The TCP slow start algorithm by itself is inefficient, so a threshold (the defau

Question

The TCP slow start algorithm by itself is inefficient, so a threshold (the default value is 64KByte) is established to improve throughput. Suppose the length of the link between host A and host B is 10,000 meters and its propagation delay is 20 msec/km. Host A is using TCP over this link to transfer a 4M byte file to host B. Assume that the TCP has been modified to allow very large windows, and so the TCP receiver window size is always set to 2M bytes. The segment/packet size is 1K bytes. Ignore the packet transmission time and handling time.

A. (8 points) Suppose no packet is lost, that is, no timeout. How many RTTs does it take to send the 4MB file?

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B. (4 points) What is the effective throughput for the file in part (A)?

C. (4 points) Suppose a timeout happens on the 10th transmission and no other timeouts occur. How many RTTs does it take to send the 4MB file?

D. (4 points) What is the effective throughput for the file in part (C)?

Explanation / Answer

A.

Time to reach to host from A to B

=> 20*10^-3 sec/km *10km=200 msec.

1 RTT=2*200 msec=400 msec.

no of RTT=Size of File/packet Size

=>4MB/1KB=4000

4000 RTT.

B.

Total Time=4000*400 msec=1600 sec

which is 26.67 minutes.

throughput=File size/time taken

=>4MB/1600 sec

=2.56 kbps

C.

Time is doubled when timeout occurs to avoid another timeout for same packet.

new RTT=old RTT*2

=>1600 sec*2=32 secs.

D.

Effective throughput= file size/Time req

=>4MB/3200 sec

=>1.28 kbps

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