s = 2 ms.
G = 250 × 0.002 = 0.5 attempts/frame time.
−1G = -0.5; e to that power ≈ 0.606530660.
S = 0.5 × 0.606530660 ≈ 0.303265330.
S/T = 0.303265330/0.002 ≈ 151.632665 frames/s.
SR = 0.303265330 × 800,000 ≈ 242612.263885 bits/s ≈ 242.612264 kbps.
Final answer: G=0.5, S≈0.303265, successful rate≈151.633 frames/s = 242.612 kbps.
A small offered load leaves many opportunities unused. A very large load creates many collisions. The best point balances those effects.
![\"\\frac{dS}{dG}=e^{-2G}(1-2G)\"](\"math/0e9a0d697e50f118.svg\")
![\"\\frac{dS}{dG}=e^{-G}(1-G)\"](\"math/e448a50c7be7dbba.svg\")
Yaad rakho: Pure optimum: G half, maximum half of 1/e. Slotted optimum: G one, maximum 1/e.
"}]},{"id":"csma","title":"CSMA, collision detection and avoidance","scope":"Unit 3 · after ALOHA","sections":[{"id":"sense","title":"Carrier sensing: why listening does not remove every collision","html":"Definition. Carrier Sense Multiple Access (CSMA) requires a station to sense the shared medium before transmitting.
Simple meaning. Pehle suno, phir bhejo; lekin door ka signal pahunchne mein time leta hai.
![\"An](\"fig/csma-propagation.df55cdec46.svg\")
Carrier sensing observes the local medium. Because signals have nonzero propagation delay, another station’s new transmission may not yet be detectable. Both stations can then start and their transmissions overlap.
Yaad rakho: Idle yahan hai; remote signal abhi raste mein ho sakta hai.
"},{"id":"persistence","title":"1-persistent, non-persistent and p-persistent CSMA","html":"Definition. Persistence specifies how a station behaves after observing that the medium is busy or becomes idle.
Simple meaning. 1 = turant; non = random wait; p = probability ke basis pe start.
![\"CSMA](\"fig/persistence.528dafcd8a.svg\")
| Rule | When idle | When busy | Trade-off |
|---|---|---|---|
| 1-persistent | Transmit immediately | Keep listening; transmit as soon as idle | Low idle delay, but waiting stations may start together |
| Non-persistent | Transmit | Wait a random time, then sense again | Less synchronized contention; may leave avoidable idle time |
| p-persistent (slotted) | Transmit with probability p; defer a slot with probability 1−p | Wait until available, then resume the probabilistic slot rule | Controls contention; needs slots and a choice of p |
| Rule | When idle | When busy | Trade-off |
|---|---|---|---|
| 1-persistent | Transmit immediately | Keep listening; transmit as soon as idle | Low idle delay, but waiting stations may start together |
| Non-persistent | Transmit | Wait a random time, then sense again | Less synchronized contention; may leave avoidable idle time |
| p-persistent (slotted) | Transmit with probability p; defer a slot with probability 1−p | Wait until available, then resume the probabilistic slot rule | Controls contention; needs slots and a choice of p |
Yaad rakho: 1 = start now; non = wait then listen; p = chance per idle slot.
"},{"id":"cd","title":"CSMA/CD: detect a collision while transmitting","html":"Definition. CSMA with Collision Detection monitors the medium during transmission, aborts a detected collision and retries after a randomized backoff.
Simple meaning. CD = takkar detect hui toh transmission rok do.
![\"CSMA/CD:](\"fig/collision-roundtrip.42358ec7bf.svg\")
![\"T_{\\mathrm{frame}}\\geq](\"math/4f522413155daf19.svg\")
![\"L_{\\min}\\geq](\"math/eae5a57f0575ef7d.svg\")
CSMA/CD senses before transmission and detects collisions during transmission. A collision causes abort/jam and a random backoff before retry. To detect the most distant collision before the frame finishes, the transmission duration must be at least the worst-case round-trip propagation delay. Thus Lmin/R ≥ 2Tp in the idealized model.
Yaad rakho: Detect karna hai toh sender tab tak bhej raha hona chahiye.
"},{"id":"minframe-1","title":"Collision detection · minimum size 1","html":"Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
R=10 Mbps; Tp=25.6 μs.
Lmin in bits and the minimum whole-byte length.
![\"L_{\\min}=\\lceil](\"math/dc84d9f4c66a26e3.svg\")
The frame must last at least a round trip. Convert Mbps to bits/s and μs to seconds before multiplying.
R=10,000,000 bits/s; Tp=0.0000256 s.
2Tp=2×25.6=51.2 μs.
Lmin=2×10,000,000×0.0000256=512 bits.
512/8=64; round upward if needed → 64 bytes.
Final answer: 512 bits, or at least 64 whole bytes.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
R=100 Mbps; Tp=2 μs.
Lmin in bits and the minimum whole-byte length.
The frame must last at least a round trip. Convert Mbps to bits/s and μs to seconds before multiplying.
R=100,000,000 bits/s; Tp=0.000002 s.
2Tp=2×2=4 μs.
Lmin=2×100,000,000×0.000002=400 bits.
400/8=50; round upward if needed → 50 bytes.
Final answer: 400 bits, or at least 50 whole bytes.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
R=10 Mbps; Tp=10 μs.
Lmin in bits and the minimum whole-byte length.
The frame must last at least a round trip. Convert Mbps to bits/s and μs to seconds before multiplying.
R=10,000,000 bits/s; Tp=0.00001 s.
2Tp=2×10=20 μs.
Lmin=2×10,000,000×0.00001=200 bits.
200/8=25; round upward if needed → 25 bytes.
Final answer: 200 bits, or at least 25 whole bytes.
Definition. Binary exponential backoff enlarges the range of possible random waiting slots after repeated collisions, reducing the chance of the same stations colliding again.
Simple meaning. Collision zyada → random wait ka range bada.
![\"k=\\min(n,10),\\qquad](\"math/0f0e79cc7f4deaac.svg\")
![\"T_{\\mathrm{backoff}}=K](\"math/b5a8dc2b6f87ba95.svg\")
A station selects a random integer K from an expanding window and waits K slot times before retrying. After n collisions, classic Ethernet uses k=min(n,10) and 0≤K≤2ᵏ−1. After three collisions the possible values are 0 through 7.
Yaad rakho: 2ᵏ choices hain, isliye last value 2ᵏ−1.
"},{"id":"backoff-1","title":"Backoff · example 1","html":"Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
n=2; K=3; Tslot=51.2 μs.
The random range and the delay for the selected K.
![\"k=\\min(n,10),\\quad](\"math/c06068d72176a7c3.svg\")
![\"T=KT_{\\mathrm{slot}}\"](\"math/9da1be02953867d3.svg\")
Count starts at zero for K. The selected K must fit the allowed range.
k=min(2,10)=2.
22=4; largest K=4−1=3. Range: 0 to 3.
3 is within 0…3.
3×51.2=153.6 μs.
Final answer: K may be 0…3; selected delay=153.6 μs.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
n=3; K=5; Tslot=51.2 μs.
The random range and the delay for the selected K.
Count starts at zero for K. The selected K must fit the allowed range.
k=min(3,10)=3.
23=8; largest K=8−1=7. Range: 0 to 7.
5 is within 0…7.
5×51.2=256 μs.
Final answer: K may be 0…7; selected delay=256 μs.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
n=5; K=17; Tslot=5.12 μs.
The random range and the delay for the selected K.
Count starts at zero for K. The selected K must fit the allowed range.
k=min(5,10)=5.
25=32; largest K=32−1=31. Range: 0 to 31.
17 is within 0…31.
17×5.12=87.04 μs.
Final answer: K may be 0…31; selected delay=87.04 μs.
Definition. CSMA with Collision Avoidance uses carrier sensing and randomized access delays, with acknowledgements and optional reservation exchanges, to reduce collisions on a wireless shared medium.
Simple meaning. CA = pehle collision avoid karne ki koshish; guarantee nahi.
![\"CSMA/CA:](\"fig/ca-sequence.566280e50f.svg\")
| Aspect | CSMA/CD | CSMA/CA |
|---|---|---|
| Main action | Detect and abort a collision during sending | Reduce collision risk before sending |
| Typical teaching context | Shared half-duplex wired Ethernet | Wi-Fi contention access |
| Mechanisms | Carrier sensing, detection/jam, backoff | Carrier sensing, random backoff, ACK; optional RTS/CTS |
Yaad rakho: CD = detect during; CA = avoid before.
"},{"id":"hidden","title":"Hidden terminals and RTS/CTS","html":"Definition. Hidden terminals are stations that cannot sense each other but whose transmissions can interfere at a common receiver.
Simple meaning. A aur C ek dusre ko nahi sunte, par dono B ko bhej rahe hain.
![\"Hidden](\"fig/hidden-terminal.1ffcf794d5.svg\")
A and C may both reach B but be unable to sense one another. Simultaneous transmissions can then collide at B. A short RTS followed by B’s CTS announces a reservation; a station hearing the CTS defers, reducing the risk of a long data-frame collision.
Yaad rakho: Hidden: senders invisible to each other, receiver common.
"}]},{"id":"switching","title":"Switching techniques","scope":"Unit 4 · Network Layer","sections":[{"id":"types","title":"Circuit, message and packet switching","html":"Definition. Switching selects how data passes through intermediate network nodes from a source to a destination.
Simple meaning. Raasta reserve karna, poora message ruk-ruk ke bhejna, ya chhote packets bhejna—yeh teen ideas hain.
![\"Three](\"fig/switching.2b51f856d9.svg\")
| Aspect | Circuit switching | Message switching | Packet switching |
|---|---|---|---|
| Unit handled | Continuous stream on established circuit | Whole message | Small packets |
| Resource use | Capacity reserved along a circuit | Shared; message stored at each intermediate node | Shared; each packet stored/forwarded in this model |
| Setup | Required before data transfer | No circuit setup required | Datagram: no connection setup; virtual circuit: setup |
| Delay | Setup delay, then relatively predictable service | Large store-and-forward delay for long messages | Variable delay; packet pipeline can reduce completion time |
| Main limitation | Reserved capacity can be idle | Large buffers; unsuitable for tight delay needs | Headers, possible queueing/loss; datagrams may reorder |
| Aspect | Circuit switching | Message switching | Packet switching |
|---|---|---|---|
| Unit handled | Continuous stream on established circuit | Whole message | Small packets |
| Resource use | Capacity reserved along a circuit | Shared; message stored at each intermediate node | Shared; each packet stored/forwarded in this model |
| Setup | Required before data transfer | No circuit setup required | Datagram: no connection setup; virtual circuit: setup |
| Delay | Setup delay, then relatively predictable service | Large store-and-forward delay for long messages | Variable delay; packet pipeline can reduce completion time |
| Main limitation | Reserved capacity can be idle | Large buffers; unsuitable for tight delay needs | Headers, possible queueing/loss; datagrams may reorder |
For ordinary bursty office data, packet switching is generally appropriate because resources are shared among active transfers. The design must still account for queueing, loss and the required service quality.
Yaad rakho: Circuit reserves; message waits whole; packets pipeline.
"},{"id":"datagram","title":"Datagram versus virtual-circuit packet switching","html":"Definition. Datagram switching routes each packet independently; virtual-circuit switching first establishes a logical path and forwarding state for a flow.
Simple meaning. Datagram mein har packet ka decision; virtual circuit mein pehle logical route set.
| Feature | Datagram | Virtual circuit |
|---|---|---|
| Setup | Usually none at network layer | Logical connection setup |
| Forwarding information | Destination address | Virtual-circuit identifier |
| Router state | No per-connection state required for basic IP forwarding | Connection/path state maintained |
| Path | May vary | Established logical path until changed |
Yaad rakho: Virtual circuit bhi packet switching ho sakta hai.
"},{"id":"pipeline","title":"Store-and-forward and pipelining","html":"Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Definition. Store-and-forward requires a node to receive the complete forwarding unit before sending it onward; pipelining allows successive units to occupy different links simultaneously.
Simple meaning. Pehla packet next link pe jaa raha hai, tab doosra packet first link pe aa sakta hai.
![\"Packet](\"fig/pipeline.a023f04605.svg\")
![\"T_{\\mathrm{message}}=hM/R+h\\tau\"](\"math/88eb2b8ca0c2e1fd.svg\")
![\"T_{\\mathrm{packet}}=(h+k-1)L/R+h\\tau\"](\"math/209da411e5cae6a2.svg\")
Packet switching can pipeline smaller units across successive links. After the first packet is forwarded, the next packet can use an earlier link while the first uses a later link. Whole-message store-and-forward must receive the entire message at each node, preventing that packet-level pipeline.
Yaad rakho: First packet h times; remaining packets one time each.
"},{"id":"delay-1","title":"Compare completion times · example 1","html":"Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
| Given | Value |
|---|---|
| M | 12000 bits |
| h | 3 |
| k | 6 |
| R | 1,000,000 bits/s |
| τ | 0.002 s/link |
| Circuit setup | 0.01 s |
Time until the final data bit reaches the destination in each method.
![\"T_m=hM/R+h\\tau\"](\"math/9a18147a7dd3f73d.svg\")
![\"T_p=(h+k-1)L/R+h\\tau\"](\"math/2fccb0c108e66b5b.svg\")
![\"T_c=T_{\\mathrm{setup}}+M/R+h\\tau\"](\"math/59666d3c7abd8bb9.svg\")
Message switching repeats whole-message serialization at every hop. Packet switching pipelines equal packets. A circuit streams bits through the established equal-rate path.
L=M/k=12000/6=2000 bits.
3×12000/1,000,000 + 3×0.002 = 0.042000 s = 42 ms.
L/R=2000/1,000,000=0.002 s.
(3+6−1)×0.002+3×0.002=0.022000 s = 22 ms.
0.01+12000/1,000,000+3×0.002=0.028000 s = 28 ms.
Final answer: Message=42 ms; packets=22 ms; circuit=28 ms.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
| Given | Value |
|---|---|
| M | 8000 bits |
| h | 2 |
| k | 4 |
| R | 1,000,000 bits/s |
| τ | 0.001 s/link |
| Circuit setup | 0.005 s |
Time until the final data bit reaches the destination in each method.
Message switching repeats whole-message serialization at every hop. Packet switching pipelines equal packets. A circuit streams bits through the established equal-rate path.
L=M/k=8000/4=2000 bits.
2×8000/1,000,000 + 2×0.001 = 0.018000 s = 18 ms.
L/R=2000/1,000,000=0.002 s.
(2+4−1)×0.002+2×0.001=0.012000 s = 12 ms.
0.005+8000/1,000,000+2×0.001=0.015000 s = 15 ms.
Final answer: Message=18 ms; packets=12 ms; circuit=15 ms.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
M=12,000 bits; payload=1,000 bits; header=100 bits; h=4; R=1,000,000 bits/s; τ=0.001 s.
k, L, total serialized bits on one link and completion time.
![\"k=M/P,\\quad](\"math/cf2d3cfa77719992.svg\")
![\"T=(h+k-1)L/R+h\\tau\"](\"math/a1dee7ec7d280062.svg\")
The payload determines packet count; payload plus header determines packet transmission time.
k=12,000/1,000=12.
L=1,000+100=1,100 bits.
12×1,100=13,200 bits, of which 1,200 are headers.
1,100/1,000,000=0.0011 s.
(4+12−1)×0.0011+4×0.001=0.0205 s=20.5 ms.
Final answer: 12 packets; 1,100 bits each; 13,200 bits on one link; completion=20.5 ms.
Definition. An IPv4 address is a 32-bit logical address, conventionally written as four decimal octets separated by dots.
Simple meaning. IP ke chaar boxes hain; har box 8 bits ka, value 0 se 255.
![\"IPv4](\"fig/ipv4-prefix.d99075bfca.svg\")
IPv4 contains 32 bits, written as four decimal octets of 8 bits each. Each octet ranges from 0 to 255. In /26 notation, 26 leading bits are the network prefix and 6 trailing bits are available within the block, giving 2⁶=64 total addresses.
Yaad rakho: 4 octets × 8 bits = 32; slash = network bits.
"},{"id":"classes","title":"Classful addressing: historical default boundaries","html":"Definition. Classful addressing divides the IPv4 address space into historical classes with fixed default network boundaries for Classes A, B and C.
Simple meaning. Class first octet se; modern subnet given slash se.
| Class | First octet | Default / purpose |
|---|---|---|
| A | 0–127; ordinary historical assignments 1–126 | /8; special ranges exist |
| B | 128–191 | /16 |
| C | 192–223 | /24 |
| D | 224–239 | Multicast |
| E | 240–255 | Reserved/experimental; special addresses exist |
Historical defaults are /8 for A, /16 for B and /24 for C. CIDR explicitly supplies the prefix length and supports variable-sized blocks. Therefore an explicit CIDR prefix takes precedence over a historical class default.
Yaad rakho: Class pehchaan sakte ho; calculation mein given slash follow karo.
"},{"id":"special","title":"Private, loopback and special IPv4 addresses","html":"Definition. Private IPv4 ranges are address blocks reserved for use within private networks rather than ordinary public Internet routing.
Simple meaning. Private IP ghar/college ke andar reuse ho sakta hai; globally unique public IP jaisa treat mat karo.
| Prefix | Range |
|---|---|
| 10.0.0.0/8 | 10.0.0.0–10.255.255.255 |
| 172.16.0.0/12 | 172.16.0.0–172.31.255.255 |
| 192.168.0.0/16 | 192.168.0.0–192.168.255.255 |
For example, 192.168.1.10 can be used in a private LAN.
Yaad rakho: 10 full; 172 only 16–31; 192.168 full.
"},{"id":"mask","title":"Subnet masks: 1 keeps a network bit; 0 marks a host bit","html":"Definition. A subnet mask is a 32-bit value with contiguous leading 1s for network-prefix positions and trailing 0s for host positions.
Simple meaning. Mask mein 1 = network part rakhna; 0 = host part zero karna.
![\"Bitwise](\"fig/mask-and.3b9a8ff9ea.svg\")
| A | B | A AND B |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
![\"\\text{Network](\"math/3a94cf83bf5040ba.svg\")
The prefix mask contains 1s in network positions and 0s in host positions. Bitwise AND between the IP address and mask preserves the network bits and clears the host bits. For 192.168.10.77/26, the mask is 255.255.255.192 and the network is 192.168.10.64.
Yaad rakho: AND: dono 1 ho tabhi output 1.
"},{"id":"mask-1","title":"Prefix to mask · example 1","html":"Prefix length p=26.
Mask, h and N.
![\"h=32-p,\\qquad](\"math/ab3d8502169db5ce.svg\")
A prefix contains p leading ones; total combinations depend on the remaining host bits.
26 ones followed by 6 zeroes: 11111111.11111111.11111111.11000000.
| Mask bits | Decimal |
|---|---|
| 11111111 | 255 |
| 11111111 | 255 |
| 11111111 | 255 |
| 11000000 | 192 |
h=32−26=6.
N=26=64.
Final answer: Mask=255.255.255.192; host bits=6; total addresses=64.
Prefix length p=20.
Mask, h and N.
A prefix contains p leading ones; total combinations depend on the remaining host bits.
20 ones followed by 12 zeroes: 11111111.11111111.11110000.00000000.
| Mask bits | Decimal |
|---|---|
| 11111111 | 255 |
| 11111111 | 255 |
| 11110000 | 240 |
| 00000000 | 0 |
h=32−20=12.
N=212=4,096.
Final answer: Mask=255.255.240.0; host bits=12; total addresses=4,096.
Prefix length p=27.
Mask, h and N.
A prefix contains p leading ones; total combinations depend on the remaining host bits.
27 ones followed by 5 zeroes: 11111111.11111111.11111111.11100000.
| Mask bits | Decimal |
|---|---|
| 11111111 | 255 |
| 11111111 | 255 |
| 11111111 | 255 |
| 11100000 | 224 |
h=32−27=5.
N=25=32.
Final answer: Mask=255.255.255.224; host bits=5; total addresses=32.
Prefix length p=23.
Mask, h and N.
A prefix contains p leading ones; total combinations depend on the remaining host bits.
23 ones followed by 9 zeroes: 11111111.11111111.11111110.00000000.
| Mask bits | Decimal |
|---|---|
| 11111111 | 255 |
| 11111111 | 255 |
| 11111110 | 254 |
| 00000000 | 0 |
h=32−23=9.
N=29=512.
Final answer: Mask=255.255.254.0; host bits=9; total addresses=512.
Definition. Subnetting divides an address block into smaller prefix-defined blocks. In an ordinary subnet, the network and directed-broadcast addresses bound its usable host range.
Simple meaning. Pehle block nikalo; phir first aur last. Uske baad hosts.
![\"A](\"fig/subnet-blocks.67a24f46fc.svg\")
![\"N=2^{32-p}\"](\"math/06561cae3cdafd6c.svg\")
![\"\\text{block](\"math/6d26c03e47eb5b6c.svg\")
![\"\\text{ordinary](\"math/0d9d5151a6681f69.svg\")
A /26 has 32−26=6 host bits, so it contains 2⁶=64 total addresses. Under ordinary IPv4 subnet rules, one is the network address and one is directed broadcast, leaving 64−2=62 usable host addresses.
Yaad rakho: Total mein sab count; usable mein boundary ke 2 minus.
"},{"id":"block-1","title":"Complete block · 192.168.10.77/26","html":"IP/prefix = 192.168.10.77/26.
The complete containing address block and its usable host range.
![\"\\text{step}=256-\\text{changing](\"math/cb1467f42fc2ae0f.svg\")
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−26=6; N=26=64.
Mask=255.255.255.192; binary=11111111.11111111.11111111.11000000.
The changing octet is number 4. Its mask value is 192, so step=256−192=64. The IP value 77 lies in 64…127.
Keep earlier octets; place 64 in octet 4; set any later octets to zero. Network=192.168.10.64.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 11000000 | 11111111 | 11000000 | 192 |
| 2 | 10101000 | 11111111 | 10101000 | 168 |
| 3 | 00001010 | 11111111 | 00001010 | 10 |
| 4 | 01001101 | 11000000 | 01000000 | 64 |
Use 127 in the changing octet and 255 in any later octets. Broadcast=192.168.10.127.
First usable=192.168.10.65; last usable=192.168.10.126; count=64−2=62.
Final answer: Network 192.168.10.64; broadcast 192.168.10.127; 64 total, 62 ordinary usable; range 192.168.10.65–192.168.10.126.
IP/prefix = 200.140.64.48/26.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−26=6; N=26=64.
Mask=255.255.255.192; binary=11111111.11111111.11111111.11000000.
The changing octet is number 4. Its mask value is 192, so step=256−192=64. The IP value 48 lies in 0…63.
Keep earlier octets; place 0 in octet 4; set any later octets to zero. Network=200.140.64.0.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 11001000 | 11111111 | 11001000 | 200 |
| 2 | 10001100 | 11111111 | 10001100 | 140 |
| 3 | 01000000 | 11111111 | 01000000 | 64 |
| 4 | 00110000 | 11000000 | 00000000 | 0 |
Use 63 in the changing octet and 255 in any later octets. Broadcast=200.140.64.63.
First usable=200.140.64.1; last usable=200.140.64.62; count=64−2=62.
Final answer: Network 200.140.64.0; broadcast 200.140.64.63; 64 total, 62 ordinary usable; range 200.140.64.1–200.140.64.62.
IP/prefix = 193.20.35.40/26.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−26=6; N=26=64.
Mask=255.255.255.192; binary=11111111.11111111.11111111.11000000.
The changing octet is number 4. Its mask value is 192, so step=256−192=64. The IP value 40 lies in 0…63.
Keep earlier octets; place 0 in octet 4; set any later octets to zero. Network=193.20.35.0.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 11000001 | 11111111 | 11000001 | 193 |
| 2 | 00010100 | 11111111 | 00010100 | 20 |
| 3 | 00100011 | 11111111 | 00100011 | 35 |
| 4 | 00101000 | 11000000 | 00000000 | 0 |
Use 63 in the changing octet and 255 in any later octets. Broadcast=193.20.35.63.
First usable=193.20.35.1; last usable=193.20.35.62; count=64−2=62.
Final answer: Network 193.20.35.0; broadcast 193.20.35.63; 64 total, 62 ordinary usable; range 193.20.35.1–193.20.35.62.
IP/prefix = 192.168.1.150/27.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−27=5; N=25=32.
Mask=255.255.255.224; binary=11111111.11111111.11111111.11100000.
The changing octet is number 4. Its mask value is 224, so step=256−224=32. The IP value 150 lies in 128…159.
Keep earlier octets; place 128 in octet 4; set any later octets to zero. Network=192.168.1.128.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 11000000 | 11111111 | 11000000 | 192 |
| 2 | 10101000 | 11111111 | 10101000 | 168 |
| 3 | 00000001 | 11111111 | 00000001 | 1 |
| 4 | 10010110 | 11100000 | 10000000 | 128 |
Use 159 in the changing octet and 255 in any later octets. Broadcast=192.168.1.159.
First usable=192.168.1.129; last usable=192.168.1.158; count=32−2=30.
Final answer: Network 192.168.1.128; broadcast 192.168.1.159; 32 total, 30 ordinary usable; range 192.168.1.129–192.168.1.158.
IP/prefix = 10.4.9.55/28.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−28=4; N=24=16.
Mask=255.255.255.240; binary=11111111.11111111.11111111.11110000.
The changing octet is number 4. Its mask value is 240, so step=256−240=16. The IP value 55 lies in 48…63.
Keep earlier octets; place 48 in octet 4; set any later octets to zero. Network=10.4.9.48.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 00001010 | 11111111 | 00001010 | 10 |
| 2 | 00000100 | 11111111 | 00000100 | 4 |
| 3 | 00001001 | 11111111 | 00001001 | 9 |
| 4 | 00110111 | 11110000 | 00110000 | 48 |
Use 63 in the changing octet and 255 in any later octets. Broadcast=10.4.9.63.
First usable=10.4.9.49; last usable=10.4.9.62; count=16−2=14.
Final answer: Network 10.4.9.48; broadcast 10.4.9.63; 16 total, 14 ordinary usable; range 10.4.9.49–10.4.9.62.
IP/prefix = 172.16.35.123/20.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−20=12; N=212=4,096.
Mask=255.255.240.0; binary=11111111.11111111.11110000.00000000.
The changing octet is number 3. Its mask value is 240, so step=256−240=16. The IP value 35 lies in 32…47.
Keep earlier octets; place 32 in octet 3; set any later octets to zero. Network=172.16.32.0.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 10101100 | 11111111 | 10101100 | 172 |
| 2 | 00010000 | 11111111 | 00010000 | 16 |
| 3 | 00100011 | 11110000 | 00100000 | 32 |
| 4 | 01111011 | 00000000 | 00000000 | 0 |
Use 47 in the changing octet and 255 in any later octets. Broadcast=172.16.47.255.
First usable=172.16.32.1; last usable=172.16.47.254; count=4096−2=4094.
Final answer: Network 172.16.32.0; broadcast 172.16.47.255; 4,096 total, 4,094 ordinary usable; range 172.16.32.1–172.16.47.254.
IP/prefix = 10.20.5.130/23.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−23=9; N=29=512.
Mask=255.255.254.0; binary=11111111.11111111.11111110.00000000.
The changing octet is number 3. Its mask value is 254, so step=256−254=2. The IP value 5 lies in 4…5.
Keep earlier octets; place 4 in octet 3; set any later octets to zero. Network=10.20.4.0.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 00001010 | 11111111 | 00001010 | 10 |
| 2 | 00010100 | 11111111 | 00010100 | 20 |
| 3 | 00000101 | 11111110 | 00000100 | 4 |
| 4 | 10000010 | 00000000 | 00000000 | 0 |
Use 5 in the changing octet and 255 in any later octets. Broadcast=10.20.5.255.
First usable=10.20.4.1; last usable=10.20.5.254; count=512−2=510.
Final answer: Network 10.20.4.0; broadcast 10.20.5.255; 512 total, 510 ordinary usable; range 10.20.4.1–10.20.5.254.
IP/prefix = 192.168.2.129/25.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−25=7; N=27=128.
Mask=255.255.255.128; binary=11111111.11111111.11111111.10000000.
The changing octet is number 4. Its mask value is 128, so step=256−128=128. The IP value 129 lies in 128…255.
Keep earlier octets; place 128 in octet 4; set any later octets to zero. Network=192.168.2.128.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 11000000 | 11111111 | 11000000 | 192 |
| 2 | 10101000 | 11111111 | 10101000 | 168 |
| 3 | 00000010 | 11111111 | 00000010 | 2 |
| 4 | 10000001 | 10000000 | 10000000 | 128 |
Use 255 in the changing octet and 255 in any later octets. Broadcast=192.168.2.255.
First usable=192.168.2.129; last usable=192.168.2.254; count=128−2=126.
Final answer: Network 192.168.2.128; broadcast 192.168.2.255; 128 total, 126 ordinary usable; range 192.168.2.129–192.168.2.254.
IP/prefix = 192.0.2.6/30.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−30=2; N=22=4.
Mask=255.255.255.252; binary=11111111.11111111.11111111.11111100.
The changing octet is number 4. Its mask value is 252, so step=256−252=4. The IP value 6 lies in 4…7.
Keep earlier octets; place 4 in octet 4; set any later octets to zero. Network=192.0.2.4.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 11000000 | 11111111 | 11000000 | 192 |
| 2 | 00000000 | 11111111 | 00000000 | 0 |
| 3 | 00000010 | 11111111 | 00000010 | 2 |
| 4 | 00000110 | 11111100 | 00000100 | 4 |
Use 7 in the changing octet and 255 in any later octets. Broadcast=192.0.2.7.
First usable=192.0.2.5; last usable=192.0.2.6; count=4−2=2.
Final answer: Network 192.0.2.4; broadcast 192.0.2.7; 4 total, 2 ordinary usable; range 192.0.2.5–192.0.2.6.
Definition. A /31 prefix can address both endpoints of a point-to-point IPv4 link under RFC 3021; a /32 identifies one IPv4 address, commonly as a host route.
Simple meaning. Minus 2 har slash pe blindly mat lagao.
RFC 3021 permits both addresses of a /31 to be used by the two endpoints of an IPv4 point-to-point link. Applying the ordinary broadcast-subnet formula would incorrectly claim there are zero usable endpoints.
Yaad rakho: /31 special link; /32 single address.
"},{"id":"hosts-1","title":"Hosts to prefix · 25 hosts","html":"Required usable hosts H=25.
The smallest sufficient subnet, expressed as /p and mask.
![\"2^h-2\\geq](\"math/9d320d9c17125391.svg\")
Host demand excludes network and broadcast, so use the usable-host inequality. Choose the smallest h that satisfies it.
With h=4, usable=24−2=14, which is less than 25.
With h=5, usable=25−2=30, which is sufficient.
p=32−5=27; mask=255.255.255.224.
30−25=5.
Final answer: /27, mask 255.255.255.224; 30 usable, 5 spare usable addresses.
Required usable hosts H=64.
The smallest sufficient subnet, expressed as /p and mask.
Host demand excludes network and broadcast, so use the usable-host inequality. Choose the smallest h that satisfies it.
With h=6, usable=26−2=62, which is less than 64.
With h=7, usable=27−2=126, which is sufficient.
p=32−7=25; mask=255.255.255.128.
126−64=62.
Final answer: /25, mask 255.255.255.128; 126 usable, 62 spare usable addresses.
Required usable hosts H=1000.
The smallest sufficient subnet, expressed as /p and mask.
Host demand excludes network and broadcast, so use the usable-host inequality. Choose the smallest h that satisfies it.
With h=9, usable=29−2=510, which is less than 1000.
With h=10, usable=210−2=1022, which is sufficient.
p=32−10=22; mask=255.255.252.0.
1022−1000=22.
Final answer: /22, mask 255.255.252.0; 1022 usable, 22 spare usable addresses.
Parent=192.168.8.0/24; required equal-size subnets=4.
Borrowed bits, new mask/prefix and complete subnet ranges.
![\"2^s\\geq](\"math/ca4fab7e8dac4ad7.svg\")
![\"p_{\\mathrm{new}}=p_{\\mathrm{old}}+s\"](\"math/3c981479a80689de.svg\")
Borrow the smallest s giving enough subnets. In modern subnetting, zero and all-ones subnets are valid; do not subtract two from the subnet count.
s=2, because 22=4 is the smallest power of two at least 4.
/24+2=/26; mask=255.255.255.192. Each subnet contains 64 total and 62 usable addresses.
| Subnet | First/network | Last/broadcast | Usable range |
|---|---|---|---|
| 192.168.8.0/26 | 192.168.8.0 | 192.168.8.63 | 192.168.8.1–192.168.8.62 |
| 192.168.8.64/26 | 192.168.8.64 | 192.168.8.127 | 192.168.8.65–192.168.8.126 |
| 192.168.8.128/26 | 192.168.8.128 | 192.168.8.191 | 192.168.8.129–192.168.8.190 |
| 192.168.8.192/26 | 192.168.8.192 | 192.168.8.255 | 192.168.8.193–192.168.8.254 |
4 available−4 required=0 extra equal-size subnet(s).
Final answer: /26; 4 subnets; 62 usable hosts each.
Parent=172.16.0.0/16; required equal-size subnets=8.
Borrowed bits, new mask/prefix and complete subnet ranges.
Borrow the smallest s giving enough subnets. In modern subnetting, zero and all-ones subnets are valid; do not subtract two from the subnet count.
s=3, because 23=8 is the smallest power of two at least 8.
/16+3=/19; mask=255.255.224.0. Each subnet contains 8192 total and 8190 usable addresses.
| Subnet | First/network | Last/broadcast | Usable range |
|---|---|---|---|
| 172.16.0.0/19 | 172.16.0.0 | 172.16.31.255 | 172.16.0.1–172.16.31.254 |
| 172.16.32.0/19 | 172.16.32.0 | 172.16.63.255 | 172.16.32.1–172.16.63.254 |
| 172.16.64.0/19 | 172.16.64.0 | 172.16.95.255 | 172.16.64.1–172.16.95.254 |
| 172.16.96.0/19 | 172.16.96.0 | 172.16.127.255 | 172.16.96.1–172.16.127.254 |
| 172.16.128.0/19 | 172.16.128.0 | 172.16.159.255 | 172.16.128.1–172.16.159.254 |
| 172.16.160.0/19 | 172.16.160.0 | 172.16.191.255 | 172.16.160.1–172.16.191.254 |
| 172.16.192.0/19 | 172.16.192.0 | 172.16.223.255 | 172.16.192.1–172.16.223.254 |
| 172.16.224.0/19 | 172.16.224.0 | 172.16.255.255 | 172.16.224.1–172.16.255.254 |
8 available−8 required=0 extra equal-size subnet(s).
Final answer: /19; 8 subnets; 8190 usable hosts each.
Parent=10.0.0.0/24; required equal-size subnets=6.
Borrowed bits, new mask/prefix and complete subnet ranges.
Borrow the smallest s giving enough subnets. In modern subnetting, zero and all-ones subnets are valid; do not subtract two from the subnet count.
s=3, because 23=8 is the smallest power of two at least 6.
/24+3=/27; mask=255.255.255.224. Each subnet contains 32 total and 30 usable addresses.
| Subnet | First/network | Last/broadcast | Usable range |
|---|---|---|---|
| 10.0.0.0/27 | 10.0.0.0 | 10.0.0.31 | 10.0.0.1–10.0.0.30 |
| 10.0.0.32/27 | 10.0.0.32 | 10.0.0.63 | 10.0.0.33–10.0.0.62 |
| 10.0.0.64/27 | 10.0.0.64 | 10.0.0.95 | 10.0.0.65–10.0.0.94 |
| 10.0.0.96/27 | 10.0.0.96 | 10.0.0.127 | 10.0.0.97–10.0.0.126 |
| 10.0.0.128/27 | 10.0.0.128 | 10.0.0.159 | 10.0.0.129–10.0.0.158 |
| 10.0.0.160/27 | 10.0.0.160 | 10.0.0.191 | 10.0.0.161–10.0.0.190 |
| 10.0.0.192/27 | 10.0.0.192 | 10.0.0.223 | 10.0.0.193–10.0.0.222 |
| 10.0.0.224/27 | 10.0.0.224 | 10.0.0.255 | 10.0.0.225–10.0.0.254 |
8 available−6 required=2 extra equal-size subnet(s).
Final answer: /27; 8 subnets; 30 usable hosts each.
Definition. CIDR expresses address blocks using an explicit prefix length; VLSM uses different subnet prefix lengths to match different size requirements within an allocated block.
Simple meaning. CIDR = slash se block; VLSM = departments ko need ke hisaab se different block sizes.
CIDR identifies a classless address block with an explicit /p prefix. VLSM permits subnets of different prefix lengths inside a larger block. Matching subnet sizes to requirements reduces waste while maintaining non-overlapping aligned allocations.
Yaad rakho: Hosts bola toh +2 allowance; addresses bola toh total count directly.
"},{"id":"vlsm-1","title":"Complete VLSM allocation · example 1","html":"| Department | Usable hosts |
|---|---|
| A | 100 |
| B | 50 |
| C | 25 |
| D | 10 |
An aligned subnet for each department, usable ranges, broadcasts and remaining addresses.
![\"2^h-2\\geq](\"math/0d2799e2eb681b5f.svg\")
Choose the smallest sufficient subnet per requirement. Align each start to its block size and never overlap a previously assigned range.
A (100) → B (50) → C (25) → D (10)
100 hosts require 27−2=126 usable; a smaller /26 would give 62. Choose /25. The next aligned block is 192.168.10.0/25; it ends at 192.168.10.127.
50 hosts require 26−2=62 usable; a smaller /27 would give 30. Choose /26. The next aligned block is 192.168.10.128/26; it ends at 192.168.10.191.
25 hosts require 25−2=30 usable; a smaller /28 would give 14. Choose /27. The next aligned block is 192.168.10.192/27; it ends at 192.168.10.223.
10 hosts require 24−2=14 usable; a smaller /29 would give 6. Choose /28. The next aligned block is 192.168.10.224/28; it ends at 192.168.10.239.
| Group | Hosts | Network/prefix | Mask | Usable range | Broadcast | Spare usable |
|---|---|---|---|---|---|---|
| A | 100 | 192.168.10.0/25 | 255.255.255.128 | 192.168.10.1–192.168.10.126 | 192.168.10.127 | 26 |
| B | 50 | 192.168.10.128/26 | 255.255.255.192 | 192.168.10.129–192.168.10.190 | 192.168.10.191 | 12 |
| C | 25 | 192.168.10.192/27 | 255.255.255.224 | 192.168.10.193–192.168.10.222 | 192.168.10.223 | 5 |
| D | 10 | 192.168.10.224/28 | 255.255.255.240 | 192.168.10.225–192.168.10.238 | 192.168.10.239 | 4 |
Reserved=128+64+32+16=240. Parent total=256. Unallocated=256−240=16.
Final answer: 240 total addresses reserved; 16 unallocated. The table gives one valid complete allocation.
| Department | Usable hosts |
|---|---|
| Research | 60 |
| Admin | 25 |
| IoT | 12 |
An aligned subnet for each department, usable ranges, broadcasts and remaining addresses.
Choose the smallest sufficient subnet per requirement. Align each start to its block size and never overlap a previously assigned range.
Research (60) → Admin (25) → IoT (12)
60 hosts require 26−2=62 usable; a smaller /27 would give 30. Choose /26. The next aligned block is 10.0.0.0/26; it ends at 10.0.0.63.
25 hosts require 25−2=30 usable; a smaller /28 would give 14. Choose /27. The next aligned block is 10.0.0.64/27; it ends at 10.0.0.95.
12 hosts require 24−2=14 usable; a smaller /29 would give 6. Choose /28. The next aligned block is 10.0.0.96/28; it ends at 10.0.0.111.
| Group | Hosts | Network/prefix | Mask | Usable range | Broadcast | Spare usable |
|---|---|---|---|---|---|---|
| Research | 60 | 10.0.0.0/26 | 255.255.255.192 | 10.0.0.1–10.0.0.62 | 10.0.0.63 | 2 |
| Admin | 25 | 10.0.0.64/27 | 255.255.255.224 | 10.0.0.65–10.0.0.94 | 10.0.0.95 | 5 |
| IoT | 12 | 10.0.0.96/28 | 255.255.255.240 | 10.0.0.97–10.0.0.110 | 10.0.0.111 | 2 |
Reserved=64+32+16=112. Parent total=256. Unallocated=256−112=144.
Final answer: 112 total addresses reserved; 144 unallocated. The table gives one valid complete allocation.
| Department | Usable hosts |
|---|---|
| Lab | 200 |
| Staff | 100 |
| Guests | 50 |
An aligned subnet for each department, usable ranges, broadcasts and remaining addresses.
Choose the smallest sufficient subnet per requirement. Align each start to its block size and never overlap a previously assigned range.
Lab (200) → Staff (100) → Guests (50)
200 hosts require 28−2=254 usable; a smaller /25 would give 126. Choose /24. The next aligned block is 172.16.8.0/24; it ends at 172.16.8.255.
100 hosts require 27−2=126 usable; a smaller /26 would give 62. Choose /25. The next aligned block is 172.16.9.0/25; it ends at 172.16.9.127.
50 hosts require 26−2=62 usable; a smaller /27 would give 30. Choose /26. The next aligned block is 172.16.9.128/26; it ends at 172.16.9.191.
| Group | Hosts | Network/prefix | Mask | Usable range | Broadcast | Spare usable |
|---|---|---|---|---|---|---|
| Lab | 200 | 172.16.8.0/24 | 255.255.255.0 | 172.16.8.1–172.16.8.254 | 172.16.8.255 | 54 |
| Staff | 100 | 172.16.9.0/25 | 255.255.255.128 | 172.16.9.1–172.16.9.126 | 172.16.9.127 | 26 |
| Guests | 50 | 172.16.9.128/26 | 255.255.255.192 | 172.16.9.129–172.16.9.190 | 172.16.9.191 | 12 |
Reserved=256+128+64=448. Parent total=512. Unallocated=512−448=64.
Final answer: 448 total addresses reserved; 64 unallocated. The table gives one valid complete allocation.
Definition. Route aggregation represents several contiguous networks by a shorter common prefix, reducing the number of advertised routing entries.
Simple meaning. Side-by-side blocks ko ek larger boundary mein summarize karo; gap chhupana nahi.
The blocks must be contiguous with no gaps and together form a power-of-two address count starting on the corresponding larger-block boundary. Their shared leading bits define the aggregate prefix.
Yaad rakho: Consecutive + power of two + aligned start.
"},{"id":"aggregate-1","title":"Exact route aggregation · example 1","html":"One prefix covering exactly the same address range.
![\"\\text{total](\"math/96879d7c45f4939b.svg\")
![\"p=32-\\log_2(N)\"](\"math/796d561e6716df45.svg\")
Adding sizes gives the candidate prefix. Contiguity and start-boundary alignment must also be satisfied.
Each input block has 256 addresses.
4×256=1024.
1024=210; prefix=32−10=/22.
Inputs cover 192.168.4.0–192.168.7.255, without gaps. 192.168.4.0/22 covers the same start and end and is correctly aligned.
Final answer: 192.168.4.0/22
One prefix covering exactly the same address range.
Adding sizes gives the candidate prefix. Contiguity and start-boundary alignment must also be satisfied.
Each input block has 128 addresses.
2×128=256.
256=28; prefix=32−8=/24.
Inputs cover 10.20.0.0–10.20.0.255, without gaps. 10.20.0.0/24 covers the same start and end and is correctly aligned.
Final answer: 10.20.0.0/24
One prefix covering exactly the same address range.
Adding sizes gives the candidate prefix. Contiguity and start-boundary alignment must also be satisfied.
Each input block has 512 addresses.
2×512=1024.
1024=210; prefix=32−10=/22.
Inputs cover 172.16.8.0–172.16.11.255, without gaps. 172.16.8.0/22 covers the same start and end and is correctly aligned.
Final answer: 172.16.8.0/22
Definition. Longest-prefix matching selects the matching forwarding-table entry with the greatest number of prefix bits.
Simple meaning. Pehle match check; matching entries mein sabse bada slash choose.
A router selects the most specific route whose prefix contains the destination. For 10.1.2.77, a matching 10.1.2.0/24 route takes precedence over 10.0.0.0/8 and 0.0.0.0/0.
Yaad rakho: Bada slash only among matches.
"},{"id":"match-1","title":"Forwarding table · example 1","html":"| Network/prefix | Next hop |
|---|---|
| 0.0.0.0/0 | R0 |
| 10.0.0.0/8 | R1 |
| 10.1.0.0/16 | R2 |
| 10.1.2.0/24 | R3 |
| 10.1.2.128/25 | R4 |
Destination=10.1.2.150
The matching entries and selected next hop.
Choose the largest prefix length among entries containing the destination.
Prefix length is considered only after confirming an address-range match.
| Route | Destination inside? |
|---|---|
| 0.0.0.0/0 | Yes |
| 10.0.0.0/8 | Yes |
| 10.1.0.0/16 | Yes |
| 10.1.2.0/24 | Yes |
| 10.1.2.128/25 | Yes |
0.0.0.0/0, 10.0.0.0/8, 10.1.0.0/16, 10.1.2.0/24, 10.1.2.128/25. The longest is 10.1.2.128/25.
10.1.2.128/25 uses R4.
Final answer: Forward via R4 using 10.1.2.128/25.
| Network/prefix | Next hop |
|---|---|
| 0.0.0.0/0 | R0 |
| 10.0.0.0/8 | R1 |
| 10.1.0.0/16 | R2 |
| 10.1.2.0/24 | R3 |
| 10.1.2.128/25 | R4 |
Destination=10.1.2.77
The matching entries and selected next hop.
Choose the largest prefix length among entries containing the destination.
Prefix length is considered only after confirming an address-range match.
| Route | Destination inside? |
|---|---|
| 0.0.0.0/0 | Yes |
| 10.0.0.0/8 | Yes |
| 10.1.0.0/16 | Yes |
| 10.1.2.0/24 | Yes |
| 10.1.2.128/25 | No |
0.0.0.0/0, 10.0.0.0/8, 10.1.0.0/16, 10.1.2.0/24. The longest is 10.1.2.0/24.
10.1.2.0/24 uses R3.
Final answer: Forward via R3 using 10.1.2.0/24.
| Network/prefix | Next hop |
|---|---|
| 0.0.0.0/0 | R0 |
| 10.0.0.0/8 | R1 |
| 10.1.0.0/16 | R2 |
| 10.1.2.0/24 | R3 |
| 10.1.2.128/25 | R4 |
Destination=192.0.2.9
The matching entries and selected next hop.
Choose the largest prefix length among entries containing the destination.
Prefix length is considered only after confirming an address-range match.
| Route | Destination inside? |
|---|---|
| 0.0.0.0/0 | Yes |
| 10.0.0.0/8 | No |
| 10.1.0.0/16 | No |
| 10.1.2.0/24 | No |
| 10.1.2.128/25 | No |
0.0.0.0/0. The longest is 0.0.0.0/0.
0.0.0.0/0 uses R0.
Final answer: Forward via R0 using 0.0.0.0/0.
Definition. IPv6 uses 128-bit addresses, conventionally represented by eight groups of four hexadecimal digits separated by colons.
Simple meaning. IPv4 mein 4 decimal boxes; IPv6 mein 8 hexadecimal groups.
![\"IPv6:](\"fig/ipv6-address.25497337ae.svg\")
| Hex | Decimal | Four bits |
|---|---|---|
| 0 | 0 | 0000 |
| 1 | 1 | 0001 |
| 2 | 2 | 0010 |
| 3 | 3 | 0011 |
| 4 | 4 | 0100 |
| 5 | 5 | 0101 |
| 6 | 6 | 0110 |
| 7 | 7 | 0111 |
| 8 | 8 | 1000 |
| 9 | 9 | 1001 |
| a | 10 | 1010 |
| b | 11 | 1011 |
| c | 12 | 1100 |
| d | 13 | 1101 |
| e | 14 | 1110 |
| f | 15 | 1111 |
An IPv6 address contains 128 bits, written as eight 16-bit hexadecimal groups. Leading zeros inside a group can be removed. One contiguous run of all-zero groups can be represented with ::, used at most once.
Yaad rakho: 8 groups × 4 hex digits × 4 bits = 128.
"},{"id":"expansion","title":"Expand :: by counting visible groups","html":"Definition. IPv6 expansion restores exactly eight groups, each written with four hexadecimal digits.
Simple meaning. Visible groups count karo; 8 mein se minus; utne zero groups :: ki jagah.
Count all explicitly written groups. Insert 8 minus this count all-zero groups at ::, then left-pad every group to four hexadecimal digits. The result must contain exactly eight groups.
Yaad rakho: Already-written 0 ko bhi count karna.
"},{"id":"expand-1","title":"Expand IPv6 · example 1","html":"Given address: 0:15::1:12:1213.
The full 128-bit address in eight-group hexadecimal notation.
![\"\\text{missing](\"math/a013f99fcbb0987c.svg\")
A written 0 is a visible group. :: represents only the omitted all-zero groups.
Left: 0, 15 (2 group(s)). Right: 1, 12, 1213 (3 group(s)).
Visible=2+3=5. Missing=8−5=3.
0:15:0:0:0:1:12:1213.
| Group number | Four-digit group |
|---|---|
| 1 | 0000 |
| 2 | 0015 |
| 3 | 0000 |
| 4 | 0000 |
| 5 | 0000 |
| 6 | 0001 |
| 7 | 0012 |
| 8 | 1213 |
0000:0015:0000:0000:0000:0001:0012:1213.
Final answer: 0000:0015:0000:0000:0000:0001:0012:1213
Given address: 2001:db8:0:1::9.
The full 128-bit address in eight-group hexadecimal notation.
A written 0 is a visible group. :: represents only the omitted all-zero groups.
Left: 2001, db8, 0, 1 (4 group(s)). Right: 9 (1 group(s)).
Visible=4+1=5. Missing=8−5=3.
2001:db8:0:1:0:0:0:9.
| Group number | Four-digit group |
|---|---|
| 1 | 2001 |
| 2 | 0db8 |
| 3 | 0000 |
| 4 | 0001 |
| 5 | 0000 |
| 6 | 0000 |
| 7 | 0000 |
| 8 | 0009 |
2001:0db8:0000:0001:0000:0000:0000:0009.
Final answer: 2001:0db8:0000:0001:0000:0000:0000:0009
Given address: ::1.
The full 128-bit address in eight-group hexadecimal notation.
A written 0 is a visible group. :: represents only the omitted all-zero groups.
Left: none (0 group(s)). Right: 1 (1 group(s)).
Visible=0+1=1. Missing=8−1=7.
0:0:0:0:0:0:0:1.
| Group number | Four-digit group |
|---|---|
| 1 | 0000 |
| 2 | 0000 |
| 3 | 0000 |
| 4 | 0000 |
| 5 | 0000 |
| 6 | 0000 |
| 7 | 0000 |
| 8 | 0001 |
0000:0000:0000:0000:0000:0000:0000:0001.
Final answer: 0000:0000:0000:0000:0000:0000:0000:0001
Given address: fe80::abcd:ef12.
The full 128-bit address in eight-group hexadecimal notation.
A written 0 is a visible group. :: represents only the omitted all-zero groups.
Left: fe80 (1 group(s)). Right: abcd, ef12 (2 group(s)).
Visible=1+2=3. Missing=8−3=5.
fe80:0:0:0:0:0:abcd:ef12.
| Group number | Four-digit group |
|---|---|
| 1 | fe80 |
| 2 | 0000 |
| 3 | 0000 |
| 4 | 0000 |
| 5 | 0000 |
| 6 | 0000 |
| 7 | abcd |
| 8 | ef12 |
fe80:0000:0000:0000:0000:0000:abcd:ef12.
Final answer: fe80:0000:0000:0000:0000:0000:abcd:ef12
Definition. Canonical IPv6 text formatting removes leading zeros and compresses the longest eligible run of consecutive zero groups.
Simple meaning. Group ke left wale zero hatao; sabse lambi zero-chain ko :: banao.
Compress the leftmost of the equal longest runs. Leading zeros in groups are removed and hexadecimal letters are lowercase. A single zero group is retained as 0 rather than compressed to :: in canonical text.
Yaad rakho: Longest first; tie hua toh leftmost.
"},{"id":"compress-1","title":"Compress IPv6 · example 1","html":"2001:0db8:0000:0000:0012:0000:0000:0001.
A canonical shortened spelling of the same address.
Remove leading zeroes; compress the longest run of at least two zero groups; choose the leftmost tied run.
These rules preserve all 128 bits while removing redundant writing.
2001:db8:0:0:12:0:0:1.
groups 3–4: 2 zero group(s); groups 6–7: 2 zero group(s).
Compress groups 3–4.
2001:db8::12:0:0:1.
Final answer: 2001:db8::12:0:0:1
2001:0db8:0000:0001:0002:0003:0004:0005.
A canonical shortened spelling of the same address.
Remove leading zeroes; compress the longest run of at least two zero groups; choose the leftmost tied run.
These rules preserve all 128 bits while removing redundant writing.
2001:db8:0:1:2:3:4:5.
groups 3–3: 1 zero group(s).
No run has two or more groups, so do not use ::.
2001:db8:0:1:2:3:4:5.
Final answer: 2001:db8:0:1:2:3:4:5
0000:0000:0000:0000:0000:0000:0000:0000.
A canonical shortened spelling of the same address.
Remove leading zeroes; compress the longest run of at least two zero groups; choose the leftmost tied run.
These rules preserve all 128 bits while removing redundant writing.
0:0:0:0:0:0:0:0.
groups 1–8: 8 zero group(s).
Compress groups 1–8.
::.
Final answer: ::
Definition. The IPv6 base header is a fixed 40-byte structure containing addressing and forwarding-control fields, followed by optional extension headers and upper-layer data.
Simple meaning. Base header fixed 40 bytes; extra information extension headers mein.
![\"IPv6](\"fig/ipv6-header.aef1bfbac7.svg\")
| Field | Bits | Purpose |
|---|---|---|
| Version | 4 | Value 6 |
| Traffic Class | 8 | Traffic treatment/congestion-related marking |
| Flow Label | 20 | Identifies a flow for appropriate handling |
| Payload Length | 16 | Bytes after the base header; includes extension headers when present |
| Next Header | 8 | Type of the next extension header or upper-layer protocol |
| Hop Limit | 8 | Decremented on forwarding; limits packet lifetime in hops |
| Source address | 128 | Sending interface address |
| Destination address | 128 | Destination address |
| Field | Bits | Purpose |
|---|---|---|
| Version | 4 | Value 6 |
| Traffic Class | 8 | Traffic treatment/congestion-related marking |
| Flow Label | 20 | Identifies a flow for appropriate handling |
| Payload Length | 16 | Bytes after the base header; includes extension headers when present |
| Next Header | 8 | Type of the next extension header or upper-layer protocol |
| Hop Limit | 8 | Decremented on forwarding; limits packet lifetime in hops |
| Source address | 128 | Sending interface address |
| Destination address | 128 | Destination address |
Yaad rakho: IPv6: 40 fixed; payload field excludes those 40.
"},{"id":"compare","title":"IPv4 versus IPv6 headers and address types","html":"Definition. IPv6 extends address size and reorganizes IP forwarding information into a fixed base header with extension headers.
Simple meaning. Sirf address bada nahi; header ka arrangement bhi badla.
| Feature | IPv4 | IPv6 |
|---|---|---|
| Address width | 32 bits | 128 bits |
| Base/header size | 20–60 bytes depending on options | 40-byte fixed base header |
| Length field | Total length includes IPv4 header | Payload Length excludes 40-byte base header |
| Lifetime field | TTL | Hop Limit |
| Following protocol | Protocol field | Next Header; may identify an extension |
| Header checksum | Present | No base-header checksum |
| Options | Inside IPv4 header | Typically extension headers |
| Fragmentation fields | In IPv4 header; routers may fragment if permitted | Fragment extension header; fragmentation by source, not routers |
| Feature | IPv4 | IPv6 |
|---|---|---|
| Address width | 32 bits | 128 bits |
| Base/header size | 20–60 bytes depending on options | 40-byte fixed base header |
| Length field | Total length includes IPv4 header | Payload Length excludes 40-byte base header |
| Lifetime field | TTL | Hop Limit |
| Following protocol | Protocol field | Next Header; may identify an extension |
| Header checksum | Present | No base-header checksum |
| Options | Inside IPv4 header | Typically extension headers |
| Fragmentation fields | In IPv4 header; routers may fragment if permitted | Fragment extension header; fragmentation by source, not routers |
Yaad rakho: v4 total length; v6 payload length.
"},{"id":"length-1","title":"IPv6 packet length · example 1","html":"Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Base=40 bytes; extensions=0 bytes; upper-layer content=1000 bytes. Ordinary non-jumbo packet.
Payload Length and total bytes.
![\"P=E+D,\\qquad](\"math/30fbafad36710670.svg\")
Payload begins immediately after the IPv6 base header. Extensions are part of that payload.
P=0+1000=1000 bytes.
T=40+1000=1040 bytes.
The Payload Length field stores 1000, not 1040.
Final answer: Payload Length=1000; total=1040 bytes.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Base=40 bytes; extensions=24 bytes; upper-layer content=1200 bytes. Ordinary non-jumbo packet.
Payload Length and total bytes.
Payload begins immediately after the IPv6 base header. Extensions are part of that payload.
P=24+1200=1224 bytes.
T=40+1224=1264 bytes.
The Payload Length field stores 1224, not 1264.
Final answer: Payload Length=1224; total=1264 bytes.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Base=40 bytes; extensions=40 bytes; upper-layer content=500 bytes. Ordinary non-jumbo packet.
Payload Length and total bytes.
Payload begins immediately after the IPv6 base header. Extensions are part of that payload.
P=40+500=540 bytes.
T=40+540=580 bytes.
The Payload Length field stores 540, not 580.
Final answer: Payload Length=540; total=580 bytes.
Definition. Address Resolution Protocol (ARP) resolves a known IPv4 address to a link-layer hardware address on the local network.
Simple meaning. IP pata hai; local delivery ke liye MAC chahiye.
![\"ARP](\"fig/arp.c8af39ddb3.svg\")
ARP maps a known local next-hop IPv4 address to a hardware address. A sender broadcasts a request when the mapping is missing; the address owner replies, and the mapping is cached. For a remote IP destination, the sender resolves its gateway’s local MAC address, then forwards the packet toward that gateway.
Yaad rakho: Remote packet ka next-hop MAC gateway ka hota hai.
"},{"id":"next-hop-1","title":"ARP decision · example 1","html":"H=192.168.1.10/24; gateway=192.168.1.1; destination=192.168.1.20.
Local/remote decision and ARP target.
Compare source and destination network prefixes using the source mask. ARP resolves the selected local next-hop IP.
The IP destination stays the final destination; the Ethernet next-hop address depends on whether that destination is local.
With /24, H is in 192.168.1.0–192.168.1.255.
192.168.1.20 is inside that block, so delivery is local.
Resolve 192.168.1.20 using ARP.
IP destination remains 192.168.1.20; frame destination is that host’s MAC.
Final answer: ARP target: 192.168.1.20.
H=192.168.1.10/24; gateway=192.168.1.1; destination=203.0.113.7.
Local/remote decision and ARP target.
Compare source and destination network prefixes using the source mask. ARP resolves the selected local next-hop IP.
The IP destination stays the final destination; the Ethernet next-hop address depends on whether that destination is local.
With /24, H is in 192.168.1.0–192.168.1.255.
203.0.113.7 is outside that block, so use a router.
Resolve 192.168.1.1 using ARP.
IP destination remains 203.0.113.7; frame destination is the gateway’s MAC.
Final answer: ARP target: 192.168.1.1.
H=192.168.1.10/24; gateway=192.168.1.1; destination=192.168.2.20.
Local/remote decision and ARP target.
Compare source and destination network prefixes using the source mask. ARP resolves the selected local next-hop IP.
The IP destination stays the final destination; the Ethernet next-hop address depends on whether that destination is local.
With /24, H is in 192.168.1.0–192.168.1.255.
192.168.2.20 is outside that block, so use a router.
Resolve 192.168.1.1 using ARP.
IP destination remains 192.168.2.20; frame destination is the gateway’s MAC.
Final answer: ARP target: 192.168.1.1.
Definition. Reverse ARP (RARP) allows a client that knows its hardware address to obtain its IPv4 address from a RARP server.
Simple meaning. ARP: IP se MAC. RARP: MAC se IP.
| Protocol | Known | Needed | Use |
|---|---|---|---|
| ARP | Local next-hop IPv4 | MAC/hardware address | Local frame delivery |
| RARP | Client hardware address | Client IPv4 | Historical boot-time assignment |
Yaad rakho: RARP reverse karta hai; DHCP configuration badhata hai.
"},{"id":"dhcp","title":"DHCP: automatic configuration through DORA","html":"Definition. Dynamic Host Configuration Protocol (DHCP) provides IP configuration parameters to clients and can lease reusable addresses for limited periods.
Simple meaning. DORA = Discover, Offer, Request, Acknowledge.
![\"DHCP](\"fig/dhcp.510e8eca42.svg\")
DHCP reduces manual configuration and supports controlled address reuse through leases.
Yaad rakho: Discover–Offer–Request–ACK; pehle D, phir O, phir R, phir A.
"},{"id":"lease","title":"DHCP leases, renewal and a relay","html":"Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Definition. A DHCP lease grants use of an address for a specified lifetime; renewal extends that permission, while a relay supports communication across subnets.
Simple meaning. Address permanent gift nahi: time-limited lease, phir renew.
A lease permits a client to use its assigned address for a limited period and may be renewed. A relay conveys DHCP messages between a client subnet and a remote DHCP server, allowing centralized configuration without placing a server on every subnet.
Yaad rakho: Lease = limited time; relay = subnet ke paar DHCP support.
"},{"id":"lease-1","title":"DHCP timer example 1","html":"Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Lease duration L=8 h; use default fractional timers.
T1 and T2 as elapsed times.
![\"T_1=0.5L,\\qquad](\"math/ca97cfd9b6235ae7.svg\")
Use these defaults only when the server/question has not specified other timer values.
T1=0.5×8=4 h.
T2=0.875×8=7 h.
4 h < 7 h < 8 h.
Final answer: T1=4 h; T2=7 h.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Lease duration L=16 h; use default fractional timers.
T1 and T2 as elapsed times.
Use these defaults only when the server/question has not specified other timer values.
T1=0.5×16=8 h.
T2=0.875×16=14 h.
8 h < 14 h < 16 h.
Final answer: T1=8 h; T2=14 h.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Lease duration L=24 h; use default fractional timers.
T1 and T2 as elapsed times.
Use these defaults only when the server/question has not specified other timer values.
T1=0.5×24=12 h.
T2=0.875×24=21 h.
12 h < 21 h < 24 h.
Final answer: T1=12 h; T2=21 h.
Definition. Routing discovers or computes paths through a network; forwarding uses a table to send an individual packet toward the selected next hop.
Simple meaning. Routing table banata hai; forwarding us table ko use karta hai.
Routing determines suitable network paths and produces routing/forwarding information. Forwarding applies that information to a particular packet. If A has learned that network X is reached through B, computing that choice is routing and sending a packet to B is forwarding.
Yaad rakho: Predecessor = pichhla node; next hop = source ke baad pehla node.
"},{"id":"relaxation","title":"Relaxation: replace an old route only when the candidate is cheaper","html":"Definition. Relaxing an edge u→v compares the existing distance to v with a candidate path that reaches u first and then crosses the edge to v.
Simple meaning. Purana cost versus u tak ka cost + u→v edge. Jo chhota, woh rakho.
![\"d[v]\\leftarrow\\min\\bigl(d[v],d[u]+w(u,v)\\bigr)\"](\"math/96f918d5ce249dc4.svg\")
Relaxation evaluates candidate=d[u]+w(u,v). If the candidate is less than d[v], it updates d[v] and records u as predecessor. Otherwise the existing route is retained.
Yaad rakho: Add path + edge, compare, then replace only if smaller.
"},{"id":"relax-1","title":"One edge relaxation · example 1","html":"d[U]=4; w(U,V)=3; old d[V]=10.
New d[V] and predecessor action.
![\"c=d[U]+w(U,V),\\quad](\"math/e662a1e6810b8636.svg\")
The candidate contains both the cost to U and the final edge to V.
4+(3)=7.
min(10,7)=7.
Set predecessor[V]=U.
Final answer: New d[V]=7.
d[U]=5; w(U,V)=4; old d[V]=7.
New d[V] and predecessor action.
The candidate contains both the cost to U and the final edge to V.
5+(4)=9.
min(7,9)=7.
Keep the existing predecessor; the candidate does not improve the route.
Final answer: New d[V]=7.
d[U]=6; w(U,V)=-2; old d[V]=9.
New d[V] and predecessor action.
The candidate contains both the cost to U and the final edge to V.
6+(-2)=4.
min(9,4)=4.
Set predecessor[V]=U.
Final answer: New d[V]=4.
Definition. Dijkstra’s algorithm finds shortest paths from a source in a graph whose edge weights are nonnegative.
Simple meaning. Sabse chhota unfinished distance select karo, phir uske neighbours improve karo.
The standard algorithm requires nonnegative edge weights.
Yaad rakho: Minimum tentative total choose; negative edges pe Dijkstra nahi.
"},{"id":"dijkstra-1","title":"Dijkstra · complete example 1","html":"| Edge | Cost |
|---|---|
| A→B | 4 |
| A→C | 2 |
| C→B | 1 |
| B→D | 3 |
| C→D | 7 |
Source=A; target=D. Edges are directed; do not add reverse edges.
The path A to D, its total cost and intermediate work.
![\"d[v]\\leftarrow\\min(d[v],d[u]+w(u,v))\"](\"math/651d18e4e3d7619d.svg\")
Use Dijkstra because all weights are nonnegative. Select the smallest unsettled total distance, not the cheapest isolated edge.
![\"Dijkstra](\"fig/dijkstra-1.973d01595e.svg\")
d[A]=0; every other distance=∞. No predecessors are known.
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Tentative |
| B | ∞ | — | Tentative |
| C | ∞ | — | Tentative |
| D | ∞ | — | Tentative |
Unsettled candidates before selection: A=0, B=∞, C=∞, D=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| A→B | ∞ | 0+(4)=4 | Improve to 4; predecessor=A |
| A→C | ∞ | 0+(2)=2 | Improve to 2; predecessor=A |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 4 | A | Tentative |
| C | 2 | A | Tentative |
| D | ∞ | — | Tentative |
Unsettled candidates before selection: B=4, C=2, D=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| C→B | 4 | 2+(1)=3 | Improve to 3; predecessor=C |
| C→D | ∞ | 2+(7)=9 | Improve to 9; predecessor=C |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | C | Tentative |
| C | 2 | A | Settled |
| D | 9 | C | Tentative |
Unsettled candidates before selection: B=3, D=9. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| B→D | 9 | 3+(3)=6 | Improve to 6; predecessor=B |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | C | Settled |
| C | 2 | A | Settled |
| D | 6 | B | Tentative |
Unsettled candidates before selection: D=6. Select the minimum; ties use alphabetical order.
No outgoing edges to relax.
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | C | Settled |
| C | 2 | A | Settled |
| D | 6 | B | Settled |
Trace backwards: D ← B ← C ← A. Reverse it: A → C → B → D.
Edge sum: 2 + 1 + 3 = 6.
Final answer: A → C → B → D; cost=6.
| Edge | Cost |
|---|---|
| A—B | 2 |
| A—C | 5 |
| B—C | 1 |
| B—D | 4 |
| C—D | 1 |
| D—E | 3 |
| C—E | 8 |
Source=A; target=E. Each edge can be traversed in either direction.
The path A to E, its total cost and intermediate work.
Use Dijkstra because all weights are nonnegative. Select the smallest unsettled total distance, not the cheapest isolated edge.
![\"Dijkstra](\"fig/dijkstra-2.170b001531.svg\")
d[A]=0; every other distance=∞. No predecessors are known.
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Tentative |
| B | ∞ | — | Tentative |
| C | ∞ | — | Tentative |
| D | ∞ | — | Tentative |
| E | ∞ | — | Tentative |
Unsettled candidates before selection: A=0, B=∞, C=∞, D=∞, E=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| A→B | ∞ | 0+(2)=2 | Improve to 2; predecessor=A |
| A→C | ∞ | 0+(5)=5 | Improve to 5; predecessor=A |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 2 | A | Tentative |
| C | 5 | A | Tentative |
| D | ∞ | — | Tentative |
| E | ∞ | — | Tentative |
Unsettled candidates before selection: B=2, C=5, D=∞, E=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| B→C | 5 | 2+(1)=3 | Improve to 3; predecessor=B |
| B→D | ∞ | 2+(4)=6 | Improve to 6; predecessor=B |
| B→A | 0 | 2+(2)=4 | Already settled; retain |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 2 | A | Settled |
| C | 3 | B | Tentative |
| D | 6 | B | Tentative |
| E | ∞ | — | Tentative |
Unsettled candidates before selection: C=3, D=6, E=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| C→D | 6 | 3+(1)=4 | Improve to 4; predecessor=C |
| C→E | ∞ | 3+(8)=11 | Improve to 11; predecessor=C |
| C→A | 0 | 3+(5)=8 | Already settled; retain |
| C→B | 2 | 3+(1)=4 | Already settled; retain |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 2 | A | Settled |
| C | 3 | B | Settled |
| D | 4 | C | Tentative |
| E | 11 | C | Tentative |
Unsettled candidates before selection: D=4, E=11. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| D→E | 11 | 4+(3)=7 | Improve to 7; predecessor=D |
| D→B | 2 | 4+(4)=8 | Already settled; retain |
| D→C | 3 | 4+(1)=5 | Already settled; retain |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 2 | A | Settled |
| C | 3 | B | Settled |
| D | 4 | C | Settled |
| E | 7 | D | Tentative |
Unsettled candidates before selection: E=7. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| E→D | 4 | 7+(3)=10 | Already settled; retain |
| E→C | 3 | 7+(8)=15 | Already settled; retain |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 2 | A | Settled |
| C | 3 | B | Settled |
| D | 4 | C | Settled |
| E | 7 | D | Settled |
Trace backwards: E ← D ← C ← B ← A. Reverse it: A → B → C → D → E.
Edge sum: 2 + 1 + 1 + 3 = 7.
Final answer: A → B → C → D → E; cost=7.
| Edge | Cost |
|---|---|
| A→B | 3 |
| A→C | 3 |
| B→D | 2 |
| C→D | 2 |
Source=A; target=E. Edges are directed; do not add reverse edges.
The path A to E, its total cost and intermediate work.
Use Dijkstra because all weights are nonnegative. Select the smallest unsettled total distance, not the cheapest isolated edge.
![\"Dijkstra](\"fig/dijkstra-3.312b4c0fd9.svg\")
d[A]=0; every other distance=∞. No predecessors are known.
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Tentative |
| B | ∞ | — | Tentative |
| C | ∞ | — | Tentative |
| D | ∞ | — | Tentative |
| E | ∞ | — | Tentative |
Unsettled candidates before selection: A=0, B=∞, C=∞, D=∞, E=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| A→B | ∞ | 0+(3)=3 | Improve to 3; predecessor=A |
| A→C | ∞ | 0+(3)=3 | Improve to 3; predecessor=A |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | A | Tentative |
| C | 3 | A | Tentative |
| D | ∞ | — | Tentative |
| E | ∞ | — | Tentative |
Unsettled candidates before selection: B=3, C=3, D=∞, E=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| B→D | ∞ | 3+(2)=5 | Improve to 5; predecessor=B |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | A | Settled |
| C | 3 | A | Tentative |
| D | 5 | B | Tentative |
| E | ∞ | — | Tentative |
Unsettled candidates before selection: C=3, D=5, E=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| C→D | 5 | 3+(2)=5 | Retain 5 (equal cost) |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | A | Settled |
| C | 3 | A | Settled |
| D | 5 | B | Tentative |
| E | ∞ | — | Tentative |
Unsettled candidates before selection: D=5, E=∞. Select the minimum; ties use alphabetical order.
No outgoing edges to relax.
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | A | Settled |
| C | 3 | A | Settled |
| D | 5 | B | Settled |
| E | ∞ | — | Tentative |
Every remaining tentative distance is ∞. No path from the source reaches these vertices.
There is no finite path to recover.
Final answer: E is unreachable from A.
Definition. Bellman–Ford computes single-source shortest paths by repeatedly relaxing every edge. It supports negative edge weights and detects reachable negative cycles.
Simple meaning. Har edge ko baar-baar relax; negative edge chalega, negative cycle problem hai.
Bellman–Ford repeatedly relaxes all edges and can handle negative weights. It uses up to V−1 passes and an additional improvement check for reachable negative cycles. Dijkstra settles the smallest tentative vertex and requires nonnegative weights for its standard correctness guarantee.
Yaad rakho: Pass = all edges once. Negative edge ≠ negative cycle.
"},{"id":"bellman-1","title":"Bellman–Ford · complete example 1","html":"| Directed edge | Weight |
|---|---|
| A→B | 4 |
| A→C | 5 |
| B→C | -2 |
| C→D | 3 |
The shortest route and cost, with complete passes and a negative-cycle check.
Bellman–Ford permits negative edges. We use the listed order with in-place updates; do not mix this with a table that reads only the previous pass.
![\"Bellman–Ford](\"fig/bellman-1.1bf4232fcf.svg\")
4 vertices → at most 3 main passes. d[A]=0; every other distance=∞. Use the listed edge order and immediate (in-place) updates.
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| A→B | ∞ | 0+(4)=4 | 4 | Update; predecessor=A |
| A→C | ∞ | 0+(5)=5 | 5 | Update; predecessor=A |
| B→C | 5 | 4+(-2)=2 | 2 | Update; predecessor=B |
| C→D | ∞ | 2+(3)=5 | 5 | Update; predecessor=C |
| Vertex | A | B | C | D |
|---|---|---|---|---|
| End-of-pass distance | 0 | 4 | 2 | 5 |
| Predecessor | — | A | B | C |
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| A→B | 4 | 0+(4)=4 | 4 | No improvement |
| A→C | 2 | 0+(5)=5 | 2 | No improvement |
| B→C | 2 | 4+(-2)=2 | 2 | No improvement |
| C→D | 5 | 2+(3)=5 | 5 | No improvement |
| Vertex | A | B | C | D |
|---|---|---|---|---|
| End-of-pass distance | 0 | 4 | 2 | 5 |
| Predecessor | — | A | B | C |
This entire pass changed no distance. Further identical relaxation passes will not improve anything.
Backward predecessors: D ← C ← B ← A. Forward route: A → B → C → D.
Cost = (4) + (-2) + (3) = 5. No edge can now improve its destination label.
Final answer: A → B → C → D; shortest cost=5.
| Directed edge | Weight |
|---|---|
| C→D | 2 |
| B→C | 1 |
| A→B | 3 |
| A→D | 12 |
The shortest route and cost, with complete passes and a negative-cycle check.
Bellman–Ford permits negative edges. We use the listed order with in-place updates; do not mix this with a table that reads only the previous pass.
![\"Bellman–Ford](\"fig/bellman-2.9bfc43d501.svg\")
4 vertices → at most 3 main passes. d[A]=0; every other distance=∞. Use the listed edge order and immediate (in-place) updates.
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| C→D | ∞ | ∞+(2)=∞ | ∞ | No improvement |
| B→C | ∞ | ∞+(1)=∞ | ∞ | No improvement |
| A→B | ∞ | 0+(3)=3 | 3 | Update; predecessor=A |
| A→D | ∞ | 0+(12)=12 | 12 | Update; predecessor=A |
| Vertex | A | B | C | D |
|---|---|---|---|---|
| End-of-pass distance | 0 | 3 | ∞ | 12 |
| Predecessor | — | A | — | A |
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| C→D | 12 | ∞+(2)=∞ | 12 | No improvement |
| B→C | ∞ | 3+(1)=4 | 4 | Update; predecessor=B |
| A→B | 3 | 0+(3)=3 | 3 | No improvement |
| A→D | 12 | 0+(12)=12 | 12 | No improvement |
| Vertex | A | B | C | D |
|---|---|---|---|---|
| End-of-pass distance | 0 | 3 | 4 | 12 |
| Predecessor | — | A | B | A |
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| C→D | 12 | 4+(2)=6 | 6 | Update; predecessor=C |
| B→C | 4 | 3+(1)=4 | 4 | No improvement |
| A→B | 3 | 0+(3)=3 | 3 | No improvement |
| A→D | 6 | 0+(12)=12 | 6 | No improvement |
| Vertex | A | B | C | D |
|---|---|---|---|---|
| End-of-pass distance | 0 | 3 | 4 | 6 |
| Predecessor | — | A | B | C |
Backward predecessors: D ← C ← B ← A. Forward route: A → B → C → D.
Cost = (3) + (1) + (2) = 6. No edge can now improve its destination label.
Final answer: A → B → C → D; shortest cost=6.
| Directed edge | Weight |
|---|---|
| A→B | 1 |
| B→C | -2 |
| C→B | -1 |
The shortest route and cost, with complete passes and a negative-cycle check.
Bellman–Ford permits negative edges. We use the listed order with in-place updates; do not mix this with a table that reads only the previous pass.
![\"Bellman–Ford](\"fig/bellman-3.b876d89f79.svg\")
3 vertices → at most 2 main passes. d[A]=0; every other distance=∞. Use the listed edge order and immediate (in-place) updates.
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| A→B | ∞ | 0+(1)=1 | 1 | Update; predecessor=A |
| B→C | ∞ | 1+(-2)=-1 | -1 | Update; predecessor=B |
| C→B | 1 | -1+(-1)=-2 | -2 | Update; predecessor=C |
| Vertex | A | B | C |
|---|---|---|---|
| End-of-pass distance | 0 | -2 | -1 |
| Predecessor | — | C | B |
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| A→B | -2 | 0+(1)=1 | -2 | No improvement |
| B→C | -1 | -2+(-2)=-4 | -4 | Update; predecessor=B |
| C→B | -2 | -4+(-1)=-5 | -5 | Update; predecessor=C |
| Vertex | A | B | C |
|---|---|---|---|
| End-of-pass distance | 0 | -5 | -4 |
| Predecessor | — | C | B |
After the allowed main passes, an edge still improves:
| Edge | Candidate | Current distance |
|---|---|---|
| B→C | -7 | -4 |
A reachable negative cycle exists. Do not present the changing distance labels as finite shortest-path answers.
Final answer: A reachable negative cycle prevents a finite shortest distance for vertices reachable from that cycle.
Definition. Distance-vector routing represents each destination by an estimated distance and next hop, updating those estimates using neighbouring routers’ advertised distances.
Simple meaning. Neighbour ke destination-cost mein apna neighbour-link cost add karo.
![\"Distance](\"fig/distance-vector.8f3c90bc87.svg\")
![\"\\text{candidate](\"math/16a3dc8e6e4edd60.svg\")
![\"D_x(y)=\\min_{v\\in](\"math/9703c520036a4b0a.svg\")
For each neighbour v, add the local link cost to that neighbour’s advertised destination cost. Select the minimum and record that neighbour as next hop. N(x) denotes the neighbours of x.
Yaad rakho: Neighbour ka distance + neighbour tak apna distance.
"},{"id":"dv-1","title":"Distance-vector update · example 1","html":"| Neighbour | X-to-neighbour cost | Advertised distance to Z |
|---|---|---|
| B | 1 | 4 |
| C | 2 | 2 |
| D | 3 | 5 |
Candidates and the minimum total route cost.
![\"C_v=c(X,v)+D_v(Z)\"](\"math/2bb516a29dc6ed94.svg\")
The advertisement starts from the neighbour, so the local link to that neighbour must be added.
| Via | Addition | Total |
|---|---|---|
| B | 1+4 | 5 |
| C | 2+2 | 4 |
| D | 3+5 | 8 |
min(5, 4, 8)=4.
C is the minimum-cost neighbour.
Final answer: Cost=4; best next hop(s): C.
| Neighbour | X-to-neighbour cost | Advertised distance to Z |
|---|---|---|
| B | 2 | 1 |
| C | 1 | 4 |
| D | 5 | 0 |
Candidates and the minimum total route cost.
The advertisement starts from the neighbour, so the local link to that neighbour must be added.
| Via | Addition | Total |
|---|---|---|
| B | 2+1 | 3 |
| C | 1+4 | 5 |
| D | 5+0 | 5 |
min(3, 5, 5)=3.
B is the minimum-cost neighbour.
Final answer: Cost=3; best next hop(s): B.
| Neighbour | X-to-neighbour cost | Advertised distance to Z |
|---|---|---|
| B | 3 | 2 |
| C | 2 | 3 |
| D | 4 | 4 |
Candidates and the minimum total route cost.
The advertisement starts from the neighbour, so the local link to that neighbour must be added.
| Via | Addition | Total |
|---|---|---|
| B | 3+2 | 5 |
| C | 2+3 | 5 |
| D | 4+4 | 8 |
min(5, 5, 8)=5.
B, C are tied; a protocol may use a defined tie rule or equal-cost multipath.
Final answer: Cost=5; best next hop(s): B, C.
Definition. Routing Information Protocol (RIP) is an interior distance-vector routing protocol that uses a bounded metric, commonly hop count, to select routes.
Simple meaning. RIP mein hops count; 15 tak reachable, 16 means unreachable.
RIP uses distance-vector routing and commonly hop count as its metric. Reachable metrics are at most 15; 16 represents infinity. Its main limitations include limited diameter, a coarse metric and relatively slow convergence with possible temporary loops.
Yaad rakho: RIP: neighbour distances; 16 is infinity.
"},{"id":"loops","title":"Count-to-infinity and loop-mitigation methods","html":"Definition. Count-to-infinity is a distance-vector failure in which routers repeatedly increase their estimates for an unreachable destination because they believe one another’s stale route advertisements.
Simple meaning. Dono neighbour ek dusre ko bol rahe: mere through jao—par actual route gaya.
Split horizon suppresses advertising a route back in the direction from which it was learned. Poison reverse explicitly advertises that route back with an infinite/unreachable metric. Both reduce two-router distance-vector loops caused by stale information.
Yaad rakho: Split: wapas mat bolo. Poison reverse: wapas bolo, par unreachable.
"},{"id":"link-state","title":"Link-state routing: share a map, calculate locally","html":"Definition. Link-state routing distributes information about routers’ local links so that routers can build a topology database and compute shortest paths independently.
Simple meaning. Distance vector mein neighbour ka answer; link state mein topology map.
![\"Link-state](\"fig/link-state.bbee3173ec.svg\")
Yaad rakho: Same map, different root, different tree.
"},{"id":"ospf","title":"OSPF: link-state shortest paths within an autonomous system","html":"Definition. Open Shortest Path First (OSPF) is an interior link-state routing protocol that selects routes using shortest-path calculations and additive link costs.
Simple meaning. OSPF = link-state map + Dijkstra + summed costs.
OSPF exchanges and maintains link-state information, builds a topology database and runs Dijkstra’s shortest-path computation rooted at the local router. It chooses a path with the lowest sum of link costs and installs the corresponding next hop. Equal-cost paths may be supported.
Yaad rakho: OSPF cost add karo; sirf hops nahi.
"},{"id":"ospf-packets","title":"The five OSPF packet types","html":"Definition. OSPFv2 uses five packet types for neighbour maintenance and reliable exchange of link-state database information.
Simple meaning. Hello → database summary → request → update → acknowledgement; har packet ka kaam alag.
![\"OSPFv2:](\"fig/ospf-packets.6093f0d178.svg\")
| Packet type | What it does | Simple example |
|---|---|---|
| Hello | Discover/maintain neighbours and exchange relationship parameters | Are we still neighbours? |
| Database Description (DD/DBD) | Summarize link-state database contents during synchronization | Here are the LSA headers I have |
| Link State Request (LSR) | Ask for specific missing/newer link-state information | Please send this LSA |
| Link State Update (LSU) | Carry one or more LSAs | Here is the requested or newly flooded link-state information |
| Link State Acknowledgment (LSAck) | Acknowledge receipt of LSAs | I received these LSAs |
| Packet type | What it does | Simple example |
|---|---|---|
| Hello | Discover/maintain neighbours and exchange relationship parameters | Are we still neighbours? |
| Database Description (DD/DBD) | Summarize link-state database contents during synchronization | Here are the LSA headers I have |
| Link State Request (LSR) | Ask for specific missing/newer link-state information | Please send this LSA |
| Link State Update (LSU) | Carry one or more LSAs | Here is the requested or newly flooded link-state information |
| Link State Acknowledgment (LSAck) | Acknowledge receipt of LSAs | I received these LSAs |
Yaad rakho: H–D–R–U–A: Hello, Description, Request, Update, Ack.
"},{"id":"comparison","title":"RIP versus OSPF; Bellman–Ford versus Dijkstra","html":"Definition. RIP and OSPF are routing protocols; Bellman–Ford and Dijkstra are algorithms underlying different route-computation approaches.
Simple meaning. Protocol message bhejta hai; algorithm route calculate karta hai.
| Feature | RIP | OSPF |
|---|---|---|
| Family | Distance vector | Link state |
| Typical metric | Hop count; 16 unreachable | Sum of configured link costs |
| Knowledge exchanged | Destination distances via neighbours | Link-state information |
| Algorithm connection | Distributed Bellman–Ford idea | Dijkstra shortest paths |
| Scale design | Small/simple networks; limited diameter | Area-based scalable design |
Yaad rakho: Distance vector → RIP; link state → OSPF.
