From: Ashish . (invent.xr@gmail.com)
Date: Sat Aug 30 2008 - 19:23:27 ART
Hi Farhad,
I understood the logic u used in this particular prob....however i'm not
able to figure out the way to use the same logic in the age old questions
a: permit all odd ip
b: permit all even ip
can you help me understand the above two in say 1.1.1.0/24 .
thanks in advance,
Ashish
On 8/30/08, Fahad Khan <fahad.khan@gmail.com> wrote:
>
> The actual game is of 3rd octat in this case.
>
> starting IP = x.x.4.0
> ending IP = x.x.15.0
>
> Four most significant bits are common in the 3rd octat i.e 0000
>
> so we have 0100 (4) and 1111(15) for our play
>
> now for calculating IP in accesslist,
>
> perform AND operation in between, 0100 and 1111
>
> 0100
> 1111
> -------
> 0100 = 4
> -------
> Hence, the IP is decided now as x.x.4.0
>
> Now for calculation of wildcard mask, perform XOR operation
>
> 0100
> 1111
> -------
> 1011 = 11
> -------
>
> Rules for XOR:
>
> - For same number of inputs output is 0 otherwise 1
> - For even number of 1s output is 0 while for odd number of 1s output is 1.
>
> hense you get the ACL as:
>
> permit x.x.4.0 0.0.11.0
>
>
> HTH
> On 8/30/08, ashhy <mail4hafiz@gmail.com> wrote:
> >
> > Hi all
> > i want to know how to get this right acl for the below mentioned question
> >
> > i can use only one acl to do this ,i am getting 16 routes of class c
> > from xxx.xxx.1.0 to xxx.xxx.16.0
> > from backbone out of which i should allow only
> >
> > xxx.xxx.4.0
> > xxx.xxx.5.0
> > xxx.xxx.6.0
> > xxx.xxx.7.0
> > xxx.xxx.12.0
> > xxx.xxx.13.0
> > xxx.xxx.14.0
> > xxx.xxx.15.0
> >
> > please guide me or else suggest if this is already disscussed.
> >
> > Thanks
> > Asshy
> >
> >
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>
> --
> Fahad Khan
>
>
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