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Subnetting
March 7 (Day); March 2 (Night)

Introduction to Network Addresses:
o
TCP/IP is the most widely used protocol in the world. The Internet uses only IP
addressing. In order for a host to access the Internet, it must have an IP address.
o
IP addresses consist of four numbers (between 0-255) separated by dots. Each
number is known as an octet.
o
Because each octet can be between 0-255, it can be represented by 8 bits.
Therefore, the entire address is represented by 32 bits.
o
In its most basic form, the IP address has two parts: a network ID and a host ID.

The network ID portion of an IP address refers to the network. Hosts on
a network can communicate directly only with devices that have the
same network ID.

Hosts might share the same physical segment, but if they have different
network IDs, they usually cannot communicate with each other – unless
there is another device that can make a connection between the networks.
This device is usually a router.
Host
Net A

Host
Net A
Router
Host
Net B
Host
Net B
Once data reaches the appropriate network, it uses the host portion of the
IP address to find the individual computer.
o
All networks have a network address. Usually, it is going to be the lowest value
in that particular network.

Classes of Network Addresses:
o
There are five classes of IP addresses (A through E). Only the first three classes
(A-C) are used regularly.
o
Summary of Network Classes:
Class
A
B
C
1st Octet
1-126
128-191
192-223
Network/
Host ID
N.H.H.H
N.N.H.H
N.N.N.H
# of
Networks
126
16,382
2,097,150
Hosts per
Network
16,777,216
65,536
256
o
In class A addresses, the first octet is the network ID and the last three octets are
host ID.

Because the last three octets are dedicated to host addresses, this means
you can have 16,777,216 hosts on a class A network.

We know this because 2^24=16,777,216 OR 256 x 256x
256=16,777,216.
o
In class B addresses, the first two octets are the network ID and the last two
octets are host ID.

Because the last two octets are dedicated to host addresses, this means
you can have 65,536 hosts on a class A network.

o
We know this because 2^16=65,536 OR 256 x 256=65,536.
In class C addresses, the first three octets are the network ID and the last octet is
the host.

Because the last octet is dedicated to host addresses, this means you can
have 256 hosts on a class A network.


We know this because 28=256 OR 256.
What is Subnetting:
o
Network administrators sometimes need to divide networks, especially large
ones, into smaller networks. These smaller divisions are called subnetworks.
o
Each subnetwork is connected to one another by a device known as a router.
o
Larger networks are broken down into subnetworks through the use of a
numbering scheme known as the subnet mask.
o
Routers use the “ANDing” process to determine if a destination host is on the
same network or not. Basically, you multiply the binary equivalent of the IP
address by the binary equivalent of the subnet mask to get the network address.
o
Here are the default subnet masks. Class A uses 255.0.0.0. Class B uses
255.255.0.0. Class C uses 255.255.255.0.
o
Let’s apply the ANDing process and the subnet mask to calculate the network
address for the host 200.1.1.5.
Decimal form
Binary form
Host address
200.1.1.5
11001000.00000001.00000001.00000101
Subnet mask
255.255.255.0
11111111.11111111.11111111.00000000
Resulting network address (binary
11001000.00000001.00000001.00000000
form):
Resulting network address (dec form): 200.1.1.0

Creating a Custom Subnet Mask for a Class C Address:
o
Let’s pretend that we want less than 256 hosts on one network with a network
address of 200.1.1.0. We want two subnets and to separate them with a router for
security reasons and to reduce traffic.
o
The first step to subnetting is to determine how many subnets we need. In this
case we need two.
o
The next step is to see how many bits we should borrow from the host portion of
the network address, begin to add the bits from the right to left until the total is
equal to or greater than the number of subnets we need.
128
64
32
16
8
4
2
1
2 + 1 = 3 which is greater than the number of subnets we need.
o
After we know how many bits to borrow, we take them from the left side of the
fourth octet of the host address.
Bits we need to borrow
128
64
32
16
8
4
2
1
1
1
0
0
0
0
0
0
Our custom subnet mask is going to be 255.255.255.192. The 192 comes
from the value of the first two bits from the left. (The 192 comes from 128 +
64).
o
Since we borrow two bits for the network ID. This means we may have 22=4
subnets.
o
Since we borrow two bits for the network ID, this leaves only six bits for the host
ID. This means that we may have 26=64 hosts on each subnet.
o
Here’s a chart that demonstrates the four subnets we can create.
Subnet No.
0
1
2
3
o
Network Address
200.1.1.0
200.1.1.64
200.1.1.128
200.1.1.192
Range of Available Addresses
200.1.1.0 – 200.1.1.63
200.1.1.64 – 200.1.1.127
200.1.1.128 – 200.1.1.191
200.1.1.192 – 200.1.1.255
Based on our example, can a host of 200.1.1.5 send data directly to a host of
200.1.1.70?
o
Based on our example, can a host of 200.1.1.5 send data directly to a host of
200.1.1.60?