7 votes 7 votes Consider TCP congestion control. Assuming RTT as 4 seconds and segment size as 3KB, if bandwidth is 500kbps, what is the smallest window size for which there is no stalling in this case? Computer Networks computer-networks tcp + – rahul sharma 5 asked Aug 16, 2017 rahul sharma 5 1.7k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply joshi_nitish commented Aug 16, 2017 reply Follow Share 252KB ?? 0 votes 0 votes sourav. commented Aug 16, 2017 reply Follow Share $651KB$? 0 votes 0 votes Please log in or register to add a comment.
4 votes 4 votes Given Data Bandwidth = 500 Kbps Segment size = 3KB Round Trip Time = 4 sec So window size will depend upon the data we can send in 1 RTT i.e in 4 sec To send 500 Kb we need 1 sec In 4 sec we can send = 500 x 4 kb Given segment size = 3 KB Therefore number of such segments we can send in 4 sec = $\frac{500 \times 4 k}{3 \times 2^{3} k}$ So in 4 sec we can send $\left \lfloor 83.33 \right \rfloor = 83$ segments Therefore smallest window size should be 83 stblue answered Aug 16, 2017 stblue comment Share Follow See all 9 Comments See all 9 9 Comments reply just_bhavana commented Aug 16, 2017 reply Follow Share why have you taken floor ? Shouldn't it be ceil, else 0.33 segments of data will be lost 1 votes 1 votes stblue commented Aug 17, 2017 reply Follow Share just to send 0.33 segment we need to do further segmentation, since 0.33 is not enough for 3KB of data, so its better to avoid 0.33 segment. 0 votes 0 votes rahul sharma 5 commented Aug 17, 2017 reply Follow Share Please clear following points 1. What does "No stalling" means in questions ? 2. If you are finidng windows size in similar way as sliding window,then answer should be 83.33+1=84.33 ? 0 votes 0 votes stblue commented Aug 17, 2017 i edited by stblue Aug 17, 2017 reply Follow Share 1. "No stall" means we don't want to stop sending packets, suppose we can send 10 packets, but we are sending only 8 of them at once, After sending those 8 packets, we are not doing any useful work for remaining time in which we can send 2 more packets, so here we put a stall of 2 packets, In other words, i think its the same thing saying we want to achieve maximum efficiency. 2. Why you are adding 1 to it ? 0 votes 0 votes rahul sharma 5 commented Aug 17, 2017 reply Follow Share I am adding + 1 because the maximum sender window size in sliding window is 2*Bandwidth*Delay+1.or Number of frames in RTT+1. 1=N/(1+a) // this formulae also if you apply,will give you 84.33 0 votes 0 votes stblue commented Aug 18, 2017 reply Follow Share Even i used to think same i.e number of frames should be RTT+1 , and logically it seem correct. But everyone is using RTT.. Here is similar discussion you can follow, read the replies on bikram's sir comment. https://www.facebook.com/groups/gateoverflow/permalink/654549848083562/ If you can provide me some known link where you read RTT+1, That will Helpful. @Arjun sir, @joshi_nitish , @just_bhavana what should be the window size RTT or RTT+1. 0 votes 0 votes rahul sharma 5 commented Aug 18, 2017 reply Follow Share Maximum window size is always number of frames in RTT+1 always. Refer tanenbaum Sec 3.4 p 233 it explained this thing beautifully 1 votes 1 votes Rajesh Panwar commented Jan 14, 2019 reply Follow Share i don't think this one is correct solution 0 votes 0 votes sachin486 commented Dec 29, 2021 reply Follow Share in Gate please be carefull to take 5Kb as 5000kb and 3KB data as 3 x 8 x 1024 bits so that u can get correct answer that is 83 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes i think window size:- 85 segments ???? ... if not ,please explain.... hs_yadav answered Aug 16, 2017 hs_yadav comment Share Follow See 1 comment See all 1 1 comment reply Rajesh Panwar commented Jan 14, 2019 reply Follow Share could you explain? 0 votes 0 votes Please log in or register to add a comment.