[lug] CIDR question

[Archer Sully]

----- Original Message -----
From: "Andrew Diederich" <andrew at NETdelivery.com>

> Ok, I've got an easy one for someone who knows the math.  I'm switching
> ISP's today and am losing my class C for a partial C.  (Boo! hiss!)  For
> routing purposes it looks like I have two contiguous /26 networks.  From a
> fancy cidr HOWTO like thing (http://www.ibm.net.il/~hank/cidr.html) that
> tells me I have 2 one-quarter segments of a class C.
>
> So, if I have the front end of a C (x.x.x.0 - x.x.x.128?) What would be my
> network and broadcast numbers for the two /26 networks?  I'd really like
to
> find the formulas while I'm at it, 'cause it's embarrassing to ask this
sort
> of question.  *grin*

There are four possible 1/4 class C nets for a given class C:
x.x.x.{0,64,128,192}/26
The netmask is always 255.255.255.192 (or 0xffffffa0 for the hex minded).
Broadcast is always all 1's in the host part (eg for 192.168.0.64/26, it
would
be 192.168.0.127).

Since you have two distinct networks you might have to multi-home a machine
to
route between them.  I've never set this situation up, so I don't know how
it works.
You might be able to just knock a bit off your netmask, which works from a
practical
standpoint, but maybe not for the router.  Check with your ISP.

What I discovered, though, is that NAT plus anonymous networks are your
friends.
I'm very glad that I went that route for my home network rather than getting
official
IP numbers for my internal machines.  Its easier to set up and a lot less
hassle.

-- archer