Computer-Networks
September 9, 2024Computer-Networks
September 9, 2024Sliding-Window-Protocol
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Question 14
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The distance between two stations M and N is L kilometers. All frames are K bits long. The propagation delay per kilometer is t seconds. Let R bits/second be the channel capacity. Assuming that processing delay is negligible, the minimum number of bits for the sequence number field in a frame for maximum utilization, when the sliding window protocol is used, is:
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⌈log2(2LtR+2K/K)⌉
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⌈log2(2LtR/K)⌉
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⌈log2(2LtR+K/K)⌉
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⌈log2(2LtR+K/2K)⌉
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Question 14 Explanation:
Maximum window size at sender side,
N = (Tk + 2Tp)/Tk
Tp = l × l sec
Tk = K/R sec
So, N = (K/R+2Lt)/(K/R) = K+2LtR/K
Note that when asked for general sliding window protocol (not GBN nor SR) then we do not care about receiver’s window size.
So, no. of bits required,
⌈log2 K+2LtR/K⌉
N = (Tk + 2Tp)/Tk
Tp = l × l sec
Tk = K/R sec
So, N = (K/R+2Lt)/(K/R) = K+2LtR/K
Note that when asked for general sliding window protocol (not GBN nor SR) then we do not care about receiver’s window size.
So, no. of bits required,
⌈log2 K+2LtR/K⌉
Correct Answer: C
Question 14 Explanation:
Maximum window size at sender side,
N = (Tk + 2Tp)/Tk
Tp = l × l sec
Tk = K/R sec
So, N = (K/R+2Lt)/(K/R) = K+2LtR/K
Note that when asked for general sliding window protocol (not GBN nor SR) then we do not care about receiver’s window size.
So, no. of bits required,
⌈log2 K+2LtR/K⌉
N = (Tk + 2Tp)/Tk
Tp = l × l sec
Tk = K/R sec
So, N = (K/R+2Lt)/(K/R) = K+2LtR/K
Note that when asked for general sliding window protocol (not GBN nor SR) then we do not care about receiver’s window size.
So, no. of bits required,
⌈log2 K+2LtR/K⌉
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