0 votes 0 votes Compute approximate optimal window size when packet size is 53 bytes, RTT is 60 microseconds and bottleneck bandwidth is 155 Mbps. 21 22 23 24 Computer Networks sliding-window computer-networks + – Amitesh Sharma asked Nov 28, 2017 Amitesh Sharma 894 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes answer is 22.....as shown down.... SHUBHAM SHASTRI answered Nov 28, 2017 • edited Nov 28, 2017 by SHUBHAM SHASTRI SHUBHAM SHASTRI comment Share Follow See all 2 Comments See all 2 2 Comments reply Amitesh Sharma commented Nov 28, 2017 reply Follow Share see Akash's ans I also got 23, but the given ans was 22 I am confused between floor and ceil 0 votes 0 votes SHUBHAM SHASTRI commented Nov 28, 2017 reply Follow Share ITS FLOOR..i also did wrong at first time 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes optimal window size = 1+2a {a=2PT/TT} TT = 53*8/155*10^6 =2.73 microsec optimal window size =1+60/2.73 =floor value[22.97] = 22 Akash Mittal answered Nov 28, 2017 Akash Mittal comment Share Follow See all 2 Comments See all 2 2 Comments reply Amitesh Sharma commented Nov 28, 2017 reply Follow Share I wasn't sure about the floor i did ceil and ans was 23. thank you 0 votes 0 votes hs_yadav commented Nov 28, 2017 reply Follow Share ....sender will send a single packet and wait for (2TP=RTT).....now the no. of packet transmitted by sender within 2TP(60mis) time(i.e 60/2.73=21.97->21) + one packet(initially transmitted) would be equal to one window........ 0 votes 0 votes Please log in or register to add a comment.
