MadeEasy Subject Test 2019: Computer Networks - Flow Control Methods

Question

consider a network connecting two nodes A and B having a propagation delay of 6*10^-4 micro sec.bandwidth of network is 150 Mbps if each frame size is 5000 bytes and both uses go back n sliding window protocal, where maximum 100 frames can be sent at a time then the maximum possible data rate is ----- [ mbps]

Comments

If possible can you please provide the solution provided?

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this is the sol

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Couple of things to be corrected in question Propagation delay is  $6*10^4$ micro and the result in Mbps

@suneetha check my answer below

Answer 1

Transmission time= $\frac{Length}{Bandwidth}$

                             = $\frac{5000*8}{150*10^6}$

                             = 266.667 micro sec

Propagation time  =$6*10^4$ micro sec

Efficiency = $\frac{Transmission time}{Transmission time+2* Propagation time}$

                  =$\frac{266.667}{266.667+120000}$

                  =$\frac{266.667}{120266.667}$

                  =$\frac{266.667}{120266.667}$

                  = $0.0022173225549522576$ *100

                  = $0.22173225549522576$

 

Effective Bandwidth = $Efficiency$*$Bandwidth$

                                    =$0.22173225549522576 * 150 $Mbps

                                    =$33.25983832428386$ Mbps

Comments

thanks Hemanth_13

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here no need of checking wheather window size is greater than 1+2a?