From: Chua, Parry (Parry.Chua@xxxxxxxxxx)
Date: Fri Nov 02 2001 - 00:44:55 GMT-3
Hi,
Sorry for the error, resend as follow
The quick way is to determine the differences.
161=10100001= 128 + 00 + 32 + 1
225=11100001= 128 + 64 + 32 + 1
============
ExR=01000000= (64) , this will be the mask, only one bit can be 0 or 1.
==
And=10100001= 161 this should be the start address
OR=11100001= 225 the higer value it can go.
To cover the above two network by one access list will be
access-list 1 [permit | deny] 161.97.37.0 64.0.0.255
What if change 225 to 226, then it will not cover the exact changes, here it
go
161=10100001 start
226=11100010 end
==============
ExR=01000011 = 67 will be the mask, there are 3 bits can be 0 or 1 to have
up to 8 combination.
And=10100000 = 160 start
OR=11100011 = 227 end
access-list 1 permit 160.97.37.0 67.0.0.255
=== ==
The above mean that start from 160 to 163 then 224 to 227 cover some extral
networks. As you
can see that the masks is still split...
Hope this help.
> Parry Chua
>
-----Original Message-----
From: Andy Cuberly [mailto:andyc@netcat.com]
Sent: Friday, November 02, 2001 5:54 AM
To: CCIE Group Study
Subject: Access List Wild Card Mask - Bit splitting
Can anyone give me a good explanation of why it is bad to split up the bits
in the wild card mask? example below:
161.97.37.0 64.0.0.255
I am trying to determine if it is possible to list these two networks with
one Access List statement.
161.97.37.0
225.97.37.0
Andy Cuberly
This archive was generated by hypermail 2.1.4 : Fri Jun 21 2002 - 06:45:01 GMT-3