GATE 2015 CS – Question 36
Suppose that the stop-and-wait protocol is used on a link with a bit rate of 64 kilobits per second and 20 milliseconds propagation delay. Assume that the transmission time for the acknowledgement and the processing time at nodes are negligible. Then the minimum frame size in bytes to achieve a link utilization of at least 50% is ________.
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Correct answer: 160
Explanation
Utilization is the fraction of time the sender spends transmitting, so for 50% the transmission time must be at least the time the sender spends waiting. Taking that waiting time as 20 ms, the frame must take at least 20 ms to send, which is $64000 \times 0.02 = 1280$ bits, or $\frac{1280}{8} = 160$ bytes. This follows the official key.