# tcp congestion avoidance

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?


Answer:

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.

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

## 1 Answer

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