Consider a simplified TCPs AIMD algorithm where thecon- gestion window size is m
ID: 3619348 • Letter: C
Question
Consider a simplified TCPs AIMD algorithm where thecon- gestion window size is measured in number of segments, not inbytes. In additive increase, the congestion window size increases by one segmentin each RTT. In mul- tiplicative decrease, the congestion window size decreases byhalf (if the result is not an integer, round down to the nearest integer). Suppose thattwo TCP connections, C1and C2, share a single congested link of speed 30 segmentsper second. Assume that both C1 and C2 are in the congestion avoidance phase.Connection C1s RTT is 100 ms and connection C2s RTT is 200 ms. Assume that when thedata rate in the link exceeds the links speed, all TCP connections experiencedata segment loss. a. If both C1 and C2 at time t0 have a congestion window of 10segments, what are their congestion window sizes after 2200 msec? b. In the long run, will these two connections get the sameshare of the bandwidth of the congested link? Explain.Consider a simplified TCPs AIMD algorithm where thecon- gestion window size is measured in number of segments, not inbytes. In additive increase, the congestion window size increases by one segmentin each RTT. In mul- tiplicative decrease, the congestion window size decreases byhalf (if the result is not an integer, round down to the nearest integer). Suppose thattwo TCP connections, C1and C2, share a single congested link of speed 30 segmentsper second. Assume that both C1 and C2 are in the congestion avoidance phase.Connection C1s RTT is 100 ms and connection C2s RTT is 200 ms. Assume that when thedata rate in the link exceeds the links speed, all TCP connections experiencedata segment loss. a. If both C1 and C2 at time t0 have a congestion window of 10segments, what are their congestion window sizes after 2200 msec? b. In the long run, will these two connections get the sameshare of the bandwidth of the congested link? Explain.
Explanation / Answer
For every 100ms (RTT for C1) , Congestion window of C1 willincrease by1. Similarly for C2, congestion window increase by 1 after 200ms (RTTfor C2). Initially Congestion window of C1 (CW1) = 10 similarly CW2 = 10 The capacity of link is 30...when Cw1 +Cw2 Exceeds 30, AIMD will beapplied. At t=100ms: CW1 = 11, CW2 =10 At t=200ms: CW1 = 12, CW2 =11 Therefore for every 200ms, Cw1 increase by 2, CW2 increases by1 At t=400ms: CW1 = 14, CW2 =12 At t=600ms: CW1 = 16, CW2 =13 At t=700ms: CW1 = 17, CW2 =13 At t=800ms: CW1 = 18, CW2 =14.....total is 32, exceeded linkcapacity.. At t=900ms: therefore there will be collisions and AIMD isapplied. Therefore Congested windows will be decreased by half CW1 = 9, CW2 = 7 At t=1000ms: CW1 = 10, CW2 =7 At t=1200ms: CW1 = 12, CW2 =8 At t=1400ms: CW1 = 14, CW2 = 9 At t=1600ms: CW1 = 16, CW2 =10 At t=1800ms: CW1 = 18, CW2 =11 At t=1900ms: CW1 = 19, CW2 =11...total 30 At t=2000ms: CW1 = 20, CW2 = 12...total exceeded 30 At t=2100ms: There will be collisions and AIMD is applied. Therefore Congested windows will be decreased by half CW1 = 10, CW2 = 6 At t=2200ms: CW1 = 11, CW2 =6 Therefore contentions windows of C1 (CW1 = 11) and C2 (CW2) = 6)after 2200ms We can clearly observe that C1 is exploiting the link more thanthat of C2, sending more segements per second rather than that ofC2 which has more RTT.