Consider an error-free 64 kbps satellite channel used to send 512-byte data frames in one direction, with very short acknowledgements coming back. What is the maximum throughput for a window size of 15?
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Correct answer: C. 64 kbps
Explanation
Frame time = 4096/64000 = 64 ms. With 270 ms one-way delay, the cycle is 64 + 540 = 604 ms. A window of 15 takes 15 × 64 = 960 ms ≥ 604 ms, so the link stays busy: 64 kbps.More Data Communication and Computer Networks questions
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