GATE | GATE-CS-2014-(Set-1) | Question 65
Consider a selective repeat sliding window protocol that uses a frame size of 1 KB to send data on a 1.5 Mbps link with a one-way latency of 50 msec. To achieve a link utilization of 60%, the minimum number of bits required to represent the sequence number field is ________.
Transmission delay = Frame Size/bandwidth = (1*8*1024)/(1.5 * 10^6)=5.33ms Propagation delay = 50ms Efficiency = Window Size/(1+2a) = .6 a = Propagation delay/Transmission delay So, window size = 11.856(approx) min sequence number = 2*window size = 23.712 bits required in Min sequence number = log2(23.712) Answer is 4.56 Ceil(4.56) = 5
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