[ale] Networking question

[Joe Knapka]

Trey Sizemore <trey at fastmail.fm> writes:

> I have a networking conceptual problem.  Suppose I plan to use the
> 192.168.0.1 network for an intranet.  With a default network mask of
> 255.255.255.0 I would have 254 host addresses available.
> 
> With a 25-bit mask (255.255.255.128) I would be able to have 2
> subnetworks and each would be able to have 126 addresses available.  The
> host ranges for each would be 192.168.1.1 - 192.168.1.127 and
> 192.168.1.129 - 192.168.1.255.
> 
> Now, with a 26-bit mask (255.255.255.192) I understand that I would be
> able to have 4 subnets and each would support 62 hosts, but what would
> the host ranges be?

The way I like to think about these things, is to insert an extra dot
in the middle of (in this case) the low-order byte.  In this case, the
extra dot goes between the top two (network) and bottom six (host)
bits of the low-order byte, so you have (last byte expressed as
binary):

192.168.1.00.000000 thru 192.168.1.00.111111
192.168.1.01.000000 thru 192.168.1.01.111111
192.168.1.10.000000 thru 192.168.1.10.111111
192.168.1.11.000000 thru 192.168.1.11.111111

Then just remove the extra dot to compute the address ranges. So the
range of the last byte of the first subnet is 00000000 to 00111111,
giving 192.168.1.0 to 192.168.1.63.  The second subnet range is
01000000 to 01111111, giving 192.168.1.64 thru 192.168.1.127. Etc.

Cheers,

-- Joe Knapka