GATE 2026 CS (CS1) – Question 45
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?
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Correct answer: (B) 21
Explanation
Given parameters:
- Frame size $L = 1000$ bits
- Bandwidth $B = 100\text{ kbps} = 100 \times 10^3 = 10^5\text{ bps}$
- Propagation delay $T_p = 100\text{ ms} = 0.1\text{ s}$
- Transmission time of one frame:
$$T_t = \frac{L}{B} = \frac{1000}{100000} = 0.01\text{ s} = 10\text{ ms}$$
The total round-trip cycle time to transmit a frame and receive its acknowledgement is:
$$\text{Cycle time} = T_t + 2T_p = 10\text{ ms} + 2(100\text{ ms}) = 210\text{ ms}$$
To achieve 100% link utilization, the sender must be able to continuously transmit frames throughout the entire cycle time without waiting for an ACK:
$$W \ge \frac{\text{Cycle time}}{T_t} = 1 + 2a = 1 + \frac{2T_p}{T_t} = 1 + \frac{200\text{ ms}}{10\text{ ms}} = 1 + 20 = 21$$
Thus, the minimum window size is $W = 21$ frames. Therefore, option (B) is correct.