GATE 2020 CS Q.55: solution and answer

Official answer: 44Source: Official answer key, GATE 2020 CS (IIT Delhi)

1. Elapsed time is $60\text{ms}$ and $\text{RTT}=6\text{ms}$, so $60/6=10$ round trips have completed, with all acknowledgements processed.

2. At $t=0$ the connection starts in slow start with $\text{cwnd}=1\text{MSS}=2\text{KB}$.

3. In slow start the window doubles every RTT until it reaches the threshold of 32 KB: $$\begin{array}{cc}\text{After RTT}& \text{cwnd (KB)}\\ 0& 2\\ 1& 4\\ 2& 8\\ 3& 16\\ 4& 32\end{array}$$

4. After RTT 4, $\text{cwnd}=32\text{KB}=\text{ssthresh}$, so slow start ends and congestion avoidance begins (additive increase).

5. In congestion avoidance the window grows by 1 MSS $=2\text{KB}$ per RTT: $$\begin{array}{cc}\text{After RTT}& \text{cwnd (KB)}\\ 5& 34\\ 6& 36\\ 7& 38\\ 8& 40\\ 9& 42\\ 10& 44\end{array}$$

6. The receiver advertised window is 50 KB and $44<50$, so the sending window is not capped by the receiver. The congestion window after 10 RTTs is $44\text{KB}$.