Re: Acl question

From: Anant Tamgole (anant.tamgole@gmail.com)
Date: Sat Aug 30 2008 - 06:36:46 ART


Hi,

Below ACL will do the job.

permit xxx.xxx.4.0 0.0.11.0

How it is derived is given below

0000 0100 - 4
0000 0101 - 5
0000 0110 - 6
0000 0111 - 7

0000 1100 - 12
0000 1101 - 13
0000 1110 - 14
0000 1111 - 15
--------------
0000 1011 - 11 (Here 1 is don't case and 0 is match exactly)

Anant

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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-- 
With Best Regards
Anant

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