From: Aabid Saleem (aabids@nesma.net.sa)
Date: Mon Apr 14 2008 - 15:03:07 ART
Thanks Shine --
now its bit clear to me how to subnet these issues, I will do practice
on it and many thanks
I read that question, but I was still step away to understand, that's
why I posted the questions
Aabid
Shine Joseph wrote:
> Abid,
>
> Not that long ago, the same question appeared in the GS. Check in the
> archive before posting questions. Most likely the question was already been
> answered. My response to the same question is as follows:
> ========================================================
>
> The key here is to convert the decimals into binary and we are interested
> only in the third octet. So,
> 5 is 0000 0101
> 10 is 0000 1010
> 13 is 0000 1101
> 14 is 0000 1110
>
> Since the question is specific about number of lines in the acl, group the
> binary in such a way that it forms 2 groups with maximum matching of bit
> positions.
>
> Line 1 and 2 mismatches in 2 positions
> Line 1 and 3 mismatches in 1 position
> Line 1 and 4 mismatches in 3 positions
>
> The minimum mismatch is among lines 1 and 3.
>
> 0000 0101
> 0000 1101
>
> The forth bit position value in decimal is 8
>
> Similarly, you can group the remaining 2 lines
>
> 0000 1010
> 0000 1110
>
> Here, the third position is the only mismatch; with its value in decimal is
> 4.
>
> The acl entries must be
>
> 192.168.5.0 0.0.8.255 and
> 192.1.168.10.0 0.0.4.255
>
> HTH
> Shine
>
>
> -----Original Message-----
> From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of Rods
> Rods2
> Sent: Friday, 22 February 2008 1:56 PM
> To: ccielab@groupstudy.com
> Subject: Wildcard mask
>
> Hello masters.
>
> I am getting confusing studying some kinds of wildcard masks. I understand
> well the tradional wildcard mask for VLSM, but others are very weird.
> I would like to know how to calculate this masks, as example:
> How to only permit routes that the third octect is 5, 10, 13, 14 using only
> two ACL. (Net 192.168.x.0) ?
>
> I got the answer from a book:
>
> access-list 10 permit 192.168.5.0 0.0.8.255
> access-list 10 permit 192.168.10.0 0.0.4.255
>
> How to get that answer? I really didn't undestand. Is that rigth ?
>
> Thanks in advance.
>
> -----Original Message-----
> From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
> Aabid Saleem
> Sent: Tuesday, 15 April 2008 1:10 AM
> To: Cisco certification
> Subject: ACL -- requirement
>
> Hi,
> i am new in this list,
> I have little confusion in creating an ACL for the network --
>
> I am receiving multiple network from BGP peer from the range 192.168.0.0/16
>
> as per task i need to filter all but allow 192.168.5.0/24,
> 192.168.10.0/24, 192.168.13.0/24 and 192.168.14.0/24 using only two
> line ACL
> catch is two line ACL,
>
> please answer it how it can be done for other scenarios, i need to
> understand ACL implementation for any other question for the same reason
>
>
> Aabid
>
>
> Pass the CCIE in six weeks, Guaranteed!
> http://www.certscience.com/CCIE
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