An 8 Mbps (about to be congested) non-buffered link is used to send a huge file

Question

An 8 Mbps (about to be congested) non-buffered link is used to send a huge file between two hosts. The receiving host has a large buffer than the congestion window. Assume the turnaround time is 22 ms, and the TCP Tahoe connection is in congestion avoidance phase. Answer the following questions.

a.) What is the maximum window size in bytes that this TCP connection can achieve?

b.) What is the maximum window size in segments, average window size in segments, and average throughput if the TCP segment size is 2,000 bytes?

Explanation / Answer

Solution:

Given is:

The RTT is 22 ms

the connection is 8 Mbps

a)

Maximum Window size = Link speed * latency / 8 = 8 * 10^6 * 22 * 10^-3 / 8 = 22*10^3 bytes

= 22000 bytes.

b)

Maximum window size in segments = 22000/2000 = 11 segements

The congetion window size varies from W to w/2

Average window size = 0.75* 11 = 9=> This is the average window size

Average throughput = (1 * 22000 * 8 / 22 * 10^-3) = 8 Mbps

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