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Authors:
James F. Kurose, Keith W. Ross
Chapter:
Transport Layer
Exercise:
Problems
Question:52 | ISBN:9780132856201 | Edition: 6

Question

Consider a modification to TCP’s congestion control algorithm. Instead of additive increase, we can use multiplicative increase. A TCP sender increases its window size by a small positive constant a (0 < a < 1) whenever it receives a valid ACK. Find the functional relationship between loss rate L and maximum congestion window W. Argue that for this modified TCP, regardless of TCP’s average throughput, a TCP connection always spends the same amount of time to increase its congestion window size from W/2 to W.

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Answer

The maximum window size is represented by w.

The total number of segments(S) sent out during the interval when TCP changes its window size from w/2 up to and includes w.

S=w/2 +(w/2)*(1+a)+(w/2)*(1+a)^2 +(w/2)*(1+a)^3 +(w/2)*(1+a)^4 + . . .+ (w/2)*(1+a)^n

Here,n=log(1+a) 2,then S=w*(2a+1)/(2a).

 

The Loss rate is derived by

L=1/S= (2a)/ (w*(2a+1)).

 

The TCP takes time to increase its window size from w/2 to w is calculated by

n*RTT = (log(1+a) 2 * RTT

This is independent of TCP’s average throughput.

TCP’s average throughput is derived by

Throughput B=MSS*S/((n+1)*RTT) =MSS/(L*(k+1)*RTT)

Note that the derived throughput is different from original throughput.

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