Subnetting Made Simple By Keith W. Noe – CCNA, CCAI Ivy Tech Community College Sellersburg,...

Subnetting Made Subnetting Made SimpleSimple

By Keith W. Noe – CCNA, CCAIIvy Tech Community College

Sellersburg, Indiana

Objective

• To subnet a Class “C” IP network.

Why Subdivide a Network

• Class A networks are designed to have more than 16 million hosts

• Think about this: 100 megabits (Mbps) of bandwidth divided by 16,000,000 hosts.

• Result: each host gets 6.25 bits per second (bps) of bandwidth.

• Gee, isn’t 19.9 kbps dial-up great?

Why Subdivide a Network

• One reason is to increase the bandwidth for the users.

• Another reason is to group users with a common purpose.

• Other reasons?

The Basics

• You will need to know to the binary number system.

• There are many websites for you to use.• About Computing & Technology• There is an easy way to do this

http://compnetworking.about.com/cs/basicnetworking/a/binaryhexnumber.htm

Basic Terminology - Decimal

• Each individual number in the decimal number system is called a digit

• Why? We have ten fingers• Each finger is called a digit

Basic Terminology - Decimal

• Each individual number in the decimal number system is called a digit

• Why? We have ten fingers• Each finger is called a digit• Therefore there are ten digits in the decimal

number system• The ten digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

Basic Terminology - Binary

• Each individual number in the binary number system is called a BIT

• Why? BIT stands for BInary digiT• There are two BITS in the binary number

system• The two bits are 0 & 1

Basics

• Technicians must be able to translate numbers between the binary and decimal number system

• Why?• In networking, this is a skill that will make

your job easier

The Basics

• Write down the following numbers. Start at the right and work your way to the left.

128 64 32 16 8 4 2 1

This is the numbers that you will use to convert between the decimal and binary number systems

Decimal to Binary Conversion

• In this example, we will be converting a decimal number to binary

• Convert 18510 to Binary.• Hint: 185 is an odd number, therefore, your

binary equivalent number will also be odd.• Remember, numbers ending in ZERO are

even and numbers ending in ONE are odd.


18510 = ____________2

128 64 32 16 8 4 2 1

Begin by subtracting the largest number from 185 without the difference being less than zero (negative)


18510 = ____________2

128 64 32 16 8 4 2 1 1

We can subtract 128 from 185. The difference is 57.


18510 = ____________2

128 64 32 16 8 4 2 1 1 0

Next, subtract 57 – 64. The result is –7. Therefore put a zero under the 64.


18510 = ____________2

128 64 32 16 8 4 2 1 1 0 1

Next, subtract 57 – 32. The difference is 25. Put a 1 under the 32.


18510 = ____________2

128 64 32 16 8 4 2 1 1 0 1 1



18510 = ____________2

128 64 32 16 8 4 2 1 1 0 1 1 1



18510 = ____________2

128 64 32 16 8 4 2 1 1 0 1 1 1 0

Next, subtract 1 – 4. The difference is -3. Put a 0 under the 4.


18510 = ____________2

128 64 32 16 8 4 2 1 1 0 1 1 1 0 0

Next, subtract 1 – 2 The difference is -1. Put a 0 under the 2.


18510 = ____________2

128 64 32 16 8 4 2 1 1 0 1 1 1 0 0 1

Next, subtract 1 – 1 The difference is 0. Put a 1 underneath the 1.


18510 = 10111012

128 64 32 16 8 4 2 1 1 0 1 1 1 0 0 1

Binary to Decimal Conversion

• Subtraction was used for converting a decimal number to binary

• Therefore, addition will be used to convert a binary number to decimal

• We will use the same number used in the last example.

• 101110012 will be converted to decimal.

Binary to Decimal conversion

• Begin with the same chart

128 84 32 16 8 4 2 1


• Write the binary number for conversion as shown

128 64 32 16 8 4 2 1 1 0 1 1 1 0 0 1


• Each column that has a one, write down the number above it.

128 64 32 16 8 4 2 1 1 0 1 1 1 0 0 1

128 32 16 8 1


• Add these numbers together

128 64 32 16 8 4 2 1 1 0 1 1 1 0 0 1

128 + 32 + 16 + 8 + 1= ?


• Add these numbers together

128 64 32 16 8 4 2 1 1 0 1 1 1 0 0 1

128 + 32 + 16 + 8 + 1= 185

Subnetting

• All IP addresses are stored as binary numbers in the computer.

• On the human side, we enter network numbers as dotted-decimal numbers.

More Terminology

• The largest binary number that can be used as part of a network address is 255

• 25510 equals 111111112

• That is; 8 binary 1s or bits.• 8 bits = 1 byte (IBM term)• A byte is also called an OCTET• Remember, the word OCTET means 8.

Address Classes & Ranges

• Class A – 1 to 126• Class B – 128 to 191• Class C – 192 to 223

Address Classes Number of Networks

• Class A – 126 Networks• Class B – 65,534 Networks• Class C – 2,097,152 Networks

Classes and Number of Hosts

• Class A – 16,777,214 hosts• Class B – 65,534 hosts• Class C – 254 hosts

Subnetting a Class C Network

• The first three octets of a Class C address identifies the network number.

