Computer-Networks
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Question 492
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Station A uses 32 byte packets to transmit messages to station B using sliding window protocol. The round trip delay between A and B is 40 milliseconds and the bottleneck bandwidth on the path between A and B is 64 kbps. The optimal window size of A is ________.
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20
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10
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30
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40
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Question 492 Explanation:
Given data,
L = 32 byte
Bandwidth = 64 kbps,
2Tp = 40 millisec
Tt = L / B = 32 *8 / 64*103 = 4 millisec
a= Tp/Tt
Optimal window size (N) = 1 + 2a
N = 1 + 2a ⇒ 1+ (40/4) ⇒ 11
The correct answer is 11. Option-B is more relevant.
L = 32 byte
Bandwidth = 64 kbps,
2Tp = 40 millisec
Tt = L / B = 32 *8 / 64*103 = 4 millisec
a= Tp/Tt
Optimal window size (N) = 1 + 2a
N = 1 + 2a ⇒ 1+ (40/4) ⇒ 11
The correct answer is 11. Option-B is more relevant.
Correct Answer: B
Question 492 Explanation:
Given data,
L = 32 byte
Bandwidth = 64 kbps,
2Tp = 40 millisec
Tt = L / B = 32 *8 / 64*103 = 4 millisec
a= Tp/Tt
Optimal window size (N) = 1 + 2a
N = 1 + 2a ⇒ 1+ (40/4) ⇒ 11
The correct answer is 11. Option-B is more relevant.
L = 32 byte
Bandwidth = 64 kbps,
2Tp = 40 millisec
Tt = L / B = 32 *8 / 64*103 = 4 millisec
a= Tp/Tt
Optimal window size (N) = 1 + 2a
N = 1 + 2a ⇒ 1+ (40/4) ⇒ 11
The correct answer is 11. Option-B is more relevant.
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