Software-Engineering
November 18, 2023Software-Engineering
November 18, 2023NTA-UGC-NET 2021 Dec & 2022 June Paper-2
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Question 22
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Consider an error free 64 kbps satellite channel used to send 512-byte data frame in one direction with very short acknowledgements coming back the other way. What is the maximum throughput for window size of 15?
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32 kbps
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48 kbps
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64 kbps
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70 kbps
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Question 22 Explanation:
Sliding window will send specified number of frames, wait for one ack for that set of frames
Given, 512 bytes x 8 bits/B = 4096 bits per frame
Transmission time to send one frame= 4096/64000 bps = 64 msec
Propagation time = 270msec
Round trip delay = (2*Propagation time)=540 msec
Transmission time + Round trip delay =604 msec
Given, 512 bytes x 8 bits/B = 4096 bits per frame
Transmission time to send one frame= 4096/64000 bps = 64 msec
Propagation time = 270msec
Round trip delay = (2*Propagation time)=540 msec
Transmission time + Round trip delay =604 msec
The maximum throughput is possible when Window size = [1+ 2Tpropagation/Ttransmission]
= (1+2*270/64)=9.43~10
= Window size =15
-The window size will give throughput as maximum bandwidth available that is 64 Kbps
Correct Answer: C
Question 22 Explanation:
Sliding window will send specified number of frames, wait for one ack for that set of frames
Given, 512 bytes x 8 bits/B = 4096 bits per frame
Transmission time to send one frame= 4096/64000 bps = 64 msec
Propagation time = 270msec
Round trip delay = (2*Propagation time)=540 msec
Transmission time + Round trip delay =604 msec
Given, 512 bytes x 8 bits/B = 4096 bits per frame
Transmission time to send one frame= 4096/64000 bps = 64 msec
Propagation time = 270msec
Round trip delay = (2*Propagation time)=540 msec
Transmission time + Round trip delay =604 msec
The maximum throughput is possible when Window size = [1+ 2Tpropagation/Ttransmission]
= (1+2*270/64)=9.43~10
= Window size =15
-The window size will give throughput as maximum bandwidth available that is 64 Kbps
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