"}]},{"id":"packet-support","title":"IP packet details · supporting practice","scope":"Unit 4 support · header comparison; fragmentation is additional practice","sections":[{"id":"ipv4-header","title":"IPv4 header fields needed for the IPv6 comparison","html":"Definition. An IPv4 datagram contains a variable-length header followed by payload. The header supplies addressing, forwarding and fragmentation information.
Simple meaning. Header delivery instructions hai; payload actual carried content.
![\"IPv4](\"fig/ipv4-header.5d41e2a297.svg\")
Total Length measures the whole IPv4 packet in bytes. IHL measures the header in 32-bit words, so header bytes=4×IHL. Protocol identifies the payload’s upper-layer protocol, such as TCP or UDP.
Yaad rakho: IHL words ×4 = bytes; total minus header = payload.
"},{"id":"ihl-1","title":"IPv4 header-length example 1","html":"Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
IHL=5; Total Length=1000 bytes.
Header bytes and payload bytes.
![\"H=4\\,\\mathrm{IHL},\\qquad](\"math/830ed738bb3329f8.svg\")
IHL counts 32-bit words; each word is 4 bytes.
H=4×5=20 bytes.
P=1000−20=980 bytes.
20+980=1000 bytes.
Final answer: Header=20 bytes; payload=980 bytes.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
IHL=6; Total Length=1200 bytes.
Header bytes and payload bytes.
IHL counts 32-bit words; each word is 4 bytes.
H=4×6=24 bytes.
P=1200−24=1176 bytes.
24+1176=1200 bytes.
Final answer: Header=24 bytes; payload=1176 bytes.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
IHL=8; Total Length=800 bytes.
Header bytes and payload bytes.
IHL counts 32-bit words; each word is 4 bytes.
H=4×8=32 bytes.
P=800−32=768 bytes.
32+768=800 bytes.
Final answer: Header=32 bytes; payload=768 bytes.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Definition. IPv4 fragmentation divides a packet that exceeds an outgoing link’s MTU into smaller fragments, when fragmentation is permitted.
Simple meaning. MTU se packet bada hai toh allowed case mein fragments; offset original payload mein position batata hai.
Fragment Offset is measured in units of 8 payload bytes. MF=1 says more fragments of the original packet follow; MF=0 marks the last fragment. Each fragment carries its own IP header.
Yaad rakho: Offset ×8 = starting payload byte.
"},{"id":"fragment-1","title":"Optional fragmentation · example 1","html":"Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Total=4000; H=20; MTU=1500 bytes; DF=0.
All fragments and their header-related values.
![\"P=T-H\"](\"math/00cfb5dffef469d2.svg\")
![\"P_{\\max}=8\\left\\lfloor\\frac{\\mathrm{MTU}-H}{8}\\right\\rfloor\"](\"math/fd129cf4e0fe3540.svg\")
![\"\\mathrm{offset}=\\mathrm{preceding\\](\"math/76b072d173a1b0cc.svg\")
Subtract the header before splitting the original payload. Round non-final fragment payload capacity down to a multiple of eight.
4000−20=3980 bytes.
MTU−H=1500−20=1480. Floor((1480)/8)×8=1480 bytes.
| Fragment | Payload bytes | Total bytes | Starting byte | Offset /8 | MF |
|---|---|---|---|---|---|
| 1 | 1480 | 1500 | 0 | 0 | 1 |
| 2 | 1480 | 1500 | 1480 | 185 | 1 |
| 3 | 1020 | 1040 | 2960 | 370 | 0 |
1480+1480+1020=3980 bytes. Every total length is at most 1500.
Final answer: 3 fragments; lengths, offsets and MF are listed in the table.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Total=2000; H=20; MTU=620 bytes; DF=0.
All fragments and their header-related values.
Subtract the header before splitting the original payload. Round non-final fragment payload capacity down to a multiple of eight.
2000−20=1980 bytes.
MTU−H=620−20=600. Floor((600)/8)×8=600 bytes.
| Fragment | Payload bytes | Total bytes | Starting byte | Offset /8 | MF |
|---|---|---|---|---|---|
| 1 | 600 | 620 | 0 | 0 | 1 |
| 2 | 600 | 620 | 600 | 75 | 1 |
| 3 | 600 | 620 | 1200 | 150 | 1 |
| 4 | 180 | 200 | 1800 | 225 | 0 |
600+600+600+180=1980 bytes. Every total length is at most 620.
Final answer: 4 fragments; lengths, offsets and MF are listed in the table.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Total=1500; H=20; MTU=600 bytes; DF=0.
All fragments and their header-related values.
Subtract the header before splitting the original payload. Round non-final fragment payload capacity down to a multiple of eight.
1500−20=1480 bytes.
MTU−H=600−20=580. Floor((580)/8)×8=576 bytes.
| Fragment | Payload bytes | Total bytes | Starting byte | Offset /8 | MF |
|---|---|---|---|---|---|
| 1 | 576 | 596 | 0 | 0 | 1 |
| 2 | 576 | 596 | 576 | 72 | 1 |
| 3 | 328 | 348 | 1152 | 144 | 0 |
576+576+328=1480 bytes. Every total length is at most 600.
Final answer: 3 fragments; lengths, offsets and MF are listed in the table.
Supporting formula collection. Use the compact TEE formula sheet first; the full calculation set below is not an observed-paper requirement.
![\"T_f=L/R\"](\"math/9e384aa108fc34e3.svg\")
L=bits, R=bits/s, Tf=seconds. Use for serialization.
d=distance, v=propagation speed. Use consistent distance units.
h equal-rate store-and-forward links, k equal packets, L total bits/packet; τ per-link propagation. No queueing or processing.
![\"G=\\lambda](\"math/777410ed8c9656d0.svg\")
λ=aggregate attempts/s; G=attempts/frame time.
![\"S_{\\mathrm{pure}}=Ge^{-2G}\"](\"math/7b496d33240552f3.svg\")
Maximum 1/(2e)≈0.18394 at G=0.5.
![\"S_{\\mathrm{slotted}}=Ge^{-G}\"](\"math/679e9767406c13ac.svg\")
Maximum 1/e≈0.367879 at G=1.
![\"F_{\\mathrm{success}}=S/T_f\"](\"math/2656297b585ffbee.svg\")
Convert normalized S into successful frames/s.
![\"B_{\\mathrm{success}}=SR\"](\"math/4164452cf4f984f6.svg\")
Successful full-frame bits/s; not necessarily application payload goodput.
![\"P(\\mathrm{success\\](\"math/a1083530dd6239fa.svg\")
For G>0. Do not confuse this probability with normalized throughput S.
Five fully worked rate questions
"},{"id":"csma","title":"Collision timing and backoff","html":"R=bits/s; Tp=one-way seconds. Theoretical collision-detection bound.
![\"k=\\min(n,10),\\quad](\"math/f340420d788785f5.svg\")
n collisions; K selected integer for classic Ethernet backoff.
![\"T_{\\mathrm{backoff}}=KT_{\\mathrm{slot}}\"](\"math/c833ce43529e0206.svg\")
Multiply selected K by slot duration, retaining units.
![\"h=32-p,\\quad](\"math/1b1ff40e2008c7a0.svg\")
p=prefix bits, h=host bits, N=total addresses.
![\"U=2^h-2\"](\"math/92ad06adcbf23a21.svg\")
Ordinary usable-host count. /31 point-to-point and /32 are exceptions.
![\"\\text{step}=256-M\"](\"math/9f548fe5bbbe07ef.svg\")
M is the first mask octet not equal to 255.
![\"\\text{network}=\\text{address](\"math/75b831766f4f99cd.svg\")
Bitwise AND keeps network bits and clears host bits.
![\"\\text{broadcast}=\\text{network}+(N-1)\"](\"math/e5e4b592ee8fd45e.svg\")
Use full address arithmetic, not an overflowing final octet.
Nine complete address-block examples
"},{"id":"design","title":"Host sizing, VLSM and aggregation","html":"![\"2^h-2\\geq](\"math/ab881b9b5531cc3b.svg\")
H requested usable hosts. Smallest sufficient h gives smallest ordinary subnet.
![\"2^s\\geq](\"math/20f9f1a5957fbd76.svg\")
K equal subnets required; s borrowed bits. Do not subtract two from modern subnet count.
For a power-of-two total address requirement, subject to boundary alignment.
Supporting formula collection. Use the compact TEE formula sheet first; the full calculation set below is not an observed-paper requirement.
![\"\\text{omitted](\"math/38a8a52c7013e253.svg\")
Count every written group, including a written 0.
![\"\\mathrm{IPv6\\](\"math/a05ef7dd84fc645e.svg\")
Payload includes extension headers, excludes the base 40 bytes.
![\"\\mathrm{IPv4\\](\"math/0abf62f351ee6ca7.svg\")
IHL is in 32-bit words.
![\"\\mathrm{IPv4\\](\"math/dfefa084951b704c.svg\")
IPv4 Total Length includes the IPv4 header.
Old destination distance versus cost to u plus edge u→v.
Distance vector: local link cost plus neighbour advertisement.
![\"\\mathrm{RIP\\](\"math/6790d72650e66c1e.svg\")
16 represents unreachable. Ordinary hop increments commonly equal 1.
The supplied ZIPs contain the 5 December 2025 final and 4 February 2026 re-exam, each with an answer scheme. This section solves the 14 question parts that match ALOHA onward through Unit 4. The original 5/10-mark values are retained; these are not presented as actual M2 papers.
| Paper | Matching questions | Recurring work |
|---|---|---|
| 5 Dec 2025 | Q1(c), Q1(d), Q2(b), Q3(b), Q4(a), Q6(b), Q7(a) | Slotted ALOHA, address block, switching, OSPF packets, IPv6, CSMA, Dijkstra |
| 4 Feb 2026 | Q1(c), Q1(d), Q2(b), Q3(b), Q4(a), Q6(b), Q7(a) | Pure ALOHA, address block, allocation, OSPF, IPv6 header, throughput rates, Bellman–Ford |
Questions on introductory models, CRC/ARQ, transport, congestion control and application protocols fall outside this stated M2 cutoff. They are omitted from the M2 question bank. These two papers show question styles; they cannot establish a reliable forecast of the next paper.
"}]},{"id":"paper-2025-final","title":"Final · 5 December 2025","scope":"Actual TEE paper · current-M2 matching parts; original marks","sections":[{"id":"q1c","title":"Q1C · Slotted ALOHA throughput","html":"Protocol = Slotted ALOHA; G = 1 attempts/frame time.
S, 100S and the maximum possible S.
Select the formula from the protocol name. The exponent is −G because Slotted ALOHA has a T vulnerable period.
−1 × 1 = -1.
S = 1 × 0.367879441 ≈ 0.367879441 successful frames/frame time.
0.367879441 × 100 ≈ 36.7879%.
The maximum is 1/(1e) ≈ 0.367879 = 36.7879%, at G=1.
Final answer: S ≈ 0.367879; percentage ≈ 36.7879%; Smax ≈ 0.367879 at G=1.
IP/prefix = 200.140.64.48/26.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−26=6; N=26=64.
Mask=255.255.255.192; binary=11111111.11111111.11111111.11000000.
The changing octet is number 4. Its mask value is 192, so step=256−192=64. The IP value 48 lies in 0…63.
Keep earlier octets; place 0 in octet 4; set any later octets to zero. Network=200.140.64.0.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 11001000 | 11111111 | 11001000 | 200 |
| 2 | 10001100 | 11111111 | 10001100 | 140 |
| 3 | 01000000 | 11111111 | 01000000 | 64 |
| 4 | 00110000 | 11000000 | 00000000 | 0 |
Use 63 in the changing octet and 255 in any later octets. Broadcast=200.140.64.63.
First usable=200.140.64.1; last usable=200.140.64.62; count=64−2=62.
Final answer: Network 200.140.64.0; broadcast 200.140.64.63; 64 total, 62 ordinary usable; range 200.140.64.1–200.140.64.62.
A setup phase reserves a circuit through intermediate switches. Data then streams through the established path until release. It offers predictable reserved service and low per-unit forwarding overhead, but setup takes time and idle periods waste reserved capacity. A traditional dedicated voice circuit is an appropriate illustration.
Each intermediate node receives and stores the entire message before forwarding it. Links can be shared and no end-to-end circuit is required, but large messages need buffers and cause substantial store-and-forward delay. It illustrates delay-tolerant whole-message delivery, rather than an efficient choice for interactive office traffic.
A message is segmented into packets, allowing shared links and pipelined forwarding. This suits bursty data and supports flexible routing. Costs include packet headers, variable queueing delay, potential packet loss and possible reordering in datagram networks.
| Aspect | Circuit switching | Message switching | Packet switching |
|---|---|---|---|
| Unit handled | Continuous stream on established circuit | Whole message | Small packets |
| Resource use | Capacity reserved along a circuit | Shared; message stored at each intermediate node | Shared; each packet stored/forwarded in this model |
| Setup | Required before data transfer | No circuit setup required | Datagram: no connection setup; virtual circuit: setup |
| Delay | Setup delay, then relatively predictable service | Large store-and-forward delay for long messages | Variable delay; packet pipeline can reduce completion time |
| Main limitation | Reserved capacity can be idle | Large buffers; unsuitable for tight delay needs | Headers, possible queueing/loss; datagrams may reorder |
For ordinary office file, web and collaboration traffic, choose packet switching, with adequate capacity and traffic management. Its statistical sharing serves bursty sources efficiently. If strict deterministic service is required, state that requirement and discuss additional service engineering rather than promising that every packet network guarantees timing.
| Packet type | What it does | Simple example |
|---|---|---|
| Hello | Discover/maintain neighbours and exchange relationship parameters | Are we still neighbours? |
| Database Description (DD/DBD) | Summarize link-state database contents during synchronization | Here are the LSA headers I have |
| Link State Request (LSR) | Ask for specific missing/newer link-state information | Please send this LSA |
| Link State Update (LSU) | Carry one or more LSAs | Here is the requested or newly flooded link-state information |
| Link State Acknowledgment (LSAck) | Acknowledge receipt of LSAs | I received these LSAs |
Database Description packets carry summaries; Link State Update packets carry LSAs themselves. Do not confuse an LSA type with an OSPF packet type.
Given address: 0:15::1:12:1213.
The full 128-bit address in eight-group hexadecimal notation.
A written 0 is a visible group. :: represents only the omitted all-zero groups.
Left: 0, 15 (2 group(s)). Right: 1, 12, 1213 (3 group(s)).
Visible=2+3=5. Missing=8−5=3.
0:15:0:0:0:1:12:1213.
| Group number | Four-digit group |
|---|---|
| 1 | 0000 |
| 2 | 0015 |
| 3 | 0000 |
| 4 | 0000 |
| 5 | 0000 |
| 6 | 0001 |
| 7 | 0012 |
| 8 | 1213 |
0000:0015:0000:0000:0000:0001:0012:1213.
Final answer: 0000:0015:0000:0000:0000:0001:0012:1213
| Feature | IPv4 | IPv6 |
|---|---|---|
| Address width | 32 bits | 128 bits |
| Base/header size | 20–60 bytes depending on options | 40-byte fixed base header |
| Length field | Total length includes IPv4 header | Payload Length excludes 40-byte base header |
| Lifetime field | TTL | Hop Limit |
| Following protocol | Protocol field | Next Header; may identify an extension |
| Header checksum | Present | No base-header checksum |
| Options | Inside IPv4 header | Typically extension headers |
| Fragmentation fields | In IPv4 header; routers may fragment if permitted | Fragment extension header; fragmentation by source, not routers |
The supplied scheme’s sentence says four missing groups, but five groups are already visible, so only three are missing. Its printed eight-group expansion is consistent with three. The count shown above is independently verified.
| Rule | When idle | When busy | Trade-off |
|---|---|---|---|
| 1-persistent | Transmit immediately | Keep listening; transmit as soon as idle | Low idle delay, but waiting stations may start together |
| Non-persistent | Transmit | Wait a random time, then sense again | Less synchronized contention; may leave avoidable idle time |
| p-persistent (slotted) | Transmit with probability p; defer a slot with probability 1−p | Wait until available, then resume the probabilistic slot rule | Controls contention; needs slots and a choice of p |
All three sense the channel; their persistence rules differ. Propagation delay means carrier sensing cannot guarantee the absence of collisions. In p-persistent CSMA, each eligible idle-slot decision uses the specified probability.
| Edge | Cost |
|---|---|
| A→B | 4 |
| A→C | 2 |
| B→C | 5 |
| B→D | 10 |
| C→D | 3 |
| C→E | 4 |
| D→E | 1 |
| D→F | 2 |
| E→F | 3 |
| F→G | 5 |
Source=A; target=G. Edges are directed; do not add reverse edges.
The path A to G, its total cost and intermediate work.
Use Dijkstra because all weights are nonnegative. Select the smallest unsettled total distance, not the cheapest isolated edge.
![\"Dijkstra](\"fig/dijkstra-pyq.ffe6efac11.svg\")
d[A]=0; every other distance=∞. No predecessors are known.
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Tentative |
| B | ∞ | — | Tentative |
| C | ∞ | — | Tentative |
| D | ∞ | — | Tentative |
| E | ∞ | — | Tentative |
| F | ∞ | — | Tentative |
| G | ∞ | — | Tentative |
Unsettled candidates before selection: A=0, B=∞, C=∞, D=∞, E=∞, F=∞, G=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| A→B | ∞ | 0+(4)=4 | Improve to 4; predecessor=A |
| A→C | ∞ | 0+(2)=2 | Improve to 2; predecessor=A |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 4 | A | Tentative |
| C | 2 | A | Tentative |
| D | ∞ | — | Tentative |
| E | ∞ | — | Tentative |
| F | ∞ | — | Tentative |
| G | ∞ | — | Tentative |
Unsettled candidates before selection: B=4, C=2, D=∞, E=∞, F=∞, G=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| C→D | ∞ | 2+(3)=5 | Improve to 5; predecessor=C |
| C→E | ∞ | 2+(4)=6 | Improve to 6; predecessor=C |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 4 | A | Tentative |
| C | 2 | A | Settled |
| D | 5 | C | Tentative |
| E | 6 | C | Tentative |
| F | ∞ | — | Tentative |
| G | ∞ | — | Tentative |
Unsettled candidates before selection: B=4, D=5, E=6, F=∞, G=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| B→C | 2 | 4+(5)=9 | Already settled; retain |
| B→D | 5 | 4+(10)=14 | Retain 5 |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 4 | A | Settled |
| C | 2 | A | Settled |
| D | 5 | C | Tentative |
| E | 6 | C | Tentative |
| F | ∞ | — | Tentative |
| G | ∞ | — | Tentative |
Unsettled candidates before selection: D=5, E=6, F=∞, G=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| D→E | 6 | 5+(1)=6 | Retain 6 (equal cost) |
| D→F | ∞ | 5+(2)=7 | Improve to 7; predecessor=D |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 4 | A | Settled |
| C | 2 | A | Settled |
| D | 5 | C | Settled |
| E | 6 | C | Tentative |
| F | 7 | D | Tentative |
| G | ∞ | — | Tentative |
Unsettled candidates before selection: E=6, F=7, G=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| E→F | 7 | 6+(3)=9 | Retain 7 |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 4 | A | Settled |
| C | 2 | A | Settled |
| D | 5 | C | Settled |
| E | 6 | C | Settled |
| F | 7 | D | Tentative |
| G | ∞ | — | Tentative |
Unsettled candidates before selection: F=7, G=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| F→G | ∞ | 7+(5)=12 | Improve to 12; predecessor=F |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 4 | A | Settled |
| C | 2 | A | Settled |
| D | 5 | C | Settled |
| E | 6 | C | Settled |
| F | 7 | D | Settled |
| G | 12 | F | Tentative |
Unsettled candidates before selection: G=12. Select the minimum; ties use alphabetical order.
No outgoing edges to relax.
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 4 | A | Settled |
| C | 2 | A | Settled |
| D | 5 | C | Settled |
| E | 6 | C | Settled |
| F | 7 | D | Settled |
| G | 12 | F | Settled |
Trace backwards: G ← F ← D ← C ← A. Reverse it: A → C → D → F → G.
Edge sum: 2 + 3 + 2 + 5 = 12.
Final answer: A → C → D → F → G; cost=12.
![\"Dijkstra](\"fig/dijkstra-pyq-solution.d92888ba14.svg\")
Protocol = Pure ALOHA; G = 0.5 attempts/frame time.
S, 100S and the maximum possible S.
Select the formula from the protocol name. The exponent is −2G because Pure ALOHA has a 2T vulnerable period.
−2 × 0.5 = -1.
S = 0.5 × 0.367879441 ≈ 0.183939721 successful frames/frame time.
0.183939721 × 100 ≈ 18.3940%.
The maximum is 1/(2e) ≈ 0.183940 = 18.3940%, at G=0.5.
Final answer: S ≈ 0.183940; percentage ≈ 18.3940%; Smax ≈ 0.183940 at G=0.5.
IP/prefix = 193.20.35.40/26.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−26=6; N=26=64.
Mask=255.255.255.192; binary=11111111.11111111.11111111.11000000.
The changing octet is number 4. Its mask value is 192, so step=256−192=64. The IP value 40 lies in 0…63.
Keep earlier octets; place 0 in octet 4; set any later octets to zero. Network=193.20.35.0.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 11000001 | 11111111 | 11000001 | 193 |
| 2 | 00010100 | 11111111 | 00010100 | 20 |
| 3 | 00100011 | 11111111 | 00100011 | 35 |
| 4 | 00101000 | 11000000 | 00000000 | 0 |
Use 63 in the changing octet and 255 in any later octets. Broadcast=193.20.35.63.
First usable=193.20.35.1; last usable=193.20.35.62; count=64−2=62.
Final answer: Network 193.20.35.0; broadcast 193.20.35.63; 64 total, 62 ordinary usable; range 193.20.35.1–193.20.35.62.
| Department | Labs | Total addresses per lab |
|---|---|---|
| A | 32 | 128 |
| B | 64 | 64 |
| C | 16 | 512 |
Parent: 172.16.0.0/16 with 65,536 addresses.
Lab prefixes, department ranges and unused total addresses.
![\"p=32-\\log_2(N_{\\mathrm{addresses\\](\"math/96f141ad847f8a75.svg\")
![\"N_{\\mathrm{department}}=N_{\\mathrm{labs}}N_{\\mathrm{addresses\\](\"math/830cf354ee16974f.svg\")
The question requests addresses, not usable hosts. Therefore do not add two before selecting each lab size.
128=2⁷ addresses/lab → /25. 32×128=4,096 addresses → department block /20. Allocate 172.16.0.0/20, spanning 172.16.0.0–172.16.15.255.
64=2⁶ → /26 per lab. 64×64=4,096 → /20 department block. Allocate 172.16.16.0/20, spanning 172.16.16.0–172.16.31.255.
512=2⁹ → /23 per lab. 16×512=8,192 → /19 department block. Allocate 172.16.32.0/19, spanning 172.16.32.0–172.16.63.255. Its start at third octet 32 is aligned for /19.
| Lab | Prefix | First address | Last address |
|---|---|---|---|
| A1 | 172.16.0.0/25 | 172.16.0.0 | 172.16.0.127 |
| A2 | 172.16.0.128/25 | 172.16.0.128 | 172.16.0.255 |
| A3 | 172.16.1.0/25 | 172.16.1.0 | 172.16.1.127 |
| A4 | 172.16.1.128/25 | 172.16.1.128 | 172.16.1.255 |
| A5 | 172.16.2.0/25 | 172.16.2.0 | 172.16.2.127 |
| A6 | 172.16.2.128/25 | 172.16.2.128 | 172.16.2.255 |
| A7 | 172.16.3.0/25 | 172.16.3.0 | 172.16.3.127 |
| A8 | 172.16.3.128/25 | 172.16.3.128 | 172.16.3.255 |
| A9 | 172.16.4.0/25 | 172.16.4.0 | 172.16.4.127 |
| A10 | 172.16.4.128/25 | 172.16.4.128 | 172.16.4.255 |
| A11 | 172.16.5.0/25 | 172.16.5.0 | 172.16.5.127 |
| A12 | 172.16.5.128/25 | 172.16.5.128 | 172.16.5.255 |
| A13 | 172.16.6.0/25 | 172.16.6.0 | 172.16.6.127 |
| A14 | 172.16.6.128/25 | 172.16.6.128 | 172.16.6.255 |
| A15 | 172.16.7.0/25 | 172.16.7.0 | 172.16.7.127 |
| A16 | 172.16.7.128/25 | 172.16.7.128 | 172.16.7.255 |
| A17 | 172.16.8.0/25 | 172.16.8.0 | 172.16.8.127 |
| A18 | 172.16.8.128/25 | 172.16.8.128 | 172.16.8.255 |
| A19 | 172.16.9.0/25 | 172.16.9.0 | 172.16.9.127 |
| A20 | 172.16.9.128/25 | 172.16.9.128 | 172.16.9.255 |
| A21 | 172.16.10.0/25 | 172.16.10.0 | 172.16.10.127 |
| A22 | 172.16.10.128/25 | 172.16.10.128 | 172.16.10.255 |
| A23 | 172.16.11.0/25 | 172.16.11.0 | 172.16.11.127 |
| A24 | 172.16.11.128/25 | 172.16.11.128 | 172.16.11.255 |
| A25 | 172.16.12.0/25 | 172.16.12.0 | 172.16.12.127 |
| A26 | 172.16.12.128/25 | 172.16.12.128 | 172.16.12.255 |
| A27 | 172.16.13.0/25 | 172.16.13.0 | 172.16.13.127 |
| A28 | 172.16.13.128/25 | 172.16.13.128 | 172.16.13.255 |
| A29 | 172.16.14.0/25 | 172.16.14.0 | 172.16.14.127 |
| A30 | 172.16.14.128/25 | 172.16.14.128 | 172.16.14.255 |
| A31 | 172.16.15.0/25 | 172.16.15.0 | 172.16.15.127 |
| A32 | 172.16.15.128/25 | 172.16.15.128 | 172.16.15.255 |
| B1 | 172.16.16.0/26 | 172.16.16.0 | 172.16.16.63 |
| B2 | 172.16.16.64/26 | 172.16.16.64 | 172.16.16.127 |
| B3 | 172.16.16.128/26 | 172.16.16.128 | 172.16.16.191 |
| B4 | 172.16.16.192/26 | 172.16.16.192 | 172.16.16.255 |
| B5 | 172.16.17.0/26 | 172.16.17.0 | 172.16.17.63 |
| B6 | 172.16.17.64/26 | 172.16.17.64 | 172.16.17.127 |
| B7 | 172.16.17.128/26 | 172.16.17.128 | 172.16.17.191 |
| B8 | 172.16.17.192/26 | 172.16.17.192 | 172.16.17.255 |
| B9 | 172.16.18.0/26 | 172.16.18.0 | 172.16.18.63 |
| B10 | 172.16.18.64/26 | 172.16.18.64 | 172.16.18.127 |
| B11 | 172.16.18.128/26 | 172.16.18.128 | 172.16.18.191 |
| B12 | 172.16.18.192/26 | 172.16.18.192 | 172.16.18.255 |
| B13 | 172.16.19.0/26 | 172.16.19.0 | 172.16.19.63 |
| B14 | 172.16.19.64/26 | 172.16.19.64 | 172.16.19.127 |
| B15 | 172.16.19.128/26 | 172.16.19.128 | 172.16.19.191 |
| B16 | 172.16.19.192/26 | 172.16.19.192 | 172.16.19.255 |
| B17 | 172.16.20.0/26 | 172.16.20.0 | 172.16.20.63 |
| B18 | 172.16.20.64/26 | 172.16.20.64 | 172.16.20.127 |
| B19 | 172.16.20.128/26 | 172.16.20.128 | 172.16.20.191 |
| B20 | 172.16.20.192/26 | 172.16.20.192 | 172.16.20.255 |
| B21 | 172.16.21.0/26 | 172.16.21.0 | 172.16.21.63 |
| B22 | 172.16.21.64/26 | 172.16.21.64 | 172.16.21.127 |
| B23 | 172.16.21.128/26 | 172.16.21.128 | 172.16.21.191 |
| B24 | 172.16.21.192/26 | 172.16.21.192 | 172.16.21.255 |
| B25 | 172.16.22.0/26 | 172.16.22.0 | 172.16.22.63 |
| B26 | 172.16.22.64/26 | 172.16.22.64 | 172.16.22.127 |
| B27 | 172.16.22.128/26 | 172.16.22.128 | 172.16.22.191 |
| B28 | 172.16.22.192/26 | 172.16.22.192 | 172.16.22.255 |
| B29 | 172.16.23.0/26 | 172.16.23.0 | 172.16.23.63 |
| B30 | 172.16.23.64/26 | 172.16.23.64 | 172.16.23.127 |
| B31 | 172.16.23.128/26 | 172.16.23.128 | 172.16.23.191 |
| B32 | 172.16.23.192/26 | 172.16.23.192 | 172.16.23.255 |
| B33 | 172.16.24.0/26 | 172.16.24.0 | 172.16.24.63 |
| B34 | 172.16.24.64/26 | 172.16.24.64 | 172.16.24.127 |
| B35 | 172.16.24.128/26 | 172.16.24.128 | 172.16.24.191 |
| B36 | 172.16.24.192/26 | 172.16.24.192 | 172.16.24.255 |
| B37 | 172.16.25.0/26 | 172.16.25.0 | 172.16.25.63 |
| B38 | 172.16.25.64/26 | 172.16.25.64 | 172.16.25.127 |
| B39 | 172.16.25.128/26 | 172.16.25.128 | 172.16.25.191 |
| B40 | 172.16.25.192/26 | 172.16.25.192 | 172.16.25.255 |
| B41 | 172.16.26.0/26 | 172.16.26.0 | 172.16.26.63 |
| B42 | 172.16.26.64/26 | 172.16.26.64 | 172.16.26.127 |
| B43 | 172.16.26.128/26 | 172.16.26.128 | 172.16.26.191 |
| B44 | 172.16.26.192/26 | 172.16.26.192 | 172.16.26.255 |
| B45 | 172.16.27.0/26 | 172.16.27.0 | 172.16.27.63 |
| B46 | 172.16.27.64/26 | 172.16.27.64 | 172.16.27.127 |
| B47 | 172.16.27.128/26 | 172.16.27.128 | 172.16.27.191 |
| B48 | 172.16.27.192/26 | 172.16.27.192 | 172.16.27.255 |
| B49 | 172.16.28.0/26 | 172.16.28.0 | 172.16.28.63 |
| B50 | 172.16.28.64/26 | 172.16.28.64 | 172.16.28.127 |
| B51 | 172.16.28.128/26 | 172.16.28.128 | 172.16.28.191 |
| B52 | 172.16.28.192/26 | 172.16.28.192 | 172.16.28.255 |
| B53 | 172.16.29.0/26 | 172.16.29.0 | 172.16.29.63 |
| B54 | 172.16.29.64/26 | 172.16.29.64 | 172.16.29.127 |
| B55 | 172.16.29.128/26 | 172.16.29.128 | 172.16.29.191 |
| B56 | 172.16.29.192/26 | 172.16.29.192 | 172.16.29.255 |
| B57 | 172.16.30.0/26 | 172.16.30.0 | 172.16.30.63 |
| B58 | 172.16.30.64/26 | 172.16.30.64 | 172.16.30.127 |
| B59 | 172.16.30.128/26 | 172.16.30.128 | 172.16.30.191 |
| B60 | 172.16.30.192/26 | 172.16.30.192 | 172.16.30.255 |
| B61 | 172.16.31.0/26 | 172.16.31.0 | 172.16.31.63 |
| B62 | 172.16.31.64/26 | 172.16.31.64 | 172.16.31.127 |
| B63 | 172.16.31.128/26 | 172.16.31.128 | 172.16.31.191 |
| B64 | 172.16.31.192/26 | 172.16.31.192 | 172.16.31.255 |
| C1 | 172.16.32.0/23 | 172.16.32.0 | 172.16.33.255 |
| C2 | 172.16.34.0/23 | 172.16.34.0 | 172.16.35.255 |
| C3 | 172.16.36.0/23 | 172.16.36.0 | 172.16.37.255 |
| C4 | 172.16.38.0/23 | 172.16.38.0 | 172.16.39.255 |
| C5 | 172.16.40.0/23 | 172.16.40.0 | 172.16.41.255 |
| C6 | 172.16.42.0/23 | 172.16.42.0 | 172.16.43.255 |
| C7 | 172.16.44.0/23 | 172.16.44.0 | 172.16.45.255 |
| C8 | 172.16.46.0/23 | 172.16.46.0 | 172.16.47.255 |
| C9 | 172.16.48.0/23 | 172.16.48.0 | 172.16.49.255 |
| C10 | 172.16.50.0/23 | 172.16.50.0 | 172.16.51.255 |
| C11 | 172.16.52.0/23 | 172.16.52.0 | 172.16.53.255 |
| C12 | 172.16.54.0/23 | 172.16.54.0 | 172.16.55.255 |
| C13 | 172.16.56.0/23 | 172.16.56.0 | 172.16.57.255 |
| C14 | 172.16.58.0/23 | 172.16.58.0 | 172.16.59.255 |
| C15 | 172.16.60.0/23 | 172.16.60.0 | 172.16.61.255 |
| C16 | 172.16.62.0/23 | 172.16.62.0 | 172.16.63.255 |
4,096+4,096+8,192=16,384 addresses.
65,536−16,384=49,152 addresses remain unallocated, from 172.16.64.0 through 172.16.255.255.
Final answer: A: /25 per lab; B: /26 per lab; C: /23 per lab. 49,152 total addresses remain unallocated.
| Field | Bits | Purpose |
|---|---|---|
| Version | 4 | Value 6 |
| Traffic Class | 8 | Traffic treatment/congestion-related marking |
| Flow Label | 20 | Identifies a flow for appropriate handling |
| Payload Length | 16 | Bytes after the base header; includes extension headers when present |
| Next Header | 8 | Type of the next extension header or upper-layer protocol |
| Hop Limit | 8 | Decremented on forwarding; limits packet lifetime in hops |
| Source address | 128 | Sending interface address |
| Destination address | 128 | Destination address |
The first two 32-bit rows contain 64 bits and the two addresses contain 256 bits, giving a 320-bit base header. Payload Length excludes these 40 bytes. Next Header can identify an extension header or upper-layer protocol. IPv6 has no base-header checksum; optional information is carried through extension headers.
Interpret the stated aggregate production rates as offered attempt rates in the standard textbook model. The following working includes normalized throughput, successful frames/s and successful frame bits/s for each case.
| Symbol | Given |
|---|---|
| L | 200 bits |
| R | 200 kbps = 200,000 bits/s |
| λ | 1000 aggregate attempts/s |
T, G, S, successful frames/s and bit/s.
First normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
T = 200/200,000 = 0.001 s = 1 ms.
G = 1000 × 0.001 = 1 attempts/frame time.
−2G = -2; e to that power ≈ 0.135335283.
S = 1 × 0.135335283 ≈ 0.135335283.
S/T = 0.135335283/0.001 ≈ 135.335283 frames/s.
SR = 0.135335283 × 200,000 ≈ 27067.056647 bits/s ≈ 27.067057 kbps.
Final answer: G=1, S≈0.135335, successful rate≈135.335 frames/s = 27.067 kbps.
| Symbol | Given |
|---|---|
| L | 200 bits |
| R | 200 kbps = 200,000 bits/s |
| λ | 500 aggregate attempts/s |
T, G, S, successful frames/s and bit/s.
First normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
T = 200/200,000 = 0.001 s = 1 ms.
G = 500 × 0.001 = 0.5 attempts/frame time.
−2G = -1; e to that power ≈ 0.367879441.
S = 0.5 × 0.367879441 ≈ 0.183939721.
S/T = 0.183939721/0.001 ≈ 183.939721 frames/s.
SR = 0.183939721 × 200,000 ≈ 36787.944117 bits/s ≈ 36.787944 kbps.
Final answer: G=0.5, S≈0.183940, successful rate≈183.940 frames/s = 36.788 kbps.
| Symbol | Given |
|---|---|
| L | 200 bits |
| R | 200 kbps = 200,000 bits/s |
| λ | 250 aggregate attempts/s |
T, G, S, successful frames/s and bit/s.
First normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
T = 200/200,000 = 0.001 s = 1 ms.
G = 250 × 0.001 = 0.25 attempts/frame time.
−2G = -0.5; e to that power ≈ 0.606530660.
S = 0.25 × 0.606530660 ≈ 0.151632665.
S/T = 0.151632665/0.001 ≈ 151.632665 frames/s.
SR = 0.151632665 × 200,000 ≈ 30326.532986 bits/s ≈ 30.326533 kbps.
Final answer: G=0.25, S≈0.151633, successful rate≈151.633 frames/s = 30.327 kbps.
| Directed edge | Weight |
|---|---|
| A→B | 7 |
| A→C | 6 |
| A→D | 5 |
| B→E | -2 |
| D→C | -2 |
| C→B | -2 |
| C→E | 1 |
| E→G | 5 |
| D→F | 3 |
| F→G | 4 |
The shortest route and cost, with complete passes and a negative-cycle check.
Bellman–Ford permits negative edges. We use the listed order with in-place updates; do not mix this with a table that reads only the previous pass.
![\"Bellman–Ford](\"fig/bellman-pyq.8245b97635.svg\")
7 vertices → at most 6 main passes. d[A]=0; every other distance=∞. Use the listed edge order and immediate (in-place) updates.
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| A→B | ∞ | 0+(7)=7 | 7 | Update; predecessor=A |
| A→C | ∞ | 0+(6)=6 | 6 | Update; predecessor=A |
| A→D | ∞ | 0+(5)=5 | 5 | Update; predecessor=A |
| B→E | ∞ | 7+(-2)=5 | 5 | Update; predecessor=B |
| D→C | 6 | 5+(-2)=3 | 3 | Update; predecessor=D |
| C→B | 7 | 3+(-2)=1 | 1 | Update; predecessor=C |
| C→E | 5 | 3+(1)=4 | 4 | Update; predecessor=C |
| E→G | ∞ | 4+(5)=9 | 9 | Update; predecessor=E |
| D→F | ∞ | 5+(3)=8 | 8 | Update; predecessor=D |
| F→G | 9 | 8+(4)=12 | 9 | No improvement |
| Vertex | A | B | C | D | E | F | G |
|---|---|---|---|---|---|---|---|
| End-of-pass distance | 0 | 1 | 3 | 5 | 4 | 8 | 9 |
| Predecessor | — | C | D | A | C | D | E |
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| A→B | 1 | 0+(7)=7 | 1 | No improvement |
| A→C | 3 | 0+(6)=6 | 3 | No improvement |
| A→D | 5 | 0+(5)=5 | 5 | No improvement |
| B→E | 4 | 1+(-2)=-1 | -1 | Update; predecessor=B |
| D→C | 3 | 5+(-2)=3 | 3 | No improvement |
| C→B | 1 | 3+(-2)=1 | 1 | No improvement |
| C→E | -1 | 3+(1)=4 | -1 | No improvement |
| E→G | 9 | -1+(5)=4 | 4 | Update; predecessor=E |
| D→F | 8 | 5+(3)=8 | 8 | No improvement |
| F→G | 4 | 8+(4)=12 | 4 | No improvement |
| Vertex | A | B | C | D | E | F | G |
|---|---|---|---|---|---|---|---|
| End-of-pass distance | 0 | 1 | 3 | 5 | -1 | 8 | 4 |
| Predecessor | — | C | D | A | B | D | E |
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| A→B | 1 | 0+(7)=7 | 1 | No improvement |
| A→C | 3 | 0+(6)=6 | 3 | No improvement |
| A→D | 5 | 0+(5)=5 | 5 | No improvement |
| B→E | -1 | 1+(-2)=-1 | -1 | No improvement |
| D→C | 3 | 5+(-2)=3 | 3 | No improvement |
| C→B | 1 | 3+(-2)=1 | 1 | No improvement |
| C→E | -1 | 3+(1)=4 | -1 | No improvement |
| E→G | 4 | -1+(5)=4 | 4 | No improvement |
| D→F | 8 | 5+(3)=8 | 8 | No improvement |
| F→G | 4 | 8+(4)=12 | 4 | No improvement |
| Vertex | A | B | C | D | E | F | G |
|---|---|---|---|---|---|---|---|
| End-of-pass distance | 0 | 1 | 3 | 5 | -1 | 8 | 4 |
| Predecessor | — | C | D | A | B | D | E |
This entire pass changed no distance. Further identical relaxation passes will not improve anything.
Backward predecessors: G ← E ← B ← C ← D ← A. Forward route: A → D → C → B → E → G.
Cost = (5) + (-2) + (-2) + (-2) + (5) = 4. No edge can now improve its destination label.
Final answer: A → D → C → B → E → G; shortest cost=4.
![\"Bellman–Ford](\"fig/bellman-pyq-solution.9ae8451e95.svg\")
Each paper contains five question parts. Q1(a) and Q1(b) are compulsory, 2 marks each. Choose any two of Q2, Q3 and Q4, 3 marks each. All five answers are supplied immediately below their questions so you can also learn the optional question you did not attempt.
These are authored M2 practice papers, not leaked or predicted questions. Choose an exam-length timer only after confirming the actual CN duration; no CN time limit was supplied.
Paper A emphasizes ALOHA, blocks and Dijkstra. Paper B includes IPv6, VLSM and Bellman–Ford. Paper C emphasizes access protocols, DHCP and route decisions.
"}]},{"id":"mock-a","title":"M2 mock A","scope":"Q1(a)+Q1(b) compulsory; choose any two of Q2/Q3/Q4 · 10 marks attempted","sections":[{"id":"q1a","title":"Q1A · Carrier sensing and collisions","html":"Carrier sensing observes the local medium. Because signals have nonzero propagation delay, another station’s new transmission may not yet be detectable. Both stations can then start and their transmissions overlap.
p=27; ordinary IPv4 broadcast subnet.
h, total and usable counts.
![\"h=32-p,\\quad](\"math/391b44ac8c0c63db.svg\")
Two boundary addresses are reserved under these ordinary rules.
Final answer: 5 host bits, 32 total, 30 usable.
| Symbol | Given |
|---|---|
| L | 500 bits |
| R | 500 kbps = 500,000 bits/s |
| λ | 500 aggregate attempts/s |
T, G, S, successful frames/s and bit/s.
First normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
T = 500/500,000 = 0.001 s = 1 ms.
G = 500 × 0.001 = 0.5 attempts/frame time.
−2G = -1; e to that power ≈ 0.367879441.
S = 0.5 × 0.367879441 ≈ 0.183939721.
S/T = 0.183939721/0.001 ≈ 183.939721 frames/s.
SR = 0.183939721 × 500,000 ≈ 91969.860293 bits/s ≈ 91.969860 kbps.
Final answer: G=0.5, S≈0.183940, successful rate≈183.940 frames/s = 91.970 kbps.
IP/prefix = 192.168.5.197/27.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−27=5; N=25=32.
Mask=255.255.255.224; binary=11111111.11111111.11111111.11100000.
The changing octet is number 4. Its mask value is 224, so step=256−224=32. The IP value 197 lies in 192…223.
Keep earlier octets; place 192 in octet 4; set any later octets to zero. Network=192.168.5.192.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 11000000 | 11111111 | 11000000 | 192 |
| 2 | 10101000 | 11111111 | 10101000 | 168 |
| 3 | 00000101 | 11111111 | 00000101 | 5 |
| 4 | 11000101 | 11100000 | 11000000 | 192 |
Use 223 in the changing octet and 255 in any later octets. Broadcast=192.168.5.223.
First usable=192.168.5.193; last usable=192.168.5.222; count=32−2=30.
Final answer: Network 192.168.5.192; broadcast 192.168.5.223; 32 total, 30 ordinary usable; range 192.168.5.193–192.168.5.222.
| Edge | Cost |
|---|---|
| A→B | 2 |
| A→C | 6 |
| B→C | 1 |
| B→D | 7 |
| C→D | 2 |
Source=A; target=D. Edges are directed; do not add reverse edges.
The path A to D, its total cost and intermediate work.
Use Dijkstra because all weights are nonnegative. Select the smallest unsettled total distance, not the cheapest isolated edge.
![\"Dijkstra](\"fig/dijkstra-mock.f988abf338.svg\")
d[A]=0; every other distance=∞. No predecessors are known.
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Tentative |
| B | ∞ | — | Tentative |
| C | ∞ | — | Tentative |
| D | ∞ | — | Tentative |
Unsettled candidates before selection: A=0, B=∞, C=∞, D=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| A→B | ∞ | 0+(2)=2 | Improve to 2; predecessor=A |
| A→C | ∞ | 0+(6)=6 | Improve to 6; predecessor=A |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 2 | A | Tentative |
| C | 6 | A | Tentative |
| D | ∞ | — | Tentative |
Unsettled candidates before selection: B=2, C=6, D=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| B→C | 6 | 2+(1)=3 | Improve to 3; predecessor=B |
| B→D | ∞ | 2+(7)=9 | Improve to 9; predecessor=B |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 2 | A | Settled |
| C | 3 | B | Tentative |
| D | 9 | B | Tentative |
Unsettled candidates before selection: C=3, D=9. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| C→D | 9 | 3+(2)=5 | Improve to 5; predecessor=C |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 2 | A | Settled |
| C | 3 | B | Settled |
| D | 5 | C | Tentative |
Unsettled candidates before selection: D=5. Select the minimum; ties use alphabetical order.
No outgoing edges to relax.
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 2 | A | Settled |
| C | 3 | B | Settled |
| D | 5 | C | Settled |
Trace backwards: D ← C ← B ← A. Reverse it: A → B → C → D.
Edge sum: 2 + 1 + 2 = 5.
Final answer: A → B → C → D; cost=5.
| Protocol | Known | Needed | Use |
|---|---|---|---|
| ARP | Local next-hop IPv4 | MAC/hardware address | Local frame delivery |
| RARP | Client hardware address | Client IPv4 | Historical boot-time assignment |
Given address: 2001:db8::10.
The full 128-bit address in eight-group hexadecimal notation.
A written 0 is a visible group. :: represents only the omitted all-zero groups.
Left: 2001, db8 (2 group(s)). Right: 10 (1 group(s)).
Visible=2+1=3. Missing=8−3=5.
2001:db8:0:0:0:0:0:10.
| Group number | Four-digit group |
|---|---|
| 1 | 2001 |
| 2 | 0db8 |
| 3 | 0000 |
| 4 | 0000 |
| 5 | 0000 |
| 6 | 0000 |
| 7 | 0000 |
| 8 | 0010 |
2001:0db8:0000:0000:0000:0000:0000:0010.
Final answer: 2001:0db8:0000:0000:0000:0000:0000:0010
| Department | Usable hosts |
|---|---|
| A | 60 |
| B | 30 |
| C | 12 |
An aligned subnet for each department, usable ranges, broadcasts and remaining addresses.
Choose the smallest sufficient subnet per requirement. Align each start to its block size and never overlap a previously assigned range.
A (60) → B (30) → C (12)
60 hosts require 26−2=62 usable; a smaller /27 would give 30. Choose /26. The next aligned block is 192.168.40.0/26; it ends at 192.168.40.63.
30 hosts require 25−2=30 usable; a smaller /28 would give 14. Choose /27. The next aligned block is 192.168.40.64/27; it ends at 192.168.40.95.
12 hosts require 24−2=14 usable; a smaller /29 would give 6. Choose /28. The next aligned block is 192.168.40.96/28; it ends at 192.168.40.111.
| Group | Hosts | Network/prefix | Mask | Usable range | Broadcast | Spare usable |
|---|---|---|---|---|---|---|
| A | 60 | 192.168.40.0/26 | 255.255.255.192 | 192.168.40.1–192.168.40.62 | 192.168.40.63 | 2 |
| B | 30 | 192.168.40.64/27 | 255.255.255.224 | 192.168.40.65–192.168.40.94 | 192.168.40.95 | 0 |
| C | 12 | 192.168.40.96/28 | 255.255.255.240 | 192.168.40.97–192.168.40.110 | 192.168.40.111 | 2 |
Reserved=64+32+16=112. Parent total=256. Unallocated=256−112=144.
Final answer: 112 total addresses reserved; 144 unallocated. The table gives one valid complete allocation.
| Directed edge | Weight |
|---|---|
| A→B | 5 |
| A→C | 8 |
| B→C | -3 |
| C→D | 4 |
| B→D | 7 |
The shortest route and cost, with complete passes and a negative-cycle check.
Bellman–Ford permits negative edges. We use the listed order with in-place updates; do not mix this with a table that reads only the previous pass.
![\"Bellman–Ford](\"fig/bellman-mock.47d822ce76.svg\")
4 vertices → at most 3 main passes. d[A]=0; every other distance=∞. Use the listed edge order and immediate (in-place) updates.
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| A→B | ∞ | 0+(5)=5 | 5 | Update; predecessor=A |
| A→C | ∞ | 0+(8)=8 | 8 | Update; predecessor=A |
| B→C | 8 | 5+(-3)=2 | 2 | Update; predecessor=B |
| C→D | ∞ | 2+(4)=6 | 6 | Update; predecessor=C |
| B→D | 6 | 5+(7)=12 | 6 | No improvement |
| Vertex | A | B | C | D |
|---|---|---|---|---|
| End-of-pass distance | 0 | 5 | 2 | 6 |
| Predecessor | — | A | B | C |
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| A→B | 5 | 0+(5)=5 | 5 | No improvement |
| A→C | 2 | 0+(8)=8 | 2 | No improvement |
| B→C | 2 | 5+(-3)=2 | 2 | No improvement |
| C→D | 6 | 2+(4)=6 | 6 | No improvement |
| B→D | 6 | 5+(7)=12 | 6 | No improvement |
| Vertex | A | B | C | D |
|---|---|---|---|---|
| End-of-pass distance | 0 | 5 | 2 | 6 |
| Predecessor | — | A | B | C |
This entire pass changed no distance. Further identical relaxation passes will not improve anything.
Backward predecessors: D ← C ← B ← A. Forward route: A → B → C → D.
Cost = (5) + (-3) + (4) = 6. No edge can now improve its destination label.
Final answer: A → B → C → D; shortest cost=6.
OSPF exchanges and maintains link-state information, builds a topology database and runs Dijkstra’s shortest-path computation rooted at the local router. It chooses a path with the lowest sum of link costs and installs the corresponding next hop. Equal-cost paths may be supported.
Five packet types: Hello, Database Description, Link State Request, Link State Update and Link State Acknowledgment.
| Feature | Pure | Slotted |
|---|---|---|
| Start time | Any time | Slot boundary only |
| Synchronization | No common slots required | Required |
| Vulnerable period | 2T | T |
| Maximum S | 1/(2e) ≈ 0.184 | 1/e ≈ 0.368 |
| G at maximum | 0.5 | 1 |
DHCP reduces manual configuration and supports controlled address reuse through leases.
R=100 Mbps; Tp=2 μs.
Lmin in bits and the minimum whole-byte length.
The frame must last at least a round trip. Convert Mbps to bits/s and μs to seconds before multiplying.
R=100,000,000 bits/s; Tp=0.000002 s.
2Tp=2×2=4 μs.
Lmin=2×100,000,000×0.000002=400 bits.
400/8=50; round upward if needed → 50 bytes.
Final answer: 400 bits, or at least 50 whole bytes.
| Prefix | Next hop |
|---|---|
| 0.0.0.0/0 | A |
| 172.16.0.0/16 | B |
| 172.16.8.0/21 | C |
Destination=172.16.10.45.
Most specific matching next hop.
Choose the largest matching prefix length; an entry that does not contain the destination is ineligible.
Final answer: next hop C via 172.16.8.0/21.
B advertises 15; C advertises 4; each local cost=1. RIP infinity=16.
Candidate metrics and usable next hop.
![\"\\text{candidate}=\\min(16,1+\\text{advertised](\"math/2010d425a3ad7305.svg\")
RIP caps its metric at 16, representing unreachable.
Final answer: route via C at metric 5; B’s route is unreachable.
Revise the methods in this sequence; keep every answer tied to its question and units.
Yaad rakho: Numerical mein units, prefix, arrow aur requested quantity pehle circle karo.
"},{"id":"traps","title":"The mistakes that change a correct method into a wrong answer","html":"| Trap | Correct action |
|---|---|
| Use 500 frames/s directly as G | First multiply by frame time |
| S and success probability treated as identical | Success probability per attempt is S/G |
| kbps converted using 1024 | Use decimal network rates unless explicitly specified otherwise |
| All subnet blocks assumed to change only last octet | Locate changing mask octet; /20 changes the third |
| First address means first usable | First address is network; first usable is network+1 |
| Subtract two from requested total addresses | Only ordinary usable-host count excludes the boundary two |
| Any consecutive networks aggregated blindly | Check power-of-two size and alignment |
| Two :: compressions in one IPv6 address | Use :: at most once |
| Dijkstra used on negative edges | Use Bellman–Ford when appropriate |
| Undirected reverse edges added to directed PYQ | Follow printed arrows only |
| All Bellman–Ford passes treated as identical conventions | State in-place or previous-pass method; do not mix |
| RIP metric 16 described as reachable | 16 means unreachable |
| ARP asked for remote server MAC | Resolve local gateway/next-hop MAC |
| Question type | Answer structure |
|---|---|
| Define protocol | One precise definition → central mechanism → one correct example |
| Compare two protocols | Two-column table with distinct comparable points → application |
| Explain ARP or DHCP | Purpose → actors → ordered messages → result |
| Explain routing algorithm | Initialization → update/selection rule → repeat/stop → path extraction |
| Draw a header | Label fields and widths → explain field meanings → length convention |
For a 2-mark answer, keep the explanation focused on the requested points. For a 3-mark algorithm/numerical, show the essential sequence, not just a memorized final number. Original 10-mark PYQs here intentionally contain longer working.
"}]}],"about":[{"id":"about","title":"Portion and references","scope":"CN M2 · source mapping","sections":[{"id":"sources","title":"Sources, exclusions and corrections","html":"Primary scope: your message, “From ALOHA to Unit 4 complete,” and the current AY 2026–27 Computer Networks syllabus, code 702AI0C023. Pattern: the confirmed 2+2 compulsory and any two 3-mark questions. No separate CN course-policy document was supplied.
Supplied materials reviewed: the CN English AnswerBook PDF/DOCX as a secondary reference; the December 2025 and February 2026 question papers and associated answer schemes extracted from the two Computer Networks ZIPs. The large answer-book contains substantial material outside this M2 cutoff; its broad predictions and extra topics do not redefine the confirmed portion.
This release covers CN M2, not Units 1–6 as a complete TEE course. The matched TEE questions are included because they test this M2 material. No video is claimed as generated or reviewed in this CN release.
"}]}]},"tee":{"read":[{"id":"tee-start","title":"Start here · CN TEE","scope":"TEE · Units 1–6 · current syllabus 702AI0C023","sections":[{"id":"plan","title":"Full CN TEE reading order","html":"| Part / actual marks | Observed family | What to practise |
|---|---|---|
| Q1(a–b) · 5+5 | Network design and reference models | Concepts, comparison and labelled diagram |
| Q1(c–d) · 5+5 | ALOHA and IPv4 address block | Formula, substitution, throughput; network/last address/count |
| Q2(a) · 10 | Block coding/Hamming distance or CRC | Explanation plus complete bit-level calculation |
| Q2(b) · 10 | Switching recommendation or subnet allocation | Compare mechanisms or show aligned allocation |
| Q3 · 5+5+10 | Network classification/topologies, OSPF, HTTP/DNS | Theory, stages and diagrams |
| Q4 · 10+10 | IPv6 and UDP/TCP | Address expansion/header diagram; protocol working |
| Q5 · 10+10 | SR/GBN and SMTP/FTP | Window/ACK diagrams; ordered application flow |
| Q6 · 5+5+10 | Media, random access, transport/congestion | Media theory; CSMA comparison or ALOHA rates; theory comparison |
| Q7 · 10+10 | Dijkstra/Bellman–Ford and buckets | Full graph iterations; bucket diagram/working/limitations |
Basis: 5 December 2025 final and 4 February 2026 re-exam, all six original pages. This is an observed pattern, not confirmation of the next paper. Required syllabus topics stay in the main course even where these two papers did not ask them.
Study all six units of the AY 2026–27 syllabus, course 702AI0C023. Begin with the small maths and vocabulary lessons; then read the layers in order. The M2 tab remains restricted to ALOHA onward through Unit 4.
| Unit | Scope | Syllabus hours |
|---|---|---|
| 1 | Networks, peer-to-peer/client-server, classifications, topologies | 2 |
| 2 | OSI/TCP-IP models and transmission media | 2 |
| 3 | Error detection/correction, block coding, Hamming distance, CRC, ARQ, random access | 7 |
| 4 | Switching, IPv4/subnetting/CIDR, IPv6, ARP/RARP/DHCP, RIP/Bellman–Ford/link state/Dijkstra/OSPF | 9 |
| 5 | Process communication, UDP/TCP/SCTP, congestion control, QoS and buckets | 6 |
| 6 | HTTP, DNS, FTP and SMTP | 4 |
The final paper dated 5 December 2025 and re-exam dated 4 February 2026, course 702AI0C007, both allow 3 hours for 100 marks. Q1 is compulsory; attempt any four of Q2–Q7. Each whole question is 20 marks. Q1 has four 5-mark parts. These are historical source facts, not confirmation of a future paper.
The current syllabus specifies TEE 100 marks scaled to 50. A separate current CN exam-policy document was not supplied. The three TEE mocks follow the supplied historical structure and have solutions for every choice.
Coverage means every named syllabus topic has an explicit teaching location. It cannot guarantee which questions will appear or how marks will be awarded. Optional supporting numericals are labelled so they do not hide the core syllabus.
Definition. A computer network is a collection of interconnected devices that exchange data using agreed communication rules called protocols.
Simple meaning. Device number bhejta hai; link usko carry karta hai; protocol batata hai kaise bhejna hai.
A frame is a data-link-layer unit for one-link delivery; an IP packet is a network-layer unit for delivery across interconnected networks. A router examines the destination IP address, selects a next hop and forwards the packet in a new link-layer frame.
Yaad rakho: Frame = ek link; packet = networks ke paar.
"},{"id":"bits","title":"Bits, bytes and rate: read the units first","html":"Definition. A bit is one binary digit. A byte contains eight bits. A data rate measures the number of bits transmitted per second.
Simple meaning. Capital B = byte; small b = bit. Size aur speed ki units same karo.
Transmission delay is L/R, where L is the number of bits and R is the bit rate. Propagation delay is d/v, where d is link length and v is propagation speed. Larger frames increase transmission delay; a longer link increases propagation delay.
Yaad rakho: Bits / bits-per-second = seconds; metres / metres-per-second = seconds.
"},{"id":"delay-1","title":"Delay example 1","html":"| Quantity | Value |
|---|---|
| Frame length | 1000 bytes |
| Rate | 2 Mbps |
| Link length | 2,000 m |
| Signal speed | 200,000,000 m/s |
The two separate delays, in seconds and microseconds.
L must be in bits and R in bits/s. Link length and signal speed must use the same distance unit.
1000 × 8 = 8,000 bits.
2 × 1,000,000 = 2,000,000 bits/s.
8,000 ÷ 2,000,000 = 0.004000 s = 4000 μs.
2,000 ÷ 200,000,000 = 0.000010 s = 10 μs.
Final answer: Transmission = 4000 μs; propagation = 10 μs.
| Quantity | Value |
|---|---|
| Frame length | 500 bytes |
| Rate | 10 Mbps |
| Link length | 10,000 m |
| Signal speed | 200,000,000 m/s |
The two separate delays, in seconds and microseconds.
L must be in bits and R in bits/s. Link length and signal speed must use the same distance unit.
500 × 8 = 4,000 bits.
10 × 1,000,000 = 10,000,000 bits/s.
4,000 ÷ 10,000,000 = 0.000400 s = 400 μs.
10,000 ÷ 200,000,000 = 0.000050 s = 50 μs.
Final answer: Transmission = 400 μs; propagation = 50 μs.
| Quantity | Value |
|---|---|
| Frame length | 1500 bytes |
| Rate | 100 Mbps |
| Link length | 100,000 m |
| Signal speed | 250,000,000 m/s |
The two separate delays, in seconds and microseconds.
L must be in bits and R in bits/s. Link length and signal speed must use the same distance unit.
1500 × 8 = 12,000 bits.
100 × 1,000,000 = 100,000,000 bits/s.
12,000 ÷ 100,000,000 = 0.000120 s = 120 μs.
100,000 ÷ 250,000,000 = 0.000400 s = 400 μs.
Final answer: Transmission = 120 μs; propagation = 400 μs.
Definition. Binary is a base-two number system. Each position represents a power of 2; its digit says whether that value is present (1) or absent (0).
Simple meaning. Right se 1 shuru karo; left jaate hi har box double: 1, 2, 4, 8…
Assign weights 128,64,32,16,8,4,2,1 from left to right. Multiply each weight by the bit in that position and add the eight products. For 11000000, the result is 128+64 = 192.
Yaad rakho: Position ka weight likho; jahan 1 hai bas woh add karo.
"},{"id":"binary-1","title":"Binary to decimal · example 1","html":"Binary octet = 00000101.
Its decimal value.
The symbol Σ means add all the terms. bₖ is the bit at position k; the rightmost position is k=0. The table expands the entire sum.
| Position from right | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |
|---|---|---|---|---|---|---|---|---|
| Place value | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
| Bit | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 |
| Place value | Bit | Product |
|---|---|---|
| 128 | 0 | 128 × 0 = 0 |
| 64 | 0 | 64 × 0 = 0 |
| 32 | 0 | 32 × 0 = 0 |
| 16 | 0 | 16 × 0 = 0 |
| 8 | 0 | 8 × 0 = 0 |
| 4 | 1 | 4 × 1 = 4 |
| 2 | 0 | 2 × 0 = 0 |
| 1 | 1 | 1 × 1 = 1 |
0 + 0 + 0 + 0 + 0 + 4 + 0 + 1 = 5.
Final answer: 5
Binary octet = 10100110.
Its decimal value.
The symbol Σ means add all the terms. bₖ is the bit at position k; the rightmost position is k=0. The table expands the entire sum.
| Position from right | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |
|---|---|---|---|---|---|---|---|---|
| Place value | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
| Bit | 1 | 0 | 1 | 0 | 0 | 1 | 1 | 0 |
| Place value | Bit | Product |
|---|---|---|
| 128 | 1 | 128 × 1 = 128 |
| 64 | 0 | 64 × 0 = 0 |
| 32 | 1 | 32 × 1 = 32 |
| 16 | 0 | 16 × 0 = 0 |
| 8 | 0 | 8 × 0 = 0 |
| 4 | 1 | 4 × 1 = 4 |
| 2 | 1 | 2 × 1 = 2 |
| 1 | 0 | 1 × 0 = 0 |
128 + 0 + 32 + 0 + 0 + 4 + 2 + 0 = 166.
Final answer: 166
Binary octet = 11100000.
Its decimal value.
The symbol Σ means add all the terms. bₖ is the bit at position k; the rightmost position is k=0. The table expands the entire sum.
| Position from right | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |
|---|---|---|---|---|---|---|---|---|
| Place value | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
| Bit | 1 | 1 | 1 | 0 | 0 | 0 | 0 | 0 |
| Place value | Bit | Product |
|---|---|---|
| 128 | 1 | 128 × 1 = 128 |
| 64 | 1 | 64 × 1 = 64 |
| 32 | 1 | 32 × 1 = 32 |
| 16 | 0 | 16 × 0 = 0 |
| 8 | 0 | 8 × 0 = 0 |
| 4 | 0 | 4 × 0 = 0 |
| 2 | 0 | 2 × 0 = 0 |
| 1 | 0 | 1 × 0 = 0 |
128 + 64 + 32 + 0 + 0 + 0 + 0 + 0 = 224.
Final answer: 224
Binary octet = 11000000.
Its decimal value.
The symbol Σ means add all the terms. bₖ is the bit at position k; the rightmost position is k=0. The table expands the entire sum.
| Position from right | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |
|---|---|---|---|---|---|---|---|---|
| Place value | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
| Bit | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
| Place value | Bit | Product |
|---|---|---|
| 128 | 1 | 128 × 1 = 128 |
| 64 | 1 | 64 × 1 = 64 |
| 32 | 0 | 32 × 0 = 0 |
| 16 | 0 | 16 × 0 = 0 |
| 8 | 0 | 8 × 0 = 0 |
| 4 | 0 | 4 × 0 = 0 |
| 2 | 0 | 2 × 0 = 0 |
| 1 | 0 | 1 × 0 = 0 |
128 + 64 + 0 + 0 + 0 + 0 + 0 + 0 = 192.
Final answer: 192
Decimal value = 13.
Eight bits, including leading zeroes.
Use the weights 128,64,32,16,8,4,2,1.
At each weight, write 1 and subtract if the remaining value is at least that weight; otherwise write 0.
| Weight | Remaining before | Bit decision | Remaining after |
|---|---|---|---|
| 128 | 13 | 0: too large | 13 |
| 64 | 13 | 0: too large | 13 |
| 32 | 13 | 0: too large | 13 |
| 16 | 13 | 0: too large | 13 |
| 8 | 13 | 1: subtract | 5 |
| 4 | 5 | 1: subtract | 1 |
| 2 | 1 | 0: too large | 1 |
| 1 | 1 | 1: subtract | 0 |
00001101.
8 + 4 + 1 = 13.
Final answer: 00001101
Decimal value = 192.
Eight bits, including leading zeroes.
Use the weights 128,64,32,16,8,4,2,1.
At each weight, write 1 and subtract if the remaining value is at least that weight; otherwise write 0.
| Weight | Remaining before | Bit decision | Remaining after |
|---|---|---|---|
| 128 | 192 | 1: subtract | 64 |
| 64 | 64 | 1: subtract | 0 |
| 32 | 0 | 0: too large | 0 |
| 16 | 0 | 0: too large | 0 |
| 8 | 0 | 0: too large | 0 |
| 4 | 0 | 0: too large | 0 |
| 2 | 0 | 0: too large | 0 |
| 1 | 0 | 0: too large | 0 |
11000000.
128 + 64 = 192.
Final answer: 11000000
Decimal value = 224.
Eight bits, including leading zeroes.
Use the weights 128,64,32,16,8,4,2,1.
At each weight, write 1 and subtract if the remaining value is at least that weight; otherwise write 0.
| Weight | Remaining before | Bit decision | Remaining after |
|---|---|---|---|
| 128 | 224 | 1: subtract | 96 |
| 64 | 96 | 1: subtract | 32 |
| 32 | 32 | 1: subtract | 0 |
| 16 | 0 | 0: too large | 0 |
| 8 | 0 | 0: too large | 0 |
| 4 | 0 | 0: too large | 0 |
| 2 | 0 | 0: too large | 0 |
| 1 | 0 | 0: too large | 0 |
11100000.
128 + 64 + 32 = 224.
Final answer: 11100000
Definition. A power describes repeated multiplication. The number e is approximately 2.718281828; e raised to a negative power is a positive number below 1.
Simple meaning. 2 ki power bits ke combinations batati hai; e ki negative power ALOHA ki collision probability mein aati hai.
| Power | Value |
|---|---|
| 20 | 1 |
| 21 | 2 |
| 22 | 4 |
| 23 | 8 |
| 24 | 16 |
| 25 | 32 |
| 26 | 64 |
| 27 | 128 |
| 28 | 256 |
| 29 | 512 |
| 210 | 1024 |
| 211 | 2048 |
| 212 | 4096 |
| 213 | 8192 |
| 214 | 16384 |
| 215 | 32768 |
| 216 | 65536 |
Each bit has two choices; six independent bit positions provide 2×2×2×2×2×2 = 64 combinations. A fraction is converted to percent by multiplying by 100, so 0.18394 corresponds to 18.394%.
Yaad rakho: Minus exponent andar hai; final exponential positive rehta hai.
"},{"id":"calculator","title":"Calculator steps for ALOHA and subnetting","html":"Use the normal calculation mode. Exact menu names vary across fx-991ES-family models; the mathematics below is the same. The eˣ function is usually the shifted function of ln. The EXP/×10ˣ key enters scientific notation and is not eˣ.
Yaad rakho: eˣ aur ×10ˣ alag keys/functions hain. Brackets se denominator safe rakho.
"},{"id":"graphs","title":"How to read a routing graph","html":"Definition. A weighted graph represents routers as vertices and links as edges labelled with costs. A directed edge may be used only in the arrow direction.
Simple meaning. Circle router hai, line link hai, number us link ka cost. Arrow ulta mat chalna.
Hop count counts links or router transitions according to the protocol. A weighted path cost adds the numerical metric assigned to each edge. The fewest-hop path need not have the lowest weighted cost; an indirect route costing 7 is cheaper than a direct link costing 10.
Yaad rakho: Shortest = smallest cost, zaroori nahi fewest arrows.
"}]},{"id":"introduction","title":"Unit 1 · Networks and topology","scope":"TEE · Unit 1 · current syllabus 702AI0C023","sections":[{"id":"network-types","title":"What a network is and how networks are classified","html":"A computer network is a set of interconnected devices that exchange data and share resources using communication protocols. A link carries signals; a protocol defines message format, meaning and timing. Hosts run applications, switches forward local frames and routers connect IP networks.
| Type | Meaning and scope | Example |
|---|---|---|
| PAN | Personal Area Network: a person’s nearby devices | Phone linked to earbuds |
| LAN | Local Area Network: a building or local campus network under local administration | Laboratory Ethernet/Wi-Fi |
| MAN | Metropolitan Area Network: interconnects sites across a town/city | City-wide institutional network |
| WAN | Wide Area Network: connects geographically separated networks over regional/global links | Branches in different cities |
| ARPANET | Historical packet-switched research network; a major predecessor of the Internet | Early interconnected research institutions; not a modern geographic size category |
Classification by geographical scope is descriptive: a fixed kilometre boundary or a guaranteed speed is not part of every definition. Networks can also be classified by wired/wireless medium or by client-server/peer-to-peer organization.
Yaad rakho: PAN personal, LAN local, MAN city, WAN wide.
"},{"id":"models","title":"Client-server and peer-to-peer","html":"In a client-server design, users’ clients contact designated service processes. In peer-to-peer (P2P), participating hosts can supply and consume resources. These are application organizations, not different physical cable types.
![\"Client-server](\"fig/tee-client-peer.a7c6f1b535.svg\")
| Point | Client-server | Peer-to-peer |
|---|---|---|
| Roles | Clients request; designated servers provide services | Peers can both request and provide resources |
| Control | Central policies, accounts and backups are easier to coordinate | Control and copies distributed among peers |
| Performance | Predictable with adequate server capacity; server may bottleneck | Resources grow with peers, but peer capacity/availability vary |
| Scalability | Scale servers, replicas, caching and load distribution | More peers can contribute capacity; coordination becomes harder |
| Reliability | One unreplicated server is a failure point; replication helps | Loss of one peer need not stop system if useful replicas remain |
| Security | Central access control and monitoring | Trust and consistent access policy are harder |
| Best fit | Managed collaboration, databases, login services | Distributed content sharing among participating peers |
Central servers can authenticate users, maintain the authoritative shared document and control versions. The organization must provision capacity, replicate data and avoid a single server failure. P2P can exchange large files or local edits among peers and distribute load, but needs discovery, conflict resolution, trust and availability strategies. A hybrid can keep account control centralized while distributing selected content.
Yaad rakho: Server mein service central; peers dono roles karte hain.
"},{"id":"topologies","title":"Network topologies and choosing a layout","html":"Topology is the arrangement of nodes and links. Physical topology describes actual links; logical topology describes how data flows. A switch-centred physical star is different from a shared collision-domain bus.
![\"Four](\"fig/tee-topologies.d9e18f041b.svg\")
| Topology | Structure | Benefit | Limitation |
|---|---|---|---|
| Bus | Devices share a common backbone | Simple small shared-medium layout | Backbone fault affects segment; collisions possible on classic shared bus |
| Star | Each device connects to a central switch/hub | Easy expansion and fault isolation | Central-device failure affects star; more cabling |
| Ring | Each node connected to two neighbours in a ring | Structured traversal; token-based variants coordinate access | Simple single-ring break can disrupt operation unless protected |
| Full mesh | Every pair has a direct link | Redundant paths and direct connectivity | Expensive ports/cabling: n(n−1)/2 undirected links |
| Tree/hybrid | Hierarchical stars or combined topologies | Scales into buildings/floors | Upper-level faults affect dependent branches |
Yaad rakho: Layout draw karo, then one strength and one failure case.
"}]},{"id":"physical","title":"Unit 2 · Models and transmission media","scope":"TEE · Unit 2 · current syllabus 702AI0C023","sections":[{"id":"osi","title":"OSI layers and encapsulation","html":"The Open Systems Interconnection (OSI) model separates network functions into seven layers. A layer offers services upward and uses services below. Peer layers communicate logically through protocol rules; actual data moves down the sender’s stack, across links and up the receiver’s stack.
![\"OSI](\"fig/tee-models.5bff0518c8.svg\")
| Layer | Task | Example |
|---|---|---|
| 7 Application | Services used by network applications | HTTP, DNS, FTP, SMTP |
| 6 Presentation | Representation, encoding and application data transformations | Character encoding, serialization; encryption as a conceptual function |
| 5 Session | Manage dialogue and synchronization/checkpoints | Session establishment and resumption concepts |
| 4 Transport | Process delivery; reliability/flow control where supplied | TCP, UDP, SCTP; ports |
| 3 Network | Logical addressing and routing across networks | IP; routers |
| 2 Data link | Framing, local link addressing, medium access and error detection | Ethernet/Wi-Fi; MAC addresses |
| 1 Physical | Transmit signal symbols/bits through a medium | Electrical, optical or radio signals |
Yaad rakho: Application se physical tak wrap; receiver par unwrap.
"},{"id":"tcpip","title":"TCP/IP layers and comparison with OSI","html":"The TCP/IP model describes the Internet protocol suite in four layers. Many teaching books split the link layer into data link and physical, making a five-layer teaching model; identify the convention before drawing.
| TCP/IP layer | Function | Examples |
|---|---|---|
| Application | Defines application message formats and exchanges for Web access, naming, email and file transfer; includes presentation/session functions as needed | HTTP, DNS, SMTP, FTP |
| Transport | End-to-end process communication | TCP, UDP, SCTP |
| Internet | IP addressing, routing and best-effort datagrams | IPv4/IPv6 |
| Link/network access | Local delivery and physical access | Ethernet/Wi-Fi and their medium |
| Point | OSI | TCP/IP |
|---|---|---|
| Layer count | Seven | Four here; five-layer teaching variant splits link |
| Upper functions | Application, presentation, session separated | Grouped under application |
| Lower functions | Data link and physical separated | Grouped under link/network access |
| Purpose | General reference architecture | Architecture of deployed Internet protocol family |
| Network service | Reference model can describe different services | Internet layer uses connectionless IP |
| Transport | Conceptual transport service | TCP reliable byte stream; UDP datagrams; SCTP messages |
Yaad rakho: OSI 7; TCP/IP four mein upper three merge, lower two merge.
"},{"id":"media","title":"Guided and unguided transmission media","html":"Guided media constrain signals to a physical path: twisted pair, coaxial cable and optical fiber. Unguided media carry electromagnetic waves through free space: radio, microwave and infrared. Wireless bandwidth and reach depend on frequency, power, antenna, obstacles and regulations; media names alone do not fix a speed.
![\"Transmission](\"fig/tee-media.aff26cd0b5.svg\")
| Medium | Two key characteristics | Limitations / use |
|---|---|---|
| Twisted pair | Electrical signal; pairs twisted to reduce interference | Low-cost LAN access; distance/noise limits depend on category and design |
| Coaxial cable | Shielded central conductor; greater shielding than basic unshielded pair | Bulkier; cable broadband and legacy shared LANs |
| Optical fiber | Light in glass/plastic; high capacity with immunity to electromagnetic interference | Transceivers and careful splicing/bend control needed; backbone links |
| Radio waves | Wireless propagation; many bands support broad/non-line-of-sight coverage | Interference, shared spectrum and security configuration matter; Wi-Fi/mobile |
| Terrestrial microwave | Directional antennas; typically line of sight | Alignment and obstructions matter; point-to-point backhaul |
| Satellite microwave | Wide coverage through relay in space | Cost and latency depend strongly on orbit/system; remote connectivity |
| Infrared | Short-range light; usually blocked by walls | Limited coverage; short-range indoor/remote-control links |
Yaad rakho: Guided = cable path; unguided = free-space waves.
"},{"id":"fiber","title":"Optical fiber: advantages and limits","html":"Transmission media are classified into guided media (twisted pair, coaxial cable, optical fiber) and unguided media (radio, microwave, infrared).
Optical fiber carries data as light in a core surrounded by cladding. It is a guided medium. Single-mode fiber supports one main propagation mode and is used for long/high-capacity links; multimode supports multiple modes and is often used over shorter distances.
| Advantages | Disadvantages / design costs |
|---|---|
| High data capacity and low attenuation over suitable links | Optical transceivers, connectors and splicing add cost/skill requirements |
| Immune to electromagnetic interference; no electrical crosstalk | Tight bends, dirty connectors and poor installation increase loss |
| Electrical isolation between buildings | Fiber is mechanically sensitive and requires appropriate protection |
| Harder to tap casually than an unprotected radio channel | It is not automatically confidential; encryption/access control remain necessary |
Yaad rakho: Fiber light carry karta hai; immunity means no EMI, not no security risk.
"},{"id":"campus","title":"Apply wired and wireless concepts to a campus","html":"| Criterion | Wired campus links | Wireless campus access |
|---|---|---|
| Speed/capacity | Dedicated switch ports and fiber backbones can provide high, predictable capacity | Capacity shared over radio; performance depends on contention, coverage and interference |
| Reliability | Stable when cables/ports and redundant uplinks are healthy | Mobility and flexible access; radio interference and coverage holes can cause variability |
| Security | Physical access control helps, but ports still need authentication/segmentation | Use authenticated encryption and managed access; radio extends beyond rooms |
| Deployment | Cabling cost and fixed access points | Supports mobile users and faster coverage changes |
| Recommended role | Servers, laboratories, inter-building backbone | Students’ phones/laptops, roaming and guest access |
Choose a hybrid: wired core/backbone and critical endpoints, with managed access points providing wireless mobility. Separate guest traffic, authenticate users and plan redundant infrastructure. Neither medium alone guarantees security or availability.
Yaad rakho: Backbone stable cable; mobile users radio; security dono mein configure karo.
"}]},{"id":"error-control","title":"Unit 3 · Errors, block coding and CRC","scope":"TEE · Unit 3 before ALOHA · current syllabus 702AI0C023","sections":[{"id":"principles","title":"Error detection, correction and block coding","html":"A transmission error changes one or more bits. A single-bit error flips one bit. A burst error affects a span from the first corrupted bit to the last; the bits in between need not all change. Detection identifies an invalid received pattern; correction identifies the intended pattern within the code’s guaranteed capability.
![\"R_c=k/n,\\quad](\"math/d973fda532c4ffe6.svg\")
Yaad rakho: Detection asks error hua? Correction asks kaunsa correct word?
"},{"id":"parity","title":"Parity and the limits of redundancy","html":"With even parity, append a bit so the total number of 1s is even. The receiver repeats the parity test. A single-bit flip toggles parity and is detected. Two flips preserve parity and can pass the test. Odd parity uses an odd total; its limitations are analogous.
| Data | Number of 1s | Even-parity bit | Codeword |
|---|---|---|---|
| 1011 | 3 | 1 | 10111 |
| 1100 | 2 | 0 | 11000 |
| 0000 | 0 | 0 | 00000 |
Parity detects every odd number of bit flips, not every burst or every nonzero error. Detection also does not reveal the error position. The unused patterns, and the minimum distance between valid patterns, determine the guaranteed capability.
Yaad rakho: Odd flips parity badalta; even flips bach sakte hain.
"},{"id":"hamming-distance","title":"Hamming distance and guaranteed capability","html":"Hamming distance d(a,b) is the number of positions where two equal-length bit strings differ. XOR the strings and count the 1s. Minimum distance dmin is the smallest distance between distinct valid codewords; do not use the largest pairwise distance.
![\"d(a,b)=\\mathrm{weight}(a\\oplus](\"math/08c67b82aac87a74.svg\")
![\"s_{\\max}=d_{\\min}-1,\\qquad](\"math/bd55e600f9b980f0.svg\")
Yaad rakho: XOR ones count; minimum minus one detect, half floor correct.
"},{"id":"distance-1","title":"Hamming-distance working · example 1","html":"| Codeword | Bits |
|---|---|
| C1 | 1001101 |
| C2 | 1110001 |
| C3 | 1010101 |
All pair distances, dmin and the maximum guaranteed numbers of errors.
![\"s=d_{\\min}-1,\\quad](\"math/b0c374242e18e388.svg\")
XOR exposes differing positions. Use the closest codeword pair, because it is the weakest separation in the code.
| Pair | First | Second | XOR | Distance |
|---|---|---|---|---|
| C1 vs C2 | 1001101 | 1110001 | 0111100 | 4 |
| C1 vs C3 | 1001101 | 1010101 | 0011000 | 2 |
| C2 vs C3 | 1110001 | 1010101 | 0100100 | 2 |
dmin = min(4, 2, 2) = 2.
Detect = 2−1 = 1; correct = floor((2−1)/2) = 0.
2 bit changes can reach another valid codeword, so only 1 are guaranteed detectable. Correction regions require 2t < 2.
Final answer: dmin=2; detect up to 1 errors; correct up to 0 errors.
| Codeword | Bits |
|---|---|
| C1 | 000 |
| C2 | 111 |
All pair distances, dmin and the maximum guaranteed numbers of errors.
XOR exposes differing positions. Use the closest codeword pair, because it is the weakest separation in the code.
| Pair | First | Second | XOR | Distance |
|---|---|---|---|---|
| C1 vs C2 | 000 | 111 | 111 | 3 |
dmin = min(3) = 3.
Detect = 3−1 = 2; correct = floor((3−1)/2) = 1.
3 bit changes can reach another valid codeword, so only 2 are guaranteed detectable. Correction regions require 2t < 3.
Final answer: dmin=3; detect up to 2 errors; correct up to 1 errors.
| Codeword | Bits |
|---|---|
| C1 | 00000 |
| C2 | 11110 |
| C3 | 00111 |
All pair distances, dmin and the maximum guaranteed numbers of errors.
XOR exposes differing positions. Use the closest codeword pair, because it is the weakest separation in the code.
| Pair | First | Second | XOR | Distance |
|---|---|---|---|---|
| C1 vs C2 | 00000 | 11110 | 11110 | 4 |
| C1 vs C3 | 00000 | 00111 | 00111 | 3 |
| C2 vs C3 | 11110 | 00111 | 11001 | 3 |
dmin = min(4, 3, 3) = 3.
Detect = 3−1 = 2; correct = floor((3−1)/2) = 1.
3 bit changes can reach another valid codeword, so only 2 are guaranteed detectable. Correction regions require 2t < 3.
Final answer: dmin=3; detect up to 2 errors; correct up to 1 errors.
A Cyclic Redundancy Check (CRC) treats a binary message as a polynomial over GF(2). Addition and subtraction are XOR: 0⊕0=0, 1⊕0=1, 0⊕1=1, 1⊕1=0. There are no carries or borrows. A degree-r generator contains r+1 bits, with leading and constant coefficients 1.
![\"T(x)=x^rM(x)\\oplus](\"math/5edc4316a910e4aa.svg\")
![\"CRC](\"fig/tee-crc.24705171b3.svg\")
A degree-r generator with nonzero constant term detects every burst of length at most r. Generators containing factor (x+1) detect every odd number of bit errors. Longer-error guarantees depend on the polynomial and frame length; CRC is not a general correction method.
Yaad rakho: CRC mein subtract bhi XOR; receiver par full frame divide.
"},{"id":"crc-1","title":"CRC worked division · example 1","html":"Message=110010; generator=1101; generator degree r=3. This is textbook systematic CRC with zero initial remainder and no reflection/final XOR.
Exactly r CRC bits, the transmitted frame and receiver syndromes.
![\"R=(M\\,2^r)\\bmod_2](\"math/3db4ba7cff91f15a.svg\")
Modulo-2 division uses aligned XOR, not ordinary decimal division.
Message bits=110010; generator bits=1101. Leading exponent is 3, so CRC length is 3.
110010 followed by 3 zeros = 110010000.
| Before | Aligned generator | After XOR |
|---|---|---|
| 110010000 | 110100000 | 000110000 |
| 000110000 | 000110100 | 000000100 |
Remainder padded to 3 bits = 100. Leading zero columns need no subtraction.
Message + CRC = 110010 + 100 = 110010100.
| Before | Aligned generator | After XOR |
|---|---|---|
| 110010100 | 110100000 | 000110100 |
| 000110100 | 000110100 | 000000000 |
Receiver remainder = 000; the uncorrupted frame passes.
Flip the last bit: 110010101. Dividing by 1101 leaves 001, which is nonzero, so this error is detected. A zero syndrome would mean no detected error, not proof of no corruption.
Final answer: CRC=100; transmitted frame=110010100; uncorrupted receiver remainder=000.
Message=1001; generator=1011; generator degree r=3. This is textbook systematic CRC with zero initial remainder and no reflection/final XOR.
Exactly r CRC bits, the transmitted frame and receiver syndromes.
Modulo-2 division uses aligned XOR, not ordinary decimal division.
Message bits=1001; generator bits=1011. Leading exponent is 3, so CRC length is 3.
1001 followed by 3 zeros = 1001000.
| Before | Aligned generator | After XOR |
|---|---|---|
| 1001000 | 1011000 | 0010000 |
| 0010000 | 0010110 | 0000110 |
Remainder padded to 3 bits = 110. Leading zero columns need no subtraction.
