1. 138.38.228.53255.255.254.0 10001010.00100110.11100010.0011010111111111.11111111.11111110.00000000 IP addressSubnet mask AND 10001010.00100110.11100010.00000000 138.38.228.0 Network address

2. 138.38.228.53255.255.254.0 IP addressSubnet mask In practice it’s usually easier and quicker, once you spot that: 255 AND 138 = And a similar thing with 0 in the subnet mask; so the only “difficult” part is if you get a byte that’s not 255 or 0 in the subnet mask – like our 254

3. 138.38.228.53255.255.254.0 10001010.00100110.11100010.0011010111111111.11111111.11111110.00000000 IP addressSubnet mask 138.38.228.53/23 CIDR notation So you count up the number of continuous bits in the mask, from the left, and put that number after a slash. Obviously you can’t do this if you have a subnet mask where the bits aren’t continuous e.g. 255.255.253.0 = 11111111.11111111.11111101.00000000

4. 255.255.254.0 11111111.11111111.11111110.00000000 Subnet mask How many hosts/networks? 23 bits = 2 23 = 8388608 possible networks with this mask 9 bits = 2 9 = 512 hosts For any network there are two special addresses: the broadcast address: all 1s, and the network address: all 0s. So it’s 512-2 = 510 hosts.

5. Classes 11111111.11111111.11111111.00000000 11111111.11111111.00000000.00000000 11111111.00000000.00000000.00000000 Subnet mask: More networks More hosts in each network How many possible hosts in a Class C?