Chapter 3 - Page 1 Infogem Institute of Technology CCNA Course IP Addressing & Subnetting IP...

Chapter 3 -

Infogem Institute of Technology CCNA Course

IP Addressing & Subnetting

IP ADDRESSING&

SUBNETTING

Chapter 3 -



TOPICS

• Numbering Systems

• What is IP Addressing?– The Hierarchical IP Addressing Scheme– Additional Classes of Networks– Who Assigns Network Addresses?

• Subnetting a Network– Implementing Subnetting– How to Implement Subnetting– Subnet Masks– Can This Be Make Easier?

Chapter 3 -



NUMBERING SYSTEMS

• Decimal – base 10– Digits 0-9– Example: 163

• Binary – base 2– Digits 0-1– Example: 1010 0011

• Hexadecimal – base 16– Digits 0-9, a-f– Example: a3

Chapter 3 -



DECIMAL NUMBERING SYSTEM

• Each position in a Decimal number represents a value

• Decimal digits – 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9

163

100 10 1

1-100 6 - 10s 3 - 1s

Chapter 3 -



BINARY NUMBERING SYSTEM

• Each position in a Binary Number represents a value

• Binary Digits – 1 and 0

1 0 1 0 0 0 1 1

128 164 32 16 8 4 2

128 + 32 + 2 + 1 =

Chapter 3 -



CONVERT BINARY TO DECIMAL

• To Convert:– Identify the place holder value– Add all the values

1 0 1 0 0 0 1 1

128 164 32 16 8 4 2

1-128 1-32 1-2 1-1

128 + 32 + 2 + 1 = 163

Chapter 3 -



CONVERT DECIMAL TO BINARY

163-128 35

128 64 32 16 8 4 2 1

Chapter 3 -



CONVERSION EXERCISES

Convert the following decimal numbers to binary

128 64 32 16 8 4 2 1

11

45

72

168

195

234

255

Convert the following binary numbers to decimal

0 0 0 0 1 0 1 00 0 0 0 1 1 1 00 0 0 1 0 1 0 00 0 1 1 0 0 1 10 1 0 1 0 1 0 11 0 1 0 1 0 1 01 1 1 1 1 1 1 0

Chapter 3 -



CONVERSION SHORTCUT CHART1 22 43 84 165 326 647 1288 2569 512

10 102411 204812 409613 819214 1638415 3276816 65536 64K17 128K18 256K19 512K20 1024K21 2048K22 4096K

IP ADDRESSING SCHEME

• Addresses are fixed length of four octets (32-bits)

– An address begins with a network number

– An address ends with a local address (host/node number)

• There are three methods for depicting an IP address

– Dotted-decimal – 130.57.30.56

– Binary – 10000010.00111001.00011110.00111000

– Hexadecimal – 82 39 1E 38

All of these examples represent the same IP address

Chapter 3 -



IP ADDRESS CLASSES

Address

Class

Leading

Bit Pattern

128 | 64 | 32

Decimal

Range

(1st Octet)

Address

Template

Number

Of Networks

Hosts

Per Network

2^ Host Bits-2

A 0 1 – 126

B 10 128 – 191

C 110 192 - 223

CLASS A IP ADDRESS

• Class A: Net.Node.Node.Host

• The first bit of the first octet is always 0

• Decimal range of first byte: 1-126• Note: Network number 127 is reserved for diagnostics

• Maximum networks: 126

• Maximum nodes per network: 16,777,216

0

Network ID Host ID

Chapter 3 -



CLASS B IP ADDRESS

• Class B: Net.Net.Node.Host

• The first two bits of the first octet are always 10

• Decimal range of first byte: 128-191

• Maximum networks: 16,384

• Maximum nodes per network: 65,534

1 0

Network ID Host ID

CLASS C IP ADDRESS

1 1 0

Network ID Host ID

• Class C: Net.Net.Net.Host

• The first three bits of the first octet are always 110

• Decimal range of first byte: 192-223

• Maximum networks: 2,097,152

• Maximum nodes per network: 254

NOTE: Class D 224 – 239 – Used for multicast

Class E 240 – 255 – Used for research purposes

Chapter 3 -



ADDRESSING GUIDELINES

• Network ID cannot be 127– 127 is reserved for loop-back diagnostics– Ping 127.0.0.1 to test protocol installation

