Why do some providers require IPv6 /64 PA space to have public whois?
owen at delong.com
Mon Dec 10 23:58:46 UTC 2012
On Dec 10, 2012, at 2:53 PM, Ian Smith <I.Smith at F5.com> wrote:
>> Quite the opposite in fact. In IPv6 a /64 is roughly equivalent to a /32 in 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 routing and host space.
> IPv6/64 is a routing endpoint and v6/128 is a host. IPv6 is a 64 bit routing space and also a 64 bit host space for each routing space, not a 128 bit combined routing and host space.
You can make a /128 a routing endpoint in IPv6 just like a /32 in IPv4 with all the same rules, restrictions, and limitations.
> Evidently, the whois requirement is for networks, not nodes, which makes sense when you think about how the entity that controls a /64 is assuming responsibility for 2^64 network nodes.
Correct (in the first part). In reality, nobody has 2^64 nodes, that's more than the square of the current host addressing available in all of IPv4. You'll never see a /64 full of hosts. For one thing, there's no concept for switching hardware that could handle that large of a MAC adjacency table, nor is there ever likely to be such.
> -----Original Message-----
> From: Doug Barton [mailto:dougb at dougbarton.us]
> Sent: Monday, December 10, 2012 5:05 PM
> To: Schiller, Heather A
> Cc: Constantine A. Murenin; nanog at 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
>> :: IPv6 /64
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