NET Blog Post 2

In this lesson we covered subnetting. A subnet refers to a network within another network. These are used to increase the overall efficiency of a network by reducing the distance in which our network needs to travel to reach its destination.

The addresses we should use when we are subnetting are as followed:

10.0.0.0 – 10.255.255.255/8

172.16.0.0 – 172.31.255.255/12

192.168.0.0 – 192.168.255.255/16

Each section of the address is known as an octet. They are 8 bits in length which equates to 28 or 256. The total number of addresses available to us is 254. An IP address is made up of four octets. The total amount of positions equates to 232 or 4,294,967,296. When we break down an octet into its binary form, it follows an 8-digit form that corresponds with the table shown below:

When we subnet, we are stealing host bits and giving it to the Net. Here is what a normal IP address would look like: 192.168.1.0 We split this into two sections, the HOST, and the NET. Below we can see a representation of this:

NET = 24bits [192.168.1] [.0] 8 Bits = HOST

If we were to take away 1 bit, this would give us two subnets leaving us with only 27 (128) subnets. When you subnet, you lose number of hosts. It costs you two bits when you create a subnet.

We also briefly covered the OSI Model, below we can see the order of the top down model:

We were then provided with some subnetting tasks. I didn’t have too much trouble with this as the concept clicked relatively quickly. I came up with a basic set of rules that seemed to work for each tasks. These helped me calculate the total number of hosts, net addresses, etc.

· Subnet Mask is based on where it is split

· Number of Addresses= 2 Number of host bits

· Number of Hosts = (2 Number of host bits) - 2

· Number of Subnets = 2 Number of stolen host bits

· Number of net addresses = Number of subnets

· Number of Broadcast Addresses = Number of subnets

Tasks:

192.168.1.0/24 Steal 1 BIT (CIDR 25)

= 255.255.255.128

11111111.11111111.11111111.10000000

CIDR 25 = 255.255.255.128

# Addresses = 128

# subnet = 2

# Host = (128-2) = 126

Net Addresses 2

Broadcast Address 2

192.168.0.0/16

xxxxxxxx.xxxxxxxx.{xxxxx}xxx.xxxxxxxx

CIDR 21 = 255.255.248.0

# subnets = (2)5 = 32

# net addressess = (2)5 = 32

# Broadcast addresses = (2)5 = 32

# Hosts = (2)11 - 2 = 2048 - 2 = 2046

Addresses = 2048

The last thing that we covered was IPV6

Hex (A Decimal)

2001:00B8:0000:0000:0000:FF00:0042:8392

NET | HOST

A Nibble is 4 bits => Quad nibble = 4x4 = 16bits

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

0 1 2 3 4 5 6 7 8 9 A B C D E F

2001:00B8:0000:0000:0000:FF00:0042:8392

The total number of bits = 16 X 8 = 128