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# File Transfer Time & Throughput

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#### Solution Preview

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Propagation speed , V = 2 x 10^8 m/sec
File size = 8 x 10^6 bits
Td = Total transmission time of a frame

Td = 2Tprop + Ttrans + Tack
a = Tprop/Ttrans
Utilization, U = 1/1 + 2a
Acknowledgement = 88 bits
Tprop means Propagation time.
Ttrans means Transmission time for frame.
Tack means Transmission time for acknowledgement.
1) Packet Size, P = 256 bits
Since there is 80 bits overhead, therefore it will have (256-80=176)bits of data from file
Therefore total number of such packet needed to transmit data = 8 x 10^6/176=45455 packets

Tprop = 1000/(2 x 10^8) sec = 5 x 10^-6 sec
T trans = 256/( 10^6) = 256 x 10^-6 sec
Tack = 88 / (10^6) = 88 x 10^-6 sec
a = Tprop/Ttrans= (5 x 10^-6) / (256 x 10^-6)=.0195
U= 1/(1+2*a)= ...

#### Solution Summary

The solution discusses file transfer time and throughput.

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