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Question

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is answer around 5?

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10.

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Oh . answer will be 10..

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getting 10?

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10 ?

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Yes ..but i did like this

for 100% efficiency Ws = 1 + 2a ==> Ws =  9.43..so it should be 9 right ?

in Tt + 2Tp we should send max ...hence [Tt + 2Tp] = Ws*L

what is wrong please correct

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@jatin khachane 1 no it should be 10, because below 9.43 the efficiency cannot be 100 %, but above 9.43 whichever integer size you take efficiency will be 100%

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@Shubhgupta @Shobhit Joshi

@Utkarsh Joshi

Efficiency(E)  = [Ws*L] / [RTT*BW]

efficiency always Less than or equal to 1

[Ws*L] / [RTT*BW] <=1

==> Ws <= [RTT*BW] / L

===>Ws <= 9.43

now how can it be 10 ..then it means efficency is greater than 1 ???

Here m taking  RTT = Tt+2Tp

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@jatin khachane 1 if you make size <= 9 efficiency will always be less than 100

when you make the size >= 10 the efficiency will be 100, it's like the it is transmitting data until one RTT i.e N*Tt > Tt+2*Tp

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Both seems right to me ... :(

But if we take Ws = 10..efficency will be > 1

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@jatin khachane 1, To get 100% utilization, the sender will keep on sending frames till he gets the first ack i.e. in the time span of 1RTT+ 1st frame transmission time=   604 ms 

Now, by simple division with frame transmission time, we obtain the number of frames as 9.43 which only implies, that the ninth frame has been partially sent and the sending is in progress. Thus 10 frames is the sender's window.

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@Shubhgupta

So when we get such case ..we should take ceil then ? everytime

@Shaik Masthan

Can you please check this

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it's 10

and Shubhgupta explanation is absolutely right

by simple division with frame transmission time, we obtain the number of frames as 9.43 which only implies, that the ninth frame has been partially sent and the sending is in progress. Thus 10 frames is the sender's window.

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thanks for correcting @Shubhgupta @Magma

only doubt is

by simple division with frame transmission time, we obtain the number of frames as 9.43 which only implies, that the ninth frame has been partially sent and the sending is in progress. Thus 10 frames is the sender's window.

should it be like ..9 frames completely send and 10th frame transmitting in progress when ack comes ??

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@jatin khachane 1 yes

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ya sorry it was 10th frame only.

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It should be 9...if it's 10 then according to your formula efficiency will be more than 1

Answer 1

Given, 

Bandwidth(B)=64 kbps=64 x $10^3$ bps

$Tp = 270msec$

Length of packet (L)=$512 Byte= 512 \ast   8  bits $

$\therefore Tt= L/B=512 \ast 8/64 \ast 10^3 = 64 msec$

For maximum utilization window size should be,

$Ws= 1+2a$

      $= 1+ 2 Tp/Tt$

      $= 1+ 2 * 270/64$

      $ = 1+ 8.44 $

      $=9.44 \cong 10$