The figure above shows 5 different subnets, each with different host requirements. The given IP address from our ISP is192.168.1.0/24.

The host requirements are:

Network A - 14 hosts

Network B - 28 hosts

Network C - 2 hosts

Network D - 7 hosts

Network E - 28 hosts

As recommended, we begin the process by subnetting for the largest host requirement first. As it seems, the largest requirements are for NetworkB and NetworkE, each with 28 hosts.

Don’t forget the cram table!

 

Let’s apply the formula: usable hosts = 2^n - 2. For networks B and E, 5 bits are borrowed from the host portion and the calculation is 2^5 = 32 - 2. Only 30 usable host addresses are available in this case due to the 2 reserved addresses. Borrowing 5 bits meets the requirement but leaves little room for future growth.

So we revert to borrowing 3 bits for subnets leaving 5 bits for the hosts. This allows 8 subnets with 30 hosts each.

We have created and will allocate addresses for networks B and E first:

Network B will use Subnet 0: 192.168.1.0/27

Host address range 1 to 30 (192.168.1.1 – 192.168.1.30) 

192.168.1.31 (broadcast address)

Network E will use Subnet 1: 192.168.1.32/27

Host address range 33 to 62 (192.168.1.33 – 192.168.1.62)

192.168.1.63 (broadcast address)

The next largest host requirement is NetworkA, followed by NetworkD.

We will borrowing another bit and subnetting  the network address 192.168.1.64 will give us the following a host range of:

Network A will use Subnet 0: 192.168.1.64/28

Host address range 65 to 78 (192.168.1.65 – 192.168.1.78)

192.168.1.79 (broadcast address)

Network D will use Subnet 1: 192.168.1.80/28

Host address range 81 to 94 (192.168.1.81 – 192.168.1.94)

192.168.1.95 (broadcast address)

This allocation supports 14 hosts on each subnet and satisfies the requirement.

*In Network C, there are only two hosts. In this case we borrow two bits to meet this requirement.

Beginning from 192.168.1.96 and borrowing 2 more bits results in subnet 192.168.1.96/30.

Network C will use Subnet 1: 192.168.1.96/30

Host address range 97 to 98 (192..168.1.97 – 192.168.1.98)

192.168.1.99 (broadcast address)

From the above illustration, we have met all requirements without wasting many possible subnets and available addresses.

In this case, bits were borrowed from addresses that had already been subnetted. As you will recall from a previous section, this method is known as Variable Length Subnet Masking, or VLSM.

*use illustration to create networks for the WAN on the network..