The distance between two stations M and N is L kilometers. All frames are K bits long. The propagation delay is t second...

Question

The distance between two stations M and N is L kilometers. All frames are K bits long. The propagation delay is t seconds per kilometer. Let B bits/second be the channel capacity. Assuming that the processing delay is negligible, the minimum number of bits required for the sequence number field in a frame for maximum utilization, when the sliding window protocol is used, is:
A. [log_2(2BtL/K)]
B. [log_2(2BtL/K + 1)]
C. [log_2(BtL/K)]
D. [log_2(2BtL/K - 1)]

Accepted Answer

Minimum bits required = [log_2(2BtL/K)]. Answer: A. [log_2(2BtL/K)] Two stations M and N are L km apart. To maximise utilisation, the sender must keep transmitting during the full round-trip time without waiting. The window size must be at least as large as the number of frames that can be sent in one RTT. The sequence number field must be large enough to uniquely label all frames in the window.