determining the netmask
kerry at gotss.net
Sun Aug 18 08:13:07 EDT 2002
If you convert 255.255.255.0 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 192.168.1.1 and a netmask of 255.255.255.0 says
that this is host 1 on the 192.168.1 (or 192.168.1.0) network.
You of course know the bottom address of a range is the network address
192.168.1.0 and the top address is broadcast 192.168.1.255.
Now if the netmask were 255.255.0.0 there would be 16-bits of network
and 16-bits of host so the machine would be the
1.1 host on the 192.168.0.0 network.
You can also subnet on any bit boundary so 255.255.255.192 gives you
6 bits of host address and 26 bits of network address.
Here is an example:
If we set the subnet mask to 255.255.255.128 (25 bits) on
a network with 192.168.1 addresses we could have to networks:
192.168.1.0 to 192.168.1.127 (126 addresses)
192.168.1.128 to 192.168.1.255 (126 addresses)
The router would know that if hte ip address was 192.168.1.1 and the netmask was
255.255.255.128 that if a packet were sent to
192.168.1.129 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.
On Sat, Aug 17, 2002 at 06:10:52PM -0500, Gregory Nowak wrote:
> Hi all.
> I know that a netmask of 255.255.255.0 is a 24-bit netmask. I also know that 8 bits is a byte. However, I don't understand how it is determined that 255.255.255.0 is a 24-bit netmask. Can someone please explain this? Thanks in advance.
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