MadeEasy Test Series 2017: Computer Networks - Congestion Control

Question

Let the size of congestion window of a TCP connection be 38 KB when a timeout occurs. The propagation time of the connection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 36 KB congestion window is ________.

Comments

Rahul... See my solution.. Why you take 16kb as threshold.. Initially Window size was 38kb when timeout occurs threshold will be half of it that is 19kb.. Therefore ca will start from 19kb onwards

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I am saying threshold is 19Kb only.

 But Slow start goes like 2kb,4kb,8,kb,16kb now next it will try to goto 32kb .But this is not allowed so now we start congestion avoidance,and goto 18Kb

See the given below solution.they will go from 16kb to 18kb then 20 and so on

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Hmm.. This is really contradictory to what I was doing all the way.. Nywz I will just cross check it.. Until I get well confirmation..

Answer 1

 If congestion is detected by time out error:

1. Set threshold value to half of the current window size where congestion is detected. New threshold = 38/2 =19 KB
2. start the slow-start phase. Hence new window size = 2 KB

Communication will be like ==>> 2KB | 4KB | 8KB | 16 KB  (now threshold reaches so Additive increase starts) | 18 KB | 20 KB | 22 KB | 24 KB | 26 KB | 28 KB | 30 KB | 32 KB| 34 KB | 36 KB

NB: Each RTT is show as | (vertical line), so total time taken = 13* 100ms ==>> 2600 msec