3 votes 3 votes Compute approx. optimal window size, when packet size is 53 B, RTT is 60 micro-sec and bottleneck bandwidth is 155Kbps. Computer Networks computer-networks sliding-window go-back-n + – Tuhin Dutta asked Dec 19, 2017 Tuhin Dutta 890 views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments joshi_nitish commented Dec 19, 2017 reply Follow Share @Ashwin, yes, it is correct. 0 votes 0 votes Tuhin Dutta commented Dec 19, 2017 reply Follow Share The given answer is 21. In 1 sec ------- 155 * 106 In 60*10-6---------155*60 bits. So optimal window size should be floor(9300/(53*8)) = 21. 0 votes 0 votes Akshay Dixit commented Jan 19, 2018 reply Follow Share Please check RTD is 60 micro seconds or 60 milliseconds ? 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes We use 1+2A only when it is given that there is full utilization of channel. If nothing is mentioned then use this, Bw * RTT / Packet size = ( 155 * 10^3 * 60 * 10^-3 ) / 53 * 8 = 21.93 = Take ceil(21.93) = 21 Yogesh Mandge answered Dec 22, 2017 Yogesh Mandge comment Share Follow See all 3 Comments See all 3 3 Comments reply Akshay Dixit commented Jan 19, 2018 reply Follow Share Any reference source ? 0 votes 0 votes Akshay Dixit commented Jan 19, 2018 reply Follow Share @Yogesh Answer is not correct by the above formula , Given RTD = 2*Tp = 60 micro seconds = 60*10^ ( -6) seconds , but in your answer you considered it as milliseconds . Ans 21 is only possible if in question RTD is 60 ms instead of 60 micro seconds 0 votes 0 votes the_bob commented Sep 3, 2018 reply Follow Share @ Yogesh Mandge It has been given in the question that we need to find the optimal window size, which I guess, corresponds to utilizing the condition for maximum efficiency... 0 votes 0 votes Please log in or register to add a comment.