Question

A channel has a bit rate of 15kbps and a propagation delay of 20ms. For what value of frame size does stop-and-wait protocol give an efficiency of 25 percent? Any help is highly appreciated.

Answer 1

Answer:

We know maximum utilization for the protocol: stop and wait is :

1/ 1 +2a

Here is the delay in bandwidth. Value of a can be found by :

a = Tp/Tt

Here Tp = Propagation delay, Tt = transmission time.

We know that Tt = Frames/ rate of data

We have given data rate = 15kbps ( convert kbps into bps)

= 15kbps = 15 x 10^3 bps

Let frames = F

Then transmission delay = F/ 15 x 10^3 bps

Propogation delay = 20ms ( let us convert mili second into second here)

= 20ms = 20 x 10^ -3

Hence Propogation delay = 20 x 10^3 s

Hence a = Tp / Tt = 20 x 10^-3 / F/ 15 x 10^3 = 20 x15 / F = 300 /F

Hence a = 300/F

We know utilisation = 1 / 1 + 2a

Here a = 300/F

Therefore utilisation = 1 / 1 + 2 ( 300/F)

But we have given that utilisation = 25% = 0.25

Hence 0.25 = 1 / 1 + 2 ( 300/F)

=> 0.25 = 1 + 600/F

=> 1 = 0.25 x (1+ 600/F)

=> 1 = 0.25 + 150/ F

=> 1 - 0.25 = 150/F

=> 0.75 = 150/F

=> F = 150/0.75

=> F = 200

Thus frame size needed to achieve 25% of efficiency using stop and wait protocol = 200 bits

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