From: perkinsr@xxxxxxxxxxxxxx
Date: Wed Jul 03 2002 - 12:19:04 GMT-3
We need to meet 2 criteria, on the given subnet and odd. We all know the
subnet criteria means we will match all bits up to a certain point as noted
by the subnet mask, in this case the first 25 bits must mach. The other
criteria forces us to look at data paterns not as familiar to us, we have to
look for something that all odd numbers have in common. All odd numbers end
in a 1 when written in binary, so we have to match the first 25 bits, then
end in a 1.
Given subnet:
172.16.80.129/25
In dotted decimal is:
172.16.80.128 255.255.255.128
With the last octet in Binary is:
172.16.80.10000000 255.255.255.10000000
Which gives us the range of IPs of:
172.16.80.10000000 172.16.80.11111111
So all IPs meeting the 2 criteria have the following pattern:
172.16.80.1xxxxxx1
Putting 0s where we want a match and 1s where we don't care gives us a mask
of:
0.0.0.01111110
Which in dotted decimal is:
0.0.0.126
So our ACL with the last octet in binary would be:
172.16.80.1xxxxxx1 0.0.0.01111110
The xs can be 1 or 0 it doesn't matter but typically people make them 0s, so
that written in dotted decimal with the bits we don't care about set to zero
we have:
172.16.80.129 0.0.0.126
-----Original Message-----
From: Shane Miles [mailto:smiles@ftdata.com]
Sent: Wednesday, July 03, 2002 7:45 AM
To: 'ccielab@groupstudy.com'
Subject: RE: odd numbered hosts access-list
I always screw these up but I'll give it a try. How about:
access-list 1 permit 172.16.80.129 0.0.0.128
-----Original Message-----
From: David Ham
To: ccielab@groupstudy.com
Sent: 7/2/02 8:32 PM
Subject: odd numbered hosts access-list
Hi all,
I have simple access-list question that I couldn't figure out.
I am trying to pass only odd numbered hosts at the token-ring interfaces
with 172.16.80.129/25
what could be the answer and the exponation??
Thanks in advance.
Regards,
David Ham
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