MadeEasy Test Series: Computer Networks - Congestion Control

Question

Assume a scenario where the size of congestion window of a TCP connection be 40 KB when a timeout occurs. The maximum segment size (MSS) be 2 KB. Let the propagation delay be 200 msec. The time taken by the TCP connection to get back to 40 KB congestion window is _________ msec.

Answer 1

lets convert everything in MSS...

Mss is of size=2Kb.....

SO given window size=40...convert into MSS=40/2=20..

So new threshold=20/2=10.

now our Aim is to reach upto MSS as 20 again...

ALGO STARTS

(slow start)1|2|4|8|9|10|(congestion avoidance phase)11|12|13|14|15|16|17|18|19|20..

reached.

total transmission are vertical lines=15.

so time required=15*RTT=15*400=6000 ms.

Comments

In the slow start phase after 8 don't you think 10 should come because we haven't reached the threshold yet so we can increment till 10.