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    File Transfer Time & Throughput

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    https://brainmass.com/computer-science/files/file-transfer-time-throughput-240344

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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
    Packet Overhead = 80 bits
    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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