GATE | GATE-CS-2006 | Question 44

Station A uses 32 byte packets to transmit messages to Station B using a sliding window protocol. The round trip delay between A and B is 80 milliseconds and the bottleneck bandwidth on the path between A and B is 128 kbps. What is the optimal window size that A should use?
(A) 20
(B) 40
(C) 160
(D) 320


Answer: (B)

Explanation:

Round Trip propagation delay = 80ms 
Frame size = 32*8 bits
Bandwidth = 128kbps
Transmission Time = 32*8/(128) ms = 2 ms

Let n be the window size.

UtiliZation = n/(1+2a) where a 
            = Propagation time / 
                transmission time
            = n/(1+80/2)

For maximum utilization: n = 41 
which is close to option (B)

Source : Question 1 of https://www.geeksforgeeks.org/computer-networks-set-11/

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