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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____________ (in msec). Assume window size at the start of slow start phase is 2 KB

 

 

ans 70 msec ....explain if anyone can?
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2 KB 

4 KB

8 KB

16 KB

18 KB (Threshold)

20 KB

22 KB

24 KB ( Time Out)

2 KB ( New Window Size = 12 KB)

4 KB

8 KB

12 KB (Threshold)

14 KB

16 KB 

18 KB

Bold counts to seven =70 ms .

correct me .... and thank you @Subhanshu sir and Pranab for pointing 

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given time out will be after 8 transmission so 2kb|4kb|8kb|16|18kb(threshold) |20kb|22kb|24 kb (8th no transmission from sender)..time out------------------>new threshold 24kb/2=12kb.

again start sending

2 kb| 4kb| 8kb| 12kb| 14kb|16kb|18kb . here first full window sent after 6 RTT's so its 60ms. but question asked  the time taken to send first full window ...which is 7RTT'S hence answer is 70ms

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