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Compute the sender utilization for this protocol in the case the roundtrip time between sender and receiver is 7 ms, the network throughput is 1Mbps and each packet has a size of 1500 bytes

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  1. 4 February, 08:42
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    Given Information:

    Round-trip time = RTT = 7 ms

    Network throughput = 1 Mbps

    Packet size = 1500 bytes

    Required Information:

    Sender utilization = U = ?

    Answer:

    Sender utilization = 63.16%

    Explanation:

    The transmission delay is given by

    T = Packet size/Network throughput

    T = 1500*8 bits/1x10⁶ bits/s

    T = 12x10⁻³ sec

    The Propagation delay is given by

    Pd = RTT/2

    Where RTT is the round trip time

    Pd = 7x10⁻³/2

    Pd = 3.5x10⁻³ sec

    The sender utilization is

    U = 1 / (1 + 2 * (Pd/T))

    U = 1 / (1 + 2 * (3.5x10⁻³ / 12x10⁻³))

    U = 0.6316

    U = 63.16%

    Therefore, the sender utilization or efficiency for this protocol is 63.16% which is basically the average traffic over this link or in other words the link has been used for 63.16% of the time.
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