GATE 2023 CS Q.50: solution and answer

Official answer: B, C, DSource: Official answer key, GATE 2023 CS (IIT Kanpur)

1. A byte-stream protocol numbers every byte. A sequence number must not be reused while an old segment carrying it can still be alive in the network, so the sequence space must not wrap around within the maximum segment lifetime (MSL).

2. Link rate: $100\text{Mbps}=12.5\times 10^6$ bytes/s. MSL $=2\text{min}=120\text{s}$.

3. Bytes that can be sent in one MSL: $12.5\times 10^6\times 120=1.5\times 10^9$ bytes. The sequence space must be at least this large, so $2^n\ge 1.5\times 10^9$. Since $2^{30}\approx 1.07\times 10^9<1.5\times 10^9\le 2^{31}\approx 2.15\times 10^9$, at least 31 bits are needed.

4. A more conservative design allows for a stale segment living one MSL and its acknowledgement another MSL, so the numbers must stay unique over $2\times$MSL: $2\times 1.5\times 10^9=3\times 10^9$ bytes. Then $2^{31}\approx 2.15\times 10^9<3\times 10^9\le 2^{32}\approx 4.29\times 10^9$, so at least 32 bits are needed.

5. Window check: the bandwidth-delay product is $12.5\times 10^6\times 0.15\approx 1.9\times 10^6$ bytes, far below $2^{n-1}$ for $n\ge 30$, so the window size does not restrict the choice.

6. 30 bits gives only $2^{30}\approx 1.07\times 10^9<1.5\times 10^9$ bytes, so the numbers wrap within a single MSL under either reading. Option A is invalid.

7. 32, 34 and 36 bits each give $2^n\ge 2^{32}\approx 4.29\times 10^9\ge 3\times 10^9$, so they satisfy even the conservative requirement. A longer field only enlarges the sequence space and never causes reuse within the MSL, so these lengths are all valid. Options B, C and D are correct.