to setup up classless subnetting, you will need a router for each network you want. you will need all of these routers to be able to talk to each other (with a small netowrk RIP is good, OSPF is better for long run). Or you can setup the Routing tables yourself manually. You could buy one big 6509 from CISCO and use it. One big router that acts as several routers, does bridging....costs a LOT.
It would be hard to explain everything you need to do to set this up as it is a TRUE networking problem and would take more than a post on this website to answer your question. you have several options to go with.
as for the problem with people stealing the ip of the server....you will have to use DHCP to help on large networks, ip conflicts are hard to avoid. (it gets even worse when no-so-smart users accidently run their own DHCP servers on accident).
to re-arrange the IPs, you need to know how many subnetworks you need. once you do this, you'll know how many bits you need off your base 172.16/24.
for instance, if you need 8 subnetworks, you would do this:
you have a class B assignment, meaning u can play with 16 bits (the last 16). If you need 8 subnetworks, 8 networks= 3 bits (0 - 7, 7 = 0b111). This means you will need 4 of the last 16 bits to make up a network which means you will have the following:
10101100.00010000.
00000000.00000000 = 172.16.0.0 - 172.16.31.255 /255.255.224.0
10101100.00010000.
00100000.00000000 = 172.16.32.0 - 172.16.63.255/255.255.224.0
10101100.00010000.
01000000.00000000 = 172.16.64.0 - 172.16.95.255/255.255.224.0
10101100.00010000.
01100000.00000000 = 172.16.96.0 - 172.16.127.255/255.255.224.0
10101100.00010000.
10000000.00000000 = 172.16.128.0 - 172.16.159.255/255.255.224.0
10101100.00010000.
10100000.00000000 = 172.16.160.0 - 173.16.191.255/255.255.224.0
10101100.00010000.
11000000.00000000 = 172.16.192.0 - 172.16.223.255/255.255.224.0
10101100.00010000.
11100000.00000000 = 172.16.224.0 - 172.16.255.255/255.255.224.0
(you cant use the 1st and last IPs in each list, explained below)
you will have 13 bits to play with on each network giving you
2^13 - 2 total hosts on each network
the -2 is because you cant use .0 or .255 because .0 specifies a network and 255 is broadcast.
if u wanted 10 subnets you'd have to use up 4 bits to specify a network and u'd have to redo that table.
if your network will really use all those IP addresses (or close). you may want to hire a network guy, or hire a consultant for such a large network. (im available for hire
)
also, for a network this large, DHCP is almost a must. and your gateways should end in .1, this is not a witten convention but .51 is just a bad number to pick for a gateway address.
