Using this network ID 192.168.128.0 and assume that you need 8 subnets with 30 hosts in each subnet. Calculate

Question

I believe the correct answer is 255.255.255.254. I need assistance regarding the concept of how I arrive at this answer.

Answer 1

255.255.255.254 would give you an unuseable network. You'd only have a network address and a broadcase address for each subnet.

The correct answer is 255.255.255.224. That will give you the following subnets & usable ranges:

192.168.128.0 (192.168.128.1 - 192.168.128.30)
192.168.128.32 (192.168.128..33 - 192.168.128.62)
192.168.128.64 (192.168.128.65 - 192.168.128.94)
192.168.128..96 (192.168.128.97 - 192.168.128.126)
192.168.128.128 (192.168.128.129 - 192.168.128.158)
192.168.128.160 (192.168.128.161 - 192.168.128.190)
192.168.128.192 (192.168.128.193 - 192.168.128.222)
192.168.128.224 (192.168.128.225 - 192.168.128.254)

More information: I got your e-mail. I can't reply to it until you verify your e-mail address on the Yahoo system. I'm trying to give you my personal e-mail address so that I can explain this to you but I won't post it here for obvious reasons. Verify your e-mail address with Yahoo and then e-mail me again if you want any additional assistance.