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Optimal window size = $\frac{B*RTT}{L}$

RTT = 2*Tp = 2*$\frac{1*10^6}{2*10^8} = 10ms.$

Window size = $\frac{128*10^3 * 10 *10^-3}{32*8}$

                     =  5.

So,optimal window size is 5.

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propagation delay =distance/propagation speed 

                          =$\frac{10^{6}}{2*10^{8}}$

                          =0.5*$10^{-2}$

                         =5ms

Roundtrip time=2*P.delay

                      =2*5=10ms

1sec------------128Kbps

10ms-------------?

=128*10

=1280.

but packet size=32B=32*8=256

SWS=1280/256

        =5.

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6

As window size=1+2a

a=propagation time/transmission time

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