8/8/2019 hw1-443-882-sol

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CE 443 Farvardin 21st, 1389 Homework 1Solutions 2

C to D:

Packet Payload(B) ID DF MF Offset1 448 x 0 1 02 448 x 0 1 563 4 x 0 0 112

D to E:

Packet Payload(B) ID DF MF Offset

1 296 x 0 1 02 152 x 0 1 373 296 x 0 1 564 152 x 0 1 935 4 x 0 0 112

3. Draw the congestion window as a function of time (Consider linear increase and multiplicativedecrease plus slow start). Assume initial window size is 1 mss (at round 0). A triple duplicateACK occurs at round 2 and 13, a packet loss happens at round 9 and 17. Draw the curve upto round 22.

4. Assume that TCP only allows window sizes up to 64KB. Suppose that you are running TCPover a 100Mbps link with a latency of 100ms to transfer a 5-MB file, and the TCP receivewindow is 64KB. If TCP sends 0.5-KB packets (assuming no congestion and no lost packets):

How many RTTs does it take until slow start opens the send window to 64KB? Solution:0.5KB 2i 64KB i = 7RTTs required.

How many RTTs does it take to send the file? Solution: For the first 7 RTTs we have(.5 + 1 + 2 + 4 + 8 + 16 + 32) = 63.5KB , and after that we have 64KB of data sent in

each RTT. Hence we have 5MB63.5KB64KB =522063.5210

64210 = 79.0078 80 we require

80 + 7 = 87RTTs to transmit a 5MB file. If the time to send the file is given by the number of required RTTs multiplied by the link

latency, what is the effective throughput for the transfer? What percentage of the linkbandwidth is utilized? Solution: 5MB887100ms =

40Mb8.7s = 4.59Mbps. Therefore

4.59100 = 4.59%

utilization.