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In the Following question, it is asked the time by  when the full window size has been sent. So, does that mean the time by which sender has received the acknowledgement for it 24KB sized window packets?
Because, it takes 90ms for sender to reach window size of 24KB from 2KB using slow start and additive increase. And additional 10ms to get the ack for the 24KB window packets sent. So total 100msec.
Is my approach correct?

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Yes this approach is correct!!!                             

1st trans 2 KB, takes 10 ms to receive ack of 2 kB

2nd trans 4 KB, takes 10 ms to receive ack of 4 KB

3rd trans 8 KB, takes 10 ms to receive ack of 8 KB

4th trans 12 KB, takes 10 ms to receive ack of 12 KB

5th trans 14 KB, takes 10 ms to receive ack of 14 KB

6th trans 16 KB, takes 10 ms to receive ack of 16 KB

7th trans 18 KB, takes 10 ms to receive ack of 18 KB

8th trans 20 KB, takes 10 ms to receive ack of 20 KB

9th trans 22 KB, takes 10 ms to receive ack of 22 KB

10th trans 24 KB, takes 10 ms to receive ack of 24 KB

so total 100 ms 

hence answer is 100 ms

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