Re: [mpls] I-D Action: draft-ietf-mpls-mna-requirements-10.txt

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Tony, et.al.,

> [WG chair hat: OFF]
> The encoding is straightforward:
> 22 octets = 176 bits

Agreed. 22 octets = 176 bits.

> If you start with a Format B LSE,


If I understand correctly it is not possible to start with Format
B LSE and continue into Format C, Format C has its own OpCode, so you can
start in the Format C LSE and  then go into the number of Format D LSEs
that is required. Only Format D is defined for additional data.

However, there is a quirk that you don't take into account, I asked  about
MUTABLE data.

A Format B LSE has no room for mutable data.

draft-ietf-mpls-mna-hdr says:

   To preserve backward compatibility, if a network action encodes
   data that will change during packet forwarding, then that data
   MUST be in the least significant 4 bits in the data field of a
   Format C LSE (Section 4.3) or the least significant 8 bits of
   a Format D LSE (Section 4.4).

I.e. 4 bit is the Format C LSE and 8 bits in Format D LSEs can be used
for mutable data.

Format C: 4 bits
Format D: 8 bits each

There is 14 Format D available 14x8=112bits, the total is 116, there is no
room for 176 bits (22 octets).

There is a caveat, I think that draft-ietf-mplsd-mna-hdr is wrong, but
I might misunderstand something. I believe that the paragraph I
quoted  above should say:

   To preserve backward compatibility, if a network action encodes
   data that will change during packet forwarding, then that data
   MUST be in the least significant 7 bits in the data field of a
   Format C LSE (Section 4.3) or the least significant 11 bits of
   a Format D LSE (Section 4.4).

Note: Authors can you check this up, it is "only" the 20 first bits of an
LSE that is mutable/immutable sensitive, right?

If I'm correct.

Format C: 7 bits
Format D: 1 bits each

There is 14 Format D available 14x11=154 bits, the total is 161, so  there
is still no room for 176 bits (22 octets). The maximum is
20 octets (160 bits).

With Stewarts example of PTP timestamp of 10 octets, 20 or 22 octets are
"reasonable" or we maybe "not excessive".

/Loa

> If you start with a Format C LSE, then you can embed 20 bits in that.
You
> then need 6 additional LSEs for the data. This seems obvious.
> This seems trivial.
> A side concern is why one needs 22 octets of precision in your
timestamps.
> Do you really gain something useful over a 4 octet NTP timestamp? T
>> On Mar 2, 2024, at 1:27 AM, Stewart Bryant <[e-mail]>
wrote:
>> 
>>> On 1 Mar 2024, at 06:43, [e-mail] wrote:
>>> Can you show me how to transport 22 octets of mutable data in the
stack? I don't think 22 octets is unreasonable.
>> I agree. Certainly when it takes 10 octets to encode a PTP timestamp,
20
>> octets does not sound excessive.
>> Stewart
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