0 0 votes Computer Networks + – Vipin Rai 1.1k views answer comment Share Follow Print See all 4 Comments 4 4 Comments reply pawan kumarln commented Jan 8, 2019 reply Follow flag 7mss 0 0 replyShare Manojkumar Susarla commented Jan 8, 2019 i reshown by Manojkumar Susarla Jan 8, 2019 reply Follow flag 7 mss sorry I misread the question 0 0 replyShare pawan kumarln commented Jan 8, 2019 reply Follow flag @Manojkumar Susarla how???? 0 0 replyShare Vipin Rai commented Jan 8, 2019 reply Follow flag I am getting 8 MSS 0 0 replyShare Please log in or register to add a comment.
Best answer 0 0 votes starting from 2 MSS we proceed 2-4-8 after third tranmission we will reach threshold 12 MSS from where additive increase will start now 12-13-14-15 during 7th transmission we get time out means threshold will be floor(15/2)=7 now start for 8 transmission with intial segment size=2mss 2-4-7-8-9-10 so at the end of 12th transmission answer is 10mss Navneet Kalra answered Jan 8, 2019 • selected Jan 8, 2019 by Vipin Rai Navneet Kalra comment Share Follow See all 2 Comments 2 2 Comments reply Vipin Rai commented Jan 8, 2019 reply Follow flag Initially will it directly increase it to 12 MSS in fourth transmission ( 2 - 4 - 8 - 12 ) ? Because in slow start the window doubles , am i correct? 0 0 replyShare Navneet Kalra commented Jan 8, 2019 reply Follow flag it will increase upto 8 mss exponentially then on next transmission we will have we will have four segments acknowledged making sender size to 12 and then additive increase will start 0 0 replyShare Please log in or register to add a comment.
