Question
Consider a simplified TCP’s AIMD algorithm where the congestion window size is measured in number of segments, not in bytes. In additive increase, the congestion window size increases by one segment in each RTT. In multiplica- tive decrease, the congestion window size decreases by half (if the result is not an integer, round down to the nearest integer). Suppose that two TCP connections, C1 and C2 , share a single congested link of speed 30 segments per second. Assume that both C1 and C2 are in the congestion avoidancephase. Connection C1 ’s RTT is 50 msec and connection C2’s RTT is 100 msec. Assume that when the data rate in the link exceeds the link’s speed, all TCP connections experience data segment loss.
a. If both C1 and C 2 at time t 0 have a congestion window of 10 segments, what are their congestion window sizes after 1000 msec?
b. In the long run, will these two connections get the same share of the band-
width of the congested link? Explain.
Answer
The difference between C1 and C2 is that RTT of C1’s is only half of the RTT of C2. Therefore, C1 adjusts its window size after 50 msec, but C2 adjusts its window size after 100 msec.
Assume that C1 receives it after 50msec and C2 receives it after 100msec, whenever a loss event happens.
The following simplified model of TCP is that after each RTT, a connection determines if it should increase window size or not.
For C1, to compute the average total sending rate in the link in the previous 50 msec. If that rate exceeds the link capacity, then we assume that C1 detects loss and reduces its window size. But for C2, we compute the average total sending rate in the link in the previous 100msec. If the rate exceeds the link capacity, then we assume that C2 detects loss and reduces its window size.
The two TCP connections, C1 and C2, share a single congested link of speed 30 segments per second.
The RTT for TCP connection of C1 is 50msec and RTT of C2 is 100msec.
|
Time in sec |
Window size = No. of segments sent in next 50 msec |
Average data sending speed = segments per second (window size/0.05) |
Window size = No. of segments sent in next 100msec |
Average data sending rate = segments per second |
|
0 |
10 |
200 |
10 |
100 |
|
50 |
5 |
100 |
|
100 |
|
100 |
2 |
40 |
5 |
50 |
|
150 |
1 |
20 |
|
50 |
|
200 |
1 |
20 |
2 |
20 |
|
250 |
1 |
20 |
|
|
|
300 |
1 |
20 |
1 |
20 |
|
350 |
2 |
40 |
|
10 |
|
400 |
1 |
20 |
1 |
10 |
|
450 |
2 |
40 |
|
10 |
|
500 |
1 |
20 |
1 |
10 |
|
550 |
2 |
40 |
|
10 |
|
600 |
1 |
20 |
1 |
10 |
|
650 |
2 |
40 |
|
10 |
|
700 |
1 |
20 |
1 |
10 |
|
750 |
2 |
40 |
|
10 |
|
800 |
1 |
20 |
1 |
10 |
|
850 |
2 |
40 |
|
10 |
|
900 |
1 |
20 |
1 |
10 |
|
950 |
2 |
40 |
|
10 |
|
1000 |
1 |
20 |
1 |
10 |
We conclude that after 1000 msec, C1’s and C2’s window sizes are 1 segment each based on the above table.
b) In the long run, these two connections will not get the same share of the band-width of the congested link because C1’s bandwidth share is roughly twice as that of C2’s, because C1 has shorter RTT, only half of that of C2, so C1 can adjust its window size twice as fast as C2. According to the above table, for every cycle has 200msec.
For example, in the cycle from 850msec to1000msec, inclusive. The sending rate of C1 is 40+20+40+20 = 120, which is thrice as large as the sending of C2 given by 10+10+10+10 = 40.