$44.44 $ ?

4 votes

Consider that the stop and wait protocol is used on a link with bit rate of 128 Kbps and 40 msec as round trip time. Assume that the transmission time for the acknowledgment and processing time at nodes are negligible. If the frame size is 512 B then the link utilization is ________. (In % upto 2 decimal places)

3 votes

**Given Data :**

bit rate = 128 Kbps

round trip time(2T_{p}) = 40 msec

Therefore T_{p }=_{ } 20 msec

length of one frame = 512 Bytes = 512*8 bits

Transmission Time = length / bandwidth = 512*8 / (128kbps) = 32msec

efficiency = (T_{t} / T_{t} + 2T_{p}) = 32 / (32 + 2*40) = 0.44444444

**Answer :**

Efficiency percentage =** 44.44 %**