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The distance between two stations M and N is L kilometers. All frames are K bits long. The propagation delay per kilometer is t seconds. Let R bits/second be the channel capacity. Assuming that the processing delay is negligible, the minimum number of bits for the sequence number field in a frame for maximum utilization, when the sliding window protocol is used, is:
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The distance between two stations M and N is L kilometers. All frames ...
Answer: C
for maximum utilization η = 1
so,

gives the number of packets that are sent. It means that it is the Sender's Window Size.
But we also know that

so, to get the minimum number of bits needed to represent the sequence numbers we should consider what protocols are in use.
if GBN ARQ:

if Selective Repeat ARQ :
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The distance between two stations M and N is L kilometers. All frames are K bits long. The propagation delay per kilometer is t seconds. Let R bits/second be the channel capacity. Assuming that the processing delay is negligible, the minimum number of bits for the sequence number field in a frame for maximum utilization, when the sliding window protocol is used, is:a)b)c)d)Correct answer is option 'C'. Can you explain this answer?
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