[158780] in North American Network Operators' Group
RE: Why do some providers require IPv6 /64 PA space to have public
daemon@ATHENA.MIT.EDU (Ian Smith)
Mon Dec 10 18:19:11 2012
From: Ian Smith <I.Smith@F5.com>
To: Doug Barton <dougb@dougbarton.us>, "Schiller, Heather A"
<heather.schiller@verizon.com>
Date: Mon, 10 Dec 2012 22:53:42 +0000
In-Reply-To: <50C65C84.6080203@dougbarton.us>
Cc: "Constantine A. Murenin" <mureninc@gmail.com>,
"nanog@nanog.org" <nanog@nanog.org>
Errors-To: nanog-bounces+nanog.discuss=bloom-picayune.mit.edu@nanog.org
>Quite the opposite in fact. In IPv6 a /64 is roughly equivalent to a /32 i=
n IPv4. As in, it's the smallest possible assignment that will allow an end=
-user host to >function under normal circumstances.
>SWIP or rwhois for a /64 seems excessive to me, FWIW.
IPv4/32 is both a routing endpoint and a host. IPv4 is a 32 bit combined r=
outing and host space.
IPv6/64 is a routing endpoint and v6/128 is a host. IPv6 is a 64 bit rout=
ing space and also a 64 bit host space for each routing space, not a 128 bi=
t combined routing and host space.
Evidently, the whois requirement is for networks, not nodes, which makes se=
nse when you think about how the entity that controls a /64 is assuming res=
ponsibility for 2^64 network nodes.
-----Original Message-----
From: Doug Barton [mailto:dougb@dougbarton.us]=20
Sent: Monday, December 10, 2012 5:05 PM
To: Schiller, Heather A
Cc: Constantine A. Murenin; nanog@nanog.org
Subject: Re: Why do some providers require IPv6 /64 PA space to have public=
whois?
On 12/10/2012 01:27 PM, Schiller, Heather A wrote:
> I think most folks would agree that, IPv4 /32 :: IPv6 /128 as IPv4 /29=20
> :: IPv6 /64
Doug
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