[131539] in North American Network Operators' Group
Re: IPv6 Routing table will be bloated?
daemon@ATHENA.MIT.EDU (Franck Martin)
Tue Oct 26 18:01:13 2010
X-Barracuda-Envelope-From: franck@genius.com
Date: Wed, 27 Oct 2010 10:00:51 +1200 (FJT)
From: Franck Martin <franck@genius.com>
To: Randy Carpenter <rcarpen@network1.net>
In-Reply-To: <1190710313.13228.1288129693692.JavaMail.root@zimbra.network1.net>
Cc: nanog@nanog.org
Errors-To: nanog-bounces+nanog.discuss=bloom-picayune.mit.edu@nanog.org
Yes indeed, but you don't have to justify much if you only ask for the minimum, if you want more you need to ask...
Also, and this I like less, your membership is calculated from the number of IPs you have... I think in short $$=max(1180x1.3(log2(Addresses in /32)-8),Feev6 = 1180x1.3(log2(Addresses in /56)-22)
http://www.apnic.net/services/apply-for-resources/check-your-eligibility/check-ipv6
http://www.apnic.net/services/become-a-member/how-much-does-it-cost
----- Original Message -----
From: "Randy Carpenter" <rcarpen@network1.net>
To: "Franck Martin" <franck@genius.com>
Cc: nanog@nanog.org, "Nick Hilliard" <nick@foobar.org>
Sent: Wednesday, 27 October, 2010 10:48:13 AM
Subject: Re: IPv6 Routing table will be bloated?
It would be nice as a start, but does not really take into consideration future expansion needs.
I would think that you could draw some parallels, though.
Something like:
v4 /16 ~ v6 /32
v4 /12 ~ v6 /28
v4 /8 ~ v6 /24
I know it we don't want to equate v4 and v6, but it may help as a guideline for the size of the customer base.
