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

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Please help, I am getting total number of transmission to get BACK TO 38 as 14 and NOT 15.

after 14th, it would have became 38, so no need to consider 15th segment right?

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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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In the question it is given in question that time required to get back to 36 KB...if we got a congestion window of 36 kb at the time when the congestion window is 37 KB why will we move further to send packets again

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yes, sorry you are correct, the question is different from what they solved.

for 38 it will be 14 na?

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how 16 to 19 becoz threshold is 19kb so 16 onward additive increase(otherwise for slow start next sender window size will be 16 to 32kb but threshold is 19kb)

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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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