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If a bandwidth of the line in a stop and wait protocol is 2.5mbps and one way propagation is 20 ms what is the link utilization if size of packet is 2kBytes is___
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For stop and wait protocol the efficiency is 1/1+2a where a =Propagation Time/ Transmission Time

The utilization of the link i.e. the bandwidth is the throughput i.e. efficiency *the given bandwidth.

So for this question :

Tt=Length of the packet in bits/bandwidth=2*8*1000/2.5*10^6=6.4ms

Tp=20ms

efficiency=1/1+2*20/6.4

and link utilization=efficiency *bandwidth
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