Re: Acl question

From: Fahad Khan (fahad.khan@gmail.com)
Date: Sat Aug 30 2008 - 07:23:35 ART

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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