Start finding the subnet.
192.168.100.32 255.255.255.224, the first address is 192.168.100.32, the number of addr. in the subnet is
.255 subtrackted with .224 equals 31, 31 added to .32 equals .63
So the first addr is .32 and the last is .63 which means you have 32 ip addr. in the subnet, but you can not use the first .32, because it’s the network addr. of the subnet. And you can not use the last, because it’s the broadcast addr. of the subnet. This means that tou can only use 30 ip addr. in the subnet.
The binary method is.:
there must NOT be any 1 after 0, that makes en invalid subnet mask eg.
255.255.255.242 – 11111111.11111111.11111111.11110010
Your subnet mask eg.
255.255.255.224 – 11111111.11111111.11111111.11100000
Try the IP sub calc. from Wildcats
which is the last addr. in the subnet
and is the net addr. of the subnet.
, the last addr. is .32 plus 31 = .63.