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Consider a sliding window protocol operating at the data link layer between two stations 1000
kilometers apart, directly connected by an error-free 1.0 Mbps link. The frame sizes used are
500 bits, of which 50 bits are header and 450 bits are data. Ack frames are 100 bits, and frame
processing time is 100 microseconds. Assume that the signal propagation delay is 10 µsec/km.
If a window size of W = 10 is used, then the efficiency of the channel is
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I ALSO GOT 24.15%...
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edited by
@Anand

I guess the steps B and C are interchanged..

Also, while calculating the time taken for transmission of ACK packet (D), why have you added 50 bits ? I am asking this since the ACK frame size is already given in the question..
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how the processing time is 10*1000 micro sec
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