Stop n wait

Question

A channel has propagation delay of 10 ms .For 320 bits frame what will be channel bit rate if stop -n-wait give efficiency of 50%?

I have solved this using 2 approaches but getting different answers(as in the picture). Can anyone plz explain where i am wrong?

Answer 1

Your first approach is not understandable. Why are you multiplying propagation delay with the Bandwidth? I think it will not give total amount of data that could be sent.
Method 2 is correct.

Comments

In below link to calculate utilization size of frame/bandwidth delay product is done..thats why i multiplied RTT with B/w ..

https://books.google.co.in/books?id=bwUNZvJbEeQC&pg=PA322&lpg=PA322&dq=in+stop+and+wait+ARQ+bandwidth+1Mbps+and+1+bit+take+20+ms+to+make+a+round+trip+if+system+data+frame+are+1000&source=bl&ots=qsr_E9SupI&sig=xQkLId_-8ZCZ8qAFWoG9D3aebSQ&hl=en&sa=X&ved=0ahUKEwj9u4qo3YPKAhVRTo4KHWQ2Bd4Q6AEILTAD#v=onepage&q=in%20stop%20and%20wait%20ARQ%20bandwidth%201Mbps%20and%201%20bit%20take%2020%20ms%20to%20make%20a%20round%20trip%20if%20system%20data%20frame%20are%201000&f=false

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actually bandwidth delay product should be 2*B*10^-2..but then also answers not same

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how second method is correct? formula used here for link utilization not for  efficiency(in efficiency we consider only RTT and do not consider other overhead).
first method is almost correct only mistake is instead of  pt*B, it should be RTT*B and answer will be 32

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isn't utilization and efficiency same?