From: Daniel Fredrick (dfredrick@gmail.com)
Date: Sun Jul 02 2006 - 11:49:44 ART
One other solution could be...
permit 51.3.0.1 0.4.0.8
permit 51.5.0.1 0.0.0.8
Just throwing out the other possiblity.
Doesn't matter I guess... as long as you get the same result.
Dan
On 7/1/06, Scott Morris <swm@emanon.com> wrote:
>
> 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:
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> >
> > ______________________________________________________________________
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