109 109 votes Consider an instance of TCP’s Additive Increase Multiplicative Decrease (AIMD) algorithm where the window size at the start of the slow start phase is $2$ MSS and the threshold at the start of the first transmission is $8$ MSS. Assume that a timeout occurs during the fifth transmission. Find the congestion window size at the end of the tenth transmission. $8$ MSS $14$ MSS $7$ MSS $12$ MSS Computer Networks gatecse-2012 computer-networks congestion-control normal + – Arjun 63.0k views answer comment Share Follow Print See all 16 Comments 16 16 Comments reply Show 13 previous comments Piyush Sharma_1 commented Sep 21, 2024 reply Follow flag initial ssthres = 8MSS start from 2 MSS so at t=1 => 2 at t=2 => 4 at t=3 => 8 (reached ssthres now it will go linear i.e +=1) at t=4 =>9 at t=5 =>10 (now the timeout occur so the new ssthres will be CWND/2 i.e 5) at t=6 => 1 at t=7 =>2 at t=8 => 4 (ssthres = 5 so it can't go 8) at t=9 =>5 at t=10 =>6 at t=11 =>7 CWND AT THE END OF 10TH TRANSMISSION now CWND is (6+1) = 7 MSS 9 9 replyShare Hira Thakur commented Jan 19 reply Follow flag go-classes-dpp-cn-aimd ( follows rfc5681) its correct. 0 0 replyShare GO Classes Support commented 3 days ago reply Follow flag Watch the Detailed Video Solution by clicking the button below..!Watch Detailed Video Solution 0 0 replyShare Please log in or register to add a comment.
0 0 votes ?si=OjZRt0bFCVkaGp6q priyangsu45 answered Oct 3 priyangsu45 comment Share Follow 0 reply Please log in or register to add a comment.
