From: Scott Morris (swm@emanon.com)
Date: Wed Oct 25 2006 - 20:51:09 ART
Well.. Let's think that through...
230 = 11100110
184 = 10111000
There are NINE 1 bits there. 2^9 = 1024 matches in just the first two
octets. So I'm thinking that you'll get a LOT more matches then you intend
to get there!
While binary math is kinda fun and challenging, not everything can be
summarized into one line.
Save yourself the thinking. Just use permit/deny any!
Scott Morris, CCIE4 (R&S/ISP-Dial/Security/Service Provider) #4713, JNCIE
#153, CISSP, et al.
CCSI/JNCI-M/JNCI-J
IPExpert VP - Curriculum Development
IPExpert Sr. Technical Instructor
smorris@ipexpert.com
http://www.ipexpert.com
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
Sergio Silva
Sent: Wednesday, October 25, 2006 4:49 PM
To: ccielab@groupstudy.com
Subject: wildcard access lists
Hi Group
Please bare with me on this one,
Would you agree with this result?
Netblocks
10.0.0.0/8
172.16.0.0/12
192.168.0.0/24
access-list 1 permit 0.0.0.0 230.184.0.0
Thanks in advance
Serg
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