We know :
Throughput(effective data rate) = Link Utilisation * Bandwidth
So for window size of 1 , we have :
Link Utilisation = T.T. / [ T.T + 4 * P.T. ]
= [(512 * 8) / (64 * 103)] / [ { (512 * 8) / (64 * 103) } + 2 * 270 * 10-3]
= 64 / (64 + 540)
= 64 / 604
= 0.106 = 10.6 %
So throughput = 0.106 * 64
= 6.7 Kbps in case window size
Now for
b) Window size of 7 , efficiency = 0.106 * 7 = 70.42 %
So throughput = 47 Kbps
c) Window size of 15 , throughput = 6.7 * 15 = 106 Kbps ..But it exceeds maximum link capacity(bandwidth) which is unrealistic.So it is limited to maximum bandwidth only..Hence throughput = 64 Kbps
d) Window size of 90 , throughput = 0.6 Mbps..Same here also..It will be 64 Kbps only..
Hence option 2) is correct..