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Let γ be the value of the Slow Start Threshold (SSThresh)
Let t = tγ be the time when TCP exits SS (i.e., tγ = log(γ) because W(tγ) = 2tγ = γ )
This is the moment that TCP exits SS and enters CA
So the second equation indeed increases congestion window W(t) by ONE per RTT seconds.
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we can derive a simple function for the segment (packet) transmission rate of TCP:
RTT
|<-------------------------------->|
W(t)
---+----------------------------------+---------
\
\
W(t) segments en route to destination
W(t)
Transmission rate = B(t) = ------ (# packets / sec)
RTT
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(This situation is similar to the example in Physics:)
RTT
ρ = ------
RTT0
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where RTT0 is some pre-determined round trip time (e.g., 25 msec).
Reno Slow Start threshold value = γ RTT TCP Hybla Slow Start threshold value = ρ * γ = ------ γ RTT0 |
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where
RTT ρ 2(ρ*t/RTT) = ----- 2(t/RTT0) . . . . . . . . (SS) RTT0 WH(t) = t - tγ,0 RTT t - tγ,0 ρ ( ρ ---------- + γ ) = ----- ( --------- + γ ) . . . . (CA) RTT RTT0 RTT0 |
Reason:
RTT WH(t) = ----- 2(t/RTT0) = ρ * γ RTT0 |
1
= ----- 2(t/RTT0) . . . . . (SS)
RTT0
WH(t)
BH(t) = -----
RTT
1 t - tγ,0
= ----- ( --------- + γ ) . . . . (CA)
RTT0 RTT0
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As we have just seen before, tγ,0 is independent of RTT
Therefore, RTT does not appear anywhere as a factor in BH(t) --- so BH(t) is independent of RTT
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A TCP Hybla connection with longer RTT needs to end up at a higher CWND to achieve the SAME throughput per time unit (second)
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Now HOW should the CWND be updated ( per each NEW ACK) so that TCP Hybla's CWND will be equal to the given function ?
What are the values for α and β for the update of CWND W:
WHi + α . . . . (SS) WHi+1 = WHi + β . . . . (CA) |
so that:
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WHi + 2ρ - 1 . . . . (SS)
WHi+1 =
ρ2
WHi + ----- . . . . (CA)
WHi
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We won't try to derive it, we will just verify it (it's a lot easier to verify the correctness of a solution than to find a solution)
TCP Hybla will only exit the SS phase after CWDN > ρ * SSThreashold
SS Phase
WH(t) = ρ 2(ρ*t/RTT) |
WHi+1 = WHi + 2ρ - 1 . . . . (SS)
WH1 = ρ
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