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

mss=2KB
–1 votes
–1 votes
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

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