Subnetting

[dt-kclug at xr7.org]

On Tue, 19 Nov 2002, Lucas Peet wrote:
> My question is:  How can I split a Class C block of IP addresses like so
> - 
> 
> Network #1:
> NetIP:		10.0.0.0
> Usable IP's:	10.0.0.1-2
> Broadcast:		10.0.0.3
> 
> Network #2:
> NetIP:		10.0.0.4
> Useable IP's:	10.0.0.5-254
> Broadcast:		10.0.0.255
> 
> Is this even possible?  What other alternatives would I have besides
> subnetting, if I want both the above networks separately routable on
> different interface cards?

This is not possible.  Look at the addresses and masks in binary to see
why.  Netmasks are always 1's through the network part of the address, and
the /24 notation means that the netmask is 24 1's followed by 8 0's
(255.255.255.0).  This is easier to describe with this notation...

There isn't anything that forces you to use the same netmask for the whole 
/24 (class C) block.  Following your example, you could have 2 /30 
networks (0-3 and 4-7).  Since these two networks together make up a /29, 
you could then have another /29 (8-15) to fill out a /28.  This repeats 
until you fill up the /24; you would have a /25 (128-255) at the top.

I know this doesn't solve your problem, since you wind up with 7 subnets
total, but it should at least explain why you can't get there from here.

Maybe the real question to help is what is the network that only needs two 
addresses?  Since these are 'real' IPs, is it a point-to-point link?  
Sometimes these can be configured without IP addresses to avoid situations 
like this.

dt

-- 
Dean Troyer
dt at xr7.org