Question:medium
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?
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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Utilization is 100% only when the window can supply frames continuously for the entire send-plus-round-trip cycle: minimum W = 1 + 2(Tp/Tt).
Updated On: Jul 22, 2026
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Show Solution
The Correct Option is B
Solution and Explanation
Step 1: Think in terms of the bandwidth-delay product.
For the link to be 100% utilized, the sender must keep transmitting continuously, which requires enough unacknowledged data "in flight" to cover the full send-to-first-ACK cycle.
Step 2: Compute the cycle time in bits, at the link's bit rate.
Bandwidth \(=100 \times 10^3\) bps. Cycle time \(= T_t + 2T_p = 10\text{ms} + 200\text{ms} = 210\text{ms} = 0.21\text{s}\).
Total bits the link could carry during this cycle time \(= 100000 \text{ bps} \times 0.21\text{ s} = 21000\) bits.
Step 3: Convert bits to number of frames.
Each frame carries 1000 bits, so the number of frames needed to supply \(21000\) bits is: \[ W = \dfrac{21000 \text{ bits}}{1000 \text{ bits/frame}} = 21 \text{ frames} \]
Step 4: Interpret the result.
This means the sender must have 21 frames' worth of data outstanding (sent but not yet acknowledged) at any instant to keep the link continuously busy for the whole 210 ms cycle -- exactly the minimum window size, matching the formula-based result \(W = 1+2a = 21\). \[ \boxed{W = 21} \]
For the link to be 100% utilized, the sender must keep transmitting continuously, which requires enough unacknowledged data "in flight" to cover the full send-to-first-ACK cycle.
Step 2: Compute the cycle time in bits, at the link's bit rate.
Bandwidth \(=100 \times 10^3\) bps. Cycle time \(= T_t + 2T_p = 10\text{ms} + 200\text{ms} = 210\text{ms} = 0.21\text{s}\).
Total bits the link could carry during this cycle time \(= 100000 \text{ bps} \times 0.21\text{ s} = 21000\) bits.
Step 3: Convert bits to number of frames.
Each frame carries 1000 bits, so the number of frames needed to supply \(21000\) bits is: \[ W = \dfrac{21000 \text{ bits}}{1000 \text{ bits/frame}} = 21 \text{ frames} \]
Step 4: Interpret the result.
This means the sender must have 21 frames' worth of data outstanding (sent but not yet acknowledged) at any instant to keep the link continuously busy for the whole 210 ms cycle -- exactly the minimum window size, matching the formula-based result \(W = 1+2a = 21\). \[ \boxed{W = 21} \]
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