1. 1 Exercise: IPv4 subnetting

2. 2 Task 1 Given is an IP network with address 194.141.0.0: Divide this network into 8 subnets.

3. 3Solution l This network is class C because the starting bits are 110. l In binary form: 11000010. 10001101.00000000.00000000 l The first 3 octets represent the network address. The fourth octet is for subnet and host address.  We need a subnet mask with 3 bits ones, because 8 = 2 3 : 11111111.11111111.11111111.11100000 l In decimal form: 255.255.255.224

4. 4Comments l We use this mask in the corresponding field of the IP configuration. l The available subnets have the following addresses: 194.141.0.0194.141.0.1194.141.0.2194.141.0.3194.141.0.4194.141.0.5194.141.0.6194.141.0.7 Can not be used. In fact we have 6 real subnets. Hosts with address 0 or 255 can not be used. Therefor each subnet contains up to 256/8 - 2 = 30 hosts.

5. 5 Task 2 Given is an IP network with address 162.251.0.0: Divide this network into 32 subnets.

6. 6Solution l The subnet mask is 11111111.11111111.11111000.00000000 l in decimal: 255.255.248.0 l 30 subnets with up to 2046 hosts each.

7. 7 Task 3 l Given is a subnet mask: 255.254.0.0 l Which class is the network? l How many real subnets? l How many real hosts may contain each subnet?

8. 8Solution l This is a subnet mask for network class A. l There are 2 7 – 2 = 126 real subnets. l Each subnet may has up to 2 17 – 2 = 131070 real hosts.