Transport layer

Question

Assume slow start mechanism is used and assume there is no congestion . The round trip time is 10 ms and maximum segment size is 24 KB . How much time is required to reach 24 KB of window size with packet size of 2 KB?

A. 50 ms

B. 30 ms

C. 120 ms

D. 40 ms

Comments

D. 40 ms

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I got it , you are almost correct Bhagyashree Mukherje
Before the 2nd transmission(i.e after the 1st RTT the window size is becoming 4kb )
Before the 3rd transmission(i.e after the 2nd RTT the window size is becoming 8kb )
Before the 4th transmission(i.e after the 3rd RTT the window size is becoming 16kb )
Before the 5th transmission(i.e after the 4th  RTT the window size is becoming 24kb )
So only 4 RTT needed to reach 24kb window size.

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One small doubt..how it started from 2kb..slow start starts from initially 1 MSS which is 24 kB..how 2 kB becomes initial window size?