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GATE | GATE-CS-2014-(Set-1) | Question 65

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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 ________.
(A) 3
(B) 4
(C) 5
(D) 6

Answer: (C)


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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Last Updated : 04 Aug, 2021
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