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• Subject: Re: [vox-tech] question about ip addresses and netmasks
• From: Dale Bewley <dale@bMAPSewley.net>
• Date: Tue, 20 Feb 2001 14:08:46 -0800
• References: 20010220134714.A32719@dirac.org

```I have a hard time explaining it to other people, but
you have to picture the binary number in your head and after a while you
memorize most of these.

To convert /30 to the mask I say to myself:
32-30=2 bits
2 bits can create a maximum of 4 in decimal
256-4=252
so the netmask is 255.255.255.252.

Once you do enough of that, you quickly see 252 and think /30.
Or see 240 and think /28. (256-240=16 which is 2^4 and 32-4=28)

255.255.254.0 would be /23 meaning 23 bits in the netmask.
If you subtract 254 from 256 you'll get 2. That tells you you have two
/24 subnetworks.

It gets more complicated when you are peeling off a /29 or something in
the middle of the range from 0-255. You have to make sure you network
starts on a boundary divisible by 8. (32-29=3bits = max of 8)

i.e. For the address 1.2.3.74/28 what is the broadcast address and what
is the subnet?
It's 1.2.3.79 and 1.2.3.72. The next subnet starts at .80 so the
broadcast is one below that.

Sorry if that wasn't much help.
If anyone has a easier way to explain that, please do.
I probably should have just kept my mouth shut. :)

On Tue, 20 Feb 2001, Peter Jay Salzman wrote:
> is there a way to see, without converting numbers to binary and performing a
> a bitwise and, that:
>
>       131.155.72.0/255.255.254.0
>
> means all addresses from 131.155.72.0 through 131.155.73.255?  is there
> short cut for this?
>
> something like "255 means that quad is firm, 254 means you can go one
> higher, and 0 means the quad can be anything"?  i know that's "loose talk",
> but is it generally true?
>
> pete
>
> --
> "It's better to be safe than assimilated."                   p@dirac.org
>                       -- Chakotay                            www.dirac.org/p
>

--
Dale Bewley - Bewley Internet Solutions Inc. http://bewley.net/
```