1 votes 1 votes Q. Compute the approximate optimal window size when packet size 53 byte and round trip time (RTT) is 60 ms and bandwidth (BW)155 mbps (in byte) using sliding window protocol. Computer Networks computer-networks + – kallu singh asked Aug 25, 2017 kallu singh 762 views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply saxena0612 commented Aug 25, 2017 reply Follow Share @joshi_nitish @just_bhavana Check it once! 2 votes 2 votes G.K.T commented Aug 25, 2017 i edited by G.K.T Aug 25, 2017 reply Follow Share For optimal window size exhaust your bandwidth in given RTT so sender in never idle (Thats the basic idea). so bandwidth = 155Mbps and RTT = 60 ms so in 1 sec we may send 155MB in RTT it should be 155 MB * 60 ms = 9300 KB packet size is 53 B win size = 9300 KB / 53 B =175.47169 K or 175471 (taking floor as what we have just calculated is the maximum limit so cant go further). 0 votes 0 votes saxena0612 commented Aug 25, 2017 reply Follow Share @G.K.T Bandwidth is given in BytesI I guess? 0 votes 0 votes G.K.T commented Aug 25, 2017 reply Follow Share Yeah thanks for pointing . 0 votes 0 votes just_bhavana commented Aug 25, 2017 reply Follow Share Yes, it's correct @saxena0612 2 votes 2 votes joshi_nitish commented Aug 25, 2017 reply Follow Share correct. 0 votes 0 votes Shubhanshu commented Aug 27, 2017 reply Follow Share why you are not using w= 1 + 2a 0 votes 0 votes Shubhanshu commented Nov 10, 2017 reply Follow Share @saxena0612 @joshi_nitish For maximum eff = 100% = 1 1 = $\frac{n*TT}{TT + 2PT}$ where n = window size. TT = 2.73 sec PT = 30 $\frac{n*2.73}{2.73 + 60}$ = 1 n = 22.97 taking floor value it is 22. If we take ceil value then it will be more than 100%. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Hope you will get this...:) sandeepjkh answered Aug 25, 2017 sandeepjkh comment Share Follow See all 0 reply Please log in or register to add a comment.