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Consider an instance of TCP’s Additive Increase Multiplicative Decrease (AIMD) algorithm where the Window size at the start of the slow phase is 2 MSS and the threshold at the start of the first transmission is 8 MSS. Assume that a timeout occurs during the fifth transmission. Find the congestion Window size at the end of the tenth transmission.
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- 8 MSS
- 14 MSS
- 7 MSS
- 12 MASS
- 8 MSS
Correct Option: C
Given threshold = 8
Time = 1, during first transmission, window size = 2
(slow start phase)
Time = 2, congestion window size = 4 (double the no. of acknowledgments)
Time = 3, congestion window size is = 8
Time = 4, congestion window size = 9, after threshold (increase by one addictive increase)
Time = 5, transmits 10 MSS, but time out occurs congestion windw size = 10
Hence threshold = (congestion window size) / 2 = 10 / 2 = 5
Time = 6, transmits 2
Time = 7, transmits 4
Time = 8, transmits 5(threshold is 5)
Time = 9, transmits 6
Time = 10, transmits 7
During 10th transmission, it transmits 7 segments hence at the end of the 10th transmission the size of congestion window is 7 MSS.
