From: Cisco Champ (cisco_champ@delhirocks.com)
Date: Thu Jul 10 2008 - 15:56:19 ART
Hello Joe,
Thanks for the reply..
The discussed solution in that attachment to match two networks. Here I have 4
networks and I can use 2 line in that ACL to match 4 networks. My outcome in
binary at 3rd octate as mentioned below.
05 --> 0000 0101
10 --> 0000 1010
13 --> 0000 1101
14 --> 0000 1110
what is the way to pick and club which network. I suppose we should pick for
close network match from Most Significant Bit side. then network 5 & 10 may
club together to match and 13 & 14 can be match for another match.
=====================================
AND for network 05 & 10
05 --> 0000 0101
10 --> 0000 1010
-----------------------
0000 0000 This means 192.168.0.0
XOR for network 05 & 10
0000 1111 This means SM : 255.255.240.0
=====================================
for other two networks 13 & 14
13 --> 0000 1101
14 --> 0000 1110
---------------------
0000 1100 This means 192.168.12.0
XOR for network 13th and 14th
0000 0011 Which means SM 255.255.252.0
=====================================
I'm sure I'm very close to the solution but may not be using the right pairs
of the network. Any suggstions will be appreciated.
Thanks!!
> From: joe@affirmedsystems.com
> To: cisco_champ@delhirocks.com; ccielab@groupstudy.com
> Subject: RE: simple query on ACL
> Date: Thu, 10 Jul 2008 13:42:16 -0400
>
> yawwwwwwwwwn
>
>
http://blog.internetworkexpert.com/2007/12/26/q-how-do-i-compute-complex-wil
> dcard-masks-for-access-lists/
>
>
>
> -----Original Message-----
> From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
> Cisco Champ
> Sent: Thursday, July 10, 2008 1:32 PM
> To: ccielab@groupstudy.com
> Subject: simple query on ACL
>
> Hi:
>
> In one of the question requirement was to match (192.168.5.0, 10.0, 13.0
and
> 14.0) network. Solution is given below, which is giving the results
> correctly.
>
> access-list 1 permit 192.168.10.0 0.0.4.255
> access-list 1 permit 192.168.5.0 0.0.8.255
>
> But my question is how can we used '192.168.5.0' as Network address for
> subnet
> mask 255.255.248.0 it should be 192.168.4.0. But when I do this I didn't
get
> the required results. Can any one tell me how this one is working.
>
> Thanks in advance!!
>
>
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This archive was generated by hypermail 2.1.4 : Mon Aug 04 2008 - 06:11:54 ART