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? $20$% $25$% $40$% $66$% Computer Networks isro2013 computer-networks stop-and-wait + – makhdoom ghaya asked Apr 27, 2016 makhdoom ghaya 7.5k views answer comment Share Follow See 1 comment See all 1 1 comment reply GO Classes Support commented Dec 15, 2023 reply Follow Share 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 0 votes 0 votes Please log in or register to add a comment.
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%. Arjun answered Apr 27, 2016 selected Jun 27, 2016 by srestha Arjun comment Share Follow See all 6 Comments See all 6 6 Comments reply dhairya commented Apr 28, 2016 reply Follow Share @arjun Why have u taken 20 as usefull data..? It should be 30ns as it is the time when some useful data was send...?? –1 votes –1 votes Arjun commented Apr 28, 2016 reply Follow Share 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) 3 votes 3 votes dhairya commented Apr 29, 2016 reply Follow Share 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 ..? –1 votes –1 votes Arjun commented Jun 8, 2016 reply Follow Share 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). 2 votes 2 votes sandeee commented Jun 28, 2016 reply Follow Share 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. –1 votes –1 votes Arjun commented Jun 28, 2016 reply Follow Share You are saying it reverse. 0 votes 0 votes Please log in or register to add a comment.
9 votes 9 votes Efficiency=1/(1+2*tp/tt) =1/(1+60/20)=1/(1+3)=25% Answer (B) srestha answered Apr 27, 2016 srestha comment Share Follow See all 0 reply Please log in or register to add a comment.
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 Ashwani Kumar 2 answered Apr 29, 2016 Ashwani Kumar 2 comment Share Follow See all 0 reply Please log in or register to add a comment.
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% Soni Bisen answered Dec 6, 2018 Soni Bisen comment Share Follow See all 0 reply Please log in or register to add a comment.