This article is the first in a series of three which has been written for those who are, or may one day be responsible for designing an IP network.

---

Sit back, take a deep breath and prepare for one of my techie articles, this time on Sub-netting

---

Introduction

IP Routing requires that each physical subnet within the Intranet (a network owned by a private organization) use’s a unique Network ID.

A large organization may have insufficient network addresses available for their needs.

IP Sub-netting creates larger numbers of smaller groups of IP Addresses called Subnets which can improve network security and increase performance by reducing the size of the broadcast domain

To sub-net, bits must be borrowed from the Host portion within a Subnet Mask.

The best way I have found to explain sub-netting to my students is to run through an example configuration

Example

Your organization has been assigned a Class C network address of 198.209.5.0 you require eleven subnets

How can this be achieved

Firstly determine the default subnet mask

Three classes of network are available:

• Class A Networks – The first octet of the IP Address is 1-126 and will use a subnet mask of 255.0.0.0

• Class B Networks – The first octet of the IP Address is 128-191 and will use a subnet mask of 255.255.0.0

• Class C Networks – The first octet of the IP Address is 192-223 and will use a subnet mask of 255.255.255.0

In this example the first octet is 198 which an example of a class C network with the default subnet mask of 255.255.255.0 this can also be shown as / 24 (as 24 bits are assigned as a network ID)

To achieve more networks (subnets) bits must be borrowed from the host portion of the subnet mask, which in this example is the forth octet

Therefore we only need to work on the forth octet, this needs to be converted from decimal format into binary

To convert binary to decimal is reasonably straight forward each binary bit within each octet has a preset decimal value.

Working from the left to right work out how many bits must be borrowed to achieve 11 subnets

You will actually end up with 14 subnets. Leaving the remaining 4 bits in the forth octet for the hosts in each subnet

Now calculate the new subnet mask, working from the right to the left, the forth octet will become 240 as shown below

Therefore the new subnet mask will become: 255.255.255.240 – The original network ID was 198.209.5.0 this can NOT be used for the subnets.

We now have 14 subnets each subnet must have a separate network ID.

Look at the lowest value bit which was borrowed to calculate the subnet mask, this will then become the increment for the subnets

The lowest value bit borrowed had a value of 16, therefore the network addresses will be:

---

NOTE: the lowest value subnet 198.209.8.0 (using all 0’s) and the highest 198.209.8.240 (using all 1’s) are discarded as these CANNOT be validated

---

Let us now look at the first subnet – The Network ID will be 198.209.5.16

The last address in each subnet is the broadcast address, in the first subnet this will be 198.209.5.31

The remaining addresses will be available for the hosts on that subnet, in the first subnet this will be 198.209.5.17 through to 198.209.5.30

---

If you want more practice then try these Practice Sub-netting Questions

---