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5 June, 18:56

In case of rdt 3.0 Stop and Wait, suppose we send a packet of 1KB through 1 Gbps link and RTT=20 msec. Find the sender utilization? Find the sender throughput?

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  1. 5 June, 19:24
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    a. Utilization = 0.00039

    b. Throughput = 50Kbps

    Explanation:

    Given dа ta:

    Packet Size = L = 1kb = 8000 bits

    Transmission Rate = R = 1 Gbps = 1 x 10⁹ bps

    RTT = 20 msec

    To Find

    a. Sender Utilization = ?

    b. Throughput = ?

    Solution

    a. Sender Utilization

    As Given

    Packet Size = L = 8000 bits

    Transmission Rate = R = 1 Gbps = 1 x 10⁹ bps

    Transmission Time = L/R = 8000 bits / 1 x 10⁹ bps = 8 micro-sec

    Utilization = Transmission Time / RTT + Transmission Time

    = 8 micro-sec / 20 msec + 8 micro-sec

    = 0.008 sec / 20.008 sec

    Utilization = 0.00039

    b. Throughput

    As Given

    Packet Size = 1kb

    RTT = 20ms = 20/100 sec = 0.02 sec

    So,

    Throughput = Packet Size/RTT = 1kb / 0.02 = 50 kbps

    So, the system has 50 kbps throughput over 1 Gbps Link.
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