Stop and wait link utilisation


#1

Find the link utilization in stop and wait protocol if the bandwidth of the line 2 kbps, round trip time is 20 second and the packet size is 2000 bytes.

Bandwidth-delay product =2×10^3×20=40000

But the system is sending only 2000*8=160000 bits

so, Link utilisation is (40000/160000)×100=25%

given ans is different .
please correct me


#2

Link utilization = sender utilization = line utilization = efficiency = Tt/(Tt+2Tp)

Now Tt=L/B = 2000*8/(2000) = 8 s

efficiency = 8/(8+20) = 8/28 = 28.57 %

Additional Notes:

Bandwidth utilization = effective bandwidth = Throughput = L/(Tt+2Tp)


#3

@Ruturaj my solution is completely wrong n?
right ans is 28.57%


#4

No your solution is not wrong. But try to multiply correctly.

Bandwidth Delay pdt = 40000

System is sending 2000*8 = 16000 bits

Efficiency = (16000/40000) *100 = 40%

In my given solution, majority of books neglect the transmission time in denominator,
That is they just calculate Tt/2Tp = 8/20 = 0.4 = 40%

So either way you can do. But to be accurate it should be 28.57%.


#5

:anguished: calculation mistake
thanks