But they ask to reach 36 kb n .. ? Then why 38

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Answer should be 2600

RTT will be 2*T(p) =200 ns

timeout is at 38 means new thresh hold will be floor(CWND/2)=19 KB

now till `19 kb we will have slow start algo so

we will have 2-4-8-16-19 .

form 19 we will have additive increase

so 19-21-23-25-27-29-31-33-35-37(roughly 36) here we will get congestion window of 36 which is asked in question

total 13 RTT'S

So 13*200=2600ms

RTT will be 2*T(p) =200 ns

timeout is at 38 means new thresh hold will be floor(CWND/2)=19 KB

now till `19 kb we will have slow start algo so

we will have 2-4-8-16-19 .

form 19 we will have additive increase

so 19-21-23-25-27-29-31-33-35-37(roughly 36) here we will get congestion window of 36 which is asked in question

total 13 RTT'S

So 13*200=2600ms

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because after 16 cwnd will increase as slow start upto 19 but after that we will have additive increase of cwnd+1/floor(cwnd) so after 19 additive increase will be there and cwnd will increase by 1 MSS(in actual practice you calculate by above formula you will get cwnd as 19.something and further cwnd will increase by 20.something...but from gate point of view we neglect point and move forward.....

these type of questions were asked by IIT Delhi but were not considered as fully correct to ask

these type of questions were asked by IIT Delhi but were not considered as fully correct to ask

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