I came across this question:

At the beginning of transmission t, a TCP connection in congestion avoidance mode
has a congestion window w = 60 segments.Packet loss is observed during transmission
rounds t, t+10, and t+20 by getting multiple ACKs. What is the congestion window 
at the end of round t, t + 10, and t + 20? 

If there's no further packet loss, when will the window of w = 60 segments be reached again?


Congestion window is halved during transmission round t, leading to w = 30. At the
beginning of transmission round t + 10, the window has increased to w = 40 but it will be
halved again during transmission round t + 10 to w = 20. Similarly, after transmission round
t + 20, the window will be w = 15. With no further packet loss, 45 transmission rounds later
at t + 65, the window will reach again the original size of w = 60.

I know that due to multiple ACKs, packet loss reduces the window size to half. So at t+10, w = 30 makes sense. However, I don't really get the rest. Why does the window size increase by 10? And at t+20, why doesn't it reduce to 10 instead of 15? If someone could explain the steps, that would be great.

  • $\begingroup$ Is that all the question or there is more information regarding the data? $\endgroup$
    – Noname
    Commented Apr 21, 2020 at 5:59
  • $\begingroup$ That is all @Noname $\endgroup$
    – x89
    Commented Apr 22, 2020 at 0:11

1 Answer 1


When we are in AC mode every rtt(round trip time) the congestion window increase by 1(MSS). Here our current congestion window(cw) is 60 and we know that during transmission round t we get 3 duplicate acks-so our window is now 30MSS,now we also know that from t to t+1 for instance our window increase by 1,so from t to t+10 our window increases by 10-which means our window is now 40. We know that during the t+10 round we will get 3 duplicate acks,so now our window will become 40/2=20. From round t+10 to t+20 our cw increase by 10 more-so now its 30,and in addition we know we will get 3 duplicate acks-which means now our window will be 30/2=15MSS


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