Discussion
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 = 103) 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?
Answer: B
The minimum window is 21 frames.
The transmission time of one frame and the round-trip time are:
\[ \begin{aligned} T_{t} &= \frac{1000}{100\times10^{3}} = 10\ \text{ms} \\ 2T_{p} &= 200\ \text{ms} \end{aligned} \]
For 100% utilisation the sender must keep transmitting until the first acknowledgement returns, which takes one frame time plus the round trip:
\[ W \ge \frac{T_{t}+2T_{p}}{T_{t}} = \frac{10+200}{10} = 21 \]
20 counts only the round-trip time and leaves the link idle for one frame time per cycle. 10 and 11 use the one-way delay instead of the round trip.
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