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In optimal window size for sender or for finding the maximum window size of sender N, do we consider N=1+2a or N=a.
Where a=(propagation time/Transmission time)?

For max efficiency: n*Tt should be equal to Tt+2Tp.

ie n=1+2a. But in some places it is just n=2a.
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Yes,

In optimal window size for sender or for finding the maximum window size of sender N, we consider N=1+2a

optimal window size is when link utilization is 100% .

(N/1+2a ) = 1  where N is window size in terms of packets
 

In case optimal window size, we can indirectly say that maximum throughput should be achieved. It means link utilization is 100% .
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it varies according to the question if u r talking about effficiency they are many formulas according to the situation

generly tt/(tt+2*pt)

for stop and wait it is as above only it can also be written as 1/1+2a

for sliding window protocol it is n/1+2a

for go back n also it is same

for csma/cd

1/1+5.44a

for throughput

it is efficiency * bandwidth
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for maxium no of window size (it means maximum utilization of channel)so for maximum utillization the sender must send frames untill ack comes back or it can be said like this the frames must be sent untill RTT and then divide it by frame size u will get the max no of frames

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