• For example, 192.1.2.0• The fourth octet identifies the host address

on this particular network.• The range of numbers for the fourth octet is

0 to 255.


• Two numbers in the fourth octet are reserved and cannot be assigned to a host (computer, printer, router, etc.)

• The two addresses are 0 and 255.• 0 identifies the network and 255 is the

broadcast address for this network.


• Therefore, for network 192.1.2.0 the possible addresses are as follows:

• 192.1.2.0 is the major network address• 192.1.2.1 to 192.1.2.254 are assignable host

addresses• 192.1.2.255 is the major broadcast address

for this network.


• First you must determine the reason you are subnetting a network.

• There are many reasons. We will choose one.

• If we put all 254 hosts on this one major network, and the bandwidth is 100 Mbps, each host will have approximately 393,700 bps of bandwidth (~394 kbps)


• Suppose that we wish to give 4 Mbps of bandwidth to each user,

• Then we will subnet the network and put a maximum of 25 users on each subnet.


• Aim: maximum of 25 users or hosts• Add 2 the the total number of users.• Write this chart. 128 64 32 16 8 4 2 1 It is the same chart we used earlier for number

conversions.


• Locate between which two numbers where 27 is located.

128 64 32 16 8 4 2 1 ^27 is located between 16 and 32.

Note: After adding 2 to the number of workstations and the result is 4, 16, 32, 64, etc, draw the line to the right of that number.


• Draw a line between 16 and 32 as shown. 128 64 32 | 16 8 4 2 1 | The three bits left of the vertical line will be

used for the subnetwork number. The five bits to the right of the line will be

used for the host address.


• Place 1s below the three bots to the left of the line

• Place 0s below the five bits to the right of the line

128 64 32 | 16 8 4 2 1 1 1 1 | 0 0 0 0 0Remember, this is the fourth octet.

Determining the Subnet Mask

• The default mask for a class C address is 255.255.255.0

• The 255.255.255 identifies the part of the subnet mask used for identifying the network portion of the IP address

• The .0 identifies the host portion of the IP address.

Determining the Subnet Mask

• Start by writing the first three octets for the subnet mask: 255.255.255.

• Now calculate the subnet mask number for the fourth octet

128 64 32 | 16 8 4 2 1 1 1 1 | 0 0 0 0 0 128 + 64 + 32 = ?

Determining the Subnet Mask• Start by writing the first three octets for the

subnet mask: 255.255.255.• Now calculate the subnet mask number for

the fourth octet 128 64 32 | 16 8 4 2 1 1 1 1 | 0 0 0 0 0 128 + 64 + 32 = 224Therefore the subnet mask is 255.255.255.224

Addresses

• The last steps.• Calculate each subnet address• Calculate the 1st host address• Calculate the last host address• Calculate the broadcast address for each

subnetwork

Addresses

• The first subnetwork address is always 0. 192.1.2.0• For this example, the subnetwork address

will increment by 32. 32 is the smallest part of the subnetwork address.

128 64 32 | 16 8 4 2 1 1 1 1 | 0 0 0 0 0

Subnet Addresses 128 64 32 | 16 8 4 2 1 1 1 1 | 0 0 0 0 0

Subnet Addresses 0- 192.1.2.0 4- 192.1.2.128 1- 192.1.2.32 5- 192.1.2.160 2- 192.1.2.64 6- 192.1.2.192 3- 192.1.2.96 7- 192.1.2.224

1st Host Addresses

• The first host address is always the subnet address plus 1.

• For example: 192.1.2.0 + 1 = 192.1.2.1• Therefore the first host address is 192.1.2.1

1st Host Addresses

128 64 32 | 16 8 4 2 1 x x x | 0 0 0 0 1

1st host Addresses 0- 192.1.2.1 4- 192.1.2.129 1- 192.1.2.33 5- 192.1.2.161 2- 192.1.2.65 6- 192.1.2.193 3- 192.1.2.97 7- 192.1.2.225

1st Host Addresses 128 64 32 | 16 8 4 2 1 x x x | 0 0 0 0 11st host Addresses 0- 192.1.2.1 4- 192.1.2.129 1- 192.1.2.33 5- 192.1.2.161 2- 192.1.2.65 6- 192.1.2.193 3- 192.1.2.97 7- 192.1.2.225

Broadcast Addresses

• The broadcast address is always 1 less than the next subnetwork address.

• For example, the broadcast address for subnet 0 can be calculated by subtracting 1 from the next subnetwork address.

• 192.1.2.32 – 1 = ?• Therefore the broadcast address for subnet 0

is 192.1.2.32 – 1 = 192.1.2.31

Broadcast Addresses 128 64 32 | 16 8 4 2 1 x x x | 1 1 1 1 1


Last Host Addresses

• The last set of addresses to calculate are the the last available host addresses.

• Using the broadcast address for each subnetwork, subtract 1 to obtain the last host address.

• For example: broadcast address for subnet 0 is 192.1.2.31

• The last host address is 192.1.2.30

Last Host Addresses 128 64 32 | 16 8 4 2 1 x x x | 1 1 1 1 0


Questions?

Subnetting Made Simple By Keith W. Noe – CCNA, CCAI Ivy Tech Community College Sellersburg,...

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