Message + CRC = 1001 + 110 = 1001110.
| Before | Aligned generator | After XOR |
|---|---|---|
| 1001110 | 1011000 | 0010110 |
| 0010110 | 0010110 | 0000000 |
Receiver remainder = 000; the uncorrupted frame passes.
Flip the last bit: 1001111. Dividing by 1011 leaves 001, which is nonzero, so this error is detected. A zero syndrome would mean no detected error, not proof of no corruption.
Final answer: CRC=110; transmitted frame=1001110; uncorrupted receiver remainder=000.
Message=101101; generator=10011; generator degree r=4. This is textbook systematic CRC with zero initial remainder and no reflection/final XOR.
Exactly r CRC bits, the transmitted frame and receiver syndromes.
Modulo-2 division uses aligned XOR, not ordinary decimal division.
Message bits=101101; generator bits=10011. Leading exponent is 4, so CRC length is 4.
101101 followed by 4 zeros = 1011010000.
| Before | Aligned generator | After XOR |
|---|---|---|
| 1011010000 | 1001100000 | 0010110000 |
| 0010110000 | 0010011000 | 0000101000 |
| 0000101000 | 0000100110 | 0000001110 |
Remainder padded to 4 bits = 1110. Leading zero columns need no subtraction.
Message + CRC = 101101 + 1110 = 1011011110.
| Before | Aligned generator | After XOR |
|---|---|---|
| 1011011110 | 1001100000 | 0010111110 |
| 0010111110 | 0010011000 | 0000100110 |
| 0000100110 | 0000100110 | 0000000000 |
Receiver remainder = 0000; the uncorrupted frame passes.
Flip the last bit: 1011011111. Dividing by 10011 leaves 0001, which is nonzero, so this error is detected. A zero syndrome would mean no detected error, not proof of no corruption.
Final answer: CRC=1110; transmitted frame=1011011110; uncorrupted receiver remainder=0000.
Message=1010; generator=1011; generator degree r=3. This is textbook systematic CRC with zero initial remainder and no reflection/final XOR.
Exactly r CRC bits, the transmitted frame and receiver syndromes.
Modulo-2 division uses aligned XOR, not ordinary decimal division.
Message bits=1010; generator bits=1011. Leading exponent is 3, so CRC length is 3.
1010 followed by 3 zeros = 1010000.
| Before | Aligned generator | After XOR |
|---|---|---|
| 1010000 | 1011000 | 0001000 |
| 0001000 | 0001011 | 0000011 |
Remainder padded to 3 bits = 011. Leading zero columns need no subtraction.
Message + CRC = 1010 + 011 = 1010011.
| Before | Aligned generator | After XOR |
|---|---|---|
| 1010011 | 1011000 | 0001011 |
| 0001011 | 0001011 | 0000000 |
Receiver remainder = 000; the uncorrupted frame passes.
Flip the last bit: 1010010. Dividing by 1011 leaves 001, which is nonzero, so this error is detected. A zero syndrome would mean no detected error, not proof of no corruption.
Final answer: CRC=011; transmitted frame=1010011; uncorrupted receiver remainder=000.
Flow control prevents a fast sender from overwhelming receiver buffers. Error control handles damaged, lost and duplicate frames so that the intended data can be delivered reliably. ARQ means Automatic Repeat reQuest: retransmission is triggered by feedback or a timer.
| Mechanism | Purpose |
|---|---|
| CRC/checksum | Detect corrupted frames; alone does not recover them |
| Sequence number | Identify data and reject duplicates |
| ACK | Confirm accepted data; state cumulative or individual convention |
| Timer | Recover when either data or its ACK disappears |
| Window | Bound the outstanding unacknowledged data |
| Buffer | Hold unacknowledged sender copies; receiver buffers depend on protocol |
Yaad rakho: Flow receiver speed; error lost/corrupt/duplicate recovery.
"},{"id":"stopwait","title":"Stop-and-Wait ARQ and lost ACKs","html":"Stop-and-Wait ARQ permits only one outstanding frame. The sender stores its frame, transmits it and starts a timer. A valid matching ACK advances the sender; a timeout retransmits the stored frame.
![\"Stop-and-Wait:](\"fig/tee-stopwait.a99281ab96.svg\")
Yaad rakho: One frame outstanding; duplicate ko acknowledge karo, deliver dobara nahi.
"},{"id":"gbn","title":"Go-Back-N with window 7: three complete cases","html":"Go-Back-N (GBN) ARQ pipelines up to W outstanding frames. For m-bit sequence numbers, textbook GBN uses W ≤ 2m−1 and receiver window 1. The receiver accepts only the next in-order frame; out-of-order frames are discarded. ACK k here cumulatively acknowledges all frames before k and means “next expected is k”.
![\"GBN](\"fig/tee-gbn.5f577a6979.svg\")
![\"GBN](\"fig/tee-gbn-normal.144a24ea80.svg\")
![\"GBN:](\"fig/tee-gbn-ack.713dc9972e.svg\")
Yaad rakho: GBN receiver one; gap ke baad discard; timeout pe outstanding suffix repeat.
"},{"id":"sr","title":"Selective Repeat: window bound and buffering","html":"Selective Repeat (SR) ARQ accepts and buffers valid out-of-order frames inside the receiver window, acknowledges them individually and retransmits only missing/timed-out frames. In this answer ACK i means “frame i received”, unlike the next-expected cumulative ACK convention used for GBN.
![\"SR](\"fig/tee-sr.5ae2b52b45.svg\")
![\"W_s=W_r\\le](\"math/454034d348f17623.svg\")
Yaad rakho: SR selective resend; half sequence space keeps old and new separate.
"},{"id":"comparison","title":"Compare all three ARQ protocols","html":"| Property | Stop-and-Wait | Go-Back-N | Selective Repeat |
|---|---|---|---|
| Outstanding sender frames | 1 | Up to 2^m−1 | Up to 2^(m−1), equal windows |
| Receiver window | 1 | 1 | Same maximum as sender in standard equal-window SR |
| ACK convention used here | Next expected | Cumulative, next expected | Individual frame accepted |
| Out-of-order data | Not delivered | Discarded | Buffered within valid window |
| After timeout | Repeat the one frame | Repeat outstanding frames from base | Repeat only timed-out frame |
| Buffer/tracking cost | Low | Sender window; simple receiver | Sender and receiver windows; per-frame tracking |
| Tradeoff | Simple but idle during RTT | Good pipelining, waste after loss | Efficient loss recovery, greater complexity |
Every comparison assumes finite sequence numbers, correct duplicate handling and the textbook protocol models. The same text “ACK 3” can mean different things under different ACK conventions; state its meaning before explaining a trace.
m=3; standard GBN and equal sender/receiver SR windows.
Sequence count and allowed windows.
![\"N=2^m,\\quad](\"math/c9f05985613df1dc.svg\")
GBN needs one unused sequence value; SR keeps old and new receive windows disjoint.
23=8, numbered 0…7.
Sender maximum=8−1=7; receiver window=1.
Equal sender and receiver maxima=8/2=4.
Final answer: 8 sequence numbers; GBN sender 7, receiver 1; SR sender/receiver 4 each.
m=4; standard GBN and equal sender/receiver SR windows.
Sequence count and allowed windows.
GBN needs one unused sequence value; SR keeps old and new receive windows disjoint.
24=16, numbered 0…15.
Sender maximum=16−1=15; receiver window=1.
Equal sender and receiver maxima=16/2=8.
Final answer: 16 sequence numbers; GBN sender 15, receiver 1; SR sender/receiver 8 each.
m=5; standard GBN and equal sender/receiver SR windows.
Sequence count and allowed windows.
GBN needs one unused sequence value; SR keeps old and new receive windows disjoint.
25=32, numbered 0…31.
Sender maximum=32−1=31; receiver window=1.
Equal sender and receiver maxima=32/2=16.
Final answer: 32 sequence numbers; GBN sender 31, receiver 1; SR sender/receiver 16 each.
Definition. Pure ALOHA is a random-access protocol in which a station transmits whenever a frame is ready and retransmits after a random delay if the transmission is unsuccessful.
Simple meaning. Frame ready → bhej do. Collision hua → random wait karke retry.
In Pure ALOHA, a station transmits a ready frame without carrier sensing. If a successful acknowledgement is not received, it waits for a random interval and retransmits. Overlapping transmissions cause collisions in the basic model. Its maximum normalized throughput is 1/(2e) ≈ 18.4%, so it is inefficient at high offered load.
Yaad rakho: Pure = kisi bhi time start; retry ka wait random.
"},{"id":"vulnerable","title":"Why Pure ALOHA has a 2T vulnerable period","html":"Definition. The vulnerable period is the interval of start times of other frames that would cause a collision with a chosen frame.
Simple meaning. Apne frame se ek frame-time pehle start hua frame bhi takra sakta hai, aur baad mein bhi.
A frame lasting T can collide with an equal-length frame that began less than T earlier, or one that begins less than T later. The two intervals together have length T+T=2T.
Yaad rakho: Pure: before T + after T = 2T.
"},{"id":"slotted","title":"Slotted ALOHA: starts only at slot boundaries","html":"Definition. Slotted ALOHA is a random-access protocol that divides time into slots of one frame duration and permits transmissions only at the beginning of a slot.
Simple meaning. Frame ready hai tab bhi next slot ki start-line ka wait karo.
| Feature | Pure | Slotted |
|---|---|---|
| Start time | Any time | Slot boundary only |
| Synchronization | No common slots required | Required |
| Vulnerable period | 2T | T |
| Maximum S | 1/(2e) ≈ 0.184 | 1/e ≈ 0.368 |
| G at maximum | 0.5 | 1 |
Yaad rakho: Slotted = aligned starts; collision risk half-window, not zero.
"},{"id":"formulas","title":"G, S and throughput: which number are you finding?","html":"Definition. Offered load G is the average number of transmission attempts per frame time. Normalized throughput S is the average number of successful frames per frame time.
Simple meaning. G = kitne attempts; S = unmein successful output per frame-time. Dono same nahi.
G denotes aggregate attempts per frame transmission time and S denotes successful frames per frame transmission time. Pure ALOHA: S=Ge⁻²ᴳ, with maximum 1/(2e) at G=0.5. Slotted ALOHA: S=Ge⁻ᴳ, with maximum 1/e at G=1. These results use the equal-length Poisson collision model.
Yaad rakho: P mein pair: −2G; S mein single: −G.
\r\nFor 200-bit frames on a 200000-bit/s channel, T = 200/200000 = 0.001 s. If 500 frames/s attempt transmission, G = 0.5 and pure-ALOHA S = 0.5e⁻¹ ≈ 0.1839397. Convert to frames/s by dividing by T: 0.1839397/0.001 ≈ 183.94 frames/s.
Do not multiply S by 500: S is successful frames per frame-time, not success probability per attempt. Some slide totals use that incorrect conversion. At 250 attempts/s the correct pure-ALOHA output is 151.63 frames/s. Use the definitions and units above.
"},{"id":"pure-1","title":"Pure ALOHA · load example 1","html":"Protocol = Pure ALOHA; G = 0.25 attempts/frame time.
S, 100S and the maximum possible S.
Select the formula from the protocol name. The exponent is −2G because Pure ALOHA has a 2T vulnerable period.
−2 × 0.25 = -0.5.
S = 0.25 × 0.606530660 ≈ 0.151632665 successful frames/frame time.
0.151632665 × 100 ≈ 15.1633%.
The maximum is 1/(2e) ≈ 0.183940 = 18.3940%, at G=0.5.
Final answer: S ≈ 0.151633; percentage ≈ 15.1633%; Smax ≈ 0.183940 at G=0.5.
Protocol = Pure ALOHA; G = 0.5 attempts/frame time.
S, 100S and the maximum possible S.
Select the formula from the protocol name. The exponent is −2G because Pure ALOHA has a 2T vulnerable period.
−2 × 0.5 = -1.
S = 0.5 × 0.367879441 ≈ 0.183939721 successful frames/frame time.
0.183939721 × 100 ≈ 18.3940%.
The maximum is 1/(2e) ≈ 0.183940 = 18.3940%, at G=0.5.
Final answer: S ≈ 0.183940; percentage ≈ 18.3940%; Smax ≈ 0.183940 at G=0.5.
Protocol = Pure ALOHA; G = 1 attempts/frame time.
S, 100S and the maximum possible S.
Select the formula from the protocol name. The exponent is −2G because Pure ALOHA has a 2T vulnerable period.
−2 × 1 = -2.
S = 1 × 0.135335283 ≈ 0.135335283 successful frames/frame time.
0.135335283 × 100 ≈ 13.5335%.
The maximum is 1/(2e) ≈ 0.183940 = 18.3940%, at G=0.5.
Final answer: S ≈ 0.135335; percentage ≈ 13.5335%; Smax ≈ 0.183940 at G=0.5.
Protocol = Slotted ALOHA; G = 0.25 attempts/frame time.
S, 100S and the maximum possible S.
Select the formula from the protocol name. The exponent is −G because Slotted ALOHA has a T vulnerable period.
−1 × 0.25 = -0.25.
S = 0.25 × 0.778800783 ≈ 0.194700196 successful frames/frame time.
0.194700196 × 100 ≈ 19.4700%.
The maximum is 1/(1e) ≈ 0.367879 = 36.7879%, at G=1.
Final answer: S ≈ 0.194700; percentage ≈ 19.4700%; Smax ≈ 0.367879 at G=1.
Protocol = Slotted ALOHA; G = 1 attempts/frame time.
S, 100S and the maximum possible S.
Select the formula from the protocol name. The exponent is −G because Slotted ALOHA has a T vulnerable period.
−1 × 1 = -1.
S = 1 × 0.367879441 ≈ 0.367879441 successful frames/frame time.
0.367879441 × 100 ≈ 36.7879%.
The maximum is 1/(1e) ≈ 0.367879 = 36.7879%, at G=1.
Final answer: S ≈ 0.367879; percentage ≈ 36.7879%; Smax ≈ 0.367879 at G=1.
Protocol = Slotted ALOHA; G = 2 attempts/frame time.
S, 100S and the maximum possible S.
Select the formula from the protocol name. The exponent is −G because Slotted ALOHA has a T vulnerable period.
−1 × 2 = -2.
S = 2 × 0.135335283 ≈ 0.270670566 successful frames/frame time.
0.270670566 × 100 ≈ 27.0671%.
The maximum is 1/(1e) ≈ 0.367879 = 36.7879%, at G=1.
Final answer: S ≈ 0.270671; percentage ≈ 27.0671%; Smax ≈ 0.367879 at G=1.
| Symbol | Given |
|---|---|
| L | 200 bits |
| R | 200 kbps = 200,000 bits/s |
| λ | 1000 aggregate attempts/s |
T, G, S, successful frames/s and bit/s.
First normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
T = 200/200,000 = 0.001 s = 1 ms.
G = 1000 × 0.001 = 1 attempts/frame time.
−2G = -2; e to that power ≈ 0.135335283.
S = 1 × 0.135335283 ≈ 0.135335283.
S/T = 0.135335283/0.001 ≈ 135.335283 frames/s.
SR = 0.135335283 × 200,000 ≈ 27067.056647 bits/s ≈ 27.067057 kbps.
Final answer: G=1, S≈0.135335, successful rate≈135.335 frames/s = 27.067 kbps.
| Symbol | Given |
|---|---|
| L | 200 bits |
| R | 200 kbps = 200,000 bits/s |
| λ | 500 aggregate attempts/s |
T, G, S, successful frames/s and bit/s.
First normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
T = 200/200,000 = 0.001 s = 1 ms.
G = 500 × 0.001 = 0.5 attempts/frame time.
−2G = -1; e to that power ≈ 0.367879441.
S = 0.5 × 0.367879441 ≈ 0.183939721.
S/T = 0.183939721/0.001 ≈ 183.939721 frames/s.
SR = 0.183939721 × 200,000 ≈ 36787.944117 bits/s ≈ 36.787944 kbps.
Final answer: G=0.5, S≈0.183940, successful rate≈183.940 frames/s = 36.788 kbps.
| Symbol | Given |
|---|---|
| L | 200 bits |
| R | 200 kbps = 200,000 bits/s |
| λ | 250 aggregate attempts/s |
T, G, S, successful frames/s and bit/s.
First normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
T = 200/200,000 = 0.001 s = 1 ms.
G = 250 × 0.001 = 0.25 attempts/frame time.
−2G = -0.5; e to that power ≈ 0.606530660.
S = 0.25 × 0.606530660 ≈ 0.151632665.
S/T = 0.151632665/0.001 ≈ 151.632665 frames/s.
SR = 0.151632665 × 200,000 ≈ 30326.532986 bits/s ≈ 30.326533 kbps.
Final answer: G=0.25, S≈0.151633, successful rate≈151.633 frames/s = 30.327 kbps.
| Symbol | Given |
|---|---|
| L | 400 bits |
| R | 200 kbps = 200,000 bits/s |
| λ | 500 aggregate attempts/s |
T, G, S, successful frames/s and bit/s.
First normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
T = 400/200,000 = 0.002 s = 2 ms.
G = 500 × 0.002 = 1 attempts/frame time.
−1G = -1; e to that power ≈ 0.367879441.
S = 1 × 0.367879441 ≈ 0.367879441.
S/T = 0.367879441/0.002 ≈ 183.939721 frames/s.
SR = 0.367879441 × 200,000 ≈ 73575.888234 bits/s ≈ 73.575888 kbps.
Final answer: G=1, S≈0.367879, successful rate≈183.940 frames/s = 73.576 kbps.
| Symbol | Given |
|---|---|
| L | 1600 bits |
| R | 800 kbps = 800,000 bits/s |
| λ | 250 aggregate attempts/s |
T, G, S, successful frames/s and bit/s.
First normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
T = 1600/800,000 = 0.002 s = 2 ms.
G = 250 × 0.002 = 0.5 attempts/frame time.
−1G = -0.5; e to that power ≈ 0.606530660.
S = 0.5 × 0.606530660 ≈ 0.303265330.
S/T = 0.303265330/0.002 ≈ 151.632665 frames/s.
SR = 0.303265330 × 800,000 ≈ 242612.263885 bits/s ≈ 242.612264 kbps.
Final answer: G=0.5, S≈0.303265, successful rate≈151.633 frames/s = 242.612 kbps.
A small offered load leaves many opportunities unused. A very large load creates many collisions. The best point balances those effects.
Yaad rakho: Pure optimum: G half, maximum half of 1/e. Slotted optimum: G one, maximum 1/e.
"}]},{"id":"csma","title":"CSMA, collision detection and avoidance","scope":"TEE · shared foundations and Units 3–4 · current syllabus 702AI0C023","sections":[{"id":"sense","title":"Carrier sensing: why listening does not remove every collision","html":"Definition. Carrier Sense Multiple Access (CSMA) requires a station to sense the shared medium before transmitting.
Simple meaning. Pehle suno, phir bhejo; lekin door ka signal pahunchne mein time leta hai.
Carrier sensing observes the local medium. Because signals have nonzero propagation delay, another station’s new transmission may not yet be detectable. Both stations can then start and their transmissions overlap.
Yaad rakho: Idle yahan hai; remote signal abhi raste mein ho sakta hai.
"},{"id":"persistence","title":"1-persistent, non-persistent and p-persistent CSMA","html":"Definition. Persistence specifies how a station behaves after observing that the medium is busy or becomes idle.
Simple meaning. 1 = turant; non = random wait; p = probability ke basis pe start.
| Rule | When idle | When busy | Trade-off |
|---|---|---|---|
| 1-persistent | Transmit immediately | Keep listening; transmit as soon as idle | Low idle delay, but waiting stations may start together |
| Non-persistent | Transmit | Wait a random time, then sense again | Less synchronized contention; may leave avoidable idle time |
| p-persistent (slotted) | Transmit with probability p; defer a slot with probability 1−p | Wait until available, then resume the probabilistic slot rule | Controls contention; needs slots and a choice of p |
| Rule | When idle | When busy | Trade-off |
|---|---|---|---|
| 1-persistent | Transmit immediately | Keep listening; transmit as soon as idle | Low idle delay, but waiting stations may start together |
| Non-persistent | Transmit | Wait a random time, then sense again | Less synchronized contention; may leave avoidable idle time |
| p-persistent (slotted) | Transmit with probability p; defer a slot with probability 1−p | Wait until available, then resume the probabilistic slot rule | Controls contention; needs slots and a choice of p |
Yaad rakho: 1 = start now; non = wait then listen; p = chance per idle slot.
"},{"id":"cd","title":"CSMA/CD: detect a collision while transmitting","html":"Definition. CSMA with Collision Detection monitors the medium during transmission, aborts a detected collision and retries after a randomized backoff.
Simple meaning. CD = takkar detect hui toh transmission rok do.
CSMA/CD senses before transmission and detects collisions during transmission. A collision causes abort/jam and a random backoff before retry. To detect the most distant collision before the frame finishes, the transmission duration must be at least the worst-case round-trip propagation delay. Thus Lmin/R ≥ 2Tp in the idealized model.
Yaad rakho: Detect karna hai toh sender tab tak bhej raha hona chahiye.
"},{"id":"backoff","title":"Binary exponential backoff","html":"Definition. Binary exponential backoff enlarges the range of possible random waiting slots after repeated collisions, reducing the chance of the same stations colliding again.
Simple meaning. Collision zyada → random wait ka range bada.
A station selects a random integer K from an expanding window and waits K slot times before retrying. After n collisions, classic Ethernet uses k=min(n,10) and 0≤K≤2ᵏ−1. After three collisions the possible values are 0 through 7.
Yaad rakho: 2ᵏ choices hain, isliye last value 2ᵏ−1.
"},{"id":"ca","title":"CSMA/CA: reduce collision risk before sending","html":"Definition. CSMA with Collision Avoidance uses carrier sensing and randomized access delays, with acknowledgements and optional reservation exchanges, to reduce collisions on a wireless shared medium.
Simple meaning. CA = pehle collision avoid karne ki koshish; guarantee nahi.
| Aspect | CSMA/CD | CSMA/CA |
|---|---|---|
| Main action | Detect and abort a collision during sending | Reduce collision risk before sending |
| Typical teaching context | Shared half-duplex wired Ethernet | Wi-Fi contention access |
| Mechanisms | Carrier sensing, detection/jam, backoff | Carrier sensing, random backoff, ACK; optional RTS/CTS |
Yaad rakho: CD = detect during; CA = avoid before.
"},{"id":"hidden","title":"Hidden terminals and RTS/CTS","html":"Definition. Hidden terminals are stations that cannot sense each other but whose transmissions can interfere at a common receiver.
Simple meaning. A aur C ek dusre ko nahi sunte, par dono B ko bhej rahe hain.
A and C may both reach B but be unable to sense one another. Simultaneous transmissions can then collide at B. A short RTS followed by B’s CTS announces a reservation; a station hearing the CTS defers, reducing the risk of a long data-frame collision.
Yaad rakho: Hidden: senders invisible to each other, receiver common.
"}]},{"id":"switching","title":"Switching techniques","scope":"TEE · shared foundations and Units 3–4 · current syllabus 702AI0C023","sections":[{"id":"types","title":"Circuit, message and packet switching","html":"Definition. Switching selects how data passes through intermediate network nodes from a source to a destination.
Simple meaning. Raasta reserve karna, poora message ruk-ruk ke bhejna, ya chhote packets bhejna—yeh teen ideas hain.
| Aspect | Circuit switching | Message switching | Packet switching |
|---|---|---|---|
| Unit handled | Continuous stream on established circuit | Whole message | Small packets |
| Resource use | Capacity reserved along a circuit | Shared; message stored at each intermediate node | Shared; each packet stored/forwarded in this model |
| Setup | Required before data transfer | No circuit setup required | Datagram: no connection setup; virtual circuit: setup |
| Delay | Setup delay, then relatively predictable service | Large store-and-forward delay for long messages | Variable delay; packet pipeline can reduce completion time |
| Main limitation | Reserved capacity can be idle | Large buffers; unsuitable for tight delay needs | Headers, possible queueing/loss; datagrams may reorder |
| Aspect | Circuit switching | Message switching | Packet switching |
|---|---|---|---|
| Unit handled | Continuous stream on established circuit | Whole message | Small packets |
| Resource use | Capacity reserved along a circuit | Shared; message stored at each intermediate node | Shared; each packet stored/forwarded in this model |
| Setup | Required before data transfer | No circuit setup required | Datagram: no connection setup; virtual circuit: setup |
| Delay | Setup delay, then relatively predictable service | Large store-and-forward delay for long messages | Variable delay; packet pipeline can reduce completion time |
| Main limitation | Reserved capacity can be idle | Large buffers; unsuitable for tight delay needs | Headers, possible queueing/loss; datagrams may reorder |
For ordinary bursty office data, packet switching is generally appropriate because resources are shared among active transfers. The design must still account for queueing, loss and the required service quality.
Yaad rakho: Circuit reserves; message waits whole; packets pipeline.
"},{"id":"datagram","title":"Datagram versus virtual-circuit packet switching","html":"Definition. Datagram switching routes each packet independently; virtual-circuit switching first establishes a logical path and forwarding state for a flow.
Simple meaning. Datagram mein har packet ka decision; virtual circuit mein pehle logical route set.
| Feature | Datagram | Virtual circuit |
|---|---|---|
| Setup | Usually none at network layer | Logical connection setup |
| Forwarding information | Destination address | Virtual-circuit identifier |
| Router state | No per-connection state required for basic IP forwarding | Connection/path state maintained |
| Path | May vary | Established logical path until changed |
Yaad rakho: Virtual circuit bhi packet switching ho sakta hai.
"}]},{"id":"ipv4","title":"IPv4: understand the address before subnetting","scope":"TEE · shared foundations and Units 3–4 · current syllabus 702AI0C023","sections":[{"id":"structure","title":"IPv4 addresses: four octets, one 32-bit value","html":"Definition. An IPv4 address is a 32-bit logical address, conventionally written as four decimal octets separated by dots.
Simple meaning. IP ke chaar boxes hain; har box 8 bits ka, value 0 se 255.
IPv4 contains 32 bits, written as four decimal octets of 8 bits each. Each octet ranges from 0 to 255. In /26 notation, 26 leading bits are the network prefix and 6 trailing bits are available within the block, giving 2⁶=64 total addresses.
Yaad rakho: 4 octets × 8 bits = 32; slash = network bits.
"},{"id":"classes","title":"Classful addressing: historical default boundaries","html":"Definition. Classful addressing divides the IPv4 address space into historical classes with fixed default network boundaries for Classes A, B and C.
Simple meaning. Class first octet se; modern subnet given slash se.
| Class | First octet | Default / purpose |
|---|---|---|
| A | 0–127; ordinary historical assignments 1–126 | /8; special ranges exist |
| B | 128–191 | /16 |
| C | 192–223 | /24 |
| D | 224–239 | Multicast |
| E | 240–255 | Reserved/experimental; special addresses exist |
Historical defaults are /8 for A, /16 for B and /24 for C. CIDR explicitly supplies the prefix length and supports variable-sized blocks. Therefore an explicit CIDR prefix takes precedence over a historical class default.
Yaad rakho: Class pehchaan sakte ho; calculation mein given slash follow karo.
"},{"id":"special","title":"Private, loopback and special IPv4 addresses","html":"Definition. Private IPv4 ranges are address blocks reserved for use within private networks rather than ordinary public Internet routing.
Simple meaning. Private IP ghar/college ke andar reuse ho sakta hai; globally unique public IP jaisa treat mat karo.
| Prefix | Range |
|---|---|
| 10.0.0.0/8 | 10.0.0.0–10.255.255.255 |
| 172.16.0.0/12 | 172.16.0.0–172.31.255.255 |
| 192.168.0.0/16 | 192.168.0.0–192.168.255.255 |
For example, 192.168.1.10 can be used in a private LAN.
Yaad rakho: 10 full; 172 only 16–31; 192.168 full.
"},{"id":"mask","title":"Subnet masks: 1 keeps a network bit; 0 marks a host bit","html":"Definition. A subnet mask is a 32-bit value with contiguous leading 1s for network-prefix positions and trailing 0s for host positions.
Simple meaning. Mask mein 1 = network part rakhna; 0 = host part zero karna.
| A | B | A AND B |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
The prefix mask contains 1s in network positions and 0s in host positions. Bitwise AND between the IP address and mask preserves the network bits and clears the host bits. For 192.168.10.77/26, the mask is 255.255.255.192 and the network is 192.168.10.64.
Yaad rakho: AND: dono 1 ho tabhi output 1.
"},{"id":"mask-1","title":"Prefix to mask · example 1","html":"Prefix length p=26.
Mask, h and N.
A prefix contains p leading ones; total combinations depend on the remaining host bits.
26 ones followed by 6 zeroes: 11111111.11111111.11111111.11000000.
| Mask bits | Decimal |
|---|---|
| 11111111 | 255 |
| 11111111 | 255 |
| 11111111 | 255 |
| 11000000 | 192 |
h=32−26=6.
N=26=64.
Final answer: Mask=255.255.255.192; host bits=6; total addresses=64.
Prefix length p=20.
Mask, h and N.
A prefix contains p leading ones; total combinations depend on the remaining host bits.
20 ones followed by 12 zeroes: 11111111.11111111.11110000.00000000.
| Mask bits | Decimal |
|---|---|
| 11111111 | 255 |
| 11111111 | 255 |
| 11110000 | 240 |
| 00000000 | 0 |
h=32−20=12.
N=212=4,096.
Final answer: Mask=255.255.240.0; host bits=12; total addresses=4,096.
Prefix length p=27.
Mask, h and N.
A prefix contains p leading ones; total combinations depend on the remaining host bits.
27 ones followed by 5 zeroes: 11111111.11111111.11111111.11100000.
| Mask bits | Decimal |
|---|---|
| 11111111 | 255 |
| 11111111 | 255 |
| 11111111 | 255 |
| 11100000 | 224 |
h=32−27=5.
N=25=32.
Final answer: Mask=255.255.255.224; host bits=5; total addresses=32.
Prefix length p=23.
Mask, h and N.
A prefix contains p leading ones; total combinations depend on the remaining host bits.
23 ones followed by 9 zeroes: 11111111.11111111.11111110.00000000.
| Mask bits | Decimal |
|---|---|
| 11111111 | 255 |
| 11111111 | 255 |
| 11111110 | 254 |
| 00000000 | 0 |
h=32−23=9.
N=29=512.
Final answer: Mask=255.255.254.0; host bits=9; total addresses=512.
Definition. Subnetting divides an address block into smaller prefix-defined blocks. In an ordinary subnet, the network and directed-broadcast addresses bound its usable host range.
Simple meaning. Pehle block nikalo; phir first aur last. Uske baad hosts.
A /26 has 32−26=6 host bits, so it contains 2⁶=64 total addresses. Under ordinary IPv4 subnet rules, one is the network address and one is directed broadcast, leaving 64−2=62 usable host addresses.
Yaad rakho: Total mein sab count; usable mein boundary ke 2 minus.
"},{"id":"block-1","title":"Complete block · 192.168.10.77/26","html":"IP/prefix = 192.168.10.77/26.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−26=6; N=26=64.
Mask=255.255.255.192; binary=11111111.11111111.11111111.11000000.
The changing octet is number 4. Its mask value is 192, so step=256−192=64. The IP value 77 lies in 64…127.
Keep earlier octets; place 64 in octet 4; set any later octets to zero. Network=192.168.10.64.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 11000000 | 11111111 | 11000000 | 192 |
| 2 | 10101000 | 11111111 | 10101000 | 168 |
| 3 | 00001010 | 11111111 | 00001010 | 10 |
| 4 | 01001101 | 11000000 | 01000000 | 64 |
Use 127 in the changing octet and 255 in any later octets. Broadcast=192.168.10.127.
First usable=192.168.10.65; last usable=192.168.10.126; count=64−2=62.
Final answer: Network 192.168.10.64; broadcast 192.168.10.127; 64 total, 62 ordinary usable; range 192.168.10.65–192.168.10.126.
IP/prefix = 200.140.64.48/26.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−26=6; N=26=64.
Mask=255.255.255.192; binary=11111111.11111111.11111111.11000000.
The changing octet is number 4. Its mask value is 192, so step=256−192=64. The IP value 48 lies in 0…63.
Keep earlier octets; place 0 in octet 4; set any later octets to zero. Network=200.140.64.0.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 11001000 | 11111111 | 11001000 | 200 |
| 2 | 10001100 | 11111111 | 10001100 | 140 |
| 3 | 01000000 | 11111111 | 01000000 | 64 |
| 4 | 00110000 | 11000000 | 00000000 | 0 |
Use 63 in the changing octet and 255 in any later octets. Broadcast=200.140.64.63.
First usable=200.140.64.1; last usable=200.140.64.62; count=64−2=62.
Final answer: Network 200.140.64.0; broadcast 200.140.64.63; 64 total, 62 ordinary usable; range 200.140.64.1–200.140.64.62.
IP/prefix = 193.20.35.40/26.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−26=6; N=26=64.
Mask=255.255.255.192; binary=11111111.11111111.11111111.11000000.
The changing octet is number 4. Its mask value is 192, so step=256−192=64. The IP value 40 lies in 0…63.
Keep earlier octets; place 0 in octet 4; set any later octets to zero. Network=193.20.35.0.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 11000001 | 11111111 | 11000001 | 193 |
| 2 | 00010100 | 11111111 | 00010100 | 20 |
| 3 | 00100011 | 11111111 | 00100011 | 35 |
| 4 | 00101000 | 11000000 | 00000000 | 0 |
Use 63 in the changing octet and 255 in any later octets. Broadcast=193.20.35.63.
First usable=193.20.35.1; last usable=193.20.35.62; count=64−2=62.
Final answer: Network 193.20.35.0; broadcast 193.20.35.63; 64 total, 62 ordinary usable; range 193.20.35.1–193.20.35.62.
IP/prefix = 192.168.1.150/27.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−27=5; N=25=32.
Mask=255.255.255.224; binary=11111111.11111111.11111111.11100000.
The changing octet is number 4. Its mask value is 224, so step=256−224=32. The IP value 150 lies in 128…159.
Keep earlier octets; place 128 in octet 4; set any later octets to zero. Network=192.168.1.128.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 11000000 | 11111111 | 11000000 | 192 |
| 2 | 10101000 | 11111111 | 10101000 | 168 |
| 3 | 00000001 | 11111111 | 00000001 | 1 |
| 4 | 10010110 | 11100000 | 10000000 | 128 |
Use 159 in the changing octet and 255 in any later octets. Broadcast=192.168.1.159.
First usable=192.168.1.129; last usable=192.168.1.158; count=32−2=30.
Final answer: Network 192.168.1.128; broadcast 192.168.1.159; 32 total, 30 ordinary usable; range 192.168.1.129–192.168.1.158.
IP/prefix = 10.4.9.55/28.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−28=4; N=24=16.
Mask=255.255.255.240; binary=11111111.11111111.11111111.11110000.
The changing octet is number 4. Its mask value is 240, so step=256−240=16. The IP value 55 lies in 48…63.
Keep earlier octets; place 48 in octet 4; set any later octets to zero. Network=10.4.9.48.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 00001010 | 11111111 | 00001010 | 10 |
| 2 | 00000100 | 11111111 | 00000100 | 4 |
| 3 | 00001001 | 11111111 | 00001001 | 9 |
| 4 | 00110111 | 11110000 | 00110000 | 48 |
Use 63 in the changing octet and 255 in any later octets. Broadcast=10.4.9.63.
First usable=10.4.9.49; last usable=10.4.9.62; count=16−2=14.
Final answer: Network 10.4.9.48; broadcast 10.4.9.63; 16 total, 14 ordinary usable; range 10.4.9.49–10.4.9.62.
IP/prefix = 172.16.35.123/20.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−20=12; N=212=4,096.
Mask=255.255.240.0; binary=11111111.11111111.11110000.00000000.
The changing octet is number 3. Its mask value is 240, so step=256−240=16. The IP value 35 lies in 32…47.
Keep earlier octets; place 32 in octet 3; set any later octets to zero. Network=172.16.32.0.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 10101100 | 11111111 | 10101100 | 172 |
| 2 | 00010000 | 11111111 | 00010000 | 16 |
| 3 | 00100011 | 11110000 | 00100000 | 32 |
| 4 | 01111011 | 00000000 | 00000000 | 0 |
Use 47 in the changing octet and 255 in any later octets. Broadcast=172.16.47.255.
First usable=172.16.32.1; last usable=172.16.47.254; count=4096−2=4094.
Final answer: Network 172.16.32.0; broadcast 172.16.47.255; 4,096 total, 4,094 ordinary usable; range 172.16.32.1–172.16.47.254.
IP/prefix = 10.20.5.130/23.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−23=9; N=29=512.
Mask=255.255.254.0; binary=11111111.11111111.11111110.00000000.
The changing octet is number 3. Its mask value is 254, so step=256−254=2. The IP value 5 lies in 4…5.
Keep earlier octets; place 4 in octet 3; set any later octets to zero. Network=10.20.4.0.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 00001010 | 11111111 | 00001010 | 10 |
| 2 | 00010100 | 11111111 | 00010100 | 20 |
| 3 | 00000101 | 11111110 | 00000100 | 4 |
| 4 | 10000010 | 00000000 | 00000000 | 0 |
Use 5 in the changing octet and 255 in any later octets. Broadcast=10.20.5.255.
First usable=10.20.4.1; last usable=10.20.5.254; count=512−2=510.
Final answer: Network 10.20.4.0; broadcast 10.20.5.255; 512 total, 510 ordinary usable; range 10.20.4.1–10.20.5.254.
IP/prefix = 192.168.2.129/25.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−25=7; N=27=128.
Mask=255.255.255.128; binary=11111111.11111111.11111111.10000000.
The changing octet is number 4. Its mask value is 128, so step=256−128=128. The IP value 129 lies in 128…255.
Keep earlier octets; place 128 in octet 4; set any later octets to zero. Network=192.168.2.128.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 11000000 | 11111111 | 11000000 | 192 |
| 2 | 10101000 | 11111111 | 10101000 | 168 |
| 3 | 00000010 | 11111111 | 00000010 | 2 |
| 4 | 10000001 | 10000000 | 10000000 | 128 |
Use 255 in the changing octet and 255 in any later octets. Broadcast=192.168.2.255.
First usable=192.168.2.129; last usable=192.168.2.254; count=128−2=126.
Final answer: Network 192.168.2.128; broadcast 192.168.2.255; 128 total, 126 ordinary usable; range 192.168.2.129–192.168.2.254.
IP/prefix = 192.0.2.6/30.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−30=2; N=22=4.
Mask=255.255.255.252; binary=11111111.11111111.11111111.11111100.
The changing octet is number 4. Its mask value is 252, so step=256−252=4. The IP value 6 lies in 4…7.
Keep earlier octets; place 4 in octet 4; set any later octets to zero. Network=192.0.2.4.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 11000000 | 11111111 | 11000000 | 192 |
| 2 | 00000000 | 11111111 | 00000000 | 0 |
| 3 | 00000010 | 11111111 | 00000010 | 2 |
| 4 | 00000110 | 11111100 | 00000100 | 4 |
Use 7 in the changing octet and 255 in any later octets. Broadcast=192.0.2.7.
First usable=192.0.2.5; last usable=192.0.2.6; count=4−2=2.
Final answer: Network 192.0.2.4; broadcast 192.0.2.7; 4 total, 2 ordinary usable; range 192.0.2.5–192.0.2.6.
Definition. A /31 prefix can address both endpoints of a point-to-point IPv4 link under RFC 3021; a /32 identifies one IPv4 address, commonly as a host route.
Simple meaning. Minus 2 har slash pe blindly mat lagao.
RFC 3021 permits both addresses of a /31 to be used by the two endpoints of an IPv4 point-to-point link. Applying the ordinary broadcast-subnet formula would incorrectly claim there are zero usable endpoints.
Yaad rakho: /31 special link; /32 single address.
"},{"id":"hosts-1","title":"Hosts to prefix · 25 hosts","html":"Required usable hosts H=25.
The smallest sufficient subnet, expressed as /p and mask.
Host demand excludes network and broadcast, so use the usable-host inequality. Choose the smallest h that satisfies it.
With h=4, usable=24−2=14, which is less than 25.
With h=5, usable=25−2=30, which is sufficient.
p=32−5=27; mask=255.255.255.224.
30−25=5.
Final answer: /27, mask 255.255.255.224; 30 usable, 5 spare usable addresses.
Required usable hosts H=64.
The smallest sufficient subnet, expressed as /p and mask.
Host demand excludes network and broadcast, so use the usable-host inequality. Choose the smallest h that satisfies it.
With h=6, usable=26−2=62, which is less than 64.
With h=7, usable=27−2=126, which is sufficient.
p=32−7=25; mask=255.255.255.128.
126−64=62.
Final answer: /25, mask 255.255.255.128; 126 usable, 62 spare usable addresses.
Required usable hosts H=1000.
The smallest sufficient subnet, expressed as /p and mask.
Host demand excludes network and broadcast, so use the usable-host inequality. Choose the smallest h that satisfies it.
With h=9, usable=29−2=510, which is less than 1000.
With h=10, usable=210−2=1022, which is sufficient.
p=32−10=22; mask=255.255.252.0.
1022−1000=22.
Final answer: /22, mask 255.255.252.0; 1022 usable, 22 spare usable addresses.
Parent=192.168.8.0/24; required equal-size subnets=4.
Borrowed bits, new mask/prefix and complete subnet ranges.
Borrow the smallest s giving enough subnets. In modern subnetting, zero and all-ones subnets are valid; do not subtract two from the subnet count.
s=2, because 22=4 is the smallest power of two at least 4.
/24+2=/26; mask=255.255.255.192. Each subnet contains 64 total and 62 usable addresses.
| Subnet | First/network | Last/broadcast | Usable range |
|---|---|---|---|
| 192.168.8.0/26 | 192.168.8.0 | 192.168.8.63 | 192.168.8.1–192.168.8.62 |
| 192.168.8.64/26 | 192.168.8.64 | 192.168.8.127 | 192.168.8.65–192.168.8.126 |
| 192.168.8.128/26 | 192.168.8.128 | 192.168.8.191 | 192.168.8.129–192.168.8.190 |
| 192.168.8.192/26 | 192.168.8.192 | 192.168.8.255 | 192.168.8.193–192.168.8.254 |
4 available−4 required=0 extra equal-size subnet(s).
Final answer: /26; 4 subnets; 62 usable hosts each.
Parent=172.16.0.0/16; required equal-size subnets=8.
Borrowed bits, new mask/prefix and complete subnet ranges.
Borrow the smallest s giving enough subnets. In modern subnetting, zero and all-ones subnets are valid; do not subtract two from the subnet count.
s=3, because 23=8 is the smallest power of two at least 8.
/16+3=/19; mask=255.255.224.0. Each subnet contains 8192 total and 8190 usable addresses.
| Subnet | First/network | Last/broadcast | Usable range |
|---|---|---|---|
| 172.16.0.0/19 | 172.16.0.0 | 172.16.31.255 | 172.16.0.1–172.16.31.254 |
| 172.16.32.0/19 | 172.16.32.0 | 172.16.63.255 | 172.16.32.1–172.16.63.254 |
| 172.16.64.0/19 | 172.16.64.0 | 172.16.95.255 | 172.16.64.1–172.16.95.254 |
| 172.16.96.0/19 | 172.16.96.0 | 172.16.127.255 | 172.16.96.1–172.16.127.254 |
| 172.16.128.0/19 | 172.16.128.0 | 172.16.159.255 | 172.16.128.1–172.16.159.254 |
| 172.16.160.0/19 | 172.16.160.0 | 172.16.191.255 | 172.16.160.1–172.16.191.254 |
| 172.16.192.0/19 | 172.16.192.0 | 172.16.223.255 | 172.16.192.1–172.16.223.254 |
| 172.16.224.0/19 | 172.16.224.0 | 172.16.255.255 | 172.16.224.1–172.16.255.254 |
8 available−8 required=0 extra equal-size subnet(s).
Final answer: /19; 8 subnets; 8190 usable hosts each.
Parent=10.0.0.0/24; required equal-size subnets=6.
Borrowed bits, new mask/prefix and complete subnet ranges.
Borrow the smallest s giving enough subnets. In modern subnetting, zero and all-ones subnets are valid; do not subtract two from the subnet count.
s=3, because 23=8 is the smallest power of two at least 6.
/24+3=/27; mask=255.255.255.224. Each subnet contains 32 total and 30 usable addresses.
| Subnet | First/network | Last/broadcast | Usable range |
|---|---|---|---|
| 10.0.0.0/27 | 10.0.0.0 | 10.0.0.31 | 10.0.0.1–10.0.0.30 |
| 10.0.0.32/27 | 10.0.0.32 | 10.0.0.63 | 10.0.0.33–10.0.0.62 |
| 10.0.0.64/27 | 10.0.0.64 | 10.0.0.95 | 10.0.0.65–10.0.0.94 |
| 10.0.0.96/27 | 10.0.0.96 | 10.0.0.127 | 10.0.0.97–10.0.0.126 |
| 10.0.0.128/27 | 10.0.0.128 | 10.0.0.159 | 10.0.0.129–10.0.0.158 |
| 10.0.0.160/27 | 10.0.0.160 | 10.0.0.191 | 10.0.0.161–10.0.0.190 |
| 10.0.0.192/27 | 10.0.0.192 | 10.0.0.223 | 10.0.0.193–10.0.0.222 |
| 10.0.0.224/27 | 10.0.0.224 | 10.0.0.255 | 10.0.0.225–10.0.0.254 |
8 available−6 required=2 extra equal-size subnet(s).
Final answer: /27; 8 subnets; 30 usable hosts each.
Definition. CIDR expresses address blocks using an explicit prefix length; VLSM uses different subnet prefix lengths to match different size requirements within an allocated block.
Simple meaning. CIDR = slash se block; VLSM = departments ko need ke hisaab se different block sizes.
CIDR identifies a classless address block with an explicit /p prefix. VLSM permits subnets of different prefix lengths inside a larger block. Matching subnet sizes to requirements reduces waste while maintaining non-overlapping aligned allocations.
Yaad rakho: Hosts bola toh +2 allowance; addresses bola toh total count directly.
"},{"id":"vlsm-1","title":"Complete VLSM allocation · example 1","html":"| Department | Usable hosts |
|---|---|
| A | 100 |
| B | 50 |
| C | 25 |
| D | 10 |
An aligned subnet for each department, usable ranges, broadcasts and remaining addresses.
Choose the smallest sufficient subnet per requirement. Align each start to its block size and never overlap a previously assigned range.
A (100) → B (50) → C (25) → D (10)
100 hosts require 27−2=126 usable; a smaller /26 would give 62. Choose /25. The next aligned block is 192.168.10.0/25; it ends at 192.168.10.127.
50 hosts require 26−2=62 usable; a smaller /27 would give 30. Choose /26. The next aligned block is 192.168.10.128/26; it ends at 192.168.10.191.
25 hosts require 25−2=30 usable; a smaller /28 would give 14. Choose /27. The next aligned block is 192.168.10.192/27; it ends at 192.168.10.223.
10 hosts require 24−2=14 usable; a smaller /29 would give 6. Choose /28. The next aligned block is 192.168.10.224/28; it ends at 192.168.10.239.
| Group | Hosts | Network/prefix | Mask | Usable range | Broadcast | Spare usable |
|---|---|---|---|---|---|---|
| A | 100 | 192.168.10.0/25 | 255.255.255.128 | 192.168.10.1–192.168.10.126 | 192.168.10.127 | 26 |
| B | 50 | 192.168.10.128/26 | 255.255.255.192 | 192.168.10.129–192.168.10.190 | 192.168.10.191 | 12 |
| C | 25 | 192.168.10.192/27 | 255.255.255.224 | 192.168.10.193–192.168.10.222 | 192.168.10.223 | 5 |
| D | 10 | 192.168.10.224/28 | 255.255.255.240 | 192.168.10.225–192.168.10.238 | 192.168.10.239 | 4 |
Reserved=128+64+32+16=240. Parent total=256. Unallocated=256−240=16.
Final answer: 240 total addresses reserved; 16 unallocated. The table gives one valid complete allocation.
| Department | Usable hosts |
|---|---|
| Research | 60 |
| Admin | 25 |
| IoT | 12 |
An aligned subnet for each department, usable ranges, broadcasts and remaining addresses.
Choose the smallest sufficient subnet per requirement. Align each start to its block size and never overlap a previously assigned range.
Research (60) → Admin (25) → IoT (12)
60 hosts require 26−2=62 usable; a smaller /27 would give 30. Choose /26. The next aligned block is 10.0.0.0/26; it ends at 10.0.0.63.
25 hosts require 25−2=30 usable; a smaller /28 would give 14. Choose /27. The next aligned block is 10.0.0.64/27; it ends at 10.0.0.95.
12 hosts require 24−2=14 usable; a smaller /29 would give 6. Choose /28. The next aligned block is 10.0.0.96/28; it ends at 10.0.0.111.
| Group | Hosts | Network/prefix | Mask | Usable range | Broadcast | Spare usable |
|---|---|---|---|---|---|---|
| Research | 60 | 10.0.0.0/26 | 255.255.255.192 | 10.0.0.1–10.0.0.62 | 10.0.0.63 | 2 |
| Admin | 25 | 10.0.0.64/27 | 255.255.255.224 | 10.0.0.65–10.0.0.94 | 10.0.0.95 | 5 |
| IoT | 12 | 10.0.0.96/28 | 255.255.255.240 | 10.0.0.97–10.0.0.110 | 10.0.0.111 | 2 |
Reserved=64+32+16=112. Parent total=256. Unallocated=256−112=144.
Final answer: 112 total addresses reserved; 144 unallocated. The table gives one valid complete allocation.
| Department | Usable hosts |
|---|---|
| Lab | 200 |
| Staff | 100 |
| Guests | 50 |
An aligned subnet for each department, usable ranges, broadcasts and remaining addresses.
Choose the smallest sufficient subnet per requirement. Align each start to its block size and never overlap a previously assigned range.
Lab (200) → Staff (100) → Guests (50)
200 hosts require 28−2=254 usable; a smaller /25 would give 126. Choose /24. The next aligned block is 172.16.8.0/24; it ends at 172.16.8.255.
100 hosts require 27−2=126 usable; a smaller /26 would give 62. Choose /25. The next aligned block is 172.16.9.0/25; it ends at 172.16.9.127.
50 hosts require 26−2=62 usable; a smaller /27 would give 30. Choose /26. The next aligned block is 172.16.9.128/26; it ends at 172.16.9.191.
| Group | Hosts | Network/prefix | Mask | Usable range | Broadcast | Spare usable |
|---|---|---|---|---|---|---|
| Lab | 200 | 172.16.8.0/24 | 255.255.255.0 | 172.16.8.1–172.16.8.254 | 172.16.8.255 | 54 |
| Staff | 100 | 172.16.9.0/25 | 255.255.255.128 | 172.16.9.1–172.16.9.126 | 172.16.9.127 | 26 |
| Guests | 50 | 172.16.9.128/26 | 255.255.255.192 | 172.16.9.129–172.16.9.190 | 172.16.9.191 | 12 |
Reserved=256+128+64=448. Parent total=512. Unallocated=512−448=64.
Final answer: 448 total addresses reserved; 64 unallocated. The table gives one valid complete allocation.
Definition. Route aggregation represents several contiguous networks by a shorter common prefix, reducing the number of advertised routing entries.
Simple meaning. Side-by-side blocks ko ek larger boundary mein summarize karo; gap chhupana nahi.
The blocks must be contiguous with no gaps and together form a power-of-two address count starting on the corresponding larger-block boundary. Their shared leading bits define the aggregate prefix.
Yaad rakho: Consecutive + power of two + aligned start.
"},{"id":"aggregate-1","title":"Exact route aggregation · example 1","html":"One prefix covering exactly the same address range.
Adding sizes gives the candidate prefix. Contiguity and start-boundary alignment must also be satisfied.
Each input block has 256 addresses.
4×256=1024.
1024=210; prefix=32−10=/22.
Inputs cover 192.168.4.0–192.168.7.255, without gaps. 192.168.4.0/22 covers the same start and end and is correctly aligned.
Final answer: 192.168.4.0/22
One prefix covering exactly the same address range.
Adding sizes gives the candidate prefix. Contiguity and start-boundary alignment must also be satisfied.
Each input block has 128 addresses.
2×128=256.
256=28; prefix=32−8=/24.
Inputs cover 10.20.0.0–10.20.0.255, without gaps. 10.20.0.0/24 covers the same start and end and is correctly aligned.
Final answer: 10.20.0.0/24
One prefix covering exactly the same address range.
Adding sizes gives the candidate prefix. Contiguity and start-boundary alignment must also be satisfied.
Each input block has 512 addresses.
2×512=1024.
1024=210; prefix=32−10=/22.
Inputs cover 172.16.8.0–172.16.11.255, without gaps. 172.16.8.0/22 covers the same start and end and is correctly aligned.
Final answer: 172.16.8.0/22
Definition. Longest-prefix matching selects the matching forwarding-table entry with the greatest number of prefix bits.
Simple meaning. Pehle match check; matching entries mein sabse bada slash choose.
A router selects the most specific route whose prefix contains the destination. For 10.1.2.77, a matching 10.1.2.0/24 route takes precedence over 10.0.0.0/8 and 0.0.0.0/0.
Yaad rakho: Bada slash only among matches.
"},{"id":"match-1","title":"Forwarding table · example 1","html":"| Network/prefix | Next hop |
|---|---|
| 0.0.0.0/0 | R0 |
| 10.0.0.0/8 | R1 |
| 10.1.0.0/16 | R2 |
| 10.1.2.0/24 | R3 |
| 10.1.2.128/25 | R4 |
Destination=10.1.2.150
The matching entries and selected next hop.
Choose the largest prefix length among entries containing the destination.
Prefix length is considered only after confirming an address-range match.
| Route | Destination inside? |
|---|---|
| 0.0.0.0/0 | Yes |
| 10.0.0.0/8 | Yes |
| 10.1.0.0/16 | Yes |
| 10.1.2.0/24 | Yes |
| 10.1.2.128/25 | Yes |
0.0.0.0/0, 10.0.0.0/8, 10.1.0.0/16, 10.1.2.0/24, 10.1.2.128/25. The longest is 10.1.2.128/25.
10.1.2.128/25 uses R4.
Final answer: Forward via R4 using 10.1.2.128/25.
| Network/prefix | Next hop |
|---|---|
| 0.0.0.0/0 | R0 |
| 10.0.0.0/8 | R1 |
| 10.1.0.0/16 | R2 |
| 10.1.2.0/24 | R3 |
| 10.1.2.128/25 | R4 |
Destination=10.1.2.77
The matching entries and selected next hop.
Choose the largest prefix length among entries containing the destination.
Prefix length is considered only after confirming an address-range match.
| Route | Destination inside? |
|---|---|
| 0.0.0.0/0 | Yes |
| 10.0.0.0/8 | Yes |
| 10.1.0.0/16 | Yes |
| 10.1.2.0/24 | Yes |
| 10.1.2.128/25 | No |
0.0.0.0/0, 10.0.0.0/8, 10.1.0.0/16, 10.1.2.0/24. The longest is 10.1.2.0/24.
10.1.2.0/24 uses R3.
Final answer: Forward via R3 using 10.1.2.0/24.
| Network/prefix | Next hop |
|---|---|
| 0.0.0.0/0 | R0 |
| 10.0.0.0/8 | R1 |
| 10.1.0.0/16 | R2 |
| 10.1.2.0/24 | R3 |
| 10.1.2.128/25 | R4 |
Destination=192.0.2.9
The matching entries and selected next hop.
Choose the largest prefix length among entries containing the destination.
Prefix length is considered only after confirming an address-range match.
| Route | Destination inside? |
|---|---|
| 0.0.0.0/0 | Yes |
| 10.0.0.0/8 | No |
| 10.1.0.0/16 | No |
| 10.1.2.0/24 | No |
| 10.1.2.128/25 | No |
0.0.0.0/0. The longest is 0.0.0.0/0.
0.0.0.0/0 uses R0.
Final answer: Forward via R0 using 0.0.0.0/0.
Definition. IPv6 uses 128-bit addresses, conventionally represented by eight groups of four hexadecimal digits separated by colons.
Simple meaning. IPv4 mein 4 decimal boxes; IPv6 mein 8 hexadecimal groups.
| Hex | Decimal | Four bits |
|---|---|---|
| 0 | 0 | 0000 |
| 1 | 1 | 0001 |
| 2 | 2 | 0010 |
| 3 | 3 | 0011 |
| 4 | 4 | 0100 |
| 5 | 5 | 0101 |
| 6 | 6 | 0110 |
| 7 | 7 | 0111 |
| 8 | 8 | 1000 |
| 9 | 9 | 1001 |
| a | 10 | 1010 |
| b | 11 | 1011 |
| c | 12 | 1100 |
| d | 13 | 1101 |
| e | 14 | 1110 |
| f | 15 | 1111 |
An IPv6 address contains 128 bits, written as eight 16-bit hexadecimal groups. Leading zeros inside a group can be removed. One contiguous run of all-zero groups can be represented with ::, used at most once.
Yaad rakho: 8 groups × 4 hex digits × 4 bits = 128.
"},{"id":"expansion","title":"Expand :: by counting visible groups","html":"Definition. IPv6 expansion restores exactly eight groups, each written with four hexadecimal digits.
Simple meaning. Visible groups count karo; 8 mein se minus; utne zero groups :: ki jagah.
Count all explicitly written groups. Insert 8 minus this count all-zero groups at ::, then left-pad every group to four hexadecimal digits. The result must contain exactly eight groups.
Yaad rakho: Already-written 0 ko bhi count karna.
"},{"id":"expand-1","title":"Expand IPv6 · example 1","html":"Given address: 0:15::1:12:1213.
The full 128-bit address in eight-group hexadecimal notation.
A written 0 is a visible group. :: represents only the omitted all-zero groups.
Left: 0, 15 (2 group(s)). Right: 1, 12, 1213 (3 group(s)).
Visible=2+3=5. Missing=8−5=3.
0:15:0:0:0:1:12:1213.
| Group number | Four-digit group |
|---|---|
| 1 | 0000 |
| 2 | 0015 |
| 3 | 0000 |
| 4 | 0000 |
| 5 | 0000 |
| 6 | 0001 |
| 7 | 0012 |
| 8 | 1213 |
0000:0015:0000:0000:0000:0001:0012:1213.
Final answer: 0000:0015:0000:0000:0000:0001:0012:1213
Given address: 2001:db8:0:1::9.
The full 128-bit address in eight-group hexadecimal notation.
A written 0 is a visible group. :: represents only the omitted all-zero groups.
Left: 2001, db8, 0, 1 (4 group(s)). Right: 9 (1 group(s)).
Visible=4+1=5. Missing=8−5=3.
2001:db8:0:1:0:0:0:9.
| Group number | Four-digit group |
|---|---|
| 1 | 2001 |
| 2 | 0db8 |
| 3 | 0000 |
| 4 | 0001 |
| 5 | 0000 |
| 6 | 0000 |
| 7 | 0000 |
| 8 | 0009 |
2001:0db8:0000:0001:0000:0000:0000:0009.
Final answer: 2001:0db8:0000:0001:0000:0000:0000:0009
Given address: ::1.
The full 128-bit address in eight-group hexadecimal notation.
A written 0 is a visible group. :: represents only the omitted all-zero groups.
Left: none (0 group(s)). Right: 1 (1 group(s)).
Visible=0+1=1. Missing=8−1=7.
0:0:0:0:0:0:0:1.
| Group number | Four-digit group |
|---|---|
| 1 | 0000 |
| 2 | 0000 |
| 3 | 0000 |
| 4 | 0000 |
| 5 | 0000 |
| 6 | 0000 |
| 7 | 0000 |
| 8 | 0001 |
0000:0000:0000:0000:0000:0000:0000:0001.
Final answer: 0000:0000:0000:0000:0000:0000:0000:0001
Given address: fe80::abcd:ef12.
The full 128-bit address in eight-group hexadecimal notation.
A written 0 is a visible group. :: represents only the omitted all-zero groups.
Left: fe80 (1 group(s)). Right: abcd, ef12 (2 group(s)).
Visible=1+2=3. Missing=8−3=5.
fe80:0:0:0:0:0:abcd:ef12.
| Group number | Four-digit group |
|---|---|
| 1 | fe80 |
| 2 | 0000 |
| 3 | 0000 |
| 4 | 0000 |
| 5 | 0000 |
| 6 | 0000 |
| 7 | abcd |
| 8 | ef12 |
fe80:0000:0000:0000:0000:0000:abcd:ef12.
Final answer: fe80:0000:0000:0000:0000:0000:abcd:ef12
Definition. Canonical IPv6 text formatting removes leading zeros and compresses the longest eligible run of consecutive zero groups.
Simple meaning. Group ke left wale zero hatao; sabse lambi zero-chain ko :: banao.
Compress the leftmost of the equal longest runs. Leading zeros in groups are removed and hexadecimal letters are lowercase. A single zero group is retained as 0 rather than compressed to :: in canonical text.
Yaad rakho: Longest first; tie hua toh leftmost.
"},{"id":"compress-1","title":"Compress IPv6 · example 1","html":"2001:0db8:0000:0000:0012:0000:0000:0001.
A canonical shortened spelling of the same address.
Remove leading zeroes; compress the longest run of at least two zero groups; choose the leftmost tied run.
These rules preserve all 128 bits while removing redundant writing.
2001:db8:0:0:12:0:0:1.
groups 3–4: 2 zero group(s); groups 6–7: 2 zero group(s).
Compress groups 3–4.
2001:db8::12:0:0:1.
Final answer: 2001:db8::12:0:0:1
2001:0db8:0000:0001:0002:0003:0004:0005.
A canonical shortened spelling of the same address.
Remove leading zeroes; compress the longest run of at least two zero groups; choose the leftmost tied run.
These rules preserve all 128 bits while removing redundant writing.
2001:db8:0:1:2:3:4:5.
groups 3–3: 1 zero group(s).
No run has two or more groups, so do not use ::.
2001:db8:0:1:2:3:4:5.
Final answer: 2001:db8:0:1:2:3:4:5
0000:0000:0000:0000:0000:0000:0000:0000.
A canonical shortened spelling of the same address.
Remove leading zeroes; compress the longest run of at least two zero groups; choose the leftmost tied run.
These rules preserve all 128 bits while removing redundant writing.
0:0:0:0:0:0:0:0.
groups 1–8: 8 zero group(s).
Compress groups 1–8.
::.
Final answer: ::
Definition. The IPv6 base header is a fixed 40-byte structure containing addressing and forwarding-control fields, followed by optional extension headers and upper-layer data.
Simple meaning. Base header fixed 40 bytes; extra information extension headers mein.
| Field | Bits | Purpose |
|---|---|---|
| Version | 4 | Value 6 |
| Traffic Class | 8 | Traffic treatment/congestion-related marking |
| Flow Label | 20 | Identifies a flow for appropriate handling |
| Payload Length | 16 | Bytes after the base header; includes extension headers when present |
| Next Header | 8 | Type of the next extension header or upper-layer protocol |
| Hop Limit | 8 | Decremented on forwarding; limits packet lifetime in hops |
| Source address | 128 | Sending interface address |
| Destination address | 128 | Destination address |
| Field | Bits | Purpose |
|---|---|---|
| Version | 4 | Value 6 |
| Traffic Class | 8 | Traffic treatment/congestion-related marking |
| Flow Label | 20 | Identifies a flow for appropriate handling |
| Payload Length | 16 | Bytes after the base header; includes extension headers when present |
| Next Header | 8 | Type of the next extension header or upper-layer protocol |
| Hop Limit | 8 | Decremented on forwarding; limits packet lifetime in hops |
| Source address | 128 | Sending interface address |
| Destination address | 128 | Destination address |
Yaad rakho: IPv6: 40 fixed; payload field excludes those 40.
"},{"id":"compare","title":"IPv4 versus IPv6 headers and address types","html":"Definition. IPv6 extends address size and reorganizes IP forwarding information into a fixed base header with extension headers.
Simple meaning. Sirf address bada nahi; header ka arrangement bhi badla.
| Feature | IPv4 | IPv6 |
|---|---|---|
| Address width | 32 bits | 128 bits |
| Base/header size | 20–60 bytes depending on options | 40-byte fixed base header |
| Length field | Total length includes IPv4 header | Payload Length excludes 40-byte base header |
| Lifetime field | TTL | Hop Limit |
| Following protocol | Protocol field | Next Header; may identify an extension |
| Header checksum | Present | No base-header checksum |
| Options | Inside IPv4 header | Typically extension headers |
| Fragmentation fields | In IPv4 header; routers may fragment if permitted | Fragment extension header; fragmentation by source, not routers |
| Feature | IPv4 | IPv6 |
|---|---|---|
| Address width | 32 bits | 128 bits |
| Base/header size | 20–60 bytes depending on options | 40-byte fixed base header |
| Length field | Total length includes IPv4 header | Payload Length excludes 40-byte base header |
| Lifetime field | TTL | Hop Limit |
| Following protocol | Protocol field | Next Header; may identify an extension |
| Header checksum | Present | No base-header checksum |
| Options | Inside IPv4 header | Typically extension headers |
| Fragmentation fields | In IPv4 header; routers may fragment if permitted | Fragment extension header; fragmentation by source, not routers |
Yaad rakho: v4 total length; v6 payload length.
"}]},{"id":"mapping","title":"ARP, RARP and DHCP","scope":"TEE · shared foundations and Units 3–4 · current syllabus 702AI0C023","sections":[{"id":"arp","title":"ARP: find the local next-hop MAC address","html":"Definition. Address Resolution Protocol (ARP) resolves a known IPv4 address to a link-layer hardware address on the local network.
Simple meaning. IP pata hai; local delivery ke liye MAC chahiye.
ARP maps a known local next-hop IPv4 address to a hardware address. A sender broadcasts a request when the mapping is missing; the address owner replies, and the mapping is cached. For a remote IP destination, the sender resolves its gateway’s local MAC address, then forwards the packet toward that gateway.
Yaad rakho: Remote packet ka next-hop MAC gateway ka hota hai.
"},{"id":"next-hop-1","title":"ARP decision · example 1","html":"H=192.168.1.10/24; gateway=192.168.1.1; destination=192.168.1.20.
Local/remote decision and ARP target.
Compare source and destination network prefixes using the source mask. ARP resolves the selected local next-hop IP.
The IP destination stays the final destination; the Ethernet next-hop address depends on whether that destination is local.
With /24, H is in 192.168.1.0–192.168.1.255.
192.168.1.20 is inside that block, so delivery is local.
Resolve 192.168.1.20 using ARP.
IP destination remains 192.168.1.20; frame destination is that host’s MAC.
Final answer: ARP target: 192.168.1.20.
H=192.168.1.10/24; gateway=192.168.1.1; destination=203.0.113.7.
Local/remote decision and ARP target.
Compare source and destination network prefixes using the source mask. ARP resolves the selected local next-hop IP.
The IP destination stays the final destination; the Ethernet next-hop address depends on whether that destination is local.
With /24, H is in 192.168.1.0–192.168.1.255.
203.0.113.7 is outside that block, so use a router.
Resolve 192.168.1.1 using ARP.
IP destination remains 203.0.113.7; frame destination is the gateway’s MAC.
Final answer: ARP target: 192.168.1.1.
H=192.168.1.10/24; gateway=192.168.1.1; destination=192.168.2.20.
Local/remote decision and ARP target.
Compare source and destination network prefixes using the source mask. ARP resolves the selected local next-hop IP.
The IP destination stays the final destination; the Ethernet next-hop address depends on whether that destination is local.
With /24, H is in 192.168.1.0–192.168.1.255.
192.168.2.20 is outside that block, so use a router.
Resolve 192.168.1.1 using ARP.
IP destination remains 192.168.2.20; frame destination is the gateway’s MAC.
Final answer: ARP target: 192.168.1.1.
Definition. Reverse ARP (RARP) allows a client that knows its hardware address to obtain its IPv4 address from a RARP server.
Simple meaning. ARP: IP se MAC. RARP: MAC se IP.
| Protocol | Known | Needed | Use |
|---|---|---|---|
| ARP | Local next-hop IPv4 | MAC/hardware address | Local frame delivery |
| RARP | Client hardware address | Client IPv4 | Historical boot-time assignment |
Yaad rakho: RARP reverse karta hai; DHCP configuration badhata hai.
"},{"id":"dhcp","title":"DHCP: automatic configuration through DORA","html":"Definition. Dynamic Host Configuration Protocol (DHCP) provides IP configuration parameters to clients and can lease reusable addresses for limited periods.
Simple meaning. DORA = Discover, Offer, Request, Acknowledge.
DHCP reduces manual configuration and supports controlled address reuse through leases.
Yaad rakho: Discover–Offer–Request–ACK; pehle D, phir O, phir R, phir A.
"}]},{"id":"routing","title":"Routing: build the shortest path one step at a time","scope":"TEE · shared foundations and Units 3–4 · current syllabus 702AI0C023","sections":[{"id":"vocabulary","title":"Routing, forwarding, distance and next hop","html":"Definition. Routing discovers or computes paths through a network; forwarding uses a table to send an individual packet toward the selected next hop.
Simple meaning. Routing table banata hai; forwarding us table ko use karta hai.
Routing determines suitable network paths and produces routing/forwarding information. Forwarding applies that information to a particular packet. If A has learned that network X is reached through B, computing that choice is routing and sending a packet to B is forwarding.
Yaad rakho: Predecessor = pichhla node; next hop = source ke baad pehla node.
"},{"id":"relaxation","title":"Relaxation: replace an old route only when the candidate is cheaper","html":"Definition. Relaxing an edge u→v compares the existing distance to v with a candidate path that reaches u first and then crosses the edge to v.
Simple meaning. Purana cost versus u tak ka cost + u→v edge. Jo chhota, woh rakho.
Relaxation evaluates candidate=d[u]+w(u,v). If the candidate is less than d[v], it updates d[v] and records u as predecessor. Otherwise the existing route is retained.
Yaad rakho: Add path + edge, compare, then replace only if smaller.
"},{"id":"relax-1","title":"One edge relaxation · example 1","html":"d[U]=4; w(U,V)=3; old d[V]=10.
New d[V] and predecessor action.
The candidate contains both the cost to U and the final edge to V.
4+(3)=7.
min(10,7)=7.
Set predecessor[V]=U.
Final answer: New d[V]=7.
d[U]=5; w(U,V)=4; old d[V]=7.
New d[V] and predecessor action.
The candidate contains both the cost to U and the final edge to V.
5+(4)=9.
min(7,9)=7.
Keep the existing predecessor; the candidate does not improve the route.
Final answer: New d[V]=7.
d[U]=6; w(U,V)=-2; old d[V]=9.
New d[V] and predecessor action.
The candidate contains both the cost to U and the final edge to V.
6+(-2)=4.
min(9,4)=4.
Set predecessor[V]=U.
Final answer: New d[V]=4.
Definition. Dijkstra’s algorithm finds shortest paths from a source in a graph whose edge weights are nonnegative.
Simple meaning. Sabse chhota unfinished distance select karo, phir uske neighbours improve karo.
The standard algorithm requires nonnegative edge weights.
Yaad rakho: Minimum tentative total choose; negative edges pe Dijkstra nahi.
"},{"id":"dijkstra-1","title":"Dijkstra · complete example 1","html":"| Edge | Cost |
|---|---|
| A→B | 4 |
| A→C | 2 |
| C→B | 1 |
| B→D | 3 |
| C→D | 7 |
Source=A; target=D. Edges are directed; do not add reverse edges.
The path A to D, its total cost and intermediate work.
Use Dijkstra because all weights are nonnegative. Select the smallest unsettled total distance, not the cheapest isolated edge.
d[A]=0; every other distance=∞. No predecessors are known.
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Tentative |
| B | ∞ | — | Tentative |
| C | ∞ | — | Tentative |
| D | ∞ | — | Tentative |
Unsettled candidates before selection: A=0, B=∞, C=∞, D=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| A→B | ∞ | 0+(4)=4 | Improve to 4; predecessor=A |
| A→C | ∞ | 0+(2)=2 | Improve to 2; predecessor=A |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 4 | A | Tentative |
| C | 2 | A | Tentative |
| D | ∞ | — | Tentative |
Unsettled candidates before selection: B=4, C=2, D=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| C→B | 4 | 2+(1)=3 | Improve to 3; predecessor=C |
| C→D | ∞ | 2+(7)=9 | Improve to 9; predecessor=C |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | C | Tentative |
| C | 2 | A | Settled |
| D | 9 | C | Tentative |
Unsettled candidates before selection: B=3, D=9. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| B→D | 9 | 3+(3)=6 | Improve to 6; predecessor=B |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | C | Settled |
| C | 2 | A | Settled |
| D | 6 | B | Tentative |
Unsettled candidates before selection: D=6. Select the minimum; ties use alphabetical order.
No outgoing edges to relax.
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | C | Settled |
| C | 2 | A | Settled |
| D | 6 | B | Settled |
Trace backwards: D ← B ← C ← A. Reverse it: A → C → B → D.
Edge sum: 2 + 1 + 3 = 6.
Final answer: A → C → B → D; cost=6.
| Edge | Cost |
|---|---|
| A—B | 2 |
| A—C | 5 |
| B—C | 1 |
| B—D | 4 |
| C—D | 1 |
| D—E | 3 |
| C—E | 8 |
Source=A; target=E. Each edge can be traversed in either direction.
The path A to E, its total cost and intermediate work.
Use Dijkstra because all weights are nonnegative. Select the smallest unsettled total distance, not the cheapest isolated edge.
d[A]=0; every other distance=∞. No predecessors are known.
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Tentative |
| B | ∞ | — | Tentative |
| C | ∞ | — | Tentative |
| D | ∞ | — | Tentative |
| E | ∞ | — | Tentative |
Unsettled candidates before selection: A=0, B=∞, C=∞, D=∞, E=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| A→B | ∞ | 0+(2)=2 | Improve to 2; predecessor=A |
| A→C | ∞ | 0+(5)=5 | Improve to 5; predecessor=A |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 2 | A | Tentative |
| C | 5 | A | Tentative |
| D | ∞ | — | Tentative |
| E | ∞ | — | Tentative |
Unsettled candidates before selection: B=2, C=5, D=∞, E=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| B→C | 5 | 2+(1)=3 | Improve to 3; predecessor=B |
| B→D | ∞ | 2+(4)=6 | Improve to 6; predecessor=B |
| B→A | 0 | 2+(2)=4 | Already settled; retain |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 2 | A | Settled |
| C | 3 | B | Tentative |
| D | 6 | B | Tentative |
| E | ∞ | — | Tentative |
Unsettled candidates before selection: C=3, D=6, E=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| C→D | 6 | 3+(1)=4 | Improve to 4; predecessor=C |
| C→E | ∞ | 3+(8)=11 | Improve to 11; predecessor=C |
| C→A | 0 | 3+(5)=8 | Already settled; retain |
| C→B | 2 | 3+(1)=4 | Already settled; retain |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 2 | A | Settled |
| C | 3 | B | Settled |
| D | 4 | C | Tentative |
| E | 11 | C | Tentative |
Unsettled candidates before selection: D=4, E=11. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| D→E | 11 | 4+(3)=7 | Improve to 7; predecessor=D |
| D→B | 2 | 4+(4)=8 | Already settled; retain |
| D→C | 3 | 4+(1)=5 | Already settled; retain |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 2 | A | Settled |
| C | 3 | B | Settled |
| D | 4 | C | Settled |
| E | 7 | D | Tentative |
Unsettled candidates before selection: E=7. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| E→D | 4 | 7+(3)=10 | Already settled; retain |
| E→C | 3 | 7+(8)=15 | Already settled; retain |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 2 | A | Settled |
| C | 3 | B | Settled |
| D | 4 | C | Settled |
| E | 7 | D | Settled |
Trace backwards: E ← D ← C ← B ← A. Reverse it: A → B → C → D → E.
Edge sum: 2 + 1 + 1 + 3 = 7.
Final answer: A → B → C → D → E; cost=7.
| Edge | Cost |
|---|---|
| A→B | 3 |
| A→C | 3 |
| B→D | 2 |
| C→D | 2 |
Source=A; target=E. Edges are directed; do not add reverse edges.
The path A to E, its total cost and intermediate work.
Use Dijkstra because all weights are nonnegative. Select the smallest unsettled total distance, not the cheapest isolated edge.
d[A]=0; every other distance=∞. No predecessors are known.
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Tentative |
| B | ∞ | — | Tentative |
| C | ∞ | — | Tentative |
| D | ∞ | — | Tentative |
| E | ∞ | — | Tentative |
Unsettled candidates before selection: A=0, B=∞, C=∞, D=∞, E=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| A→B | ∞ | 0+(3)=3 | Improve to 3; predecessor=A |
| A→C | ∞ | 0+(3)=3 | Improve to 3; predecessor=A |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | A | Tentative |
| C | 3 | A | Tentative |
| D | ∞ | — | Tentative |
| E | ∞ | — | Tentative |
Unsettled candidates before selection: B=3, C=3, D=∞, E=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| B→D | ∞ | 3+(2)=5 | Improve to 5; predecessor=B |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | A | Settled |
| C | 3 | A | Tentative |
| D | 5 | B | Tentative |
| E | ∞ | — | Tentative |
Unsettled candidates before selection: C=3, D=5, E=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| C→D | 5 | 3+(2)=5 | Retain 5 (equal cost) |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | A | Settled |
| C | 3 | A | Settled |
| D | 5 | B | Tentative |
| E | ∞ | — | Tentative |
Unsettled candidates before selection: D=5, E=∞. Select the minimum; ties use alphabetical order.
No outgoing edges to relax.
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | A | Settled |
| C | 3 | A | Settled |
| D | 5 | B | Settled |
| E | ∞ | — | Tentative |
Every remaining tentative distance is ∞. No path from the source reaches these vertices.
There is no finite path to recover.
Final answer: E is unreachable from A.
Definition. Bellman–Ford computes single-source shortest paths by repeatedly relaxing every edge. It supports negative edge weights and detects reachable negative cycles.
Simple meaning. Har edge ko baar-baar relax; negative edge chalega, negative cycle problem hai.
Bellman–Ford repeatedly relaxes all edges and can handle negative weights. It uses up to V−1 passes and an additional improvement check for reachable negative cycles. Dijkstra settles the smallest tentative vertex and requires nonnegative weights for its standard correctness guarantee.
Yaad rakho: Pass = all edges once. Negative edge ≠ negative cycle.
"},{"id":"bellman-1","title":"Bellman–Ford · complete example 1","html":"| Directed edge | Weight |
|---|---|
| A→B | 4 |
| A→C | 5 |
| B→C | -2 |
| C→D | 3 |
The shortest route and cost, with complete passes and a negative-cycle check.
Bellman–Ford permits negative edges. We use the listed order with in-place updates; do not mix this with a table that reads only the previous pass.
4 vertices → at most 3 main passes. d[A]=0; every other distance=∞. Use the listed edge order and immediate (in-place) updates.
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| A→B | ∞ | 0+(4)=4 | 4 | Update; predecessor=A |
| A→C | ∞ | 0+(5)=5 | 5 | Update; predecessor=A |
| B→C | 5 | 4+(-2)=2 | 2 | Update; predecessor=B |
| C→D | ∞ | 2+(3)=5 | 5 | Update; predecessor=C |
| Vertex | A | B | C | D |
|---|---|---|---|---|
| End-of-pass distance | 0 | 4 | 2 | 5 |
| Predecessor | — | A | B | C |
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| A→B | 4 | 0+(4)=4 | 4 | No improvement |
| A→C | 2 | 0+(5)=5 | 2 | No improvement |
| B→C | 2 | 4+(-2)=2 | 2 | No improvement |
| C→D | 5 | 2+(3)=5 | 5 | No improvement |
| Vertex | A | B | C | D |
|---|---|---|---|---|
| End-of-pass distance | 0 | 4 | 2 | 5 |
| Predecessor | — | A | B | C |
This entire pass changed no distance. Further identical relaxation passes will not improve anything.
Backward predecessors: D ← C ← B ← A. Forward route: A → B → C → D.
Cost = (4) + (-2) + (3) = 5. No edge can now improve its destination label.
Final answer: A → B → C → D; shortest cost=5.
| Directed edge | Weight |
|---|---|
| C→D | 2 |
| B→C | 1 |
| A→B | 3 |
| A→D | 12 |
The shortest route and cost, with complete passes and a negative-cycle check.
Bellman–Ford permits negative edges. We use the listed order with in-place updates; do not mix this with a table that reads only the previous pass.
4 vertices → at most 3 main passes. d[A]=0; every other distance=∞. Use the listed edge order and immediate (in-place) updates.
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| C→D | ∞ | ∞+(2)=∞ | ∞ | No improvement |
| B→C | ∞ | ∞+(1)=∞ | ∞ | No improvement |
| A→B | ∞ | 0+(3)=3 | 3 | Update; predecessor=A |
| A→D | ∞ | 0+(12)=12 | 12 | Update; predecessor=A |
| Vertex | A | B | C | D |
|---|---|---|---|---|
| End-of-pass distance | 0 | 3 | ∞ | 12 |
| Predecessor | — | A | — | A |
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| C→D | 12 | ∞+(2)=∞ | 12 | No improvement |
| B→C | ∞ | 3+(1)=4 | 4 | Update; predecessor=B |
| A→B | 3 | 0+(3)=3 | 3 | No improvement |
| A→D | 12 | 0+(12)=12 | 12 | No improvement |
| Vertex | A | B | C | D |
|---|---|---|---|---|
| End-of-pass distance | 0 | 3 | 4 | 12 |
| Predecessor | — | A | B | A |
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| C→D | 12 | 4+(2)=6 | 6 | Update; predecessor=C |
| B→C | 4 | 3+(1)=4 | 4 | No improvement |
| A→B | 3 | 0+(3)=3 | 3 | No improvement |
| A→D | 6 | 0+(12)=12 | 6 | No improvement |
| Vertex | A | B | C | D |
|---|---|---|---|---|
| End-of-pass distance | 0 | 3 | 4 | 6 |
| Predecessor | — | A | B | C |
Backward predecessors: D ← C ← B ← A. Forward route: A → B → C → D.
Cost = (3) + (1) + (2) = 6. No edge can now improve its destination label.
Final answer: A → B → C → D; shortest cost=6.
| Directed edge | Weight |
|---|---|
| A→B | 1 |
| B→C | -2 |
| C→B | -1 |
The shortest route and cost, with complete passes and a negative-cycle check.
Bellman–Ford permits negative edges. We use the listed order with in-place updates; do not mix this with a table that reads only the previous pass.
3 vertices → at most 2 main passes. d[A]=0; every other distance=∞. Use the listed edge order and immediate (in-place) updates.
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| A→B | ∞ | 0+(1)=1 | 1 | Update; predecessor=A |
| B→C | ∞ | 1+(-2)=-1 | -1 | Update; predecessor=B |
| C→B | 1 | -1+(-1)=-2 | -2 | Update; predecessor=C |
| Vertex | A | B | C |
|---|---|---|---|
| End-of-pass distance | 0 | -2 | -1 |
| Predecessor | — | C | B |
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| A→B | -2 | 0+(1)=1 | -2 | No improvement |
| B→C | -1 | -2+(-2)=-4 | -4 | Update; predecessor=B |
| C→B | -2 | -4+(-1)=-5 | -5 | Update; predecessor=C |
| Vertex | A | B | C |
|---|---|---|---|
| End-of-pass distance | 0 | -5 | -4 |
| Predecessor | — | C | B |
After the allowed main passes, an edge still improves:
| Edge | Candidate | Current distance |
|---|---|---|
| B→C | -7 | -4 |
A reachable negative cycle exists. Do not present the changing distance labels as finite shortest-path answers.
Final answer: A reachable negative cycle prevents a finite shortest distance for vertices reachable from that cycle.
Definition. Distance-vector routing represents each destination by an estimated distance and next hop, updating those estimates using neighbouring routers’ advertised distances.
Simple meaning. Neighbour ke destination-cost mein apna neighbour-link cost add karo.
For each neighbour v, add the local link cost to that neighbour’s advertised destination cost. Select the minimum and record that neighbour as next hop. N(x) denotes the neighbours of x.
Yaad rakho: Neighbour ka distance + neighbour tak apna distance.
"},{"id":"dv-1","title":"Distance-vector update · example 1","html":"| Neighbour | X-to-neighbour cost | Advertised distance to Z |
|---|---|---|
| B | 1 | 4 |
| C | 2 | 2 |
| D | 3 | 5 |
Candidates and the minimum total route cost.
The advertisement starts from the neighbour, so the local link to that neighbour must be added.
| Via | Addition | Total |
|---|---|---|
| B | 1+4 | 5 |
| C | 2+2 | 4 |
| D | 3+5 | 8 |
min(5, 4, 8)=4.
C is the minimum-cost neighbour.
Final answer: Cost=4; best next hop(s): C.
| Neighbour | X-to-neighbour cost | Advertised distance to Z |
|---|---|---|
| B | 2 | 1 |
| C | 1 | 4 |
| D | 5 | 0 |
Candidates and the minimum total route cost.
The advertisement starts from the neighbour, so the local link to that neighbour must be added.
| Via | Addition | Total |
|---|---|---|
| B | 2+1 | 3 |
| C | 1+4 | 5 |
| D | 5+0 | 5 |
min(3, 5, 5)=3.
B is the minimum-cost neighbour.
