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?

Question

Tripp

Computers & Technology

23 November, 02:30

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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Answers (1)

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Ashleigh · Answer 1

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