ISRO-2013-41

Question

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?

• A. $20$%
• B. $25$%
• C. $40$%
• D. $66$%

Comments

GO to the Root of Concept, Video Explanation with timestamp 15 Questions on Stop Wait Protocol GATE PYQs MIT Berkeley Efficiency and Throughput With NOTES

Answer 1

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%.

Comments

transmission time is transmitting useful data.

propagation time is carrying the data to the other end.

Ideally we want the transmission to continue without any stopage (for 100% efficiency).

---

Dear that is propagation delay it means to fill the channel with the data, but the actual time is 20ns which will be taken by the frame to travel towards the destination/receiver.Thanks.

---

You are saying it reverse.

Answer 2

Efficiency=1/(1+2*t_p/t_t) =1/(1+60/20)=1/(1+3)=25%

Answer (B)

Answer 3

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

Answer 4

25% is answer

Tt=20ns, tp=30ns

Efficiency= Tt/Tt+2Tp

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

                                  Efficiency = 25%

Answer 5

i think ans===25% option (B)correct

ɳ=1/1+2*a

where a=propagation delay/transmission delay

Answer 6

i think ans===25% option (B)correct

ɳ=Tt/Tt+2*Tp

where a=propagation delay/transmission delay