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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.
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bandwidth=64 kbps =64*10^3 bps

data length=512 bytes=512*8 =2^12 bits

transmission time =L/B =2^12/(64*10^3) =64 msec

propagation time=270 mesc

(a)window size(N) =1715

efficiency =N/(1+2a) where a=PT/TT  =270/64

efficiency=1715/(1+2*(270/64)) =181.72

throughput =efficiency*Bandwidth =181.72*64*10^3 =11630.19 kbps

(b)window size=127

efficiency=127/(1+2*(270/64))=13.456

throughput=efficiency*bandwidth=13.456*64*10^3 =861.24 kbps
answered by Active (3.4k points)


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