An error-free satellite channel carries 512-byte data frames at 64 kbps using a sliding-window protocol with a fixed sender window of 15 frames. The one-way propagation delay is 250 ms in each direction. The sender always has data available, and the receiver acknowledges each frame immediately after receiving it completely. Acknowledgements travel on a separate reverse channel; their transmission time is negligible. Ignore frame overhead, processing delays and interframe gaps, and take 1 kbps = 1000 bits/s. What is the maximum steady-state data throughput?
64 kbps
Step 1 — Frame and acknowledgement timing:
\(T_f=\dfrac{512\times8}{64000}=0.064\text{ s}.\)
An acknowledgement returns 0.064 + 2(0.250) = 0.564 s after the corresponding frame starts transmission.
Step 2 — Window utilization:
The sender takes 15(0.064) = 0.960 s to transmit a full window. Since acknowledgements begin returning before that window is exhausted, the sender can transmit continuously.
\(S=\min\!\left(64000,\dfrac{15\times4096}{0.564}\right)=64000\text{ bits/s}=64\text{ kbps}.\)
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