Question 2 | Stop and wait ARQ

Question

In stop and wait protocol , If the packet size is 1 KB and propogation tme is 15ms the chanel capacity is 10⁹ bits / sec then find :

1. transmission time
2. channel utilisation
3. sender utilisation
4. link utilisation
5. line utilisation
6. What do you account about the difference in channel utilisation and sender utilisation

Comments

Channel Capacity is (C) = Bandwidth X Propogation Delay

So Bandwidth becommes 66.67Gbps

Answer 1

1) transmission time (1⨉1024⨉8)/10⁹ =8 micro seconds

2) channel utilization =$\frac{1}{1+2(15000/8)}$

                             =1/3751⨉100

                            =0.0266%

link and line utilization same as channel utilization , as it is for stop and wait

link :here

Comments

@srestha,As I know sender utilization=link utilization=line utilization in case of unidirectional transfer.But if transfer is bidirectional and both side sender and receiver are present than sender utilization will be different from link utilization.

verify it.

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how sender utilization can be bidirectional?

how link utilization can be unidirectional?

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@srestha, what I want to say is,In case of bidirectional data transfer, when piggybacked ack is used,then sender utilization will decrease because in this case Rtt=Tt(sendert)+2*Tp+Tt(piggyback ack).

But link utilization will increase becz now data is transferring from both sides.

I think you got my point.