0 votes 0 votes Let the size of congestion window of a TCP connection be 32KB , when the time out occurs , maximum segment size used is 2KB , if the time taken by TCP connection to get 32KB congestion window is 440 msec then the RTT of connection is ??? Lokesh Arya asked Aug 9, 2018 Lokesh Arya 332 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes When timeout occurs, in TCP's Slow Start algorithm, threshold is reduced to half which is 16KB or 8MSS. Also, slow start phase begins where congestion window is increased twice. So from 1MSS to 8 MSS window size will grow exponentially. Congestion window becomes 2MSS after one RTT and becomes 4MSS after 2 RTTs and 8MSS after 3 RTTs. At 8MSS, threshold is reached and congestion avoidance phase begins. In congestion avoidance phase, window is increased linearly. So to cover from 8MSS to 16MSS, it needs 8 RTTs Together, 11RTTs are needed (3 in slow start phase and 8 in congestion avoidance phase). Time taken in one RTT = 440/11 = 44msec Priyansh Singh answered Aug 25, 2018 Priyansh Singh comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes TILL THE TIMEOUT WE WILL APPLY SLOW START ALGORITHM WHICH INCREASES EXPONENTIALLY. SO UPTO 32KB, IT WILL TAKE 5*RTT. ACCORDING TO QUESTION TOTAL TIME = 440 MSEC SO 5*RTT = 440 RTT = 88 MSEC sunnykg answered Aug 12, 2018 sunnykg comment Share Follow See all 0 reply Please log in or register to add a comment.