MadeEasy Test Series 2017: Computer Networks - Congestion Control

Question

Consider the effect of using slow start on a line with 10 msec round trip time. The receiver window and the size of congestion window are set to 38 KB and 36 KB respectively. Sender side threshold is set to 18 KB. After 8 transmission a time-out occurs, after time out, the time taken to send first full window of 18 KB is______ Assume window size at the start of slow start phase is 2 KB.

Comments

And i think 65 should be answer.

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how are you assuming mss = 2kb

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I have made easy test series and question has stated that. Question here is incomplete but in ME test it is given.

Answer 1

congestion at 18Kb

let's take 8 transmissions

2->4->8->16->18->20->22->24 here time out occurs

 so new threshold=1/2(24)=12 KB

now to pass the full window of 18 Kb transmissions will be

2->4->8->12->14->16->18

7 Transmissions

so total time =7*10=70ms

so 70 ms is correct answer

Comments

@Tuhin Dutta

we increase the size of congestion window by 1 for each ack till the threshold ...when threshold has been reached we increase window size by 1 when we get acks for whole window ...

2->4->8->12->14->16->18"

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Here 1MSS = 2 KB. Since by exponential increase it is getting over the threshold (8+8 = 16 KB) we would increase by 2 KB till it reaches threshold and beyond that too until timeout or 3 dup ACKs occurs. Why jump from 8 to 12?

If that is the case, then for the first case : 

2->4->8->16->18->20->22->24 here time out occurs.

why is it not going 16->24? 

Please explain.

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Because for first case threshold is 18KB

Answer 2

ans is 70msec (after time out it takes 7 R.T.T)

mss=2KB

Comments

how did you assume mss=2KB

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@Rajnish Kumar

why we have taken MSS = 2KB?

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Because it's given in the question that assume window size at the start of slow start is 2kb then take 2kb is equals to 1mss.

Answer 3

i think answer should be 5ms

as max capacity of sender is18KB

and  max capacity of reciever is38KB and congestion window is 36KB

so full window of sender can be send in Tp=RTT/2

=10/2=5ms