Multiplicative decrease of data center TCP

Multiplicative decrease of data center TCP

by Shengyu Huang -
Number of replies: 2

I don't quite get why we set the multiplicative decrease to (1 - p/2) when the estimated congestion rate is p. Say if p = 10%, it means 10% of my packets are considered lost. This implies I should send only 90% of my packets (which should be 1-p), right? But by the formula, I will be still sending 95% of the packets. Did I misunderstand what p means here?



In reply to Shengyu Huang

Re: Multiplicative decrease of data center TCP

by Seyed Mohammadhossein Tabatabaee -
Hello,

Thank you for your question.

First, observe that p is the not probability of loss, but rather is the probability of congestion which is an estimation of the fraction of packets that are CE marked; these packets are not lost but they inform senders of possible congestion in a router and that is why senders should decrease their rates to avoid congestion hence packet loss. Second, when p is small (low congestion), with DHTCP, rates are decreased very slightly; however, in the worst case when p=1 (high congestion), the multiplicative decrease of DHTCP becomes 1/2 which means DHTCP behaves as TCP.

I hope this answers your question, otherwise do not hesitate to ask follow-up questions.

Best,
Hossein
In reply to Shengyu Huang

Re: Multiplicative decrease of data center TCP

by Joseph Abboud -
Hello,

p is an estimation of the network congestion: In other words, every fixed period of time (e.g. 1 RTT), p is updated by looking at the fraction of packets that were marked during that window of time. The window is reduced by p/2 instead of p to avoid dramatical drops (if p is 1, the congestion window becomes 0, which is unwanted obviously).

This makes sure that at worst case, the congestion window is cut by half, which becomes similar to TCP

Hope this answers your question :)