ISRO CS 2025 Q.40: solution and answer

Official answer: DSource: ISRO's final answer key of 15 May 2026

1. To keep a link fully busy, the sender must have unacknowledged data in flight equal to the bandwidth-delay product (BDP): $\text{window}=\text{bandwidth}\times \text{delay}$.

2. The question gives a single latency figure, $32\text{ms}$, so it is taken as the delay the window must cover (the round-trip time, the time before the first acknowledgement returns).

3. Bits in flight: $8192000\times 0.032=262144\text{bits}$, since $8192000\times 32=262144000$ and dividing by $1000$ gives $262144$.

4. Convert to bytes: $\frac{262144}{8}=32768\text{bytes}$, which is $2^{15}$.

5. So the ideal window is $32768$ bytes, option (d). A window of $65536$ bytes would be twice what is needed, and $16384$ or $8192$ bytes would leave the link idle for part of each round trip.