Hi
You need to look at in binary really. Using 192.168.5.0 255.255.255.0 as the example network
one octet in binary =
128 64 32 16 8 4 2 1
Because we are dealing with a Class C network the first 3 octets in the subnet mask will be 255 ie. 255.255.255.0 so we will concentrate only on the last octet.
So first you need 100 hosts. Well the only number larger than 100 is 128 so your first subnet is
128 64 32 16 8 4 2 1
1 0 0 0 0 0 0 0 = .128
So the first network is
192.168.5.0 255.255.255.128
which includes host 192.168.5.1 -> 192.168.5.126 with 192.168.5.127 as the broadcast address.
Next you need a network with 34 hosts. Unluckily for you this is just above 32 in the binary above so you need to select 64 ie.
128 64 32 16 8 4 2 1
1 1 0 0 0 0 0 0 = .192
(remember subnet masks must be continguous)
So carrying on from last network
192.168.5.128 255.255.255.192
hosts 192.168.5.129 -> 190
broadcast 192.168.5.191
Next you need a network with 29 hosts
128 64 32 16 8 4 2 1
1 1 1 0 0 0 0 0 = 224
192.168.5.192 255.255.255.224
hosts = 192.168.5.193 -> 222
broadcast 192.168.5.223
7 hosts next. You need to be careful with this one. Seven hosts would suggest you just need to mark out the "8" in the octet but
128 64 32 16 8 4 2 1
1 1 1 1 1 0 0 0 = 248
192.168.5.224 255.255.255.248
hosts 192.168.5.225 -> 230
broadcast 192.168.5.231
You only get 6 hosts because of the broadcast and the the subnet mask so you need a "16" in the octet ie.
128 64 32 16 8 4 2 1
1 1 1 1 0 0 0 0 = 240
192.168.5.224 255.255.255.240
host 192.168.5.225 -> 192.168.5 238
broadcast 192.168.5.239
and finally 2 hosts
128 64 32 16 8 4 2 1
1 1 1 1 1 1 1 0 = 252
192.168.5.240 255.255.255.252
hosts 192.168.5.241 & 242
broadcast 192.168.5.243
As one of the other posters said you should start with the largest requirement first.
** Edit - apologies, the octet values don't quite line up in the examples but hopefully it still makes sense **
HTH
Jon
