GATE | GATE-CS-2014-(Set-1) | Question 65
Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of the connection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________.
(A) 1100 to 1300
(B) 800 to 1000
(C) 1400 to 1600
(D) 1500 to 1700
Current size of congestion window in terms of number of segments = (Size in Bytes)/(Maximum Segment Size) = 32KB / 2KB = 16 MSS 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).
My Personal Notes arrow_drop_up
Please Login to comment...