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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)
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Given Data :

 bit rate  =  128 Kbps 

 round trip time(2Tp) = 40 msec 

Therefore Tp =  20 msec

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

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

efficiency  = (Tt / Tt + 2Tp) = 32 / (32 + 2*40)  = 0.44444444

Answer :

Efficiency percentage = 44.44 %

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first find transmission time by using formula ( L / B )

find propagation time ( RTT / 2 )

to find efficiency use formula 1 / (1+2a) ; where a=(propagation time / transmission time)

multiply it by 100 to get answer in %

that's it

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