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GATE | GATE-CS-2009 | Question 58

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  • Difficulty Level : Easy
  • Last Updated : 06 Nov, 2019
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Consider the data of previous question. Suppose that the sliding window protocol is used with the sender window size of 2^i where is the number of bits identified in the previous question and acknowledgments are always piggybacked. After sending 2^i frames, what is the minimum time the sender will have to wait before starting transmission of the next frame? (Identify the closest choice ignoring the frame processing time.)
(A) 16ms
(B) 18ms
(C) 20ms
(D) 22ms


Answer: (C)

Explanation: Size of sliding window = 2^5 = 32
Transmission time for a frame = 1ms
Total time taken for 32 frames = 32ms
Total time = 2tx + 2tp = 2+50 = 52ms
After sending 32 frames, the minimum time the sender will have to wait before starting transmission of the next frame = 52 – 32 = 20

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