determining the netmask

[Gregory Nowak]

Very interesting. Thanks to all those who responded. I get it now.
Greg


On Sun, Aug 18, 2002 at 02:28:41PM +1000, Aaron Howell wrote:
> Let me try to make it a bit clearer.
> There are a total of 4 bytes (32 bits) in a dotted quad (an ip address).
> In every netmask specification, there is a left and a right hand side.
> the right hand side specifies how many machines are referenced by the netmask, while the left represents the remaining bits.
> Netmasks, in cidr notation are represented as ip/mask
> E.G, what used to be a class c address, that is, 192.168.1.xx/255.255.255.0
> is now represented as 192.168.1.0/24 (there are 8 bits, or 255 possible hosts on the right hand side, leaving 24 bits of unused mask on the left).
> If you wanted a total of 512 possible hosts, then it would be /23, for 1024 it'd be /22.
> The original class b (255*255 hosts) is /16
> While the original class a (255*255*255 hosts) is /8.
> Traditionally, bite boundaries were used for allcation of addresses, meaning you could only have a /24, /16 or /8 allocation.
> With the introduction of cidr, its now possible to bundle several /24s into a /22 or /21 for example and have them all  routed to the one place
> (this concept is known as supernetting)
> and similarly its possible to route smaller lots of addresses to (e.g clients),
> such as a /28 or /29 (16 or 8 ips) this concept is subnetting.
> This is explained in painfully complex detail in any good book on tcp/ip.
> Regards
> Aaron