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Computers and TechnologyComputers and Technology
8 March, 10:59

Suppose packet size is 4KB (i. e. 4000 bytes), bandwidth is 8Mbps, and one-way propagation delay is 20 msec. Assume there is no packet corruption and packet loss.

(a) Suppose sender window size is 5, will the sender be kept busy? If yes, explain why. If not, what is the effective throughput?

(b) What is the minimum sender window size to achieve full utilization? Then how many bits would be needed for the sequence number field?

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  1. 8 March, 13:16
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    1) The sender will not be kept busy

    Throughput = 4 Mbps

    2) Minimum sender window size to achieve full utilization = 10

    Number of bits = 4

    Explanation:

    1) Bandwidth = 8 Mbps

    Bandwidth = 8 * 10^6 bps

    Packet size = 4000 bytes

    Propagation delay = 20 msec

    Sender window size (i. e. number of packets) = 5

    Size of data sent in bits = (packet size) * (number of packets) * (Number of bits)

    Size of data sent = 4000 * 5 * 8

    Size of data sent = 160000 bits

    Size of data that can be sent = 2 * (Propagation delay) * (Bandwidth)

    Size of data that can be sent = 2 * (20 * 10^-3) * (8 * 10^6)

    Size of data that can be sent = 320000 bits

    Efficiency = (Size of data sent) / (Size of data that can be sent)

    Efficiency = 160000/320000

    Efficiency = 0.5

    Throuput = Efficiency * Bandwidth

    Throuput = 0.5 * 8 * 10^6

    Throuput = 4 * 10^6

    Throughput = 4 Mbps

    Since there is a throughput of 4 Mbps, the sender will not be kept busy.

    2) Minimum sender window size to achieve full utilization

    Full utilization means that efficiency = 1

    Efficiency = (Size of data sent) / (Size of data that can be sent)

    Size of data that can be sent = 320000 bits

    Size of data sent in bits = (packet size) * (number of packets) * (Number of bits)

    Size of data sent = 4000 * n * 8

    Size of data sent = 32000n

    1 = (32000n) / 320000

    320000 = 32000n

    n = 10 packets

    The minimum sender window size = 10

    Number of bits needed for the sequence number field

    2^3 = 8

    3 bits will not be sufficient because 8 is not up to 19

    4 bits will be sufficient because 2^4 = 16 which is greater than 10

    2^3 = 8

    Therefore, the number of bits needed = 4
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Home » Computers and Technology » Suppose packet size is 4KB (i. e. 4000 bytes), bandwidth is 8Mbps, and one-way propagation delay is 20 msec. Assume there is no packet corruption and packet loss. (a) Suppose sender window size is 5, will the sender be kept busy? If yes, explain why.
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