[ale] Networking question

[Trey Sizemore]

On Fri Jul 22, 2005 10:26PM, Joe Knapka wrote:
> 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

This is a good way to think about it.  Much easier to think about it this
way.

-- 
Cheers,
Trey
----
 
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