From: Larry Metzger (larrymetzger@sbcglobal.net)
Date: Fri Sep 03 2004 - 12:27:22 GMT-3
Dustin:
You can't do it in a single line. Your acl will catch 4 multicast
groups (226.0.0.0-226.255.255.255, 230.0.0.0-230.255.255.255,
234.0.0.0-234.255.255.255, 238.0.0.0-238.255.255.255). The acl says to
allow bits 3 & 4 to vary which means you will get 2^2 or 4 possible
options.
Do things in binary and it will be much easier to see. If you're not
really good at this type of thing then I would suggest taking the time
to write all of the options you want to get (226, 227, 228, 229...) and
convert to binary (use a calculator if you need to). Then look at how
you can group them together. As shown in the solution, it is sometimes
easier to deny some and then allow a bigger range.
Larry
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
dusth@comcast.net
Sent: Friday, September 03, 2004 8:13 AM
To: ccielab@groupstudy.com
Subject: multicast access-list question
Hi all,
I have a question on multicast acl. Could any one explain for me
differences b/t these 2 acl.
Solution book:
access-list 51deny 239.0.0.0 0.255.255.255
access-list 51 deny 224.0.0.0 1.255.255.255
access-list 51 permit 224.0.0.0 15.255.255.255
Can I use as follow acl configuration instead? If no, then why can't I?
Though 224, 225, and 239 subnet will be implicitly deny, or because this
acl is multicast acl so have to be configure the deny statement.
access-list 51 permit 226.0.0.0 12.255.255.255
Thanks in advance
Dustin
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