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+1 vote
Compute approx optimal window size when packet size is 1000 bits and propogation delay is 250msec and bottleneck bandwith is 50kbps
asked in Computer Networks by Veteran (15k points) | 104 views
Optimal window size (N)would be 1+2a // 100%efficiency


PT=propagation time and TT=Transmission time
yes so the calculation of 2a gives 25
and 1+2a=26
so answer should be 26
but many places they give 25 as the answer
can u just check it
They didn't considered transmission time...

actually it's Bandwidth delay product..

(RTT*BW)/Data size = maximum # of packets that can be send in one go...

(RTT*BW)/Data size

this is only the calculation for 2a

why is it not 1+2a

thats my doubt


@Rupendra Choudhary how to identify when to use 1+2a and when to use (RTT*BW)/Data size.

Is there any keyword in the question which gives us hint to differentiate these formulas.

yea exactly @subhanshu that is what i wanted to know
Lets clear it from one simple example.

Lets suppose you're(A) at your home and you have to reach office and come back. lets suppose you need 2 second to get ready , after getting ready you leave for office and in 2 seconds you reach your office. Now from office next moment you're coming back (now you don't need any time to get ready) and you take same 2 second to come back..

What does MWS(max window size)means? it means how many maximum persons can leave their room and start the journey for office before you comes back (ACK for 'A' actually) so take same time scenario for every person..

so what happen when you reach office another person B can get ready in those 2 seconds and after you reach  back at your room , B would have reach office in those 2 seconds and another person 'C' would have get ready and just departed for office tell me how many maximum perosons can be departed from their room , in time you depart and comes back to your room...3 right?

It's 3 actually...

so i will go with 1+2a ....

nicely explained

maximum persons can leave their room and start the journey for office before you comes back 

but this is 2 right( B and C)


window size is maximum packets that can be transmitted before coming ACK for 1st packet...

so i suppose me must include A too...A , B ,C transmitted before A's ACK reach...
So in any case it should be 1+2a

and for the above one the ans should be 26?

@arjunsir can u please confirm it  and help us out

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