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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 ???
asked in Computer Networks by (133 points) | 48 views

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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
answered by (241 points)
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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

answered by (263 points)

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