# MadeEasy Test Series 2018: Computer Networks - Stop And Wait

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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)

edited
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$44.44$ ?
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yes it is 44.44 %

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

Efficiency percentage = 44.44 %

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I used following formula and i got 80%

Link capacity: (Packet size) / (Channel capacity)

channel capacity : 2 * tp * Bandwidth = RTT * Bandwidth

Where im wrong?? tell me .. if we can't use this formula then tell me why??

and where we can use this formula??

Thanks ,
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http://nptel.ac.in/courses/106105080/pdf/M3L3.pdf

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Nice

Comment if u have more such links it would be helpful to all :p
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Thank you so much..
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Hey, were you able to understand, when and when not to use this formula?
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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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Thanks bro... @Rahul_Rathod_

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