From: Lab Rat #109385382 (techlist01@gmail.com)
Date: Mon Nov 27 2006 - 20:51:13 ART
Awesome, thanks. That was it.address is ANDed.
So, for the below, do you guys get:
Access-list 100 permit ip 199.1.68.0 0.0.169.0 any?
Thanks,
Ed
From: Nick Griffin [mailto:nick.jon.griffin@gmail.com]
Sent: Monday, November 27, 2006 6:04 AM
To: Lab Rat #109385382
Cc: cisco@groupstudy.com; Cisco certification; security@groupstudy.com
Subject: Re: ACL Discontiguous Network Matching Question
IE has a good link for this:
http://www.internetworkexpert.com/resources/01700370.htm
On 11/27/06, Lab Rat #109385382 <techlist01@gmail.com> wrote:
I know how to derive the inverse mask for matching an ACL to discontiguous
subnets, but how does one derive the network address again? I must have
forgotten this part of the procedure.
For example, if I'm supposed to match the following networks using a single
ACL:
199.1.101.0/24
199.1.109.0/24
199.1.197.0/24
199.1.204.0/24 <http://199.1.204.0/24>
What is the network statement to use in the ACL? More importantly, why? If
there is a Doc CD location that explains this logic in detail, that would be
great.
Thanks,
Ed
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