From: Faryar Zabihi \(fzabihi\) (fzabihi@cisco.com)
Date: Sat Jul 01 2006 - 08:27:34 ART
What if I have,
192.172.1.0
192.172.2.0
192.172.3.0
192.172.4.0
192.172.5.0
192.172.6.0
192.172.7.0
192.172.8.0
192.172.9.0
192.172.10.0
192.172.11.0
192.172.12.0
192.172.13.0
192.172.14.0
192.172.15.0
150.16.1.0
I need to keep
192.172.8.0
192.172.9.0
192.172.10.0
192.172.11.0
150.16.1.0
Routing protocol RIP. Catch being ONLY ONE line ACL
First I tried to grab the networks I need but the 150.16.1.0 I could not
include with one line in an acl. Then I tried to get the ones I need to
filter out and offset by 15. my question is how do I match
192.172.1.0
192.172.2.0
192.172.3.0
192.172.4.0
192.172.5.0
192.172.6.0
192.172.7.0
192.172.12.0
192.172.13.0
192.172.14.0
192.172.15.0
Thanks in advance,
Faryar
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
ELDHO PAUL
Sent: Saturday, July 01, 2006 6:20 AM
To: Sami
Cc: ccielab@groupstudy.com
Subject: Re: Access list
A more currect answer seems to be
51.1.0.1 0.0.0.8
We will get the first portion 51.1.0.1 by doing an and opration of all
the four ip addresses and the wildcard portion by doing the XOR
operation of all the four ip addresses.
On 7/1/06, ELDHO PAUL <cciein2006@gmail.com> wrote:
>
> I think we can match it with 2 access lists
> 51.3.0.1 0.6.0.0
> 51.3.0.9 0.6.0.0
>
>
> On 7/1/06, Sami <sy1977@gmail.com> wrote:
> >
> > Group,
> >
> > One of the task say use minimum amount of line necessary to comple
> > this task.
> >
> > 51.3.0.1
> > 51.5.0.1
> > 51.7.0.1
> > 51.3.0.9
> > 51.5.0.9
> > 51.7.0.9
> >
> > How can I combine in one access list ?
> >
> > Thanks
> >
> > ____________________________________________________________________
> > ___ Subscription information may be found at:
> > http://www.groupstudy.com/list/CCIELab.html
This archive was generated by hypermail 2.1.4 : Tue Aug 01 2006 - 07:13:46 ART