Final answer: Cost=3; best next hop(s): B.
| Neighbour | X-to-neighbour cost | Advertised distance to Z |
|---|---|---|
| B | 3 | 2 |
| C | 2 | 3 |
| D | 4 | 4 |
Candidates and the minimum total route cost.
The advertisement starts from the neighbour, so the local link to that neighbour must be added.
| Via | Addition | Total |
|---|---|---|
| B | 3+2 | 5 |
| C | 2+3 | 5 |
| D | 4+4 | 8 |
min(5, 5, 8)=5.
B, C are tied; a protocol may use a defined tie rule or equal-cost multipath.
Final answer: Cost=5; best next hop(s): B, C.
Definition. Routing Information Protocol (RIP) is an interior distance-vector routing protocol that uses a bounded metric, commonly hop count, to select routes.
Simple meaning. RIP mein hops count; 15 tak reachable, 16 means unreachable.
RIP uses distance-vector routing and commonly hop count as its metric. Reachable metrics are at most 15; 16 represents infinity. Its main limitations include limited diameter, a coarse metric and relatively slow convergence with possible temporary loops.
Yaad rakho: RIP: neighbour distances; 16 is infinity.
"},{"id":"loops","title":"Count-to-infinity and loop-mitigation methods","html":"Definition. Count-to-infinity is a distance-vector failure in which routers repeatedly increase their estimates for an unreachable destination because they believe one another’s stale route advertisements.
Simple meaning. Dono neighbour ek dusre ko bol rahe: mere through jao—par actual route gaya.
Split horizon suppresses advertising a route back in the direction from which it was learned. Poison reverse explicitly advertises that route back with an infinite/unreachable metric. Both reduce two-router distance-vector loops caused by stale information.
Yaad rakho: Split: wapas mat bolo. Poison reverse: wapas bolo, par unreachable.
"},{"id":"link-state","title":"Link-state routing: share a map, calculate locally","html":"Definition. Link-state routing distributes information about routers’ local links so that routers can build a topology database and compute shortest paths independently.
Simple meaning. Distance vector mein neighbour ka answer; link state mein topology map.
Yaad rakho: Same map, different root, different tree.
"},{"id":"ospf","title":"OSPF: link-state shortest paths within an autonomous system","html":"Definition. Open Shortest Path First (OSPF) is an interior link-state routing protocol that selects routes using shortest-path calculations and additive link costs.
Simple meaning. OSPF = link-state map + Dijkstra + summed costs.
OSPF exchanges and maintains link-state information, builds a topology database and runs Dijkstra’s shortest-path computation rooted at the local router. It chooses a path with the lowest sum of link costs and installs the corresponding next hop. Equal-cost paths may be supported.
Yaad rakho: OSPF cost add karo; sirf hops nahi.
"},{"id":"ospf-packets","title":"The five OSPF packet types","html":"Definition. OSPFv2 uses five packet types for neighbour maintenance and reliable exchange of link-state database information.
Simple meaning. Hello → database summary → request → update → acknowledgement; har packet ka kaam alag.
| Packet type | What it does | Simple example |
|---|---|---|
| Hello | Discover/maintain neighbours and exchange relationship parameters | Are we still neighbours? |
| Database Description (DD/DBD) | Summarize link-state database contents during synchronization | Here are the LSA headers I have |
| Link State Request (LSR) | Ask for specific missing/newer link-state information | Please send this LSA |
| Link State Update (LSU) | Carry one or more LSAs | Here is the requested or newly flooded link-state information |
| Link State Acknowledgment (LSAck) | Acknowledge receipt of LSAs | I received these LSAs |
| Packet type | What it does | Simple example |
|---|---|---|
| Hello | Discover/maintain neighbours and exchange relationship parameters | Are we still neighbours? |
| Database Description (DD/DBD) | Summarize link-state database contents during synchronization | Here are the LSA headers I have |
| Link State Request (LSR) | Ask for specific missing/newer link-state information | Please send this LSA |
| Link State Update (LSU) | Carry one or more LSAs | Here is the requested or newly flooded link-state information |
| Link State Acknowledgment (LSAck) | Acknowledge receipt of LSAs | I received these LSAs |
Yaad rakho: H–D–R–U–A: Hello, Description, Request, Update, Ack.
"},{"id":"comparison","title":"RIP versus OSPF; Bellman–Ford versus Dijkstra","html":"Definition. RIP and OSPF are routing protocols; Bellman–Ford and Dijkstra are algorithms underlying different route-computation approaches.
Simple meaning. Protocol message bhejta hai; algorithm route calculate karta hai.
| Feature | RIP | OSPF |
|---|---|---|
| Family | Distance vector | Link state |
| Typical metric | Hop count; 16 unreachable | Sum of configured link costs |
| Knowledge exchanged | Destination distances via neighbours | Link-state information |
| Algorithm connection | Distributed Bellman–Ford idea | Dijkstra shortest paths |
| Scale design | Small/simple networks; limited diameter | Area-based scalable design |
Yaad rakho: Distance vector → RIP; link state → OSPF.
"}]},{"id":"packet-support","title":"IP packet details · supporting practice","scope":"TEE · shared foundations and Units 3–4 · current syllabus 702AI0C023","sections":[{"id":"ipv4-header","title":"IPv4 header fields needed for the IPv6 comparison","html":"Definition. An IPv4 datagram contains a variable-length header followed by payload. The header supplies addressing, forwarding and fragmentation information.
Simple meaning. Header delivery instructions hai; payload actual carried content.
Total Length measures the whole IPv4 packet in bytes. IHL measures the header in 32-bit words, so header bytes=4×IHL. Protocol identifies the payload’s upper-layer protocol, such as TCP or UDP.
Yaad rakho: IHL words ×4 = bytes; total minus header = payload.
"}]},{"id":"transport","title":"Unit 5 · Process delivery, UDP, TCP and SCTP","scope":"TEE · Unit 5 · current syllabus 702AI0C023","sections":[{"id":"process","title":"Process-to-process communication, ports and sockets","html":"IP delivers a packet toward a host interface. The transport layer delivers application data to a process endpoint, identified using a port. A socket endpoint is commonly described by IP address and port; the protocol also matters. TCP distinguishes connections by source IP/port and destination IP/port.
![\"Transport:](\"fig/tee-ports.f496c69028.svg\")
| Identifier | Purpose |
|---|---|
| MAC address | Delivery on a local link |
| IP address | Network-layer interface/addressing |
| Port number | Transport-layer application endpoint |
| Domain name | Application-level naming resolved through DNS |
Yaad rakho: IP host tak; port process tak.
"},{"id":"udp","title":"UDP services, operation and header","html":"User Datagram Protocol (UDP) provides connectionless, message-oriented process delivery. One send creates a datagram; message boundaries are retained. UDP has an 8-byte header, carries ports and length, and offers checksum-based corruption detection. It does not itself guarantee delivery, ordering, duplicate suppression, retransmission, flow control or congestion control.
![\"UDP](\"fig/tee-udp.b14cba3c28.svg\")
| UDP field | Width | Purpose |
|---|---|---|
| Source port | 16 bits | Sending endpoint; a zero value can indicate no reply port when unused |
| Destination port | 16 bits | Receiving application endpoint |
| Length | 16 bits | UDP header plus UDP payload; minimum 8 bytes |
| Checksum | 16 bits | Error detection over UDP and an IP pseudo-header; not retransmission |
Yaad rakho: UDP datagram preserve; reliability automatic nahi.
"},{"id":"tcp","title":"TCP features and segment structure","html":"Transmission Control Protocol (TCP) provides reliable ordered byte-stream communication between two endpoints over an unreliable IP service. It repairs many network losses while the connection remains viable; a failed connection still must be handled by the application.
![\"TCP](\"fig/tee-tcp.81a8dfb693.svg\")
| Field | Meaning |
|---|---|
| Source/destination ports (16 each) | Identify process endpoints |
| Sequence number (32) | First data-byte number in this segment; SYN/FIN also occupy sequence space |
| Acknowledgement number (32) | Next sequence number expected when ACK flag is set |
| Data offset (4) | TCP header length in 32-bit words; minimum 5 → 20 bytes |
| Flags | SYN establishes; ACK validates acknowledgement; FIN closes a sending direction; RST aborts; others have defined purposes |
| Window (16 base field) | Advertised receive window; scaling can be negotiated |
| Checksum (16) | Covers TCP header/data and pseudo-header |
| Urgent pointer (16) | Used with URG for the urgent indication |
| Options/padding | Optional features such as MSS; align header to 32-bit words |
Yaad rakho: TCP bytes count karta hai; writes ke borders preserve nahi.
"},{"id":"handshake","title":"TCP connection, data and termination","html":"A TCP connection establishes each endpoint’s initial sequence number and confirms that both can receive. Use client initial sequence x=1000 and server initial sequence y=5000; assume no payload in the handshake.
![\"TCP:](\"fig/tee-handshake.e053246177.svg\")
| Step | Segment | Meaning |
|---|---|---|
| 1 Client → server | SYN, Seq=1000 | Client proposes its starting sequence |
| 2 Server → client | SYN+ACK, Seq=5000, Ack=1001 | Server acknowledges client SYN and supplies its own start |
| 3 Client → server | ACK, Seq=1001, Ack=5001 | Client acknowledges server SYN; connection is established |
Client sends 200 bytes beginning at Seq=1001, so byte numbers are 1001…1200. The receiver acknowledges Ack=1201, the next byte expected. If bytes starting at 1201 are missing but later bytes arrive, the cumulative ACK remains 1201 until the gap is repaired. TCP may buffer out-of-order bytes; it delivers the byte stream in order. Each direction counts independently.
Yaad rakho: SYN plus one, FIN plus one; ACK next expected byte.
"},{"id":"comparison","title":"UDP versus TCP versus SCTP","html":"| Point | UDP | TCP | SCTP |
|---|---|---|---|
| Service unit | Datagram; preserves messages | Byte stream; no message boundaries | Messages carried in chunks; preserves boundaries |
| Connection setup | No protocol handshake | Connection; usual three-way SYN exchange | Association; four-message cookie exchange |
| Reliability | No delivery/order guarantee | Reliable ordered stream | Reliable messages; ordered per stream by default, unordered option |
| Ordering scope | None supplied | One ordered byte stream per direction | Multiple streams; per-stream order avoids blocking every stream on one stream’s ordering gap |
| Flow control | Not supplied by UDP | Receiver-advertised window | Receiver flow control |
| Congestion control | Not supplied by UDP | Yes | Yes |
| Multihoming | No built-in association failover | Classic TCP uses one address pair | Association can bind multiple addresses; built-in path failover |
| Header | 8 bytes | Minimum 20 bytes | 12-byte common header plus chunks |
| Typical use | DNS, real-time applications with their own controls | Web over HTTP/1.1 or HTTP/2, reliable byte transfers | Telecom signalling and multi-stream message applications |
| Performance claim | Low protocol overhead, not automatically fastest overall | Recovery/ordering have costs and benefits | Extra capabilities; no universal speed ranking |
The correct protocol depends on application semantics and network conditions. “UDP is always faster” and “SCTP is always slower than TCP” are not defensible general rules. All three normally run above IP; security/encryption is a separate requirement.
Yaad rakho: UDP messages simple; TCP byte stream; SCTP reliable message streams.
"},{"id":"sctp","title":"SCTP association, chunks and multiple streams","html":"Stream Control Transmission Protocol (SCTP) is a reliable message-oriented transport. A relationship between endpoints is an association. Its packets contain a common header followed by chunks, which carry data or control information.
![\"SCTP](\"fig/tee-sctp.55048562bb.svg\")
Yaad rakho: SCTP association: streams separate order, addresses provide failover.
"}]},{"id":"congestion","title":"Unit 5 · Congestion control and QoS","scope":"TEE · Unit 5 · current syllabus 702AI0C023","sections":[{"id":"policies","title":"Open-loop and closed-loop congestion control","html":"Congestion occurs when offered traffic overloads network resources such as link capacity, router buffers or processing. Queueing delay grows, packets may be dropped, and excessive retransmission can make the overload worse. Flow control protects a receiver; congestion control protects the network.
![\"Two](\"fig/tee-congestion.c27c2303bb.svg\")
| Approach | Policy / mechanism | How it helps |
|---|---|---|
| Open loop: prevent congestion without runtime congestion feedback | Retransmission policy | Carefully chosen timers/selective retransmission avoid needless extra traffic |
| Open loop | Window policy | Limit outstanding data and avoid excessive retransmission after loss |
| Open loop | Acknowledgement policy | Avoid unnecessarily frequent ACK traffic while retaining correct recovery |
| Open loop | Admission policy | Refuse new load when reserved capacity is insufficient |
| Closed loop: measure and react to congestion | Backpressure | A congested node tells upstream nodes to reduce/stop sending, possibly propagating hop by hop |
| Closed loop | Choke/control signalling | A congested node sends explicit feedback to the source |
| Closed loop | Implicit signalling | A sender infers congestion from loss/delay and changes its sending behaviour |
| Closed loop | Explicit marking | Network marks congestion (for example ECN); endpoints react to returned information |
For an answer asking for two of each: explain retransmission and admission policies under open loop, then backpressure and explicit signalling under closed loop, with the mechanism and its effect. These are general congestion-control classifications used in the course; they are not all mandatory algorithms implemented inside TCP. TCP commonly uses end-to-end feedback such as loss or ECN.
Yaad rakho: Open loop pehle prevent; closed loop feedback se react.
"},{"id":"tcp-control","title":"TCP congestion window and classic control phases","html":"A TCP sender is limited by receiver capacity and network capacity. The receive window rwnd comes from the receiver; the congestion window cwnd is maintained by the sender. Roughly, allowed outstanding data is min(rwnd,cwnd), with the existing bytes in flight subtracted before sending new bytes.
![\"\\mathrm{new\\](\"math/e04fbf2fea600eab.svg\")
Quality of Service (QoS) describes delivery characteristics important to an application: throughput/bandwidth, delay, delay variation (jitter) and loss/reliability. Different applications value them differently; a voice call is sensitive to delay/jitter, while a file transfer strongly needs complete data.
| Technique | Mechanism | Tradeoff |
|---|---|---|
| Scheduling | FIFO, priority or fair/weighted queues choose which packet transmits next | Strict priority may starve lower classes; fair scheduling balances shares |
| Traffic shaping | Delay departures to match a profile | Smooths load but adds queueing delay |
| Policing | Measure compliance; mark/drop nonconforming packets according to policy | Controls excess traffic but can lose packets |
| Admission control | Accept new flows only when resources can support them | Protects existing service by refusing excess demand |
| Resource allocation/reservation | Assign bandwidth/buffer/service resources | Requires capacity planning and coordination |
| Classification/marking | Identify traffic classes and carry service markings | Marking alone does not create capacity or guarantees |
Yaad rakho: Bandwidth, delay, jitter, loss; requirement ke hisaab se policy.
"},{"id":"leaky","title":"Leaky-bucket operation and tradeoffs","html":"In the textbook leaky-bucket queue/shaper model, arriving packets enter a finite buffer. The output sends at a configured rate whenever data is queued. When the queue empties, it cannot send nonexistent packets; when the buffer is full, excess arrivals are dropped in this finite-buffer model.
![\"Leaky-bucket](\"fig/tee-leaky.77bdbcf957.svg\")
| Advantage | Limitation |
|---|---|
| Simple bound on output rate | Bursts may overflow or suffer delay |
| Reduces short-term output variation | Unused idle service cannot be saved |
| Useful for predictable traffic shaping | No end-to-end delay/loss guarantee by itself |
The term “leaky bucket” can also describe a conformance meter in standards. This course’s diagrams and numerical questions explicitly use the finite data-queue model, so its state is queued data rather than tokens.
No algorithm is universally better. The leaky data-queue model is useful for a smooth output profile; token bucket is preferable when controlled bursts should be allowed. State the application’s delay, burst and rate needs before recommending one.
Yaad rakho: Leaky bucket stores data; idle service save nahi hota.
"},{"id":"token","title":"Token-bucket operation and limitations","html":"A token bucket generates tokens at rate r up to capacity B. Sending one unit of data consumes the corresponding tokens. Tokens accumulate during idle periods only until the bucket is full. Data can therefore burst when saved tokens exist, subject to the physical link rate.
![\"Token](\"fig/tee-token.c3ff1dbf76.svg\")
![\"\\mathrm{admitted\\](\"math/750b0148fd39654a.svg\")
| Limitation | Reason |
|---|---|
| Bursts can still congest downstream queues | Local conformance does not ensure every later link has capacity |
| Parameters must match traffic and link requirements | A small B delays/drops legitimate packets; a large B permits disruptive bursts |
| No complete QoS guarantee alone | Admission, scheduling and end-to-end capacity may still be needed |
Yaad rakho: Token permission store hota hai; actual packet alag queue mein.
"},{"id":"bucket-compare","title":"Leaky bucket versus token bucket","html":"| Point | Leaky data queue | Token bucket |
|---|---|---|
| Stored state | Queued data | Tokens representing permission |
| Idle period | Empty queue; unused service lost | Tokens accumulate up to B |
| Output burst | Smoothed by configured service rate | Allowed if enough tokens and physical capacity exist |
| Overflow/shortage | Finite queue can overflow | Shaper waits; policer may mark/drop |
| Main parameters | Queue capacity and drain rate | Token capacity and generation rate |
| Useful when | A smoother output profile is required | Average rate plus burst allowance is required |
The World Wide Web (WWW) is an information system of linked resources identified by URLs and accessed using Web clients and servers. A browser is a client; a Web server hosts or generates representations such as HTML, images or JSON. HTTP defines request/response semantics between these components. The Web is one application of the Internet, not a synonym for the entire Internet.
![\"Opening](\"fig/tee-http.de65bc75dd.svg\")
| Message component | Purpose |
|---|---|
| Request method / target | Requested action and resource; GET retrieves a representation, POST submits data for processing |
| Headers | Metadata, routing context, representation preferences and cache/connection information |
| Status code | Result class: 2xx success, 3xx redirection, 4xx client error, 5xx server error |
| Body | Representation or submitted content when present |
HTTP semantics are stateless: each request has its own meaning. Cookies/tokens can let an application maintain a login session; stateless does not mean “users cannot log in”. HTTP/1.1 and HTTP/2 commonly use TCP; HTTP/3 uses QUIC over UDP, so “HTTP always uses TCP” is not a universal statement.
Yaad rakho: DNS address dhoondta; HTTP resource mangta; browser render karta.
"},{"id":"http-state","title":"HTTP methods, status, persistence and HTTPS","html":"| Concept | Explanation | Example |
|---|---|---|
| GET | Retrieve the selected representation | Read a page |
| POST | Ask the resource to process submitted content | Submit a form |
| PUT / DELETE | Replace target representation / request removal according to server semantics | Update/delete an API resource |
| 200 / 301 / 404 / 500 | Success / permanent redirect / not found / server error | Read status before interpreting body |
| Persistent connection | Reuse a connection for multiple exchanges | Avoid repeated TCP setup for each image |
| Cookie/session token | Carry an identifier so the application can relate requests | Remain signed in |
| HTTPS | HTTP protected by authenticated encryption through TLS in common deployments | Protect credentials and content in transit |
A secure transport protects the connection; it does not prove that every server’s content is truthful. HTTP method semantics and application authorization still matter. For exam comparisons, separate HTTP semantics, connection lifetime and application session state.
The Domain Name System (DNS) is a distributed hierarchical naming system. Its namespace is a tree with the root at the top. A complete name is the ordered path of labels from a node back to the root. An absolute name may be written with a trailing dot, for example www.example.com.
![\"DNS](\"fig/tee-dns.f5345fe9fb.svg\")
| Level | Example | Meaning |
|---|---|---|
| Root | . | Top of the namespace |
| Top-level domain (TLD) | com, org, in | A child zone below the root |
| Registered/delegated domain | example.com | Name under a parent such as com |
| Subdomain/host label | www.example.com | A label below example.com; a name may own different record types |
Yaad rakho: Full path unique; www label akela unique nahi.
"},{"id":"dns-records","title":"DNS records, recursive/iterative queries and caching","html":"| Record | Purpose | Example meaning |
|---|---|---|
| A | IPv4 address | Name → IPv4 address |
| AAAA | IPv6 address | Name → IPv6 address |
| NS | Authoritative server names for a zone/delegation | Which nameservers serve the zone |
| CNAME | Alias to another canonical name | Alias → target name; not an IP address |
| MX | Mail exchanger preference and hostname | Where email for a domain should be routed |
| PTR | Reverse-mapping pointer | Address-related reverse name → domain name |
| SOA | Zone administrative/version/timing information | Start of authority for zone data |
The time to live (TTL) is a field attached to resource-record data. It states how many seconds the data may normally remain cached. A, AAAA and MX are record types; TTL is not another type alongside them.
In a recursive query, the queried resolver is asked to obtain a final answer or error for the client. With iterative resolution, a server may return the best information it has, often a referral, and the querying resolver follows it. A caching resolver reduces delay/load but may return an older still-valid record until its TTL expires.
DNS commonly uses UDP port 53 for small exchanges and TCP port 53 for situations such as zone transfers or responses requiring TCP. TCP is supported beyond zone transfer, so do not say DNS is UDP-only. Encrypted DNS transports exist but are outside the named core topic.
File Transfer Protocol (FTP) transfers files between a client and server using TCP and separates commands from file data. This separation is its central design feature. Its control connection remains available across multiple operations, while data connections carry file bytes or directory listings.
![\"FTP:](\"fig/tee-ftp.1b305fa997.svg\")
| Connection | Function | Lifetime / ports |
|---|---|---|
| Control | Commands and replies: USER, PASS, CWD, TYPE, LIST, RETR, STOR, QUIT | Client connects to server TCP port 21; normally kept for the session |
| Data | File contents and directory listing data | Opened as needed; separate from command channel |
FTP over TLS (FTPS) protects FTP connections. SFTP is a separate SSH-based file-transfer protocol, not “FTP using an SSH port”. These names should not be treated as interchangeable.
Yaad rakho: Control commands, data file; passive mein client dono connect karta.
"},{"id":"smtp","title":"SMTP phases, commands and email flow","html":"Simple Mail Transfer Protocol (SMTP) transfers email from a sending client/server to a receiving mail server using a reliable transport, normally TCP. It is a push/store-and-forward transfer protocol. User agents compose/read mail; mail-transfer agents relay it, consulting DNS MX records to find recipient-domain mail servers.
![\"SMTP:](\"fig/tee-smtp.c6f99877c4.svg\")
| Phase | Typical exchange | Purpose |
|---|---|---|
| Connection and greeting | Server 220; client EHLO (HELO for basic SMTP); server 250 capabilities | Identify peers and learn supported extensions |
| Envelope transfer | MAIL FROM:<sender>; RCPT TO:<recipient>; 250 replies | Declare return path and one or more delivery recipients |
| Message transfer | DATA; server 354; headers + blank line + body; single dot line; server 250 | Transmit the message after server permission; completion reply accepts responsibility |
| Termination | QUIT; server 221 | Close the SMTP session cleanly |
Yaad rakho: SMTP send/relay; inbox read IMAP/POP/webmail.
"},{"id":"protocol-summary","title":"Compare the four application protocols","html":"| Protocol | Purpose | Communication pattern | Common transport/port |
|---|---|---|---|
| HTTP | Transfer Web resource representations | Client requests; server responds | Plain HTTP/1.1 TCP 80; HTTPS commonly 443; version/transport matters |
| DNS | Resolve names and other naming records | Queries, referrals, answers and caching | UDP/TCP 53 in classic DNS |
| FTP | Transfer files and listings | Persistent command channel plus separate data connection | TCP control 21; active data traditionally 20, passive chosen port |
| SMTP | Submit/relay email | Greeting, envelope, DATA, termination; queued relay | TCP 25 relay; 587 submission |
Ports are conventional defaults, not a guarantee that every service runs only on that number. A port can be explicitly configured. For exam questions, identify the protocol’s purpose and message flow before memorizing a port.
λ=aggregate attempts/s; G=attempts/frame time.
Maximum 1/(2e)≈0.18394 at G=0.5.
Maximum 1/e≈0.367879 at G=1.
Convert normalized S into successful frames/s.
Successful full-frame bits/s; not necessarily application payload goodput.
For G>0. Do not confuse this probability with normalized throughput S.
Five fully worked rate questions
"},{"id":"ipv4","title":"Masks and address blocks","html":"p=prefix bits, h=host bits, N=total addresses.
Ordinary usable-host count. /31 point-to-point and /32 are exceptions.
M is the first mask octet not equal to 255.
Bitwise AND keeps network bits and clears host bits.
Use full address arithmetic, not an overflowing final octet.
Nine complete address-block examples
"},{"id":"design","title":"Host sizing, VLSM and aggregation","html":"H requested usable hosts. Smallest sufficient h gives smallest ordinary subnet.
K equal subnets required; s borrowed bits. Do not subtract two from modern subnet count.
For a power-of-two total address requirement, subject to boundary alignment.
Old destination distance versus cost to u plus edge u→v.
Distance vector: local link cost plus neighbour advertisement.
16 represents unreachable. Ordinary hop increments commonly equal 1.
| Term | Full form | Meaning |
|---|---|---|
| OSI | Open Systems Interconnection | Seven-layer reference model |
| TCP/IP | Transmission Control Protocol / Internet Protocol | Internet protocol suite |
| LAN / MAN / PAN / WAN | Local / Metropolitan / Personal / Wide Area Network | Network-scope classifications |
| ARPANET | Advanced Research Projects Agency Network | Historical packet network |
| CRC | Cyclic Redundancy Check | Polynomial error detection |
| ARQ | Automatic Repeat reQuest | Feedback/timer-based retransmission |
| GBN / SR | Go-Back-N / Selective Repeat | Sliding-window ARQ strategies |
| FEC | Forward Error Correction | Recover using coding redundancy |
| CSMA | Carrier Sense Multiple Access | Sense medium before access |
| CD / CA | Collision Detection / Collision Avoidance | Two different medium-access strategies |
| CIDR | Classless Inter-Domain Routing | Prefix-based allocation and aggregation |
| VLSM | Variable Length Subnet Mask | Different subnet sizes within a block |
| ARP / RARP | Address Resolution Protocol / Reverse ARP | Local IPv4-to-hardware mapping / historical reverse mapping |
| DHCP | Dynamic Host Configuration Protocol | Leased network configuration |
| RIP | Routing Information Protocol | Distance-vector protocol using hop metric |
| OSPF | Open Shortest Path First | Link-state interior routing protocol |
| LSA | Link State Advertisement | Topology information advertised in link-state routing |
| UDP | User Datagram Protocol | Connectionless datagram transport |
| SCTP | Stream Control Transmission Protocol | Reliable message-oriented multi-stream transport |
| QoS | Quality of Service | Bandwidth, delay, jitter, loss properties |
| RTT / MSS | Round-Trip Time / Maximum Segment Size | Round-trip delay / TCP data-size parameter |
| HTTP / WWW | Hypertext Transfer Protocol / World Wide Web | Web request-response protocol / linked-resource system |
| DNS | Domain Name System | Distributed hierarchical naming |
| FTP | File Transfer Protocol | Separate file-control and data connections |
| SMTP | Simple Mail Transfer Protocol | Mail submission/relay |
| TTL | Time To Live | Meaning depends on context: IP lifetime field or DNS cache duration |
| ACK / SYN / FIN | Acknowledgement / Synchronize / Finish | TCP control concepts; SYN and FIN consume sequence space |
Do not mix uses of the same acronym: DNS TTL is a cache time; IPv4 TTL is decremented at routers. ALOHA is a protocol name, not an acronym that requires an invented expansion.
| Topic | Exact fact |
|---|---|
| OSI / TCP-IP | Seven OSI layers; four TCP/IP layers here; five-layer teaching model splits link |
| CRC | Zero syndrome means no error detected, not certainty of no errors |
| GBN / SR ACK | This course: GBN ACK k=next expected; SR ACK k=individual frame accepted |
| IPv4 / IPv6 | 32-/128-bit addresses; IPv6 base header 40 bytes; no IPv6 broadcast |
| UDP / TCP / SCTP | UDP 8-byte header; TCP minimum 20; SCTP common 12 plus chunks |
| TCP data | Bytes counted; ACK next expected; SYN and FIN each consume one; pure ACK none |
| RIP | Metric 16 means unreachable; 15 is highest reachable hop metric |
| OSPF | Hello, Database Description, Link State Request, Link State Update, Link State Acknowledgment |
| HTTP | Default plain HTTP port 80; HTTPS commonly 443; HTTP/3 uses QUIC/UDP |
| DNS | Classic DNS uses UDP and TCP 53, not UDP only |
| FTP | Control TCP 21; active data traditionally server 20; passive port negotiated |
| SMTP | Server relay TCP 25; submission commonly 587; SMTP does not retrieve inboxes |
| Token shortage | Shaper delays; policer applies mark/drop policy; behaviour must be specified |
Question families present in the supplied papers: final Q2(a), re-exam Q2(a).
![\"r=n-k,\\quad](\"math/beca5466928acbfc.svg\")
k data bits, n codeword bits, r redundant bits; coding rate is dimensionless.
![\"d(a,b)=\\operatorname{weight}(a\\oplus](\"math/15901a0538322dc0.svg\")
Count differing bit positions in equal-length words.
Separate maximum guaranteed detection and correction counts; all distinct pairs define dmin.
![\"R(x)=x^rM(x)\\bmod](\"math/08451acce5e9e980.svg\")
Generator degree r; append r zeros and use modulo-2 XOR division. Receiver divides T directly.
"},{"id":"exam-arq","title":"Paper-backed ARQ window formulas","html":"Question families present in the supplied papers: final Q5(a); re-exam Q5(a) explains GBN.
![\"N=2^m,\\quad](\"math/17ff91ae191f528e.svg\")
m sequence bits; sequence numbers 0…N−1. GBN receiver accepts only next in order.
![\"W_{S,SR}=W_{R,SR}\\le](\"math/413c503dfd6c2a68.svg\")
Standard equal-window SR bound; prevent successive windows overlapping.
"},{"id":"exam-units","title":"Frame time for ALOHA","html":"L is frame length in bits and R is channel rate in bits/s. Divide to obtain seconds per frame. This conversion is used in re-exam Q6(b); then G=attempts per second × frame time.
"},{"id":"exam-headers","title":"IPv6 expansion and header facts","html":"![\"\\mathrm{omitted\\](\"math/e61302b7dd479191.svg\")
Count every written hexadecimal group, including a written 0. Expand :: once and pad every group to four hexadecimal digits. Final Q4(a) asks expansion and header comparison; re-exam Q4(a) asks the IPv6 packet format.
For diagrams: IPv6 has a fixed 40-byte base header; its payload-length field excludes that base header. IPv4 uses IHL in 32-bit words. Extra packet-length calculations are in optional practice.
"}]}],"papers":[{"id":"tee-pyq-guide","title":"Past papers: questions, marks and full solutions","scope":"TEE · Latest papers first; older variants mapped to current syllabus","sections":[{"id":"pattern","title":"Read the original TEE instructions first","html":"Both supplied original question papers contain three pages, are for 100 marks and allow three hours. Q1 is compulsory; attempt any four of Q2–Q7. Every whole question is 20 marks. All 18 parts of each paper are solved here, including the choices you would not attempt in one sitting.
| Whole question | Part marks | Total |
|---|---|---|
| Q1 | 5+5+5+5 | 20 |
| Q2 | 10+10 | 20 |
| Q3 | 5+5+10 | 20 |
| Q4 | 10+10 | 20 |
| Q5 | 10+10 | 20 |
| Q6 | 5+5+10 | 20 |
| Q7 | 10+10 | 20 |
The original instructions also say to start each new question on a fresh page, that right-side figures indicate marks and that suitable data may be assumed if necessary. Assume only what is genuinely missing and state it. Do not replace supplied values or directed edges.
These papers use the older course code 702AI0C007. They are aligned topic by topic to the current 702AI0C023 syllabus; they are evidence for practice, not a promise of the next examination’s pattern. The current syllabus independently confirms 100 TEE marks scaled to 50.
The question map also includes recovered older AI-only and combined-programme variants. Do not apply the latest part-split table to every older paper: use its own printed marks. One AI-only November2023 scan is missing page6 and its incomplete question is explicitly flagged. Same-date variants are different question sets.
"},{"id":"corrections","title":"Source answer-key corrections that affect exam working","html":"| Source | Problem | Verified treatment |
|---|---|---|
| Final Q4(a), IPv6 | Scheme prose says four omitted groups although five groups are visible | :: inserts three groups; full address 0000:0015:0000:0000:0000:0001:0012:1213 |
| Re-exam Q2(a), CRC | Scheme gives CRC 100 but once prints transmitted frame 110010111 | CRC is 100 and frame is 110010100; receiver division is zero |
| Re-exam Q6(b), ALOHA | Scheme multiplies offered frames/s by normalized S for the 500 and 250 cases | Successful frames/s = S/T = offered rate × e^(−2G); results approximately 183.94 and 151.63 frames/s |
| Re-exam Q6(c), transport comparison | Scheme implies a universal UDP/TCP/SCTP speed ranking | Explain overhead/features; real performance depends on workload and network |
| Bucket descriptions | Output described as constant regardless of whether data exists | A data queue drains at configured rate only while nonempty; no packets are created |
| Current syllabus Unit 3 | Printed CDMA/CA in the random-access list | Teach standard CSMA/CA and disclose the apparent typo; CDMA is a separate method |
The question paper fixes the question. An inconsistent scheme is not copied into a solved answer. Arithmetic checks and protocol references support the corrections.
"},{"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: 10, Unit 2: 10, Unit 3: 30, Unit 4: 40, Unit 5: 30, Unit 6: 20. Unspecified internal task marks are not split or guessed.
\r\n| Question | Marks | Current unit and topic | Type | Answer |
|---|---|---|---|---|
| Q1(a) | 5 | Unit 1 · Client-server and peer-to-peer models | Theory | Full solution |
| Q1(b) | 5 | Unit 2 · TCP/IP reference model | Theory | Full solution |
| Q1(c) | 5 | Unit 3 · Pure ALOHA throughput | Mixed | Full solution |
| Q1(d) | 5 | Unit 4 · IPv4 CIDR address block | Numerical | Full solution |
| Q2(a) | 10 | Unit 3 · CRC polynomial division | Numerical | Full solution |
| Q2(b) | 10 | Unit 4 · VLSM subnet allocation | Numerical | Full solution |
| Q3(a) | 5 | Unit 1 · Network definition and topology | Theory | Full solution |
| Q3(b) | 5 | Unit 4 · OSPF shortest-path calculation | Theory | Full solution |
| Q3(c) | 10 | Unit 6 · DNS namespace and resolution | Theory | Full solution |
| Q4(a) | 10 | Unit 4 · IPv6 packet header | Theory | Full solution |
| Q4(b) | 10 | Unit 5 · TCP connection lifecycle | Theory | Full solution |
| Q5(a) | 10 | Unit 3 · Go-Back-N normal, loss and ACK cases | Mixed | Full solution |
| Q5(b) | 10 | Unit 6 · FTP control/data connections | Theory | Full solution |
| Q6(a) | 5 | Unit 2 · Guided/unguided media and optical fibre | Theory | Full solution |
| Q6(b) | 5 | Unit 3 · Pure ALOHA at three offered loads | Numerical | Full solution |
| Q6(c) | 10 | Unit 5 · UDP, TCP and SCTP comparison | Theory | Full solution |
| Q7(a) | 10 | Unit 4 · Bellman-Ford shortest path | Numerical | Full solution |
| Q7(b) | 10 | Unit 5 · Leaky bucket versus token bucket | 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: 5, Unit 2: 15, Unit 3: 30, Unit 4: 40, Unit 5: 30, Unit 6: 20. Unspecified internal task marks are not split or guessed.
\r\n| Question | Marks | Current unit and topic | Type | Answer |
|---|---|---|---|---|
| Q1(a) | 5 | Unit 2 · Wired and wireless media | Case | Full solution |
| Q1(b) | 5 | Unit 2 · OSI and TCP/IP models | Theory | Full solution |
| Q1(c) | 5 | Unit 3 · Slotted ALOHA throughput | Mixed | Full solution |
| Q1(d) | 5 | Unit 4 · IPv4 CIDR address block | Numerical | Full solution |
| Q2(a) | 10 i: 2; ii: 2; iii: 2; iv: 4 | Unit 3 · Block coding and Hamming distance | Mixed | Full solution |
| Q2(b) | 10 | Unit 4 · Circuit, message and packet switching | Case | Full solution |
| Q3(a) | 5 | Unit 1 · Network classifications and ARPANET | Theory | Full solution |
| Q3(b) | 5 | Unit 4 · OSPF packet types | Theory | Full solution |
| Q3(c) | 10 | Unit 6 · WWW and HTTP | Theory | Full solution |
| Q4(a) | 10 | Unit 4 · IPv6 expansion and header | Mixed | Full solution |
| Q4(b) | 10 | Unit 5 · UDP and TCP | Theory | Full solution |
| Q5(a) | 10 | Unit 3 · Selective Repeat sequence space and windows | Mixed | Full solution |
| Q5(b) | 10 | Unit 6 · SMTP phases | Theory | Full solution |
| Q6(a) | 5 | Unit 2 · Unguided transmission media | Theory | Full solution |
| Q6(b) | 5 | Unit 3 · CSMA persistence | Theory | Full solution |
| Q6(c) | 10 | Unit 5 · Open-loop and closed-loop congestion control | Theory | Full solution |
| Q7(a) | 10 | Unit 4 · Dijkstra shortest path | Numerical | Full solution |
| Q7(b) | 10 | Unit 5 · Token bucket | Theory | Full solution |
100 marks to attempt; 140 marks offered. Q1 compulsory; attempt any four of Q2–Q7. All whole questions carry 20 marks. 2022-23
\r\nOffered marks by current unit: Unit 1: 20, Unit 3: 30, Unit 4: 40, Unit 5: 25, Unit 6: 10. Unspecified internal task marks are not split or guessed.
\r\n| Question | Marks | Current unit and topic | Type | Answer |
|---|---|---|---|---|
| Q1(a) | 5 | Unit 1 · Network definition | Theory | Full solution |
| Q1(b) | 5 | Unit 3 · Error control | Theory | Full solution |
| Q1(c) | 5 | Unit 4 · Classful IPv4 | Theory | Full solution |
| Q1(d) | 5 | Unit 5 · UDP features | Theory | Full solution |
| Q2(a) | 10 | Unit 1 · Network topologies | Theory | Full solution |
| Q2(b) | 10 | Unit 3 · Stop-and-Wait versus Go-Back-N | Theory | Full solution |
| Q3(a) | 10 | FDMA and TDMA outside current syllabus | Theory | excluded: outside current syllabus |
| Q3(b) | 10 i: 5; ii: 5 | Unit 4 · Classful net/host IDs and /26 host | Numerical | Full solution |
| Q4(a) | 10 | Unit 3 · CSMA persistence and CSMA/CD | Theory | Full solution |
| Q4(b) | 10 | Unit 4 · OSPF and link types | Theory | Full solution |
| Q5(a) | 10 | Unit 4 · Dijkstra algorithm | Mixed | Full solution |
| Q5(b) | 10 | Unit 5 · Congestion control | Theory | Full solution |
| Q6(a) | 10 | Unit 5 · TCP phases | Theory | Full solution |
| Q6(b) | 10 | Unit 6 · DNS hierarchy and resolution | Theory | Full solution |
| Q7(a) | 5 | Unit 1 · Network classifications | Theory | Full solution |
| Q7(b) | 5 | Unit 3 · Block coding | Mixed | Full solution |
| Q7(c) | 5 | Unit 4 · ARP | Theory | Full solution |
| Q7(d) | 5 | Unit 6 · SMTP and SNMP partly current syllabus | Theory | Full solution |
100 marks to attempt; 140 marks offered. Q1 compulsory; attempt any four of Q2–Q7. All whole questions carry 20 marks. 2023-24 / 2024-25
\r\nOffered marks by current unit: Unit 1: 10, Unit 2: 10, Unit 3: 35, Unit 4: 40, Unit 5: 25, Unit 6: 20. Unspecified internal task marks are not split or guessed.
\r\n| Question | Marks | Current unit and topic | Type | Answer |
|---|---|---|---|---|
| Q1(a) | 5 | Unit 2 · TCP/IP network and transport layers | Theory | Full solution |
| Q1(b) | 5 | Unit 3 · Stop-and-Wait losses | Theory | Full solution |
| Q1(c) | 5 | Unit 5 · SCTP services | Theory | Full solution |
| Q1(d) | 5 | Unit 4 · Message versus packet switching | Theory | Full solution |
| Q2(a) | 10 | Unit 1 · Star and mesh | Theory | Full solution |
| Q2(b) | 10 | Unit 3 · CSMA/CA and RTS/CTS | Theory | Full solution |
| Q3(a) | 10 | Unit 5 · TCP connection lifecycle | Theory | Full solution |
| Q3(b) | 10 | Unit 4 · VLSM customer allocation | Numerical | Full solution |
| Q4(a) | 10 | Unit 6 · DNS resolution | Theory | Full solution |
| Q4(b) | 10 | Unit 4 · RIP route updates | Mixed | Full solution |
| Q5(a) | 10 | Unit 3 · Pure versus slotted ALOHA | Theory | Full solution |
| Q5(b) | 10 | Unit 5 · Congestion policies | Theory | Full solution |
| Q6(a) | 10 | Unit 4 · Dijkstra six-node graph | Numerical | Full solution |
| Q6(b) | 10 | Unit 3 · CRC | Numerical | Full solution |
| Q7(a) | 5 | Unit 2 · Guided versus unguided media | Theory | Full solution |
| Q7(b) | 5 | Unit 4 · IPv4 /26 block | Numerical | Full solution |
| Q7(c) | 10 | Unit 6 · FTP connections | Theory | Full solution |
Source qualification: The PDF has four pages; the fourth is blank. The handwritten date ends in 2025 over a printed 2024 template.
\r\n100 marks to attempt; 140 marks offered. Q1 compulsory; attempt any four of Q2–Q7. All whole questions carry 20 marks. Final 2024-25 / Re-exam 2023-24
\r\nOffered marks by current unit: Unit 1: 10, Unit 2: 10, Unit 3: 35, Unit 4: 40, Unit 5: 25, Unit 6: 20. Unspecified internal task marks are not split or guessed.
\r\n| Question | Marks | Current unit and topic | Type | Answer |
|---|---|---|---|---|
| Q1(a) | 5 | Unit 1 · LAN MAN WAN | Theory | Full solution |
| Q1(b) | 5 | Unit 3 · Selective Repeat versus Go-Back-N | Theory | Full solution |
| Q1(c) | 5 | Unit 5 · TCP versus UDP | Theory | Full solution |
| Q1(d) | 5 | Unit 4 · DHCP | Theory | Full solution |
| Q2(a) | 10 | Unit 2 · OSI model | Theory | Full solution |
| Q2(b) | 10 | Unit 3 · CSMA/CD | Theory | Full solution |
| Q3(a) | 10 | Unit 6 · SMTP phases | Theory | Full solution |
| Q3(b) | 10 | Unit 4 · OSPF and link types | Theory | Full solution |
| Q4(a) | 10 | Unit 3 · CRC and corrupted frame | Numerical | Full solution |
| Q4(b) | 10 | Unit 5 · UDP and applications | Theory | Full solution |
| Q5(a) | 5 | Unit 1 · Network models | Theory | Full solution |
| Q5(b) | 5 | Unit 6 · DNS hierarchy | Theory | Full solution |
| Q5(c) | 10 | Unit 4 · Bellman-Ford example | Mixed | Full solution |
| Q6(a) | 10 | Unit 3 · Pure ALOHA rates | Numerical | Full solution |
| Q6(b) | 10 | Unit 4 · VLSM groups | Numerical | Full solution |
| Q7(a) | 10 | Unit 5 · Leaky versus token bucket | Theory | Full solution |
| Q7(b) | 5 | Unit 6 · HTTP | Theory | Full solution |
| Q7(c) | 5 | Unit 4 · Distance vector versus link state | Theory | Full solution |
100 marks to attempt; 140 marks offered. Q1 compulsory; attempt any four remaining complete questions. AI-only Sem V variant
\r\nOffered marks by current unit: Unit 3: 25, Unit 4: 50, Unit 5: 10, Unit 6: 25. Unspecified internal task marks are not split or guessed.
\r\n| Question | Marks | Current unit and topic | Type | Answer |
|---|---|---|---|---|
| Q1(a) | 5 | Unit 4 · Mission-critical switching choice | Case | Full solution |
| Q1(b) | 5 | Unit 4 · Store-and-forward delay | Numerical | Full solution |
| Q1(c) | 5 | Unit 4 · Classful versus classless addressing | Theory | Full solution |
| Q1(d) | 5 | BGP AS-PATH outside current syllabus | Theory | excluded: outside current syllabus |
| Q2(a) | 10 | Unit 6 · Webmail and POP3 protocol journey | Theory | Full solution |
| Q2(b) | 5 | CDN Bring Home outside current syllabus | Theory | excluded: outside current syllabus |
| Q2(c) | 5 | Unit 4 · ARP response destination | Theory | Full solution |
| Q3(a) | 10 | Unit 4 · Store-and-forward caravan | Numerical | Full solution |
| Q3(b) | 5 | Unit 4 · Count to infinity | Theory | Full solution |
| Q3(c) | 5 | Unit 5 · Reliable application over UDP | Theory | Full solution |
| Q4(a) | 5 | Unit 3 · CRC codeword | Numerical | Full solution |
| Q4(b) | 5 | Unit 5 · Internet checksum | Numerical | Full solution |
| Q4(c) | 10 | Unit 3 · Hamming generation and correction | Numerical | Full solution |
| Q5(a) | 10 | Unit 6 · Persistent HTTP cache delay | Numerical | Full solution |
| Q5(b) | 5 | Unit 6 · Iterative DNS resolution | Theory | Full solution |
| Q5(c) | 5 | Unit 4 · Longest-prefix matching | Numerical | Full solution |
| Q6(a) | 5 | Unit 4 · Classful network and host portions | Numerical | Full solution |
| Q6(b) | 5 | Unit 3 · CSMA without collision detection | Numerical | Full solution |
| Q6(c) | 5 | Unit 4 · Recovering unknown routing weights | Numerical | Full solution |
| Q6(d) | 5 | Unit 3 · ALOHA comparison | Theory | Full solution |
| Q7(a) | 10 | Learning-switch tables, mislabeled ARP outside current syllabus | Theory | excluded: outside current syllabus |
| Q7(b) | 10 | NAT four packet stages outside current syllabus | Theory | excluded: outside current syllabus |
100 marks to attempt; 140 marks offered. Q1 compulsory; attempt any four of Q2–Q7. All whole questions carry 20 marks. 2022-23
\r\nOffered marks by current unit: Unit 1: 10, Unit 2: 10, Unit 3: 35, Unit 4: 40, Unit 5: 25, Unit 6: 20. Unspecified internal task marks are not split or guessed.
\r\n| Question | Marks | Current unit and topic | Type | Answer |
|---|---|---|---|---|
| Q1(a) | 5 | Unit 5 · TCP versus UDP | Theory | Full solution |
| Q1(b) | 5 | Unit 2 · Guided media | Theory | Full solution |
| Q1(c) | 5 | Unit 3 · CSMA persistence | Theory | Full solution |
| Q1(d) | 5 | Unit 4 · Circuit versus packet switching | Theory | Full solution |
| Q2(a) | 10 | Unit 5 · QoS and scheduling | Theory | Full solution |
| Q2(b) | 10 | Unit 3 · Slotted ALOHA rates | Numerical | Full solution |
| Q3(a) | 10 | Unit 3 · CSMA/CA | Theory | Full solution |
| Q3(b) | 10 | Unit 4 · RIP and timers | Theory | Full solution |
| Q4(a) | 10 | Unit 5 · TCP lifecycle | Theory | Full solution |
| Q4(b) | 10 | Unit 6 · DNS resolution | Theory | Full solution |
| Q5(a) | 10 | Unit 3 · Go-Back-N scenarios | Theory | Full solution |
| Q5(b) | 5 | Unit 6 · FTP connections | Theory | Full solution |
| Q5(c) | 5 | Unit 4 · DHCP dynamic allocation | Theory | Full solution |
| Q6(a) | 10 | Unit 4 · VLSM and noncanonical block address | Numerical | Full solution |
| Q6(b) | 5 | Unit 6 · SMTP | Theory | Full solution |
| Q6(c) | 5 | Unit 2 · TCP/IP layers | Theory | Full solution |
| Q7(a) | 10 | Unit 4 · Dijkstra graph | Numerical | Full solution |
| Q7(b) | 10 | Unit 1 · Four network topologies | Theory | Full solution |
100 marks to attempt; 140 marks offered. Q1 compulsory; attempt any four remaining complete questions. AI-only Sem V variant
\r\nOffered marks by current unit: Unit 3: 35, Unit 4: 40, Unit 5: 15, Unit 6: 15. Unspecified internal task marks are not split or guessed.
\r\n| Question | Marks | Current unit and topic | Type | Answer |
|---|---|---|---|---|
| Q1(a) | 5 | Unit 4 · Subnet mask | Theory | Full solution |
| Q1(b) | 5 | Unit 5 · Additional shared bottleneck | Numerical | Full solution |
| Q1(c) | 5 | Unit 6 · HTTP/1.1 delay | Numerical | Full solution |
| Q1(d) | 5 | Confidentiality versus integrity outside current syllabus | Theory | excluded: outside current syllabus |
| Q2(a) | 10 | Unit 4 · Dijkstra from w | Numerical | Full solution |
| Q2(b) | 5 | Unit 5 · Telnet connection ports | Theory | Full solution |
| Q2(c) | 5 | SDN routing application outside current syllabus | Theory | excluded: outside current syllabus |
| Q3(a) | 5 | Unit 3 · rdt3.0 error recovery | Theory | Full solution |
| Q3(b) | 5 | Unit 6 · DNS cache popularity | Theory | Full solution |
| Q3(c) | 5 | TLS impersonation outside current syllabus | Theory | excluded: outside current syllabus |
| Q3(d) | 5 | TLS nonces outside current syllabus | Theory | excluded: outside current syllabus |
| Q4(a) | 10 | Unit 4 · Distance vector at z | Numerical | Full solution |
| Q4(b) | 5 | Cryptographic attack models outside current syllabus | Theory | excluded: outside current syllabus |
| Q4(c) | 5 | CDN design outside current syllabus | Theory | excluded: outside current syllabus |
| Q5(a) | 10 | Unit 3 · Hamming error detection | Numerical | Full solution |
| Q5(b) | 5 | Unit 3 · CRC and damaged message | Numerical | Full solution |
| Q5(c) | 5 | Unit 5 · 8-bit checksum | Numerical | Full solution |
| Q6(a) | 5 | Unit 3 · Pure ALOHA timeline | Numerical | Full solution |
| Q6(b) | 5 | Unit 4 · Distance-vector link change | Numerical | Full solution |
| Q6(c) | 5 | Polling and token passing outside current syllabus | Theory | excluded: outside current syllabus |
| Q6(d) | 5 | Unit 4 · Classful and IPv6 addresses | Mixed | Full solution |
| Q7(a) | 5 | Unit 4 · Network delay components | Theory | Full solution |
| Q7(b) | 5 | Unit 6 · Webmail to IMAP recipient | Theory | Full solution |
| Q7(c) | 10 | Unit 3 · Selective Repeat implementation | Theory | Full solution |
100 marks to attempt; 140 marks offered. Q1 compulsory; attempt any four of Q2–Q7. All whole questions carry 20 marks. Final 2023-24 / Re-exam 2022-23
\r\nOffered marks by current unit: Unit 1: 10, Unit 2: 10, Unit 3: 35, Unit 4: 40, Unit 5: 25, Unit 6: 20. Unspecified internal task marks are not split or guessed.
\r\n| Question | Marks | Current unit and topic | Type | Answer |
|---|---|---|---|---|
| Q1(a) | 5 | Unit 1 · Network classification | Theory | Full solution |
| Q1(b) | 5 | Unit 3 · ALOHA and vulnerable periods | Theory | Full solution |
| Q1(c) | 5 | Unit 4 · Classful addressing | Theory | Full solution |
| Q1(d) | 5 | Unit 5 · UDP applications | Theory | Full solution |
| Q2(a) | 10 | Unit 6 · SMTP phases | Theory | Full solution |
| Q2(b) | 10 | Unit 2 · OSI layers | Theory | Full solution |
| Q3(a) | 5 | Unit 1 · Network models | Theory | Full solution |
| Q3(b) | 10 | Unit 3 · CSMA/CD | Theory | Full solution |
| Q3(c) | 5 | Unit 4 · OSPF routing | Theory | Full solution |
| Q4(a) | 10 | Unit 5 · Traffic shaping buckets | Theory | Full solution |
| Q4(b) | 10 | Unit 3 · Selective Repeat scenarios | Theory | Full solution |
| Q5(a) | 10 | Unit 4 · VLSM two groups | Numerical | Full solution |
| Q5(b) | 5 | Unit 6 · FTP | Theory | Full solution |
| Q5(c) | 5 | Unit 4 · ARP | Theory | Full solution |
| Q6(a) | 10 | Unit 3 · CRC | Numerical | Full solution |
| Q6(b) | 5 | Unit 4 · IPv4 /26 block | Numerical | Full solution |
| Q6(c) | 5 | Unit 6 · DNS hierarchy | Theory | Full solution |
| Q7(a) | 10 | Unit 4 · Bellman-Ford directed graph | Numerical | Full solution |
| Q7(b) | 10 | Unit 5 · Congestion control | Theory | Full solution |
Source qualification: Header says 02 pages, but the paper continues through printed Page 3 of 3; all three are included.
\r\n100 marks to attempt; 105 printed marks recovered; complete offered total unknown. Q1 compulsory; attempt any four remaining complete questions. AI-only Sem V variant; incomplete recovered scan
\r\nOffered marks by current unit: Unit 3: 25, Unit 4: 20, Unit 5: 10, Unit 6: 10. Unspecified internal task marks are not split or guessed.
\r\n| Question | Marks | Current unit and topic | Type | Answer |
|---|---|---|---|---|
| Q1(a) | 5 | Unit 4 · Subnetting and masks | Theory | Full solution |
| Q1(b) | 5 | Unit 5 · Shared bottleneck throughput | Numerical | Full solution |
| Q1(c) | 5 | Unit 6 · HTTP/1.0 delay | Numerical | Full solution |
| Q1(d) | 5 | Cryptographic hash versus checksum outside current syllabus | Theory | excluded: outside current syllabus |
| Q2(a) | 10 | Unit 3 · Hamming error correction | Numerical | Full solution |
| Q2(b) | 5 | Unit 3 · CRC encoding | Numerical | Full solution |
| Q2(c) | 5 | Unit 3 · Framing | Theory | Full solution |
| Q3(a) | 5 | Unit 3 · Alternating-bit state transitions | Numerical | Full solution |
| Q3(b) | 5 | Stateful packet filtering outside current syllabus | Theory | excluded: outside current syllabus |
| Q3(c) | 5 | Unit 6 · DNS cache inference | Theory | Full solution |
| Q3(d) | 5 | Unit 4 · IPv4 address classes | Theory | Full solution |
| Q4(a) | 10 | Unit 4 · Dijkstra from u | Numerical | Full solution |
| Q4(b) | 5 | Unit 5 · TCP socket port numbers | Theory | Full solution |
| Q4(c) | 5 | SDN control-plane layers outside current syllabus | Theory | excluded: outside current syllabus |
| Q5(a) | 10 | BGP topology views outside current syllabus | Theory | excluded: outside current syllabus |
| Q5(b) | 5 | Symmetric and public-key cryptography outside current syllabus | Theory | excluded: outside current syllabus |
| Q5(c) | 5 | CDN placement and server selection outside current syllabus | Theory | excluded: outside current syllabus |
| Q6(a) | 5 | Unit 3 · Incomplete multiple-access question source incomplete | Numerical | unavailable: missing source continuation |
Source qualification: Header states six pages, but the recovered PDF has only pages 1–5. Q6(a) stops before the requested protocol/tasks; page 6 and any subsequent questions are unavailable. No missing text or marks have been invented.
\r\n100 marks to attempt; 140 marks offered. Q1 compulsory; attempt any four remaining complete questions. AI-only Sem V
\r\nOffered marks by current unit: Unit 3: 39, Unit 4: 20, Unit 5: 26, Unit 6: 25. Unspecified internal task marks are not split or guessed.
\r\n| Question | Marks | Current unit and topic | Type | Answer |
|---|---|---|---|---|
| Q1(a) | 5 | Unit 5 · Shared bottleneck throughput | Numerical | Full solution |
| Q1(b) | 5 | Unit 3 · Ideal broadcast channel and slotted ALOHA | Theory | Full solution |
| Q1(c) | 5 | Unit 5 · Round-robin scheduling | Numerical | Full solution |
| Q1(d) | 5 | Unit 5 · TCP congestion control | Theory | Full solution |
| Q2(a) | 10 | Unit 6 · Persistent HTTP delay | Numerical | Full solution |
| Q2(b)(i) | 5 | Unit 3 · Block-code received words | Numerical | Full solution |
| Q2(b)(ii) | 5 | Unit 3 · CRC and syndrome | Numerical | Full solution |
| Q3(a) | 10 | Unit 4 · Dijkstra from u | Numerical | Full solution |
| Q3(b)(i) | 4 | Unit 3 · Why rdt3.0 needs a timer | Theory | Full solution |
| Q3(b)(ii) | 6 | Unit 5 · Internet checksum | Numerical | Full solution |
| Q4(a) | 10 | Unit 3 · Alternating-bit data and ACK sequence | Numerical | Full solution |
| Q4(b) | 10 | Unit 6 · HTTP/2 multiplexing versus HTTP/1.1 | Numerical | Full solution |
| Q5(a) | 10 | BGP export policy outside current syllabus | Theory | excluded: outside current syllabus |
| Q5(b)(i) | 5 | HFC access outside current syllabus | Theory | excluded: outside current syllabus |
| Q5(b)(ii) | 5 | Wireless access technologies outside current syllabus | Theory | excluded: outside current syllabus |
| Q6(a) | 10 | Unit 3 · Slotted ALOHA arrival timeline | Numerical | Full solution |
| Q6(b) | 10 | Unit 4 · ISP VLSM allocation | Numerical | Full solution |
| Q7(a) | 5 | TDM versus FDM outside current syllabus | Theory | excluded: outside current syllabus |
| Q7(b) | 5 | Unit 6 · HTTP state | Theory | Full solution |
| Q7(c) | 5 | Unit 5 · TCP server sockets | Theory | Full solution |
| Q7(d) | 5 | NAT translated flow outside current syllabus | Theory | excluded: outside current syllabus |
100 marks to attempt; 140 marks offered. Q1 compulsory; attempt any four remaining complete questions. AI-only Sem V
\r\nOffered marks by current unit: Unit 2: 5, Unit 3: 35, Unit 4: 35, Unit 5: 20, Unit 6: 15. Unspecified internal task marks are not split or guessed.
\r\n| Question | Marks | Current unit and topic | Type | Answer |
|---|---|---|---|---|
| Q1(a) | 5 | Unit 4 · Store-and-forward delay | Numerical | Full solution |
| Q1(b) | 5 | Token passing and ideal broadcast access outside current syllabus | Theory | excluded: outside current syllabus |
| Q1(c) | 5 | Unit 5 · Priority scheduling | Numerical | Full solution |
| Q1(d) | 5 | Unit 5 · TCP flow control | Theory | Full solution |
| Q2(a) | 10 | Unit 6 · HTTP delay comparison | Numerical | Full solution |
| Q2(b)(i) | 4 | Unit 3 · Block coding capacity | Numerical | Full solution |
| Q2(b)(ii) | 6 | Unit 3 · CRC encoder and decoder | Theory | Full solution |
| Q3(a) | 10 | Unit 4 · Dijkstra from t | Numerical | Full solution |
| Q3(b)(i) | 5 | Unit 5 · UDP checksum limitations | Theory | Full solution |
| Q3(b)(ii) | 5 | Unit 5 · TCP demultiplexing | Theory | Full solution |
| Q4(a) | 10 | Unit 3 · rdt3.0 global transition sequence | Numerical | Full solution |
| Q4(b)(i) | 5 | CDN placement outside current syllabus | Theory | excluded: outside current syllabus |
| Q4(b)(ii) | 5 | CDN server selection outside current syllabus | Theory | excluded: outside current syllabus |
| Q5(a) | 10 | BGP path views outside current syllabus | Theory | excluded: outside current syllabus |
| Q5(b)(i) | 5 | Access technologies outside current syllabus | Theory | excluded: outside current syllabus |
| Q5(b)(ii) | 5 | Unit 2 · Ethernet physical media | Theory | Full solution |
| Q6(a) | 10 | Unit 3 · CSMA/CD arrival timeline | Numerical | Full solution |
| Q6(b)(i) | 5 | Unit 4 · Classful addresses and masks | Theory | Full solution |
| Q6(b)(ii) | 5 | Unit 4 · CIDR block | Numerical | Full solution |
| Q7(a) | 5 | Unit 4 · Circuit switching advantage | Theory | Full solution |
| Q7(b) | 5 | Unit 6 · HTTP cookies | Theory | Full solution |
| Q7(c) | 5 | Unit 3 · Selective Repeat | Theory | Full solution |
| Q7(d) | 5 | Unit 4 · Destination-based forwarding | Theory | Full solution |
Prepare 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":"tee-paper-final","title":"Final · 5 December 2025","scope":"TEE · Actual TEE paper · all questions · current syllabus 702AI0C023","sections":[{"id":"q1a","title":"Q1A · Wired and wireless campus","html":"| Criterion | Wired campus links | Wireless campus access |
|---|---|---|
| Speed/capacity | Dedicated switch ports and fiber backbones can provide high, predictable capacity | Capacity shared over radio; performance depends on contention, coverage and interference |
| Reliability | Stable when cables/ports and redundant uplinks are healthy | Mobility and flexible access; radio interference and coverage holes can cause variability |
| Security | Physical access control helps, but ports still need authentication/segmentation | Use authenticated encryption and managed access; radio extends beyond rooms |
| Deployment | Cabling cost and fixed access points | Supports mobile users and faster coverage changes |
| Recommended role | Servers, laboratories, inter-building backbone | Students’ phones/laptops, roaming and guest access |
Choose a hybrid: wired core/backbone and critical endpoints, with managed access points providing wireless mobility. Separate guest traffic, authenticate users and plan redundant infrastructure. Neither medium alone guarantees security or availability.
Source: Final · 5 December 2025, original question paper page 1. Original marks retained; solution independently authored.
"},{"id":"q1b","title":"Q1B · OSI and TCP/IP comparison","html":"The TCP/IP model describes the Internet protocol suite in four layers. Many teaching books split the link layer into data link and physical, making a five-layer teaching model; identify the convention before drawing.
| TCP/IP layer | Function | Examples |
|---|---|---|
| Application | Defines application message formats and exchanges for Web access, naming, email and file transfer; includes presentation/session functions as needed | HTTP, DNS, SMTP, FTP |
| Transport | End-to-end process communication | TCP, UDP, SCTP |
| Internet | IP addressing, routing and best-effort datagrams | IPv4/IPv6 |
| Link/network access | Local delivery and physical access | Ethernet/Wi-Fi and their medium |
| Point | OSI | TCP/IP |
|---|---|---|
| Layer count | Seven | Four here; five-layer teaching variant splits link |
| Upper functions | Application, presentation, session separated | Grouped under application |
| Lower functions | Data link and physical separated | Grouped under link/network access |
| Purpose | General reference architecture | Architecture of deployed Internet protocol family |
| Network service | Reference model can describe different services | Internet layer uses connectionless IP |
| Transport | Conceptual transport service | TCP reliable byte stream; UDP datagrams; SCTP messages |
Source: Final · 5 December 2025, original question paper page 1. Original marks retained; solution independently authored.
"},{"id":"q1c","title":"Q1C · Slotted ALOHA","html":"Protocol = Slotted ALOHA; G = 1 attempts/frame time.
S, 100S and the maximum possible S.
Select the formula from the protocol name. The exponent is −G because Slotted ALOHA has a T vulnerable period.
−1 × 1 = -1.
S = 1 × 0.367879441 ≈ 0.367879441 successful frames/frame time.
0.367879441 × 100 ≈ 36.7879%.
The maximum is 1/(1e) ≈ 0.367879 = 36.7879%, at G=1.
Final answer: S ≈ 0.367879; percentage ≈ 36.7879%; Smax ≈ 0.367879 at G=1.
Source: Final · 5 December 2025, original question paper page 1. Original marks retained; solution independently authored.
"},{"id":"q1d","title":"Q1D · IPv4 address block","html":"IP/prefix = 200.140.64.48/26.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−26=6; N=26=64.
Mask=255.255.255.192; binary=11111111.11111111.11111111.11000000.
The changing octet is number 4. Its mask value is 192, so step=256−192=64. The IP value 48 lies in 0…63.
Keep earlier octets; place 0 in octet 4; set any later octets to zero. Network=200.140.64.0.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 11001000 | 11111111 | 11001000 | 200 |
| 2 | 10001100 | 11111111 | 10001100 | 140 |
| 3 | 01000000 | 11111111 | 01000000 | 64 |
| 4 | 00110000 | 11000000 | 00000000 | 0 |
Use 63 in the changing octet and 255 in any later octets. Broadcast=200.140.64.63.
First usable=200.140.64.1; last usable=200.140.64.62; count=64−2=62.
Final answer: Network 200.140.64.0; broadcast 200.140.64.63; 64 total, 62 ordinary usable; range 200.140.64.1–200.140.64.62.
Source: Final · 5 December 2025, original question paper page 1. Original marks retained; solution independently authored.
"},{"id":"q2a","title":"Q2A · Block coding and all pairwise distances","html":"(i) Block coding. Map each k-bit dataword to a distinct n-bit codeword by adding n−k redundant bits. Only the valid codeword subset represents data. A received invalid pattern detects an error; sufficiently separated codewords permit unique correction within the guaranteed radius. For a chosen code, the minimum Hamming distance measures that separation.
| Codeword | Bits |
|---|---|
| C1 | 1001101 |
| C2 | 1110001 |
| C3 | 1010101 |
All pair distances, dmin and the maximum guaranteed numbers of errors.
XOR exposes differing positions. Use the closest codeword pair, because it is the weakest separation in the code.
| Pair | First | Second | XOR | Distance |
|---|---|---|---|---|
| C1 vs C2 | 1001101 | 1110001 | 0111100 | 4 |
| C1 vs C3 | 1001101 | 1010101 | 0011000 | 2 |
| C2 vs C3 | 1110001 | 1010101 | 0100100 | 2 |
dmin = min(4, 2, 2) = 2.
Detect = 2−1 = 1; correct = floor((2−1)/2) = 0.
2 bit changes can reach another valid codeword, so only 1 are guaranteed detectable. Correction regions require 2t < 2.
Final answer: dmin=2; detect up to 1 errors; correct up to 0 errors.
Source: Final · 5 December 2025, original question paper page 1. Original marks retained; solution independently authored.
"},{"id":"q2b","title":"Q2B · Choose a switching technique","html":"A setup phase reserves a circuit through intermediate switches. Data then streams through the established path until release. It offers predictable reserved service and low per-unit forwarding overhead, but setup takes time and idle periods waste reserved capacity. A traditional dedicated voice circuit is an appropriate illustration.
Each intermediate node receives and stores the entire message before forwarding it. Links can be shared and no end-to-end circuit is required, but large messages need buffers and cause substantial store-and-forward delay. It illustrates delay-tolerant whole-message delivery, rather than an efficient choice for interactive office traffic.
A message is segmented into packets, allowing shared links and pipelined forwarding. This suits bursty data and supports flexible routing. Costs include packet headers, variable queueing delay, potential packet loss and possible reordering in datagram networks.
| Aspect | Circuit switching | Message switching | Packet switching |
|---|---|---|---|
| Unit handled | Continuous stream on established circuit | Whole message | Small packets |
| Resource use | Capacity reserved along a circuit | Shared; message stored at each intermediate node | Shared; each packet stored/forwarded in this model |
| Setup | Required before data transfer | No circuit setup required | Datagram: no connection setup; virtual circuit: setup |
| Delay | Setup delay, then relatively predictable service | Large store-and-forward delay for long messages | Variable delay; packet pipeline can reduce completion time |
| Main limitation | Reserved capacity can be idle | Large buffers; unsuitable for tight delay needs | Headers, possible queueing/loss; datagrams may reorder |
For ordinary office file, web and collaboration traffic, choose packet switching, with adequate capacity and traffic management. Its statistical sharing serves bursty sources efficiently. If strict deterministic service is required, state that requirement and discuss additional service engineering rather than promising that every packet network guarantees timing.
Source: Final · 5 December 2025, original question paper page 2. Original marks retained; solution independently authored.
"},{"id":"q3a","title":"Q3A · LAN, MAN, PAN, WAN, ARPANET","html":"| Type | Meaning and scope | Example |
|---|---|---|
| PAN | Personal Area Network: a person’s nearby devices | Phone linked to earbuds |
| LAN | Local Area Network: a building or local campus network under local administration | Laboratory Ethernet/Wi-Fi |
| MAN | Metropolitan Area Network: interconnects sites across a town/city | City-wide institutional network |
| WAN | Wide Area Network: connects geographically separated networks over regional/global links | Branches in different cities |
| ARPANET | Historical packet-switched research network; a major predecessor of the Internet | Early interconnected research institutions; not a modern geographic size category |
Source: Final · 5 December 2025, original question paper page 2. Original marks retained; solution independently authored.
"},{"id":"q3b","title":"Q3B · OSPF packet types","html":"| Packet type | What it does | Simple example |
|---|---|---|
| Hello | Discover/maintain neighbours and exchange relationship parameters | Are we still neighbours? |
| Database Description (DD/DBD) | Summarize link-state database contents during synchronization | Here are the LSA headers I have |
| Link State Request (LSR) | Ask for specific missing/newer link-state information | Please send this LSA |
| Link State Update (LSU) | Carry one or more LSAs | Here is the requested or newly flooded link-state information |
| Link State Acknowledgment (LSAck) | Acknowledge receipt of LSAs | I received these LSAs |
Database Description packets carry summaries; Link State Update packets carry LSAs themselves. Do not confuse an LSA type with an OSPF packet type.
Source: Final · 5 December 2025, original question paper page 2. Original marks retained; solution independently authored.
"},{"id":"q3c","title":"Q3C · WWW and HTTP","html":"The World Wide Web (WWW) is an information system of linked resources identified by URLs and accessed using Web clients and servers. A browser is a client; a Web server hosts or generates representations such as HTML, images or JSON. HTTP defines request/response semantics between these components. The Web is one application of the Internet, not a synonym for the entire Internet.
| Message component | Purpose |
|---|---|
| Request method / target | Requested action and resource; GET retrieves a representation, POST submits data for processing |
| Headers | Metadata, routing context, representation preferences and cache/connection information |
| Status code | Result class: 2xx success, 3xx redirection, 4xx client error, 5xx server error |
| Body | Representation or submitted content when present |
HTTP semantics are stateless: each request has its own meaning. Cookies/tokens can let an application maintain a login session; stateless does not mean “users cannot log in”. HTTP/1.1 and HTTP/2 commonly use TCP; HTTP/3 uses QUIC over UDP, so “HTTP always uses TCP” is not a universal statement.
Source: Final · 5 December 2025, original question paper page 2. Original marks retained; solution independently authored.
"},{"id":"q4a","title":"Q4A · IPv6 expansion and headers","html":"Given address: 0:15::1:12:1213.
The full 128-bit address in eight-group hexadecimal notation.
A written 0 is a visible group. :: represents only the omitted all-zero groups.
Left: 0, 15 (2 group(s)). Right: 1, 12, 1213 (3 group(s)).
Visible=2+3=5. Missing=8−5=3.
0:15:0:0:0:1:12:1213.
| Group number | Four-digit group |
|---|---|
| 1 | 0000 |
| 2 | 0015 |
| 3 | 0000 |
| 4 | 0000 |
| 5 | 0000 |
| 6 | 0001 |
| 7 | 0012 |
| 8 | 1213 |
0000:0015:0000:0000:0000:0001:0012:1213.
Final answer: 0000:0015:0000:0000:0000:0001:0012:1213
| Feature | IPv4 | IPv6 |
|---|---|---|
| Address width | 32 bits | 128 bits |
| Base/header size | 20–60 bytes depending on options | 40-byte fixed base header |
| Length field | Total length includes IPv4 header | Payload Length excludes 40-byte base header |
| Lifetime field | TTL | Hop Limit |
| Following protocol | Protocol field | Next Header; may identify an extension |
| Header checksum | Present | No base-header checksum |
| Options | Inside IPv4 header | Typically extension headers |
| Fragmentation fields | In IPv4 header; routers may fragment if permitted | Fragment extension header; fragmentation by source, not routers |
The supplied scheme’s sentence says four missing groups, but five groups are already visible, so only three are missing. Its printed eight-group expansion is consistent with three. The count shown above is independently verified.
Source: Final · 5 December 2025, original question paper page 2. Original marks retained; solution independently authored.
"},{"id":"q4b","title":"Q4B · UDP and TCP","html":"User Datagram Protocol (UDP) provides connectionless, message-oriented process delivery. One send creates a datagram; message boundaries are retained. UDP has an 8-byte header, carries ports and length, and offers checksum-based corruption detection. It does not itself guarantee delivery, ordering, duplicate suppression, retransmission, flow control or congestion control.
| UDP field | Width | Purpose |
|---|---|---|
| Source port | 16 bits | Sending endpoint; a zero value can indicate no reply port when unused |
| Destination port | 16 bits | Receiving application endpoint |
| Length | 16 bits | UDP header plus UDP payload; minimum 8 bytes |
| Checksum | 16 bits | Error detection over UDP and an IP pseudo-header; not retransmission |
| Point | UDP | TCP |
|---|---|---|
| Setup | Connectionless | Three-way establishment |
| Data | Message/datagram boundaries | Ordered byte stream |
| Recovery | No built-in retransmission | ACKs, timers and retransmission |
| Control | No built-in flow/congestion control | Receive window and congestion window |
| Header | 8 bytes | At least 20 bytes |
| Use | Applications tolerant of loss or managing their own recovery | Reliable ordered byte transfer |
Source: Final · 5 December 2025, original question paper page 2. Original marks retained; solution independently authored.
"},{"id":"q5a","title":"Q5A · Selective Repeat with three bits","html":"Selective Repeat (SR) ARQ accepts and buffers valid out-of-order frames inside the receiver window, acknowledges them individually and retransmits only missing/timed-out frames. In this answer ACK i means “frame i received”, unlike the next-expected cumulative ACK convention used for GBN.
Source: Final · 5 December 2025, original question paper page 2. Original marks retained; solution independently authored.
"},{"id":"q5b","title":"Q5B · SMTP phases","html":"Simple Mail Transfer Protocol (SMTP) transfers email from a sending client/server to a receiving mail server using a reliable transport, normally TCP. It is a push/store-and-forward transfer protocol. User agents compose/read mail; mail-transfer agents relay it, consulting DNS MX records to find recipient-domain mail servers.
| Phase | Typical exchange | Purpose |
|---|---|---|
| Connection and greeting | Server 220; client EHLO (HELO for basic SMTP); server 250 capabilities | Identify peers and learn supported extensions |
| Envelope transfer | MAIL FROM:<sender>; RCPT TO:<recipient>; 250 replies | Declare return path and one or more delivery recipients |
| Message transfer | DATA; server 354; headers + blank line + body; single dot line; server 250 | Transmit the message after server permission; completion reply accepts responsibility |
| Termination | QUIT; server 221 | Close the SMTP session cleanly |
Source: Final · 5 December 2025, original question paper page 2. Original marks retained; solution independently authored.
"},{"id":"q6a","title":"Q6A · Unguided media","html":"| Type | Characteristic 1 | Characteristic 2 |
|---|---|---|
| Radio | Can provide broad-area/mobility support depending on band | Shared spectrum/interference and obstacles affect reception |
| Terrestrial microwave | Directional, usually line of sight | Needs antenna alignment; obstructions/weather can matter |
| Satellite microwave | Relay covers large or remote areas | Delay/capacity depend strongly on orbit/system |
| Infrared | Short range and generally blocked by opaque walls | Limited room-scale coverage reduces inter-room interference |
Many classifications group satellite links under microwave rather than making a separate spectral category. State that grouping consistently.
Source: Final · 5 December 2025, original question paper page 2. Original marks retained; solution independently authored.
"},{"id":"q6b","title":"Q6B · CSMA persistence","html":"| Rule | When idle | When busy | Trade-off |
|---|---|---|---|
| 1-persistent | Transmit immediately | Keep listening; transmit as soon as idle | Low idle delay, but waiting stations may start together |
| Non-persistent | Transmit | Wait a random time, then sense again | Less synchronized contention; may leave avoidable idle time |
| p-persistent (slotted) | Transmit with probability p; defer a slot with probability 1−p | Wait until available, then resume the probabilistic slot rule | Controls contention; needs slots and a choice of p |
All three sense the channel; their persistence rules differ. Propagation delay means carrier sensing cannot guarantee the absence of collisions. In p-persistent CSMA, each eligible idle-slot decision uses the specified probability.
Source: Final · 5 December 2025, original question paper page 2. Original marks retained; solution independently authored.
"},{"id":"q6c","title":"Q6C · Open-loop and closed-loop control","html":"Congestion occurs when offered traffic overloads network resources such as link capacity, router buffers or processing. Queueing delay grows, packets may be dropped, and excessive retransmission can make the overload worse. Flow control protects a receiver; congestion control protects the network.
| Approach | Policy / mechanism | How it helps |
|---|---|---|
| Open loop: prevent congestion without runtime congestion feedback | Retransmission policy | Carefully chosen timers/selective retransmission avoid needless extra traffic |
| Open loop | Window policy | Limit outstanding data and avoid excessive retransmission after loss |
| Open loop | Acknowledgement policy | Avoid unnecessarily frequent ACK traffic while retaining correct recovery |
| Open loop | Admission policy | Refuse new load when reserved capacity is insufficient |
| Closed loop: measure and react to congestion | Backpressure | A congested node tells upstream nodes to reduce/stop sending, possibly propagating hop by hop |
| Closed loop | Choke/control signalling | A congested node sends explicit feedback to the source |
| Closed loop | Implicit signalling | A sender infers congestion from loss/delay and changes its sending behaviour |
| Closed loop | Explicit marking | Network marks congestion (for example ECN); endpoints react to returned information |
For an answer asking for two of each: explain retransmission and admission policies under open loop, then backpressure and explicit signalling under closed loop, with the mechanism and its effect. These are general congestion-control classifications used in the course; they are not all mandatory algorithms implemented inside TCP. TCP commonly uses end-to-end feedback such as loss or ECN.
Source: Final · 5 December 2025, original question paper page 2. Original marks retained; solution independently authored.
"},{"id":"q7a","title":"Q7A · Dijkstra with all intermediate steps","html":"| Edge | Cost |
|---|---|
| A→B | 4 |
| A→C | 2 |
| B→C | 5 |
| B→D | 10 |
| C→D | 3 |
| C→E | 4 |
| D→E | 1 |
| D→F | 2 |
| E→F | 3 |
| F→G | 5 |
Source=A; target=G. Edges are directed; do not add reverse edges.
The path A to G, its total cost and intermediate work.
Use Dijkstra because all weights are nonnegative. Select the smallest unsettled total distance, not the cheapest isolated edge.
d[A]=0; every other distance=∞. No predecessors are known.
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Tentative |
| B | ∞ | — | Tentative |
| C | ∞ | — | Tentative |
| D | ∞ | — | Tentative |
| E | ∞ | — | Tentative |
| F | ∞ | — | Tentative |
| G | ∞ | — | Tentative |
Unsettled candidates before selection: A=0, B=∞, C=∞, D=∞, E=∞, F=∞, G=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| A→B | ∞ | 0+(4)=4 | Improve to 4; predecessor=A |
| A→C | ∞ | 0+(2)=2 | Improve to 2; predecessor=A |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 4 | A | Tentative |
| C | 2 | A | Tentative |
| D | ∞ | — | Tentative |
| E | ∞ | — | Tentative |
| F | ∞ | — | Tentative |
| G | ∞ | — | Tentative |
Unsettled candidates before selection: B=4, C=2, D=∞, E=∞, F=∞, G=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| C→D | ∞ | 2+(3)=5 | Improve to 5; predecessor=C |
| C→E | ∞ | 2+(4)=6 | Improve to 6; predecessor=C |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 4 | A | Tentative |
| C | 2 | A | Settled |
| D | 5 | C | Tentative |
| E | 6 | C | Tentative |
| F | ∞ | — | Tentative |
| G | ∞ | — | Tentative |
Unsettled candidates before selection: B=4, D=5, E=6, F=∞, G=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| B→C | 2 | 4+(5)=9 | Already settled; retain |
| B→D | 5 | 4+(10)=14 | Retain 5 |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 4 | A | Settled |
| C | 2 | A | Settled |
| D | 5 | C | Tentative |
| E | 6 | C | Tentative |
| F | ∞ | — | Tentative |
| G | ∞ | — | Tentative |
Unsettled candidates before selection: D=5, E=6, F=∞, G=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| D→E | 6 | 5+(1)=6 | Retain 6 (equal cost) |
| D→F | ∞ | 5+(2)=7 | Improve to 7; predecessor=D |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 4 | A | Settled |
| C | 2 | A | Settled |
| D | 5 | C | Settled |
| E | 6 | C | Tentative |
| F | 7 | D | Tentative |
| G | ∞ | — | Tentative |
Unsettled candidates before selection: E=6, F=7, G=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| E→F | 7 | 6+(3)=9 | Retain 7 |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 4 | A | Settled |
| C | 2 | A | Settled |
| D | 5 | C | Settled |
| E | 6 | C | Settled |
| F | 7 | D | Tentative |
| G | ∞ | — | Tentative |
Unsettled candidates before selection: F=7, G=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| F→G | ∞ | 7+(5)=12 | Improve to 12; predecessor=F |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 4 | A | Settled |
| C | 2 | A | Settled |
| D | 5 | C | Settled |
| E | 6 | C | Settled |
| F | 7 | D | Settled |
| G | 12 | F | Tentative |
Unsettled candidates before selection: G=12. Select the minimum; ties use alphabetical order.
No outgoing edges to relax.
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 4 | A | Settled |
| C | 2 | A | Settled |
| D | 5 | C | Settled |
| E | 6 | C | Settled |
| F | 7 | D | Settled |
| G | 12 | F | Settled |
Trace backwards: G ← F ← D ← C ← A. Reverse it: A → C → D → F → G.
Edge sum: 2 + 3 + 2 + 5 = 12.
Final answer: A → C → D → F → G; cost=12.
Source: Final · 5 December 2025, original question paper page 3. Original marks retained; solution independently authored.
"},{"id":"q7b","title":"Q7B · Token bucket","html":"A token bucket generates tokens at rate r up to capacity B. Sending one unit of data consumes the corresponding tokens. Tokens accumulate during idle periods only until the bucket is full. Data can therefore burst when saved tokens exist, subject to the physical link rate.
| Limitation | Reason |
|---|---|
| Bursts can still congest downstream queues | Local conformance does not ensure every later link has capacity |
| Parameters must match traffic and link requirements | A small B delays/drops legitimate packets; a large B permits disruptive bursts |
| No complete QoS guarantee alone | Admission, scheduling and end-to-end capacity may still be needed |
Source: Final · 5 December 2025, original question paper page 3. Original marks retained; solution independently authored.
"}]},{"id":"tee-paper-reexam","title":"Re-exam · 4 February 2026","scope":"TEE · Actual TEE paper · all questions · current syllabus 702AI0C023","sections":[{"id":"q1a","title":"Q1A · Client-server and peer-to-peer","html":"In a client-server design, users’ clients contact designated service processes. In peer-to-peer (P2P), participating hosts can supply and consume resources. These are application organizations, not different physical cable types.
| Point | Client-server | Peer-to-peer |
|---|---|---|
| Roles | Clients request; designated servers provide services | Peers can both request and provide resources |
| Control | Central policies, accounts and backups are easier to coordinate | Control and copies distributed among peers |
| Performance | Predictable with adequate server capacity; server may bottleneck | Resources grow with peers, but peer capacity/availability vary |
| Scalability | Scale servers, replicas, caching and load distribution | More peers can contribute capacity; coordination becomes harder |
| Reliability | One unreplicated server is a failure point; replication helps | Loss of one peer need not stop system if useful replicas remain |
| Security | Central access control and monitoring | Trust and consistent access policy are harder |
| Best fit | Managed collaboration, databases, login services | Distributed content sharing among participating peers |
Central servers can authenticate users, maintain the authoritative shared document and control versions. The organization must provision capacity, replicate data and avoid a single server failure. P2P can exchange large files or local edits among peers and distribute load, but needs discovery, conflict resolution, trust and availability strategies. A hybrid can keep account control centralized while distributing selected content.
Source: Re-exam · 4 February 2026, original question paper page 1. Original marks retained; solution independently authored.
"},{"id":"q1b","title":"Q1B · TCP/IP reference model","html":"| Layer | Task | Examples |
|---|---|---|
| Application | Defines application message formats and exchanges for Web access, naming, email and file transfer; includes presentation/session functions as needed | HTTP, DNS, SMTP, FTP |
| Transport | End-to-end process communication | TCP, UDP, SCTP |
| Internet | IP addressing, routing and best-effort datagrams | IPv4/IPv6 |
| Link/network access | Local delivery and physical access | Ethernet/Wi-Fi and their medium |
Here the four-layer convention includes physical access within the link layer. A five-layer textbook diagram may split physical from data link; label which convention is being used.
