How to consolidate two networks on one access-list?

Unanswered Question
Jul 14th, 2009

Imagine that I need to write an access-list to allow traffic to

228.0.0.0/8 and

231.0.0.0/8


Then I wrote down the binary for both of them and tried to find a pattern. If I do 228.0.0.0 3.255.255.255 I also include "229.0-255.0-255.0-255" and that does not work though.


11100100 = 228

11100111 = 231



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Edison Ortiz Tue, 07/14/2009 - 11:16

That's a multicast range so what exactly are you trying to do?


The subnet as you stated should work but it depends on the use of this filtering.


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Edison.

news2010a Tue, 07/14/2009 - 11:22

The respective exercise asked to 'advertise a given router as the RP for 228.0.0.0/8 and 231.0.0.0/8'.


So I would use 'ip pim send-rp-announce loopback0 scope 20 group-list .


So I was trying to consolidate the respective access-list on one line if possible but I so far I can't see how to do that...


Marvin Rhoads Tue, 07/14/2009 - 11:26

So does your access list read:


access list 1 allow any 228.0.0.0 3.255.255.255


and then, under the relevant interface:


ip access-group 1 in


?


Are their any other access lists applied to the interface(s)?

news2010a Tue, 07/14/2009 - 11:51

In this case, it would be applied on a router part of the multicast RP configuration as shown below (I separated the access-lists since it is nto clear to me if it is possible to write this using just one access-list entry).


config t

#ip pim send-rp-announce loopback0 scope 20 group-list 1


access-list 1 permit 228.0.0.0 0.255.255.255

access-list 1 permit 231.0.0.0 0.255.255.255



Edison Ortiz Tue, 07/14/2009 - 12:06

That's the right syntax. The requirements is to advertise 228.0.0.0/8 and 231.0.0.0/8 not from 228.0.0.0/8 to 231.0.0.0/8 - if I understood your post correctly.


It seems you are studying for a CCIE and reading the requirements correctly is a fundamental part of the exam.


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Edison.

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