From: Bhaskar Sivanesan (bas_bharath@yahoo.com)
Date: Mon Jun 04 2007 - 10:21:13 ART
HI Prahsant
This is the logic. ACL wildcard mask defines 0 as MATCH and 1 as dont care.. so for the ranges you have listed, the third octet in binary is as follows.
00000001 - 1
00000010 - 2
00000011 - 3
00000100 - 4
00000101 - 5
If you observe, all the even octets have 0 and the odd ones have 1 in the last bit. So if we are to match a even octet, the last has bit has to be 0 and the remaining seven bit can be anything. So the corresponding wildcard mask will be 11111110 which corresponds to 254.
Thus the inverse mask becomes 0.0.254.255
HTH
Bhaskar
prashant sansare <prashant.sansare@gmail.com> wrote:
Hi
Can some one please help me to understand the ACL
Scenerio:Route filtering
R1-----se0/0-----R2
Networks on R2
201.1.1.1/24
201.1.2.1/24
201.1.3.1/24
201.1.4.1/24
201.1.5.1/24
EIgrp is the routing protocol running R1 needs to receive the routers with
201.1.x.0 range that have even number in the third octet
can some one pls explain me how do we filter the third octet evn
numbers,please explain me the logic of filter the even numbers
i know how acl works and stuff and how to do router filtering
access-list 2 deny 201.1.1.0 0.0.254.255
access-list 2 permit any
router eigrp 123
network 201.1.1.0
network 201.1.2.0
network 201.1.3.0
network 201.1.4.0
network 201.1.5.0
network 201.1.6.0
distribute-list 2 out se 0/0
can any one please me explain me the logic of filtring the even from the acl
how to play with wild card mask to acheive our requirment
Regards
Prashant
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