Question

Computer Networks

Suppose that the congestion window of a TCP connection starts from W/2 bytes, grows as specified in the congestion avoidance state until reaching W bytes, and then resets to W/2 bytes upon detecting a lost segment. Assume each segment contains MSS bytes of data, and each round takes RTT seconds. a) Give the number of rounds it takes for the window to grow from W/2 to W. b) Give the total number of bytes transmitted in a period, i.e., the period from the time the window is W/2 to just before the next time it becomes W/2 again Give the loss rate (fraction of lost segments in the long run). c) d) Show that for W >> MSS (: much larger than), the average throughput is approximately 1.22 MSS RTT - VL

Answer 1

a) In congestion avoidance until you get detection of lost segment we get increment of 1 packet for each acknowledgement received. So it takes total time of (W - (W/2))*RTT

b) W/2 + (W/2 + 1) + ...... + W = 3/8 W^2 + 3/4 W .It has to grow from W/2 to W. So we sum it up.

c)If we take the value W/2 + (W/2 + 1) + ...... + W = 3/8 W^2 + 3/4 W = K, then for every K packets we will observe a loss So we get the value of the loss rate L = 1/K

d)Let the first loss occur when the window size is W. Then the window size becomes W/2. So next loss occur again at W. No.of packets transmitted :
W/2 + (W/2 + 1) + ...... + W = 3/8 W^2 + 3/4 W = K
Then loss rate L = 1/K
For larger values of W , we have W^2 term dominating over W.
So loss rate becomes L = 8/(3W^2) which gives W = $\sqrt{\frac{8}{3*L}}$

The standard formula for the Average Throughtput = 0.75 (Max bytes transferred in a period)/RTT.

Max bytes transferred is when window size is max = W * MSS. So using these two equation we get the following answer.

Substituting the above found value of W in terms of loss rate we get the required equation:
$\frac{1.22*MSS}{RTT*\sqrt{L}}$ which is required answer.