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Consider sliding window protocol for a 10MBps channel with the propagation delay of 300 µs. If packet size is 1kB then what is the maximum link utilization for window size of 127?

 

a. 100%

b. 75%

c. 2.8%

d. 43%
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Answer: 100%

Tp=Propagation Delay= 300 x $10^-6$ sec

Tt=Transmission Delay= Length/BW =1000/$10^7$ =$10^-4$ sec

a=Tp/Tt= 3

N=7 (I think GateForum misprinted 127 instead of 7, as i have seen this same question at a diffrent place with same options but 7 instead of 127)

Utilization = N/(1+2a) = 7/(1+2x3) = 7/7 = 1 = 100%
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