Source: Re-exam · 4 February 2026, original question paper page 1. Original marks retained; solution independently authored.
"},{"id":"q1c","title":"Q1C · Pure ALOHA","html":"Protocol = Pure ALOHA; G = 0.5 attempts/frame time.
S, 100S and the maximum possible S.
Select the formula from the protocol name. The exponent is −2G because Pure ALOHA has a 2T vulnerable period.
−2 × 0.5 = -1.
S = 0.5 × 0.367879441 ≈ 0.183939721 successful frames/frame time.
0.183939721 × 100 ≈ 18.3940%.
The maximum is 1/(2e) ≈ 0.183940 = 18.3940%, at G=0.5.
Final answer: S ≈ 0.183940; percentage ≈ 18.3940%; Smax ≈ 0.183940 at G=0.5.
Source: Re-exam · 4 February 2026, original question paper page 1. Original marks retained; solution independently authored.
"},{"id":"q1d","title":"Q1D · IPv4 address block","html":"IP/prefix = 193.20.35.40/26.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−26=6; N=26=64.
Mask=255.255.255.192; binary=11111111.11111111.11111111.11000000.
The changing octet is number 4. Its mask value is 192, so step=256−192=64. The IP value 40 lies in 0…63.
Keep earlier octets; place 0 in octet 4; set any later octets to zero. Network=193.20.35.0.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 11000001 | 11111111 | 11000001 | 193 |
| 2 | 00010100 | 11111111 | 00010100 | 20 |
| 3 | 00100011 | 11111111 | 00100011 | 35 |
| 4 | 00101000 | 11000000 | 00000000 | 0 |
Use 63 in the changing octet and 255 in any later octets. Broadcast=193.20.35.63.
First usable=193.20.35.1; last usable=193.20.35.62; count=64−2=62.
Final answer: Network 193.20.35.0; broadcast 193.20.35.63; 64 total, 62 ordinary usable; range 193.20.35.1–193.20.35.62.
Source: Re-exam · 4 February 2026, original question paper page 1. Original marks retained; solution independently authored.
"},{"id":"q2a","title":"Q2A · CRC polynomial division","html":"Message=110010; generator=1101; generator degree r=3. This is textbook systematic CRC with zero initial remainder and no reflection/final XOR.
Exactly r CRC bits, the transmitted frame and receiver syndromes.
Modulo-2 division uses aligned XOR, not ordinary decimal division.
Message bits=110010; generator bits=1101. Leading exponent is 3, so CRC length is 3.
110010 followed by 3 zeros = 110010000.
| Before | Aligned generator | After XOR |
|---|---|---|
| 110010000 | 110100000 | 000110000 |
| 000110000 | 000110100 | 000000100 |
Remainder padded to 3 bits = 100. Leading zero columns need no subtraction.
Message + CRC = 110010 + 100 = 110010100.
| Before | Aligned generator | After XOR |
|---|---|---|
| 110010100 | 110100000 | 000110100 |
| 000110100 | 000110100 | 000000000 |
Receiver remainder = 000; the uncorrupted frame passes.
Flip the last bit: 110010101. Dividing by 1101 leaves 001, which is nonzero, so this error is detected. A zero syndrome would mean no detected error, not proof of no corruption.
Final answer: CRC=100; transmitted frame=110010100; uncorrupted receiver remainder=000.
Scheme correction: the remainder 100 is correct, but one scheme line prints 110010111. The correctly appended frame is 110010100; the receiver division shown above verifies it.
Source: Re-exam · 4 February 2026, original question paper page 1. Original marks retained; solution independently authored.
"},{"id":"q2b","title":"Q2B · Department subnet allocation","html":"| Department | Labs | Total addresses per lab |
|---|---|---|
| A | 32 | 128 |
| B | 64 | 64 |
| C | 16 | 512 |
Parent: 172.16.0.0/16 with 65,536 addresses.
Lab prefixes, department ranges and unused total addresses.
The question requests addresses, not usable hosts. Therefore do not add two before selecting each lab size.
128=2⁷ addresses/lab → /25. 32×128=4,096 addresses → department block /20. Allocate 172.16.0.0/20, spanning 172.16.0.0–172.16.15.255.
64=2⁶ → /26 per lab. 64×64=4,096 → /20 department block. Allocate 172.16.16.0/20, spanning 172.16.16.0–172.16.31.255.
512=2⁹ → /23 per lab. 16×512=8,192 → /19 department block. Allocate 172.16.32.0/19, spanning 172.16.32.0–172.16.63.255. Its start at third octet 32 is aligned for /19.
| Lab | Prefix | First address | Last address |
|---|---|---|---|
| A1 | 172.16.0.0/25 | 172.16.0.0 | 172.16.0.127 |
| A2 | 172.16.0.128/25 | 172.16.0.128 | 172.16.0.255 |
| A3 | 172.16.1.0/25 | 172.16.1.0 | 172.16.1.127 |
| A4 | 172.16.1.128/25 | 172.16.1.128 | 172.16.1.255 |
| A5 | 172.16.2.0/25 | 172.16.2.0 | 172.16.2.127 |
| A6 | 172.16.2.128/25 | 172.16.2.128 | 172.16.2.255 |
| A7 | 172.16.3.0/25 | 172.16.3.0 | 172.16.3.127 |
| A8 | 172.16.3.128/25 | 172.16.3.128 | 172.16.3.255 |
| A9 | 172.16.4.0/25 | 172.16.4.0 | 172.16.4.127 |
| A10 | 172.16.4.128/25 | 172.16.4.128 | 172.16.4.255 |
| A11 | 172.16.5.0/25 | 172.16.5.0 | 172.16.5.127 |
| A12 | 172.16.5.128/25 | 172.16.5.128 | 172.16.5.255 |
| A13 | 172.16.6.0/25 | 172.16.6.0 | 172.16.6.127 |
| A14 | 172.16.6.128/25 | 172.16.6.128 | 172.16.6.255 |
| A15 | 172.16.7.0/25 | 172.16.7.0 | 172.16.7.127 |
| A16 | 172.16.7.128/25 | 172.16.7.128 | 172.16.7.255 |
| A17 | 172.16.8.0/25 | 172.16.8.0 | 172.16.8.127 |
| A18 | 172.16.8.128/25 | 172.16.8.128 | 172.16.8.255 |
| A19 | 172.16.9.0/25 | 172.16.9.0 | 172.16.9.127 |
| A20 | 172.16.9.128/25 | 172.16.9.128 | 172.16.9.255 |
| A21 | 172.16.10.0/25 | 172.16.10.0 | 172.16.10.127 |
| A22 | 172.16.10.128/25 | 172.16.10.128 | 172.16.10.255 |
| A23 | 172.16.11.0/25 | 172.16.11.0 | 172.16.11.127 |
| A24 | 172.16.11.128/25 | 172.16.11.128 | 172.16.11.255 |
| A25 | 172.16.12.0/25 | 172.16.12.0 | 172.16.12.127 |
| A26 | 172.16.12.128/25 | 172.16.12.128 | 172.16.12.255 |
| A27 | 172.16.13.0/25 | 172.16.13.0 | 172.16.13.127 |
| A28 | 172.16.13.128/25 | 172.16.13.128 | 172.16.13.255 |
| A29 | 172.16.14.0/25 | 172.16.14.0 | 172.16.14.127 |
| A30 | 172.16.14.128/25 | 172.16.14.128 | 172.16.14.255 |
| A31 | 172.16.15.0/25 | 172.16.15.0 | 172.16.15.127 |
| A32 | 172.16.15.128/25 | 172.16.15.128 | 172.16.15.255 |
| B1 | 172.16.16.0/26 | 172.16.16.0 | 172.16.16.63 |
| B2 | 172.16.16.64/26 | 172.16.16.64 | 172.16.16.127 |
| B3 | 172.16.16.128/26 | 172.16.16.128 | 172.16.16.191 |
| B4 | 172.16.16.192/26 | 172.16.16.192 | 172.16.16.255 |
| B5 | 172.16.17.0/26 | 172.16.17.0 | 172.16.17.63 |
| B6 | 172.16.17.64/26 | 172.16.17.64 | 172.16.17.127 |
| B7 | 172.16.17.128/26 | 172.16.17.128 | 172.16.17.191 |
| B8 | 172.16.17.192/26 | 172.16.17.192 | 172.16.17.255 |
| B9 | 172.16.18.0/26 | 172.16.18.0 | 172.16.18.63 |
| B10 | 172.16.18.64/26 | 172.16.18.64 | 172.16.18.127 |
| B11 | 172.16.18.128/26 | 172.16.18.128 | 172.16.18.191 |
| B12 | 172.16.18.192/26 | 172.16.18.192 | 172.16.18.255 |
| B13 | 172.16.19.0/26 | 172.16.19.0 | 172.16.19.63 |
| B14 | 172.16.19.64/26 | 172.16.19.64 | 172.16.19.127 |
| B15 | 172.16.19.128/26 | 172.16.19.128 | 172.16.19.191 |
| B16 | 172.16.19.192/26 | 172.16.19.192 | 172.16.19.255 |
| B17 | 172.16.20.0/26 | 172.16.20.0 | 172.16.20.63 |
| B18 | 172.16.20.64/26 | 172.16.20.64 | 172.16.20.127 |
| B19 | 172.16.20.128/26 | 172.16.20.128 | 172.16.20.191 |
| B20 | 172.16.20.192/26 | 172.16.20.192 | 172.16.20.255 |
| B21 | 172.16.21.0/26 | 172.16.21.0 | 172.16.21.63 |
| B22 | 172.16.21.64/26 | 172.16.21.64 | 172.16.21.127 |
| B23 | 172.16.21.128/26 | 172.16.21.128 | 172.16.21.191 |
| B24 | 172.16.21.192/26 | 172.16.21.192 | 172.16.21.255 |
| B25 | 172.16.22.0/26 | 172.16.22.0 | 172.16.22.63 |
| B26 | 172.16.22.64/26 | 172.16.22.64 | 172.16.22.127 |
| B27 | 172.16.22.128/26 | 172.16.22.128 | 172.16.22.191 |
| B28 | 172.16.22.192/26 | 172.16.22.192 | 172.16.22.255 |
| B29 | 172.16.23.0/26 | 172.16.23.0 | 172.16.23.63 |
| B30 | 172.16.23.64/26 | 172.16.23.64 | 172.16.23.127 |
| B31 | 172.16.23.128/26 | 172.16.23.128 | 172.16.23.191 |
| B32 | 172.16.23.192/26 | 172.16.23.192 | 172.16.23.255 |
| B33 | 172.16.24.0/26 | 172.16.24.0 | 172.16.24.63 |
| B34 | 172.16.24.64/26 | 172.16.24.64 | 172.16.24.127 |
| B35 | 172.16.24.128/26 | 172.16.24.128 | 172.16.24.191 |
| B36 | 172.16.24.192/26 | 172.16.24.192 | 172.16.24.255 |
| B37 | 172.16.25.0/26 | 172.16.25.0 | 172.16.25.63 |
| B38 | 172.16.25.64/26 | 172.16.25.64 | 172.16.25.127 |
| B39 | 172.16.25.128/26 | 172.16.25.128 | 172.16.25.191 |
| B40 | 172.16.25.192/26 | 172.16.25.192 | 172.16.25.255 |
| B41 | 172.16.26.0/26 | 172.16.26.0 | 172.16.26.63 |
| B42 | 172.16.26.64/26 | 172.16.26.64 | 172.16.26.127 |
| B43 | 172.16.26.128/26 | 172.16.26.128 | 172.16.26.191 |
| B44 | 172.16.26.192/26 | 172.16.26.192 | 172.16.26.255 |
| B45 | 172.16.27.0/26 | 172.16.27.0 | 172.16.27.63 |
| B46 | 172.16.27.64/26 | 172.16.27.64 | 172.16.27.127 |
| B47 | 172.16.27.128/26 | 172.16.27.128 | 172.16.27.191 |
| B48 | 172.16.27.192/26 | 172.16.27.192 | 172.16.27.255 |
| B49 | 172.16.28.0/26 | 172.16.28.0 | 172.16.28.63 |
| B50 | 172.16.28.64/26 | 172.16.28.64 | 172.16.28.127 |
| B51 | 172.16.28.128/26 | 172.16.28.128 | 172.16.28.191 |
| B52 | 172.16.28.192/26 | 172.16.28.192 | 172.16.28.255 |
| B53 | 172.16.29.0/26 | 172.16.29.0 | 172.16.29.63 |
| B54 | 172.16.29.64/26 | 172.16.29.64 | 172.16.29.127 |
| B55 | 172.16.29.128/26 | 172.16.29.128 | 172.16.29.191 |
| B56 | 172.16.29.192/26 | 172.16.29.192 | 172.16.29.255 |
| B57 | 172.16.30.0/26 | 172.16.30.0 | 172.16.30.63 |
| B58 | 172.16.30.64/26 | 172.16.30.64 | 172.16.30.127 |
| B59 | 172.16.30.128/26 | 172.16.30.128 | 172.16.30.191 |
| B60 | 172.16.30.192/26 | 172.16.30.192 | 172.16.30.255 |
| B61 | 172.16.31.0/26 | 172.16.31.0 | 172.16.31.63 |
| B62 | 172.16.31.64/26 | 172.16.31.64 | 172.16.31.127 |
| B63 | 172.16.31.128/26 | 172.16.31.128 | 172.16.31.191 |
| B64 | 172.16.31.192/26 | 172.16.31.192 | 172.16.31.255 |
| C1 | 172.16.32.0/23 | 172.16.32.0 | 172.16.33.255 |
| C2 | 172.16.34.0/23 | 172.16.34.0 | 172.16.35.255 |
| C3 | 172.16.36.0/23 | 172.16.36.0 | 172.16.37.255 |
| C4 | 172.16.38.0/23 | 172.16.38.0 | 172.16.39.255 |
| C5 | 172.16.40.0/23 | 172.16.40.0 | 172.16.41.255 |
| C6 | 172.16.42.0/23 | 172.16.42.0 | 172.16.43.255 |
| C7 | 172.16.44.0/23 | 172.16.44.0 | 172.16.45.255 |
| C8 | 172.16.46.0/23 | 172.16.46.0 | 172.16.47.255 |
| C9 | 172.16.48.0/23 | 172.16.48.0 | 172.16.49.255 |
| C10 | 172.16.50.0/23 | 172.16.50.0 | 172.16.51.255 |
| C11 | 172.16.52.0/23 | 172.16.52.0 | 172.16.53.255 |
| C12 | 172.16.54.0/23 | 172.16.54.0 | 172.16.55.255 |
| C13 | 172.16.56.0/23 | 172.16.56.0 | 172.16.57.255 |
| C14 | 172.16.58.0/23 | 172.16.58.0 | 172.16.59.255 |
| C15 | 172.16.60.0/23 | 172.16.60.0 | 172.16.61.255 |
| C16 | 172.16.62.0/23 | 172.16.62.0 | 172.16.63.255 |
4,096+4,096+8,192=16,384 addresses.
65,536−16,384=49,152 addresses remain unallocated, from 172.16.64.0 through 172.16.255.255.
Final answer: A: /25 per lab; B: /26 per lab; C: /23 per lab. 49,152 total addresses remain unallocated.
Source: Re-exam · 4 February 2026, original question paper page 1. Original marks retained; solution independently authored.
"},{"id":"q3a","title":"Q3A · Network definition and topologies","html":"A computer network is a set of interconnected devices exchanging data through protocols. Topology describes the arrangement of nodes/links. Two examples are star and bus.
Source: Re-exam · 4 February 2026, original question paper page 2. Original marks retained; solution independently authored.
"},{"id":"q3b","title":"Q3B · OSPF best path","html":"Source: Re-exam · 4 February 2026, original question paper page 2. Original marks retained; solution independently authored.
"},{"id":"q3c","title":"Q3C · DNS namespace","html":"The Domain Name System (DNS) is a distributed hierarchical naming system. Its namespace is a tree with the root at the top. A complete name is the ordered path of labels from a node back to the root. An absolute name may be written with a trailing dot, for example www.example.com.
| Level | Example | Meaning |
|---|---|---|
| Root | . | Top of the namespace |
| Top-level domain (TLD) | com, org, in | A child zone below the root |
| Registered/delegated domain | example.com | Name under a parent such as com |
| Subdomain/host label | www.example.com | A label below example.com; a name may own different record types |
Source: Re-exam · 4 February 2026, original question paper page 2. Original marks retained; solution independently authored.
"},{"id":"q4a","title":"Q4A · IPv6 packet format","html":"| Field | Bits | Purpose |
|---|---|---|
| Version | 4 | Value 6 |
| Traffic Class | 8 | Traffic treatment/congestion-related marking |
| Flow Label | 20 | Identifies a flow for appropriate handling |
| Payload Length | 16 | Bytes after the base header; includes extension headers when present |
| Next Header | 8 | Type of the next extension header or upper-layer protocol |
| Hop Limit | 8 | Decremented on forwarding; limits packet lifetime in hops |
| Source address | 128 | Sending interface address |
| Destination address | 128 | Destination address |
The first two 32-bit rows contain 64 bits and the two addresses contain 256 bits, giving a 320-bit base header. Payload Length excludes these 40 bytes. Next Header can identify an extension header or upper-layer protocol. IPv6 has no base-header checksum; optional information is carried through extension headers.
Source: Re-exam · 4 February 2026, original question paper page 2. Original marks retained; solution independently authored.
"},{"id":"q4b","title":"Q4B · TCP connection lifecycle","html":"A TCP connection establishes each endpoint’s initial sequence number and confirms that both can receive. Use client initial sequence x=1000 and server initial sequence y=5000; assume no payload in the handshake.
| Step | Segment | Meaning |
|---|---|---|
| 1 Client → server | SYN, Seq=1000 | Client proposes its starting sequence |
| 2 Server → client | SYN+ACK, Seq=5000, Ack=1001 | Server acknowledges client SYN and supplies its own start |
| 3 Client → server | ACK, Seq=1001, Ack=5001 | Client acknowledges server SYN; connection is established |
Client sends 200 bytes beginning at Seq=1001, so byte numbers are 1001…1200. The receiver acknowledges Ack=1201, the next byte expected. If bytes starting at 1201 are missing but later bytes arrive, the cumulative ACK remains 1201 until the gap is repaired. TCP may buffer out-of-order bytes; it delivers the byte stream in order. Each direction counts independently.
Source: Re-exam · 4 February 2026, original question paper page 2. Original marks retained; solution independently authored.
"},{"id":"q5a","title":"Q5A · Go-Back-N three cases","html":"Go-Back-N (GBN) ARQ pipelines up to W outstanding frames. For m-bit sequence numbers, textbook GBN uses W ≤ 2m−1 and receiver window 1. The receiver accepts only the next in-order frame; out-of-order frames are discarded. ACK k here cumulatively acknowledges all frames before k and means “next expected is k”.
Source: Re-exam · 4 February 2026, original question paper page 2. Original marks retained; solution independently authored.
"},{"id":"q5b","title":"Q5B · FTP connections","html":"File Transfer Protocol (FTP) transfers files between a client and server using TCP and separates commands from file data. This separation is its central design feature. Its control connection remains available across multiple operations, while data connections carry file bytes or directory listings.
| Connection | Function | Lifetime / ports |
|---|---|---|
| Control | Commands and replies: USER, PASS, CWD, TYPE, LIST, RETR, STOR, QUIT | Client connects to server TCP port 21; normally kept for the session |
| Data | File contents and directory listing data | Opened as needed; separate from command channel |
FTP over TLS (FTPS) protects FTP connections. SFTP is a separate SSH-based file-transfer protocol, not “FTP using an SSH port”. These names should not be treated as interchangeable.
Source: Re-exam · 4 February 2026, original question paper page 2. Original marks retained; solution independently authored.
"},{"id":"q6a","title":"Q6A · Media classification and fiber","html":"Transmission media are classified into guided media (twisted pair, coaxial cable, optical fiber) and unguided media (radio, microwave, infrared).
Optical fiber carries data as light in a core surrounded by cladding. It is a guided medium. Single-mode fiber supports one main propagation mode and is used for long/high-capacity links; multimode supports multiple modes and is often used over shorter distances.
| Advantages | Disadvantages / design costs |
|---|---|
| High data capacity and low attenuation over suitable links | Optical transceivers, connectors and splicing add cost/skill requirements |
| Immune to electromagnetic interference; no electrical crosstalk | Tight bends, dirty connectors and poor installation increase loss |
| Electrical isolation between buildings | Fiber is mechanically sensitive and requires appropriate protection |
| Harder to tap casually than an unprotected radio channel | It is not automatically confidential; encryption/access control remain necessary |
Source: Re-exam · 4 February 2026, original question paper page 2. Original marks retained; solution independently authored.
"},{"id":"q6b","title":"Q6B · ALOHA at three rates","html":"Interpret the stated aggregate production rates as offered attempt rates in the standard textbook model. The following working includes normalized throughput, successful frames/s and successful frame bits/s for each case.
| Symbol | Given |
|---|---|
| L | 200 bits |
| R | 200 kbps = 200,000 bits/s |
| λ | 1000 aggregate attempts/s |
T, G, S, successful frames/s and bit/s.
First normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
T = 200/200,000 = 0.001 s = 1 ms.
G = 1000 × 0.001 = 1 attempts/frame time.
−2G = -2; e to that power ≈ 0.135335283.
S = 1 × 0.135335283 ≈ 0.135335283.
S/T = 0.135335283/0.001 ≈ 135.335283 frames/s.
SR = 0.135335283 × 200,000 ≈ 27067.056647 bits/s ≈ 27.067057 kbps.
Final answer: G=1, S≈0.135335, successful rate≈135.335 frames/s = 27.067 kbps.
| Symbol | Given |
|---|---|
| L | 200 bits |
| R | 200 kbps = 200,000 bits/s |
| λ | 500 aggregate attempts/s |
T, G, S, successful frames/s and bit/s.
First normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
T = 200/200,000 = 0.001 s = 1 ms.
G = 500 × 0.001 = 0.5 attempts/frame time.
−2G = -1; e to that power ≈ 0.367879441.
S = 0.5 × 0.367879441 ≈ 0.183939721.
S/T = 0.183939721/0.001 ≈ 183.939721 frames/s.
SR = 0.183939721 × 200,000 ≈ 36787.944117 bits/s ≈ 36.787944 kbps.
Final answer: G=0.5, S≈0.183940, successful rate≈183.940 frames/s = 36.788 kbps.
| Symbol | Given |
|---|---|
| L | 200 bits |
| R | 200 kbps = 200,000 bits/s |
| λ | 250 aggregate attempts/s |
T, G, S, successful frames/s and bit/s.
First normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
T = 200/200,000 = 0.001 s = 1 ms.
G = 250 × 0.001 = 0.25 attempts/frame time.
−2G = -0.5; e to that power ≈ 0.606530660.
S = 0.25 × 0.606530660 ≈ 0.151632665.
S/T = 0.151632665/0.001 ≈ 151.632665 frames/s.
SR = 0.151632665 × 200,000 ≈ 30326.532986 bits/s ≈ 30.326533 kbps.
Final answer: G=0.25, S≈0.151633, successful rate≈151.633 frames/s = 30.327 kbps.
Source: Re-exam · 4 February 2026, original question paper page 2. Original marks retained; solution independently authored.
"},{"id":"q6c","title":"Q6C · UDP TCP SCTP comparison","html":"| Point | UDP | TCP | SCTP |
|---|---|---|---|
| Service unit | Datagram; preserves messages | Byte stream; no message boundaries | Messages carried in chunks; preserves boundaries |
| Connection setup | No protocol handshake | Connection; usual three-way SYN exchange | Association; four-message cookie exchange |
| Reliability | No delivery/order guarantee | Reliable ordered stream | Reliable messages; ordered per stream by default, unordered option |
| Ordering scope | None supplied | One ordered byte stream per direction | Multiple streams; per-stream order avoids blocking every stream on one stream’s ordering gap |
| Flow control | Not supplied by UDP | Receiver-advertised window | Receiver flow control |
| Congestion control | Not supplied by UDP | Yes | Yes |
| Multihoming | No built-in association failover | Classic TCP uses one address pair | Association can bind multiple addresses; built-in path failover |
| Header | 8 bytes | Minimum 20 bytes | 12-byte common header plus chunks |
| Typical use | DNS, real-time applications with their own controls | Web over HTTP/1.1 or HTTP/2, reliable byte transfers | Telecom signalling and multi-stream message applications |
| Performance claim | Low protocol overhead, not automatically fastest overall | Recovery/ordering have costs and benefits | Extra capabilities; no universal speed ranking |
The correct protocol depends on application semantics and network conditions. “UDP is always faster” and “SCTP is always slower than TCP” are not defensible general rules. All three normally run above IP; security/encryption is a separate requirement.
Source: Re-exam · 4 February 2026, original question paper page 2. Original marks retained; solution independently authored.
"},{"id":"q7a","title":"Q7A · Bellman–Ford intermediate steps","html":"| Directed edge | Weight |
|---|---|
| A→B | 7 |
| A→C | 6 |
| A→D | 5 |
| B→E | -2 |
| D→C | -2 |
| C→B | -2 |
| C→E | 1 |
| E→G | 5 |
| D→F | 3 |
| F→G | 4 |
The shortest route and cost, with complete passes and a negative-cycle check.
Bellman–Ford permits negative edges. We use the listed order with in-place updates; do not mix this with a table that reads only the previous pass.
7 vertices → at most 6 main passes. d[A]=0; every other distance=∞. Use the listed edge order and immediate (in-place) updates.
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| A→B | ∞ | 0+(7)=7 | 7 | Update; predecessor=A |
| A→C | ∞ | 0+(6)=6 | 6 | Update; predecessor=A |
| A→D | ∞ | 0+(5)=5 | 5 | Update; predecessor=A |
| B→E | ∞ | 7+(-2)=5 | 5 | Update; predecessor=B |
| D→C | 6 | 5+(-2)=3 | 3 | Update; predecessor=D |
| C→B | 7 | 3+(-2)=1 | 1 | Update; predecessor=C |
| C→E | 5 | 3+(1)=4 | 4 | Update; predecessor=C |
| E→G | ∞ | 4+(5)=9 | 9 | Update; predecessor=E |
| D→F | ∞ | 5+(3)=8 | 8 | Update; predecessor=D |
| F→G | 9 | 8+(4)=12 | 9 | No improvement |
| Vertex | A | B | C | D | E | F | G |
|---|---|---|---|---|---|---|---|
| End-of-pass distance | 0 | 1 | 3 | 5 | 4 | 8 | 9 |
| Predecessor | — | C | D | A | C | D | E |
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| A→B | 1 | 0+(7)=7 | 1 | No improvement |
| A→C | 3 | 0+(6)=6 | 3 | No improvement |
| A→D | 5 | 0+(5)=5 | 5 | No improvement |
| B→E | 4 | 1+(-2)=-1 | -1 | Update; predecessor=B |
| D→C | 3 | 5+(-2)=3 | 3 | No improvement |
| C→B | 1 | 3+(-2)=1 | 1 | No improvement |
| C→E | -1 | 3+(1)=4 | -1 | No improvement |
| E→G | 9 | -1+(5)=4 | 4 | Update; predecessor=E |
| D→F | 8 | 5+(3)=8 | 8 | No improvement |
| F→G | 4 | 8+(4)=12 | 4 | No improvement |
| Vertex | A | B | C | D | E | F | G |
|---|---|---|---|---|---|---|---|
| End-of-pass distance | 0 | 1 | 3 | 5 | -1 | 8 | 4 |
| Predecessor | — | C | D | A | B | D | E |
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| A→B | 1 | 0+(7)=7 | 1 | No improvement |
| A→C | 3 | 0+(6)=6 | 3 | No improvement |
| A→D | 5 | 0+(5)=5 | 5 | No improvement |
| B→E | -1 | 1+(-2)=-1 | -1 | No improvement |
| D→C | 3 | 5+(-2)=3 | 3 | No improvement |
| C→B | 1 | 3+(-2)=1 | 1 | No improvement |
| C→E | -1 | 3+(1)=4 | -1 | No improvement |
| E→G | 4 | -1+(5)=4 | 4 | No improvement |
| D→F | 8 | 5+(3)=8 | 8 | No improvement |
| F→G | 4 | 8+(4)=12 | 4 | No improvement |
| Vertex | A | B | C | D | E | F | G |
|---|---|---|---|---|---|---|---|
| End-of-pass distance | 0 | 1 | 3 | 5 | -1 | 8 | 4 |
| Predecessor | — | C | D | A | B | D | E |
This entire pass changed no distance. Further identical relaxation passes will not improve anything.
Backward predecessors: G ← E ← B ← C ← D ← A. Forward route: A → D → C → B → E → G.
Cost = (5) + (-2) + (-2) + (-2) + (5) = 4. No edge can now improve its destination label.
Final answer: A → D → C → B → E → G; shortest cost=4.
Source: Re-exam · 4 February 2026, original question paper page 3. Original marks retained; solution independently authored.
"},{"id":"q7b","title":"Q7B · Leaky bucket and comparison","html":"In the textbook leaky-bucket queue/shaper model, arriving packets enter a finite buffer. The output sends at a configured rate whenever data is queued. When the queue empties, it cannot send nonexistent packets; when the buffer is full, excess arrivals are dropped in this finite-buffer model.
| Advantage | Limitation |
|---|---|
| Simple bound on output rate | Bursts may overflow or suffer delay |
| Reduces short-term output variation | Unused idle service cannot be saved |
| Useful for predictable traffic shaping | No end-to-end delay/loss guarantee by itself |
The term “leaky bucket” can also describe a conformance meter in standards. This course’s diagrams and numerical questions explicitly use the finite data-queue model, so its state is queued data rather than tokens.
| Point | Leaky data queue | Token bucket |
|---|---|---|
| Stored state | Queued data | Tokens representing permission |
| Idle period | Empty queue; unused service lost | Tokens accumulate up to B |
| Output burst | Smoothed by configured service rate | Allowed if enough tokens and physical capacity exist |
| Overflow/shortage | Finite queue can overflow | Shaper waits; policer may mark/drop |
| Main parameters | Queue capacity and drain rate | Token capacity and generation rate |
| Useful when | A smoother output profile is required | Average rate plus burst allowance is required |
Neither is always advantageous. Choose leaky queue shaping for smooth departures; choose token bucket when controlled bursts after idle time are desirable.
Source: Re-exam · 4 February 2026, original question paper page 3. Original marks retained; solution independently authored.
"}]},{"id":"paper-2025-legacy-reexam","title":"2025-12-05 · 2022-23","unit":"Historical PYQ","scope":"Actual paper; 100 marks attempted. Current-syllabus alignment is stated per question.","sections":[{"id":"q1a","title":"Q1(a) · Network definition","html":"A computer network is a set of interconnected devices that exchange data and share resources using communication protocols. A link carries signals; a protocol defines message format, meaning and timing. Hosts run applications, switches forward local frames and routers connect IP networks.
| Type | Meaning and scope | Example |
|---|---|---|
| PAN | Personal Area Network: a person’s nearby devices | Phone linked to earbuds |
| LAN | Local Area Network: a building or local campus network under local administration | Laboratory Ethernet/Wi-Fi |
| MAN | Metropolitan Area Network: interconnects sites across a town/city | City-wide institutional network |
| WAN | Wide Area Network: connects geographically separated networks over regional/global links | Branches in different cities |
| ARPANET | Historical packet-switched research network; a major predecessor of the Internet | Early interconnected research institutions; not a modern geographic size category |
Classification by geographical scope is descriptive: a fixed kilometre boundary or a guaranteed speed is not part of every definition. Networks can also be classified by wired/wireless medium or by client-server/peer-to-peer organization.
Yaad rakho: PAN personal, LAN local, MAN city, WAN wide.
Host A ─┐\r\n ├─ Switch ─ Router ─ Other network\r\nHost B ─┘
The hosts exchange application data; the switch forwards local frames; the router connects different IP networks.
A transmission error changes one or more bits. A single-bit error flips one bit. A burst error affects a span from the first corrupted bit to the last; the bits in between need not all change. Detection identifies an invalid received pattern; correction identifies the intended pattern within the code’s guaranteed capability.
Yaad rakho: Detection asks error hua? Correction asks kaunsa correct word?
An IPv4 address identifies an IP interface. The network prefix selects the destination network; remaining bits identify an address within it. Classful addressing is historical; modern routing uses explicit CIDR prefixes.
| Class | First octet | Default mask | Network / host split | Purpose |
|---|---|---|---|---|
| A | 1–126 (ordinary unicast) | 255.0.0.0 (/8) | 8 / 24 bits | Large historical unicast networks; 0 and 127 reserved |
| B | 128–191 | 255.255.0.0 (/16) | 16 / 16 bits | Medium historical unicast networks |
| C | 192–223 | 255.255.255.0 (/24) | 24 / 8 bits | Small historical unicast networks |
| D | 224–239 | No unicast default mask | Not network/host class split | Multicast |
| E | 240–255 | No unicast default mask | Not normal unicast allocation | Reserved/experimental; special addresses exist |
Example: 114.34.2.8 is class A with network 114.0.0.0 and host portion 34.2.8. Class does not override an explicitly supplied CIDR prefix.
User Datagram Protocol (UDP) provides connectionless, message-oriented process delivery. One send creates a datagram; message boundaries are retained. UDP has an 8-byte header, carries ports and length, and offers checksum-based corruption detection. It does not itself guarantee delivery, ordering, duplicate suppression, retransmission, flow control or congestion control.
| UDP field | Width | Purpose |
|---|---|---|
| Source port | 16 bits | Sending endpoint; a zero value can indicate no reply port when unused |
| Destination port | 16 bits | Receiving application endpoint |
| Length | 16 bits | UDP header plus UDP payload; minimum 8 bytes |
| Checksum | 16 bits | Error detection over UDP and an IP pseudo-header; not retransmission |
Yaad rakho: UDP datagram preserve; reliability automatic nahi.
Topology is the arrangement of nodes and links. Physical topology describes actual links; logical topology describes how data flows. A switch-centred physical star is different from a shared collision-domain bus.
| Topology | Structure | Benefit | Limitation |
|---|---|---|---|
| Bus | Devices share a common backbone | Simple small shared-medium layout | Backbone fault affects segment; collisions possible on classic shared bus |
| Star | Each device connects to a central switch/hub | Easy expansion and fault isolation | Central-device failure affects star; more cabling |
| Ring | Each node connected to two neighbours in a ring | Structured traversal; token-based variants coordinate access | Simple single-ring break can disrupt operation unless protected |
| Full mesh | Every pair has a direct link | Redundant paths and direct connectivity | Expensive ports/cabling: n(n−1)/2 undirected links |
| Tree/hybrid | Hierarchical stars or combined topologies | Scales into buildings/floors | Upper-level faults affect dependent branches |
Yaad rakho: Layout draw karo, then one strength and one failure case.
| Topology / example | Advantage 1 | Advantage 2 | Disadvantage 1 | Disadvantage 2 |
|---|---|---|---|---|
| Star: switched computer laboratory | A device can be added on a spare central port without rebuilding the whole topology | A single access-cable fault normally isolates only its attached device | Central-switch failure affects all attached devices unless redundant | A separate cable and switch port are needed per device |
| Bus: legacy coaxial Ethernet | Uses little cable in a small linear layout | No central switch is required in the simple shared-bus design | A backbone fault can disable the segment | Shared bandwidth, contention and fault diagnosis become difficult as it grows |
| Ring: token-based ring | Controlled token access avoids data collisions among compliant stations | Token rotation gives stations an organized opportunity to transmit | A break can interrupt an unprotected single ring | Adding/removing a node can disrupt a simple ring |
| Mesh: redundant router backbone | Alternate links preserve connectivity after some link failures | Direct pairwise links can carry separate exchanges without a shared central bottleneck | A full mesh needs n(n−1)/2 links and many ports | Configuration, cabling and maintenance become complex |
The ring benefits above assume token-controlled access; physical ring wiring alone does not guarantee fairness or collision avoidance. The failure disadvantages describe simple layouts without protection/redundancy.
Stop-and-Wait ARQ permits only one outstanding frame. The sender stores its frame, transmits it and starts a timer. A valid matching ACK advances the sender; a timeout retransmits the stored frame.
Yaad rakho: One frame outstanding; duplicate ko acknowledge karo, deliver dobara nahi.
\r\nGo-Back-N (GBN) ARQ pipelines up to W outstanding frames. For m-bit sequence numbers, textbook GBN uses W ≤ 2m−1 and receiver window 1. The receiver accepts only the next in-order frame; out-of-order frames are discarded. ACK k here cumulatively acknowledges all frames before k and means “next expected is k”.
Yaad rakho: GBN receiver one; gap ke baad discard; timeout pe outstanding suffix repeat.
\r\n| Property | Stop-and-Wait | Go-Back-N | Selective Repeat |
|---|---|---|---|
| Outstanding sender frames | 1 | Up to 2^m−1 | Up to 2^(m−1), equal windows |
| Receiver window | 1 | 1 | Same maximum as sender in standard equal-window SR |
| ACK convention used here | Next expected | Cumulative, next expected | Individual frame accepted |
| Out-of-order data | Not delivered | Discarded | Buffered within valid window |
| After timeout | Repeat the one frame | Repeat outstanding frames from base | Repeat only timed-out frame |
| Buffer/tracking cost | Low | Sender window; simple receiver | Sender and receiver windows; per-frame tracking |
| Tradeoff | Simple but idle during RTT | Good pipelining, waste after loss | Efficient loss recovery, greater complexity |
Every comparison assumes finite sequence numbers, correct duplicate handling and the textbook protocol models. The same text “ACK 3” can mean different things under different ACK conventions; state its meaning before explaining a trace.
Go-Back-N is usually superior in link utilization when propagation delay is significant: it sends several frames before waiting, whereas Stop-and-Wait leaves the link idle during much of the round trip. Its cumulative ACKs also simplify receiver tracking. This is not an unconditional reliability ranking: GBN needs sender buffering and may retransmit correctly received later frames after one loss. Stop-and-Wait is simpler and can suffice for low-rate, short-delay exchanges. Both recover loss correctly when their timers and sequence-number rules are followed.
This historical question is outside the named current syllabus. It is retained in the map for a truthful paper inventory but excluded from the current exam-preparation sequence.
An IPv4 address identifies an IP interface. The network prefix selects the destination network; remaining bits identify an address within it. Classful addressing is historical; modern routing uses explicit CIDR prefixes.
| Class | First octet | Default mask | Network / host split | Purpose |
|---|---|---|---|---|
| A | 1–126 (ordinary unicast) | 255.0.0.0 (/8) | 8 / 24 bits | Large historical unicast networks; 0 and 127 reserved |
| B | 128–191 | 255.255.0.0 (/16) | 16 / 16 bits | Medium historical unicast networks |
| C | 192–223 | 255.255.255.0 (/24) | 24 / 8 bits | Small historical unicast networks |
| D | 224–239 | No unicast default mask | Not network/host class split | Multicast |
| E | 240–255 | No unicast default mask | Not normal unicast allocation | Reserved/experimental; special addresses exist |
Example: 114.34.2.8 is class A with network 114.0.0.0 and host portion 34.2.8. Class does not override an explicitly supplied CIDR prefix.
| Address | Class and mask | Network ID | Host ID |
|---|---|---|---|
| 114.34.2.8 | A /8 | 114.0.0.0 | 34.2.8 (24-bit host portion) |
| 208.34.54.12 | C /24 | 208.34.54.0 | 12 (8-bit host portion) |
165.21.80.128/26
First and last addresses and total block size.
p is the prefix length. The remaining host bits vary through every combination. The network address is the first address; setting all host bits to one gives the last address.
32 − 26 = 6. N = 64.
Mask = 255.255.255.192. The first varying octet is octet 4; its block width is 64. The supplied value 128 lies in the interval 128–191. Keep earlier octets fixed. Set any later octets to 0 for the first address and 255 for the last address.
First = 165.21.80.128; last = 165.21.80.191; total = 64 addresses. Usual usable host range: 165.21.80.129–165.21.80.190.
Asked host answer: 165.21.80.190; .191 is broadcast, not a usable host.
Definition. Persistence specifies how a station behaves after observing that the medium is busy or becomes idle.
Simple meaning. 1 = turant; non = random wait; p = probability ke basis pe start.
| Rule | When idle | When busy | Trade-off |
|---|---|---|---|
| 1-persistent | Transmit immediately | Keep listening; transmit as soon as idle | Low idle delay, but waiting stations may start together |
| Non-persistent | Transmit | Wait a random time, then sense again | Less synchronized contention; may leave avoidable idle time |
| p-persistent (slotted) | Transmit with probability p; defer a slot with probability 1−p | Wait until available, then resume the probabilistic slot rule | Controls contention; needs slots and a choice of p |
Yaad rakho: 1 = start now; non = wait then listen; p = chance per idle slot.
\r\nDefinition. CSMA with Collision Detection monitors the medium during transmission, aborts a detected collision and retries after a randomized backoff.
Simple meaning. CD = takkar detect hui toh transmission rok do.
CSMA/CD senses before transmission and detects collisions during transmission. A collision causes abort/jam and a random backoff before retry. To detect the most distant collision before the frame finishes, the transmission duration must be at least the worst-case round-trip propagation delay. Thus Lmin/R ≥ 2Tp in the idealized model.
Yaad rakho: Detect karna hai toh sender tab tak bhej raha hona chahiye.
1-persistent:\r\nReady → sense → busy? yes → sense again\r\n no → transmit immediately\r\n\r\np-persistent (slotted medium):\r\nReady → sense → busy? yes → wait and sense\r\n no → draw random decision\r\n probability p → transmit\r\n probability 1−p → wait one slot → sense again
Non-persistent CSMA instead waits a random time after a busy observation before sensing again. A collision leads to the protocol's retry/backoff procedure; the persistence rule alone does not guarantee success.
Definition. Open Shortest Path First (OSPF) is an interior link-state routing protocol that selects routes using shortest-path calculations and additive link costs.
Simple meaning. OSPF = link-state map + Dijkstra + summed costs.
OSPF exchanges and maintains link-state information, builds a topology database and runs Dijkstra’s shortest-path computation rooted at the local router. It chooses a path with the lowest sum of link costs and installs the corresponding next hop. Equal-cost paths may be supported.
Yaad rakho: OSPF cost add karo; sirf hops nahi.
| Type | Meaning | Diagram |
|---|---|---|
| Point-to-point | Direct connection between two routers | R1 — R2 |
| Transit network | Multiaccess network through which routers communicate; a designated router helps represent it | R1 — LAN — R2 (also R3) |
| Stub network | Network reachable through one router, with no onward transit via another router | R1 — end-host LAN |
| Virtual link | Logical OSPF backbone connection through a transit area | Area 0 — R1 ⋯ transit area ⋯ R2 — Area 0 |
Do not confuse these topology/link descriptions with the five OSPF packet types. Hello discovers/maintains neighbours; database description, link-state request, link-state update and acknowledgement synchronize link-state information.
Definition. Dijkstra’s algorithm finds shortest paths from a source in a graph whose edge weights are nonnegative.
Simple meaning. Sabse chhota unfinished distance select karo, phir uske neighbours improve karo.
The standard algorithm requires nonnegative edge weights.
Yaad rakho: Minimum tentative total choose; negative edges pe Dijkstra nahi.
\r\n| Edge | Cost |
|---|---|
| A→B | 4 |
| A→C | 2 |
| C→B | 1 |
| B→D | 3 |
| C→D | 7 |
Source=A; target=D. Edges are directed; do not add reverse edges.
The path A to D, its total cost and intermediate work.
Use Dijkstra because all weights are nonnegative. Select the smallest unsettled total distance, not the cheapest isolated edge.
d[A]=0; every other distance=∞. No predecessors are known.
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Tentative |
| B | ∞ | — | Tentative |
| C | ∞ | — | Tentative |
| D | ∞ | — | Tentative |
Unsettled candidates before selection: A=0, B=∞, C=∞, D=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| A→B | ∞ | 0+(4)=4 | Improve to 4; predecessor=A |
| A→C | ∞ | 0+(2)=2 | Improve to 2; predecessor=A |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 4 | A | Tentative |
| C | 2 | A | Tentative |
| D | ∞ | — | Tentative |
Unsettled candidates before selection: B=4, C=2, D=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| C→B | 4 | 2+(1)=3 | Improve to 3; predecessor=C |
| C→D | ∞ | 2+(7)=9 | Improve to 9; predecessor=C |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | C | Tentative |
| C | 2 | A | Settled |
| D | 9 | C | Tentative |
Unsettled candidates before selection: B=3, D=9. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| B→D | 9 | 3+(3)=6 | Improve to 6; predecessor=B |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | C | Settled |
| C | 2 | A | Settled |
| D | 6 | B | Tentative |
Unsettled candidates before selection: D=6. Select the minimum; ties use alphabetical order.
No outgoing edges to relax.
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | C | Settled |
| C | 2 | A | Settled |
| D | 6 | B | Settled |
Trace backwards: D ← B ← C ← A. Reverse it: A → C → B → D.
Edge sum: 2 + 1 + 3 = 6.
Final answer: A → C → B → D; cost=6.
Congestion occurs when offered traffic overloads network resources such as link capacity, router buffers or processing. Queueing delay grows, packets may be dropped, and excessive retransmission can make the overload worse. Flow control protects a receiver; congestion control protects the network.
| Approach | Policy / mechanism | How it helps |
|---|---|---|
| Open loop: prevent congestion without runtime congestion feedback | Retransmission policy | Carefully chosen timers/selective retransmission avoid needless extra traffic |
| Open loop | Window policy | Limit outstanding data and avoid excessive retransmission after loss |
| Open loop | Acknowledgement policy | Avoid unnecessarily frequent ACK traffic while retaining correct recovery |
| Open loop | Admission policy | Refuse new load when reserved capacity is insufficient |
| Closed loop: measure and react to congestion | Backpressure | A congested node tells upstream nodes to reduce/stop sending, possibly propagating hop by hop |
| Closed loop | Choke/control signalling | A congested node sends explicit feedback to the source |
| Closed loop | Implicit signalling | A sender infers congestion from loss/delay and changes its sending behaviour |
| Closed loop | Explicit marking | Network marks congestion (for example ECN); endpoints react to returned information |
For an answer asking for two of each: explain retransmission and admission policies under open loop, then backpressure and explicit signalling under closed loop, with the mechanism and its effect. These are general congestion-control classifications used in the course; they are not all mandatory algorithms implemented inside TCP. TCP commonly uses end-to-end feedback such as loss or ECN.
Yaad rakho: Open loop pehle prevent; closed loop feedback se react.
A TCP connection establishes each endpoint’s initial sequence number and confirms that both can receive. Use client initial sequence x=1000 and server initial sequence y=5000; assume no payload in the handshake.
| Step | Segment | Meaning |
|---|---|---|
| 1 Client → server | SYN, Seq=1000 | Client proposes its starting sequence |
| 2 Server → client | SYN+ACK, Seq=5000, Ack=1001 | Server acknowledges client SYN and supplies its own start |
| 3 Client → server | ACK, Seq=1001, Ack=5001 | Client acknowledges server SYN; connection is established |
Client sends 200 bytes beginning at Seq=1001, so byte numbers are 1001…1200. The receiver acknowledges Ack=1201, the next byte expected. If bytes starting at 1201 are missing but later bytes arrive, the cumulative ACK remains 1201 until the gap is repaired. TCP may buffer out-of-order bytes; it delivers the byte stream in order. Each direction counts independently.
Yaad rakho: SYN plus one, FIN plus one; ACK next expected byte.
The Domain Name System (DNS) is a distributed hierarchical naming system. Its namespace is a tree with the root at the top. A complete name is the ordered path of labels from a node back to the root. An absolute name may be written with a trailing dot, for example www.example.com.
| Level | Example | Meaning |
|---|---|---|
| Root | . | Top of the namespace |
| Top-level domain (TLD) | com, org, in | A child zone below the root |
| Registered/delegated domain | example.com | Name under a parent such as com |
| Subdomain/host label | www.example.com | A label below example.com; a name may own different record types |
Yaad rakho: Full path unique; www label akela unique nahi.
Client → recursive resolver → root (.)\r\n → .com TLD\r\n → authoritative server for xyz.com\r\nClient ← final A/AAAA answer ← resolver
Assume a cache miss and that the name exists. (1) The client asks its resolver for www.xyz.com. (2) A root server refers the resolver to .com servers. (3) A .com server refers it to authoritative servers for xyz.com. (4) An authoritative server returns the applicable A/AAAA record, or a CNAME that must be followed. (5) The resolver returns the answer and caches it according to TTL. No real IP address for this domain is assumed. Root and TLD servers normally supply referrals here, not the final host address.
A computer network is a set of interconnected devices that exchange data and share resources using communication protocols. A link carries signals; a protocol defines message format, meaning and timing. Hosts run applications, switches forward local frames and routers connect IP networks.
| Type | Meaning and scope | Example |
|---|---|---|
| PAN | Personal Area Network: a person’s nearby devices | Phone linked to earbuds |
| LAN | Local Area Network: a building or local campus network under local administration | Laboratory Ethernet/Wi-Fi |
| MAN | Metropolitan Area Network: interconnects sites across a town/city | City-wide institutional network |
| WAN | Wide Area Network: connects geographically separated networks over regional/global links | Branches in different cities |
| ARPANET | Historical packet-switched research network; a major predecessor of the Internet | Early interconnected research institutions; not a modern geographic size category |
Classification by geographical scope is descriptive: a fixed kilometre boundary or a guaranteed speed is not part of every definition. Networks can also be classified by wired/wireless medium or by client-server/peer-to-peer organization.
Yaad rakho: PAN personal, LAN local, MAN city, WAN wide.
A transmission error changes one or more bits. A single-bit error flips one bit. A burst error affects a span from the first corrupted bit to the last; the bits in between need not all change. Detection identifies an invalid received pattern; correction identifies the intended pattern within the code’s guaranteed capability.
Yaad rakho: Detection asks error hua? Correction asks kaunsa correct word?
\r\nHamming distance d(a,b) is the number of positions where two equal-length bit strings differ. XOR the strings and count the 1s. Minimum distance dmin is the smallest distance between distinct valid codewords; do not use the largest pairwise distance.
Yaad rakho: XOR ones count; minimum minus one detect, half floor correct.
Definition. Address Resolution Protocol (ARP) resolves a known IPv4 address to a link-layer hardware address on the local network.
Simple meaning. IP pata hai; local delivery ke liye MAC chahiye.
ARP maps a known local next-hop IPv4 address to a hardware address. A sender broadcasts a request when the mapping is missing; the address owner replies, and the mapping is cached. For a remote IP destination, the sender resolves its gateway’s local MAC address, then forwards the packet toward that gateway.
Yaad rakho: Remote packet ka next-hop MAC gateway ka hota hai.
Simple Mail Transfer Protocol (SMTP) transfers email from a sending client/server to a receiving mail server using a reliable transport, normally TCP. It is a push/store-and-forward transfer protocol. User agents compose/read mail; mail-transfer agents relay it, consulting DNS MX records to find recipient-domain mail servers.
| Phase | Typical exchange | Purpose |
|---|---|---|
| Connection and greeting | Server 220; client EHLO (HELO for basic SMTP); server 250 capabilities | Identify peers and learn supported extensions |
| Envelope transfer | MAIL FROM:<sender>; RCPT TO:<recipient>; 250 replies | Declare return path and one or more delivery recipients |
| Message transfer | DATA; server 354; headers + blank line + body; single dot line; server 250 | Transmit the message after server permission; completion reply accepts responsibility |
| Termination | QUIT; server 221 | Close the SMTP session cleanly |
Yaad rakho: SMTP send/relay; inbox read IMAP/POP/webmail.
SNMP is a network-management protocol (manager/agent, management information, requests/responses and notifications). It is not named in the current syllabus. This question’s combined 5 marks cannot be split into SMTP and SNMP without a printed allocation.
A computer network is a set of interconnected devices that exchange data and share resources using communication protocols. A link carries signals; a protocol defines message format, meaning and timing. Hosts run applications, switches forward local frames and routers connect IP networks.
| Type | Meaning and scope | Example |
|---|---|---|
| PAN | Personal Area Network: a person’s nearby devices | Phone linked to earbuds |
| LAN | Local Area Network: a building or local campus network under local administration | Laboratory Ethernet/Wi-Fi |
| MAN | Metropolitan Area Network: interconnects sites across a town/city | City-wide institutional network |
| WAN | Wide Area Network: connects geographically separated networks over regional/global links | Branches in different cities |
| ARPANET | Historical packet-switched research network; a major predecessor of the Internet | Early interconnected research institutions; not a modern geographic size category |
Classification by geographical scope is descriptive: a fixed kilometre boundary or a guaranteed speed is not part of every definition. Networks can also be classified by wired/wireless medium or by client-server/peer-to-peer organization.
Yaad rakho: PAN personal, LAN local, MAN city, WAN wide.
| Property | Stop-and-Wait | Go-Back-N | Selective Repeat |
|---|---|---|---|
| Outstanding sender frames | 1 | Up to 2^m−1 | Up to 2^(m−1), equal windows |
| Receiver window | 1 | 1 | Same maximum as sender in standard equal-window SR |
| ACK convention used here | Next expected | Cumulative, next expected | Individual frame accepted |
| Out-of-order data | Not delivered | Discarded | Buffered within valid window |
| After timeout | Repeat the one frame | Repeat outstanding frames from base | Repeat only timed-out frame |
| Buffer/tracking cost | Low | Sender window; simple receiver | Sender and receiver windows; per-frame tracking |
| Tradeoff | Simple but idle during RTT | Good pipelining, waste after loss | Efficient loss recovery, greater complexity |
Every comparison assumes finite sequence numbers, correct duplicate handling and the textbook protocol models. The same text “ACK 3” can mean different things under different ACK conventions; state its meaning before explaining a trace.
| Point | UDP | TCP | SCTP |
|---|---|---|---|
| Service unit | Datagram; preserves messages | Byte stream; no message boundaries | Messages carried in chunks; preserves boundaries |
| Connection setup | No protocol handshake | Connection; usual three-way SYN exchange | Association; four-message cookie exchange |
| Reliability | No delivery/order guarantee | Reliable ordered stream | Reliable messages; ordered per stream by default, unordered option |
| Ordering scope | None supplied | One ordered byte stream per direction | Multiple streams; per-stream order avoids blocking every stream on one stream’s ordering gap |
| Flow control | Not supplied by UDP | Receiver-advertised window | Receiver flow control |
| Congestion control | Not supplied by UDP | Yes | Yes |
| Multihoming | No built-in association failover | Classic TCP uses one address pair | Association can bind multiple addresses; built-in path failover |
| Header | 8 bytes | Minimum 20 bytes | 12-byte common header plus chunks |
| Typical use | DNS, real-time applications with their own controls | Web over HTTP/1.1 or HTTP/2, reliable byte transfers | Telecom signalling and multi-stream message applications |
| Performance claim | Low protocol overhead, not automatically fastest overall | Recovery/ordering have costs and benefits | Extra capabilities; no universal speed ranking |
The correct protocol depends on application semantics and network conditions. “UDP is always faster” and “SCTP is always slower than TCP” are not defensible general rules. All three normally run above IP; security/encryption is a separate requirement.
Yaad rakho: UDP messages simple; TCP byte stream; SCTP reliable message streams.
Definition. Dynamic Host Configuration Protocol (DHCP) provides IP configuration parameters to clients and can lease reusable addresses for limited periods.
Simple meaning. DORA = Discover, Offer, Request, Acknowledge.
DHCP reduces manual configuration and supports controlled address reuse through leases.
Yaad rakho: Discover–Offer–Request–ACK; pehle D, phir O, phir R, phir A.
The Open Systems Interconnection (OSI) model separates network functions into seven layers. A layer offers services upward and uses services below. Peer layers communicate logically through protocol rules; actual data moves down the sender’s stack, across links and up the receiver’s stack.
| Layer | Task | Example |
|---|---|---|
| 7 Application | Services used by network applications | HTTP, DNS, FTP, SMTP |
| 6 Presentation | Representation, encoding and application data transformations | Character encoding, serialization; encryption as a conceptual function |
| 5 Session | Manage dialogue and synchronization/checkpoints | Session establishment and resumption concepts |
| 4 Transport | Process delivery; reliability/flow control where supplied | TCP, UDP, SCTP; ports |
| 3 Network | Logical addressing and routing across networks | IP; routers |
| 2 Data link | Framing, local link addressing, medium access and error detection | Ethernet/Wi-Fi; MAC addresses |
| 1 Physical | Transmit signal symbols/bits through a medium | Electrical, optical or radio signals |
Yaad rakho: Application se physical tak wrap; receiver par unwrap.
| Layer | Function 1 | Function 2 |
|---|---|---|
| Application | Provide application-level network services, such as file transfer | Define application exchanges for services such as email or name resolution |
| Presentation | Translate/serialize data into an agreed representation | Compress data or encrypt/decrypt it where these transformations are required |
| Session | Establish, manage and end a dialogue/session | Provide synchronization/checkpoints for resuming a dialogue |
| Transport | Identify sending/receiving processes using ports and segment/reassemble data | Provide end-to-end reliability and flow control when the selected protocol supports them |
| Network | Provide logical addressing across networks | Select routes and forward packets towards the destination |
| Data link | Frame the data and identify local interfaces | Control shared-medium access and detect link-frame errors |
| Physical | Represent bits as electrical, optical or radio signal symbols | Specify physical-interface/signalling properties such as timing, connectors and transmission rate |
These are reference-model responsibilities; an actual protocol stack may combine them. UDP, for example, does not supply TCP-style reliable delivery merely because both are transport protocols.
Definition. CSMA with Collision Detection monitors the medium during transmission, aborts a detected collision and retries after a randomized backoff.
Simple meaning. CD = takkar detect hui toh transmission rok do.
CSMA/CD senses before transmission and detects collisions during transmission. A collision causes abort/jam and a random backoff before retry. To detect the most distant collision before the frame finishes, the transmission duration must be at least the worst-case round-trip propagation delay. Thus Lmin/R ≥ 2Tp in the idealized model.
Yaad rakho: Detect karna hai toh sender tab tak bhej raha hona chahiye.
Frame ready → sense channel\r\n │\r\n Busy? yes → keep sensing (1-persistent case)\r\n no\r\n ↓\r\n Transmit and monitor\r\n │\r\n Collision? no → frame finished? → success\r\n yes\r\n ↓\r\n Jam indication → abort frame → increment attempt count\r\n │\r\n Retry limit reached? yes → report failure\r\n no\r\n ↓\r\n Random binary-exponential backoff → sense again
For Ethernet, the backoff range grows with collisions, capped after the specified stage; transmission is abandoned at the retry limit. Carrier sensing must still be obeyed after the random delay.
Source note: The scan says connection transfer; the third SMTP phase is connection termination.
\r\nSimple Mail Transfer Protocol (SMTP) transfers email from a sending client/server to a receiving mail server using a reliable transport, normally TCP. It is a push/store-and-forward transfer protocol. User agents compose/read mail; mail-transfer agents relay it, consulting DNS MX records to find recipient-domain mail servers.
| Phase | Typical exchange | Purpose |
|---|---|---|
| Connection and greeting | Server 220; client EHLO (HELO for basic SMTP); server 250 capabilities | Identify peers and learn supported extensions |
| Envelope transfer | MAIL FROM:<sender>; RCPT TO:<recipient>; 250 replies | Declare return path and one or more delivery recipients |
| Message transfer | DATA; server 354; headers + blank line + body; single dot line; server 250 | Transmit the message after server permission; completion reply accepts responsibility |
| Termination | QUIT; server 221 | Close the SMTP session cleanly |
Yaad rakho: SMTP send/relay; inbox read IMAP/POP/webmail.
| Phase | Commands / replies | Meaning |
|---|---|---|
| 1. Connection establishment | TCP connection; 220 greeting; EHLO/HELO; 250 reply | Open the connection and identify the SMTP peers |
| 2. Mail transfer | MAIL FROM, RCPT TO, DATA; 354; message; single dot; 250 | Transfer the envelope and then the message content |
| 3. Connection termination | QUIT; 221; close TCP | Finish the SMTP session |
Envelope transfer and message-body transfer are substeps of the single mail-transfer phase. Thus the more detailed table above still fits the standard three-phase description.
Definition. Open Shortest Path First (OSPF) is an interior link-state routing protocol that selects routes using shortest-path calculations and additive link costs.
Simple meaning. OSPF = link-state map + Dijkstra + summed costs.
OSPF exchanges and maintains link-state information, builds a topology database and runs Dijkstra’s shortest-path computation rooted at the local router. It chooses a path with the lowest sum of link costs and installs the corresponding next hop. Equal-cost paths may be supported.
Yaad rakho: OSPF cost add karo; sirf hops nahi.
| Type | Meaning | Diagram |
|---|---|---|
| Point-to-point | Direct connection between two routers | R1 — R2 |
| Transit network | Multiaccess network through which routers communicate; a designated router helps represent it | R1 — LAN — R2 (also R3) |
| Stub network | Network reachable through one router, with no onward transit via another router | R1 — end-host LAN |
| Virtual link | Logical OSPF backbone connection through a transit area | Area 0 — R1 ⋯ transit area ⋯ R2 — Area 0 |
Do not confuse these topology/link descriptions with the five OSPF packet types. Hello discovers/maintains neighbours; database description, link-state request, link-state update and acknowledgement synchronize link-state information.
Data = 1001100; generator bits = 1101; degree r = 3.
CRC remainder, complete transmitted frame and receiver check; then detect the specified bit inversion.
Append r zeros, divide using XOR (modulo 2), and replace those zeros with the r-bit remainder. XOR has no borrowing or carrying. A valid received codeword has remainder zero; the receiver does not append new zeros.
1001100 → 1001100000
| Leading position | Before XOR | Aligned generator | After XOR |
|---|---|---|---|
| 1 | 1001100000 | 1101000000 | 0100100000 |
| 2 | 0100100000 | 0110100000 | 0010000000 |
| 3 | 0010000000 | 0011010000 | 0001010000 |
| 4 | 0001010000 | 0001101000 | 0000111000 |
| 5 | 0000111000 | 0000110100 | 0000001100 |
| 7 | 0000001100 | 0000001101 | 0000000001 |
Remainder = 001. Transmit 1001100001.
| Position | Before | Generator | After |
|---|---|---|---|
| 1 | 1001100001 | 1101000000 | 0100100001 |
| 2 | 0100100001 | 0110100000 | 0010000001 |
| 3 | 0010000001 | 0011010000 | 0001010001 |
| 4 | 0001010001 | 0001101000 | 0000111001 |
| 5 | 0000111001 | 0000110100 | 0000001101 |
| 7 | 0000001101 | 0000001101 | 0000000000 |
Receiver remainder = 000. No error is detected; this is not a guarantee that every possible corruption is detectable.
1001100001 becomes 1101100001.
| Position | Before | Generator | After |
|---|---|---|---|
| 1 | 1101100001 | 1101000000 | 0000100001 |
| 5 | 0000100001 | 0000110100 | 0000010101 |
| 6 | 0000010101 | 0000011010 | 0000001111 |
| 7 | 0000001111 | 0000001101 | 0000000010 |
Remainder = 010, which is nonzero: reject the frame and request recovery according to the error-control protocol.
User Datagram Protocol (UDP) provides connectionless, message-oriented process delivery. One send creates a datagram; message boundaries are retained. UDP has an 8-byte header, carries ports and length, and offers checksum-based corruption detection. It does not itself guarantee delivery, ordering, duplicate suppression, retransmission, flow control or congestion control.
| UDP field | Width | Purpose |
|---|---|---|
| Source port | 16 bits | Sending endpoint; a zero value can indicate no reply port when unused |
| Destination port | 16 bits | Receiving application endpoint |
| Length | 16 bits | UDP header plus UDP payload; minimum 8 bytes |
| Checksum | 16 bits | Error detection over UDP and an IP pseudo-header; not retransmission |
Yaad rakho: UDP datagram preserve; reliability automatic nahi.
| Application | Why UDP can suit it |
|---|---|
| DNS queries | Small request/reply; application retries if necessary. DNS also supports TCP. |
| DHCP | A host without an assigned address can use UDP broadcasts to obtain configuration. |
| Real-time audio/video using RTP | Late packets are often less useful than timely packets; application handles loss/jitter. |
| Online game state updates | A newer position can replace a stale one; application-specific reliability can protect critical events. |
| NTP | Small timestamp exchanges; a fresh later measurement can replace a lost exchange. |
These are examples of applications that can use UDP, not a claim that every implementation of each application always uses UDP. Applications must still manage congestion responsibly.
In a client-server design, users’ clients contact designated service processes. In peer-to-peer (P2P), participating hosts can supply and consume resources. These are application organizations, not different physical cable types.
| Point | Client-server | Peer-to-peer |
|---|---|---|
| Roles | Clients request; designated servers provide services | Peers can both request and provide resources |
| Control | Central policies, accounts and backups are easier to coordinate | Control and copies distributed among peers |
| Performance | Predictable with adequate server capacity; server may bottleneck | Resources grow with peers, but peer capacity/availability vary |
| Scalability | Scale servers, replicas, caching and load distribution | More peers can contribute capacity; coordination becomes harder |
| Reliability | One unreplicated server is a failure point; replication helps | Loss of one peer need not stop system if useful replicas remain |
| Security | Central access control and monitoring | Trust and consistent access policy are harder |
| Best fit | Managed collaboration, databases, login services | Distributed content sharing among participating peers |
Central servers can authenticate users, maintain the authoritative shared document and control versions. The organization must provision capacity, replicate data and avoid a single server failure. P2P can exchange large files or local edits among peers and distribute load, but needs discovery, conflict resolution, trust and availability strategies. A hybrid can keep account control centralized while distributing selected content.
Yaad rakho: Server mein service central; peers dono roles karte hain.
The Domain Name System (DNS) is a distributed hierarchical naming system. Its namespace is a tree with the root at the top. A complete name is the ordered path of labels from a node back to the root. An absolute name may be written with a trailing dot, for example www.example.com.
| Level | Example | Meaning |
|---|---|---|
| Root | . | Top of the namespace |
| Top-level domain (TLD) | com, org, in | A child zone below the root |
| Registered/delegated domain | example.com | Name under a parent such as com |
| Subdomain/host label | www.example.com | A label below example.com; a name may own different record types |
Yaad rakho: Full path unique; www label akela unique nahi.
Client → recursive resolver → root (.)\r\n → .com TLD\r\n → authoritative server for xyz.com\r\nClient ← final A/AAAA answer ← resolver
Assume a cache miss and that the name exists. (1) The client asks its resolver for www.xyz.com. (2) A root server refers the resolver to .com servers. (3) A .com server refers it to authoritative servers for xyz.com. (4) An authoritative server returns the applicable A/AAAA record, or a CNAME that must be followed. (5) The resolver returns the answer and caches it according to TTL. No real IP address for this domain is assumed. Root and TLD servers normally supply referrals here, not the final host address.
Definition. Bellman–Ford computes single-source shortest paths by repeatedly relaxing every edge. It supports negative edge weights and detects reachable negative cycles.
Simple meaning. Har edge ko baar-baar relax; negative edge chalega, negative cycle problem hai.
Bellman–Ford repeatedly relaxes all edges and can handle negative weights. It uses up to V−1 passes and an additional improvement check for reachable negative cycles. Dijkstra settles the smallest tentative vertex and requires nonnegative weights for its standard correctness guarantee.
Yaad rakho: Pass = all edges once. Negative edge ≠ negative cycle.
\r\n| Directed edge | Weight |
|---|---|
| A→B | 4 |
| A→C | 5 |
| B→C | -2 |
| C→D | 3 |
The shortest route and cost, with complete passes and a negative-cycle check.
Bellman–Ford permits negative edges. We use the listed order with in-place updates; do not mix this with a table that reads only the previous pass.
4 vertices → at most 3 main passes. d[A]=0; every other distance=∞. Use the listed edge order and immediate (in-place) updates.
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| A→B | ∞ | 0+(4)=4 | 4 | Update; predecessor=A |
| A→C | ∞ | 0+(5)=5 | 5 | Update; predecessor=A |
| B→C | 5 | 4+(-2)=2 | 2 | Update; predecessor=B |
| C→D | ∞ | 2+(3)=5 | 5 | Update; predecessor=C |
| Vertex | A | B | C | D |
|---|---|---|---|---|
| End-of-pass distance | 0 | 4 | 2 | 5 |
| Predecessor | — | A | B | C |
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| A→B | 4 | 0+(4)=4 | 4 | No improvement |
| A→C | 2 | 0+(5)=5 | 2 | No improvement |
| B→C | 2 | 4+(-2)=2 | 2 | No improvement |
| C→D | 5 | 2+(3)=5 | 5 | No improvement |
This entire pass changed no distance. Further identical relaxation passes will not improve anything.
Backward predecessors: D ← C ← B ← A. Forward route: A → B → C → D.
Cost = (4) + (-2) + (3) = 5. No edge can now improve its destination label.
Final answer: A → B → C → D; shortest cost=5.
Frame length L = 400 bits; link rate R = 100,000 bit/s; attempted rates λ = 1200, 600, 300 frames/s.
Normalized throughput S and successful frames per second for each offered load.
![\"T_f=L/R,\\quad](\"math/043fb89e4a6dc8db.svg\")
G counts attempted frames per frame time, including retransmissions if present. S is not measured in frames/s. Multiply by the frame rate capacity 1/Tf to obtain surviving frames/s.
Tf = 400/100000 = 0.004 s. Capacity = 1/Tf = 250 frames/s.
G = 1200 × 0.004 = 4.8.
e−2G = e−9.6 = 6.77287365e-05.
S = 4.8 × 6.77287365e-05 = 0.000325097935.
Successful frames/s = 1200 × 6.77287365e-05 = 0.0812744838. Successful bit/s = 0.0812744838 × 400 = 32.5097935.
G = 600 × 0.004 = 2.4.
e−2G = e−4.8 = 0.00822974705.
S = 2.4 × 0.00822974705 = 0.0197513929.
Successful frames/s = 600 × 0.00822974705 = 4.93784823. Successful bit/s = 4.93784823 × 400 = 1975.13929.
G = 300 × 0.004 = 1.2.
e−2G = e−2.4 = 0.0907179533.
S = 1.2 × 0.0907179533 = 0.108861544.
Successful frames/s = 300 × 0.0907179533 = 27.215386. Successful bit/s = 27.215386 × 400 = 10886.1544.
188.15.0.0/16; Group 1: 64 customers × 256 addresses; Group 2: 128 customers × 128 addresses; Group 3: 64 customers × 128 addresses
Each customer prefix, every group range, first/last customer blocks, and unallocated addresses.
![\"p=32-\\log_2(\\text{addresses](\"math/2cb46cf5eb7fdc4d.svg\")
These are requested total addresses, not usable-host counts. Allocate customer blocks on their prefix boundary. A group with a non-power-of-two customer count need not be forced into one larger, wasteful CIDR aggregate.
| Group | Customers | Addresses each | Customer prefix | Used | Group range |
|---|---|---|---|---|---|
| 1 | 64 | 256 | /24 | 16384 | 188.15.0.0–188.15.63.255 |
| 2 | 128 | 128 | /25 | 16384 | 188.15.64.0–188.15.127.255 |
| 3 | 64 | 128 | /25 | 8192 | 188.15.128.0–188.15.159.255 |
| Group | First customer | Last customer |
|---|---|---|
| 1 | 188.15.0.0–188.15.0.255 | 188.15.63.0–188.15.63.255 |
| 2 | 188.15.64.0–188.15.64.127 | 188.15.127.128–188.15.127.255 |
| 3 | 188.15.128.0–188.15.128.127 | 188.15.159.128–188.15.159.255 |
Allocated = 16384 + 16384 + 8192 = 40960. Unused = 65536 − 40960 = 24576.
Unused range: 188.15.160.0–188.15.255.255.
In the textbook leaky-bucket queue/shaper model, arriving packets enter a finite buffer. The output sends at a configured rate whenever data is queued. When the queue empties, it cannot send nonexistent packets; when the buffer is full, excess arrivals are dropped in this finite-buffer model.
| Advantage | Limitation |
|---|---|
| Simple bound on output rate | Bursts may overflow or suffer delay |
| Reduces short-term output variation | Unused idle service cannot be saved |
| Useful for predictable traffic shaping | No end-to-end delay/loss guarantee by itself |
The term “leaky bucket” can also describe a conformance meter in standards. This course’s diagrams and numerical questions explicitly use the finite data-queue model, so its state is queued data rather than tokens.
No algorithm is universally better. The leaky data-queue model is useful for a smooth output profile; token bucket is preferable when controlled bursts should be allowed. State the application’s delay, burst and rate needs before recommending one.
Yaad rakho: Leaky bucket stores data; idle service save nahi hota.
\r\nA token bucket generates tokens at rate r up to capacity B. Sending one unit of data consumes the corresponding tokens. Tokens accumulate during idle periods only until the bucket is full. Data can therefore burst when saved tokens exist, subject to the physical link rate.
| Limitation | Reason |
|---|---|
| Bursts can still congest downstream queues | Local conformance does not ensure every later link has capacity |
| Parameters must match traffic and link requirements | A small B delays/drops legitimate packets; a large B permits disruptive bursts |
| No complete QoS guarantee alone | Admission, scheduling and end-to-end capacity may still be needed |
Yaad rakho: Token permission store hota hai; actual packet alag queue mein.
\r\n| Point | Leaky data queue | Token bucket |
|---|---|---|
| Stored state | Queued data | Tokens representing permission |
| Idle period | Empty queue; unused service lost | Tokens accumulate up to B |
| Output burst | Smoothed by configured service rate | Allowed if enough tokens and physical capacity exist |
| Overflow/shortage | Finite queue can overflow | Shaper waits; policer may mark/drop |
| Main parameters | Queue capacity and drain rate | Token capacity and generation rate |
| Useful when | A smoother output profile is required | Average rate plus burst allowance is required |
The World Wide Web (WWW) is an information system of linked resources identified by URLs and accessed using Web clients and servers. A browser is a client; a Web server hosts or generates representations such as HTML, images or JSON. HTTP defines request/response semantics between these components. The Web is one application of the Internet, not a synonym for the entire Internet.
| Message component | Purpose |
|---|---|
| Request method / target | Requested action and resource; GET retrieves a representation, POST submits data for processing |
| Headers | Metadata, routing context, representation preferences and cache/connection information |
| Status code | Result class: 2xx success, 3xx redirection, 4xx client error, 5xx server error |
| Body | Representation or submitted content when present |
HTTP semantics are stateless: each request has its own meaning. Cookies/tokens can let an application maintain a login session; stateless does not mean “users cannot log in”. HTTP/1.1 and HTTP/2 commonly use TCP; HTTP/3 uses QUIC over UDP, so “HTTP always uses TCP” is not a universal statement.
Yaad rakho: DNS address dhoondta; HTTP resource mangta; browser render karta.
Definition. RIP and OSPF are routing protocols; Bellman–Ford and Dijkstra are algorithms underlying different route-computation approaches.
Simple meaning. Protocol message bhejta hai; algorithm route calculate karta hai.
| Feature | RIP | OSPF |
|---|---|---|
| Family | Distance vector | Link state |
| Typical metric | Hop count; 16 unreachable | Sum of configured link costs |
| Knowledge exchanged | Destination distances via neighbours | Link-state information |
| Algorithm connection | Distributed Bellman–Ford idea | Dijkstra shortest paths |
| Scale design | Small/simple networks; limited diameter | Area-based scalable design |
Yaad rakho: Distance vector → RIP; link state → OSPF.
| Point | Distance vector | Link state |
|---|---|---|
| Information exchanged | Destination-distance estimates and route information | Descriptions of local links, neighbours and link costs |
| Recipient / distribution | Neighbour-to-neighbour exchange | Flooding across the routing domain or area |
| Topology knowledge | Ordinarily no complete topology map; relies on neighbours' advertised distances | Builds a link-state database representing the area topology |
| Computation | Distributed Bellman–Ford style minimum via neighbours | Dijkstra shortest-path tree from the router itself |
| Convergence / resource tradeoff | Simple advertisements, but failures can cause count-to-infinity and transient loops | More database/computation and flooding overhead; topology changes can be propagated promptly |
RIP and OSPF illustrate these families, respectively. Exact convergence time depends on timers, topology and implementation, so neither family has one universal time bound.
The TCP/IP model describes the Internet protocol suite in four layers. Many teaching books split the link layer into data link and physical, making a five-layer teaching model; identify the convention before drawing.
| TCP/IP layer | Function | Examples |
|---|---|---|
| Application | Defines application message formats and exchanges for Web access, naming, email and file transfer; includes presentation/session functions as needed | HTTP, DNS, SMTP, FTP |
| Transport | End-to-end process communication | TCP, UDP, SCTP |
| Internet | IP addressing, routing and best-effort datagrams | IPv4/IPv6 |
| Link/network access | Local delivery and physical access | Ethernet/Wi-Fi and their medium |
| Point | OSI | TCP/IP |
|---|---|---|
| Layer count | Seven | Four here; five-layer teaching variant splits link |
| Upper functions | Application, presentation, session separated | Grouped under application |
| Lower functions | Data link and physical separated | Grouped under link/network access |
| Purpose | General reference architecture | Architecture of deployed Internet protocol family |
| Network service | Reference model can describe different services | Internet layer uses connectionless IP |
| Transport | Conceptual transport service | TCP reliable byte stream; UDP datagrams; SCTP messages |
Yaad rakho: OSI 7; TCP/IP four mein upper three merge, lower two merge.
| Layer | Protocol | Role |
|---|---|---|
| Internet | IPv4 / IPv6 | Address and deliver best-effort packets across networks |
| Internet | ICMP / ICMPv6 | Report selected IP errors and operational information; used by diagnostic exchanges |
| Transport | TCP | Reliable, ordered byte stream with connection establishment, flow and congestion control |
| Transport | UDP | Connectionless message delivery using ports, without built-in retransmission or ordering |
| Transport | SCTP | Reliable message-oriented association with multiple streams and multihoming support |
Stop-and-Wait ARQ permits only one outstanding frame. The sender stores its frame, transmits it and starts a timer. A valid matching ACK advances the sender; a timeout retransmits the stored frame.
Yaad rakho: One frame outstanding; duplicate ko acknowledge karo, deliver dobara nahi.
Stream Control Transmission Protocol (SCTP) is a reliable message-oriented transport. A relationship between endpoints is an association. Its packets contain a common header followed by chunks, which carry data or control information.
Yaad rakho: SCTP association: streams separate order, addresses provide failover.
| Service | Explanation |
|---|---|
| Process-to-process delivery | Port numbers identify the sending and receiving applications |
| Reliable message delivery | Checks, sequence tracking, acknowledgements and retransmission recover missing chunks while preserving messages |
| Multistreaming | Separate ordering within streams reduces cross-stream ordering blockage |
| Multihoming | Multiple endpoint addresses permit an alternate path if the primary fails |
| Flow and congestion control | Receiver flow control limits receiver overload; congestion control adapts traffic to network conditions |
Definition. Switching selects how data passes through intermediate network nodes from a source to a destination.
Simple meaning. Raasta reserve karna, poora message ruk-ruk ke bhejna, ya chhote packets bhejna—yeh teen ideas hain.
| Aspect | Circuit switching | Message switching | Packet switching |
|---|---|---|---|
| Unit handled | Continuous stream on established circuit | Whole message | Small packets |
| Resource use | Capacity reserved along a circuit | Shared; message stored at each intermediate node | Shared; each packet stored/forwarded in this model |
| Setup | Required before data transfer | No circuit setup required | Datagram: no connection setup; virtual circuit: setup |
| Delay | Setup delay, then relatively predictable service | Large store-and-forward delay for long messages | Variable delay; packet pipeline can reduce completion time |
| Main limitation | Reserved capacity can be idle | Large buffers; unsuitable for tight delay needs | Headers, possible queueing/loss; datagrams may reorder |
For ordinary bursty office data, packet switching is generally appropriate because resources are shared among active transfers. The design must still account for queueing, loss and the required service quality.
Yaad rakho: Circuit reserves; message waits whole; packets pipeline.
| Method | Advantage 1 | Advantage 2 | Disadvantage 1 | Disadvantage 2 |
|---|---|---|---|---|
| Message switching | Shares link capacity without a permanently reserved circuit | Intermediate storage lets a message wait for a currently busy outgoing link | The entire message must arrive before forwarding, giving long delay | Large messages require large buffers and can hold up other messages |
| Packet switching | Statistical sharing suits bursty data from many users | Packets can be pipelined across successive links, reducing completion time | Every packet needs header/control overhead | Queueing, loss and (for datagrams) reordering need endpoint or network handling |
A packet-switched network may use datagrams or virtual circuits. The possibility of independent routes and reordering refers to datagrams, not every packet-switching implementation.
Topology is the arrangement of nodes and links. Physical topology describes actual links; logical topology describes how data flows. A switch-centred physical star is different from a shared collision-domain bus.
| Topology | Structure | Benefit | Limitation |
|---|---|---|---|
| Bus | Devices share a common backbone | Simple small shared-medium layout | Backbone fault affects segment; collisions possible on classic shared bus |
| Star | Each device connects to a central switch/hub | Easy expansion and fault isolation | Central-device failure affects star; more cabling |
| Ring | Each node connected to two neighbours in a ring | Structured traversal; token-based variants coordinate access | Simple single-ring break can disrupt operation unless protected |
| Full mesh | Every pair has a direct link | Redundant paths and direct connectivity | Expensive ports/cabling: n(n−1)/2 undirected links |
| Tree/hybrid | Hierarchical stars or combined topologies | Scales into buildings/floors | Upper-level faults affect dependent branches |
Yaad rakho: Layout draw karo, then one strength and one failure case.
| Topology / example | Advantage 1 | Advantage 2 | Disadvantage 1 | Disadvantage 2 |
|---|---|---|---|---|
| Star: switched computer laboratory | A device can be added on a spare central port without rebuilding the whole topology | A single access-cable fault normally isolates only its attached device | Central-switch failure affects all attached devices unless redundant | A separate cable and switch port are needed per device |
| Bus: legacy coaxial Ethernet | Uses little cable in a small linear layout | No central switch is required in the simple shared-bus design | A backbone fault can disable the segment | Shared bandwidth, contention and fault diagnosis become difficult as it grows |
| Ring: token-based ring | Controlled token access avoids data collisions among compliant stations | Token rotation gives stations an organized opportunity to transmit | A break can interrupt an unprotected single ring | Adding/removing a node can disrupt a simple ring |
| Mesh: redundant router backbone | Alternate links preserve connectivity after some link failures | Direct pairwise links can carry separate exchanges without a shared central bottleneck | A full mesh needs n(n−1)/2 links and many ports | Configuration, cabling and maintenance become complex |
The ring benefits above assume token-controlled access; physical ring wiring alone does not guarantee fairness or collision avoidance. The failure disadvantages describe simple layouts without protection/redundancy.
Definition. CSMA with Collision Avoidance uses carrier sensing and randomized access delays, with acknowledgements and optional reservation exchanges, to reduce collisions on a wireless shared medium.
Simple meaning. CA = pehle collision avoid karne ki koshish; guarantee nahi.
| Aspect | CSMA/CD | CSMA/CA |
|---|---|---|
| Main action | Detect and abort a collision during sending | Reduce collision risk before sending |
| Typical teaching context | Shared half-duplex wired Ethernet | Wi-Fi contention access |
| Mechanisms | Carrier sensing, detection/jam, backoff | Carrier sensing, random backoff, ACK; optional RTS/CTS |
Yaad rakho: CD = detect during; CA = avoid before.
\r\nDefinition. Hidden terminals are stations that cannot sense each other but whose transmissions can interfere at a common receiver.
Simple meaning. A aur C ek dusre ko nahi sunte, par dono B ko bhej rahe hain.
A and C may both reach B but be unable to sense one another. Simultaneous transmissions can then collide at B. A short RTS followed by B’s CTS announces a reservation; a station hearing the CTS defers, reducing the risk of a long data-frame collision.
Yaad rakho: Hidden: senders invisible to each other, receiver common.
Frame ready → sense medium\r\nBusy? yes → defer until eligible idle interval\r\n ↓\r\nChoose a random backoff → count down eligible idle slots\r\nMedium becomes busy? → freeze counter; resume after required idle wait\r\n ↓ counter reaches zero\r\nOptional RTS → CTS received? no → retry/back off\r\n ↓ yes (or RTS/CTS not used)\r\nSend DATA → ACK received? yes → success\r\n no → retry/back off or fail at retry limit
A station hearing a duration reservation updates its network allocation vector and defers. RTS/CTS is optional; loss of CTS or ACK triggers recovery but does not by itself prove a collision, because channel errors can also cause loss.
A TCP connection establishes each endpoint’s initial sequence number and confirms that both can receive. Use client initial sequence x=1000 and server initial sequence y=5000; assume no payload in the handshake.
| Step | Segment | Meaning |
|---|---|---|
| 1 Client → server | SYN, Seq=1000 | Client proposes its starting sequence |
| 2 Server → client | SYN+ACK, Seq=5000, Ack=1001 | Server acknowledges client SYN and supplies its own start |
| 3 Client → server | ACK, Seq=1001, Ack=5001 | Client acknowledges server SYN; connection is established |
Client sends 200 bytes beginning at Seq=1001, so byte numbers are 1001…1200. The receiver acknowledges Ack=1201, the next byte expected. If bytes starting at 1201 are missing but later bytes arrive, the cumulative ACK remains 1201 until the gap is repaired. TCP may buffer out-of-order bytes; it delivers the byte stream in order. Each direction counts independently.
