The distance between two stations M and N is L kilometers. All frames are K…
2007
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 processing delay is negligible, the minimum number of bits for the sequence number field in a frame for maximum utilization, when the sliding window protocolis used, is:
- A.
A
- B.
B
- C.
C
- D.
D
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Correct answer: C
Goal: find the minimum number of sequence-number bits n so that the sliding-window sender can achieve full utilization.
One-way propagation delay = L·t, so round-trip propagation delay = 2Lt.
Transmission time per frame = K/R.
To keep the sender busy (utilization = 1), the sender must be able to send W frames in the time between sending a frame and receiving its ACK. Therefore:
W · (K/R) ≥ (K/R) + 2Lt
Rearrange to get the required window size:
W ≥ 1 + 2LtR / K
So the minimum integer window is Wmin = ceil(1 + 2LtR / K).
The sequence-number field of n bits provides 2^n distinct sequence numbers, so we need
2^n ≥ Wmin = 1 + 2LtR / K
Hence
n = ceil(log2(1 + 2LtR / K)) = ceil(log2((2LtR + K)/K)).
Therefore the minimum number of bits in the sequence-number field is ceil(log2((2LtR + K)/K)).
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