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In a Stop and WaIT ARQ protcol given Bandwidth = 1Mbps and 1bit RTT = 20ms ,Frame size is 1000bits whats the % Utilization of the link?

Soln. My approach is :-

Bandwidth delay pdt = 1x10^6 x 20 x10^-3 = 20000bits

And system sends 1000bits at a time so 1000/20000  = 5%

Is this approach Correct?

Other way :- U = Tr/(Tr + RTT)

but by this answer doesnt matches

Plz Correct me
asked in Computer Networks by Loyal (5.6k points) | 75 views

1 Answer

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assuming zero processing delay at both sender and receiver and delays for acknowledgement is zero

$T_f$= $10^3$/$10^6$ = 1 ms

$T_f$ = RTT/2 =10 ms

total time for sending 1 frame T = $T_f$ + 2$T_f$ = 21 ms

in T=21 ms time we could have send =21 ms * 1Mbps bits = 21 Kilobits = 21 packets

but in 21 ms we are sending only 1 packet

hence efficiency = 1/21 = 0.0476
answered by Active (1.9k points)


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