Re: Access list

From: Petr Lapukhov (petr@internetworkexpert.com)
Date: Sat Jul 01 2006 - 09:44:02 ART

Sami,

how about this one:

10 deny 51.1.0.1 0.0.0.8
20 permit 51.1.0.1 0.6.0.8

You see, they all work well, in sense that
they all fulfill requirement to have minimum
number of lines. As long as you get the result
and dont violate any constrains - it's okay :)

HTH 

-- 
Petr Lapukhov, CCIE #16379
petr@internetworkexpert.com

Internetwork Expert, Inc.
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2006/7/1, Sami <sy1977@gmail.com>:
>
> 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
> > > >
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