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I was learning TCP slow-start exponential increase congestion control mechanism from Forouzan's book. It says following:

The size of congestion window (cwnd) starts with one maximum segment size (MSS), but increases one MSS each time an acknowledgement arrives as follows:

Start         -> cwnd = 1 -> 2^0
After 1 RTT   -> cwnd = cwnd + 1 = 1 + 1 = 2 -> 2^1
After 2 RTT   -> cwnd = cwnd + 2 = 2 + 2 = 4 -> 2^2
After 3 RTT   -> cwnd = cwnd + 4 = 4 + 4 = 8 -> 2^3

It further says:

To mention that the slow-start strategy is slower in the case of delayed acknowledgements. Remember, for each ACk, the cwnd is increased by only 1. Hence, if two segments are acknowledged cumulatively, the size of the cwnd increases by only 1, not 2. The growth is still exponential, but it is not a power of 2. With one ACK for every two segments, it is a power of 1.5.

I was not able to get how it is a power of 1.5. I tried out following:

Start         -> cwnd = 1 -> 2^0
After 2 RTT   -> cwnd = cwnd + 1 = 1 + 1 = 2 -> 2^1
After 4 RTT   -> cwnd = cwnd + 2 = 2 + 2 = 4 -> 2^2
After 6 RTT   -> cwnd = cwnd + 4 = 4 + 4 = 8 -> 2^3
After 8 RTT   -> cwnd = cwnd + 8 = 8 + 8 = 16-> 2^4

I am not able to see how this is exponential with the power of 1.5. (Notice that the only difference in above calculations and the one given in the book is interval between consecutive RTTs considered)

PS: This is more sort of mathematical question in the context of networks. Sorry for that. :(

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It's actually more like the power of 1.4: an increase in each of x segments means 2x, an increase in every 2nd segment means 2(x/2) or √2 x.

| improve this answer | |
  • Great!!! Thats exactly what it is. – anir Oct 31 '19 at 5:35

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