With a netmask of .240 then your last block is any block of 16 IPs on a 16 block boundary, ie. 192.168.1.0 thru 192.168.1.15 or 192.168.1.16 thru 192.168.1.31 or 192.168.1.32 thru 192.168.1.47 or 192.168.1.48 thru 192.168.1.63 etc. You get the idea, always remembering that the first and last IP in the block you define is unusable as a host address. In the example you quote then your block is wrong because the address 192.168.0.15 would be in the first 16 address subnet whereas the address 192.168.0.30 would be in the next 16 address subnet, so your range would be: 192.169.0.16 - network address 192.168.0.17 thru 192.168.0.30 - useable host adresses 192.168.0.31 - broadcast address You have to subnet on boundaries that are a power of 2. -- Howard. ______________________________________________________ LANNet Computing Associates <http://www.lannet.com.au> On Mon, 17 Jul 2000, Dean Hamstead wrote: > > > > > > > And its that simple, in a subnet with a mask of 255.255.255.240, you have > > > 15 available addresses. BUT. Remever, with IP addressing the lowest > > > address (all 0's in the host space) is the network address (identifies the > > > subnet) and the highest address (all 1's in the host space) is the > > > broadcast address. So you only really have 13 addresses that you can use > > > for your actual hosts in the subnet. > > > > You actually have 16 addresses in this range, with 14 useable. IP > > numbering starts from ZERO - in other words, ZERO is a valid IP address - > > so your range is from 0 to 15 - with 0 being the network address, and 15 > > being the broadcast address. > > when you use a subnet like 255.255.255.240 which 14 address's can be > used? > with 255.255.255.0 is all 255 in the last byte, eg 192.168.0.x > > with 255.255.255.240 is it the last 14 address's? > can i use it to take a chunk from the middle? > eg. 192.168.0.15 - 30? > > Dean > > > -- SLUG - Sydney Linux User Group Mailing List - http://slug.org.au/ More Info: http://slug.org.au/lists/listinfo/slug
