From: Anshuk Kesarwani (anshuk.ccie@gmail.com)
Date: Sun Feb 24 2008 - 06:08:33 ARST
Hi ,
The key for understanding the wildcard masks in the binary maths . Where a 0
means have to match and a 1 means either is aceptable.
now see covnert the wild card in your example
192.168.5.0 0.0.8.255
here 5 will be converted to 00000101
and wild card 8 will be 00001000
now as per the rule the forth bit can be either 0 or 1 as wildcard is 1
here. So this entry matches 5 and 5+8(13)
For the other one matches 10 and 14.
Hope that helps.
Anshuk
On 2/22/08, 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.
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