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From: Joseph Touch <touch@strayalpha.com>
In-Reply-To: <FC3BA2A2-0052-418E-A4D0-F9DC9FD5C2A2@apple.com>
Date: Tue, 26 Jan 2021 20:21:42 -0800
Cc: tcpm IETF list <tcpm@ietf.org>,
 "Scheffenegger, Richard" <Richard.Scheffenegger@netapp.com>,
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To: Vidhi Goel <vidhi_goel=40apple.com@dmarc.ietf.org>
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Archived-At: <https://mailarchive.ietf.org/arch/msg/tcpm/9-MsOzQqb6dEN5WOSLLlS2Q2dBE>
Subject: Re: [tcpm] AccECN field order
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> On Jan 26, 2021, at 7:47 PM, Vidhi Goel =
<vidhi_goel=3D40apple.com@dmarc.ietf.org> wrote:
>=20
>> Please see my note from 12/3/20, which shows how a single codepoints =
would suffice without increasing length.=20
>=20
> If the TCP option code points is not a scarce resource,

They are. Very much so.

> then why not just keep it simple with "two different codepoints=E2=80=9D=
? Is there a reason that you think we should use only one code point?

That=E2=80=99s the core of the discussion we=E2=80=99ve been having.

> Regarding your suggestion of using 1 bit from the 24-bit counter =
field, do you mean specify ordering once before the first ECN counter =
field or specify before every counter field?
> For example, if all three counters are present, are you suggesting, a) =
or b)? To me b) would make more sense but that makes the implementation =
complex as the fields have different length.
>=20
> a) Kind | Length | 1-bit ordering | 23-bit EE1B | 1-bit ordering | =
23-bit ECEB | 1-bit ordering | 23-bit EE0B
> b) Kind | Length | 1-bit ordering | 23-bit EE1B | 24-bit ECEB | 24-bit =
EE0B

The point of the bit is to encode the different kinds. (b) does this =
just fine. If we want all three counters to be the same length, simply =
do so and require the other two =E2=80=9Cfirst bits=E2=80=9D to be zero.

Joe

