Hello & I have a question, I have an original IP range of 192.168.2.x, an original subnet mask of 255.255.255.0, and I need 6 subnets. -Now- I know that i need a subnet bit of 224, and that my new subnet mask will be 255.255.255.224 and I know that my increment range will be in increments of 32, but this will not work! could please explain to me in detail whats wrong and what my true range IP values will be?

Go to this site to get the free IP Subnetter Actually, there is nothing wrong with that. You will get 7 subnets because the first one is 192.168.2.0 Here is the break down. I'll just use the last numbers for ease. subnet valid range broadcast 0 1 thru 30 31 32 33 thru 62 63 64 65 thru 94 95 96 97 thru 126 127 128 129 thru 158 159 160 161 thru 190 191 192 193 thru 222 223 224 225 thru 254 255

Sorry, you get 8 subnets not 7 :)

Why won't it work? There would actually be 8 subnets, but 2 would be invalid. So you would end up with 6 usable subnets with 30 hosts per subnet. Your first range of addresses would start at 192.168.2.33. (192.168.2.32 would be the subnet ID, not a valid IP address.) That range would end at 192.168.2.62. 192.168.2.63 would be the broadcast address for that subnet, not a valid IP address for a client. 192.168.2.64 would be the next subnet ID so the next subnet range would start at 192.168.2.65 - 192.168.2.94. etc. That subnets would go like this: 1.) 192.168.2.33 - 192.168.2.62 2.) 192.168.2.65 - 192.168.2.94 3.) 192.168.2.97 - 192.168.2.126 4.) 192.168.2.129 - 192.168.2.158 5.) 192.168.2.161 - 192.168.2.190 6.) 192.168.2.193 - 192.168.2.222 That's it. Hope that helps.

Hello Glen, Why do you say that 2 of the subnets are not valid?

Well, to be technical, it depends a bit on your hardware and whether you are allowing a zero subnet. Often, the 0 subnet is not supported. Based on his question, and him mentioning the 6 subnets, I assumed he was not wanting the zero subnet. It's hard to explain without drawing a picture but I'll try. With a 224 mask the first 3 bits in the last octet are reserved for the network ID. They can not be all 0's or all 1's in binary. Either can the host ID which in this case is the last 6 bits. 224 in binary is 11100000. This means we use the last 6 bits for host ID. So, since the first 3 bits can not be all 0's or all 1's and the host ID, the last 6 bits can not be all 0's or all 1's, the lowest number in binary would be 00100001. (The network ID can not be all 0's since this would indicate the whole subnet.) 00100001 = 33 - The first valid IP address. Now, the last first 3 bits in the last octet can not be all ones either so the first 3 bits in binary would have to be 110 - and the host ID can not be all 1's so the highest number in the last 6 bits would be 111110. Put all 8 bits together, 11011110 = 222. The last valid IP address. Hope that helps clear things up. Dude, your post is not necessarily wrong, I was just referencing the fact that he said that the mask he was using (which is correct) and everything DRS mentioned seemed correct so I was just showing him how it could correct.

Well done Glen. Also an address which finishes in all 1's is the broadcast address of that subnet.