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If two stations A and B uses copper twisted pair cable having capacity 200 bits/sec. if A transmitted 20 frames to B each of size 50 bits, Assume RTT is neglected. What is the time (in Sec) required to transmit the single frame from A to B if pipelining has been used?
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Please spread some light on this question.
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250 msec = 0.25 sec( which is the transmission time for a single frame) will be the  time taken  to transfer 1 frame from A to B. In the question it is mentioned that round trip time is neglected.
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Can you please share the solution. It will be a great help in getting clearly what you mean.
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Bandwidth= 200 bits/sec

Frame size= 50 bits

Transmission time = 50/200= 0.25sec

Total time to transfer a single frame from A to B = transmission time of one frame = 0.25sec [as RTT is neglected, we need to consider only the transmission time]

Correct me if i m wrong

## 1 Answer

+1 vote
250 ms .
answered by (343 points)
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HOW PLZ EXPLAIN
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I don't have answer. But can you please explain

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