GATE CS 2026 Set 1 Q.45: Data Link Layer: Framing, Error Detection and Sliding Window

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 ( $1\text{k} = 10^3$ ) 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?

Official answer and solution

Official answer: B
From: Official GATE 2026 answer key

Solution

Compute frame transmission time $T_t$ , propagation ratio $a = T_p / T_t$ , and use $W = 1 + 2a$ for $100\%$ utilization.

Frame transmission time $T_t$ : $T_t = \frac{1000\text{ bits}}{100 \times 10^3\text{ bits/s}} = 0.01\text{ s} = 10\text{ ms}$
Round-trip time: $T_t + 2T_p = 10\text{ ms} + 2(100\text{ ms}) = 210\text{ ms}$
Minimum window size $W$ for 100% utilization: $W = \frac{T_t + 2T_p}{T_t} = 1 + 2a = 1 + \frac{200\text{ ms}}{10\text{ ms}} = 21\text{ frames}$

Final answer

Option (B)

Common trap

Forgetting to add 1 to $2a = 20$ (or forgetting the factor of 2 for round-trip propagation delay).

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2026 Computer Science and Information Technology - Session 1 (CS1)
Question: 45 of 65
Section: Computer Science and Information Technology

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