Yaad rakho: SYN plus one, FIN plus one; ACK next expected byte.
180.50.0.0/16; Group 1: 80 customers × 256 addresses; Group 2: 150 customers × 256 addresses; Group 3: 32 customers × 128 addresses
Each customer prefix, every group range, first/last customer blocks, and unallocated addresses.
These are requested total addresses, not usable-host counts. Allocate customer blocks on their prefix boundary. A group with a non-power-of-two customer count need not be forced into one larger, wasteful CIDR aggregate.
| Group | Customers | Addresses each | Customer prefix | Used | Group range |
|---|---|---|---|---|---|
| 1 | 80 | 256 | /24 | 20480 | 180.50.0.0–180.50.79.255 |
| 2 | 150 | 256 | /24 | 38400 | 180.50.80.0–180.50.229.255 |
| 3 | 32 | 128 | /25 | 4096 | 180.50.230.0–180.50.245.255 |
| Group | First customer | Last customer |
|---|---|---|
| 1 | 180.50.0.0–180.50.0.255 | 180.50.79.0–180.50.79.255 |
| 2 | 180.50.80.0–180.50.80.255 | 180.50.229.0–180.50.229.255 |
| 3 | 180.50.230.0–180.50.230.127 | 180.50.245.128–180.50.245.255 |
Allocated = 20480 + 38400 + 4096 = 62976. Unused = 65536 − 62976 = 2560.
Unused range: 180.50.246.0–180.50.255.255.
The Domain Name System (DNS) is a distributed hierarchical naming system. Its namespace is a tree with the root at the top. A complete name is the ordered path of labels from a node back to the root. An absolute name may be written with a trailing dot, for example www.example.com.
| Level | Example | Meaning |
|---|---|---|
| Root | . | Top of the namespace |
| Top-level domain (TLD) | com, org, in | A child zone below the root |
| Registered/delegated domain | example.com | Name under a parent such as com |
| Subdomain/host label | www.example.com | A label below example.com; a name may own different record types |
Yaad rakho: Full path unique; www label akela unique nahi.
Client → recursive resolver → root (.)\r\n → .com TLD\r\n → authoritative server for xyz.com\r\nClient ← final A/AAAA answer ← resolver
Assume a cache miss and that the name exists. (1) The client asks its resolver for www.xyz.com. (2) A root server refers the resolver to .com servers. (3) A .com server refers it to authoritative servers for xyz.com. (4) An authoritative server returns the applicable A/AAAA record, or a CNAME that must be followed. (5) The resolver returns the answer and caches it according to TTL. No real IP address for this domain is assumed. Root and TLD servers normally supply referrals here, not the final host address.
Definition. Routing Information Protocol (RIP) is an interior distance-vector routing protocol that uses a bounded metric, commonly hop count, to select routes.
Simple meaning. RIP mein hops count; 15 tak reachable, 16 means unreachable.
RIP uses distance-vector routing and commonly hop count as its metric. Reachable metrics are at most 15; 16 represents infinity. Its main limitations include limited diameter, a coarse metric and relatively slow convergence with possible temporary loops.
Yaad rakho: RIP: neighbour distances; 16 is infinity.
Let router A have neighbours B and C, each one hop away. B advertises destination network N at 2 hops; C advertises N at 4 hops. A adds its local one-hop cost to each advertisement.
| Candidate | Calculation | Result |
|---|---|---|
| Via B | 1 + 2 | 3 hops |
| Via C | 1 + 4 | 5 hops |
A chooses B and records (destination N, next hop B, metric 3). If B subsequently advertises metric 16, that route is unreachable. If the advertisement from C is still valid, A can select C at metric 5 instead. Updates can propagate to other routers; split horizon/poisoned reverse and triggered updates help limit loops and speed convergence, but do not make every transient loop impossible.
Definition. Pure ALOHA is a random-access protocol in which a station transmits whenever a frame is ready and retransmits after a random delay if the transmission is unsuccessful.
Simple meaning. Frame ready → bhej do. Collision hua → random wait karke retry.
In Pure ALOHA, a station transmits a ready frame without carrier sensing. If a successful acknowledgement is not received, it waits for a random interval and retransmits. Overlapping transmissions cause collisions in the basic model. Its maximum normalized throughput is 1/(2e) ≈ 18.4%, so it is inefficient at high offered load.
Yaad rakho: Pure = kisi bhi time start; retry ka wait random.
\r\nDefinition. The vulnerable period is the interval of start times of other frames that would cause a collision with a chosen frame.
Simple meaning. Apne frame se ek frame-time pehle start hua frame bhi takra sakta hai, aur baad mein bhi.
A frame lasting T can collide with an equal-length frame that began less than T earlier, or one that begins less than T later. The two intervals together have length T+T=2T.
Yaad rakho: Pure: before T + after T = 2T.
\r\nDefinition. Slotted ALOHA is a random-access protocol that divides time into slots of one frame duration and permits transmissions only at the beginning of a slot.
Simple meaning. Frame ready hai tab bhi next slot ki start-line ka wait karo.
| Feature | Pure | Slotted |
|---|---|---|
| Start time | Any time | Slot boundary only |
| Synchronization | No common slots required | Required |
| Vulnerable period | 2T | T |
| Maximum S | 1/(2e) ≈ 0.184 | 1/e ≈ 0.368 |
| G at maximum | 0.5 | 1 |
Yaad rakho: Slotted = aligned starts; collision risk half-window, not zero.
Congestion occurs when offered traffic overloads network resources such as link capacity, router buffers or processing. Queueing delay grows, packets may be dropped, and excessive retransmission can make the overload worse. Flow control protects a receiver; congestion control protects the network.
| Approach | Policy / mechanism | How it helps |
|---|---|---|
| Open loop: prevent congestion without runtime congestion feedback | Retransmission policy | Carefully chosen timers/selective retransmission avoid needless extra traffic |
| Open loop | Window policy | Limit outstanding data and avoid excessive retransmission after loss |
| Open loop | Acknowledgement policy | Avoid unnecessarily frequent ACK traffic while retaining correct recovery |
| Open loop | Admission policy | Refuse new load when reserved capacity is insufficient |
| Closed loop: measure and react to congestion | Backpressure | A congested node tells upstream nodes to reduce/stop sending, possibly propagating hop by hop |
| Closed loop | Choke/control signalling | A congested node sends explicit feedback to the source |
| Closed loop | Implicit signalling | A sender infers congestion from loss/delay and changes its sending behaviour |
| Closed loop | Explicit marking | Network marks congestion (for example ECN); endpoints react to returned information |
For an answer asking for two of each: explain retransmission and admission policies under open loop, then backpressure and explicit signalling under closed loop, with the mechanism and its effect. These are general congestion-control classifications used in the course; they are not all mandatory algorithms implemented inside TCP. TCP commonly uses end-to-end feedback such as loss or ECN.
Yaad rakho: Open loop pehle prevent; closed loop feedback se react.
| From | To | Cost |
|---|---|---|
| A | B | 3 |
| A | D | 7 |
| B | D | 5 |
| B | E | 5 |
| D | E | 4 |
| D | F | 2 |
| E | F | 1 |
| E | C | 8 |
| F | C | 3 |
The original diagram is undirected. All edge labels are reproduced below. Source = A.
Swipe across the diagram to read its labels ↔![\"dijkstra](\"fig/old-dijkstra-2025.c24a5141a1.svg\")
| From | To | Cost |
|---|---|---|
| A | B | 3 |
| A | D | 7 |
| B | D | 5 |
| B | E | 5 |
| D | E | 4 |
| D | F | 2 |
| E | F | 1 |
| E | C | 8 |
| F | C | 3 |
Shortest distance, complete route and predecessor for every node from A.
![\"d(v)\\leftarrow\\min(d(v),d(u)+w(u,v))\"](\"math/e31cde29472d2c2d.svg\")
Relaxation replaces a tentative route when going through u is cheaper. Dijkstra settles the unvisited node of least tentative distance. Nonnegative edge costs make the settled distance final. Equal-distance nodes may be settled in either order.
| Pass / settled node | Strict improvements | A | B | C | D | E | F |
|---|---|---|---|---|---|---|---|
| 0 | Initial | 0 | ∞ | ∞ | ∞ | ∞ | ∞ |
| A | B←A: 0+3=3; D←A: 0+7=7 | 0 | 3 | ∞ | 7 | ∞ | ∞ |
| B | E←B: 3+5=8 | 0 | 3 | ∞ | 7 | 8 | ∞ |
| D | F←D: 7+2=9 | 0 | 3 | ∞ | 7 | 8 | 9 |
| E | C←E: 8+8=16 | 0 | 3 | 16 | 7 | 8 | 9 |
| F | C←F: 9+3=12 | 0 | 3 | 12 | 7 | 8 | 9 |
| C | No change | 0 | 3 | 12 | 7 | 8 | 9 |
| Destination | Distance | Previous | Route |
|---|---|---|---|
| A | 0 | — | A |
| B | 3 | A | A → B |
| C | 12 | F | A → D → F → C |
| D | 7 | A | A → D |
| E | 8 | B | A → B → E |
| F | 9 | D | A → D → F |
A tie between two equally short paths is valid; this table retains the first predecessor on equal distances.
Data = 1010011110; generator bits = 1011; degree r = 3.
CRC remainder, complete transmitted frame and receiver check.
Append r zeros, divide using XOR (modulo 2), and replace those zeros with the r-bit remainder. XOR has no borrowing or carrying. A valid received codeword has remainder zero; the receiver does not append new zeros.
1010011110 → 1010011110000
| Leading position | Before XOR | Aligned generator | After XOR |
|---|---|---|---|
| 1 | 1010011110000 | 1011000000000 | 0001011110000 |
| 4 | 0001011110000 | 0001011000000 | 0000000110000 |
| 8 | 0000000110000 | 0000000101100 | 0000000011100 |
| 9 | 0000000011100 | 0000000010110 | 0000000001010 |
| 10 | 0000000001010 | 0000000001011 | 0000000000001 |
Remainder = 001. Transmit 1010011110001.
| Position | Before | Generator | After |
|---|---|---|---|
| 1 | 1010011110001 | 1011000000000 | 0001011110001 |
| 4 | 0001011110001 | 0001011000000 | 0000000110001 |
| 8 | 0000000110001 | 0000000101100 | 0000000011101 |
| 9 | 0000000011101 | 0000000010110 | 0000000001011 |
| 10 | 0000000001011 | 0000000001011 | 0000000000000 |
Receiver remainder = 000. No error is detected; this is not a guarantee that every possible corruption is detectable.
Guided media constrain signals to a physical path: twisted pair, coaxial cable and optical fiber. Unguided media carry electromagnetic waves through free space: radio, microwave and infrared. Wireless bandwidth and reach depend on frequency, power, antenna, obstacles and regulations; media names alone do not fix a speed.
| Medium | Two key characteristics | Limitations / use |
|---|---|---|
| Twisted pair | Electrical signal; pairs twisted to reduce interference | Low-cost LAN access; distance/noise limits depend on category and design |
| Coaxial cable | Shielded central conductor; greater shielding than basic unshielded pair | Bulkier; cable broadband and legacy shared LANs |
| Optical fiber | Light in glass/plastic; high capacity with immunity to electromagnetic interference | Transceivers and careful splicing/bend control needed; backbone links |
| Radio waves | Wireless propagation; many bands support broad/non-line-of-sight coverage | Interference, shared spectrum and security configuration matter; Wi-Fi/mobile |
| Terrestrial microwave | Directional antennas; typically line of sight | Alignment and obstructions matter; point-to-point backhaul |
| Satellite microwave | Wide coverage through relay in space | Cost and latency depend strongly on orbit/system; remote connectivity |
| Infrared | Short-range light; usually blocked by walls | Limited coverage; short-range indoor/remote-control links |
Yaad rakho: Guided = cable path; unguided = free-space waves.
\r\nTransmission media are classified into guided media (twisted pair, coaxial cable, optical fiber) and unguided media (radio, microwave, infrared).
Optical fiber carries data as light in a core surrounded by cladding. It is a guided medium. Single-mode fiber supports one main propagation mode and is used for long/high-capacity links; multimode supports multiple modes and is often used over shorter distances.
| Advantages | Disadvantages / design costs |
|---|---|
| High data capacity and low attenuation over suitable links | Optical transceivers, connectors and splicing add cost/skill requirements |
| Immune to electromagnetic interference; no electrical crosstalk | Tight bends, dirty connectors and poor installation increase loss |
| Electrical isolation between buildings | Fiber is mechanically sensitive and requires appropriate protection |
| Harder to tap casually than an unprotected radio channel | It is not automatically confidential; encryption/access control remain necessary |
Yaad rakho: Fiber light carry karta hai; immunity means no EMI, not no security risk.
| Medium / working principle | Advantages | Disadvantages |
|---|---|---|
| Twisted pair: voltage signals travel on two insulated copper conductors twisted together | Low purchase/installation cost; flexible and easy to terminate for local access | More susceptible to electromagnetic interference than fiber; distance and capacity are limited by cable category and signalling design |
| Optical fiber: light pulses travel through a core guided by its refractive-index structure | High capacity and long reach; immune to electromagnetic interference | Optical equipment and splicing can cost more; tight bends, damage and dirty connectors degrade the link |
These comparisons describe typical engineering tradeoffs, not a guarantee that every fiber installation is faster or more expensive than every copper installation.
210.16.36.56/26
First and last addresses and total block size.
p is the prefix length. The remaining host bits vary through every combination. The network address is the first address; setting all host bits to one gives the last address.
32 − 26 = 6. N = 64.
Mask = 255.255.255.192. The first varying octet is octet 4; its block width is 64. The supplied value 56 lies in the interval 0–63. Keep earlier octets fixed. Set any later octets to 0 for the first address and 255 for the last address.
First = 210.16.36.0; last = 210.16.36.63; total = 64 addresses. Usual usable host range: 210.16.36.1–210.16.36.62.
File Transfer Protocol (FTP) transfers files between a client and server using TCP and separates commands from file data. This separation is its central design feature. Its control connection remains available across multiple operations, while data connections carry file bytes or directory listings.
| Connection | Function | Lifetime / ports |
|---|---|---|
| Control | Commands and replies: USER, PASS, CWD, TYPE, LIST, RETR, STOR, QUIT | Client connects to server TCP port 21; normally kept for the session |
| Data | File contents and directory listing data | Opened as needed; separate from command channel |
FTP over TLS (FTPS) protects FTP connections. SFTP is a separate SSH-based file-transfer protocol, not “FTP using an SSH port”. These names should not be treated as interchangeable.
Yaad rakho: Control commands, data file; passive mein client dono connect karta.
A computer network is a set of interconnected devices that exchange data and share resources using communication protocols. A link carries signals; a protocol defines message format, meaning and timing. Hosts run applications, switches forward local frames and routers connect IP networks.
| Type | Meaning and scope | Example |
|---|---|---|
| PAN | Personal Area Network: a person’s nearby devices | Phone linked to earbuds |
| LAN | Local Area Network: a building or local campus network under local administration | Laboratory Ethernet/Wi-Fi |
| MAN | Metropolitan Area Network: interconnects sites across a town/city | City-wide institutional network |
| WAN | Wide Area Network: connects geographically separated networks over regional/global links | Branches in different cities |
| ARPANET | Historical packet-switched research network; a major predecessor of the Internet | Early interconnected research institutions; not a modern geographic size category |
Classification by geographical scope is descriptive: a fixed kilometre boundary or a guaranteed speed is not part of every definition. Networks can also be classified by wired/wireless medium or by client-server/peer-to-peer organization.
Yaad rakho: PAN personal, LAN local, MAN city, WAN wide.
Definition. Pure ALOHA is a random-access protocol in which a station transmits whenever a frame is ready and retransmits after a random delay if the transmission is unsuccessful.
Simple meaning. Frame ready → bhej do. Collision hua → random wait karke retry.
In Pure ALOHA, a station transmits a ready frame without carrier sensing. If a successful acknowledgement is not received, it waits for a random interval and retransmits. Overlapping transmissions cause collisions in the basic model. Its maximum normalized throughput is 1/(2e) ≈ 18.4%, so it is inefficient at high offered load.
Yaad rakho: Pure = kisi bhi time start; retry ka wait random.
\r\nDefinition. The vulnerable period is the interval of start times of other frames that would cause a collision with a chosen frame.
Simple meaning. Apne frame se ek frame-time pehle start hua frame bhi takra sakta hai, aur baad mein bhi.
A frame lasting T can collide with an equal-length frame that began less than T earlier, or one that begins less than T later. The two intervals together have length T+T=2T.
Yaad rakho: Pure: before T + after T = 2T.
\r\nDefinition. Slotted ALOHA is a random-access protocol that divides time into slots of one frame duration and permits transmissions only at the beginning of a slot.
Simple meaning. Frame ready hai tab bhi next slot ki start-line ka wait karo.
| Feature | Pure | Slotted |
|---|---|---|
| Start time | Any time | Slot boundary only |
| Synchronization | No common slots required | Required |
| Vulnerable period | 2T | T |
| Maximum S | 1/(2e) ≈ 0.184 | 1/e ≈ 0.368 |
| G at maximum | 0.5 | 1 |
Yaad rakho: Slotted = aligned starts; collision risk half-window, not zero.
An IPv4 address identifies an IP interface. The network prefix selects the destination network; remaining bits identify an address within it. Classful addressing is historical; modern routing uses explicit CIDR prefixes.
| Class | First octet | Default mask | Network / host split | Purpose |
|---|---|---|---|---|
| A | 1–126 (ordinary unicast) | 255.0.0.0 (/8) | 8 / 24 bits | Large historical unicast networks; 0 and 127 reserved |
| B | 128–191 | 255.255.0.0 (/16) | 16 / 16 bits | Medium historical unicast networks |
| C | 192–223 | 255.255.255.0 (/24) | 24 / 8 bits | Small historical unicast networks |
| D | 224–239 | No unicast default mask | Not network/host class split | Multicast |
| E | 240–255 | No unicast default mask | Not normal unicast allocation | Reserved/experimental; special addresses exist |
Example: 114.34.2.8 is class A with network 114.0.0.0 and host portion 34.2.8. Class does not override an explicitly supplied CIDR prefix.
User Datagram Protocol (UDP) provides connectionless, message-oriented process delivery. One send creates a datagram; message boundaries are retained. UDP has an 8-byte header, carries ports and length, and offers checksum-based corruption detection. It does not itself guarantee delivery, ordering, duplicate suppression, retransmission, flow control or congestion control.
| UDP field | Width | Purpose |
|---|---|---|
| Source port | 16 bits | Sending endpoint; a zero value can indicate no reply port when unused |
| Destination port | 16 bits | Receiving application endpoint |
| Length | 16 bits | UDP header plus UDP payload; minimum 8 bytes |
| Checksum | 16 bits | Error detection over UDP and an IP pseudo-header; not retransmission |
Yaad rakho: UDP datagram preserve; reliability automatic nahi.
| Application | Why UDP can suit it |
|---|---|
| DNS queries | Small request/reply; application retries if necessary. DNS also supports TCP. |
| DHCP | A host without an assigned address can use UDP broadcasts to obtain configuration. |
| Real-time audio/video using RTP | Late packets are often less useful than timely packets; application handles loss/jitter. |
| Online game state updates | A newer position can replace a stale one; application-specific reliability can protect critical events. |
| NTP | Small timestamp exchanges; a fresh later measurement can replace a lost exchange. |
These are examples of applications that can use UDP, not a claim that every implementation of each application always uses UDP. Applications must still manage congestion responsibly.
Simple Mail Transfer Protocol (SMTP) transfers email from a sending client/server to a receiving mail server using a reliable transport, normally TCP. It is a push/store-and-forward transfer protocol. User agents compose/read mail; mail-transfer agents relay it, consulting DNS MX records to find recipient-domain mail servers.
| Phase | Typical exchange | Purpose |
|---|---|---|
| Connection and greeting | Server 220; client EHLO (HELO for basic SMTP); server 250 capabilities | Identify peers and learn supported extensions |
| Envelope transfer | MAIL FROM:<sender>; RCPT TO:<recipient>; 250 replies | Declare return path and one or more delivery recipients |
| Message transfer | DATA; server 354; headers + blank line + body; single dot line; server 250 | Transmit the message after server permission; completion reply accepts responsibility |
| Termination | QUIT; server 221 | Close the SMTP session cleanly |
Yaad rakho: SMTP send/relay; inbox read IMAP/POP/webmail.
| Phase | Commands / replies | Meaning |
|---|---|---|
| 1. Connection establishment | TCP connection; 220 greeting; EHLO/HELO; 250 reply | Open the connection and identify the SMTP peers |
| 2. Mail transfer | MAIL FROM, RCPT TO, DATA; 354; message; single dot; 250 | Transfer the envelope and then the message content |
| 3. Connection termination | QUIT; 221; close TCP | Finish the SMTP session |
Envelope transfer and message-body transfer are substeps of the single mail-transfer phase. Thus the more detailed table above still fits the standard three-phase description.
The Open Systems Interconnection (OSI) model separates network functions into seven layers. A layer offers services upward and uses services below. Peer layers communicate logically through protocol rules; actual data moves down the sender’s stack, across links and up the receiver’s stack.
| Layer | Task | Example |
|---|---|---|
| 7 Application | Services used by network applications | HTTP, DNS, FTP, SMTP |
| 6 Presentation | Representation, encoding and application data transformations | Character encoding, serialization; encryption as a conceptual function |
| 5 Session | Manage dialogue and synchronization/checkpoints | Session establishment and resumption concepts |
| 4 Transport | Process delivery; reliability/flow control where supplied | TCP, UDP, SCTP; ports |
| 3 Network | Logical addressing and routing across networks | IP; routers |
| 2 Data link | Framing, local link addressing, medium access and error detection | Ethernet/Wi-Fi; MAC addresses |
| 1 Physical | Transmit signal symbols/bits through a medium | Electrical, optical or radio signals |
Yaad rakho: Application se physical tak wrap; receiver par unwrap.
| Layer | Function 1 | Function 2 |
|---|---|---|
| Application | Provide application-level network services, such as file transfer | Define application exchanges for services such as email or name resolution |
| Presentation | Translate/serialize data into an agreed representation | Compress data or encrypt/decrypt it where these transformations are required |
| Session | Establish, manage and end a dialogue/session | Provide synchronization/checkpoints for resuming a dialogue |
| Transport | Identify sending/receiving processes using ports and segment/reassemble data | Provide end-to-end reliability and flow control when the selected protocol supports them |
| Network | Provide logical addressing across networks | Select routes and forward packets towards the destination |
| Data link | Frame the data and identify local interfaces | Control shared-medium access and detect link-frame errors |
| Physical | Represent bits as electrical, optical or radio signal symbols | Specify physical-interface/signalling properties such as timing, connectors and transmission rate |
These are reference-model responsibilities; an actual protocol stack may combine them. UDP, for example, does not supply TCP-style reliable delivery merely because both are transport protocols.
In a client-server design, users’ clients contact designated service processes. In peer-to-peer (P2P), participating hosts can supply and consume resources. These are application organizations, not different physical cable types.
| Point | Client-server | Peer-to-peer |
|---|---|---|
| Roles | Clients request; designated servers provide services | Peers can both request and provide resources |
| Control | Central policies, accounts and backups are easier to coordinate | Control and copies distributed among peers |
| Performance | Predictable with adequate server capacity; server may bottleneck | Resources grow with peers, but peer capacity/availability vary |
| Scalability | Scale servers, replicas, caching and load distribution | More peers can contribute capacity; coordination becomes harder |
| Reliability | One unreplicated server is a failure point; replication helps | Loss of one peer need not stop system if useful replicas remain |
| Security | Central access control and monitoring | Trust and consistent access policy are harder |
| Best fit | Managed collaboration, databases, login services | Distributed content sharing among participating peers |
Central servers can authenticate users, maintain the authoritative shared document and control versions. The organization must provision capacity, replicate data and avoid a single server failure. P2P can exchange large files or local edits among peers and distribute load, but needs discovery, conflict resolution, trust and availability strategies. A hybrid can keep account control centralized while distributing selected content.
Yaad rakho: Server mein service central; peers dono roles karte hain.
Definition. CSMA with Collision Detection monitors the medium during transmission, aborts a detected collision and retries after a randomized backoff.
Simple meaning. CD = takkar detect hui toh transmission rok do.
CSMA/CD senses before transmission and detects collisions during transmission. A collision causes abort/jam and a random backoff before retry. To detect the most distant collision before the frame finishes, the transmission duration must be at least the worst-case round-trip propagation delay. Thus Lmin/R ≥ 2Tp in the idealized model.
Yaad rakho: Detect karna hai toh sender tab tak bhej raha hona chahiye.
Frame ready → sense channel\r\n │\r\n Busy? yes → keep sensing (1-persistent case)\r\n no\r\n ↓\r\n Transmit and monitor\r\n │\r\n Collision? no → frame finished? → success\r\n yes\r\n ↓\r\n Jam indication → abort frame → increment attempt count\r\n │\r\n Retry limit reached? yes → report failure\r\n no\r\n ↓\r\n Random binary-exponential backoff → sense again
For Ethernet, the backoff range grows with collisions, capped after the specified stage; transmission is abandoned at the retry limit. Carrier sensing must still be obeyed after the random delay.
Definition. Open Shortest Path First (OSPF) is an interior link-state routing protocol that selects routes using shortest-path calculations and additive link costs.
Simple meaning. OSPF = link-state map + Dijkstra + summed costs.
OSPF exchanges and maintains link-state information, builds a topology database and runs Dijkstra’s shortest-path computation rooted at the local router. It chooses a path with the lowest sum of link costs and installs the corresponding next hop. Equal-cost paths may be supported.
Yaad rakho: OSPF cost add karo; sirf hops nahi.
In the textbook leaky-bucket queue/shaper model, arriving packets enter a finite buffer. The output sends at a configured rate whenever data is queued. When the queue empties, it cannot send nonexistent packets; when the buffer is full, excess arrivals are dropped in this finite-buffer model.
| Advantage | Limitation |
|---|---|
| Simple bound on output rate | Bursts may overflow or suffer delay |
| Reduces short-term output variation | Unused idle service cannot be saved |
| Useful for predictable traffic shaping | No end-to-end delay/loss guarantee by itself |
The term “leaky bucket” can also describe a conformance meter in standards. This course’s diagrams and numerical questions explicitly use the finite data-queue model, so its state is queued data rather than tokens.
No algorithm is universally better. The leaky data-queue model is useful for a smooth output profile; token bucket is preferable when controlled bursts should be allowed. State the application’s delay, burst and rate needs before recommending one.
Yaad rakho: Leaky bucket stores data; idle service save nahi hota.
\r\nA token bucket generates tokens at rate r up to capacity B. Sending one unit of data consumes the corresponding tokens. Tokens accumulate during idle periods only until the bucket is full. Data can therefore burst when saved tokens exist, subject to the physical link rate.
| Limitation | Reason |
|---|---|
| Bursts can still congest downstream queues | Local conformance does not ensure every later link has capacity |
| Parameters must match traffic and link requirements | A small B delays/drops legitimate packets; a large B permits disruptive bursts |
| No complete QoS guarantee alone | Admission, scheduling and end-to-end capacity may still be needed |
Yaad rakho: Token permission store hota hai; actual packet alag queue mein.
\r\n| Point | Leaky data queue | Token bucket |
|---|---|---|
| Stored state | Queued data | Tokens representing permission |
| Idle period | Empty queue; unused service lost | Tokens accumulate up to B |
| Output burst | Smoothed by configured service rate | Allowed if enough tokens and physical capacity exist |
| Overflow/shortage | Finite queue can overflow | Shaper waits; policer may mark/drop |
| Main parameters | Queue capacity and drain rate | Token capacity and generation rate |
| Useful when | A smoother output profile is required | Average rate plus burst allowance is required |
Selective Repeat (SR) ARQ accepts and buffers valid out-of-order frames inside the receiver window, acknowledges them individually and retransmits only missing/timed-out frames. In this answer ACK i means “frame i received”, unlike the next-expected cumulative ACK convention used for GBN.
Yaad rakho: SR selective resend; half sequence space keeps old and new separate.
| Case | Receiver action | Sender action |
|---|---|---|
| Normal transmission | Accept valid frames in its window; deliver the contiguous in-order sequence; ACK each frame | Send within the window and advance its base when the earliest outstanding frames are acknowledged |
| Lost data frame | Buffer later valid frames and ACK them, leaving the missing gap | On the missing frame's timeout, retransmit only that frame |
| Lost ACK | If an already accepted frame is repeated, acknowledge it again without delivering it twice | The individual timer may expire and repeat that one frame; the repeated ACK ends recovery |
Advantage: selective retransmission saves link capacity after isolated loss. Disadvantage: buffering out-of-order frames and tracking individual timers/ACKs make the receiver and sender more complex.
191.10.0.0/16; Group 1: 128 customers × 256 addresses; Group 2: 64 customers × 128 addresses
Each customer prefix, every group range, first/last customer blocks, and unallocated addresses.
These are requested total addresses, not usable-host counts. Allocate customer blocks on their prefix boundary. A group with a non-power-of-two customer count need not be forced into one larger, wasteful CIDR aggregate.
| Group | Customers | Addresses each | Customer prefix | Used | Group range |
|---|---|---|---|---|---|
| 1 | 128 | 256 | /24 | 32768 | 191.10.0.0–191.10.127.255 |
| 2 | 64 | 128 | /25 | 8192 | 191.10.128.0–191.10.159.255 |
| Group | First customer | Last customer |
|---|---|---|
| 1 | 191.10.0.0–191.10.0.255 | 191.10.127.0–191.10.127.255 |
| 2 | 191.10.128.0–191.10.128.127 | 191.10.159.128–191.10.159.255 |
Allocated = 32768 + 8192 = 40960. Unused = 65536 − 40960 = 24576.
Unused range: 191.10.160.0–191.10.255.255.
File Transfer Protocol (FTP) transfers files between a client and server using TCP and separates commands from file data. This separation is its central design feature. Its control connection remains available across multiple operations, while data connections carry file bytes or directory listings.
| Connection | Function | Lifetime / ports |
|---|---|---|
| Control | Commands and replies: USER, PASS, CWD, TYPE, LIST, RETR, STOR, QUIT | Client connects to server TCP port 21; normally kept for the session |
| Data | File contents and directory listing data | Opened as needed; separate from command channel |
FTP over TLS (FTPS) protects FTP connections. SFTP is a separate SSH-based file-transfer protocol, not “FTP using an SSH port”. These names should not be treated as interchangeable.
Yaad rakho: Control commands, data file; passive mein client dono connect karta.
Definition. Address Resolution Protocol (ARP) resolves a known IPv4 address to a link-layer hardware address on the local network.
Simple meaning. IP pata hai; local delivery ke liye MAC chahiye.
ARP maps a known local next-hop IPv4 address to a hardware address. A sender broadcasts a request when the mapping is missing; the address owner replies, and the mapping is cached. For a remote IP destination, the sender resolves its gateway’s local MAC address, then forwards the packet toward that gateway.
Yaad rakho: Remote packet ka next-hop MAC gateway ka hota hai.
Data = 110101; generator bits = 1011; degree r = 3.
CRC remainder, complete transmitted frame and receiver check.
Append r zeros, divide using XOR (modulo 2), and replace those zeros with the r-bit remainder. XOR has no borrowing or carrying. A valid received codeword has remainder zero; the receiver does not append new zeros.
110101 → 110101000
| Leading position | Before XOR | Aligned generator | After XOR |
|---|---|---|---|
| 1 | 110101000 | 101100000 | 011001000 |
| 2 | 011001000 | 010110000 | 001111000 |
| 3 | 001111000 | 001011000 | 000100000 |
| 4 | 000100000 | 000101100 | 000001100 |
| 6 | 000001100 | 000001011 | 000000111 |
Remainder = 111. Transmit 110101111.
| Position | Before | Generator | After |
|---|---|---|---|
| 1 | 110101111 | 101100000 | 011001111 |
| 2 | 011001111 | 010110000 | 001111111 |
| 3 | 001111111 | 001011000 | 000100111 |
| 4 | 000100111 | 000101100 | 000001011 |
| 6 | 000001011 | 000001011 | 000000000 |
Receiver remainder = 000. No error is detected; this is not a guarantee that every possible corruption is detectable.
211.15.35.40/26
First and last addresses and total block size.
p is the prefix length. The remaining host bits vary through every combination. The network address is the first address; setting all host bits to one gives the last address.
32 − 26 = 6. N = 64.
Mask = 255.255.255.192. The first varying octet is octet 4; its block width is 64. The supplied value 40 lies in the interval 0–63. Keep earlier octets fixed. Set any later octets to 0 for the first address and 255 for the last address.
First = 211.15.35.0; last = 211.15.35.63; total = 64 addresses. Usual usable host range: 211.15.35.1–211.15.35.62.
The Domain Name System (DNS) is a distributed hierarchical naming system. Its namespace is a tree with the root at the top. A complete name is the ordered path of labels from a node back to the root. An absolute name may be written with a trailing dot, for example www.example.com.
| Level | Example | Meaning |
|---|---|---|
| Root | . | Top of the namespace |
| Top-level domain (TLD) | com, org, in | A child zone below the root |
| Registered/delegated domain | example.com | Name under a parent such as com |
| Subdomain/host label | www.example.com | A label below example.com; a name may own different record types |
Yaad rakho: Full path unique; www label akela unique nahi.
Client → recursive resolver → root (.)\r\n → .com TLD\r\n → authoritative server for xyz.com\r\nClient ← final A/AAAA answer ← resolver
Assume a cache miss and that the name exists. (1) The client asks its resolver for www.xyz.com. (2) A root server refers the resolver to .com servers. (3) A .com server refers it to authoritative servers for xyz.com. (4) An authoritative server returns the applicable A/AAAA record, or a CNAME that must be followed. (5) The resolver returns the answer and caches it according to TTL. No real IP address for this domain is assumed. Root and TLD servers normally supply referrals here, not the final host address.
| From | To | Cost |
|---|---|---|
| A | B | 4 |
| A | C | 3 |
| A | D | 6 |
| C | B | -1 |
| D | C | -4 |
| B | E | -1 |
| C | E | 2 |
| D | F | -2 |
| E | F | 3 |
The original diagram is directed. All edge labels are reproduced below. Source = A.
Swipe across the diagram to read its labels ↔![\"bellman](\"fig/old-bellman-2023.5222c69942.svg\")
| From | To | Cost |
|---|---|---|
| A | B | 4 |
| A | C | 3 |
| A | D | 6 |
| C | B | -1 |
| D | C | -4 |
| B | E | -1 |
| C | E | 2 |
| D | F | -2 |
| E | F | 3 |
Shortest distance, complete route and predecessor for every node from A.
Relaxation replaces a tentative route when going through u is cheaper. Bellman–Ford permits negative edge weights. This table uses synchronous passes: pass k uses distances at the end of pass k−1, so it covers paths of at most k edges.
| Pass / settled node | Strict improvements | A | B | C | D | E | F |
|---|---|---|---|---|---|---|---|
| 0 | Initial | 0 | ∞ | ∞ | ∞ | ∞ | ∞ |
| 1 | B←A: 0+(4)=4; C←A: 0+(3)=3; D←A: 0+(6)=6 | 0 | 4 | 3 | 6 | ∞ | ∞ |
| 2 | B←C: 3+(-1)=2; C←D: 6+(-4)=2; E←B: 4+(-1)=3; F←D: 6+(-2)=4 | 0 | 2 | 2 | 6 | 3 | 4 |
| 3 | B←C: 2+(-1)=1; E←B: 2+(-1)=1 | 0 | 1 | 2 | 6 | 1 | 4 |
| 4 | E←B: 1+(-1)=0 | 0 | 1 | 2 | 6 | 0 | 4 |
| 5 | F←E: 0+(3)=3 | 0 | 1 | 2 | 6 | 0 | 3 |
| Destination | Distance | Previous | Route |
|---|---|---|---|
| A | 0 | — | A |
| B | 1 | C | A → D → C → B |
| C | 2 | D | A → D → C |
| D | 6 | A | A → D |
| E | 0 | B | A → D → C → B → E |
| F | 3 | E | A → D → C → B → E → F |
| Edge | Candidate d[u] + weight | Current d[v] | Improvement? |
|---|---|---|---|
| A → B | 0 + (4) = 4 | 1 | No |
| A → C | 0 + (3) = 3 | 2 | No |
| A → D | 0 + (6) = 6 | 6 | No |
| C → B | 2 + (-1) = 1 | 1 | No |
| D → C | 6 + (-4) = 2 | 2 | No |
| B → E | 1 + (-1) = 0 | 0 | No |
| C → E | 2 + (2) = 4 | 0 | No |
| D → F | 6 + (-2) = 4 | 3 | No |
| E → F | 0 + (3) = 3 | 3 | No |
No edge improves its destination on the extra pass. Therefore no negative cycle reachable from the source is detected.
Congestion occurs when offered traffic overloads network resources such as link capacity, router buffers or processing. Queueing delay grows, packets may be dropped, and excessive retransmission can make the overload worse. Flow control protects a receiver; congestion control protects the network.
| Approach | Policy / mechanism | How it helps |
|---|---|---|
| Open loop: prevent congestion without runtime congestion feedback | Retransmission policy | Carefully chosen timers/selective retransmission avoid needless extra traffic |
| Open loop | Window policy | Limit outstanding data and avoid excessive retransmission after loss |
| Open loop | Acknowledgement policy | Avoid unnecessarily frequent ACK traffic while retaining correct recovery |
| Open loop | Admission policy | Refuse new load when reserved capacity is insufficient |
| Closed loop: measure and react to congestion | Backpressure | A congested node tells upstream nodes to reduce/stop sending, possibly propagating hop by hop |
| Closed loop | Choke/control signalling | A congested node sends explicit feedback to the source |
| Closed loop | Implicit signalling | A sender infers congestion from loss/delay and changes its sending behaviour |
| Closed loop | Explicit marking | Network marks congestion (for example ECN); endpoints react to returned information |
For an answer asking for two of each: explain retransmission and admission policies under open loop, then backpressure and explicit signalling under closed loop, with the mechanism and its effect. These are general congestion-control classifications used in the course; they are not all mandatory algorithms implemented inside TCP. TCP commonly uses end-to-end feedback such as loss or ECN.
Yaad rakho: Open loop pehle prevent; closed loop feedback se react.
| Point | UDP | TCP | SCTP |
|---|---|---|---|
| Service unit | Datagram; preserves messages | Byte stream; no message boundaries | Messages carried in chunks; preserves boundaries |
| Connection setup | No protocol handshake | Connection; usual three-way SYN exchange | Association; four-message cookie exchange |
| Reliability | No delivery/order guarantee | Reliable ordered stream | Reliable messages; ordered per stream by default, unordered option |
| Ordering scope | None supplied | One ordered byte stream per direction | Multiple streams; per-stream order avoids blocking every stream on one stream’s ordering gap |
| Flow control | Not supplied by UDP | Receiver-advertised window | Receiver flow control |
| Congestion control | Not supplied by UDP | Yes | Yes |
| Multihoming | No built-in association failover | Classic TCP uses one address pair | Association can bind multiple addresses; built-in path failover |
| Header | 8 bytes | Minimum 20 bytes | 12-byte common header plus chunks |
| Typical use | DNS, real-time applications with their own controls | Web over HTTP/1.1 or HTTP/2, reliable byte transfers | Telecom signalling and multi-stream message applications |
| Performance claim | Low protocol overhead, not automatically fastest overall | Recovery/ordering have costs and benefits | Extra capabilities; no universal speed ranking |
The correct protocol depends on application semantics and network conditions. “UDP is always faster” and “SCTP is always slower than TCP” are not defensible general rules. All three normally run above IP; security/encryption is a separate requirement.
Yaad rakho: UDP messages simple; TCP byte stream; SCTP reliable message streams.
Guided media constrain signals to a physical path: twisted pair, coaxial cable and optical fiber. Unguided media carry electromagnetic waves through free space: radio, microwave and infrared. Wireless bandwidth and reach depend on frequency, power, antenna, obstacles and regulations; media names alone do not fix a speed.
| Medium | Two key characteristics | Limitations / use |
|---|---|---|
| Twisted pair | Electrical signal; pairs twisted to reduce interference | Low-cost LAN access; distance/noise limits depend on category and design |
| Coaxial cable | Shielded central conductor; greater shielding than basic unshielded pair | Bulkier; cable broadband and legacy shared LANs |
| Optical fiber | Light in glass/plastic; high capacity with immunity to electromagnetic interference | Transceivers and careful splicing/bend control needed; backbone links |
| Radio waves | Wireless propagation; many bands support broad/non-line-of-sight coverage | Interference, shared spectrum and security configuration matter; Wi-Fi/mobile |
| Terrestrial microwave | Directional antennas; typically line of sight | Alignment and obstructions matter; point-to-point backhaul |
| Satellite microwave | Wide coverage through relay in space | Cost and latency depend strongly on orbit/system; remote connectivity |
| Infrared | Short-range light; usually blocked by walls | Limited coverage; short-range indoor/remote-control links |
Yaad rakho: Guided = cable path; unguided = free-space waves.
\r\nTransmission media are classified into guided media (twisted pair, coaxial cable, optical fiber) and unguided media (radio, microwave, infrared).
Optical fiber carries data as light in a core surrounded by cladding. It is a guided medium. Single-mode fiber supports one main propagation mode and is used for long/high-capacity links; multimode supports multiple modes and is often used over shorter distances.
| Advantages | Disadvantages / design costs |
|---|---|
| High data capacity and low attenuation over suitable links | Optical transceivers, connectors and splicing add cost/skill requirements |
| Immune to electromagnetic interference; no electrical crosstalk | Tight bends, dirty connectors and poor installation increase loss |
| Electrical isolation between buildings | Fiber is mechanically sensitive and requires appropriate protection |
| Harder to tap casually than an unprotected radio channel | It is not automatically confidential; encryption/access control remain necessary |
Yaad rakho: Fiber light carry karta hai; immunity means no EMI, not no security risk.
| Medium / working principle | Advantages | Disadvantages |
|---|---|---|
| Twisted pair: voltage signals travel on two insulated copper conductors twisted together | Low purchase/installation cost; flexible and easy to terminate for local access | More susceptible to electromagnetic interference than fiber; distance and capacity are limited by cable category and signalling design |
| Optical fiber: light pulses travel through a core guided by its refractive-index structure | High capacity and long reach; immune to electromagnetic interference | Optical equipment and splicing can cost more; tight bends, damage and dirty connectors degrade the link |
These comparisons describe typical engineering tradeoffs, not a guarantee that every fiber installation is faster or more expensive than every copper installation.
Definition. Persistence specifies how a station behaves after observing that the medium is busy or becomes idle.
Simple meaning. 1 = turant; non = random wait; p = probability ke basis pe start.
| Rule | When idle | When busy | Trade-off |
|---|---|---|---|
| 1-persistent | Transmit immediately | Keep listening; transmit as soon as idle | Low idle delay, but waiting stations may start together |
| Non-persistent | Transmit | Wait a random time, then sense again | Less synchronized contention; may leave avoidable idle time |
| p-persistent (slotted) | Transmit with probability p; defer a slot with probability 1−p | Wait until available, then resume the probabilistic slot rule | Controls contention; needs slots and a choice of p |
Yaad rakho: 1 = start now; non = wait then listen; p = chance per idle slot.
\r\nDefinition. CSMA with Collision Detection monitors the medium during transmission, aborts a detected collision and retries after a randomized backoff.
Simple meaning. CD = takkar detect hui toh transmission rok do.
CSMA/CD senses before transmission and detects collisions during transmission. A collision causes abort/jam and a random backoff before retry. To detect the most distant collision before the frame finishes, the transmission duration must be at least the worst-case round-trip propagation delay. Thus Lmin/R ≥ 2Tp in the idealized model.
Yaad rakho: Detect karna hai toh sender tab tak bhej raha hona chahiye.
1-persistent:\r\nReady → sense → busy? yes → sense again\r\n no → transmit immediately\r\n\r\np-persistent (slotted medium):\r\nReady → sense → busy? yes → wait and sense\r\n no → draw random decision\r\n probability p → transmit\r\n probability 1−p → wait one slot → sense again
Non-persistent CSMA instead waits a random time after a busy observation before sensing again. A collision leads to the protocol's retry/backoff procedure; the persistence rule alone does not guarantee success.
Definition. Switching selects how data passes through intermediate network nodes from a source to a destination.
Simple meaning. Raasta reserve karna, poora message ruk-ruk ke bhejna, ya chhote packets bhejna—yeh teen ideas hain.
| Aspect | Circuit switching | Message switching | Packet switching |
|---|---|---|---|
| Unit handled | Continuous stream on established circuit | Whole message | Small packets |
| Resource use | Capacity reserved along a circuit | Shared; message stored at each intermediate node | Shared; each packet stored/forwarded in this model |
| Setup | Required before data transfer | No circuit setup required | Datagram: no connection setup; virtual circuit: setup |
| Delay | Setup delay, then relatively predictable service | Large store-and-forward delay for long messages | Variable delay; packet pipeline can reduce completion time |
| Main limitation | Reserved capacity can be idle | Large buffers; unsuitable for tight delay needs | Headers, possible queueing/loss; datagrams may reorder |
For ordinary bursty office data, packet switching is generally appropriate because resources are shared among active transfers. The design must still account for queueing, loss and the required service quality.
Yaad rakho: Circuit reserves; message waits whole; packets pipeline.
| Method | Advantage 1 | Advantage 2 | Disadvantage 1 | Disadvantage 2 |
|---|---|---|---|---|
| Message switching | Shares link capacity without a permanently reserved circuit | Intermediate storage lets a message wait for a currently busy outgoing link | The entire message must arrive before forwarding, giving long delay | Large messages require large buffers and can hold up other messages |
| Packet switching | Statistical sharing suits bursty data from many users | Packets can be pipelined across successive links, reducing completion time | Every packet needs header/control overhead | Queueing, loss and (for datagrams) reordering need endpoint or network handling |
A packet-switched network may use datagrams or virtual circuits. The possibility of independent routes and reordering refers to datagrams, not every packet-switching implementation.
Quality of Service (QoS) describes delivery characteristics important to an application: throughput/bandwidth, delay, delay variation (jitter) and loss/reliability. Different applications value them differently; a voice call is sensitive to delay/jitter, while a file transfer strongly needs complete data.
| Technique | Mechanism | Tradeoff |
|---|---|---|
| Scheduling | FIFO, priority or fair/weighted queues choose which packet transmits next | Strict priority may starve lower classes; fair scheduling balances shares |
| Traffic shaping | Delay departures to match a profile | Smooths load but adds queueing delay |
| Policing | Measure compliance; mark/drop nonconforming packets according to policy | Controls excess traffic but can lose packets |
| Admission control | Accept new flows only when resources can support them | Protects existing service by refusing excess demand |
| Resource allocation/reservation | Assign bandwidth/buffer/service resources | Requires capacity planning and coordination |
| Classification/marking | Identify traffic classes and carry service markings | Marking alone does not create capacity or guarantees |
Yaad rakho: Bandwidth, delay, jitter, loss; requirement ke hisaab se policy.
FIFO: arrivals → [one queue] → link\r\nPriority: high → [queue H] ─┐\r\n low → [queue L] ─┴→ choose H first → link\r\nWeighted fair: Q1, Q2, Q3 → scheduler shares 1:2:1 → link
FIFO sends in arrival order, so a long burst can delay urgent traffic. Strict priority chooses the highest nonempty class and can starve lower classes. Fair/weighted fair scheduling shares service among queues according to equal or configured weights, preventing one flow from taking every opportunity. The four QoS dimensions are bandwidth/throughput, delay, jitter and loss/reliability.
Frame length L = 150 bits; link rate R = 400,000 bit/s; attempted rates λ = 800, 400 frames/s.
Normalized throughput S and successful frames per second for each offered load.
![\"T_f=L/R,\\quad](\"math/094350050f7316f3.svg\")
G counts attempted frames per frame time, including retransmissions if present. S is not measured in frames/s. Multiply by the frame rate capacity 1/Tf to obtain surviving frames/s.
Tf = 150/400000 = 0.000375 s. Capacity = 1/Tf = 2666.66667 frames/s.
G = 800 × 0.000375 = 0.3.
e−1G = e−0.3 = 0.740818221.
S = 0.3 × 0.740818221 = 0.222245466.
Successful frames/s = 800 × 0.740818221 = 592.654577. Successful bit/s = 592.654577 × 150 = 88898.1865.
G = 400 × 0.000375 = 0.15.
e−1G = e−0.15 = 0.860707976.
S = 0.15 × 0.860707976 = 0.129106196.
Successful frames/s = 400 × 0.860707976 = 344.283191. Successful bit/s = 344.283191 × 150 = 51642.4786.
Definition. CSMA with Collision Avoidance uses carrier sensing and randomized access delays, with acknowledgements and optional reservation exchanges, to reduce collisions on a wireless shared medium.
Simple meaning. CA = pehle collision avoid karne ki koshish; guarantee nahi.
| Aspect | CSMA/CD | CSMA/CA |
|---|---|---|
| Main action | Detect and abort a collision during sending | Reduce collision risk before sending |
| Typical teaching context | Shared half-duplex wired Ethernet | Wi-Fi contention access |
| Mechanisms | Carrier sensing, detection/jam, backoff | Carrier sensing, random backoff, ACK; optional RTS/CTS |
Yaad rakho: CD = detect during; CA = avoid before.
\r\nDefinition. Hidden terminals are stations that cannot sense each other but whose transmissions can interfere at a common receiver.
Simple meaning. A aur C ek dusre ko nahi sunte, par dono B ko bhej rahe hain.
A and C may both reach B but be unable to sense one another. Simultaneous transmissions can then collide at B. A short RTS followed by B’s CTS announces a reservation; a station hearing the CTS defers, reducing the risk of a long data-frame collision.
Yaad rakho: Hidden: senders invisible to each other, receiver common.
Frame ready → sense medium\r\nBusy? yes → defer until eligible idle interval\r\n ↓\r\nChoose a random backoff → count down eligible idle slots\r\nMedium becomes busy? → freeze counter; resume after required idle wait\r\n ↓ counter reaches zero\r\nOptional RTS → CTS received? no → retry/back off\r\n ↓ yes (or RTS/CTS not used)\r\nSend DATA → ACK received? yes → success\r\n no → retry/back off or fail at retry limit
A station hearing a duration reservation updates its network allocation vector and defers. RTS/CTS is optional; loss of CTS or ACK triggers recovery but does not by itself prove a collision, because channel errors can also cause loss.
Definition. Routing Information Protocol (RIP) is an interior distance-vector routing protocol that uses a bounded metric, commonly hop count, to select routes.
Simple meaning. RIP mein hops count; 15 tak reachable, 16 means unreachable.
RIP uses distance-vector routing and commonly hop count as its metric. Reachable metrics are at most 15; 16 represents infinity. Its main limitations include limited diameter, a coarse metric and relatively slow convergence with possible temporary loops.
Yaad rakho: RIP: neighbour distances; 16 is infinity.
| Timer | Usual RFC value | Action |
|---|---|---|
| Periodic update | Approximately 30 seconds, with randomized variation | Send routing information to neighbours |
| Route timeout | 180 seconds without a valid refresh | Declare the route invalid and set metric 16 |
| Garbage collection | 120 seconds after invalidation | Keep advertising the unreachable route before deleting it |
Some vendor courses additionally describe an invalid/hold-down/flush timer set (often 180/180/240 seconds). Those are implementation conventions and must not be mixed with the RFC timeout-plus-garbage sequence as if they were identical clocks.
A TCP connection establishes each endpoint’s initial sequence number and confirms that both can receive. Use client initial sequence x=1000 and server initial sequence y=5000; assume no payload in the handshake.
| Step | Segment | Meaning |
|---|---|---|
| 1 Client → server | SYN, Seq=1000 | Client proposes its starting sequence |
| 2 Server → client | SYN+ACK, Seq=5000, Ack=1001 | Server acknowledges client SYN and supplies its own start |
| 3 Client → server | ACK, Seq=1001, Ack=5001 | Client acknowledges server SYN; connection is established |
Client sends 200 bytes beginning at Seq=1001, so byte numbers are 1001…1200. The receiver acknowledges Ack=1201, the next byte expected. If bytes starting at 1201 are missing but later bytes arrive, the cumulative ACK remains 1201 until the gap is repaired. TCP may buffer out-of-order bytes; it delivers the byte stream in order. Each direction counts independently.
Yaad rakho: SYN plus one, FIN plus one; ACK next expected byte.
The Domain Name System (DNS) is a distributed hierarchical naming system. Its namespace is a tree with the root at the top. A complete name is the ordered path of labels from a node back to the root. An absolute name may be written with a trailing dot, for example www.example.com.
| Level | Example | Meaning |
|---|---|---|
| Root | . | Top of the namespace |
| Top-level domain (TLD) | com, org, in | A child zone below the root |
| Registered/delegated domain | example.com | Name under a parent such as com |
| Subdomain/host label | www.example.com | A label below example.com; a name may own different record types |
Yaad rakho: Full path unique; www label akela unique nahi.
Client → recursive resolver → root (.)\r\n → .com TLD\r\n → authoritative server for xyz.com\r\nClient ← final A/AAAA answer ← resolver
Assume a cache miss and that the name exists. (1) The client asks its resolver for www.xyz.com. (2) A root server refers the resolver to .com servers. (3) A .com server refers it to authoritative servers for xyz.com. (4) An authoritative server returns the applicable A/AAAA record, or a CNAME that must be followed. (5) The resolver returns the answer and caches it according to TTL. No real IP address for this domain is assumed. Root and TLD servers normally supply referrals here, not the final host address.
Go-Back-N (GBN) ARQ pipelines up to W outstanding frames. For m-bit sequence numbers, textbook GBN uses W ≤ 2m−1 and receiver window 1. The receiver accepts only the next in-order frame; out-of-order frames are discarded. ACK k here cumulatively acknowledges all frames before k and means “next expected is k”.
Yaad rakho: GBN receiver one; gap ke baad discard; timeout pe outstanding suffix repeat.
Advantage: pipelining improves utilization over Stop-and-Wait while retaining a simple one-frame receiver window. Disadvantage: after a gap, the receiver discards later frames, so a timeout can force retransmission of several frames that arrived correctly.
File Transfer Protocol (FTP) transfers files between a client and server using TCP and separates commands from file data. This separation is its central design feature. Its control connection remains available across multiple operations, while data connections carry file bytes or directory listings.
| Connection | Function | Lifetime / ports |
|---|---|---|
| Control | Commands and replies: USER, PASS, CWD, TYPE, LIST, RETR, STOR, QUIT | Client connects to server TCP port 21; normally kept for the session |
| Data | File contents and directory listing data | Opened as needed; separate from command channel |
FTP over TLS (FTPS) protects FTP connections. SFTP is a separate SSH-based file-transfer protocol, not “FTP using an SSH port”. These names should not be treated as interchangeable.
Yaad rakho: Control commands, data file; passive mein client dono connect karta.
Definition. Dynamic Host Configuration Protocol (DHCP) provides IP configuration parameters to clients and can lease reusable addresses for limited periods.
Simple meaning. DORA = Discover, Offer, Request, Acknowledge.
DHCP reduces manual configuration and supports controlled address reuse through leases.
Yaad rakho: Discover–Offer–Request–ACK; pehle D, phir O, phir R, phir A.
Source note: 190.20.10.0 is inside 190.20.0.0/16, not the network address. Solution must disclose canonical network interpretation.
190.20.10.0/16; Group 1: 128 customers × 256 addresses; Group 2: 128 customers × 128 addresses
Source correction/assumption: The written address is a host address inside 190.20.0.0/16. A /16 cannot begin at third octet 10. Use the canonical 190.20.0.0/16 block; do not silently start unaligned subnets at 190.20.10.0.
Each customer prefix, every group range, first/last customer blocks, and unallocated addresses.
These are requested total addresses, not usable-host counts. Allocate customer blocks on their prefix boundary. A group with a non-power-of-two customer count need not be forced into one larger, wasteful CIDR aggregate.
| Group | Customers | Addresses each | Customer prefix | Used | Group range |
|---|---|---|---|---|---|
| 1 | 128 | 256 | /24 | 32768 | 190.20.0.0–190.20.127.255 |
| 2 | 128 | 128 | /25 | 16384 | 190.20.128.0–190.20.191.255 |
| Group | First customer | Last customer |
|---|---|---|
| 1 | 190.20.0.0–190.20.0.255 | 190.20.127.0–190.20.127.255 |
| 2 | 190.20.128.0–190.20.128.127 | 190.20.191.128–190.20.191.255 |
Allocated = 32768 + 16384 = 49152. Unused = 65536 − 49152 = 16384.
Unused range: 190.20.192.0–190.20.255.255.
Simple Mail Transfer Protocol (SMTP) transfers email from a sending client/server to a receiving mail server using a reliable transport, normally TCP. It is a push/store-and-forward transfer protocol. User agents compose/read mail; mail-transfer agents relay it, consulting DNS MX records to find recipient-domain mail servers.
| Phase | Typical exchange | Purpose |
|---|---|---|
| Connection and greeting | Server 220; client EHLO (HELO for basic SMTP); server 250 capabilities | Identify peers and learn supported extensions |
| Envelope transfer | MAIL FROM:<sender>; RCPT TO:<recipient>; 250 replies | Declare return path and one or more delivery recipients |
| Message transfer | DATA; server 354; headers + blank line + body; single dot line; server 250 | Transmit the message after server permission; completion reply accepts responsibility |
| Termination | QUIT; server 221 | Close the SMTP session cleanly |
Yaad rakho: SMTP send/relay; inbox read IMAP/POP/webmail.
| Phase | Commands / replies | Meaning |
|---|---|---|
| 1. Connection establishment | TCP connection; 220 greeting; EHLO/HELO; 250 reply | Open the connection and identify the SMTP peers |
| 2. Mail transfer | MAIL FROM, RCPT TO, DATA; 354; message; single dot; 250 | Transfer the envelope and then the message content |
| 3. Connection termination | QUIT; 221; close TCP | Finish the SMTP session |
Envelope transfer and message-body transfer are substeps of the single mail-transfer phase. Thus the more detailed table above still fits the standard three-phase description.
The TCP/IP model describes the Internet protocol suite in four layers. Many teaching books split the link layer into data link and physical, making a five-layer teaching model; identify the convention before drawing.
| TCP/IP layer | Function | Examples |
|---|---|---|
| Application | Defines application message formats and exchanges for Web access, naming, email and file transfer; includes presentation/session functions as needed | HTTP, DNS, SMTP, FTP |
| Transport | End-to-end process communication | TCP, UDP, SCTP |
| Internet | IP addressing, routing and best-effort datagrams | IPv4/IPv6 |
| Link/network access | Local delivery and physical access | Ethernet/Wi-Fi and their medium |
| Point | OSI | TCP/IP |
|---|---|---|
| Layer count | Seven | Four here; five-layer teaching variant splits link |
| Upper functions | Application, presentation, session separated | Grouped under application |
| Lower functions | Data link and physical separated | Grouped under link/network access |
| Purpose | General reference architecture | Architecture of deployed Internet protocol family |
| Network service | Reference model can describe different services | Internet layer uses connectionless IP |
| Transport | Conceptual transport service | TCP reliable byte stream; UDP datagrams; SCTP messages |
Yaad rakho: OSI 7; TCP/IP four mein upper three merge, lower two merge.
| Layer | Protocol | Role |
|---|---|---|
| Internet | IPv4 / IPv6 | Address and deliver best-effort packets across networks |
| Internet | ICMP / ICMPv6 | Report selected IP errors and operational information; used by diagnostic exchanges |
| Transport | TCP | Reliable, ordered byte stream with connection establishment, flow and congestion control |
| Transport | UDP | Connectionless message delivery using ports, without built-in retransmission or ordering |
| Transport | SCTP | Reliable message-oriented association with multiple streams and multihoming support |
| From | To | Cost |
|---|---|---|
| A | B | 3 |
| A | C | 5 |
| B | C | 9 |
| B | D | 8 |
| B | E | 6 |
| C | E | 3 |
| D | E | 10 |
| D | F | 3 |
| E | F | 5 |
The original diagram is undirected. All edge labels are reproduced below. Source = A.
Swipe across the diagram to read its labels ↔![\"dijkstra](\"fig/old-dijkstra-2024.b1cf42a96f.svg\")
| From | To | Cost |
|---|---|---|
| A | B | 3 |
| A | C | 5 |
| B | C | 9 |
| B | D | 8 |
| B | E | 6 |
| C | E | 3 |
| D | E | 10 |
| D | F | 3 |
| E | F | 5 |
Shortest distance, complete route and predecessor for every node from A.
Relaxation replaces a tentative route when going through u is cheaper. Dijkstra settles the unvisited node of least tentative distance. Nonnegative edge costs make the settled distance final. Equal-distance nodes may be settled in either order.
| Pass / settled node | Strict improvements | A | B | C | D | E | F |
|---|---|---|---|---|---|---|---|
| 0 | Initial | 0 | ∞ | ∞ | ∞ | ∞ | ∞ |
| A | B←A: 0+3=3; C←A: 0+5=5 | 0 | 3 | 5 | ∞ | ∞ | ∞ |
| B | D←B: 3+8=11; E←B: 3+6=9 | 0 | 3 | 5 | 11 | 9 | ∞ |
| C | E←C: 5+3=8 | 0 | 3 | 5 | 11 | 8 | ∞ |
| E | F←E: 8+5=13 | 0 | 3 | 5 | 11 | 8 | 13 |
| D | No change | 0 | 3 | 5 | 11 | 8 | 13 |
| F | No change | 0 | 3 | 5 | 11 | 8 | 13 |
| Destination | Distance | Previous | Route |
|---|---|---|---|
| A | 0 | — | A |
| B | 3 | A | A → B |
| C | 5 | A | A → C |
| D | 11 | B | A → B → D |
| E | 8 | C | A → C → E |
| F | 13 | E | A → C → E → F |
A tie between two equally short paths is valid; this table retains the first predecessor on equal distances.
Topology is the arrangement of nodes and links. Physical topology describes actual links; logical topology describes how data flows. A switch-centred physical star is different from a shared collision-domain bus.
| Topology | Structure | Benefit | Limitation |
|---|---|---|---|
| Bus | Devices share a common backbone | Simple small shared-medium layout | Backbone fault affects segment; collisions possible on classic shared bus |
| Star | Each device connects to a central switch/hub | Easy expansion and fault isolation | Central-device failure affects star; more cabling |
| Ring | Each node connected to two neighbours in a ring | Structured traversal; token-based variants coordinate access | Simple single-ring break can disrupt operation unless protected |
| Full mesh | Every pair has a direct link | Redundant paths and direct connectivity | Expensive ports/cabling: n(n−1)/2 undirected links |
| Tree/hybrid | Hierarchical stars or combined topologies | Scales into buildings/floors | Upper-level faults affect dependent branches |
Yaad rakho: Layout draw karo, then one strength and one failure case.
| Topology / example | Advantage 1 | Advantage 2 | Disadvantage 1 | Disadvantage 2 |
|---|---|---|---|---|
| Star: switched computer laboratory | A device can be added on a spare central port without rebuilding the whole topology | A single access-cable fault normally isolates only its attached device | Central-switch failure affects all attached devices unless redundant | A separate cable and switch port are needed per device |
| Bus: legacy coaxial Ethernet | Uses little cable in a small linear layout | No central switch is required in the simple shared-bus design | A backbone fault can disable the segment | Shared bandwidth, contention and fault diagnosis become difficult as it grows |
| Ring: token-based ring | Controlled token access avoids data collisions among compliant stations | Token rotation gives stations an organized opportunity to transmit | A break can interrupt an unprotected single ring | Adding/removing a node can disrupt a simple ring |
| Mesh: redundant router backbone | Alternate links preserve connectivity after some link failures | Direct pairwise links can carry separate exchanges without a shared central bottleneck | A full mesh needs n(n−1)/2 links and many ports | Configuration, cabling and maintenance become complex |
The ring benefits above assume token-controlled access; physical ring wiring alone does not guarantee fairness or collision avoidance. The failure disadvantages describe simple layouts without protection/redundancy.
Definition. Subnetting divides an address block into smaller prefix-defined blocks. In an ordinary subnet, the network and directed-broadcast addresses bound its usable host range.
Simple meaning. Pehle block nikalo; phir first aur last. Uske baad hosts.
A /26 has 32−26=6 host bits, so it contains 2⁶=64 total addresses. Under ordinary IPv4 subnet rules, one is the network address and one is directed broadcast, leaving 64−2=62 usable host addresses.
Yaad rakho: Total mein sab count; usable mein boundary ke 2 minus.
An IPv4 address identifies an IP interface. The network prefix selects the destination network; remaining bits identify an address within it. Classful addressing is historical; modern routing uses explicit CIDR prefixes.
| Class | First octet | Default mask | Network / host split | Purpose |
|---|---|---|---|---|
| A | 1–126 (ordinary unicast) | 255.0.0.0 (/8) | 8 / 24 bits | Large historical unicast networks; 0 and 127 reserved |
| B | 128–191 | 255.255.0.0 (/16) | 16 / 16 bits | Medium historical unicast networks |
| C | 192–223 | 255.255.255.0 (/24) | 24 / 8 bits | Small historical unicast networks |
| D | 224–239 | No unicast default mask | Not network/host class split | Multicast |
| E | 240–255 | No unicast default mask | Not normal unicast allocation | Reserved/experimental; special addresses exist |
Example: 114.34.2.8 is class A with network 114.0.0.0 and host portion 34.2.8. Class does not override an explicitly supplied CIDR prefix.
Example: 192.168.1.70 AND255.255.255.192 gives network192.168.1.64. The mask distinguishes the network prefix from the host portion. CIDR masks need not equal historical default class masks.
4 flows, server links 400 Mbps, client links 300 Mbps, shared link 1000 Mbps.
Maximum throughput of each flow.
![\"T=\\min(R_s,R_c,R/n)\"](\"math/26062ae8bce5ae5e.svg\")
A flow cannot exceed any link on its path; fair sharing divides the common capacity among n active flows.
1000 Mbps ÷ 4 = 250 Mbps per flow.
T=min(400,300,250)=250 Mbps.
Total delivered rate=4×250=1000 Mbps.
Known IP; new initial TCP connection; zero GET transmission; each object transmission=0.5 RTT.
Time to obtain both the base HTML and its referenced JPEG.
![\"T_{object}=T_{setup}+RTT+T_{transmission}\"](\"math/174653a5aa0dead0.svg\")
A referenced object cannot be requested until the HTML reveals it. HTTP/1.0 here uses a fresh connection per object; persistent HTTP/1.1 reuses one.
1 RTT for TCP setup + 1 RTT for GET/first response bit + 0.5 RTT to transmit = 2.5 RTT.
1 RTT for another TCP setup + 1 RTT for GET/response + 0.5 RTT transmission = 2.5 RTT.
Total = 2.5+2.5=5 RTT. DNS adds zero here because the IP address is already known.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
Received word 1011011; even parity. Number positions 7,6,5,4,3,2,1 from left to right, so position 1 is the rightmost bit.
Syndrome, erroneous position, and corrected word.
![\"s_p=\\bigoplus_{i\\,\\&\\,p\\ne0}b_i,\\quad](\"math/915c318ac72ec415.svg\")
Each parity check covers positions whose binary position number contains that parity bit. For even parity, XOR must be zero; a 1 identifies a failed group.
| Position | 7 | 6 | 5 | 4 | 3 | 2 | 1 |
|---|---|---|---|---|---|---|---|
| Received bit | 1 | 0 | 1 | 1 | 0 | 1 | 1 |
| Parity | Covered positions | Bits added | Sum | Sum mod 2 |
|---|---|---|---|---|
| 1 | 1, 3, 5, 7 | 1 + 0 + 1 + 1 | 3 | 1 |
| 2 | 2, 3, 6, 7 | 1 + 0 + 0 + 1 | 2 | 0 |
| 4 | 4, 5, 6, 7 | 1 + 1 + 0 + 1 | 3 | 1 |
Syndrome s4s2s1 = 101; decimal = 5. Under the single-bit-error assumption, flip position 5 from the right. Corrected word = 1001011.
Ordinary Hamming(7,4) corrects one bit. A multi-bit corruption may produce a misleading nonzero syndrome, so do not claim guaranteed correction of arbitrary errors.
Data = 1011011010; generator bits = 1011; degree r = 3.
CRC remainder, complete transmitted frame and receiver check.
Append r zeros, divide using XOR (modulo 2), and replace those zeros with the r-bit remainder. XOR has no borrowing or carrying. A valid received codeword has remainder zero; the receiver does not append new zeros.
1011011010 → 1011011010000
| Leading position | Before XOR | Aligned generator | After XOR |
|---|---|---|---|
| 1 | 1011011010000 | 1011000000000 | 0000011010000 |
| 6 | 0000011010000 | 0000010110000 | 0000001100000 |
| 7 | 0000001100000 | 0000001011000 | 0000000111000 |
| 8 | 0000000111000 | 0000000101100 | 0000000010100 |
| 9 | 0000000010100 | 0000000010110 | 0000000000010 |
Remainder = 010. Transmit 1011011010010.
| Position | Before | Generator | After |
|---|---|---|---|
| 1 | 1011011010010 | 1011000000000 | 0000011010010 |
| 6 | 0000011010010 | 0000010110000 | 0000001100010 |
| 7 | 0000001100010 | 0000001011000 | 0000000111010 |
| 8 | 0000000111010 | 0000000101100 | 0000000010110 |
| 9 | 0000000010110 | 0000000010110 | 0000000000000 |
Receiver remainder = 000. No error is detected; this is not a guarantee that every possible corruption is detectable.
Framing separates a continuous bit stream into data-link units so the receiver knows where each frame begins and ends.
| Method | Operation | Key issue |
|---|---|---|
| Fixed size | Every N bytes forms one frame | No delimiter needed, but padding may waste capacity |
| Length/character count | Header gives frame length | A damaged count can lose synchronization |
| Byte stuffing | Flag bytes delimit frames; escape any flag/escape inside payload | Receiver removes inserted escape bytes |
| Bit stuffing | Flag01111110; insert0 after five consecutive data1s | Receiver removes the stuffed0 |
| Coding violation | Reserve physical symbols unavailable to data | Depends on the line code |
Variable-length frames fit different packet sizes, reducing padding and often using bandwidth efficiently. They need boundary/length handling and variable buffering; fixed-size frames give simple predictable processing. Example: a20-byte payload need not occupy an entire500-byte fixed frame when a suitable short variable frame is allowed.
Two messages; no DATA or ACK errors; original FSM transition labels in the question.
The global interleaved transition sequence, including the final ACK.
Send DATA → receive and acknowledge DATA → receive matching ACK → advance the sequence bit.
Each successful ACK permits the sender to send the next new message.
The sequence bit labels the outstanding DATA packet. ACK0 confirms data0 and ACK1 confirms data1 in this printed FSM; these are not TCP cumulative byte numbers. An ACK loss does not advance the sender. A duplicate data packet is acknowledged again but not delivered twice.
| Send0 | → | ACK0 | → | Send1 | → | ACK1 |
| Order | Transition | Action |
|---|---|---|
| 1 | S0 | Send data0; start timer |
| 2 | R0 | Accept data0; deliver; send ACK0 |
| 3 | S3 | Accept ACK0; stop timer |
| 4 | S5 | Send data1; start timer |
| 5 | R2 | Accept data1; deliver; send ACK1 |
| 6 | S8 | Accept ACK1; stop timer |
Required sequence: S0, R0, S3, S5, R2, S8. For the four placeholders: t1=S3, t2=S5, t3=R2, t4=S8.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
A rough method is to query the local resolver for a site and compare response time or remaining TTL with a likely cold lookup. A cached answer can be returned without an external lookup, so a fast response or an aged TTL may suggest a previous resolution.
This is not proof of a visit a few seconds ago: cached records may come from another user, background requests, prefetching or older use; TTLs differ and network timing is noisy. A visit may also use an already cached browser/OS answer without contacting that DNS server. State “possible weak inference”, not a certain browsing history.
An IPv4 address identifies an IP interface. The network prefix selects the destination network; remaining bits identify an address within it. Classful addressing is historical; modern routing uses explicit CIDR prefixes.
| Class | First octet | Default mask | Network / host split | Purpose |
|---|---|---|---|---|
| A | 1–126 (ordinary unicast) | 255.0.0.0 (/8) | 8 / 24 bits | Large historical unicast networks; 0 and 127 reserved |
| B | 128–191 | 255.255.0.0 (/16) | 16 / 16 bits | Medium historical unicast networks |
| C | 192–223 | 255.255.255.0 (/24) | 24 / 8 bits | Small historical unicast networks |
| D | 224–239 | No unicast default mask | Not network/host class split | Multicast |
| E | 240–255 | No unicast default mask | Not normal unicast allocation | Reserved/experimental; special addresses exist |
Example: 114.34.2.8 is class A with network 114.0.0.0 and host portion 34.2.8. Class does not override an explicitly supplied CIDR prefix.
| Edge endpoint1 | Endpoint2 | Cost |
|---|---|---|
| z | x | 8 |
| z | y | 12 |
| x | y | 6 |
| x | v | 3 |
| x | w | 6 |
| y | v | 8 |
| y | t | 7 |
| v | t | 4 |
| v | u | 3 |
| v | w | 4 |
| w | u | 3 |
| t | u | 2 |
The original diagram is undirected. All edge labels are reproduced below. Source = u.
Swipe across the diagram to read its labels ↔![\"Original](\"fig/old-ai-cn-2023-11-22-ai-q4a.23319ec754.svg\")
| From | To | Cost |
|---|---|---|
| z | x | 8 |
| z | y | 12 |
| x | y | 6 |
| x | v | 3 |
| x | w | 6 |
| y | v | 8 |
| y | t | 7 |
| v | t | 4 |
| v | u | 3 |
| v | w | 4 |
| w | u | 3 |
| t | u | 2 |
Shortest distance, complete route and predecessor for every node from u.
Relaxation replaces a tentative route when going through u is cheaper. Dijkstra settles the unvisited node of least tentative distance. Nonnegative edge costs make the settled distance final. Equal-distance nodes may be settled in either order.
| Pass / settled node | Strict improvements | t | u | v | w | x | y | z |
|---|---|---|---|---|---|---|---|---|
| 0 | Initial | ∞ | 0 | ∞ | ∞ | ∞ | ∞ | ∞ |
| u | v←u: 0+3=3; w←u: 0+3=3; t←u: 0+2=2 | 2 | 0 | 3 | 3 | ∞ | ∞ | ∞ |
| t | y←t: 2+7=9 | 2 | 0 | 3 | 3 | ∞ | 9 | ∞ |
| v | x←v: 3+3=6 | 2 | 0 | 3 | 3 | 6 | 9 | ∞ |
| w | No change | 2 | 0 | 3 | 3 | 6 | 9 | ∞ |
| x | z←x: 6+8=14 | 2 | 0 | 3 | 3 | 6 | 9 | 14 |
| y | No change | 2 | 0 | 3 | 3 | 6 | 9 | 14 |
| z | No change | 2 | 0 | 3 | 3 | 6 | 9 | 14 |
| Destination | Distance | Previous | Route |
|---|---|---|---|
| t | 2 | u | u → t |
| u | 0 | — | u |
| v | 3 | u | u → v |
| w | 3 | u | u → w |
| x | 6 | v | u → v → x |
| y | 9 | t | u → t → y |
| z | 14 | x | u → v → x → z |
A tie between two equally short paths is valid; this table retains the first predecessor on equal distances.
IP delivers a packet toward a host interface. The transport layer delivers application data to a process endpoint, identified using a port. A socket endpoint is commonly described by IP address and port; the protocol also matters. TCP distinguishes connections by source IP/port and destination IP/port.
| Identifier | Purpose |
|---|---|
| MAC address | Delivery on a local link |
| IP address | Network-layer interface/addressing |
| Port number | Transport-layer application endpoint |
| Domain name | Application-level naming resolved through DNS |
Yaad rakho: IP host tak; port process tak.
| Direction | Source port | Destination port |
|---|---|---|
| Client→server | Example49152 | 23 (Telnet) |
| Server→client | 23 | 49152 |
Different simultaneous connections from the same client IP to the same server IP and port23 require different client ports, e.g.49152 and49153. Otherwise the TCP four-tuples would be identical.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
The recovered page gives arrivals but stops before the protocol and requested task. Page6 is needed for an exact answer. Guessing ALOHA or CSMA from another paper would silently change this question. The complete, separately identified older papers contain solved examples of each protocol.
One packet, L=1000 bits;3 links;rate2000000 bits/s;propagation1ms/link;store-and-forward.
Total time until the final packet bit reaches the destination.
![\"T=\\sum_{i=1}^{3}(L/R_i+d_{prop,i})\"](\"math/73c28ce2abb77cac.svg\")
Each router must receive the complete packet before sending it onward, so all three serialization times count.
1000/2000000 seconds=0.0005s=0.5ms.
0.5+1=1.5ms.
3×1.5=4.5ms. Queuing/processing are ignored as instructed.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
Arrival groups in the question; one-unit service; slot-boundary starts; FIFO inside each queue.
Departure order at t=1 through7.
At each slot start, consider only packets that have already arrived.
Strict priority chooses the high queue whenever it is nonempty. A running packet is not preempted.
| Service interval | Departure packet | Reason |
|---|---|---|
| [0,1) | 1 | Only1 is available. |
| [1,2) | 2 | High-priority2 is oldest. |
| [2,3) | 3 | High-priority3 precedes low-priority4. |
| [3,4) | 5 | High-priority5 has arrived. |
| [4,5) | 7 | High-priority7 precedes low-priority4 and6. |
| [5,6) | 4 | No high-priority packet remains; send oldest low4. |
| [6,7) | 6 | Send remaining low6. |
1 2 3 5 7 4 6
Transmission Control Protocol (TCP) provides reliable ordered byte-stream communication between two endpoints over an unreliable IP service. It repairs many network losses while the connection remains viable; a failed connection still must be handled by the application.
| Field | Meaning |
|---|---|
| Source/destination ports (16 each) | Identify process endpoints |
| Sequence number (32) | First data-byte number in this segment; SYN/FIN also occupy sequence space |
| Acknowledgement number (32) | Next sequence number expected when ACK flag is set |
| Data offset (4) | TCP header length in 32-bit words; minimum 5 → 20 bytes |
| Flags | SYN establishes; ACK validates acknowledgement; FIN closes a sending direction; RST aborts; others have defined purposes |
| Window (16 base field) | Advertised receive window; scaling can be negotiated |
| Checksum (16) | Covers TCP header/data and pseudo-header |
| Urgent pointer (16) | Used with URG for the urgent indication |
| Options/padding | Optional features such as MSS; align header to 32-bit words |
Yaad rakho: TCP bytes count karta hai; writes ke borders preserve nahi.
Mechanism: the receiver advertises free buffer space as rwnd. The sender limits outstanding unacknowledged bytes to the advertised window, subject also to congestion control. ACKs move the window; a zero window pauses ordinary sending, with persist probes to learn when space reopens. For example, with4000 free bytes and1000 already outstanding, at most3000 more bytes can be outstanding under flow control. Flow control protects the receiver; congestion control protects the network.
Known IP; new initial TCP connection; zero GET transmission; each object transmission=0.5 RTT.
Time to obtain the base object and then the pair of objects.
A referenced object cannot be requested until the HTML reveals it. HTTP/1.0 here uses a fresh connection per object; persistent HTTP/1.1 reuses one.
1 RTT for TCP setup + 1 RTT for GET/first response bit + 0.5 RTT to transmit = 2.5 RTT.
1 RTT for another TCP setup + 1 RTT for GET/response + 0.5 RTT transmission = 2.5 RTT.
Total = 2.5+2.5=5 RTT. DNS adds zero here because the IP address is already known.
Source note: The phrase “successfully received” does not specify a channel/error event. The answer distinguishes the 16 valid encoded words from the 32 possible received 5-bit patterns.
k=4 data bits; n=5 coded bits.
Dataword count, possible received words, valid codewords.
![\"N_{data}=2^k,\\quad](\"math/6ac9b03072926d25.svg\")
A one-to-one encoder maps every dataword to exactly one valid codeword. The remaining n-bit patterns are not valid codewords.
2⁴=16.
2⁵=32 possible received patterns; only16 are selected valid codewords.
All16 distinct datawords can be represented and recovered when received correctly. The question gives no channel outcomes, so an actual count of successful transmissions cannot be calculated. Invalid patterns indicate an error; an error that changes one valid word to another can go undetected.
A Cyclic Redundancy Check (CRC) treats a binary message as a polynomial over GF(2). Addition and subtraction are XOR: 0⊕0=0, 1⊕0=1, 0⊕1=1, 1⊕1=0. There are no carries or borrows. A degree-r generator contains r+1 bits, with leading and constant coefficients 1.
A degree-r generator with nonzero constant term detects every burst of length at most r. Generators containing factor (x+1) detect every odd number of bit errors. Longer-error guarantees depend on the polynomial and frame length; CRC is not a general correction method.
Yaad rakho: CRC mein subtract bhi XOR; receiver par full frame divide.
| Data + r zero bits | → | Modulo-2 divide by generator | → | Append r-bit remainder | → | Receiver divides full codeword | → | Zero remainder: no detected error |
| Edge endpoint1 | Endpoint2 | Cost |
|---|---|---|
| z | x | 8 |
| z | y | 12 |
| x | y | 6 |
| x | v | 3 |
| x | w | 6 |
| y | v | 8 |
| y | t | 7 |
| v | t | 4 |
| v | u | 3 |
| v | w | 4 |
| w | u | 3 |
| t | u | 2 |
The original diagram is undirected. All edge labels are reproduced below. Source = t.
Swipe across the diagram to read its labels ↔![\"Original](\"fig/old-ai-cn-2022-11-16-ai-q3a.503e410dd7.svg\")
| From | To | Cost |
|---|---|---|
| z | x | 8 |
| z | y | 12 |
| x | y | 6 |
| x | v | 3 |
| x | w | 6 |
| y | v | 8 |
| y | t | 7 |
| v | t | 4 |
| v | u | 3 |
| v | w | 4 |
| w | u | 3 |
| t | u | 2 |
Shortest distance, complete route and predecessor for every node from t.
Relaxation replaces a tentative route when going through u is cheaper. Dijkstra settles the unvisited node of least tentative distance. Nonnegative edge costs make the settled distance final. Equal-distance nodes may be settled in either order.
| Pass / settled node | Strict improvements | t | u | v | w | x | y | z |
|---|---|---|---|---|---|---|---|---|
| 0 | Initial | 0 | ∞ | ∞ | ∞ | ∞ | ∞ | ∞ |
| t | u←t: 0+2=2; y←t: 0+7=7; v←t: 0+4=4 | 0 | 2 | 4 | ∞ | ∞ | 7 | ∞ |
| u | w←u: 2+3=5 | 0 | 2 | 4 | 5 | ∞ | 7 | ∞ |
| v | x←v: 4+3=7 | 0 | 2 | 4 | 5 | 7 | 7 | ∞ |
| w | No change | 0 | 2 | 4 | 5 | 7 | 7 | ∞ |
| x | z←x: 7+8=15 | 0 | 2 | 4 | 5 | 7 | 7 | 15 |
| y | No change | 0 | 2 | 4 | 5 | 7 | 7 | 15 |
| z | No change | 0 | 2 | 4 | 5 | 7 | 7 | 15 |
| Destination | Distance | Previous | Route |
|---|---|---|---|
| t | 0 | — | t |
| u | 2 | t | t → u |
| v | 4 | t | t → v |
| w | 5 | u | t → u → w |
| x | 7 | v | t → v → x |
| y | 7 | t | t → y |
| z | 15 | x | t → v → x → z |
A tie between two equally short paths is valid; this table retains the first predecessor on equal distances.
No. An Internet checksum compresses a long message into16 bits, so different messages can have the same sum. Compensating changes can leave the one’s-complement sum unchanged. For example, word pair0001,0002 and0000,0003 (shown in hexadecimal) both sum to0003 and have the same complement checksum. A matching checksum means “no error detected”; it does not guarantee bit-for-bit identity. UDP does not automatically repair errors or retransmit lost data.
IP delivers a packet toward a host interface. The transport layer delivers application data to a process endpoint, identified using a port. A socket endpoint is commonly described by IP address and port; the protocol also matters. TCP distinguishes connections by source IP/port and destination IP/port.
| Identifier | Purpose |
|---|---|
| MAC address | Delivery on a local link |
| IP address | Network-layer interface/addressing |
| Port number | Transport-layer application endpoint |
| Domain name | Application-level naming resolved through DNS |
Yaad rakho: IP host tak; port process tak.
An established TCP connection is identified by source IP, source port, destination IP, destination port. IP addresses come from the IP header; port numbers come from the TCP header. The transport protocol distinguishes TCP from UDP.
A web server’s listening socket accepts connections on port80. Accepted connections from A and B use distinct connection sockets/state, although both server-side local ports remain80. Different remote IP/port pairs distinguish them. Persistent connections reuse their own established socket for subsequent requests.
Two messages; no DATA or ACK errors; original FSM transition labels in the question.
The global interleaved transition sequence, including the final ACK.
Send DATA → receive and acknowledge DATA → receive matching ACK → advance the sequence bit.
Each successful ACK permits the sender to send the next new message.
