7,487 views
7 votes
7 votes

What will be the efficiency of a Stop and Wait protocol, if the transmission time for a frame is $20$ns and the propagation time is $30$ns?

  1. $20$%
  2. $25$%
  3. $40$%
  4. $66$%

6 Answers

Best answer
15 votes
15 votes
Transmission time = 20ns.

Time for ACK to reach back $= TT + PD + TT_{ack} + PD \\= 20 + 30 + 0 + 30 = 80ns.$

Efficiency is the ratio of time where useful data is sent to the total time (or it is the ratio of the amount of useful data sent to the maximum amount of data that could be sent)= 20/80 = 25%.
selected by
9 votes
9 votes

Efficiency=1/(1+2*tp/tt) =1/(1+60/20)=1/(1+3)=25%

Answer (B)

5 votes
5 votes

Efficiency in Stop and Wait Protocol is given by 

Efficiency =  T,transmission / (T,transmission + 2* T,propagation)

Efficiency= 20 / (20+2*30)

               = 20/80 =0.25

For percentage = 0.25 * 100 =25%  Option (B) is correct

1 votes
1 votes
25% is answer

Tt=20ns, tp=30ns

Efficiency= Tt/Tt+2Tp

20ns/(20+30*2)ns=20/80=0. 25*100

                                  Efficiency = 25%
Answer:

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