Question

A byte-oriented stop and wait protocol uses frames of 143 bytes and acknowledgements of 4 bytes. The processing delays at the sender and receiver are 2.6 μs. The physical layer operates at 27 Mbit/s over a cable of length 14.0 km, on which signals travel at two-thirds the speed of light. What is the cycle time (the time from the start of transmission of a frame to the start of transmission of the next frame)? Give your answer in μs, rounded to one decimal place.

Answer 1

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