GATE 2026 CS1 Q.44: solution and answer

Official answer: BSource: Official answer key, GATE 2026 CS1 (IIT Guwahati)

1. The threshold is 10000 segments, far above 2000, so the sender stays in slow start. With no loss, the congestion window starts at 1 MSS and doubles every round-trip time (RTT = 1 ms).

2. Take round $k$ (for $k=1,2,\dots$) to begin at $t=(k-1)$ ms. It sends $2^{k-1}$ segments.

3. Segments sent in rounds $1$ to $k$ in total: $1+2+4+\cdots +2^{k-1}=2^k-1$.

4. $$\begin{array}{ccc}\text{Round}& \text{Starts at (ms)}& \text{Segments sent}\\ 1& 0& 1\text{ (segment 1)}\\ 2& 1& 2\text{ (segments 2 to 3)}\\ 10& 9& 512\text{ (segments 512 to 1023)}\\ 11& 10& 1024\text{ (segments 1024 to 2047)}\end{array}$$

5. After round 10, $2^{10}-1=1023$ segments have been sent. The window in round 11 is $1024<10000$, so it is still slow start.

6. Segment 2000 satisfies $1024\le 2000\le 2047$, so it is sent in round 11, which begins at $t=10$ ms.

7. The segments of round 11 go out as ACKs return during that one RTT, before round 12 begins at $t=11$ ms. Hence $10\le t<11$, which is option (B).