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Consider two computers A and B are
connected through a network of 30 Mbps. Assume the distance between them is 3000km and the signal propagation speed is same as the speed of light and the packet size is 12 KB.
What is the minimum number of bits required for window to achieve 100% of utilization during GoBack-N and selective repeat protocol?
A . 5 and 6
B. 6 and 7
C. 6 and 6
D. 7 and 8
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BW = 30Mbps, Distance = 3000Km, Speed = 3*10^8 m/s, Data Length = 12*10^3*8bits

Transmission Delay (Tt) = Data Length / BW

                                = (12 * 10^3* 8) / (30 * 10^6) = 3.2 ms

Propagation Delay (Tp) = Distance / Speed

                              = (3000 * 10^3) / (3 * 10^8) = 10 ms

Efficiency = N / (1 + 2a), where N = Window Size, and a = Tp / Tt = 10 / 3.2 = 3.125

            1  = N / (1 + 2 *3.125)

            N = ceil(7.25) = 8

For GB-N, N should be 8(Sender) + 1(Receiver) = 9, No. of bits required is 4

For Selective Repeat, N should be 8(Sender) +8(Receiver) = 16, No. of bits required is 4
Please Correct me if I am wrong

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Why sender and receiver window size are same in selective repeat?