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. Regards, Kerry. 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. > Greg > > > _______________________________________________ > Speakup mailing list > Speakup at braille.uwo.ca > http://speech.braille.uwo.ca/mailman/listinfo/speakup > > -- Kerry Hoath: kerry at gotss.net kerry at gotss.eu.org or kerry at gotss.spice.net.au ICQ: 8226547 msn: kerry at gotss.net Yahoo: kerryhoath at yahoo.com.au
