4. 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 protocol is used, is:
Ans: option (c)
Explanation:
Given,
Bandwidth = R bps, Frame Size = K bits, distance between stations = L kms
Propagation delay = t seconds per km
= Lt seconds
We know, Throughput =1 window/ RTT
Since link utilization is maximum => Throughput = R
RTT = (2 x propagation delay) + Transmission Time
= 2Lt + (K/R)
1 window in terms of bits = Throughput x RTT
= R {2Lt + (K/R)} = 2LtR + K
1 window in terms of frames = [2LtR + K ]/K
Sequence numbers required: 2n = [2LtR + K ]/K
{where n is the number of bits for the sequence number field}
Therefore n = log2 [(2LtR + K)/K]