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Consider host A uses 32B packets to transmit to host B using SWP. The distance between host A and B is 1000000 meters and the propagation delays is 2 * 100000000 m/s. The data rate is 128 kbps. The optimal window size that A should use will be .................?
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here we have used sliding window protocol.....

tx=L/B .....that is 32*8/(128*10^3)= 2 mili second....(all are in BITS )

now they given in question that   signal travels 2*10^8 meters in 1 sec hence to travel the distance of 10^6 how much time we need=

                                                Tp=10^6/(2*10^8) sec that is = 5 mili second...

now for SLIDING WINDOW PROTOCOL we have formula as 1+2a......(a=Tp/Tx)

           by putting all values .....1+2*(5)/2 .....(mili /mili cancel)......by solving that you will get it as 6......
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Window size is 6