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Consider a TCP connection with 10 ms round-trip time, max segment size = 2 KB, receiver window = 64 KB. Suppose packet #4 timed out and all other transmissions were successful. RTO value is 50 ms. Calculate the time required to reach the receiver window size. Assume RTT is fixed.
posted Aug 25, 2016 in Computer Networks 368 views
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2 Comments

 

Solution
Initially, cwnd = 2KB,
ssthresh = 64 KB
ACK #1 at RTT 1, cwnd = 4 KB,
ACK #2 at RTT 2, cwnd = 8 KB,
ACK #3 at RTT 3, cwnd = 16 KB
#4 is lost at RTT3+RTO = 30 + 50 =80 ms
ssthresh = 16/2 = 8 KB, cwnd = 2 KB
retransmit #4
ACK #4 at RTT 1, cwnd = 4 KB
ACK #5 at RTT 2, cwnd = 8 KB
ACK #6 at RTT 3, cwnd = 8 KB
Since we reached sstresh, switch to
additive increase:
ACK #6 at RTT 3, cwnd = 9 KB
..
ACK #x at RTT x-3, cwnd = x+3 KB
..
ACK #61 at RTT 58, cwnd = 64 KB
RTT 58 = 580 + 80 = 660 ms
 

@Aantra in question it is given 4th packet is time out...and 4th packet is availble at RTT2
...