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2 Answers

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when a timeout occurs, the sender will go to slow start phase

New threshold=$\frac{cwnd}{2}=\frac{18}{2}$=9 KB

cwnd=1 MSS=1 KB

first transmission(after timeout)=1 KB

Second transmission=2 KB

Third transmission=4 KB

Fourth transmission=8 KB

After 4 successful transmissions cwnd=8 KB
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Slow start threshold = 18 MSS / 2 = 9 MSS

Congestion window size = 1 MSS

Window size at the start of 1st transmission = 1 MSS

Window size at the start of 2nd transmission = 2 MSS

Window size at the start of 3rd transmission = 4 MSS

Window size at the start of 4th transmission = 8 MSS

Window size at the start of 5th transmission = 9 MSS

Thus, after 4 successful transmissions, window size will be 9 MSS or 9 KB.

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ajaysoni1924 asked Mar 18, 2019
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