determining the netmask

Kerry Hoath kerry at
Sun Aug 18 08:13:07 EDT 2002

If you convert to binary you get
11111111 11111111 11111111 00000000
the left-hand side has 24 ones an d then 8 zeros.
Oversimplified t his means that the left-hand 24 bits represent the network portion
of the ip address and the right-hand section is the host portion.
So an ip address of and a netmask of says
that this is host 1 on the 192.168.1 (or network.
You of course know the bottom address of a range is the network address and the top address is broadcast

Now if the netmask were there would be 16-bits of network
and 16-bits of host so the machine would be the
1.1 host on the network.
You can also subnet on any bit boundary so gives you
6 bits of host address and 26 bits of network address.

Here is an example:
If we set the subnet mask to (25 bits) on
a network with 192.168.1 addresses we could have to networks: to (126 addresses) to (126 addresses)
The router would know that if hte ip address was and the netmask was that if a packet were sent to it would be considered to not be on the local network.

Any questions? If this is unclear I can rephrase i used to teach this stuff.

Regards, Kerry.
On Sat, Aug 17, 2002 at 06:10:52PM -0500, Gregory Nowak wrote:
> Hi all.
> I know that a netmask of is a 24-bit netmask. I also know that 8 bits is a byte. However, I don't understand how it is determined that is a 24-bit netmask. Can someone please explain this? Thanks in advance.
> Greg
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Kerry Hoath:  kerry at kerry at or  kerry at
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