Ipv4 Addressing Tutorial

By Miragetek

IPv4 addresses are 32 bit binary numbers that allow devices to communicate with each other using the TCP/IP suite of protocols. A typical scenario of two IP devices communicating with each other would be your computer talking to a web server on the internet in order to display a webpage contained on the web server. An IP address is made up of 4 octets, each octet contains 8 bits, thus totalling 32 bits. This article relates to class C IP addresses only, which are the bread and butter of computer networking.

This is what an IP address looks like: 192.168.1.126

The Binary numeral system uses 0 and 1, so each bit in the address is represented as either 0 or 1.

The 192.168.1.126 address written in binary looks like: 11000000.10101000.00000001.01111110

How do we get 11000000.10101000.00000001.01111110 from 192.168.1.126? It´s actually quite easy.

Here´s a table that shows you the decimal value of each bit in an octet:

128 | 64 | 32 | 16 | 8 | 4 | 2 | 1

We can tell by using this table as a reference that:

The 1st bit in an octet is worth 128
The 2nd bit in an octet is worth 64
The 3rd bit in an octet is worth 32
The 4th bit in an octet is worth 16
The 5th bit in an octet is worth 8
The 6th bit in an octet is worth 4
The 7th bit in an octet is worth 2
The 8th bit in an octet is worth 1

Let´s use the first octet in the 192.168.1.126 address as an example (the decimal value of the first octet is 192), we will convert it from decimal to binary. In order to do this, we look at the table and find out which bits when added together will add up to exactly 192.

By doing some simple maths, we know that 128 + 64 = 192, so we know that the bits relating to 128 and 64 should be represented with a 1 and the remaining bits should be represented with a 0. This gives us the following binary number: 11000000

Let´s do another example, but this time using the second octet of the 192.168.1.126 address. We know we have to get 168 by adding up a certain combination of bits. If we add 128 and 32 and 8 together, we get 168, which means that the bits corresponding to 128 and 32 and 8 will be represented by a 1 and the remaining bits will be represented by a 0. This gives us the following binary number: 10101000

One last example using the thrid octet of the 192.168.1.126 address. We have to get 1, we know that the 8th bit has a value of 1, so all the bits other than the 8th bit will have a value of 0. This gives us the following binary number: 00000001

You should now understand the relationship between 192.168.1.126 and 11000000.10101000.00000001.01111110

You now have the foundation needed to understand more advance IP addressing concepts. Please read my Subnet Mask Tutorial in order to understand how to subnet an IP address.

Article Source: http://interpret.zar.vg

About The Author
My name is Jason Woodruff and I work for Miragetek Global Resources Ltd. I work within the IT industry, here's a link to our website:

www.miragetekshop.co.uk