Keith:

> Hello all, I need the input of a network guru:

That's me.  :)

> I seem to recall that some of the fundamental rules at the IP layer is
> that:
> *	Every node on a subnet must have all it's network bits set appropriately
> in its IP address.
> *	Node must be to the right of the LSB of the network address

> I've been told (by IT in my org) to use: 12.147.70.88/27 for a little
> embedded device.  This doesn't make sense to me...
> 	12.147.70.88  = 12.147.70.01011000
> network	12.147.70.224 = 12.147.70.11100000
> (last octets in bin, duh..)

I have no idea where you got the .224 for your network address.  That is
the last octet of the netmask.  Your network address is .64 .

In this case, here are the four parameters for your connection to the IP
network:

network address: 	12.147.70.64
IP address:		12.147.70.88
broadcast address:	12.147.70.95
netmask:		255.255.255.224 (/27)

In a /27 network the .88 address is a valid IP address.  There is no
reason it shouldn't work if you are properly setting your broadcast
address (.95), and if your route contains the correct network address (.64).

> On my development network, I can recreate the problem, and when I switch
> the IP address to include the network bits - everything works AOK.

> So, my questions are:
> A)  Am I way off base here?  Am I thinking circa 80's IP rules and things
> are different now?

Things are the same except for the class business.  The 12 address space
is a "Class A" but you are treating it as not a Class A address space.  Cisco
routers need to be told "ip classless" to work properly.  The Linux kernel
(AFAIK) defaults to classless IP addressing and does not need any special
configuration to work properly.

> B)  Apparently an implementation has changed in the linux kernel since my
> notebook doesn't complain - is this true?

The Linux kernel now adds the network route in its routing table based on
the IP address, netmask, and broadcast address supplied when ifconfigging
the interface.

--
Jim Ockers (ockers@ockers.net)
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