The sequence bit labels the outstanding DATA packet. ACK0 confirms data0 and ACK1 confirms data1 in this printed FSM; these are not TCP cumulative byte numbers. An ACK loss does not advance the sender. A duplicate data packet is acknowledged again but not delivered twice.
| Send0 | → | ACK0 | → | Send1 | → | ACK1 |
| Order | Transition | Action |
|---|---|---|
| 1 | S0 | Send data0; start timer |
| 2 | R0 | Accept data0; deliver; send ACK0 |
| 3 | S3 | Accept ACK0; stop timer |
| 4 | S5 | Send data1; start timer |
| 5 | R2 | Accept data1; deliver; send ACK1 |
| 6 | S8 | Accept ACK1; stop timer |
Required sequence: S0, R0, S3, S5, R2, S8. For the four placeholders: t1=S3, t2=S5, t3=R2, t4=S8.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
Guided media constrain signals to a physical path: twisted pair, coaxial cable and optical fiber. Unguided media carry electromagnetic waves through free space: radio, microwave and infrared. Wireless bandwidth and reach depend on frequency, power, antenna, obstacles and regulations; media names alone do not fix a speed.
| Medium | Two key characteristics | Limitations / use |
|---|---|---|
| Twisted pair | Electrical signal; pairs twisted to reduce interference | Low-cost LAN access; distance/noise limits depend on category and design |
| Coaxial cable | Shielded central conductor; greater shielding than basic unshielded pair | Bulkier; cable broadband and legacy shared LANs |
| Optical fiber | Light in glass/plastic; high capacity with immunity to electromagnetic interference | Transceivers and careful splicing/bend control needed; backbone links |
| Radio waves | Wireless propagation; many bands support broad/non-line-of-sight coverage | Interference, shared spectrum and security configuration matter; Wi-Fi/mobile |
| Terrestrial microwave | Directional antennas; typically line of sight | Alignment and obstructions matter; point-to-point backhaul |
| Satellite microwave | Wide coverage through relay in space | Cost and latency depend strongly on orbit/system; remote connectivity |
| Infrared | Short-range light; usually blocked by walls | Limited coverage; short-range indoor/remote-control links |
Yaad rakho: Guided = cable path; unguided = free-space waves.
\r\nTransmission media are classified into guided media (twisted pair, coaxial cable, optical fiber) and unguided media (radio, microwave, infrared).
Optical fiber carries data as light in a core surrounded by cladding. It is a guided medium. Single-mode fiber supports one main propagation mode and is used for long/high-capacity links; multimode supports multiple modes and is often used over shorter distances.
| Advantages | Disadvantages / design costs |
|---|---|
| High data capacity and low attenuation over suitable links | Optical transceivers, connectors and splicing add cost/skill requirements |
| Immune to electromagnetic interference; no electrical crosstalk | Tight bends, dirty connectors and poor installation increase loss |
| Electrical isolation between buildings | Fiber is mechanically sensitive and requires appropriate protection |
| Harder to tap casually than an unprotected radio channel | It is not automatically confidential; encryption/access control remain necessary |
Yaad rakho: Fiber light carry karta hai; immunity means no EMI, not no security risk.
Ethernet can use balanced twisted-pair copper (e.g.100BASE-TX/1000BASE-T), optical fiber (e.g.1000BASE-SX/LX), and historical coaxial cable (10BASE2/10BASE5). Twisted pair is economical for typical building runs; fiber supports longer reach and immunity to electromagnetic interference. The exact reach and rate depend on the Ethernet standard, not simply the word “fiber”.
Arrivals0.3,1.7,1.8,2.5,4.2,4.6; one-unit frames; no retries before time5.
Successful packet numbers and each relevant start/collision decision.
Pure ALOHA sends immediately. Slotted ALOHA waits for the next slot. CSMA/CD senses only signals that have had time to propagate.
Check intervals, not just whether arrival times differ. Any collision corrupts a frame; collision detection shortens the wasted transmission but does not make that frame successful.
| Packet | Transmission / slot | Reason | Outcome |
|---|---|---|---|
| 1 | 0.3–1.3 | Signal clears other nodes at1.5 | Success |
| 2 | 1.7–2.0 | At1.7 channel idle; detects3 at1.8+0.2=2.0 | Collision |
| 3 | 1.8–1.9 | Cannot sense2 until1.9; then stops | Collision |
| 4 | 2.5–3.5 | Last collision signal clears by2.2, so idle | Success |
| 5 | 4.2–5.2 | Previous signal cleared by3.7 | Success |
| 6 | No transmission | At4.6 senses5, whose signal arrives4.4 | Busy; waits |
1, 4, 5.
An IPv4 address identifies an IP interface. The network prefix selects the destination network; remaining bits identify an address within it. Classful addressing is historical; modern routing uses explicit CIDR prefixes.
| Class | First octet | Default mask | Network / host split | Purpose |
|---|---|---|---|---|
| A | 1–126 (ordinary unicast) | 255.0.0.0 (/8) | 8 / 24 bits | Large historical unicast networks; 0 and 127 reserved |
| B | 128–191 | 255.255.0.0 (/16) | 16 / 16 bits | Medium historical unicast networks |
| C | 192–223 | 255.255.255.0 (/24) | 24 / 8 bits | Small historical unicast networks |
| D | 224–239 | No unicast default mask | Not network/host class split | Multicast |
| E | 240–255 | No unicast default mask | Not normal unicast allocation | Reserved/experimental; special addresses exist |
Example: 114.34.2.8 is class A with network 114.0.0.0 and host portion 34.2.8. Class does not override an explicitly supplied CIDR prefix.
210.90.12.7/20
First and last addresses and total block size.
p is the prefix length. The remaining host bits vary through every combination. The network address is the first address; setting all host bits to one gives the last address.
32 − 20 = 12. N = 4096.
Mask = 255.255.240.0. The first varying octet is octet 3; its block width is 16. The supplied value 12 lies in the interval 0–15. Keep earlier octets fixed. Set any later octets to 0 for the first address and 255 for the last address.
First = 210.90.0.0; last = 210.90.15.255; total = 4096 addresses. Usual usable host range: 210.90.0.1–210.90.15.254.
Definition. Switching selects how data passes through intermediate network nodes from a source to a destination.
Simple meaning. Raasta reserve karna, poora message ruk-ruk ke bhejna, ya chhote packets bhejna—yeh teen ideas hain.
| Aspect | Circuit switching | Message switching | Packet switching |
|---|---|---|---|
| Unit handled | Continuous stream on established circuit | Whole message | Small packets |
| Resource use | Capacity reserved along a circuit | Shared; message stored at each intermediate node | Shared; each packet stored/forwarded in this model |
| Setup | Required before data transfer | No circuit setup required | Datagram: no connection setup; virtual circuit: setup |
| Delay | Setup delay, then relatively predictable service | Large store-and-forward delay for long messages | Variable delay; packet pipeline can reduce completion time |
| Main limitation | Reserved capacity can be idle | Large buffers; unsuitable for tight delay needs | Headers, possible queueing/loss; datagrams may reorder |
For ordinary bursty office data, packet switching is generally appropriate because resources are shared among active transfers. The design must still account for queueing, loss and the required service quality.
Yaad rakho: Circuit reserves; message waits whole; packets pipeline.
Choice: circuit switching reserves a path and capacity before communication, giving predictable service and avoiding packet-queue contention after setup. A dedicated voice circuit is a standard example. It sacrifices statistical sharing and may waste reserved capacity when silent.
| Concept | Explanation | Example |
|---|---|---|
| GET | Retrieve the selected representation | Read a page |
| POST | Ask the resource to process submitted content | Submit a form |
| PUT / DELETE | Replace target representation / request removal according to server semantics | Update/delete an API resource |
| 200 / 301 / 404 / 500 | Success / permanent redirect / not found / server error | Read status before interpreting body |
| Persistent connection | Reuse a connection for multiple exchanges | Avoid repeated TCP setup for each image |
| Cookie/session token | Carry an identifier so the application can relate requests | Remain signed in |
| HTTPS | HTTP protected by authenticated encryption through TLS in common deployments | Protect credentials and content in transit |
A secure transport protects the connection; it does not prove that every server’s content is truthful. HTTP method semantics and application authorization still matter. For exam comparisons, separate HTTP semantics, connection lifetime and application session state.
| First visit | → | Server creates customer/session ID | → | Set-Cookie header | → | Browser stores cookie | → | Later Cookie header | → | Server retrieves purchase record |
Keep the purchase record in the server database keyed by a suitable identifier; the browser need only carry that identifier. Example: cookie customer=R81 links later visits to R81’s stored orders. HTTP requests remain independent at the protocol level; the application supplies continuity. Account login can restore the record when a cookie is absent.
Selective Repeat (SR) ARQ accepts and buffers valid out-of-order frames inside the receiver window, acknowledges them individually and retransmits only missing/timed-out frames. In this answer ACK i means “frame i received”, unlike the next-expected cumulative ACK convention used for GBN.
Yaad rakho: SR selective resend; half sequence space keeps old and new separate.
| Destination | A output interface |
|---|---|
| H3 (its destination address/prefix) | 3 |
For part(ii), an ordinary destination-only forwarding lookup sees the same destination H3 for both packets, so it cannot make the requested distinction by sender. It would need policy/source-aware forwarding or a more general match-action rule. A simple table sending H3 to3 sends both H1 andH2 that way. The alternate path through4 exists physically but is not selected by source under the stated rule.
10 flows, server links 25 Mbps, client links 50 Mbps, shared link 200 Mbps.
Maximum throughput of each flow.
A flow cannot exceed any link on its path; fair sharing divides the common capacity among n active flows.
200 Mbps ÷ 10 = 20 Mbps per flow.
T=min(25,50,20)=20 Mbps.
Total delivered rate=10×20=200 Mbps.
| Ideal characteristic | Slotted ALOHA |
|---|---|
| One active node uses rate R | Yes, at slot boundaries with no collisions; an initial wait may occur |
| M active nodes each achieve R/M | Not in general: collisions and empty slots waste capacity |
| Decentralized without master or clock synchronization | No master is needed, but all nodes must synchronize slots |
| Simple | Yes: transmit at a slot boundary, then retry after random backoff |
The standard independent-attempt throughput maximum approaches1/e≈0.368 of the channel as the contender count becomes large, so it does not deliver ideal full utilization under contention.
Arrival groups in the question; one-unit service; slot-boundary starts; FIFO inside each queue.
Departure order at t=1 through7.
At each slot start, consider only packets that have already arrived.
Round robin alternates nonempty queues, skipping an empty queue.
| Service interval | Departure packet | Reason |
|---|---|---|
| [0,1) | 1 | Only packet1 has arrived. |
| [1,2) | 2 | Use set2: oldest is2. |
| [2,3) | 4 | Use set1: oldest is4. |
| [3,4) | 3 | Use set2: oldest is3. |
| [4,5) | 6 | Use set1:6 has arrived. |
| [5,6) | 5 | Use set2:oldest is5. |
| [6,7) | 7 | Set1 empty; set2 supplies7. |
1 2 4 3 6 5 7
A TCP sender is limited by receiver capacity and network capacity. The receive window rwnd comes from the receiver; the congestion window cwnd is maintained by the sender. Roughly, allowed outstanding data is min(rwnd,cwnd), with the existing bytes in flight subtracted before sending new bytes.
Known IP; new initial TCP connection; zero GET transmission; each object transmission=0.5 RTT.
Time to obtain the base object and then the pair of objects.
A referenced object cannot be requested until the HTML reveals it. HTTP/1.0 here uses a fresh connection per object; persistent HTTP/1.1 reuses one.
1 RTT for TCP setup + 1 RTT for GET/first response bit + 0.5 RTT to transmit = 2.5 RTT.
Existing TCP connection: 0 setup + 1 RTT for GET/response + 0.5 RTT transmission = 1.5 RTT.
Total = 2.5+1.5=4 RTT. DNS adds zero here because the IP address is already known.
Sent data01, encoded011; valid codewords000,011,101,110.
Receiver outcome for111 and000.
A receiver first checks membership in the valid-codeword set, then decodes a valid word.
This even-parity code detects any single-bit error but cannot identify a unique bit to correct.
111 has odd parity and is not a valid word. Relative to011, one bit differs. The receiver detects an error and rejects/requests retransmission; it cannot uniquely infer the original data.
000 is a valid word and decodes to00. Relative to the actual sent011, two bits changed. Therefore this is an undetected error: the receiver accepts the wrong dataword00.
Data = 1001; generator bits = 1011; degree r = 3.
CRC remainder, complete transmitted frame and receiver check.
Append r zeros, divide using XOR (modulo 2), and replace those zeros with the r-bit remainder. XOR has no borrowing or carrying. A valid received codeword has remainder zero; the receiver does not append new zeros.
1001 → 1001000
| Leading position | Before XOR | Aligned generator | After XOR |
|---|---|---|---|
| 1 | 1001000 | 1011000 | 0010000 |
| 3 | 0010000 | 0010110 | 0000110 |
Remainder = 110. Transmit 1001110.
| Position | Before | Generator | After |
|---|---|---|---|
| 1 | 1001110 | 1011000 | 0010110 |
| 3 | 0010110 | 0010110 | 0000000 |
Receiver remainder = 000. No error is detected; this is not a guarantee that every possible corruption is detectable.
| Edge endpoint1 | Endpoint2 | Cost |
|---|---|---|
| u | v | 4 |
| u | w | 3 |
| u | x | 5 |
| v | w | 1 |
| v | x | 8 |
| x | w | 7 |
| x | y | 7 |
| w | y | 6 |
| w | z | 7 |
| y | z | 7 |
The original diagram is undirected. All edge labels are reproduced below. Source = u.
Swipe across the diagram to read its labels ↔![\"Original](\"fig/old-ai-cn-2023-01-27-ai-q3a.e2a89263bf.svg\")
| From | To | Cost |
|---|---|---|
| u | v | 4 |
| u | w | 3 |
| u | x | 5 |
| v | w | 1 |
| v | x | 8 |
| x | w | 7 |
| x | y | 7 |
| w | y | 6 |
| w | z | 7 |
| y | z | 7 |
Shortest distance, complete route and predecessor for every node from u.
Relaxation replaces a tentative route when going through u is cheaper. Dijkstra settles the unvisited node of least tentative distance. Nonnegative edge costs make the settled distance final. Equal-distance nodes may be settled in either order.
| Pass / settled node | Strict improvements | u | v | w | x | y | z |
|---|---|---|---|---|---|---|---|
| 0 | Initial | 0 | ∞ | ∞ | ∞ | ∞ | ∞ |
| u | v←u: 0+4=4; w←u: 0+3=3; x←u: 0+5=5 | 0 | 4 | 3 | 5 | ∞ | ∞ |
| w | y←w: 3+6=9; z←w: 3+7=10 | 0 | 4 | 3 | 5 | 9 | 10 |
| v | No change | 0 | 4 | 3 | 5 | 9 | 10 |
| x | No change | 0 | 4 | 3 | 5 | 9 | 10 |
| y | No change | 0 | 4 | 3 | 5 | 9 | 10 |
| z | No change | 0 | 4 | 3 | 5 | 9 | 10 |
| Destination | Distance | Previous | Route |
|---|---|---|---|
| u | 0 | — | u |
| v | 4 | u | u → v |
| w | 3 | u | u → w |
| x | 5 | u | u → x |
| y | 9 | w | u → w → y |
| z | 10 | w | u → w → z |
A tie between two equally short paths is valid; this table retains the first predecessor on equal distances.
Yes. A known RTT tells the sender when a response should arrive, but a lost DATA or ACK produces no arrival event. A timer is required to detect that absence and trigger retransmission. It can be set using the known RTT plus appropriate transmission/processing allowance. Example: DATA0 is lost; without a timer the sender could wait forever forACK0. A timeout repeats DATA0, while the sequence bit prevents duplicate delivery if only the ACK was lost.
1011010001000110, 1100100001101110; word width = 16 bits.
Sender checksum and receiver verification.
![\"S=\\operatorname{endAroundCarry}(\\text{sum](\"math/c15ebbb2cede607a.svg\")
One’s-complement addition returns an overflow carry to the least-significant end. Complement only after all words have been added.
| Word | Decimal | Raw addition | Folded addition | Running binary sum |
|---|---|---|---|---|
| 1011010001000110 | 46150 | 46150 | 46150 + carry 0 = 46150 | 1011010001000110 |
| 1100100001101110 | 51310 | 97460 | 31924 + carry 1 = 31925 | 0111110010110101 |
Running sum = 0111110010110101. Change every 0 to 1 and every 1 to 0: checksum = 1000001101001010 (decimal 33610). In the requested two-byte format: 10000011 01001010.
31925 + 33610 = 65535 = 1111111111111111. All bits are 1; complement gives all zeros. Therefore this intact message passes the checksum check. A matching checksum detects no error; it does not prove that every possible corruption is absent.
The complete timeline in the question, starting with data0 at t=0.
Six interleaved DATA/ACK numbers.
Advance DATA sequence only after a correct matching ACK. On timeout resend the same sequence; on duplicate or corrupt data repeat the last successful ACK.
This keeps lost ACKs from causing duplicate application delivery.
The sequence bit labels the outstanding DATA packet. ACK0 confirms data0 and ACK1 confirms data1 in this printed FSM; these are not TCP cumulative byte numbers. An ACK loss does not advance the sender. A duplicate data packet is acknowledged again but not delivered twice.
| Event | Number | Reason |
|---|---|---|
| t2 data | 1 | ACK0 has arrived; next DATA is1 |
| t3 ACK | 1 | Receiver accepted data1 |
| after t4 timeout data | 1 | ACK1 lost; sender resends1 |
| t5 ACK | 1 | Duplicate data1: re-ACK1, do not redeliver |
| t6 data | 0 | ACK1 now arrives; wrap to0 |
| t7 ACK | 1 | Data0 corrupt: receiver still expects0 and repeatsACK1 |
Answer: 1, 1, 1, 1, 0, 1. At t8 ACK1 is wrong for outstanding data0, so S1 ignores it; the next timeout retransmits0.
RTT4; frame time1; O1/O2/O3/O4 have10/1/2/1 frames; established connection.
Every object completion time and average delay.
![\"T_j=RTT+\\text{position](\"math/20bdb51428a09b4e.svg\")
Requests need2 units to reach the server and final frame bits need2 to return. Thus add4 to each last-frame transmission count; do not add a new RTT per frame.
| Method | Frame order | O1 completion | O2 completion | O3 completion | O4 completion | Average |
|---|---|---|---|---|---|---|
| HTTP/2 | O1, O2, O3, O4, O1, O3, O1, O1, O1, O1, O1, O1, O1, O1 | 18 | 6 | 10 | 8 | 10.5 |
| HTTP/1.1 | O1, O1, O1, O1, O1, O1, O1, O1, O1, O1, O2, O3, O3, O4 | 14 | 15 | 17 | 18 | 16.0 |
HTTP/2: (18+6+10+8)/4=42/4=10.5. HTTP/1.1: (14+15+17+18)/4=64/4=16. All objects finish by18 in both cases; multiplexing makes short objects finish earlier.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
Arrivals0.3,1.7,1.8,2.5,4.2,4.6; one-unit frames; no retries before time5.
Successful packet numbers and each relevant start/collision decision.
Pure ALOHA sends immediately. Slotted ALOHA waits for the next slot. CSMA/CD senses only signals that have had time to propagate.
Check intervals, not just whether arrival times differ. Any collision corrupts a frame; collision detection shortens the wasted transmission but does not make that frame successful.
| Packet | Transmission / slot | Reason | Outcome |
|---|---|---|---|
| 1 | 1–2 | Alone in slot1 | Success |
| 2 | 2–3 | Same slot as3 | Collision |
| 3 | 2–3 | Same slot as2 | Collision |
| 4 | 3–4 | Alone in slot3 | Success |
| 5 | 5–6 | Same slot as6 | Collision |
| 6 | 5–6 | Same slot as5 | Collision |
1, 4. The slot beginning at5 is the initial attempt of packets5 and6, not a retransmission.
190.100.0.0/16; Group 1: 64 customers × 256 addresses; Group 2: 128 customers × 128 addresses; Group 3: 128 customers × 64 addresses
Source correction/assumption: All requests are total addresses, not usable hosts; therefore no extra two addresses are added.
Each customer prefix, every group range, first/last customer blocks, and unallocated addresses.
These are requested total addresses, not usable-host counts. Allocate customer blocks on their prefix boundary. A group with a non-power-of-two customer count need not be forced into one larger, wasteful CIDR aggregate.
| Group | Customers | Addresses each | Customer prefix | Used | Group range |
|---|---|---|---|---|---|
| 1 | 64 | 256 | /24 | 16384 | 190.100.0.0–190.100.63.255 |
| 2 | 128 | 128 | /25 | 16384 | 190.100.64.0–190.100.127.255 |
| 3 | 128 | 64 | /26 | 8192 | 190.100.128.0–190.100.159.255 |
| Group | First customer | Last customer |
|---|---|---|
| 1 | 190.100.0.0–190.100.0.255 | 190.100.63.0–190.100.63.255 |
| 2 | 190.100.64.0–190.100.64.127 | 190.100.127.128–190.100.127.255 |
| 3 | 190.100.128.0–190.100.128.63 | 190.100.159.192–190.100.159.255 |
Allocated = 16384 + 16384 + 8192 = 40960. Unused = 65536 − 40960 = 24576.
Unused range: 190.100.160.0–190.100.255.255.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
| Concept | Explanation | Example |
|---|---|---|
| GET | Retrieve the selected representation | Read a page |
| POST | Ask the resource to process submitted content | Submit a form |
| PUT / DELETE | Replace target representation / request removal according to server semantics | Update/delete an API resource |
| 200 / 301 / 404 / 500 | Success / permanent redirect / not found / server error | Read status before interpreting body |
| Persistent connection | Reuse a connection for multiple exchanges | Avoid repeated TCP setup for each image |
| Cookie/session token | Carry an identifier so the application can relate requests | Remain signed in |
| HTTPS | HTTP protected by authenticated encryption through TLS in common deployments | Protect credentials and content in transit |
A secure transport protects the connection; it does not prove that every server’s content is truthful. HTTP method semantics and application authorization still matter. For exam comparisons, separate HTTP semantics, connection lifetime and application session state.
| First visit | → | Server creates customer/session ID | → | Set-Cookie header | → | Browser stores cookie | → | Later Cookie header | → | Server retrieves purchase record |
Keep the purchase record in the server database keyed by a suitable identifier; the browser need only carry that identifier. Example: cookie customer=R81 links later visits to R81’s stored orders. HTTP requests remain independent at the protocol level; the application supplies continuity. Account login can restore the record when a cookie is absent.
IP delivers a packet toward a host interface. The transport layer delivers application data to a process endpoint, identified using a port. A socket endpoint is commonly described by IP address and port; the protocol also matters. TCP distinguishes connections by source IP/port and destination IP/port.
| Identifier | Purpose |
|---|---|
| MAC address | Delivery on a local link |
| IP address | Network-layer interface/addressing |
| Port number | Transport-layer application endpoint |
| Domain name | Application-level naming resolved through DNS |
Yaad rakho: IP host tak; port process tak.
An established TCP connection is identified by source IP, source port, destination IP, destination port. IP addresses come from the IP header; port numbers come from the TCP header. The transport protocol distinguishes TCP from UDP.
A web server’s listening socket accepts connections on port80. Accepted connections from A and B use distinct connection sockets/state, although both server-side local ports remain80. Different remote IP/port pairs distinguish them. Persistent connections reuse their own established socket for subsequent requests.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
Definition. Subnetting divides an address block into smaller prefix-defined blocks. In an ordinary subnet, the network and directed-broadcast addresses bound its usable host range.
Simple meaning. Pehle block nikalo; phir first aur last. Uske baad hosts.
A /26 has 32−26=6 host bits, so it contains 2⁶=64 total addresses. Under ordinary IPv4 subnet rules, one is the network address and one is directed broadcast, leaving 64−2=62 usable host addresses.
Yaad rakho: Total mein sab count; usable mein boundary ke 2 minus.
An IPv4 address identifies an IP interface. The network prefix selects the destination network; remaining bits identify an address within it. Classful addressing is historical; modern routing uses explicit CIDR prefixes.
| Class | First octet | Default mask | Network / host split | Purpose |
|---|---|---|---|---|
| A | 1–126 (ordinary unicast) | 255.0.0.0 (/8) | 8 / 24 bits | Large historical unicast networks; 0 and 127 reserved |
| B | 128–191 | 255.255.0.0 (/16) | 16 / 16 bits | Medium historical unicast networks |
| C | 192–223 | 255.255.255.0 (/24) | 24 / 8 bits | Small historical unicast networks |
| D | 224–239 | No unicast default mask | Not network/host class split | Multicast |
| E | 240–255 | No unicast default mask | Not normal unicast allocation | Reserved/experimental; special addresses exist |
Example: 114.34.2.8 is class A with network 114.0.0.0 and host portion 34.2.8. Class does not override an explicitly supplied CIDR prefix.
Example: 192.168.1.70 AND255.255.255.192 gives network192.168.1.64. The mask distinguishes the network prefix from the host portion. CIDR masks need not equal historical default class masks.
4 flows, server links 400 Mbps, client links 300 Mbps, shared link 1000 Mbps.
Minimum extra shared link Re.
A flow cannot exceed any link on its path; fair sharing divides the common capacity among n active flows.
1000 Mbps ÷ 4 = 250 Mbps per flow.
T=min(400,300,250)=250 Mbps.
Each of Hosts1 and2 requires 250 Mbps. Re must carry their sum: Re≥500 Mbps. Rc=300 Mbps is already at least 250 Mbps.
Known IP; new initial TCP connection; zero GET transmission; each object transmission=0.5 RTT.
Time to obtain both the base HTML and its referenced JPEG.
A referenced object cannot be requested until the HTML reveals it. HTTP/1.0 here uses a fresh connection per object; persistent HTTP/1.1 reuses one.
1 RTT for TCP setup + 1 RTT for GET/first response bit + 0.5 RTT to transmit = 2.5 RTT.
Existing TCP connection: 0 setup + 1 RTT for GET/response + 0.5 RTT transmission = 1.5 RTT.
Total = 2.5+1.5=4 RTT. DNS adds zero here because the IP address is already known.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
| Edge endpoint1 | Endpoint2 | Cost |
|---|---|---|
| z | x | 8 |
| z | y | 12 |
| x | y | 6 |
| x | v | 3 |
| x | w | 6 |
| y | v | 8 |
| y | t | 7 |
| v | t | 4 |
| v | u | 3 |
| v | w | 4 |
| w | u | 3 |
| t | u | 2 |
The original diagram is undirected. All edge labels are reproduced below. Source = w.
Swipe across the diagram to read its labels ↔![\"Original](\"fig/old-ai-cn-2024-02-06-ai-q2a.60ea5488ef.svg\")
| From | To | Cost |
|---|---|---|
| z | x | 8 |
| z | y | 12 |
| x | y | 6 |
| x | v | 3 |
| x | w | 6 |
| y | v | 8 |
| y | t | 7 |
| v | t | 4 |
| v | u | 3 |
| v | w | 4 |
| w | u | 3 |
| t | u | 2 |
Shortest distance, complete route and predecessor for every node from w.
Relaxation replaces a tentative route when going through u is cheaper. Dijkstra settles the unvisited node of least tentative distance. Nonnegative edge costs make the settled distance final. Equal-distance nodes may be settled in either order.
| Pass / settled node | Strict improvements | t | u | v | w | x | y | z |
|---|---|---|---|---|---|---|---|---|
| 0 | Initial | ∞ | ∞ | ∞ | 0 | ∞ | ∞ | ∞ |
| w | u←w: 0+3=3; x←w: 0+6=6; v←w: 0+4=4 | ∞ | 3 | 4 | 0 | 6 | ∞ | ∞ |
| u | t←u: 3+2=5 | 5 | 3 | 4 | 0 | 6 | ∞ | ∞ |
| v | y←v: 4+8=12 | 5 | 3 | 4 | 0 | 6 | 12 | ∞ |
| t | No change | 5 | 3 | 4 | 0 | 6 | 12 | ∞ |
| x | z←x: 6+8=14 | 5 | 3 | 4 | 0 | 6 | 12 | 14 |
| y | No change | 5 | 3 | 4 | 0 | 6 | 12 | 14 |
| z | No change | 5 | 3 | 4 | 0 | 6 | 12 | 14 |
| Destination | Distance | Previous | Route |
|---|---|---|---|
| t | 5 | u | w → u → t |
| u | 3 | w | w → u |
| v | 4 | w | w → v |
| w | 0 | — | w |
| x | 6 | w | w → x |
| y | 12 | v | w → v → y |
| z | 14 | x | w → x → z |
A tie between two equally short paths is valid; this table retains the first predecessor on equal distances.
IP delivers a packet toward a host interface. The transport layer delivers application data to a process endpoint, identified using a port. A socket endpoint is commonly described by IP address and port; the protocol also matters. TCP distinguishes connections by source IP/port and destination IP/port.
| Identifier | Purpose |
|---|---|
| MAC address | Delivery on a local link |
| IP address | Network-layer interface/addressing |
| Port number | Transport-layer application endpoint |
| Domain name | Application-level naming resolved through DNS |
Yaad rakho: IP host tak; port process tak.
| Direction | Source port | Destination port |
|---|---|---|
| Client→server | Example49152 | 23 (Telnet) |
| Server→client | 23 | 49152 |
Different client hosts may both use49152: their different source IP addresses distinguish the four-tuples. The server keeps separate connection state.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
The sequence bit labels the outstanding DATA packet. ACK0 confirms data0 and ACK1 confirms data1 in this printed FSM; these are not TCP cumulative byte numbers. An ACK loss does not advance the sender. A duplicate data packet is acknowledged again but not delivered twice.
| Case | Possible ordered transitions | Explanation |
|---|---|---|
| No error | S0,R0,S3 | DATA0 and ACK0 both arrive |
| Corrupt data0 | S0,R3,S1,S2,R0,S3 | Receiver repeatsACK1; sender ignores it, times out, resends0 |
| Lost data0 | S0,S2,R0,S3 | No receiver event for the lost packet |
| Corrupt ACK0 | S0,R0,S1,S2,R1,S3 | Receiver has delivered0; duplicate0 getsACK0 without redelivery |
| Lost ACK0 | S0,R0,S2,R1,S3 | No sender event for a lost ACK; timeout triggers retry |
After recovery, S5 sends data1 and the symmetric transitions S7/R2/R3/S8 apply as appropriate. These are representative possible traces; repeated losses/corruptions can repeat the retry loops arbitrarily often. No single finite list can enumerate every possible error history.
Take repeated authorized snapshots of cache entries or resolution counters and group names by web service. A record frequently present or refreshed after expiry suggests repeated resolutions; query logs/counters, if available, provide a better frequency estimate than a single cache snapshot.
Do not equate DNS queries with page views. TTL duration, client/browser caches, shared CDN names, prefetching and multiple domains per page bias the result. A long-TTL unpopular name may remain cached longer than a popular short-TTL name. The result is only a rough popularity indicator, not a count of users or visits.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
At z, neighbor costs c(z,v)=6 and c(z,x)=2; all original links are stated in the question.
Distance table at z over synchronous exchanges.
![\"D_z(d)=\\min\\{6+D_v(d),\\](\"math/b9f38cdb21f5d397.svg\")
Add the local link cost to each neighbor’s advertised distance. Round0 below uses the neighbors’ initialized direct-link vectors; next rounds use their previous completed exchange.
| Exchange input round | Destination | Via v | Via x | Minimum |
|---|---|---|---|---|
| 0 | u | 7 | ∞ | 7 |
| 0 | v | 6 | 5 | 5 |
| 0 | x | 9 | 2 | 2 |
| 0 | y | ∞ | 5 | 5 |
| 1 | u | 7 | 6 | 6 |
| 1 | v | 6 | 5 | 5 |
| 1 | x | 9 | 2 | 2 |
| 1 | y | 9 | 5 | 5 |
| 2 | u | 7 | 6 | 6 |
| 2 | v | 6 | 5 | 5 |
| 2 | x | 9 | 2 | 2 |
| 2 | y | 9 | 5 | 5 |
| 3 | u | 7 | 6 | 6 |
| 3 | v | 6 | 5 | 5 |
| 3 | x | 9 | 2 | 2 |
| 3 | y | 9 | 5 | 5 |
From z: u=6 viax→v→u; v=5 viax→v; x=2 direct; y=5 viax→y; z=0. The route to u throughx→y→u costs7 and is therefore not the minimum.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
Received word 1011001; even parity. Number positions 7,6,5,4,3,2,1 from left to right, so position 1 is the rightmost bit.
Syndrome, erroneous position, and corrected word.
Each parity check covers positions whose binary position number contains that parity bit. For even parity, XOR must be zero; a 1 identifies a failed group.
| Position | 7 | 6 | 5 | 4 | 3 | 2 | 1 |
|---|---|---|---|---|---|---|---|
| Received bit | 1 | 0 | 1 | 1 | 0 | 0 | 1 |
| Parity | Covered positions | Bits added | Sum | Sum mod 2 |
|---|---|---|---|---|
| 1 | 1, 3, 5, 7 | 1 + 0 + 1 + 1 | 3 | 1 |
| 2 | 2, 3, 6, 7 | 0 + 0 + 0 + 1 | 1 | 1 |
| 4 | 4, 5, 6, 7 | 1 + 1 + 0 + 1 | 3 | 1 |
Syndrome s4s2s1 = 111; decimal = 7. Under the single-bit-error assumption, flip position 7 from the right. Corrected word = 0011001.
Ordinary Hamming(7,4) corrects one bit. A multi-bit corruption may produce a misleading nonzero syndrome, so do not claim guaranteed correction of arbitrary errors.
Data = 11010110101; generator bits = 10101; degree r = 4.
CRC remainder, complete transmitted frame and receiver check; then detect the specified bit inversion.
Append r zeros, divide using XOR (modulo 2), and replace those zeros with the r-bit remainder. XOR has no borrowing or carrying. A valid received codeword has remainder zero; the receiver does not append new zeros.
11010110101 → 110101101010000
| Leading position | Before XOR | Aligned generator | After XOR |
|---|---|---|---|
| 1 | 110101101010000 | 101010000000000 | 011111101010000 |
| 2 | 011111101010000 | 010101000000000 | 001010101010000 |
| 3 | 001010101010000 | 001010100000000 | 000000001010000 |
| 9 | 000000001010000 | 000000001010100 | 000000000000100 |
Remainder = 0100. Transmit 110101101010100.
| Position | Before | Generator | After |
|---|---|---|---|
| 1 | 110101101010100 | 101010000000000 | 011111101010100 |
| 2 | 011111101010100 | 010101000000000 | 001010101010100 |
| 3 | 001010101010100 | 001010100000000 | 000000001010100 |
| 9 | 000000001010100 | 000000001010100 | 000000000000000 |
Receiver remainder = 0000. No error is detected; this is not a guarantee that every possible corruption is detectable.
110101101010100 becomes 110101111010100.
| Position | Before | Generator | After |
|---|---|---|---|
| 1 | 110101111010100 | 101010000000000 | 011111111010100 |
| 2 | 011111111010100 | 010101000000000 | 001010111010100 |
| 3 | 001010111010100 | 001010100000000 | 000000011010100 |
| 8 | 000000011010100 | 000000010101000 | 000000001111100 |
| 9 | 000000001111100 | 000000001010100 | 000000000101000 |
| 10 | 000000000101000 | 000000000101010 | 000000000000010 |
Remainder = 0010, which is nonzero: reject the frame and request recovery according to the error-control protocol.
10101001, 00111001; word width = 8 bits.
Sender checksum and receiver verification.
One’s-complement addition returns an overflow carry to the least-significant end. Complement only after all words have been added.
| Word | Decimal | Raw addition | Folded addition | Running binary sum |
|---|---|---|---|---|
| 10101001 | 169 | 169 | 169 + carry 0 = 169 | 10101001 |
| 00111001 | 57 | 226 | 226 + carry 0 = 226 | 11100010 |
Running sum = 11100010. Change every 0 to 1 and every 1 to 0: checksum = 00011101 (decimal 29).
226 + 29 = 255 = 11111111. All bits are 1; complement gives all zeros. Therefore this intact message passes the checksum check. A matching checksum detects no error; it does not prove that every possible corruption is absent.
Arrivals0.3,1.7,1.8,2.5,4.2,4.6; one-unit frames; no retries before time5.
Successful packet numbers and each relevant start/collision decision.
Pure ALOHA sends immediately. Slotted ALOHA waits for the next slot. CSMA/CD senses only signals that have had time to propagate.
Check intervals, not just whether arrival times differ. Any collision corrupts a frame; collision detection shortens the wasted transmission but does not make that frame successful.
| Packet | Transmission / slot | Reason | Outcome |
|---|---|---|---|
| 1 | 0.3–1.3 | No overlap | Success |
| 2 | 1.7–2.7 | Overlaps3 and4 | Collision |
| 3 | 1.8–2.8 | Overlaps2 and4 | Collision |
| 4 | 2.5–3.5 | Overlaps2 and3 | Collision |
| 5 | 4.2–5.2 | Overlaps6 | Collision |
| 6 | 4.6–5.6 | Overlaps5 | Collision |
1.
x–w2,x–y5,w–y2; Dw(u)=5,Dy(u)=6.
Dx(w),Dx(y),Dx(u) and a link change leaving Dx(u) unchanged.
![\"D_x(u)=\\min(2+5,5+6)\"](\"math/a570618da07d1f4e.svg\")
Compare complete candidate route costs, not the last advertised value alone.
Dx(w)=2 direct. Dx(y)=min(5,2+2)=4 viaw.
Dx(u)=min(7,11)=7 viaw. Vector for(w,y,u) is (2,4,7).
Increase c(x,y) from5 to6. Route viaw remains2+5=7; route viay becomes6+6=12. The minimum cost to u is unchanged, so x has no changed u-distance to advertise. Other destination entries must be considered separately.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
An IPv4 address identifies an IP interface. The network prefix selects the destination network; remaining bits identify an address within it. Classful addressing is historical; modern routing uses explicit CIDR prefixes.
| Class | First octet | Default mask | Network / host split | Purpose |
|---|---|---|---|---|
| A | 1–126 (ordinary unicast) | 255.0.0.0 (/8) | 8 / 24 bits | Large historical unicast networks; 0 and 127 reserved |
| B | 128–191 | 255.255.0.0 (/16) | 16 / 16 bits | Medium historical unicast networks |
| C | 192–223 | 255.255.255.0 (/24) | 24 / 8 bits | Small historical unicast networks |
| D | 224–239 | No unicast default mask | Not network/host class split | Multicast |
| E | 240–255 | No unicast default mask | Not normal unicast allocation | Reserved/experimental; special addresses exist |
Example: 114.34.2.8 is class A with network 114.0.0.0 and host portion 34.2.8. Class does not override an explicitly supplied CIDR prefix.
11000011₂=128+64+2+1=195: ClassC. 129 lies in128–191: ClassB.
Definition. IPv6 extends address size and reorganizes IP forwarding information into a fixed base header with extension headers.
Simple meaning. Sirf address bada nahi; header ka arrangement bhi badla.
| Feature | IPv4 | IPv6 |
|---|---|---|
| Address width | 32 bits | 128 bits |
| Base/header size | 20–60 bytes depending on options | 40-byte fixed base header |
| Length field | Total length includes IPv4 header | Payload Length excludes 40-byte base header |
| Lifetime field | TTL | Hop Limit |
| Following protocol | Protocol field | Next Header; may identify an extension |
| Header checksum | Present | No base-header checksum |
| Options | Inside IPv4 header | Typically extension headers |
| Fragmentation fields | In IPv4 header; routers may fragment if permitted | Fragment extension header; fragmentation by source, not routers |
Yaad rakho: v4 total length; v6 payload length.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Definition. Store-and-forward requires a node to receive the complete forwarding unit before sending it onward; pipelining allows successive units to occupy different links simultaneously.
Simple meaning. Pehla packet next link pe jaa raha hai, tab doosra packet first link pe aa sakta hai.
Packet switching can pipeline smaller units across successive links. After the first packet is forwarded, the next packet can use an earlier link while the first uses a later link. Whole-message store-and-forward must receive the entire message at each node, preventing that packet-level pipeline.
Yaad rakho: First packet h times; remaining packets one time each.
| Delay | Formula / cause | Constant or variable on a fixed route |
|---|---|---|
| Processing | Header/error/lookup work | Often modeled constant, can vary with implementation/load |
| Queuing | Waiting behind other packets | Variable with traffic |
| Transmission | L/R | Constant only when packet sizeL and link rateR are fixed |
| Propagation | Distance/speed | Approximately constant for a fixed physical link |
Sum these components over the traversed links/nodes; a fixed route alone does not make queueing delay constant.
| Alice browser | → | HTTP(S) webmail service | → | SMTP between mail servers | → | Bob mailbox | → | IMAP client |
| Stage | Protocol and role | Transport / conventional ports |
|---|---|---|
| Browser submits message | HTTP POST over HTTPS; application uploads/pushes the message | HTTPS:TCP443; historical HTTP:TCP80 |
| Sender mail server→recipient mail server | SMTP pushes the mail to recipient domain | TCP25 for server relay; client submission commonly587 |
| Bob retrieves/synchronizes | IMAP pulls/synchronizes mailbox state | TCP143 (IMAP) /993 with implicit TLS |
DNS may first locate the recipient domain’s MX record and mail-host IP (typicallyUDP53, withTCP53 when needed). HTTP is a request/response protocol rather than a pure push-only protocol; classify this particular POST as submission. The named IMAP/POP3 stage is historical email context; SMTP is the explicitly listed current application topic.
Selective Repeat (SR) ARQ accepts and buffers valid out-of-order frames inside the receiver window, acknowledges them individually and retransmits only missing/timed-out frames. In this answer ACK i means “frame i received”, unlike the next-expected cumulative ACK convention used for GBN.
Yaad rakho: SR selective resend; half sequence space keeps old and new separate.
Definition. Switching selects how data passes through intermediate network nodes from a source to a destination.
Simple meaning. Raasta reserve karna, poora message ruk-ruk ke bhejna, ya chhote packets bhejna—yeh teen ideas hain.
| Aspect | Circuit switching | Message switching | Packet switching |
|---|---|---|---|
| Unit handled | Continuous stream on established circuit | Whole message | Small packets |
| Resource use | Capacity reserved along a circuit | Shared; message stored at each intermediate node | Shared; each packet stored/forwarded in this model |
| Setup | Required before data transfer | No circuit setup required | Datagram: no connection setup; virtual circuit: setup |
| Delay | Setup delay, then relatively predictable service | Large store-and-forward delay for long messages | Variable delay; packet pipeline can reduce completion time |
| Main limitation | Reserved capacity can be idle | Large buffers; unsuitable for tight delay needs | Headers, possible queueing/loss; datagrams may reorder |
For ordinary bursty office data, packet switching is generally appropriate because resources are shared among active transfers. The design must still account for queueing, loss and the required service quality.
Yaad rakho: Circuit reserves; message waits whole; packets pipeline.
Choice: circuit switching reserves a path and capacity before communication, giving predictable service and avoiding packet-queue contention after setup. A dedicated voice circuit is a standard example. It sacrifices statistical sharing and may waste reserved capacity when silent.
For the rocket case, reservation fits the stated timing need. However, no switching technique alone guarantees absolute delivery: link failures, bit errors and equipment faults still require redundancy, error control and fault-tolerant system design. State the reservation advantage without promising impossible certainty.
One packet, L=2000 bits;3 links;rate1000000 bits/s;propagation2ms/link;store-and-forward.
Total time until the final packet bit reaches the destination.
Each router must receive the complete packet before sending it onward, so all three serialization times count.
2000/1000000 seconds=0.002s=2ms.
2+2=4ms.
3×4=12ms. Queuing/processing are ignored as instructed.
An IPv4 address identifies an IP interface. The network prefix selects the destination network; remaining bits identify an address within it. Classful addressing is historical; modern routing uses explicit CIDR prefixes.
| Class | First octet | Default mask | Network / host split | Purpose |
|---|---|---|---|---|
| A | 1–126 (ordinary unicast) | 255.0.0.0 (/8) | 8 / 24 bits | Large historical unicast networks; 0 and 127 reserved |
| B | 128–191 | 255.255.0.0 (/16) | 16 / 16 bits | Medium historical unicast networks |
| C | 192–223 | 255.255.255.0 (/24) | 24 / 8 bits | Small historical unicast networks |
| D | 224–239 | No unicast default mask | Not network/host class split | Multicast |
| E | 240–255 | No unicast default mask | Not normal unicast allocation | Reserved/experimental; special addresses exist |
Example: 114.34.2.8 is class A with network 114.0.0.0 and host portion 34.2.8. Class does not override an explicitly supplied CIDR prefix.
| Classful | CIDR/classless |
|---|---|
| Fixed A/B/C boundaries /8,/16,/24 | Prefix length explicitly states any suitable network boundary |
| Simple historical classification | Flexible allocation and aggregation |
| Wastes addresses when demand lies between fixed classes | Matches sizes more closely, reducing waste |
| Network determined partly by leading bits | Network determined by prefix mask |
Example: an organization needing roughly1000 addresses can receive a /22 block (1024 addresses) rather than an entire old ClassB /16. Classless routing relies on the prefix length; the first octet alone is insufficient. Advantages of CIDR are flexible allocations, less address wastage and route aggregation. Disadvantages are more involved prefix/mask planning and longest-prefix lookup; badly fragmented allocations may prevent aggregation.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
| Alice browser | → | HTTP(S) webmail service | → | SMTP between mail servers | → | Bob mailbox | → | POP3 client |
| Stage | Protocol and role | Transport / conventional ports |
|---|---|---|
| Browser submits message | HTTP POST over HTTPS; application uploads/pushes the message | HTTPS:TCP443; historical HTTP:TCP80 |
| Sender mail server→recipient mail server | SMTP pushes the mail to recipient domain | TCP25 for server relay; client submission commonly587 |
| Bob retrieves/synchronizes | POP3 pulls messages for download | TCP110 (POP3) /995 with implicit TLS |
DNS may first locate the recipient domain’s MX record and mail-host IP (typicallyUDP53, withTCP53 when needed). HTTP is a request/response protocol rather than a pure push-only protocol; classify this particular POST as submission. The named IMAP/POP3 stage is historical email context; SMTP is the explicitly listed current application topic.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
Definition. Address Resolution Protocol (ARP) resolves a known IPv4 address to a link-layer hardware address on the local network.
Simple meaning. IP pata hai; local delivery ke liye MAC chahiye.
ARP maps a known local next-hop IPv4 address to a hardware address. A sender broadcasts a request when the mapping is missing; the address owner replies, and the mapping is cached. For a remote IP destination, the sender resolves its gateway’s local MAC address, then forwards the packet toward that gateway.
Yaad rakho: Remote packet ka next-hop MAC gateway ka hota hai.
The request already contains the requester’s MAC address. The owner of the requested IP can therefore address its reply directly to that MAC. Unicast avoids making every LAN host process a reply intended for one requester and reduces unnecessary traffic. Ordinary request is broadcast because the destination MAC was unknown; ordinary response is unicast because it is now known. Gratuitous/announcement ARP is a different case and can be broadcast.
10 cars; service1car/s; road400km; speed10km/s; each booth waits for all10 cars.
All seven timing/overlap questions.
![\"T_{service}=N/r,\\quad](\"math/d9a16d08c4aea6f6.svg\")
Transmission corresponds to serializing the whole caravan; propagation is the road journey of each individual car.
Car1 leaves at1s; car10 leaves at10s. Travel=400/10=40s. Thus first arrival at booth2 is41s and last arrival is50s. Booth2 starts service only at50s.
| Part | Answer | Reason |
|---|---|---|
| 1 | 1 second | One car per second |
| 2 | 10 seconds | 10/1 |
| 3 | 40 seconds | Measured from first car departure |
| 4 | 40 seconds | Last car travels the same road at the same speed |
| 5 | 49 seconds | From first departure t1 to next service t50:50−1 |
| 6 | No | First booth service0–10; second50–60 |
| 7 | Yes | Neither is servicing during(10,50); in particular before first arrival, (10,41) |
Common mistake: part5 is not50s because its clock starts when the first car leaves, one second after the first service began.
Definition. Count-to-infinity is a distance-vector failure in which routers repeatedly increase their estimates for an unreachable destination because they believe one another’s stale route advertisements.
Simple meaning. Dono neighbour ek dusre ko bol rahe: mere through jao—par actual route gaya.
Split horizon suppresses advertising a route back in the direction from which it was learned. Poison reverse explicitly advertises that route back with an infinite/unreachable metric. Both reduce two-router distance-vector loops caused by stale information.
Yaad rakho: Split: wapas mat bolo. Poison reverse: wapas bolo, par unreachable.
User Datagram Protocol (UDP) provides connectionless, message-oriented process delivery. One send creates a datagram; message boundaries are retained. UDP has an 8-byte header, carries ports and length, and offers checksum-based corruption detection. It does not itself guarantee delivery, ordering, duplicate suppression, retransmission, flow control or congestion control.
| UDP field | Width | Purpose |
|---|---|---|
| Source port | 16 bits | Sending endpoint; a zero value can indicate no reply port when unused |
| Destination port | 16 bits | Receiving application endpoint |
| Length | 16 bits | UDP header plus UDP payload; minimum 8 bytes |
| Checksum | 16 bits | Error detection over UDP and an IP pseudo-header; not retransmission |
Yaad rakho: UDP datagram preserve; reliability automatic nahi.
Yes, reliability can be implemented above UDP. Add sequence numbers, acknowledgments, checksums, retransmission timers, duplicate suppression, ordering buffers and flow/congestion handling as required. If packet5 is lost, an application-level timeout retransmits5; a receiver seeing a duplicate5 acknowledges it without delivering it twice. UDP itself still supplies no delivery guarantee: the application protocol supplies these mechanisms and may eventually report failure rather than guarantee delivery across a permanent outage.
Data = 10011101; generator bits = 1001; degree r = 3.
CRC remainder, complete transmitted frame and receiver check.
Append r zeros, divide using XOR (modulo 2), and replace those zeros with the r-bit remainder. XOR has no borrowing or carrying. A valid received codeword has remainder zero; the receiver does not append new zeros.
10011101 → 10011101000
| Leading position | Before XOR | Aligned generator | After XOR |
|---|---|---|---|
| 1 | 10011101000 | 10010000000 | 00001101000 |
| 5 | 00001101000 | 00001001000 | 00000100000 |
| 6 | 00000100000 | 00000100100 | 00000000100 |
Remainder = 100. Transmit 10011101100.
| Position | Before | Generator | After |
|---|---|---|---|
| 1 | 10011101100 | 10010000000 | 00001101100 |
| 5 | 00001101100 | 00001001000 | 00000100100 |
| 6 | 00000100100 | 00000100100 | 00000000000 |
Receiver remainder = 000. No error is detected; this is not a guarantee that every possible corruption is detectable.
10110001, 10101011, 00110101, 10100001; word width = 8 bits.
Sender checksum and receiver verification.
One’s-complement addition returns an overflow carry to the least-significant end. Complement only after all words have been added.
| Word | Decimal | Raw addition | Folded addition | Running binary sum |
|---|---|---|---|---|
| 10110001 | 177 | 177 | 177 + carry 0 = 177 | 10110001 |
| 10101011 | 171 | 348 | 92 + carry 1 = 93 | 01011101 |
| 00110101 | 53 | 146 | 146 + carry 0 = 146 | 10010010 |
| 10100001 | 161 | 307 | 51 + carry 1 = 52 | 00110100 |
Running sum = 00110100. Change every 0 to 1 and every 1 to 0: checksum = 11001011 (decimal 203).
52 + 203 = 255 = 11111111. All bits are 1; complement gives all zeros. Therefore this intact message passes the checksum check. A matching checksum detects no error; it does not prove that every possible corruption is absent.
Source note: The paper does not state bit-position direction. The solution numbers positions from the right, matching current faculty slides; position 7 means the seventh from the right throughout.
Data 10101111; even parity; one error at position 7 from the right.
Codeword, received word, syndrome, corrected data.
![\"2^r\\ge](\"math/ced24dba71859ec0.svg\")
Use the smallest r satisfying the inequality. Four checks label positions up to 12; parity positions are powers of two.
r=3 gives 8 < 8+3+1=12, so it is insufficient. r=4 gives16 ≥8+4+1=13. Total length =12; reserve positions1,2,4,8.
| Position (right-to-left index) | 12 | 11 | 10 | 9 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Role | data | data | data | data | p8 | data | data | data | p4 | data | p2 | p1 |
| Data/parity | 1 | 0 | 1 | 0 | ? | 1 | 1 | 1 | ? | 1 | ? | ? |
| Parity | Other covered positions | Other bits | Parity bit |
|---|---|---|---|
| 1 | 3, 5, 7, 9, 11 | 1 + 1 + 1 + 0 + 0 | 1 |
| 2 | 3, 6, 7, 10, 11 | 1 + 1 + 1 + 1 + 0 | 0 |
| 4 | 5, 6, 7, 12 | 1 + 1 + 1 + 1 | 0 |
| 8 | 9, 10, 11, 12 | 0 + 1 + 0 + 1 | 0 |
Transmitted codeword = 101001110101.
Position7 changes 1→0. Received word = 101000110101.
| Parity check | Positions | XOR |
|---|---|---|
| 1 | 1,3,5,7,9,11 | 1 |
| 2 | 2,3,6,7,10,11 | 1 |
| 4 | 4,5,6,7,12 | 1 |
| 8 | 8,9,10,11,12 | 0 |
s8s4s2s1 = 0111 = 7. Flip position7 back; all four checks become0. Remove positions1,2,4,8 to recover 10101111.
Source note: The paper does not explicitly state whether the origin TCP connection already exists. Both the established persistent-connection convention and one new shared connection are distinguished.
200 sequential requests;50%hits;hit delay0;miss RTT1s and object transmission4s.
Total time, making connection setup assumption explicit.
![\"T=N_{hit}T_{hit}+N_{miss}(RTT+T_{object})+T_{setup}\"](\"math/277a57e2fbd98438.svg\")
Sequential requests cannot overlap, and without conditional validation a cache hit needs no origin round trip.
Hits=200×0.5=100; misses=100.
Each miss costs1+4=5s. Total response time=100×0+100×5=500s.
With an already-established persistent origin connection, answer500s. If the model instead includes one new reusable origin TCP connection, add1RTT once:501s. The paper does not state that initial condition; do not add100 separate handshakes to a persistent connection or silently assume one connection per object. Multiple origin connections would require their setup count.
The Domain Name System (DNS) is a distributed hierarchical naming system. Its namespace is a tree with the root at the top. A complete name is the ordered path of labels from a node back to the root. An absolute name may be written with a trailing dot, for example www.example.com.
| Level | Example | Meaning |
|---|---|---|
| Root | . | Top of the namespace |
| Top-level domain (TLD) | com, org, in | A child zone below the root |
| Registered/delegated domain | example.com | Name under a parent such as com |
| Subdomain/host label | www.example.com | A label below example.com; a name may own different record types |
Yaad rakho: Full path unique; www label akela unique nahi.
| Part | Answer | Explanation |
|---|---|---|
| 1 | DNS Local | Check the local resolver cache first |
| 2 | DNS TLD | Root supplies a referral toward .edu TLD servers; the response packet itself goes back to local DNS |
| 3 | DNS Authoritative | TLD supplies a referral toward the responsible authoritative servers; response goes back to local DNS |
| 4 | A record | Contains an IPv4 address; AAAA would contain IPv6 |
| 5 | Iterative between the resolver and hierarchy; recursive client-to-resolver | Neither is universally best in every role. Typical deployment combines the two |
The wording “where does the response link” is read as the referral target, not the packet’s network destination. This distinction prevents confusing local DNS with the next server category.
Prefixes01→1,10→2,011→3,100→4,001→5,default→6.
Output interface for each destination.
Among matching prefixes choose the one with the most bits; use default only when none match.
The leading bits must match from the first bit; a substring appearing later does not count.
| Destination | Matching prefixes | Longest | Interface |
|---|---|---|---|
| 00111101 | 001 | 001 | 5 |
| 01010101 | 01 (but not011) | 01 | 1 |
| 11001001 | None | Default | 6 |
5, 1, 6.
An IPv4 address identifies an IP interface. The network prefix selects the destination network; remaining bits identify an address within it. Classful addressing is historical; modern routing uses explicit CIDR prefixes.
| Class | First octet | Default mask | Network / host split | Purpose |
|---|---|---|---|---|
| A | 1–126 (ordinary unicast) | 255.0.0.0 (/8) | 8 / 24 bits | Large historical unicast networks; 0 and 127 reserved |
| B | 128–191 | 255.255.0.0 (/16) | 16 / 16 bits | Medium historical unicast networks |
| C | 192–223 | 255.255.255.0 (/24) | 24 / 8 bits | Small historical unicast networks |
| D | 224–239 | No unicast default mask | Not network/host class split | Multicast |
| E | 240–255 | No unicast default mask | Not normal unicast allocation | Reserved/experimental; special addresses exist |
Example: 114.34.2.8 is class A with network 114.0.0.0 and host portion 34.2.8. Class does not override an explicitly supplied CIDR prefix.
191.5.8.9 and214.25.6.3; historical default class masks.
Class, mask, network, broadcast and host portion.
Network = address AND mask; broadcast = network with all host bits1.
Without a custom prefix the question’s classful default determines the split.
| Address | Class | Mask | Network | Broadcast | Host portion |
|---|---|---|---|---|---|
| 191.5.8.9 | B | 255.255.0.0 (/16) | 191.5.0.0 | 191.5.255.255 | 8.9 (16 bits) |
| 214.25.6.3 | C | 255.255.255.0 (/24) | 214.25.6.0 | 214.25.6.255 | 3 (8 bits) |
The supplied address itself is a host address; “host portion” means the suffix left after removing the network bits.
Arrival times0.3,0.6,0.9,1.6,2.7,3.3,4.3;frame1;propagation0.4.
Starts/busy decisions and successfully received packets.
A transmission starting at a is sensed by another node during [a+0.4,a+1.4]. Without collision detection, it continues for its full one-unit duration.
A node may start before an earlier distant transmission has arrived. Such starts can collide even though each node initially sensed idle.
| Packet | Arrival | Decision | Reason |
|---|---|---|---|
| 1 | 0.3 | 0.3 | No prior signal |
| 2 | 0.6 | 0.6 | Signal1 first arrives0.7 |
| 3 | 0.9 | s | Signal1 busy[0.7,1.7] |
| 4 | 1.6 | s | Signals1 and2 are busy |
| 5 | 2.7 | 2.7 | Both prior signals cleared by2.0 |
| 6 | 3.3 | s | Signal5 busy[3.1,4.1] |
| 7 | 4.3 | 4.3 | Signal5 cleared by4.1 |
Frames1 and2 overlap and collide. Frames5 and7 have no overlapping reception intervals at another node. Starts:0.3,0.6,s,s,2.7,s,4.3. Successes:5,7.
dV(U)=6 with predecessorV; dV(nodeX)=4 with predecessorV. Unknown linkX is U–V; linkY is U–nodeX.
Unknown edge costs or n/a.
![\"d(U)=d(\\operatorname{pred}(U))+c(\\operatorname{pred}(U),U)\"](\"math/af3d37f6202e501e.svg\")
A predecessor edge lies on the reported shortest path, so its cost is fixed by the distance difference. An unused edge is usually constrained, not uniquely determined.
U’s predecessor isV:6=0+X, therefore X=6.
The alternate routeV→nodeX→U costs4+Y. It must not be shorter than6, henceY≥2 (assuming nonnegative costs). Many values satisfy this; ifY=2 there is a tie and predecessorV remains valid. Therefore Y=n/a (not uniquely determined).
Definition. Pure ALOHA is a random-access protocol in which a station transmits whenever a frame is ready and retransmits after a random delay if the transmission is unsuccessful.
Simple meaning. Frame ready → bhej do. Collision hua → random wait karke retry.
In Pure ALOHA, a station transmits a ready frame without carrier sensing. If a successful acknowledgement is not received, it waits for a random interval and retransmits. Overlapping transmissions cause collisions in the basic model. Its maximum normalized throughput is 1/(2e) ≈ 18.4%, so it is inefficient at high offered load.
Yaad rakho: Pure = kisi bhi time start; retry ka wait random.
\r\nDefinition. The vulnerable period is the interval of start times of other frames that would cause a collision with a chosen frame.
Simple meaning. Apne frame se ek frame-time pehle start hua frame bhi takra sakta hai, aur baad mein bhi.
A frame lasting T can collide with an equal-length frame that began less than T earlier, or one that begins less than T later. The two intervals together have length T+T=2T.
Yaad rakho: Pure: before T + after T = 2T.
\r\nDefinition. Slotted ALOHA is a random-access protocol that divides time into slots of one frame duration and permits transmissions only at the beginning of a slot.
Simple meaning. Frame ready hai tab bhi next slot ki start-line ka wait karo.
| Feature | Pure | Slotted |
|---|---|---|
| Start time | Any time | Slot boundary only |
| Synchronization | No common slots required | Required |
| Vulnerable period | 2T | T |
| Maximum S | 1/(2e) ≈ 0.184 | 1/e ≈ 0.368 |
| G at maximum | 0.5 | 1 |
Yaad rakho: Slotted = aligned starts; collision risk half-window, not zero.
Source note: This is MAC learning, not an ARP cache. The diagram’s example (A,2) conflicts with the drawn A interface1. No unqualified ARP solution is presented.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
This historical topic is not named in the current six-unit syllabus. It is recorded for a complete paper inventory, but excluded from the required preparation and current-unit mark totals.
Historical pattern used for these compiled mocks: 3 hours, 100 marks. Q1 is compulsory; attempt any four whole questions from Q2–Q7. Every whole question is 20 marks. These are practice papers, not predicted examination papers. They combine lesson questions, reused actual-paper questions and newly varied numerical problems; the actual-paper tab preserves the original papers separately.
Attempt Q1(a–d), then choose four complete 20-mark question groups. The solutions follow each part so that every choice is fully reviewable. For a closed-book attempt, write the question in your notebook before reading its answer.
| Planning suggestion | Time |
|---|---|
| Read and select four question groups | 10 minutes |
| Write five 20-mark questions including Q1 | About 32 minutes each |
| Check units, diagrams, sequence numbers and chosen parts | 10 minutes |
Timing is a suggestion based on the historical three-hour papers. It is not a current instructor instruction. Mock numerical assumptions are explicit; use those assumptions in your answer.
Each mock part follows a question family present in the originals. Q6(a) is transmission-media theory; TCP sequence-number drills are optional, not mock questions. Reused questions and fresh numerical variants are practice, with mock marks rather than a claim that the new data appeared in an exam.
"}]},{"id":"tee-mock-a","title":"TEE mock A · full paper","scope":"TEE · 100 attempted marks · solutions for every choice · current syllabus 702AI0C023","sections":[{"id":"q1a","title":"Q1A · Networks and design","html":"| Criterion | Wired campus links | Wireless campus access |
|---|---|---|
| Speed/capacity | Dedicated switch ports and fiber backbones can provide high, predictable capacity | Capacity shared over radio; performance depends on contention, coverage and interference |
| Reliability | Stable when cables/ports and redundant uplinks are healthy | Mobility and flexible access; radio interference and coverage holes can cause variability |
| Security | Physical access control helps, but ports still need authentication/segmentation | Use authenticated encryption and managed access; radio extends beyond rooms |
| Deployment | Cabling cost and fixed access points | Supports mobile users and faster coverage changes |
| Recommended role | Servers, laboratories, inter-building backbone | Students’ phones/laptops, roaming and guest access |
Choose a hybrid: wired core/backbone and critical endpoints, with managed access points providing wireless mobility. Separate guest traffic, authenticate users and plan redundant infrastructure. Neither medium alone guarantees security or availability.
The TCP/IP model describes the Internet protocol suite in four layers. Many teaching books split the link layer into data link and physical, making a five-layer teaching model; identify the convention before drawing.
| TCP/IP layer | Function | Examples |
|---|---|---|
| Application | Defines application message formats and exchanges for Web access, naming, email and file transfer; includes presentation/session functions as needed | HTTP, DNS, SMTP, FTP |
| Transport | End-to-end process communication | TCP, UDP, SCTP |
| Internet | IP addressing, routing and best-effort datagrams | IPv4/IPv6 |
| Link/network access | Local delivery and physical access | Ethernet/Wi-Fi and their medium |
| Point | OSI | TCP/IP |
|---|---|---|
| Layer count | Seven | Four here; five-layer teaching variant splits link |
| Upper functions | Application, presentation, session separated | Grouped under application |
| Lower functions | Data link and physical separated | Grouped under link/network access |
| Purpose | General reference architecture | Architecture of deployed Internet protocol family |
| Network service | Reference model can describe different services | Internet layer uses connectionless IP |
| Transport | Conceptual transport service | TCP reliable byte stream; UDP datagrams; SCTP messages |
Protocol = Pure ALOHA; G = 0.25 attempts/frame time.
S, 100S and the maximum possible S.
Select the formula from the protocol name. The exponent is −2G because Pure ALOHA has a 2T vulnerable period.
−2 × 0.25 = -0.5.
S = 0.25 × 0.606530660 ≈ 0.151632665 successful frames/frame time.
0.151632665 × 100 ≈ 15.1633%.
The maximum is 1/(2e) ≈ 0.183940 = 18.3940%, at G=0.5.
Final answer: S ≈ 0.151633; percentage ≈ 15.1633%; Smax ≈ 0.183940 at G=0.5.
IP/prefix = 192.168.8.173/27.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−27=5; N=25=32.
Mask=255.255.255.224; binary=11111111.11111111.11111111.11100000.
The changing octet is number 4. Its mask value is 224, so step=256−224=32. The IP value 173 lies in 160…191.
Keep earlier octets; place 160 in octet 4; set any later octets to zero. Network=192.168.8.160.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 11000000 | 11111111 | 11000000 | 192 |
| 2 | 10101000 | 11111111 | 10101000 | 168 |
| 3 | 00001000 | 11111111 | 00001000 | 8 |
| 4 | 10101101 | 11100000 | 10100000 | 160 |
Use 191 in the changing octet and 255 in any later octets. Broadcast=192.168.8.191.
First usable=192.168.8.161; last usable=192.168.8.190; count=32−2=30.
Final answer: Network 192.168.8.160; broadcast 192.168.8.191; 32 total, 30 ordinary usable; range 192.168.8.161–192.168.8.190.
Block coding maps each k-bit dataword to an n-bit codeword, adding n−k redundant bits. Only selected codewords are valid; their separation determines error-detection/correction capability.
| Codeword | Bits |
|---|---|
| C1 | 0000000 |
| C2 | 1111000 |
| C3 | 0011110 |
All pair distances, dmin and the maximum guaranteed numbers of errors.
XOR exposes differing positions. Use the closest codeword pair, because it is the weakest separation in the code.
| Pair | First | Second | XOR | Distance |
|---|---|---|---|---|
| C1 vs C2 | 0000000 | 1111000 | 1111000 | 4 |
| C1 vs C3 | 0000000 | 0011110 | 0011110 | 4 |
| C2 vs C3 | 1111000 | 0011110 | 1100110 | 4 |
dmin = min(4, 4, 4) = 4.
Detect = 4−1 = 3; correct = floor((4−1)/2) = 1.
4 bit changes can reach another valid codeword, so only 3 are guaranteed detectable. Correction regions require 2t < 4.
Final answer: dmin=4; detect up to 3 errors; correct up to 1 errors.
A setup phase reserves a circuit through intermediate switches. Data then streams through the established path until release. It offers predictable reserved service and low per-unit forwarding overhead, but setup takes time and idle periods waste reserved capacity. A traditional dedicated voice circuit is an appropriate illustration.
Each intermediate node receives and stores the entire message before forwarding it. Links can be shared and no end-to-end circuit is required, but large messages need buffers and cause substantial store-and-forward delay. It illustrates delay-tolerant whole-message delivery, rather than an efficient choice for interactive office traffic.
A message is segmented into packets, allowing shared links and pipelined forwarding. This suits bursty data and supports flexible routing. Costs include packet headers, variable queueing delay, potential packet loss and possible reordering in datagram networks.
| Aspect | Circuit switching | Message switching | Packet switching |
|---|---|---|---|
| Unit handled | Continuous stream on established circuit | Whole message | Small packets |
| Resource use | Capacity reserved along a circuit | Shared; message stored at each intermediate node | Shared; each packet stored/forwarded in this model |
| Setup | Required before data transfer | No circuit setup required | Datagram: no connection setup; virtual circuit: setup |
| Delay | Setup delay, then relatively predictable service | Large store-and-forward delay for long messages | Variable delay; packet pipeline can reduce completion time |
| Main limitation | Reserved capacity can be idle | Large buffers; unsuitable for tight delay needs | Headers, possible queueing/loss; datagrams may reorder |
For ordinary office file, web and collaboration traffic, choose packet switching, with adequate capacity and traffic management. Its statistical sharing serves bursty sources efficiently. If strict deterministic service is required, state that requirement and discuss additional service engineering rather than promising that every packet network guarantees timing.
A computer network is a set of interconnected devices that exchange data and share resources using communication protocols. A link carries signals; a protocol defines message format, meaning and timing. Hosts run applications, switches forward local frames and routers connect IP networks.
| Type | Meaning and scope | Example |
|---|---|---|
| PAN | Personal Area Network: a person’s nearby devices | Phone linked to earbuds |
| LAN | Local Area Network: a building or local campus network under local administration | Laboratory Ethernet/Wi-Fi |
| MAN | Metropolitan Area Network: interconnects sites across a town/city | City-wide institutional network |
| WAN | Wide Area Network: connects geographically separated networks over regional/global links | Branches in different cities |
| ARPANET | Historical packet-switched research network; a major predecessor of the Internet | Early interconnected research institutions; not a modern geographic size category |
Classification by geographical scope is descriptive: a fixed kilometre boundary or a guaranteed speed is not part of every definition. Networks can also be classified by wired/wireless medium or by client-server/peer-to-peer organization.
| Packet type | What it does | Simple example |
|---|---|---|
| Hello | Discover/maintain neighbours and exchange relationship parameters | Are we still neighbours? |
| Database Description (DD/DBD) | Summarize link-state database contents during synchronization | Here are the LSA headers I have |
| Link State Request (LSR) | Ask for specific missing/newer link-state information | Please send this LSA |
| Link State Update (LSU) | Carry one or more LSAs | Here is the requested or newly flooded link-state information |
| Link State Acknowledgment (LSAck) | Acknowledge receipt of LSAs | I received these LSAs |
Database Description packets carry summaries; Link State Update packets carry LSAs themselves. Do not confuse an LSA type with an OSPF packet type.
The World Wide Web (WWW) is an information system of linked resources identified by URLs and accessed using Web clients and servers. A browser is a client; a Web server hosts or generates representations such as HTML, images or JSON. HTTP defines request/response semantics between these components. The Web is one application of the Internet, not a synonym for the entire Internet.
| Message component | Purpose |
|---|---|
| Request method / target | Requested action and resource; GET retrieves a representation, POST submits data for processing |
| Headers | Metadata, routing context, representation preferences and cache/connection information |
| Status code | Result class: 2xx success, 3xx redirection, 4xx client error, 5xx server error |
| Body | Representation or submitted content when present |
HTTP semantics are stateless: each request has its own meaning. Cookies/tokens can let an application maintain a login session; stateless does not mean “users cannot log in”. HTTP/1.1 and HTTP/2 commonly use TCP; HTTP/3 uses QUIC over UDP, so “HTTP always uses TCP” is not a universal statement.
Given address: 0:15::1:12:1213.
The full 128-bit address in eight-group hexadecimal notation.
A written 0 is a visible group. :: represents only the omitted all-zero groups.
Left: 0, 15 (2 group(s)). Right: 1, 12, 1213 (3 group(s)).
Visible=2+3=5. Missing=8−5=3.
0:15:0:0:0:1:12:1213.
| Group number | Four-digit group |
|---|---|
| 1 | 0000 |
| 2 | 0015 |
| 3 | 0000 |
| 4 | 0000 |
| 5 | 0000 |
| 6 | 0001 |
| 7 | 0012 |
| 8 | 1213 |
0000:0015:0000:0000:0000:0001:0012:1213.
Final answer: 0000:0015:0000:0000:0000:0001:0012:1213
| Feature | IPv4 | IPv6 |
|---|---|---|
| Address width | 32 bits | 128 bits |
| Base/header size | 20–60 bytes depending on options | 40-byte fixed base header |
| Length field | Total length includes IPv4 header | Payload Length excludes 40-byte base header |
| Lifetime field | TTL | Hop Limit |
| Following protocol | Protocol field | Next Header; may identify an extension |
| Header checksum | Present | No base-header checksum |
| Options | Inside IPv4 header | Typically extension headers |
| Fragmentation fields | In IPv4 header; routers may fragment if permitted | Fragment extension header; fragmentation by source, not routers |
The supplied scheme’s sentence says four missing groups, but five groups are already visible, so only three are missing. Its printed eight-group expansion is consistent with three. The count shown above is independently verified.
A TCP connection establishes each endpoint’s initial sequence number and confirms that both can receive. Use client initial sequence x=1000 and server initial sequence y=5000; assume no payload in the handshake.
| Step | Segment | Meaning |
|---|---|---|
| 1 Client → server | SYN, Seq=1000 | Client proposes its starting sequence |
| 2 Server → client | SYN+ACK, Seq=5000, Ack=1001 | Server acknowledges client SYN and supplies its own start |
| 3 Client → server | ACK, Seq=1001, Ack=5001 | Client acknowledges server SYN; connection is established |
Client sends 200 bytes beginning at Seq=1001, so byte numbers are 1001…1200. The receiver acknowledges Ack=1201, the next byte expected. If bytes starting at 1201 are missing but later bytes arrive, the cumulative ACK remains 1201 until the gap is repaired. TCP may buffer out-of-order bytes; it delivers the byte stream in order. Each direction counts independently.
Go-Back-N (GBN) ARQ pipelines up to W outstanding frames. For m-bit sequence numbers, textbook GBN uses W ≤ 2m−1 and receiver window 1. The receiver accepts only the next in-order frame; out-of-order frames are discarded. ACK k here cumulatively acknowledges all frames before k and means “next expected is k”.
File Transfer Protocol (FTP) transfers files between a client and server using TCP and separates commands from file data. This separation is its central design feature. Its control connection remains available across multiple operations, while data connections carry file bytes or directory listings.
| Connection | Function | Lifetime / ports |
|---|---|---|
| Control | Commands and replies: USER, PASS, CWD, TYPE, LIST, RETR, STOR, QUIT | Client connects to server TCP port 21; normally kept for the session |
| Data | File contents and directory listing data | Opened as needed; separate from command channel |
FTP over TLS (FTPS) protects FTP connections. SFTP is a separate SSH-based file-transfer protocol, not “FTP using an SSH port”. These names should not be treated as interchangeable.
| Type | Characteristic 1 | Characteristic 2 |
|---|---|---|
| Radio | Can provide broad-area/mobility support depending on band | Shared spectrum/interference and obstacles affect reception |
| Terrestrial microwave | Directional, usually line of sight | Needs antenna alignment; obstructions/weather can matter |
| Satellite microwave | Relay covers large or remote areas | Delay/capacity depend strongly on orbit/system |
| Infrared | Short range and generally blocked by opaque walls | Limited room-scale coverage reduces inter-room interference |
Many classifications group satellite links under microwave rather than making a separate spectral category. State that grouping consistently.
| Rule | When idle | When busy | Trade-off |
|---|---|---|---|
| 1-persistent | Transmit immediately | Keep listening; transmit as soon as idle | Low idle delay, but waiting stations may start together |
| Non-persistent | Transmit | Wait a random time, then sense again | Less synchronized contention; may leave avoidable idle time |
| p-persistent (slotted) | Transmit with probability p; defer a slot with probability 1−p | Wait until available, then resume the probabilistic slot rule | Controls contention; needs slots and a choice of p |
All three sense the channel; their persistence rules differ. Propagation delay means carrier sensing cannot guarantee the absence of collisions. In p-persistent CSMA, each eligible idle-slot decision uses the specified probability.
Congestion occurs when offered traffic overloads network resources such as link capacity, router buffers or processing. Queueing delay grows, packets may be dropped, and excessive retransmission can make the overload worse. Flow control protects a receiver; congestion control protects the network.
| Approach | Policy / mechanism | How it helps |
|---|---|---|
| Open loop: prevent congestion without runtime congestion feedback | Retransmission policy | Carefully chosen timers/selective retransmission avoid needless extra traffic |
| Open loop | Window policy | Limit outstanding data and avoid excessive retransmission after loss |
| Open loop | Acknowledgement policy | Avoid unnecessarily frequent ACK traffic while retaining correct recovery |
| Open loop | Admission policy | Refuse new load when reserved capacity is insufficient |
| Closed loop: measure and react to congestion | Backpressure | A congested node tells upstream nodes to reduce/stop sending, possibly propagating hop by hop |
| Closed loop | Choke/control signalling | A congested node sends explicit feedback to the source |
| Closed loop | Implicit signalling | A sender infers congestion from loss/delay and changes its sending behaviour |
| Closed loop | Explicit marking | Network marks congestion (for example ECN); endpoints react to returned information |
For an answer asking for two of each: explain retransmission and admission policies under open loop, then backpressure and explicit signalling under closed loop, with the mechanism and its effect. These are general congestion-control classifications used in the course; they are not all mandatory algorithms implemented inside TCP. TCP commonly uses end-to-end feedback such as loss or ECN.
| Edge | Cost |
|---|---|
| A→B | 3 |
| A→C | 2 |
| B→D | 4 |
| C→B | 1 |
| C→D | 7 |
| D→E | 2 |
| C→E | 10 |
Source=A; target=E. Edges are directed; do not add reverse edges.
The path A to E, its total cost and intermediate work.
Use Dijkstra because all weights are nonnegative. Select the smallest unsettled total distance, not the cheapest isolated edge.
![\"Dijkstra](\"fig/tee-mock-a-route.6f748058c9.svg\")
d[A]=0; every other distance=∞. No predecessors are known.
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Tentative |
| B | ∞ | — | Tentative |
| C | ∞ | — | Tentative |
| D | ∞ | — | Tentative |
| E | ∞ | — | Tentative |
Unsettled candidates before selection: A=0, B=∞, C=∞, D=∞, E=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| A→B | ∞ | 0+(3)=3 | Improve to 3; predecessor=A |
| A→C | ∞ | 0+(2)=2 | Improve to 2; predecessor=A |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | A | Tentative |
| C | 2 | A | Tentative |
| D | ∞ | — | Tentative |
| E | ∞ | — | Tentative |
Unsettled candidates before selection: B=3, C=2, D=∞, E=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| C→B | 3 | 2+(1)=3 | Retain 3 (equal cost) |
| C→D | ∞ | 2+(7)=9 | Improve to 9; predecessor=C |
| C→E | ∞ | 2+(10)=12 | Improve to 12; predecessor=C |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | A | Tentative |
| C | 2 | A | Settled |
| D | 9 | C | Tentative |
| E | 12 | C | Tentative |
Unsettled candidates before selection: B=3, D=9, E=12. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| B→D | 9 | 3+(4)=7 | Improve to 7; predecessor=B |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | A | Settled |
| C | 2 | A | Settled |
| D | 7 | B | Tentative |
| E | 12 | C | Tentative |
Unsettled candidates before selection: D=7, E=12. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| D→E | 12 | 7+(2)=9 | Improve to 9; predecessor=D |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | A | Settled |
| C | 2 | A | Settled |
| D | 7 | B | Settled |
| E | 9 | D | Tentative |
Unsettled candidates before selection: E=9. Select the minimum; ties use alphabetical order.
No outgoing edges to relax.
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | A | Settled |
| C | 2 | A | Settled |
| D | 7 | B | Settled |
| E | 9 | D | Settled |
Trace backwards: E ← D ← B ← A. Reverse it: A → B → D → E.
Edge sum: 3 + 4 + 2 = 9.
Final answer: A → B → D → E; cost=9.
![\"Dijkstra](\"fig/tee-mock-a-route-solution.c1a801d55e.svg\")
A token bucket generates tokens at rate r up to capacity B. Sending one unit of data consumes the corresponding tokens. Tokens accumulate during idle periods only until the bucket is full. Data can therefore burst when saved tokens exist, subject to the physical link rate.
| Limitation | Reason |
|---|---|
| Bursts can still congest downstream queues | Local conformance does not ensure every later link has capacity |
| Parameters must match traffic and link requirements | A small B delays/drops legitimate packets; a large B permits disruptive bursts |
| No complete QoS guarantee alone | Admission, scheduling and end-to-end capacity may still be needed |
In a client-server design, users’ clients contact designated service processes. In peer-to-peer (P2P), participating hosts can supply and consume resources. These are application organizations, not different physical cable types.
| Point | Client-server | Peer-to-peer |
|---|---|---|
| Roles | Clients request; designated servers provide services | Peers can both request and provide resources |
| Control | Central policies, accounts and backups are easier to coordinate | Control and copies distributed among peers |
| Performance | Predictable with adequate server capacity; server may bottleneck | Resources grow with peers, but peer capacity/availability vary |
| Scalability | Scale servers, replicas, caching and load distribution | More peers can contribute capacity; coordination becomes harder |
| Reliability | One unreplicated server is a failure point; replication helps | Loss of one peer need not stop system if useful replicas remain |
| Security | Central access control and monitoring | Trust and consistent access policy are harder |
| Best fit | Managed collaboration, databases, login services | Distributed content sharing among participating peers |
Central servers can authenticate users, maintain the authoritative shared document and control versions. The organization must provision capacity, replicate data and avoid a single server failure. P2P can exchange large files or local edits among peers and distribute load, but needs discovery, conflict resolution, trust and availability strategies. A hybrid can keep account control centralized while distributing selected content.
The TCP/IP model describes the Internet protocol suite in four layers. Many teaching books split the link layer into data link and physical, making a five-layer teaching model; identify the convention before drawing.
| TCP/IP layer | Function | Examples |
|---|---|---|
| Application | Defines application message formats and exchanges for Web access, naming, email and file transfer; includes presentation/session functions as needed | HTTP, DNS, SMTP, FTP |
| Transport | End-to-end process communication | TCP, UDP, SCTP |
| Internet | IP addressing, routing and best-effort datagrams | IPv4/IPv6 |
| Link/network access | Local delivery and physical access | Ethernet/Wi-Fi and their medium |
| Point | OSI | TCP/IP |
|---|---|---|
| Layer count | Seven | Four here; five-layer teaching variant splits link |
| Upper functions | Application, presentation, session separated | Grouped under application |
| Lower functions | Data link and physical separated | Grouped under link/network access |
| Purpose | General reference architecture | Architecture of deployed Internet protocol family |
| Network service | Reference model can describe different services | Internet layer uses connectionless IP |
| Transport | Conceptual transport service | TCP reliable byte stream; UDP datagrams; SCTP messages |
Protocol = Slotted ALOHA; G = 0.6 attempts/frame time.
S, 100S and the maximum possible S.
Select the formula from the protocol name. The exponent is −G because Slotted ALOHA has a T vulnerable period.
−1 × 0.6 = -0.6.
![\"e^{-0.6}](\"math/aa9f70bc612091c4.svg\")
S = 0.6 × 0.548811636 ≈ 0.329286982 successful frames/frame time.
0.329286982 × 100 ≈ 32.9287%.
The maximum is 1/(1e) ≈ 0.367879 = 36.7879%, at G=1.
Final answer: S ≈ 0.329287; percentage ≈ 32.9287%; Smax ≈ 0.367879 at G=1.
IP/prefix = 10.20.45.200/20.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−20=12; N=212=4,096.
Mask=255.255.240.0; binary=11111111.11111111.11110000.00000000.
The changing octet is number 3. Its mask value is 240, so step=256−240=16. The IP value 45 lies in 32…47.
Keep earlier octets; place 32 in octet 3; set any later octets to zero. Network=10.20.32.0.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 00001010 | 11111111 | 00001010 | 10 |
| 2 | 00010100 | 11111111 | 00010100 | 20 |
| 3 | 00101101 | 11110000 | 00100000 | 32 |
| 4 | 11001000 | 00000000 | 00000000 | 0 |
Use 47 in the changing octet and 255 in any later octets. Broadcast=10.20.47.255.
First usable=10.20.32.1; last usable=10.20.47.254; count=4096−2=4094.
Final answer: Network 10.20.32.0; broadcast 10.20.47.255; 4,096 total, 4,094 ordinary usable; range 10.20.32.1–10.20.47.254.
Message=101101; generator=10011; generator degree r=4. This is textbook systematic CRC with zero initial remainder and no reflection/final XOR.
Exactly r CRC bits, the transmitted frame and receiver syndromes.
Modulo-2 division uses aligned XOR, not ordinary decimal division.
Message bits=101101; generator bits=10011. Leading exponent is 4, so CRC length is 4.
101101 followed by 4 zeros = 1011010000.
| Before | Aligned generator | After XOR |
|---|---|---|
| 1011010000 | 1001100000 | 0010110000 |
| 0010110000 | 0010011000 | 0000101000 |
| 0000101000 | 0000100110 | 0000001110 |
Remainder padded to 4 bits = 1110. Leading zero columns need no subtraction.
Message + CRC = 101101 + 1110 = 1011011110.
| Before | Aligned generator | After XOR |
|---|---|---|
| 1011011110 | 1001100000 | 0010111110 |
| 0010111110 | 0010011000 | 0000100110 |
| 0000100110 | 0000100110 | 0000000000 |
Receiver remainder = 0000; the uncorrupted frame passes.
Flip the last bit: 1011011111. Dividing by 10011 leaves 0001, which is nonzero, so this error is detected. A zero syndrome would mean no detected error, not proof of no corruption.
