GATE 2024 CS2 Q.54: solution and answer

Official answer: 4Source: Official answer key, GATE 2024 CS2 (IISc Bengaluru)

1. A timeout is a severe loss signal. TCP sets the slow-start threshold to half the window at the time of loss: $\text{ssthresh}=12/2=6$ MSS.

2. The congestion window is reset to $\text{cwnd}=1$ MSS and TCP re-enters slow start.

3. First RTT after the timeout: the window is $1$ MSS. The single segment is acknowledged, so the window doubles to $2$ MSS for the next RTT.

4. Second RTT: the window is $2$ MSS. Both segments are acknowledged, so the window doubles to $4$ MSS for the next RTT.

5. Third RTT: the window is $4$ MSS. This is below $\text{ssthresh}=6$, so TCP is still in slow start and there is no linear growth yet.

6. $$\text{cwnd}:12\stackrel{\text{timeout}}{\to }1\to 2\to 4$$

7. So the congestion window during the third RTT is $4$ MSS.