• Network ID cannot be all 1s or all 0s– All 1s indicates “Broadcast”– All 0s indicates “Any Address”

• Host IDs cannot be all 1s or all 0s– All 1s indicates “Network Broadcast”– All 0s indicates “The network only…”

• Private IP address (Private Internets)– Used for internal IP addressing– Will never be used on internet– 10.0.0.0 – 10.255.255.255– 172.16.0.0 – 172.31.255.255– 192.168.0.0 – 192.168.255.255

• Class D Multicast– 224 – 239

• Class E Experimental or Scientific purposes– 240 - 247

Chapter 3 -



IP ADDRESSING EXERCISE

IP Address Class Network ID Host ID

10.226.30.8

100.100.3.2

150.50.5.145

201.2.10.25

168.16.30.98

126.63.120.135

ETHERNET INTERFACE SETUP

207.201.1.0

P2R1

P2R3P2R2

e0

s1 s0

e0s1

s0

e0

s1

s0

P1R1

P1R3P1R2

e0

s1 s0

e0s1

s0

e0

s1

s0207.201.2.0

207.201.3.0

Chapter 3 -



• Divide a large network into smaller logical networks connected together by routers

• Uses bits from the host (node) portion of the address to create a subnet address

• Benefits of subnetting:

– Reduced network traffic

– Optimized network performance

– Facilitate spanning of large geographical distances

SUBNETTING A NETWORK

Chapter 3 -



SUBNET MASKS

• Distinguishes the network address from the host address– 1’s represent the positions in the IP address that refer to the network or subnet address– 0’s represent the positions in the IP address that refer to the host address– When coming from host end, remember to jump over one for subnet mask

255 . 255 . 0 . 0

11111111 11111111 00000000 00000000

NetworkMask

HostMask

Chapter 3 -



DEFAULT SUBNET MASKS

• Networks that have not been divided into subnets will use the default network mask for that class of address

– Class A 255 . 0 . 0 . 0

(binary) 11111111 . 00000000 . 00000000 . 00000000

– Class B 255 . 255 . 0 . 0

(binary) 11111111 . 11111111 . 00000000 . 00000000

– Class C 255 . 255 . 255 . 0

(binary) 11111111 . 11111111 . 11111111 . 00000000

DEFAULT SUBNET MASKS

# of Subnets = 2N – 2

# of Hosts/subnetwork = 2H – 2

N=number of bits used for subnetting

H = number of bits used for the host address

Chapter 3 -



CUSTOM SUBNET MASKS

• Maximum number of subnets

2 (number of masked bits) – 2 24 = (16) – 2 = 14

• Maximum number of hosts

2 (number of unmasked bits) –2 212 (4096) – 2 = 4094

255 . 255 . 240 . 0

11111111 11111111 11110000 00000000

NetworkMask

HostMask

SubnetMask

Chapter 3 -



HOW TO IMPLEMENT SUBNETTING

• An IP address can be given a subnet address by reducing the number bits designated as the host portion of the address

255 . 255 . 240 . 0

11111111 11111111 11110000 00000000

NetworkMask

HostMask

SubnetMask

Subnet Mask

150 . 19 . 16 . 15

SubnetAddress

10010110 00010011 00010000 00000001

Network Address Host AddressClass B Network Address

HOW TO IMPLEMENT SUBNETTING

• Determine the number of required Network Ids

– One for each subnet

– One for each Wide Area Network connection

150.50.0.0255.255.0.0

Chapter 3 -



NETWORK ADDRESSING

• Determine the number of required host Ids per subnet– One for each TCP/IP host– One for each router interface

