In this example allow for the following:

Network address: 192.168.10.0

The Perth router has to support 60 hosts. In this case, a minimum of six bits are needed in the host portion of the address. Six bits will yield 62 possible host addresses, 2⁶ = 64 – 2 = 62, so the division was 192.168.10.0/26.

The Sydney and Singapore routers have to support 12 hosts each. In these cases, a minimum of four bits are needed in the host portion of the address. Four bits will yield 14 possible host addresses, 2⁴ = 16 – 2 = 14, so the division is 192.168.10.96/28 for Sydney and 192.168.10.112/28 for Singapore.

The Kuala Lumpur router requires 28 hosts. In this case, a minimum of five bits are needed in the host portion of the address. Five bits will yield 30 possible host addresses, 2⁵ = 32 – 2 = 30, so the division here is 192.168.10.64/27.

The following are the point-to-point connections:

• Perth to Kuala Lumpur 192.168.10.128/30 – Since only two addresses are required, a minimum of two bits are needed in the host portion of the address. Two bits will yield two possible host addresses (2² = 4 – 2 = 2) so the division here is 192.168.10.128/30.
• Sydney to Kuala Lumpur 192.168.10.132/30 – Since only two addresses are required, a minimum of two bits are needed in the host portion of the address. Two bits will yield two possible host addresses (2² = 4 – 2 = 2) so the division here is 192.168.10.132/30.
• Singapore to Kuala Lumpur 192.168.10.136/30 – Since only two addresses are required, a minimum of two bits are needed in the host portion of the address. Two bits will yield two possible host addresses (2² = 4 – 2 = 2) so the division here is 192.168.10.136/30.

There is sufficient host address space for two host endpoints on a point-to-point serial link. The example for Singapore to Kuala Lumpur is configured as follows: