Re: [v6ops] draft-ietf-v6ops-ula-usage-recommendations - work or abandon?

[Mark Smith <[e-mail]>]

On 9 November 2015 at 10:08, Philip Homburg <[e-mail]> wrote:
> In your letter dated Sun, 8 Nov 2015 22:06:14 +0000 you wrote:
>>On 08/11/2015 19:47, Philip Homburg wrote:
>>> Call me old fashioned, but what NAT breaks is not what is traditionally
>>> called a protocol layering violation.
>>
>>When you code layer 3 information at what we usually refer to as layer 7,
>>this is layering violation.
>
> I disagree. Absent any session, presentation, and security layers,
> transport layer identifiers are the application layer identifiers. There is
> nothing else.
>
> It would be different if we actually had stable identifiers that were
> in actual use. But we don't. There is hardly any use of LISP.

<snip>

>
> Again, we have to use transport layer identifiers, because the IP stack
> doesn't define anything else.
>

So I was thinking about this from the perspective of MPTCP.

I think MPTCP is separating the locator and identifier semantics of IP
addresses, by creating a temporary per-session host identifier, which
is the 32 bit MPTCP session token.

Assuming that the application is oblivious to MPTCP because it should
be hidden within the TCP layer, there is still an issue with NATs.
Although MPTCP is now not using IP addresses as identifiers, it is
still trying to use them as locators, as a host announces to its MPTCP
peer what its local IP addresses are, so that additional MPTCP
subflows can come up if the addresses are reachable by the MPTCP peer.

So a NAT would still cause problems with MPTCP unless the NAT was
MPTCP aware, and translated the addresses that are in the MPTCP TCP
option sent to the MPTCP peer, which won't be able to be done if
end-to-end IPsec was in use.

So does that mean that the transport layer shouldn't be using IP
addresses either, even if they're now only locators in an MPTCP
scenario? Is this really saying that layers above the network layer in
hosts shouldn't be aware at all of or be able to find out of the
host's network location?

Mark.