[172281] in North American Network Operators' Group
Re: Getting pretty close to default IPv4 route maximum for 6500/7600
daemon@ATHENA.MIT.EDU (=?utf-8?Q?=C5=81ukasz_Bromirski?=)
Tue Jun 10 14:07:44 2014
X-Original-To: nanog@nanog.org
From: =?utf-8?Q?=C5=81ukasz_Bromirski?= <lukasz@bromirski.net>
In-Reply-To: <539742E4.70108@ispn.net>
Date: Tue, 10 Jun 2014 20:07:35 +0200
To: Blake Hudson <blake@ispn.net>
Cc: nanog@nanog.org
Errors-To: nanog-bounces@nanog.org
On 10 Jun 2014, at 19:39, Blake Hudson <blake@ispn.net> wrote:
>> And yes, you=E2=80=99re right - no matter how many neighbors you =
have, the FIB
>> will only contain best paths, so it will be closer to 500k entries in
>> total rather than N times number of neighbours.
> Please correct me if I'm wrong, but if the BGP table contains ~500k =
prefixes, which are then summarized into ~300k routes (RIB), and the FIB =
contains only the "best path" entries from the RIB, wouldn't the FIB be =
at or below 300k?
Because you need to do your own summarization or ask your upstreams
to do it for you. Until then, most of transit accepts loosely
prefixes in exact length but also longer (i.e. /24 but also both /25s).
You=E2=80=99ll see more and more deaggregation with the rise of smaller
entities fighting for chance to do some traffic engineering, so be
prepared to constant rise of prefixes overall.
--=20
"There's no sense in being precise when | =C5=81ukasz =
Bromirski
you don't know what you're talking | jid:lbromirski@jabber.org
about." John von Neumann | http://lukasz.bromirski.net
