From: Scott Morris (smorris@ipexpert.com)
Date: Sat Feb 23 2008 - 22:51:08 ARST
What's the binary tell you?
5 = 0000 0101
10= 0000 1010
13= 0000 1101
14= 0000 1110
Off the cuff, lots of differences there.... Notepad and cut/paste is your
friend.
5 = 0000 0101
13= 0000 1101
These only have one bit of difference (the 8 bit).
10= 0000 1010
14= 0000 1110
These only have one bit of difference (the 4 bit).
In the mask, a 0 bit means the value must be the same, a 1 bit means it can
be any value.
So what I'd end up with then is:
access-list permit 192.168.5.0 0.0.8.x
access-list permit 192.168.10.0 0.0.4.x
HTH,
Scott Morris, CCIE4 (R&S/ISP-Dial/Security/Service Provider) #4713, JNCIE-M
#153, JNCIS-ER, CISSP, et al.
CCSI/JNCI-M/JNCI-ER
VP - Technical Training - IPexpert, Inc.
IPexpert Sr. Technical Instructor
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-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
Snyder, Daniel P
Sent: Friday, February 22, 2008 5:32 AM
To: ccielab@groupstudy.com
Subject: FW: question about ACL on a filter
Say I have the following networks...
192.168.0.0/24 - 192.168.15.0/24
I need to make a filter on routes coming in. The router can only accept the
following routes:
192.168.5.0
192.168.10.0
192.168.13.0
192.168.14.0
The thing is it can only be a 2 line acl... Is this even possible??
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