150.50.0.0255.255.0.0

NETWORK ADDRESSING

• One subnet mask for your entire network

• A unique subnet ID for each physical segment

• A range of host IDs for each subnet

Subnet A

Host

Host

Host

Host

Subnet C

Subnet B

Network 150.19.0.0Subnet Mask 255.255.0.0

Chapter 3 -



IMPLEMENTING SUBNETTING

The subnet mask will determine how many bits are used to create new subnetworks and

how many bits are available for use as Host Ids

• Requirements Analysis– Determine number of subnets required– Determine number of hosts on largest subnet

• Compute Subnet Mask

• Determine Subnet Ranges

• Determine Assignable Host Addresses

Subnet A150.19.96.0

Host Host

Host

Subnet C150.19.64.0

Subnet B150.19.32.0

Network 150.19.0.0Subnet Mask 255.255.224.0

150.19.64.16150.19.96.6

Host

150.19.96.5

150.19.64.125

SUBNET PROCEDURES

Chapter 3 -



SUBNET REQUIREMENTS ANALYSIS

• How many subnets are required?

• How many hosts per subnet?

14 PC’s

12 PC’s

10 VAC’s

2 Main-frames

• Add 2 subnet and host fields. These are for the reserved addresses (all 0s and all 1s)– 5 + 2 = 7 Subnet– 15 + 2 = 17 Hosts

• Round Subnet and Host up to next power of 2– 128 64 32 16 8 4 2

– Subnets = 7 rounded up to 23 (8) –2 = 6 subnets– Hosts = 23 rounded up to 25 (32) – 2 = 30 Subnets

COMPUTE SUBNET MASKS

Chapter 3 -



COMPUTE SUBNET MASK

• 23 – the 3 represents the number subnet bits which are masked and are indicated with 1’s

• 25 – the 5 represents the number of host bits which are not masked and are indicated with 0’s

• The appropriate Subnet Mask is:

128 64 32 16 8 4 2 1

1 1 1 0 0 0 0 0 = 224

23 25

Net ID Host ID

DETERMINE SUBNETS

224 = 1 1 1 0 0 0 0 0reserved = 0 0 0 0 0 0 0 0

32 = 0 0 1 0 0 0 0 064 = 0 1 0 0 0 0 0 096 = 0 1 1 0 0 0 0 0

128 = 1 0 0 0 0 0 0 0160 = 1 0 1 0 0 0 0 0192 = 1 1 0 0 0 0 0 0

reserved = 1 1 1 0 0 0 0 0

FieldHost

MaskSubnet

Determine the range of available subnets

Chapter 3 -



SUBNET EXAMPLE – CLASS C

• Valid hosts in subnet 255.255.255.224

Mask = 111 0 0000

Subnet

32 001 0 0001 33 (start)0 1110 62 (end)

64 010 0 0001 65 (start)1 1110 94 (end)

96 011 0 0001 97 (start)1 1110 126 (end)

128 100 0 0001 129 (start)1 1110 158 (end)

160 101 0 0001 (start)1 1110 (end)

192 110 0 0001 (start)1 1110 (end)

Binary Hosts

Chapter 3 -



SUBNET REQUIREMENTS ANALYSIS

58 PC’s, 2 Servers

34 PC’s, 1 Server

• How many subnets are required?

•How many hosts per subnet?

Chapter 3 -



SUBNET PROCEDURES

• Requirements Analysis– Determine number of subnets required: ______________– Determine number of hosts on largest subnet: ______________

• Compute Subnet Mask: ________________________– Subnet multiple: ______________– Hosts/subnet: _______________

• Subnet Net IDs Host Address Range_______________ __________________________________________ __________________________________________ __________________________________________ ___________________________

SUBNETTING A NETWORK ADDRESS

150 . 50 . 1 . 0

P2R1

P2R3P2R2

e0

s1 s0

e0s1

s0

e0

s1

s0

P1R1

P1R3P1R2

e0

s1 s0

e0s1

s0

e0

s1

s0

Assigned by Pod 2

Assigned by Pod 1

150.50.1.2

200.20.2.0 100.10.1.0

150.50.1.3

Chapter 3 -



Lab – Page 158Review Questions

Page 160

Homework: - Written Lab- Review 3-10 & 13-20

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