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Consider an error-free 64-kbps satellite channel used to send 512-byte data frames in
one direction, with very short acknowledgements coming back the other way. What is
the maximum throughput for window sizes of 1, 7, 15, and 127? The earth-satellite
propagation time is 270 msec.
in Computer Networks by Boss (10.5k points) | 30 views

1 Answer

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Bandwidth=64 Kbits/sec

Frame size=512 Bytes

Transmission Delay($T_{t}$)=$\frac{512*8}{64*10^3}$=64 ms

Propagation Delay($T_{t}$)=270 ms

a=$\frac{T_{p}}{T_{t}}$

Throughput or Bandwidth utilization=Efficiency*Bandwidth=$\frac{N}{1+2a}*Bandwidth=\frac{N}{1+2a}*64 Kbits/sec$

 

    N      Efficiency         Throughput
    1       10.59%       6.78 Kbits/sec  
    7       74.17%      47,470 bits/sec
    9       95.36%      61,033 bits/sec

 

For window size 10 or greater the channel utilization will be 100% and throughput will be 64 Kbits/sec

by Active (5k points)

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