On Friday 12 April 2002 18:10, Ewan Leith wrote:
You are right, ive not been paying attention to this thread, but though 2 IPs with those subnet masks are on the same 9 bit subnet.
The whole subnet will consist of 10.0.0.0-10.0.1.255
Exactly, although it's a supernet rather than a subnet; you're making the host portion of the address larger rather than smaller.
What he wants is a normal Class C subnet.
Classes are Bad. Classes are old, and largely superseded thanks to classless routing protocols. Although knowing them is good, thinking of everything in a class-based system can lead to some really bad mistakes. As such, it's better to avoid calling things 'class C/B/A', and just go on the length of the CIDR mask; a traditional Class A is a /8, a Class B /16, and a Class C /24. If you talk of a 256-host block as a /24, it means that you can do the sums more easily, since a /23 comprises twice as many addresses as a /24, and a /22 comprises four /24s and a /21 comprises eight /24s (although you have to remember to subtract two addresses per network).
On Fri, 2002-04-12 at 18:54, John Higgins wrote:
First network 10.0.0.0 255.255.254.0 Second network 10.0.1.0 255.255.254.0
In no way do I mean to sound rude - but isn't that completely wrong?
Doesn't the above subnet mask mean that those addresses are on the same subnet? If the subnet mask ends at the 23rd binary digit then 10.0.0.x and 10.0.1.x are in the same subnet - the first 23 binary digits are 00001010 00000000 0000000 - and doesn't that mean they both exist within the same host area of the last 9 binary digits?
Exactly. It tells the network devices that you have one /23 area. If you wanted to look at it as a /24 with a 1-bit supernet, then fine; you still end up with two effective /24s to play with. (And we're all assuming that Michael only needs to add a maximum of 254 more hosts...) The range of valid addresses is 10.0.0.1-10.0.1.254 - to do this, you need a /23 netmask.
Wouldn't you need a 255.255.255.0 subnet mask in this case? This looks awfully similar to 2 class C networks and all class C's have 255.255.255.0 as the subnet mask.
Gah. Brainfart on my part. Although the whole area is 10.0.0.0 /23, the networks themselves are 10.0.0.0 /24 and 10.0.1.0 /24. Sorry about that. And yes, if you used the /23 netmask, you'd just end up with one big (510-host) subnet. Of course, depending on how Michael actually wants to do things, this might be more useful than having two separate networks. If he wants traffic to pass freely between the two networks, then he would be advised to aggregate the two networks into one; if not, then two networks will suit him better. A little addendum - if you're using classful addressing, 10.0.0.0 is most definitely *not* a Class C address; because classful addresses must obey the first octet rules so that network devices can grok their class (i.e. Class A (0-127) - first octet starts 0 Class B (128-191) first octet starts 10 Class C (192-223) first octet starts 110 Class D (224-239) first octet starts 111) As such, the RFC 1918 address combined with the /24 mask means that it *can't* be classful unless there's a 16-bit subnet in the network; which is somewhat unlikely. cheers, Gideon.