Final answer: CRC=1110; transmitted frame=1011011110; uncorrupted receiver remainder=0000.
| Department | Labs | Total addresses per lab |
|---|---|---|
| A | 32 | 128 |
| B | 64 | 64 |
| C | 16 | 512 |
Parent: 172.16.0.0/16 with 65,536 addresses.
Lab prefixes, department ranges and unused total addresses.
The question requests addresses, not usable hosts. Therefore do not add two before selecting each lab size.
128=2⁷ addresses/lab → /25. 32×128=4,096 addresses → department block /20. Allocate 172.16.0.0/20, spanning 172.16.0.0–172.16.15.255.
64=2⁶ → /26 per lab. 64×64=4,096 → /20 department block. Allocate 172.16.16.0/20, spanning 172.16.16.0–172.16.31.255.
512=2⁹ → /23 per lab. 16×512=8,192 → /19 department block. Allocate 172.16.32.0/19, spanning 172.16.32.0–172.16.63.255. Its start at third octet 32 is aligned for /19.
| Lab | Prefix | First address | Last address |
|---|---|---|---|
| A1 | 172.16.0.0/25 | 172.16.0.0 | 172.16.0.127 |
| A2 | 172.16.0.128/25 | 172.16.0.128 | 172.16.0.255 |
| A3 | 172.16.1.0/25 | 172.16.1.0 | 172.16.1.127 |
| A4 | 172.16.1.128/25 | 172.16.1.128 | 172.16.1.255 |
| A5 | 172.16.2.0/25 | 172.16.2.0 | 172.16.2.127 |
| A6 | 172.16.2.128/25 | 172.16.2.128 | 172.16.2.255 |
| A7 | 172.16.3.0/25 | 172.16.3.0 | 172.16.3.127 |
| A8 | 172.16.3.128/25 | 172.16.3.128 | 172.16.3.255 |
| A9 | 172.16.4.0/25 | 172.16.4.0 | 172.16.4.127 |
| A10 | 172.16.4.128/25 | 172.16.4.128 | 172.16.4.255 |
| A11 | 172.16.5.0/25 | 172.16.5.0 | 172.16.5.127 |
| A12 | 172.16.5.128/25 | 172.16.5.128 | 172.16.5.255 |
| A13 | 172.16.6.0/25 | 172.16.6.0 | 172.16.6.127 |
| A14 | 172.16.6.128/25 | 172.16.6.128 | 172.16.6.255 |
| A15 | 172.16.7.0/25 | 172.16.7.0 | 172.16.7.127 |
| A16 | 172.16.7.128/25 | 172.16.7.128 | 172.16.7.255 |
| A17 | 172.16.8.0/25 | 172.16.8.0 | 172.16.8.127 |
| A18 | 172.16.8.128/25 | 172.16.8.128 | 172.16.8.255 |
| A19 | 172.16.9.0/25 | 172.16.9.0 | 172.16.9.127 |
| A20 | 172.16.9.128/25 | 172.16.9.128 | 172.16.9.255 |
| A21 | 172.16.10.0/25 | 172.16.10.0 | 172.16.10.127 |
| A22 | 172.16.10.128/25 | 172.16.10.128 | 172.16.10.255 |
| A23 | 172.16.11.0/25 | 172.16.11.0 | 172.16.11.127 |
| A24 | 172.16.11.128/25 | 172.16.11.128 | 172.16.11.255 |
| A25 | 172.16.12.0/25 | 172.16.12.0 | 172.16.12.127 |
| A26 | 172.16.12.128/25 | 172.16.12.128 | 172.16.12.255 |
| A27 | 172.16.13.0/25 | 172.16.13.0 | 172.16.13.127 |
| A28 | 172.16.13.128/25 | 172.16.13.128 | 172.16.13.255 |
| A29 | 172.16.14.0/25 | 172.16.14.0 | 172.16.14.127 |
| A30 | 172.16.14.128/25 | 172.16.14.128 | 172.16.14.255 |
| A31 | 172.16.15.0/25 | 172.16.15.0 | 172.16.15.127 |
| A32 | 172.16.15.128/25 | 172.16.15.128 | 172.16.15.255 |
| B1 | 172.16.16.0/26 | 172.16.16.0 | 172.16.16.63 |
| B2 | 172.16.16.64/26 | 172.16.16.64 | 172.16.16.127 |
| B3 | 172.16.16.128/26 | 172.16.16.128 | 172.16.16.191 |
| B4 | 172.16.16.192/26 | 172.16.16.192 | 172.16.16.255 |
| B5 | 172.16.17.0/26 | 172.16.17.0 | 172.16.17.63 |
| B6 | 172.16.17.64/26 | 172.16.17.64 | 172.16.17.127 |
| B7 | 172.16.17.128/26 | 172.16.17.128 | 172.16.17.191 |
| B8 | 172.16.17.192/26 | 172.16.17.192 | 172.16.17.255 |
| B9 | 172.16.18.0/26 | 172.16.18.0 | 172.16.18.63 |
| B10 | 172.16.18.64/26 | 172.16.18.64 | 172.16.18.127 |
| B11 | 172.16.18.128/26 | 172.16.18.128 | 172.16.18.191 |
| B12 | 172.16.18.192/26 | 172.16.18.192 | 172.16.18.255 |
| B13 | 172.16.19.0/26 | 172.16.19.0 | 172.16.19.63 |
| B14 | 172.16.19.64/26 | 172.16.19.64 | 172.16.19.127 |
| B15 | 172.16.19.128/26 | 172.16.19.128 | 172.16.19.191 |
| B16 | 172.16.19.192/26 | 172.16.19.192 | 172.16.19.255 |
| B17 | 172.16.20.0/26 | 172.16.20.0 | 172.16.20.63 |
| B18 | 172.16.20.64/26 | 172.16.20.64 | 172.16.20.127 |
| B19 | 172.16.20.128/26 | 172.16.20.128 | 172.16.20.191 |
| B20 | 172.16.20.192/26 | 172.16.20.192 | 172.16.20.255 |
| B21 | 172.16.21.0/26 | 172.16.21.0 | 172.16.21.63 |
| B22 | 172.16.21.64/26 | 172.16.21.64 | 172.16.21.127 |
| B23 | 172.16.21.128/26 | 172.16.21.128 | 172.16.21.191 |
| B24 | 172.16.21.192/26 | 172.16.21.192 | 172.16.21.255 |
| B25 | 172.16.22.0/26 | 172.16.22.0 | 172.16.22.63 |
| B26 | 172.16.22.64/26 | 172.16.22.64 | 172.16.22.127 |
| B27 | 172.16.22.128/26 | 172.16.22.128 | 172.16.22.191 |
| B28 | 172.16.22.192/26 | 172.16.22.192 | 172.16.22.255 |
| B29 | 172.16.23.0/26 | 172.16.23.0 | 172.16.23.63 |
| B30 | 172.16.23.64/26 | 172.16.23.64 | 172.16.23.127 |
| B31 | 172.16.23.128/26 | 172.16.23.128 | 172.16.23.191 |
| B32 | 172.16.23.192/26 | 172.16.23.192 | 172.16.23.255 |
| B33 | 172.16.24.0/26 | 172.16.24.0 | 172.16.24.63 |
| B34 | 172.16.24.64/26 | 172.16.24.64 | 172.16.24.127 |
| B35 | 172.16.24.128/26 | 172.16.24.128 | 172.16.24.191 |
| B36 | 172.16.24.192/26 | 172.16.24.192 | 172.16.24.255 |
| B37 | 172.16.25.0/26 | 172.16.25.0 | 172.16.25.63 |
| B38 | 172.16.25.64/26 | 172.16.25.64 | 172.16.25.127 |
| B39 | 172.16.25.128/26 | 172.16.25.128 | 172.16.25.191 |
| B40 | 172.16.25.192/26 | 172.16.25.192 | 172.16.25.255 |
| B41 | 172.16.26.0/26 | 172.16.26.0 | 172.16.26.63 |
| B42 | 172.16.26.64/26 | 172.16.26.64 | 172.16.26.127 |
| B43 | 172.16.26.128/26 | 172.16.26.128 | 172.16.26.191 |
| B44 | 172.16.26.192/26 | 172.16.26.192 | 172.16.26.255 |
| B45 | 172.16.27.0/26 | 172.16.27.0 | 172.16.27.63 |
| B46 | 172.16.27.64/26 | 172.16.27.64 | 172.16.27.127 |
| B47 | 172.16.27.128/26 | 172.16.27.128 | 172.16.27.191 |
| B48 | 172.16.27.192/26 | 172.16.27.192 | 172.16.27.255 |
| B49 | 172.16.28.0/26 | 172.16.28.0 | 172.16.28.63 |
| B50 | 172.16.28.64/26 | 172.16.28.64 | 172.16.28.127 |
| B51 | 172.16.28.128/26 | 172.16.28.128 | 172.16.28.191 |
| B52 | 172.16.28.192/26 | 172.16.28.192 | 172.16.28.255 |
| B53 | 172.16.29.0/26 | 172.16.29.0 | 172.16.29.63 |
| B54 | 172.16.29.64/26 | 172.16.29.64 | 172.16.29.127 |
| B55 | 172.16.29.128/26 | 172.16.29.128 | 172.16.29.191 |
| B56 | 172.16.29.192/26 | 172.16.29.192 | 172.16.29.255 |
| B57 | 172.16.30.0/26 | 172.16.30.0 | 172.16.30.63 |
| B58 | 172.16.30.64/26 | 172.16.30.64 | 172.16.30.127 |
| B59 | 172.16.30.128/26 | 172.16.30.128 | 172.16.30.191 |
| B60 | 172.16.30.192/26 | 172.16.30.192 | 172.16.30.255 |
| B61 | 172.16.31.0/26 | 172.16.31.0 | 172.16.31.63 |
| B62 | 172.16.31.64/26 | 172.16.31.64 | 172.16.31.127 |
| B63 | 172.16.31.128/26 | 172.16.31.128 | 172.16.31.191 |
| B64 | 172.16.31.192/26 | 172.16.31.192 | 172.16.31.255 |
| C1 | 172.16.32.0/23 | 172.16.32.0 | 172.16.33.255 |
| C2 | 172.16.34.0/23 | 172.16.34.0 | 172.16.35.255 |
| C3 | 172.16.36.0/23 | 172.16.36.0 | 172.16.37.255 |
| C4 | 172.16.38.0/23 | 172.16.38.0 | 172.16.39.255 |
| C5 | 172.16.40.0/23 | 172.16.40.0 | 172.16.41.255 |
| C6 | 172.16.42.0/23 | 172.16.42.0 | 172.16.43.255 |
| C7 | 172.16.44.0/23 | 172.16.44.0 | 172.16.45.255 |
| C8 | 172.16.46.0/23 | 172.16.46.0 | 172.16.47.255 |
| C9 | 172.16.48.0/23 | 172.16.48.0 | 172.16.49.255 |
| C10 | 172.16.50.0/23 | 172.16.50.0 | 172.16.51.255 |
| C11 | 172.16.52.0/23 | 172.16.52.0 | 172.16.53.255 |
| C12 | 172.16.54.0/23 | 172.16.54.0 | 172.16.55.255 |
| C13 | 172.16.56.0/23 | 172.16.56.0 | 172.16.57.255 |
| C14 | 172.16.58.0/23 | 172.16.58.0 | 172.16.59.255 |
| C15 | 172.16.60.0/23 | 172.16.60.0 | 172.16.61.255 |
| C16 | 172.16.62.0/23 | 172.16.62.0 | 172.16.63.255 |
4,096+4,096+8,192=16,384 addresses.
65,536−16,384=49,152 addresses remain unallocated, from 172.16.64.0 through 172.16.255.255.
Final answer: A: /25 per lab; B: /26 per lab; C: /23 per lab. 49,152 total addresses remain unallocated.
Topology is the arrangement of nodes and links. Physical topology describes actual links; logical topology describes how data flows. A switch-centred physical star is different from a shared collision-domain bus.
| Topology | Structure | Benefit | Limitation |
|---|---|---|---|
| Bus | Devices share a common backbone | Simple small shared-medium layout | Backbone fault affects segment; collisions possible on classic shared bus |
| Star | Each device connects to a central switch/hub | Easy expansion and fault isolation | Central-device failure affects star; more cabling |
| Ring | Each node connected to two neighbours in a ring | Structured traversal; token-based variants coordinate access | Simple single-ring break can disrupt operation unless protected |
| Full mesh | Every pair has a direct link | Redundant paths and direct connectivity | Expensive ports/cabling: n(n−1)/2 undirected links |
| Tree/hybrid | Hierarchical stars or combined topologies | Scales into buildings/floors | Upper-level faults affect dependent branches |
The Domain Name System (DNS) is a distributed hierarchical naming system. Its namespace is a tree with the root at the top. A complete name is the ordered path of labels from a node back to the root. An absolute name may be written with a trailing dot, for example www.example.com.
| Level | Example | Meaning |
|---|---|---|
| Root | . | Top of the namespace |
| Top-level domain (TLD) | com, org, in | A child zone below the root |
| Registered/delegated domain | example.com | Name under a parent such as com |
| Subdomain/host label | www.example.com | A label below example.com; a name may own different record types |
| Field | Bits | Purpose |
|---|---|---|
| Version | 4 | Value 6 |
| Traffic Class | 8 | Traffic treatment/congestion-related marking |
| Flow Label | 20 | Identifies a flow for appropriate handling |
| Payload Length | 16 | Bytes after the base header; includes extension headers when present |
| Next Header | 8 | Type of the next extension header or upper-layer protocol |
| Hop Limit | 8 | Decremented on forwarding; limits packet lifetime in hops |
| Source address | 128 | Sending interface address |
| Destination address | 128 | Destination address |
The first two 32-bit rows contain 64 bits and the two addresses contain 256 bits, giving a 320-bit base header. Payload Length excludes these 40 bytes. Next Header can identify an extension header or upper-layer protocol. IPv6 has no base-header checksum; optional information is carried through extension headers.
User Datagram Protocol (UDP) provides connectionless, message-oriented process delivery. One send creates a datagram; message boundaries are retained. UDP has an 8-byte header, carries ports and length, and offers checksum-based corruption detection. It does not itself guarantee delivery, ordering, duplicate suppression, retransmission, flow control or congestion control.
| UDP field | Width | Purpose |
|---|---|---|
| Source port | 16 bits | Sending endpoint; a zero value can indicate no reply port when unused |
| Destination port | 16 bits | Receiving application endpoint |
| Length | 16 bits | UDP header plus UDP payload; minimum 8 bytes |
| Checksum | 16 bits | Error detection over UDP and an IP pseudo-header; not retransmission |
Selective Repeat (SR) ARQ accepts and buffers valid out-of-order frames inside the receiver window, acknowledges them individually and retransmits only missing/timed-out frames. In this answer ACK i means “frame i received”, unlike the next-expected cumulative ACK convention used for GBN.
Simple Mail Transfer Protocol (SMTP) transfers email from a sending client/server to a receiving mail server using a reliable transport, normally TCP. It is a push/store-and-forward transfer protocol. User agents compose/read mail; mail-transfer agents relay it, consulting DNS MX records to find recipient-domain mail servers.
| Phase | Typical exchange | Purpose |
|---|---|---|
| Connection and greeting | Server 220; client EHLO (HELO for basic SMTP); server 250 capabilities | Identify peers and learn supported extensions |
| Envelope transfer | MAIL FROM:<sender>; RCPT TO:<recipient>; 250 replies | Declare return path and one or more delivery recipients |
| Message transfer | DATA; server 354; headers + blank line + body; single dot line; server 250 | Transmit the message after server permission; completion reply accepts responsibility |
| Termination | QUIT; server 221 | Close the SMTP session cleanly |
Transmission media are classified into guided media (twisted pair, coaxial cable, optical fiber) and unguided media (radio, microwave, infrared).
Optical fiber carries data as light in a core surrounded by cladding. It is a guided medium. Single-mode fiber supports one main propagation mode and is used for long/high-capacity links; multimode supports multiple modes and is often used over shorter distances.
| Advantages | Disadvantages / design costs |
|---|---|
| High data capacity and low attenuation over suitable links | Optical transceivers, connectors and splicing add cost/skill requirements |
| Immune to electromagnetic interference; no electrical crosstalk | Tight bends, dirty connectors and poor installation increase loss |
| Electrical isolation between buildings | Fiber is mechanically sensitive and requires appropriate protection |
| Harder to tap casually than an unprotected radio channel | It is not automatically confidential; encryption/access control remain necessary |
Interpret the stated aggregate production rates as offered attempt rates in the standard textbook model. The following working includes normalized throughput, successful frames/s and successful frame bits/s for each case.
| Symbol | Given |
|---|---|
| L | 200 bits |
| R | 200 kbps = 200,000 bits/s |
| λ | 1000 aggregate attempts/s |
T, G, S, successful frames/s and bit/s.
First normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
T = 200/200,000 = 0.001 s = 1 ms.
G = 1000 × 0.001 = 1 attempts/frame time.
−2G = -2; e to that power ≈ 0.135335283.
S = 1 × 0.135335283 ≈ 0.135335283.
S/T = 0.135335283/0.001 ≈ 135.335283 frames/s.
SR = 0.135335283 × 200,000 ≈ 27067.056647 bits/s ≈ 27.067057 kbps.
Final answer: G=1, S≈0.135335, successful rate≈135.335 frames/s = 27.067 kbps.
| Symbol | Given |
|---|---|
| L | 200 bits |
| R | 200 kbps = 200,000 bits/s |
| λ | 500 aggregate attempts/s |
T, G, S, successful frames/s and bit/s.
First normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
T = 200/200,000 = 0.001 s = 1 ms.
G = 500 × 0.001 = 0.5 attempts/frame time.
−2G = -1; e to that power ≈ 0.367879441.
S = 0.5 × 0.367879441 ≈ 0.183939721.
S/T = 0.183939721/0.001 ≈ 183.939721 frames/s.
SR = 0.183939721 × 200,000 ≈ 36787.944117 bits/s ≈ 36.787944 kbps.
Final answer: G=0.5, S≈0.183940, successful rate≈183.940 frames/s = 36.788 kbps.
| Symbol | Given |
|---|---|
| L | 200 bits |
| R | 200 kbps = 200,000 bits/s |
| λ | 250 aggregate attempts/s |
T, G, S, successful frames/s and bit/s.
First normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
T = 200/200,000 = 0.001 s = 1 ms.
G = 250 × 0.001 = 0.25 attempts/frame time.
−2G = -0.5; e to that power ≈ 0.606530660.
S = 0.25 × 0.606530660 ≈ 0.151632665.
S/T = 0.151632665/0.001 ≈ 151.632665 frames/s.
SR = 0.151632665 × 200,000 ≈ 30326.532986 bits/s ≈ 30.326533 kbps.
Final answer: G=0.25, S≈0.151633, successful rate≈151.633 frames/s = 30.327 kbps.
| Point | UDP | TCP | SCTP |
|---|---|---|---|
| Service unit | Datagram; preserves messages | Byte stream; no message boundaries | Messages carried in chunks; preserves boundaries |
| Connection setup | No protocol handshake | Connection; usual three-way SYN exchange | Association; four-message cookie exchange |
| Reliability | No delivery/order guarantee | Reliable ordered stream | Reliable messages; ordered per stream by default, unordered option |
| Ordering scope | None supplied | One ordered byte stream per direction | Multiple streams; per-stream order avoids blocking every stream on one stream’s ordering gap |
| Flow control | Not supplied by UDP | Receiver-advertised window | Receiver flow control |
| Congestion control | Not supplied by UDP | Yes | Yes |
| Multihoming | No built-in association failover | Classic TCP uses one address pair | Association can bind multiple addresses; built-in path failover |
| Header | 8 bytes | Minimum 20 bytes | 12-byte common header plus chunks |
| Typical use | DNS, real-time applications with their own controls | Web over HTTP/1.1 or HTTP/2, reliable byte transfers | Telecom signalling and multi-stream message applications |
| Performance claim | Low protocol overhead, not automatically fastest overall | Recovery/ordering have costs and benefits | Extra capabilities; no universal speed ranking |
The correct protocol depends on application semantics and network conditions. “UDP is always faster” and “SCTP is always slower than TCP” are not defensible general rules. All three normally run above IP; security/encryption is a separate requirement.
| Directed edge | Weight |
|---|---|
| A→B | 4 |
| A→C | 5 |
| B→C | -2 |
| B→D | 6 |
| C→D | 2 |
| D→E | 3 |
| C→E | 8 |
The shortest route and cost, with complete passes and a negative-cycle check.
Bellman–Ford permits negative edges. We use the listed order with in-place updates; do not mix this with a table that reads only the previous pass.
![\"Bellman–Ford](\"fig/tee-mock-b-route.720ccc32b6.svg\")
5 vertices → at most 4 main passes. d[A]=0; every other distance=∞. Use the listed edge order and immediate (in-place) updates.
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| A→B | ∞ | 0+(4)=4 | 4 | Update; predecessor=A |
| A→C | ∞ | 0+(5)=5 | 5 | Update; predecessor=A |
| B→C | 5 | 4+(-2)=2 | 2 | Update; predecessor=B |
| B→D | ∞ | 4+(6)=10 | 10 | Update; predecessor=B |
| C→D | 10 | 2+(2)=4 | 4 | Update; predecessor=C |
| D→E | ∞ | 4+(3)=7 | 7 | Update; predecessor=D |
| C→E | 7 | 2+(8)=10 | 7 | No improvement |
| Vertex | A | B | C | D | E |
|---|---|---|---|---|---|
| End-of-pass distance | 0 | 4 | 2 | 4 | 7 |
| Predecessor | — | A | B | C | D |
| Edge | Old d[target] | Candidate via source | New d[target] | Decision |
|---|---|---|---|---|
| A→B | 4 | 0+(4)=4 | 4 | No improvement |
| A→C | 2 | 0+(5)=5 | 2 | No improvement |
| B→C | 2 | 4+(-2)=2 | 2 | No improvement |
| B→D | 4 | 4+(6)=10 | 4 | No improvement |
| C→D | 4 | 2+(2)=4 | 4 | No improvement |
| D→E | 7 | 4+(3)=7 | 7 | No improvement |
| C→E | 7 | 2+(8)=10 | 7 | No improvement |
| Vertex | A | B | C | D | E |
|---|---|---|---|---|---|
| End-of-pass distance | 0 | 4 | 2 | 4 | 7 |
| Predecessor | — | A | B | C | D |
This entire pass changed no distance. Further identical relaxation passes will not improve anything.
Backward predecessors: E ← D ← C ← B ← A. Forward route: A → B → C → D → E.
Cost = (4) + (-2) + (2) + (3) = 7. No edge can now improve its destination label.
Final answer: A → B → C → D → E; shortest cost=7.
![\"Bellman–Ford](\"fig/tee-mock-b-route-solution.f6bae119c4.svg\")
In the textbook leaky-bucket queue/shaper model, arriving packets enter a finite buffer. The output sends at a configured rate whenever data is queued. When the queue empties, it cannot send nonexistent packets; when the buffer is full, excess arrivals are dropped in this finite-buffer model.
| Advantage | Limitation |
|---|---|
| Simple bound on output rate | Bursts may overflow or suffer delay |
| Reduces short-term output variation | Unused idle service cannot be saved |
| Useful for predictable traffic shaping | No end-to-end delay/loss guarantee by itself |
The term “leaky bucket” can also describe a conformance meter in standards. This course’s diagrams and numerical questions explicitly use the finite data-queue model, so its state is queued data rather than tokens.
No algorithm is universally better. The leaky data-queue model is useful for a smooth output profile; token bucket is preferable when controlled bursts should be allowed. State the application’s delay, burst and rate needs before recommending one.
| Criterion | Wired campus links | Wireless campus access |
|---|---|---|
| Speed/capacity | Dedicated switch ports and fiber backbones can provide high, predictable capacity | Capacity shared over radio; performance depends on contention, coverage and interference |
| Reliability | Stable when cables/ports and redundant uplinks are healthy | Mobility and flexible access; radio interference and coverage holes can cause variability |
| Security | Physical access control helps, but ports still need authentication/segmentation | Use authenticated encryption and managed access; radio extends beyond rooms |
| Deployment | Cabling cost and fixed access points | Supports mobile users and faster coverage changes |
| Recommended role | Servers, laboratories, inter-building backbone | Students’ phones/laptops, roaming and guest access |
Choose a hybrid: wired core/backbone and critical endpoints, with managed access points providing wireless mobility. Separate guest traffic, authenticate users and plan redundant infrastructure. Neither medium alone guarantees security or availability.
The Open Systems Interconnection (OSI) model separates network functions into seven layers. A layer offers services upward and uses services below. Peer layers communicate logically through protocol rules; actual data moves down the sender’s stack, across links and up the receiver’s stack.
| Layer | Task | Example |
|---|---|---|
| 7 Application | Services used by network applications | HTTP, DNS, FTP, SMTP |
| 6 Presentation | Representation, encoding and application data transformations | Character encoding, serialization; encryption as a conceptual function |
| 5 Session | Manage dialogue and synchronization/checkpoints | Session establishment and resumption concepts |
| 4 Transport | Process delivery; reliability/flow control where supplied | TCP, UDP, SCTP; ports |
| 3 Network | Logical addressing and routing across networks | IP; routers |
| 2 Data link | Framing, local link addressing, medium access and error detection | Ethernet/Wi-Fi; MAC addresses |
| 1 Physical | Transmit signal symbols/bits through a medium | Electrical, optical or radio signals |
Protocol = Pure ALOHA; G = 0.8 attempts/frame time.
S, 100S and the maximum possible S.
Select the formula from the protocol name. The exponent is −2G because Pure ALOHA has a 2T vulnerable period.
−2 × 0.8 = -1.6.
![\"e^{-1.6}](\"math/e7173ebbd6128f2f.svg\")
S = 0.8 × 0.201896518 ≈ 0.161517214 successful frames/frame time.
0.161517214 × 100 ≈ 16.1517%.
The maximum is 1/(2e) ≈ 0.183940 = 18.3940%, at G=0.5.
Final answer: S ≈ 0.161517; percentage ≈ 16.1517%; Smax ≈ 0.183940 at G=0.5.
IP/prefix = 172.18.7.33/23.
The complete containing address block and its usable host range.
Use the explicit prefix. We show both the interval method and bitwise AND so the boundary can be understood.
h=32−23=9; N=29=512.
Mask=255.255.254.0; binary=11111111.11111111.11111110.00000000.
The changing octet is number 3. Its mask value is 254, so step=256−254=2. The IP value 7 lies in 6…7.
Keep earlier octets; place 6 in octet 3; set any later octets to zero. Network=172.18.6.0.
| Octet | IP bits | Mask bits | AND result | Decimal |
|---|---|---|---|---|
| 1 | 10101100 | 11111111 | 10101100 | 172 |
| 2 | 00010010 | 11111111 | 00010010 | 18 |
| 3 | 00000111 | 11111110 | 00000110 | 6 |
| 4 | 00100001 | 00000000 | 00000000 | 0 |
Use 7 in the changing octet and 255 in any later octets. Broadcast=172.18.7.255.
First usable=172.18.6.1; last usable=172.18.7.254; count=512−2=510.
Final answer: Network 172.18.6.0; broadcast 172.18.7.255; 512 total, 510 ordinary usable; range 172.18.6.1–172.18.7.254.
Block coding maps each k-bit dataword to an n-bit codeword, adding n−k redundant bits. Only selected codewords are valid; their separation determines error-detection/correction capability.
| Codeword | Bits |
|---|---|
| C1 | 0000000 |
| C2 | 1110000 |
| C3 | 0011100 |
All pair distances, dmin and the maximum guaranteed numbers of errors.
XOR exposes differing positions. Use the closest codeword pair, because it is the weakest separation in the code.
| Pair | First | Second | XOR | Distance |
|---|---|---|---|---|
| C1 vs C2 | 0000000 | 1110000 | 1110000 | 3 |
| C1 vs C3 | 0000000 | 0011100 | 0011100 | 3 |
| C2 vs C3 | 1110000 | 0011100 | 1101100 | 4 |
dmin = min(3, 3, 4) = 3.
Detect = 3−1 = 2; correct = floor((3−1)/2) = 1.
3 bit changes can reach another valid codeword, so only 2 are guaranteed detectable. Correction regions require 2t < 3.
Final answer: dmin=3; detect up to 2 errors; correct up to 1 errors.
A setup phase reserves a circuit through intermediate switches. Data then streams through the established path until release. It offers predictable reserved service and low per-unit forwarding overhead, but setup takes time and idle periods waste reserved capacity. A traditional dedicated voice circuit is an appropriate illustration.
Each intermediate node receives and stores the entire message before forwarding it. Links can be shared and no end-to-end circuit is required, but large messages need buffers and cause substantial store-and-forward delay. It illustrates delay-tolerant whole-message delivery, rather than an efficient choice for interactive office traffic.
A message is segmented into packets, allowing shared links and pipelined forwarding. This suits bursty data and supports flexible routing. Costs include packet headers, variable queueing delay, potential packet loss and possible reordering in datagram networks.
| Aspect | Circuit switching | Message switching | Packet switching |
|---|---|---|---|
| Unit handled | Continuous stream on established circuit | Whole message | Small packets |
| Resource use | Capacity reserved along a circuit | Shared; message stored at each intermediate node | Shared; each packet stored/forwarded in this model |
| Setup | Required before data transfer | No circuit setup required | Datagram: no connection setup; virtual circuit: setup |
| Delay | Setup delay, then relatively predictable service | Large store-and-forward delay for long messages | Variable delay; packet pipeline can reduce completion time |
| Main limitation | Reserved capacity can be idle | Large buffers; unsuitable for tight delay needs | Headers, possible queueing/loss; datagrams may reorder |
For ordinary office file, web and collaboration traffic, choose packet switching, with adequate capacity and traffic management. Its statistical sharing serves bursty sources efficiently. If strict deterministic service is required, state that requirement and discuss additional service engineering rather than promising that every packet network guarantees timing.
A computer network is a set of interconnected devices that exchange data and share resources using communication protocols. A link carries signals; a protocol defines message format, meaning and timing. Hosts run applications, switches forward local frames and routers connect IP networks.
| Type | Meaning and scope | Example |
|---|---|---|
| PAN | Personal Area Network: a person’s nearby devices | Phone linked to earbuds |
| LAN | Local Area Network: a building or local campus network under local administration | Laboratory Ethernet/Wi-Fi |
| MAN | Metropolitan Area Network: interconnects sites across a town/city | City-wide institutional network |
| WAN | Wide Area Network: connects geographically separated networks over regional/global links | Branches in different cities |
| ARPANET | Historical packet-switched research network; a major predecessor of the Internet | Early interconnected research institutions; not a modern geographic size category |
Classification by geographical scope is descriptive: a fixed kilometre boundary or a guaranteed speed is not part of every definition. Networks can also be classified by wired/wireless medium or by client-server/peer-to-peer organization.
| Packet type | What it does | Simple example |
|---|---|---|
| Hello | Discover/maintain neighbours and exchange relationship parameters | Are we still neighbours? |
| Database Description (DD/DBD) | Summarize link-state database contents during synchronization | Here are the LSA headers I have |
| Link State Request (LSR) | Ask for specific missing/newer link-state information | Please send this LSA |
| Link State Update (LSU) | Carry one or more LSAs | Here is the requested or newly flooded link-state information |
| Link State Acknowledgment (LSAck) | Acknowledge receipt of LSAs | I received these LSAs |
Database Description packets carry summaries; Link State Update packets carry LSAs themselves. Do not confuse an LSA type with an OSPF packet type.
Simple Mail Transfer Protocol (SMTP) transfers email from a sending client/server to a receiving mail server using a reliable transport, normally TCP. It is a push/store-and-forward transfer protocol. User agents compose/read mail; mail-transfer agents relay it, consulting DNS MX records to find recipient-domain mail servers.
| Phase | Typical exchange | Purpose |
|---|---|---|
| Connection and greeting | Server 220; client EHLO (HELO for basic SMTP); server 250 capabilities | Identify peers and learn supported extensions |
| Envelope transfer | MAIL FROM:<sender>; RCPT TO:<recipient>; 250 replies | Declare return path and one or more delivery recipients |
| Message transfer | DATA; server 354; headers + blank line + body; single dot line; server 250 | Transmit the message after server permission; completion reply accepts responsibility |
| Termination | QUIT; server 221 | Close the SMTP session cleanly |
Given address: 0:15::1:12:1213.
The full 128-bit address in eight-group hexadecimal notation.
A written 0 is a visible group. :: represents only the omitted all-zero groups.
Left: 0, 15 (2 group(s)). Right: 1, 12, 1213 (3 group(s)).
Visible=2+3=5. Missing=8−5=3.
0:15:0:0:0:1:12:1213.
| Group number | Four-digit group |
|---|---|
| 1 | 0000 |
| 2 | 0015 |
| 3 | 0000 |
| 4 | 0000 |
| 5 | 0000 |
| 6 | 0001 |
| 7 | 0012 |
| 8 | 1213 |
0000:0015:0000:0000:0000:0001:0012:1213.
Final answer: 0000:0015:0000:0000:0000:0001:0012:1213
| Feature | IPv4 | IPv6 |
|---|---|---|
| Address width | 32 bits | 128 bits |
| Base/header size | 20–60 bytes depending on options | 40-byte fixed base header |
| Length field | Total length includes IPv4 header | Payload Length excludes 40-byte base header |
| Lifetime field | TTL | Hop Limit |
| Following protocol | Protocol field | Next Header; may identify an extension |
| Header checksum | Present | No base-header checksum |
| Options | Inside IPv4 header | Typically extension headers |
| Fragmentation fields | In IPv4 header; routers may fragment if permitted | Fragment extension header; fragmentation by source, not routers |
The supplied scheme’s sentence says four missing groups, but five groups are already visible, so only three are missing. Its printed eight-group expansion is consistent with three. The count shown above is independently verified.
| Point | UDP | TCP | SCTP |
|---|---|---|---|
| Service unit | Datagram; preserves messages | Byte stream; no message boundaries | Messages carried in chunks; preserves boundaries |
| Connection setup | No protocol handshake | Connection; usual three-way SYN exchange | Association; four-message cookie exchange |
| Reliability | No delivery/order guarantee | Reliable ordered stream | Reliable messages; ordered per stream by default, unordered option |
| Ordering scope | None supplied | One ordered byte stream per direction | Multiple streams; per-stream order avoids blocking every stream on one stream’s ordering gap |
| Flow control | Not supplied by UDP | Receiver-advertised window | Receiver flow control |
| Congestion control | Not supplied by UDP | Yes | Yes |
| Multihoming | No built-in association failover | Classic TCP uses one address pair | Association can bind multiple addresses; built-in path failover |
| Header | 8 bytes | Minimum 20 bytes | 12-byte common header plus chunks |
| Typical use | DNS, real-time applications with their own controls | Web over HTTP/1.1 or HTTP/2, reliable byte transfers | Telecom signalling and multi-stream message applications |
| Performance claim | Low protocol overhead, not automatically fastest overall | Recovery/ordering have costs and benefits | Extra capabilities; no universal speed ranking |
The correct protocol depends on application semantics and network conditions. “UDP is always faster” and “SCTP is always slower than TCP” are not defensible general rules. All three normally run above IP; security/encryption is a separate requirement.
Go-Back-N (GBN) ARQ pipelines up to W outstanding frames. For m-bit sequence numbers, textbook GBN uses W ≤ 2m−1 and receiver window 1. The receiver accepts only the next in-order frame; out-of-order frames are discarded. ACK k here cumulatively acknowledges all frames before k and means “next expected is k”.
The Domain Name System (DNS) is a distributed hierarchical naming system. Its namespace is a tree with the root at the top. A complete name is the ordered path of labels from a node back to the root. An absolute name may be written with a trailing dot, for example www.example.com.
| Level | Example | Meaning |
|---|---|---|
| Root | . | Top of the namespace |
| Top-level domain (TLD) | com, org, in | A child zone below the root |
| Registered/delegated domain | example.com | Name under a parent such as com |
| Subdomain/host label | www.example.com | A label below example.com; a name may own different record types |
| Type | Characteristic 1 | Characteristic 2 |
|---|---|---|
| Radio | Can provide broad-area/mobility support depending on band | Shared spectrum/interference and obstacles affect reception |
| Terrestrial microwave | Directional, usually line of sight | Needs antenna alignment; obstructions/weather can matter |
| Satellite microwave | Relay covers large or remote areas | Delay/capacity depend strongly on orbit/system |
| Infrared | Short range and generally blocked by opaque walls | Limited room-scale coverage reduces inter-room interference |
Many classifications group satellite links under microwave rather than making a separate spectral category. State that grouping consistently.
| Rule | When idle | When busy | Trade-off |
|---|---|---|---|
| 1-persistent | Transmit immediately | Keep listening; transmit as soon as idle | Low idle delay, but waiting stations may start together |
| Non-persistent | Transmit | Wait a random time, then sense again | Less synchronized contention; may leave avoidable idle time |
| p-persistent (slotted) | Transmit with probability p; defer a slot with probability 1−p | Wait until available, then resume the probabilistic slot rule | Controls contention; needs slots and a choice of p |
All three sense the channel; their persistence rules differ. Propagation delay means carrier sensing cannot guarantee the absence of collisions. In p-persistent CSMA, each eligible idle-slot decision uses the specified probability.
Congestion occurs when offered traffic overloads network resources such as link capacity, router buffers or processing. Queueing delay grows, packets may be dropped, and excessive retransmission can make the overload worse. Flow control protects a receiver; congestion control protects the network.
| Approach | Policy / mechanism | How it helps |
|---|---|---|
| Open loop: prevent congestion without runtime congestion feedback | Retransmission policy | Carefully chosen timers/selective retransmission avoid needless extra traffic |
| Open loop | Window policy | Limit outstanding data and avoid excessive retransmission after loss |
| Open loop | Acknowledgement policy | Avoid unnecessarily frequent ACK traffic while retaining correct recovery |
| Open loop | Admission policy | Refuse new load when reserved capacity is insufficient |
| Closed loop: measure and react to congestion | Backpressure | A congested node tells upstream nodes to reduce/stop sending, possibly propagating hop by hop |
| Closed loop | Choke/control signalling | A congested node sends explicit feedback to the source |
| Closed loop | Implicit signalling | A sender infers congestion from loss/delay and changes its sending behaviour |
| Closed loop | Explicit marking | Network marks congestion (for example ECN); endpoints react to returned information |
For an answer asking for two of each: explain retransmission and admission policies under open loop, then backpressure and explicit signalling under closed loop, with the mechanism and its effect. These are general congestion-control classifications used in the course; they are not all mandatory algorithms implemented inside TCP. TCP commonly uses end-to-end feedback such as loss or ECN.
| Edge | Cost |
|---|---|
| A→B | 5 |
| A→C | 2 |
| B→D | 4 |
| C→B | 1 |
| C→D | 7 |
| D→E | 2 |
| C→E | 10 |
Source=A; target=E. Edges are directed; do not add reverse edges.
The path A to E, its total cost and intermediate work.
Use Dijkstra because all weights are nonnegative. Select the smallest unsettled total distance, not the cheapest isolated edge.
![\"Dijkstra](\"fig/tee-mock-c-route.93b735aa12.svg\")
d[A]=0; every other distance=∞. No predecessors are known.
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Tentative |
| B | ∞ | — | Tentative |
| C | ∞ | — | Tentative |
| D | ∞ | — | Tentative |
| E | ∞ | — | Tentative |
Unsettled candidates before selection: A=0, B=∞, C=∞, D=∞, E=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| A→B | ∞ | 0+(5)=5 | Improve to 5; predecessor=A |
| A→C | ∞ | 0+(2)=2 | Improve to 2; predecessor=A |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 5 | A | Tentative |
| C | 2 | A | Tentative |
| D | ∞ | — | Tentative |
| E | ∞ | — | Tentative |
Unsettled candidates before selection: B=5, C=2, D=∞, E=∞. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| C→B | 5 | 2+(1)=3 | Improve to 3; predecessor=C |
| C→D | ∞ | 2+(7)=9 | Improve to 9; predecessor=C |
| C→E | ∞ | 2+(10)=12 | Improve to 12; predecessor=C |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | C | Tentative |
| C | 2 | A | Settled |
| D | 9 | C | Tentative |
| E | 12 | C | Tentative |
Unsettled candidates before selection: B=3, D=9, E=12. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| B→D | 9 | 3+(4)=7 | Improve to 7; predecessor=B |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | C | Settled |
| C | 2 | A | Settled |
| D | 7 | B | Tentative |
| E | 12 | C | Tentative |
Unsettled candidates before selection: D=7, E=12. Select the minimum; ties use alphabetical order.
| Edge | Old distance | Candidate | Decision |
|---|---|---|---|
| D→E | 12 | 7+(2)=9 | Improve to 9; predecessor=D |
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | C | Settled |
| C | 2 | A | Settled |
| D | 7 | B | Settled |
| E | 9 | D | Tentative |
Unsettled candidates before selection: E=9. Select the minimum; ties use alphabetical order.
No outgoing edges to relax.
| Vertex | Distance from A | Predecessor | State |
|---|---|---|---|
| A | 0 | — | Settled |
| B | 3 | C | Settled |
| C | 2 | A | Settled |
| D | 7 | B | Settled |
| E | 9 | D | Settled |
Trace backwards: E ← D ← B ← C ← A. Reverse it: A → C → B → D → E.
Edge sum: 2 + 1 + 4 + 2 = 9.
Final answer: A → C → B → D → E; cost=9.
![\"Dijkstra](\"fig/tee-mock-c-route-solution.94d22e6eb0.svg\")
A token bucket generates tokens at rate r up to capacity B. Sending one unit of data consumes the corresponding tokens. Tokens accumulate during idle periods only until the bucket is full. Data can therefore burst when saved tokens exist, subject to the physical link rate.
| Limitation | Reason |
|---|---|
| Bursts can still congest downstream queues | Local conformance does not ensure every later link has capacity |
| Parameters must match traffic and link requirements | A small B delays/drops legitimate packets; a large B permits disruptive bursts |
| No complete QoS guarantee alone | Admission, scheduling and end-to-end capacity may still be needed |
If time is short, use the source coverage map to find weak topics. Syllabus hours and repeated questions indicate emphasis, but they do not justify skipping a unit or promising a future question.
Yaad rakho: Layer ka role samjho; numerical mein units aur assumptions pehle.
"},{"id":"u12","title":"Units 1–2: five-mark answer frames","html":"| Prompt | Write this structure |
|---|---|
| Compare models | Define each → matched rows for performance/scaling/reliability → one scenario recommendation |
| Explain topology | Definition → labelled diagram → operation → advantage → failure/limitation |
| OSI / TCP-IP | Layer order → one distinct task each → protocol examples → encapsulation |
| Transmission media | Guided versus unguided → named types → two characteristics each → suitable application |
| Campus design | Wired backbone and endpoints + wireless mobility → capacity/reliability/security tradeoffs |
Avoid absolute statements such as “wireless is always slow”, “fiber cannot be tapped”, “P2P has no failures” or “a server always fails the entire system”. Explain the conditions and redundancy.
"},{"id":"u3","title":"Unit 3: error and access traps","html":"| Trap | Correction |
|---|---|
| Hamming distance calculated by decimal subtraction | XOR equal-length strings and count 1s |
| Pick largest pairwise distance | Find the minimum of every distinct pair |
| Detect and correct counts assumed identical | Detect dmin−1; correct floor((dmin−1)/2) |
| CRC generator has four bits, so append four zeros | Append degree r, one fewer than generator bit count |
| CRC receiver appends more zeros | Divide the received complete frame directly |
| Every lost ACK forces immediate retransmission | A later cumulative ACK may acknowledge the missing earlier ACK in GBN |
| SR sender skips a missing base after a later ACK | Window base cannot pass its earliest unacknowledged frame |
| GBN and SR ACK labels interpreted identically | State cumulative next-expected versus individual-frame convention |
| Use S as probability per attempted frame | S is successful frames per frame time; success probability is S/G |
| Carrier sense guarantees no collisions | Propagation delay creates an uncertainty interval |
| CSMA/CA called collision-free | Avoidance reduces risk; collisions remain possible |
For the actual final graph the path is A→C→D→F→G, cost 12. For the actual re-exam graph it is A→D→C→B→E→G, cost 4. Memorizing those numbers is not enough for a changed graph: show the table.
"},{"id":"u5","title":"Unit 5: distinguish the quantities","html":"| Pair | Difference |
|---|---|
| IP address vs port | Interface/network delivery vs process endpoint |
| TCP vs UDP | Ordered byte stream with recovery vs datagrams without built-in delivery recovery |
| rwnd vs cwnd | Receiver capacity vs network congestion allowance |
| Flow control vs congestion control | Receiver protection vs shared network protection |
| Open loop vs closed loop | Prevention without runtime congestion feedback vs measurement/feedback/reaction |
| Delay vs jitter | Time taken vs variation in time taken |
| Leaky queue vs token bucket | Stores data to smooth output vs stores permission to allow bounded bursts |
| Shaper vs policer | Delays nonconforming traffic vs measures and applies mark/drop policy |
| SCTP streams vs multihoming | Separate message-ordering streams vs alternate endpoint addresses/paths |
For the observed TCP question, draw establishment, data transfer and termination, and explain the arrows. Separate sequence-number calculation drills are optional; they were not asked in the supplied papers.
"},{"id":"u6","title":"Unit 6: redraw four message flows","html":"| Protocol | Draw and explain |
|---|---|
| HTTP | Browser → DNS → server connection → request(method,target,headers) → response(status,headers,body) → render/extra objects |
| DNS | Root → TLD → domain/subdomain; stub → recursive resolver → referrals → authoritative record; explain full-name uniqueness |
| FTP | Control TCP21 persists; separate data connection; active server initiation versus passive client initiation |
| SMTP | 220/EHLO → MAIL FROM → RCPT TO → DATA/354 → message/dot/250 → QUIT/221; recipient later retrieves via another protocol |
A good 10-mark answer has labelled actors and arrow directions, the ordered mechanism, important fields/commands and one practical example. Keep security and version qualifications precise without replacing the main mechanism.
"}]}],"about":[{"id":"tee-about","title":"TEE coverage and source notes","scope":"TEE · Evidence and scope limits · current syllabus 702AI0C023","sections":[{"id":"coverage","title":"Every named syllabus topic has a teaching location","html":"| Unit | Current syllabus topic | Read |
|---|---|---|
| 1 | Computer network | Open lesson |
| 1 | Peer-to-peer and client-server communication | Open lesson |
| 1 | Classifications of computer networks | Open lesson |
| 1 | Network topologies | Open lesson |
| 2 | OSI model | Open lesson |
| 2 | TCP/IP model | Open lesson |
| 2 | Transmission media | Open lesson |
| 3 | Fundamentals of error detection and correction | Open lesson |
| 3 | Block coding | Open lesson |
| 3 | Hamming distance | Open lesson |
| 3 | Flow control and error control | Open lesson |
| 3 | CRC | Open lesson |
| 3 | Stop and Wait ARQ | Open lesson |
| 3 | Go-Back-N ARQ | Open lesson |
| 3 | Selective Repeat ARQ | Open lesson |
| 3 | Pure ALOHA | Open lesson |
| 3 | Slotted ALOHA | Open lesson |
| 3 | CSMA/CD | Open lesson |
| 3 | Printed CDMA/CA, interpreted CSMA/CA | Open lesson |
| 4 | Switching techniques | Open lesson |
| 4 | IPv4 addressing | Open lesson |
| 4 | Subnet mask | Open lesson |
| 4 | CIDR | Open lesson |
| 4 | IPv6 | Open lesson |
| 4 | ARP | Open lesson |
| 4 | RARP | Open lesson |
| 4 | DHCP | Open lesson |
| 4 | Shortest-path routing | Open lesson |
| 4 | RIP | Open lesson |
| 4 | Bellman–Ford | Open lesson |
| 4 | Link-state routing | Open lesson |
| 4 | Dijkstra | Open lesson |
| 4 | OSPF | Open lesson |
| 5 | Process-to-process communication | Open lesson |
| 5 | UDP services and operation | Open lesson |
| 5 | TCP features | Open lesson |
| 5 | TCP three-way handshaking | Open lesson |
| 5 | UDP/TCP comparison | Open lesson |
| 5 | SCTP | Open lesson |
| 5 | Open-loop and closed-loop congestion control | Open lesson |
| 5 | Quality of Service | Open lesson |
| 5 | QoS improving techniques | Open lesson |
| 5 | Leaky bucket | Open lesson |
| 5 | Token bucket | Open lesson |
| 6 | HTTP | Open lesson |
| 6 | DNS | Open lesson |
| 6 | FTP | Open lesson |
| 6 | SMTP | Open lesson |
This matrix checks explicit named-topic coverage, not guaranteed examination outcomes. Supporting material includes Hamming-code construction, mesh counting, ideal ARQ utilization, TCP sequence arithmetic, bucket calculations, HTTP timing and IPv4 fragmentation. They develop the core concepts and are labelled where assumptions or scope go beyond the named syllabus.
Each actual paper contains 18 solved parts, all retained here. Each authored TEE mock presents the historical seven-question choice structure: 140 marks of offered questions, of which 100 are attempted by following Q1 plus four choices.
"},{"id":"sources","title":"Sources, verification and limitations","html":"Primary scope: Syllabus_BTech&MBATech AI_SEM V_AY 26-27_AI subjects_9.6.26_NEW CODE.pdf, CN pages 1–2, course 702AI0C023. It lists six units and 100 TEE marks scaled to 50. Unit durations sum to 30 hours.
Primary questions: QP_Final-Exam_Computer Networks(702AI0C007)_Semester V_2025-2026.pdf, dated 5 December 2025; QP_Re-Exam_Computer Networks(702AI0C007)_Semester V_2025-2026.pdf, dated 4 February 2026; associated answer schemes from the supplied Computer Networks archives. Question wording is normalized for legible typography, with original data, parts and marks retained.
Secondary material: the supplied CN Master/English answer-book is background reference. Its size, predictions and topics outside the syllabus do not establish scope or guarantee marks. The course does not claim that every page of that large secondary collection was independently checked.
Checks: each named syllabus item has a route; both actual papers have all 18 parts; each mock has seven 20-mark groups. Arithmetic is checked separately using bit-string/polynomial routines, integer address boundaries, graph oracles and independent queue/token balances. Rendered assets, links and HTML structure are checked. These are academic and technical checks, not instructor approval.
Read the explicit answer-key corrections
The historical papers’ course code and dates differ from the current syllabus. The current future-paper time, choices and unit weighting require the current instructor’s instructions; no separate current CN policy was supplied. The standard CSMA/CA interpretation of printed CDMA/CA is disclosed. No claim of guaranteed marks, predicted questions or complete validation of every possible network edge case is made.
These exercises explain related mechanisms but their calculation formats do not occur in the two supplied CN papers. They are kept separate from the main TEE reading order and mocks. Use them only after the syllabus explanations and actual papers, or when a specific concept needs another example.
Hamming distance, CRC, ALOHA, address blocks/subnets, IPv6 expansion, SR windows and shortest paths do have numerical support in the papers. Hamming-code construction, HTTP timing and TCP sequence arithmetic are different question types; one does not prove the other was asked.
"}]},{"id":"introduction","title":"Unit 1 · Networks and topology · optional practice","scope":"Optional extension · no matching calculation in the two supplied papers","sections":[{"id":"mesh-1","title":"Topology counting · example 1","html":"Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
n=4; links are undirected and no self-links exist.
Number of distinct links and per-node ports.
![\"L=\\frac{n(n-1)}{2},\\qquad](\"math/5de641c579576dbf.svg\")
Each node connects to n−1 others; counting endpoints counts each link twice.
4−1=3 ports.
4×3÷2=6 links.
6×2=12 link endpoints, equal to 4 nodes × 3 ports.
Final answer: 6 links; 3 ports per node.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
n=6; links are undirected and no self-links exist.
Number of distinct links and per-node ports.
Each node connects to n−1 others; counting endpoints counts each link twice.
6−1=5 ports.
6×5÷2=15 links.
15×2=30 link endpoints, equal to 6 nodes × 5 ports.
Final answer: 15 links; 5 ports per node.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
n=8; links are undirected and no self-links exist.
Number of distinct links and per-node ports.
Each node connects to n−1 others; counting endpoints counts each link twice.
8−1=7 ports.
8×7÷2=28 links.
28×2=56 link endpoints, equal to 8 nodes × 7 ports.
Final answer: 28 links; 7 ports per node.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
A basic binary Hamming code uses parity checks whose outcomes identify a single incorrect bit. For m data bits and r check bits, choose the smallest r satisfying 2r ≥ m+r+1. The syndrome must represent each of the m+r possible single-error positions and the no-error state.
![\"2^r\\ge](\"math/ecdbdeb0433d59c5.svg\")
![\"Hamming](\"fig/tee-hamming.509c933bba.svg\")
Yaad rakho: Parity powers of two; syndrome binary tells the position.
"},{"id":"hamming-code-1","title":"Single-error correction · example 1","html":"Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Data=1011; even parity; error position=5; only one bit flips.
Redundant bits, complete codeword and syndrome correction.
![\"\\mathrm{syndrome}=s_1+2s_2+4s_4+\\cdots\"](\"math/a38011b407dfa547.svg\")
Parity positions correspond to binary index bits, so a single flipped position changes precisely its index checks.
m=4; r=3: 23=8 ≥ 4+3+1=8; one fewer parity bit fails.
| Check | Covered positions (including check) | Parity bit |
|---|---|---|
| p1 | 1, 3, 5, 7 | 0 |
| p2 | 2, 3, 6, 7 | 1 |
| p4 | 4, 5, 6, 7 | 0 |
| Position | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|
| Encoded bit | 0 | 1 | 1 | 0 | 0 | 1 | 1 |
Received = 0110111
| Syndrome check | Received-bit XOR | Result |
|---|---|---|
| s1 | 0 XOR 1 XOR 1 XOR 1 | 1 |
| s2 | 1 XOR 1 XOR 1 XOR 1 | 0 |
| s4 | 0 XOR 1 XOR 1 XOR 1 | 1 |
Syndrome integer=5. Flip position 5; recovered codeword=0110011; extracted data=1011.
Final answer: Codeword 0110011; syndrome 5; corrected data 1011.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Data=0110; even parity; error position=3; only one bit flips.
Redundant bits, complete codeword and syndrome correction.
Parity positions correspond to binary index bits, so a single flipped position changes precisely its index checks.
m=4; r=3: 23=8 ≥ 4+3+1=8; one fewer parity bit fails.
| Check | Covered positions (including check) | Parity bit |
|---|---|---|
| p1 | 1, 3, 5, 7 | 1 |
| p2 | 2, 3, 6, 7 | 1 |
| p4 | 4, 5, 6, 7 | 0 |
| Position | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|
| Encoded bit | 1 | 1 | 0 | 0 | 1 | 1 | 0 |
Received = 1110110
| Syndrome check | Received-bit XOR | Result |
|---|---|---|
| s1 | 1 XOR 1 XOR 1 XOR 0 | 1 |
| s2 | 1 XOR 1 XOR 1 XOR 0 | 1 |
| s4 | 0 XOR 1 XOR 1 XOR 0 | 0 |
Syndrome integer=3. Flip position 3; recovered codeword=1100110; extracted data=0110.
Final answer: Codeword 1100110; syndrome 3; corrected data 0110.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Data=1101; even parity; error position=7; only one bit flips.
Redundant bits, complete codeword and syndrome correction.
Parity positions correspond to binary index bits, so a single flipped position changes precisely its index checks.
m=4; r=3: 23=8 ≥ 4+3+1=8; one fewer parity bit fails.
| Check | Covered positions (including check) | Parity bit |
|---|---|---|
| p1 | 1, 3, 5, 7 | 1 |
| p2 | 2, 3, 6, 7 | 0 |
| p4 | 4, 5, 6, 7 | 0 |
| Position | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|
| Encoded bit | 1 | 0 | 1 | 0 | 1 | 0 | 1 |
Received = 1010100
| Syndrome check | Received-bit XOR | Result |
|---|---|---|
| s1 | 1 XOR 1 XOR 1 XOR 0 | 1 |
| s2 | 0 XOR 1 XOR 0 XOR 0 | 1 |
| s4 | 0 XOR 1 XOR 0 XOR 0 | 1 |
Syndrome integer=7. Flip position 7; recovered codeword=1010101; extracted data=1101.
Final answer: Codeword 1010101; syndrome 7; corrected data 1101.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Assume an error-free link, constant data-frame transmission time Tt, one-way propagation Tp, ACK transmission Ta, negligible processing and no queueing. One stop-and-wait cycle takes Tt+2Tp+Ta. Useful data occupies the forward link for Tt of that cycle.
![\"U_{SW}=\\frac{T_t}{T_t+2T_p+T_a}\"](\"math/0ed523dd39640c27.svg\")
![\"U_W\\approx\\min\\left(1,\\frac{WT_t}{T_t+2T_p+T_a}\\right)\"](\"math/041cbb2263043e29.svg\")
![\"W_{\\mathrm{required}}=\\left\\lceil\\frac{T_t+2T_p+T_a}{T_t}\\right\\rceil\"](\"math/c54f515ee0c2f8e8.svg\")
These are ideal teaching bounds for equal frames and a sufficiently fast independent ACK path. Loss, ACK scheduling, headers and a limited sequence space reduce actual performance. A calculated window must also satisfy the selected ARQ protocol’s sequence-number bound.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Tt=1 ms; Tp=5 ms; Ta=0 ms; W=4; no processing/queueing delay.
Two utilizations and the required window.
![\"C=T_t+2T_p+T_a,\\quad](\"math/46d5773e25e12379.svg\")
Both propagation directions matter. All durations already share the same unit.
C=1+2×5+0=11 ms.
U=1/11=0.090909=9.09%.
U=min(1,4×1/11)=0.363636=36.36%.
Wrequired=ceil(11/1)=11; check the sequence-space bound separately.
Final answer: Stop-and-Wait=9.09%; window-4=36.36%; ideal required window=11.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Tt=2 ms; Tp=3 ms; Ta=0 ms; W=4; no processing/queueing delay.
Two utilizations and the required window.
Both propagation directions matter. All durations already share the same unit.
C=2+2×3+0=8 ms.
U=2/8=0.250000=25.00%.
U=min(1,4×2/8)=1.000000=100.00%.
Wrequired=ceil(8/2)=4; check the sequence-space bound separately.
Final answer: Stop-and-Wait=25.00%; window-4=100.00%; ideal required window=4.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Tt=4 ms; Tp=8 ms; Ta=1 ms; W=6; no processing/queueing delay.
Two utilizations and the required window.
Both propagation directions matter. All durations already share the same unit.
C=4+2×8+1=21 ms.
U=4/21=0.190476=19.05%.
U=min(1,6×4/21)=1.000000=100.00%.
Wrequired=ceil(21/4)=6; check the sequence-space bound separately.
Final answer: Stop-and-Wait=19.05%; window-6=100.00%; ideal required window=6.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
R=10 Mbps; Tp=25.6 μs.
Lmin in bits and the minimum whole-byte length.
The frame must last at least a round trip. Convert Mbps to bits/s and μs to seconds before multiplying.
R=10,000,000 bits/s; Tp=0.0000256 s.
2Tp=2×25.6=51.2 μs.
Lmin=2×10,000,000×0.0000256=512 bits.
512/8=64; round upward if needed → 64 bytes.
Final answer: 512 bits, or at least 64 whole bytes.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
R=100 Mbps; Tp=2 μs.
Lmin in bits and the minimum whole-byte length.
The frame must last at least a round trip. Convert Mbps to bits/s and μs to seconds before multiplying.
R=100,000,000 bits/s; Tp=0.000002 s.
2Tp=2×2=4 μs.
Lmin=2×100,000,000×0.000002=400 bits.
400/8=50; round upward if needed → 50 bytes.
Final answer: 400 bits, or at least 50 whole bytes.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
R=10 Mbps; Tp=10 μs.
Lmin in bits and the minimum whole-byte length.
The frame must last at least a round trip. Convert Mbps to bits/s and μs to seconds before multiplying.
R=10,000,000 bits/s; Tp=0.00001 s.
2Tp=2×10=20 μs.
Lmin=2×10,000,000×0.00001=200 bits.
200/8=25; round upward if needed → 25 bytes.
Final answer: 200 bits, or at least 25 whole bytes.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
n=2; K=3; Tslot=51.2 μs.
The random range and the delay for the selected K.
Count starts at zero for K. The selected K must fit the allowed range.
k=min(2,10)=2.
22=4; largest K=4−1=3. Range: 0 to 3.
3 is within 0…3.
3×51.2=153.6 μs.
Final answer: K may be 0…3; selected delay=153.6 μs.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
n=3; K=5; Tslot=51.2 μs.
The random range and the delay for the selected K.
Count starts at zero for K. The selected K must fit the allowed range.
k=min(3,10)=3.
23=8; largest K=8−1=7. Range: 0 to 7.
5 is within 0…7.
5×51.2=256 μs.
Final answer: K may be 0…7; selected delay=256 μs.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
n=5; K=17; Tslot=5.12 μs.
The random range and the delay for the selected K.
Count starts at zero for K. The selected K must fit the allowed range.
k=min(5,10)=5.
25=32; largest K=32−1=31. Range: 0 to 31.
17 is within 0…31.
17×5.12=87.04 μs.
Final answer: K may be 0…31; selected delay=87.04 μs.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Definition. Store-and-forward requires a node to receive the complete forwarding unit before sending it onward; pipelining allows successive units to occupy different links simultaneously.
Simple meaning. Pehla packet next link pe jaa raha hai, tab doosra packet first link pe aa sakta hai.
Packet switching can pipeline smaller units across successive links. After the first packet is forwarded, the next packet can use an earlier link while the first uses a later link. Whole-message store-and-forward must receive the entire message at each node, preventing that packet-level pipeline.
Yaad rakho: First packet h times; remaining packets one time each.
"},{"id":"delay-1","title":"Compare completion times · example 1","html":"Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
| Given | Value |
|---|---|
| M | 12000 bits |
| h | 3 |
| k | 6 |
| R | 1,000,000 bits/s |
| τ | 0.002 s/link |
| Circuit setup | 0.01 s |
Time until the final data bit reaches the destination in each method.
Message switching repeats whole-message serialization at every hop. Packet switching pipelines equal packets. A circuit streams bits through the established equal-rate path.
L=M/k=12000/6=2000 bits.
3×12000/1,000,000 + 3×0.002 = 0.042000 s = 42 ms.
L/R=2000/1,000,000=0.002 s.
(3+6−1)×0.002+3×0.002=0.022000 s = 22 ms.
0.01+12000/1,000,000+3×0.002=0.028000 s = 28 ms.
Final answer: Message=42 ms; packets=22 ms; circuit=28 ms.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
| Given | Value |
|---|---|
| M | 8000 bits |
| h | 2 |
| k | 4 |
| R | 1,000,000 bits/s |
| τ | 0.001 s/link |
| Circuit setup | 0.005 s |
Time until the final data bit reaches the destination in each method.
Message switching repeats whole-message serialization at every hop. Packet switching pipelines equal packets. A circuit streams bits through the established equal-rate path.
L=M/k=8000/4=2000 bits.
2×8000/1,000,000 + 2×0.001 = 0.018000 s = 18 ms.
L/R=2000/1,000,000=0.002 s.
(2+4−1)×0.002+2×0.001=0.012000 s = 12 ms.
0.005+8000/1,000,000+2×0.001=0.015000 s = 15 ms.
Final answer: Message=18 ms; packets=12 ms; circuit=15 ms.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
M=12,000 bits; payload=1,000 bits; header=100 bits; h=4; R=1,000,000 bits/s; τ=0.001 s.
k, L, total serialized bits on one link and completion time.
The payload determines packet count; payload plus header determines packet transmission time.
k=12,000/1,000=12.
L=1,000+100=1,100 bits.
12×1,100=13,200 bits, of which 1,200 are headers.
1,100/1,000,000=0.0011 s.
(4+12−1)×0.0011+4×0.001=0.0205 s=20.5 ms.
Final answer: 12 packets; 1,100 bits each; 13,200 bits on one link; completion=20.5 ms.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Base=40 bytes; extensions=0 bytes; upper-layer content=1000 bytes. Ordinary non-jumbo packet.
Payload Length and total bytes.
Payload begins immediately after the IPv6 base header. Extensions are part of that payload.
P=0+1000=1000 bytes.
T=40+1000=1040 bytes.
The Payload Length field stores 1000, not 1040.
Final answer: Payload Length=1000; total=1040 bytes.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Base=40 bytes; extensions=24 bytes; upper-layer content=1200 bytes. Ordinary non-jumbo packet.
Payload Length and total bytes.
Payload begins immediately after the IPv6 base header. Extensions are part of that payload.
P=24+1200=1224 bytes.
T=40+1224=1264 bytes.
The Payload Length field stores 1224, not 1264.
Final answer: Payload Length=1224; total=1264 bytes.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Base=40 bytes; extensions=40 bytes; upper-layer content=500 bytes. Ordinary non-jumbo packet.
Payload Length and total bytes.
Payload begins immediately after the IPv6 base header. Extensions are part of that payload.
P=40+500=540 bytes.
T=40+540=580 bytes.
The Payload Length field stores 540, not 580.
Final answer: Payload Length=540; total=580 bytes.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Definition. A DHCP lease grants use of an address for a specified lifetime; renewal extends that permission, while a relay supports communication across subnets.
Simple meaning. Address permanent gift nahi: time-limited lease, phir renew.
A lease permits a client to use its assigned address for a limited period and may be renewed. A relay conveys DHCP messages between a client subnet and a remote DHCP server, allowing centralized configuration without placing a server on every subnet.
Yaad rakho: Lease = limited time; relay = subnet ke paar DHCP support.
"},{"id":"lease-1","title":"DHCP timer example 1","html":"Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Lease duration L=8 h; use default fractional timers.
T1 and T2 as elapsed times.
Use these defaults only when the server/question has not specified other timer values.
T1=0.5×8=4 h.
T2=0.875×8=7 h.
4 h < 7 h < 8 h.
Final answer: T1=4 h; T2=7 h.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Lease duration L=16 h; use default fractional timers.
T1 and T2 as elapsed times.
Use these defaults only when the server/question has not specified other timer values.
T1=0.5×16=8 h.
T2=0.875×16=14 h.
8 h < 14 h < 16 h.
Final answer: T1=8 h; T2=14 h.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Lease duration L=24 h; use default fractional timers.
T1 and T2 as elapsed times.
Use these defaults only when the server/question has not specified other timer values.
T1=0.5×24=12 h.
T2=0.875×24=21 h.
12 h < 21 h < 24 h.
Final answer: T1=12 h; T2=21 h.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
IHL=5; Total Length=1000 bytes.
Header bytes and payload bytes.
IHL counts 32-bit words; each word is 4 bytes.
H=4×5=20 bytes.
P=1000−20=980 bytes.
20+980=1000 bytes.
Final answer: Header=20 bytes; payload=980 bytes.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
IHL=6; Total Length=1200 bytes.
Header bytes and payload bytes.
IHL counts 32-bit words; each word is 4 bytes.
H=4×6=24 bytes.
P=1200−24=1176 bytes.
24+1176=1200 bytes.
Final answer: Header=24 bytes; payload=1176 bytes.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
IHL=8; Total Length=800 bytes.
Header bytes and payload bytes.
IHL counts 32-bit words; each word is 4 bytes.
H=4×8=32 bytes.
P=800−32=768 bytes.
32+768=800 bytes.
Final answer: Header=32 bytes; payload=768 bytes.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Definition. IPv4 fragmentation divides a packet that exceeds an outgoing link’s MTU into smaller fragments, when fragmentation is permitted.
Simple meaning. MTU se packet bada hai toh allowed case mein fragments; offset original payload mein position batata hai.
Fragment Offset is measured in units of 8 payload bytes. MF=1 says more fragments of the original packet follow; MF=0 marks the last fragment. Each fragment carries its own IP header.
Yaad rakho: Offset ×8 = starting payload byte.
"},{"id":"fragment-1","title":"Optional fragmentation · example 1","html":"Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Total=4000; H=20; MTU=1500 bytes; DF=0.
All fragments and their header-related values.
Subtract the header before splitting the original payload. Round non-final fragment payload capacity down to a multiple of eight.
4000−20=3980 bytes.
MTU−H=1500−20=1480. Floor((1480)/8)×8=1480 bytes.
| Fragment | Payload bytes | Total bytes | Starting byte | Offset /8 | MF |
|---|---|---|---|---|---|
| 1 | 1480 | 1500 | 0 | 0 | 1 |
| 2 | 1480 | 1500 | 1480 | 185 | 1 |
| 3 | 1020 | 1040 | 2960 | 370 | 0 |
1480+1480+1020=3980 bytes. Every total length is at most 1500.
Final answer: 3 fragments; lengths, offsets and MF are listed in the table.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Total=2000; H=20; MTU=620 bytes; DF=0.
All fragments and their header-related values.
Subtract the header before splitting the original payload. Round non-final fragment payload capacity down to a multiple of eight.
2000−20=1980 bytes.
MTU−H=620−20=600. Floor((600)/8)×8=600 bytes.
| Fragment | Payload bytes | Total bytes | Starting byte | Offset /8 | MF |
|---|---|---|---|---|---|
| 1 | 600 | 620 | 0 | 0 | 1 |
| 2 | 600 | 620 | 600 | 75 | 1 |
| 3 | 600 | 620 | 1200 | 150 | 1 |
| 4 | 180 | 200 | 1800 | 225 | 0 |
600+600+600+180=1980 bytes. Every total length is at most 620.
Final answer: 4 fragments; lengths, offsets and MF are listed in the table.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Total=1500; H=20; MTU=600 bytes; DF=0.
All fragments and their header-related values.
Subtract the header before splitting the original payload. Round non-final fragment payload capacity down to a multiple of eight.
1500−20=1480 bytes.
MTU−H=600−20=580. Floor((580)/8)×8=576 bytes.
| Fragment | Payload bytes | Total bytes | Starting byte | Offset /8 | MF |
|---|---|---|---|---|---|
| 1 | 576 | 596 | 0 | 0 | 1 |
| 2 | 576 | 596 | 576 | 72 | 1 |
| 3 | 328 | 348 | 1152 | 144 | 0 |
576+576+328=1480 bytes. Every total length is at most 600.
Final answer: 3 fragments; lengths, offsets and MF are listed in the table.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
UDP payload=100 B; UDP fixed header=8 B; IP header=20 B.
UDP Length field and total packet bytes.
![\"L_{UDP}=8+P,\\quad](\"math/b72774caf03e04db.svg\")
UDP Length includes its own header, but does not include the IP header.
100+8=108 bytes.
108+20=128 bytes.
20 IP + 8 UDP + 100 data = 128 bytes.
Final answer: UDP Length=108 bytes; IP packet=128 bytes.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
UDP payload=512 B; UDP fixed header=8 B; IP header=20 B.
UDP Length field and total packet bytes.
UDP Length includes its own header, but does not include the IP header.
512+8=520 bytes.
520+20=540 bytes.
20 IP + 8 UDP + 512 data = 540 bytes.
Final answer: UDP Length=520 bytes; IP packet=540 bytes.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
UDP payload=1200 B; UDP fixed header=8 B; IP header=40 B.
UDP Length field and total packet bytes.
UDP Length includes its own header, but does not include the IP header.
1200+8=1208 bytes.
1208+40=1248 bytes.
40 IP + 8 UDP + 1200 data = 1248 bytes.
Final answer: UDP Length=1208 bytes; IP packet=1248 bytes.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Client ISN=1000; server ISN=5000; data=200 B; no sequence wrap or loss.
Each sequence/acknowledgement boundary.
![\"\\mathrm{first}=ISN+1,\\quad](\"math/e95efbc97b48115e.svg\")
SYN and FIN consume one sequence value each; data consumes one value per byte; an empty ACK consumes none.
Server ACK=1001; client ACK=5001.
First byte=1001; last=1001+200−1=1200.
Next expected byte=1200+1=1201; receiver sends ACK=1201.
Client FIN Seq=1201; FIN ACK=1202.
Final answer: Handshake ACKs 1001 and 5001; data 1001…1200; data ACK 1201; FIN ACK 1202.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Client ISN=99; server ISN=700; data=500 B; no sequence wrap or loss.
Each sequence/acknowledgement boundary.
SYN and FIN consume one sequence value each; data consumes one value per byte; an empty ACK consumes none.
Server ACK=100; client ACK=701.
First byte=100; last=100+500−1=599.
Next expected byte=599+1=600; receiver sends ACK=600.
Client FIN Seq=600; FIN ACK=601.
Final answer: Handshake ACKs 100 and 701; data 100…599; data ACK 600; FIN ACK 601.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
Client ISN=4000; server ISN=9000; data=1024 B; no sequence wrap or loss.
Each sequence/acknowledgement boundary.
SYN and FIN consume one sequence value each; data consumes one value per byte; an empty ACK consumes none.
Server ACK=4001; client ACK=9001.
First byte=4001; last=4001+1024−1=5024.
Next expected byte=5024+1=5025; receiver sends ACK=5025.
Client FIN Seq=5025; FIN ACK=5026.
Final answer: Handshake ACKs 4001 and 9001; data 4001…5024; data ACK 5025; FIN ACK 5026.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
| Time (s) | Arrival packets |
|---|---|
| 0 | 8 |
| 2 | 9 |
| 5 | 3 |
Queue after each burst, dropped packets and time the queue finally empties.
![\"q_{before}=\\max(0,q_{previous}-r\\Delta](\"math/b431990cb4021a43.svg\")
![\"a_{accepted}=\\min(a,B-q_{before})\"](\"math/b022488f5ef91a76.svg\")
This explicit event order separates elapsed-time service from instantaneous admission.
q=0 at t=0 before the first arrival; no negative-time service.
| t | Arrive | Drained | Before | Admit | Drop | After |
|---|---|---|---|---|---|---|
| 0 | 8 | 0 | 0 | 8 | 0 | 8 |
| 2 | 9 | 4 | 4 | 6 | 3 | 10 |
| 5 | 3 | 6 | 4 | 3 | 0 | 7 |
Total dropped=3 packets.
Last event t=5; queue=7. Time needed=7/2=3.5 s; final empty time=8.5 s.
Final answer: 3 packets dropped; queue finally empty at t=8.5 s.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
| Time (s) | Arrival packets |
|---|---|
| 0 | 12 |
| 1 | 7 |
| 3 | 4 |
Queue after each burst, dropped packets and time the queue finally empties.
This explicit event order separates elapsed-time service from instantaneous admission.
q=0 at t=0 before the first arrival; no negative-time service.
| t | Arrive | Drained | Before | Admit | Drop | After |
|---|---|---|---|---|---|---|
| 0 | 12 | 0 | 0 | 12 | 0 | 12 |
| 1 | 7 | 3 | 9 | 3 | 4 | 12 |
| 3 | 4 | 6 | 6 | 4 | 0 | 10 |
Total dropped=4 packets.
Last event t=3; queue=10. Time needed=10/3=3.33333 s; final empty time=6.33333 s.
Final answer: 4 packets dropped; queue finally empty at t=6.33333 s.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
| Time (s) | Arrival packets |
|---|---|
| 0 | 10 |
| 1 | 18 |
| 4 | 6 |
Queue after each burst, dropped packets and time the queue finally empties.
This explicit event order separates elapsed-time service from instantaneous admission.
q=0 at t=0 before the first arrival; no negative-time service.
| t | Arrive | Drained | Before | Admit | Drop | After |
|---|---|---|---|---|---|---|
| 0 | 10 | 0 | 0 | 10 | 0 | 10 |
| 1 | 18 | 5 | 5 | 15 | 3 | 20 |
| 4 | 6 | 15 | 5 | 6 | 0 | 11 |
Total dropped=3 packets.
Last event t=4; queue=11. Time needed=11/5=2.2 s; final empty time=6.2 s.
Final answer: 3 packets dropped; queue finally empty at t=6.2 s.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
B=10000; r=2000; initial tokens=3000; L=7000 bytes; L≤B.
Token deficit and eligibility wait.
![\"\\Delta=\\max(0,L-T),\\quad](\"math/b2d0e2858bb8f8cc.svg\")
Each byte needs one token. This is a shaper, so it waits for missing tokens; the packet fits within token capacity.
Δ=max(0,7000−3000)=4000.
Wait=4000/2000=2 s.
3000+2000×2=7000 tokens, at most capacity 10000. Sending consumes 7000, leaving 0.
Final answer: The packet is eligible after 2 s; physical transmission takes additional time.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
B=8000; r=1000; initial tokens=500; L=2500 bytes; L≤B.
Token deficit and eligibility wait.
Each byte needs one token. This is a shaper, so it waits for missing tokens; the packet fits within token capacity.
Δ=max(0,2500−500)=2000.
Wait=2000/1000=2 s.
500+1000×2=2500 tokens, at most capacity 8000. Sending consumes 2500, leaving 0.
Final answer: The packet is eligible after 2 s; physical transmission takes additional time.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
B=12000; r=3000; initial tokens=4000; L=10000 bytes; L≤B.
Token deficit and eligibility wait.
Each byte needs one token. This is a shaper, so it waits for missing tokens; the packet fits within token capacity.
Δ=max(0,10000−4000)=6000.
Wait=6000/3000=2 s.
4000+3000×2=10000 tokens, at most capacity 12000. Sending consumes 10000, leaving 0.
Final answer: The packet is eligible after 2 s; physical transmission takes additional time.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
B=10000 bytes; r=2000 B/s; P=12000 B/s; continuously backlogged fluid traffic.
Time until tokens exhaust and bytes sent at the peak rate.
![\"t=\\frac{B}{P-r},\\quad](\"math/931f9e3ab8969d1f.svg\")
Tokens are consumed at P and simultaneously replenished at r, so net depletion is P−r.
12000−2000=10000 tokens/s.
t=10000/10000=1 s.
D=12000×1=12000 bytes. Check B+rt=10000+2000×1=12000 bytes.
Final answer: Peak burst lasts 1 s and sends 12000 bytes.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
B=6000 bytes; r=3000 B/s; P=9000 B/s; continuously backlogged fluid traffic.
Time until tokens exhaust and bytes sent at the peak rate.
Tokens are consumed at P and simultaneously replenished at r, so net depletion is P−r.
9000−3000=6000 tokens/s.
t=6000/6000=1 s.
D=9000×1=9000 bytes. Check B+rt=6000+3000×1=9000 bytes.
Final answer: Peak burst lasts 1 s and sends 9000 bytes.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
B=16000 bytes; r=4000 B/s; P=8000 B/s; continuously backlogged fluid traffic.
Time until tokens exhaust and bytes sent at the peak rate.
Tokens are consumed at P and simultaneously replenished at r, so net depletion is P−r.
8000−4000=4000 tokens/s.
t=16000/4000=4 s.
D=8000×4=32000 bytes. Check B+rt=16000+4000×4=32000 bytes.
Final answer: Peak burst lasts 4 s and sends 32000 bytes.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
n=3; RTT=50 ms; each payload=10000 B; rate=1000000 bit/s; ideal HTTP/1.x timing model.
Total time under each stated connection policy.
![\"T_{non}=n(2RTT+T_{object})\"](\"math/d5549e76446992ce.svg\")
![\"T_{persistent}=RTT+n(RTT+T_{object})\"](\"math/257c101efa25bb1f.svg\")
Nonpersistent uses one setup RTT plus one request/first-response RTT per object. Persistent pays setup once, then one sequential request RTT per object.
Tobject=10000×8/1000000 seconds = 80 ms.
Tnon=3×(2×50+80)=540 ms.
Tpersistent=50+3×(50+80)=440 ms.
540−440=100 ms = (3−1)×50; exactly the removed setup round trips.
Final answer: Nonpersistent=540 ms; persistent=440 ms.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
n=5; RTT=20 ms; each payload=20000 B; rate=2000000 bit/s; ideal HTTP/1.x timing model.
Total time under each stated connection policy.
Nonpersistent uses one setup RTT plus one request/first-response RTT per object. Persistent pays setup once, then one sequential request RTT per object.
Tobject=20000×8/2000000 seconds = 80 ms.
Tnon=5×(2×20+80)=600 ms.
Tpersistent=20+5×(20+80)=520 ms.
600−520=80 ms = (5−1)×20; exactly the removed setup round trips.
Final answer: Nonpersistent=600 ms; persistent=520 ms.
Optional worked extension. This calculation format was not found in the supplied December 2025 final or February 2026 re-exam. The numbers and practice marks are authored. Study the core concept and actual-paper questions first; this is extra explanation, not a TEE prediction.
n=2; RTT=100 ms; each payload=5000 B; rate=500000 bit/s; ideal HTTP/1.x timing model.
Total time under each stated connection policy.
Nonpersistent uses one setup RTT plus one request/first-response RTT per object. Persistent pays setup once, then one sequential request RTT per object.
Tobject=5000×8/500000 seconds = 80 ms.
Tnon=2×(2×100+80)=560 ms.
Tpersistent=100+2×(100+80)=460 ms.
560−460=100 ms = (2−1)×100; exactly the removed setup round trips.
Final answer: Nonpersistent=560 ms; persistent=460 ms.
Supporting formula collection. Use the compact TEE formula sheet first; the full calculation set below is not an observed-paper requirement.
L=bits, R=bits/s, Tf=seconds. Use for serialization.
d=distance, v=propagation speed. Use consistent distance units.
h equal-rate store-and-forward links, k equal packets, L total bits/packet; τ per-link propagation. No queueing or processing.
R=bits/s; Tp=one-way seconds. Theoretical collision-detection bound.
n collisions; K selected integer for classic Ethernet backoff.
Multiply selected K by slot duration, retaining units.
Supporting formula collection. Use the compact TEE formula sheet first; the full calculation set below is not an observed-paper requirement.
Count every written group, including a written 0.
Payload includes extension headers, excludes the base 40 bytes.
IHL is in 32-bit words.
IPv4 Total Length includes the IPv4 header.
k data bits, n codeword bits, r redundant bits; coding rate is dimensionless.
Count differing bit positions in equal-length words.
Separate maximum guaranteed detection and correction counts; all distinct pairs define dmin.
Smallest r for m data bits in single-error Hamming construction; parity positions 1,2,4,….
Generator degree r; append r zeros and use modulo-2 XOR division. Receiver divides T directly.
CRC and receiver-check working
"},{"id":"arq","title":"ARQ windows and utilization","html":"m sequence bits; sequence numbers 0…N−1. GBN receiver accepts only next in order.
Standard equal-window SR bound; prevent successive windows overlapping.
Ideal error-free Stop-and-Wait; Tt data transmission, Tp one-way propagation, Ta ACK transmission.
![\"U_W=\\min(1,WT_t/(T_t+2T_p+T_a))\"](\"math/73d665b598b96583.svg\")
Supporting ideal pipeline bound; no queueing/processing, equal frames.
![\"W_{need}=\\lceil(T_t+2T_p+T_a)/T_t\\rceil\"](\"math/acc548bc5c4e4387.svg\")
Round upward for a full pipeline, then verify numbering constraints.
"},{"id":"transport","title":"UDP and TCP arithmetic","html":"![\"L_{UDP}=8+L_{data}\"](\"math/940bfc5f08006c8a.svg\")
UDP length includes its own eight-byte header, not the IP header.
![\"H_{TCP}=4\\times\\mathrm{data\\](\"math/4b6e848089433565.svg\")
Offset counts 32-bit words; minimum 20-byte TCP header.
![\"ACK=\\mathrm{first\\](\"math/8d1628ee7de8ff31.svg\")
After receiving all L consecutive data bytes, acknowledge the next expected byte. SYN and FIN each add one.
![\"W_{new}=\\max(0,\\min(rwnd,cwnd)-F)\"](\"math/13bf9aaf2ed709a3.svg\")
Supporting simplified sending allowance; F bytes in flight, all quantities in the same units.
"},{"id":"buckets","title":"Leaky queue and token bucket","html":"![\"q_b=\\max(0,q-r\\Delta](\"math/b9d2a7d69c31ca42.svg\")
Leaky data-queue state before the next arrival; service cannot exceed queued data.
![\"a_{ok}=\\min(a,B-q_b),\\quad](\"math/e3b5b2a12bccb581.svg\")
B finite data capacity; admit instantaneous arrivals after interval service.
![\"T_{new}=\\min(B,T+r\\Delta](\"math/017cf4ec943dd0f5.svg\")
Token refill before the next admission, capped by token capacity.
![\"t_{wait}=\\max(0,L-T)/r\"](\"math/f9ae1e0794f3ac3b.svg\")
Single packet of L≤B tokens, no other queued traffic, shaper not policer.
![\"D(t)\\le](\"math/6a390266ecc7c5fa.svg\")
Full token-bucket allowance; physical link rate also bounds actual departure.
![\"t_{burst}=B/(P-r),\\quad](\"math/e637610146069ae8.svg\")
Fluid peak-rate burst, initially full and continuously backlogged. Net depletion is P−r.
Three sustained-burst examples
"},{"id":"support","title":"Topology and HTTP timing · supporting examples","html":"![\"L_{mesh}=n(n-1)/2\"](\"math/1aaf8e0c5d7c5d53.svg\")
Undirected full mesh without self-links; per-node ports n−1.
![\"T_{non}=n(2RTT+T_o)\"](\"math/21c4c4bee22560eb.svg\")
Sequential nonpersistent HTTP/1.x idealization; no DNS/TLS/loss/parallelism.
![\"T_{persistent}=RTT+n(RTT+T_o)\"](\"math/e9189d93499eafbc.svg\")
One initial TCP handshake then sequential requests without pipelining; To is one object’s serialization.
"}]}]}}}