From: Sean.Zimmerman@clubcorp.com
Date: Thu Jun 28 2007 - 13:45:26 ART
Without going to 1's and 0's, there is a trick. The applicable masks for
each octet are 255 (8 bits),254 (7 bits),252 (6 bits),248 (5 bits),240 (4
bits) ,224 (3 bits),192 (2 bits),128 (1 bit) and, of course, 0. If you
subtract each of these from 256 (excluding 255 and 0), you get the base
subnet number. For instance, for the /18, you would subtract 192 from 256,
which gives you 64. The applicable subnets for that range would be
X.X.0.0/18, X.X.64.0/18, X.X.128/18 and X.X.192.0/18. As you can tell,
using 224 (/19) gives you a base of 32, which wouldn't cover the entire
range. 128 (/17) with a base of 128 is to broad.
Sean
"Eric Dobyns" <eric_dobyns@yahoo.com>
Sent by: nobody@groupstudy.com
06/28/2007 11:11 AM
Please respond to
"Eric Dobyns" <eric_dobyns@yahoo.com>
To
"'Douglas M Todd, Jr'" <dtodd@PARTNERS.ORG>
cc
<ccielab@groupstudy.com>
Subject
RE: Is there a trick to this?
That's a very useful link! I knew about doing the binary math, but
thought
there might be a way someone knew of to avoid wasting so much IP space.
Thanks, folks.
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
Douglas M Todd, Jr
Sent: Thursday, June 28, 2007 9:23 AM
To: Eric Dobyns
Cc: ccielab@groupstudy.com
Subject: Re: Is there a trick to this?
Yip, there is a way to do this.
Best to read up on access-lists and "and" "or" operations.
You will need to find with the "and" operation the network.
You will need to find with the "or" operation the ""mask""
This will create a one line access-list which will cover the whole range.
This may help you out.
http://www.internetworkexpert.com/resources/01700370.htm
Eric Dobyns wrote:
> Take the following 4 loopback interfaces:
>
>
>
>
>
> 10.1.22.1 255.255.255.0
>
> 10.1.34.1 255.255.255.0
>
> 10.1.37.1 255.255.255.0
>
> 10.1.47.1 255.255.255.0
>
>
>
> Redistribute them into IGP as a single advertisement while wasting the
least
> amount of address space.
>
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