Re: [tsvwg] ECN encapsulation draft - proposed resolution

["De Schepper, Koen (Nokia - BE/Antwerp)" <[e-mail]>]

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-----Original Message-----
From: Jonathan Morton <[e-mail]> 
Sent: Wednesday, June 9, 2021 9:37 PM
To: De Schepper, Koen (Nokia - BE/Antwerp) <[e-mail]>
Cc: Bob Briscoe <[e-mail]>; [e-mail]; Markku Kojo <[e-mail]>
Subject: Re: [tsvwg] ECN encapsulation draft - proposed resolution

> On 9 Jun, 2021, at 7:07 pm, De Schepper, Koen (Nokia - BE/Antwerp) <[e-mail]> wrote:
> 
> So for me following simple algorithms would be ok to use:
> - only mark if the first (even partial) segment of the reassembled packet is marked (ignore marks on any other segments)

This would mean that any mark applied to a segment that covered only the middle or tail of a packet would reliably have no effect.  I firmly disagree with that approach, due to the possibility of accidentally triggering pathological conditions where a long series of marks is lost.

[K] It could as well mean that one outer packet marks 2 inner packets. Note that AQMs best apply random marking and that any pathological conditions will be quickly broken by the closed loop.

> - apply a random probability on the packet based on the ratio of marked bytes in the packet and full packet size

I can see where you're going with that, but I maintain - based on my analysis a couple of posts ago - that maintaining the number of marked bytes is simply the wrong approach.

The interval *between* marks, measured in either bytes or time, is the important property to preserve; this is actually easier to achieve and is natural to perform statelessly.  If the AQM device can rely on that behaviour, it can be designed to perform byte-mode or time-domain marking to match, which dovetails nicely with all existing transport implementations (whether with high-fidelity or conventional congestion control).

[K] These algorithm preserve the marking probability (whether bytes or packets), which boils down to exactly preserving the interval between marks. Not sure what the difference is?? The interval i=1/p with p the probability. So preserving the probability, preserves the interval... not?

Koen.

 - Jonathan Morton