From: Scott Morris (swm@emanon.com)
Date: Sat Jul 01 2006 - 15:04:53 ART
See my previous post with the three ways to do it. Make sure you ONLY match
those things listed, and it doesn't have to be one line only. The answer
they gave IS absolutely correct, it's just not the only way to do it.
HTH,
Scott Morris, CCIE4 (R&S/ISP-Dial/Security/Service Provider) #4713, JNCIE
#153, CISSP, et al.
CCSI/JNCI
IPExpert CCIE Program Manager
IPExpert Sr. Technical Instructor
smorris@ipexpert.com
http://www.ipexpert.com
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of Sami
Sent: Saturday, July 01, 2006 8:04 AM
To: Faryar Zabihi (fzabihi)
Cc: ELDHO PAUL; ccielab@groupstudy.com
Subject: Re: Access list
All,
I am totally confused..which one is correct..everyone has different
answer..IE lab 20 section 9.1 solution guide says
51.3.0.1 0.0.0.8
51.5.0.1 0.2.0.8
Thanks
On 7/1/06, Faryar Zabihi (fzabihi) <fzabihi@cisco.com> wrote:
>
>
> 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
>
> ______________________________________________________________________
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