GATE 2026 CS1 Q.45: solution and answer

GATE 2026 · CS1Q.45 · 2 marks · MCQ

Consider the implementation of sliding window protocol over a lossless link, with a window size of $W$ frames, where each frame is of size 1000 bits (including header). The bandwidth of the link is 100 kbps (1k = $1 0^{3}$ ) and the one-way propagation delay is 100 milliseconds. Assume that processing times at the sender and receiver are zero and the transmission time of acknowledgements is also zero. Which one of the following options gives the minimum size of $W$ (in number of frames) required to achieve 100% link utilization?

(A)
10
(B)
21
(C)
20
(D)
11

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Answer and solution

ANSWER

The minimum window size is $W = 21$ frames, option (B).

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

Frame transmission time: $T_{t} = \frac{1000 bits}{100 \times 1 0 ^{3} bps} = 10 ms$ .
One-way propagation delay is $T_{p} = 100 ms$ , so the round trip is $2 T_{p} = 200 ms$ . Acknowledgements take zero time to transmit and nothing is spent on processing.
The sender starts the first frame at time 0. Its last bit leaves at $10 ms$ , reaches the receiver at $110 ms$ , and the acknowledgement is back at $210 ms$ . One full cycle is $T_{t} + 2 T_{p} = 210 ms$ .
For 100% utilization the sender must keep transmitting for the whole cycle, so the window must cover the cycle: $W \times T_{t} \geq T_{t} + 2 T_{p}$ .
$W \geq \frac{10 + 200}{10} = 21$ .
Check: with $W = 20$ the sender is busy for $200 ms$ and then idles for $10 ms$ until the first acknowledgement arrives at $210 ms$ , so utilization is $\frac{20}{21} < 100%$ . With $W = 21$ the first acknowledgement arrives just as the 21st frame finishes, so the sender never idles.
Minimum $W = 21$ frames, option (B).

KEY IDEA

For 100% utilization the window must cover one full cycle: $W = ⌈(T_{t} + 2 T_{p}) / T_{t} ⌉ = 1 + 2 T_{p} / T_{t}$ . With $T_{t} = 10$ ms and $2 T_{p} = 200$ ms this gives 21.

WATCH OUT FOR

Using $W = 2 T_{p} / T_{t} = 20$ and forgetting the one frame time $T_{t}$ needed to send the first frame, which gives option (C). The cycle is $T_{t} + 2 T_{p}$ , not $2 T_{p}$ alone.

From the GATE 2026 CS1 paper. Every question of this paper · Download the full paper · All 7 Performance questions, with notes

GATE 2026 CS1 Q.45: solution and answer

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