3.2k views

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. $66$%

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
–1
@arjun

Why have u taken 20 as usefull data..?

It should be 30ns as it is the time when some useful data was send...??
+2
Transmission time is the time where data is sent. You can also consider it like the ratio of amount of data sent to the amount of data that could have been sent. (Data is sent in a pipelined manner)
–1
transmission time is the time where data is put on the transmission channel...?

and propagation time is when data is taken from transmitter to receiver...? right na...??

so propogation is the time where actual data is send ..?
+2
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).
–1
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.
0
You are saying it reverse.

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

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

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

ɳ=1/1+2*a

where a=propagation delay/transmission delay
i think ans===25% option (B)correct

ɳ=Tt/Tt+2*Tp

where a=propagation delay/transmission delay

Tt=20ns, tp=30ns

Efficiency= Tt/Tt+2Tp

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

Efficiency = 25%

1
2