0 votes 0 votes Consider the size of congestion window of a TCP connection be 36 KB when a timeout occurs. The round trip time of the connection is 200 μsec and the maximum segment size used is 2 KB. What is the time taken by the TCP connection to get back to 36 KB congestion window? ben10 asked Aug 4, 2018 ben10 201 views answer comment Share Follow See 1 comment See all 1 1 comment reply Mk Utkarsh commented Aug 4, 2018 reply Follow Share $2 | 4 | 8 | 16 | 18 | 20 | 22 | 24 | 26 | 28 | 30 | 32 | 34 | 36$ $13 \times 200 = 2600μsec$ 1 votes 1 votes Please log in or register to add a comment.
0 votes 0 votes WHEN TIME OUT OCCUR SENDER WINDOW SIZE = 1 MSS = 2KB THRESHOLD VALUE = 9 MSS = 18 KB SO 1MSS,2MSS,4MSS,8MSS,9MSS,10MSS,11MSS,12MSS,13MSS,14MSS,15MSS,16MSS,17MSS,18MSS. -> TOTAL TIME = 14*200 = 2800 MICOSEC sunnykg answered Aug 12, 2018 sunnykg comment Share Follow See all 0 reply Please log in or register to add a comment.