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/amp/
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