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Question

In selective repeat ARQ, packet size is 2000 bytes transmission time for one packet is 1ms. If distance between hosts is 10km and signal speed is 4ms per km (4ms/km) and frame sequence number are 6 bit long in frame format then the throughput (in Mbps) is

Answer 1

The Length of the packet = 2000B.

Transmission time (Tt) = 1ms

Tt = Length of Packet / Bandwidth

0.001 sec = 2000B / Bandwidth

Bandwidth = 2 x 10^6 B/sec = 2MB/sec = 16Mbps  [1B=8-bit]

Distance between hosts = 10km and signal speed is 4ms/km

So, total propagation time (Tp) = 10 km x 4ms /km = 40ms

As the protocol given is selective repeat ARQ, and we are given total 6-bits for frame sequence number. With 6 bits we can generate total 64 different combinations.

In Selective repeat ARQ the widow size of sender and receiver is equal.

So, Window size of receiver = Window size of Sender = 32

Now,

Efficiency = (Useful Time / Total Time) x (Window Size)

                  = {Tt / (Tt+2Tp)} x Ws

                  = (1ms / 1ms + 2x40ms) x 32

                  = (1/81)x32 = 0.395061

Throughput = Efficiency x Bandwidth

                      = 0.395061 x 16 Mbps = 6.320987 Mbps