SWP OPTIMAL WINDOW SIZE

Question

Comments

here we will take the optimal window size as (1+2a) or The bytes which we could send in a RTT.?

Answer 1

Propagation Time (Tp)= d/v = (10^6)/((2/3)*3*10^8) =5msec

No. of bits transmitted in 1 sec = 128*10^3

No. of bits transmitted in 10msec = 1280 bits

Window SIze = 1280/(32*8) = 5

Hence 5 is the correct answer

Comments

Isn't the window size will be 1+5 as 1 packet is propagating and other 5 can be transmitted in that time. please correct me if i am wrong

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@ aman_singhr  I think you are correct,but in some text question 1 is not added and in some question it is added.

@ Habibkhan Please look at this.I think we should add 1 in WS as it is also the part of window.

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It should be 1+2a.  I believe this question being taken in ACE test series. Don't trust there answers completely. I have seen lot of questions solution they are telling wrong.

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@Ashwini Kumar 2, here why you are taking the RTT to be only $2*T_p$. Isn't RTT the total time from the data sent to the time ack is received. In that case it should be $T_t+2T_p$, $T_t$ for ack is ignored, then answer would be 6, i guess

Answer 2

1+2a 

Optimal window size is 6