From: Rich Collins (nilsi2002@gmail.com)
Date: Sun Feb 24 2008 - 01:24:51 ARST
I just answered the same question on this formun.
Third octet in binary - X is the wildcard 8 and 4 respectively.
5 and 13 X101
10 and 14 1X10
You have to look for the common denominator which is the 5 and 10. Then you
have the wildcard 8 and 4. The wildcard means either a 0 (giving you a 5 or
10) or if it's set to a 1 binary (means adding an 8 or 4 respectively) to
bring you up to the 13 and 14 by adding to the base of 5 or 10.
You can search this forum to find more explanations and examples.
-Rich
On 2/21/08, Rods Rods2 <rods1234@hotmail.com> wrote:
>
> 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.
> _________________________________________________________________
> Express yourself instantly with MSN Messenger! Download today it's FREE!
> http://messenger.msn.click-url.com/go/onm00200471ave/direct/01/
This archive was generated by hypermail 2.1.4 : Sat Mar 01 2008 - 16:54:49 ARST