Basic Sub Netting

8/8/2019 Basic Sub Netting

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Subnetting BasicsSubnetting Basics

Michael D. MannMichael D. Mann

Copyright 2009 by : Michael D. Mann Copyright 2009 by : Michael D. Mann


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Prerequisite KnowledgePrerequisite Knowledge

Class 1st OctetRange

Valid NetworkNumbers

TotalNetworks

for this

class

Number ofhosts per

network

AN.H.H.H

1 -126 1.0.0.0 to

126.0.0.0

126 224 2

16,777,214

B

N.N.H.H

128 191 128.0.0.0 to

191.255.0.0

16,384 216 2

65,534

CN.N.N.H

192 223 192.0.0.0 -

223.255.255.0

2,097,152 28 2

254

N =

networkoctet H=

host octet


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DefinitionsDefinitions

Network ID : all host bits are set to 0

192.168.32.0000 0000Ex : 192.168.32.0 /24

Broadcast ID : all host bits are set to 1

192.168.32.1111 1111

Ex : 192.168.32.255 /24

IP Address : any value in between

192.168.32.0000 00011111 1110

Ex : 192.168.32.1 - .254 /24


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Binary Number SystemBinary Number System

1 BYTE = 8 BITS : 1010 0001

Each bit position has an associated PLACE VALUE as indicated :

27 26 25 24 23 22 21 20

1 0 1 0 0 0 0 1

Add up the place values wherever you have a 1 bit27 + 25 + 20 = 16110

1100 00102 = 19410?

This is how you convert from binary to decimal


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A subnet mask gives meaning to a network address space

Class A default subnet mask : 255.0.0.0 ( /8 )

Class B default subnet mask : 255.255.0.0 ( /16 )

Class C default subnet mask : 255.255.255.0 ( /24 )


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A subnet mask is logically ANDed with a network address spacein order to return the network and subnetwork space

The logical AND Truth Table

AND

1

1 0

0

1 0

00

Click once to see the result of 1 AND 1


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A subnet mask is logically ANDed with a network address space

in order to return the network and subnetwork space

Ex1 : 192.168.32.145 /24 (no subnetting)

1100 0000. 1010 1000. 0010 0000. 1001 0001

255.255.255.0

Default mask :

1111 1111. 1111 1111. 1111 1111. 0000 0000

AND Result : 1100 0000. 1010 1000. 0010 0000. 0000 0000

192. 168. 32. 0

Notice how only the network address is returned (192.168.32.0)


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Subnetting BasicsSubnetting BasicsExample

Concept :

Math :

Let x represent the number of bits required.

Let y represent the number of subnets required.

Find out how manybits to borrow from the host octet to create

Equations : 2x >=y and 2x < y * 2, substitute 4 for y :

Notes :

The value for y is usually given in the problem.

Find a value for x that satisfiesBOTH

equations.

Class C 192.168.32.0 /24 :

subnet the address space to create 4 subnets.

the number ofsubnets you need.

Equations : 2x >=4 and 2x < 8


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Subnetting BasicsSubnetting BasicsExample

Math :

Let x represent the number of bits required for4 subnets.

Let y represent the number of subnets required.

Equations : 2x >=y and 2x < y * 2, substitute 4 for y :

The value of x that satisfies BOTHequations =

Class C 192.168.32.0 /24 :


Equations : 2x >=4 and 2x < 8

27 =

128 26 =

64 25 =

3

2 24 =

16 23 =

8 22 =

4 21 =

2 20 =

1

Powers of base 2 :

Solution :

?2

Note : x = log y / log 2


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Ex : 192.168.3

2.145 /26

(subnetted)1100 0000. 1010 1000. 0010 0000. 1001 0001

255.255.255.192

Custom mask :

1111 1111. 1111 1111. 1111 1111. 1100 0000

AND Result : 1100 0000. 1010 1000. 0010 0000. 1000 0000

192. 168. 32. 128

Notice the networkandsubnet address is returned (192.168.32.128)

What is the host ID ? 10 010001 (17)

Example

Class C 192.168.32.145 /24 :



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Subnetting BasicsSubnetting BasicsEquationsEquations

To find the number of subnets use : 2x = number of subnets

Where x = number of bits required

To find the number of hosts use : 2x 2 = number of hosts

Where x = number of bits required

You must subtract 2 to account for the 2 IPS that cannot

be used for hosts : host bits =all 0s and host bits =all 1s

(See slide 3)


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Subnetting is the process of subSubnetting is the process of sub--dividing a networkdividing a network

Network Space : 192.168.32.0 /24

192.168.32.0 /24

This blue box represents the whole network space

28 - 2 = 254 total hosts


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Subnetting BasicsSubnetting BasicsSubnetting is the process of subSubnetting is the process of sub--dividing a networkdividing a network

Network Space : 192.168.32.00 000000 /26

Custom created mask /26 = 255.255.255.11 000000

Class C default mask /24 = 255.255.255.0000 0000

First Subnet =00 000000 Second Subnet =01 000000

Third Subnet =10 000000 Fourth Subnet =11 000000

Subnet Bits and Host Bits


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Subnetting BasicsSubnetting BasicsSubnetting is the process of subSubnetting is the process of sub--dividing a networkdividing a network

Network Space : 192.168.3

2.00 000000 /26

.0

00

26 2 =

62hosts

.64

01

26 2 =

62hosts

.128

10

26 2 =

62hosts

.192

11

26 2 =

62hosts

This blue box represents the sub-divided network space


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The new subThe new sub--divided network :divided network :

.0

00

26 2 =

62hosts

.64

01

26 2 =

62hosts

.128

10

26 2 =

62hosts

.192

11

26 2 =

62hosts

Subnet 1 : 192.168.32.00 /26

Subnet 2 : 192.168.32.6464 /26

Subnet 3 : 192.168.32.128128 /26Subnet 4 : 192.168.32.192192 /26

Notice how the subnet addressessubnet addresses are multiples of6464


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The MULTIPLIER :The MULTIPLIER :

Subnet 1 : 192.168.32.00 /26 Subnet 2 : 192.168.32.6464 /26

Subnet 3 : 192.168.32.128128 /26 Subnet 4 : 192.168.32.192192 /26

Notice how the subnet addressessubnet addresses are multiples of6464

The number6464 is called the MultiplierMultiplier.

Custom created subnet mask : 255.255.255.192

Shortcut to find the Multiplier :256

192 =6464

Concept :192 in Binary =11 000000 ( /26 )

Find the Least Significant Subnet Bit : 11 000000

Take its Place Value : 26 =6464


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.0

00

26 2 =

62hosts

.64

01

26 2 =

62hosts

.128

10

26 2 =

62hosts

.192

11

26 2 =

62hosts

Subnet 1 Subnet Address :

Subnet 1 Minimum Host :

Subnet 1 Maximum Host :

Subnet 1 Broadcast :

192.168.32.0/26 > 00

192.168.32.1 /26 > 00

192.168.32.62 /26 > 00

192.168.32.63 /26 > 00

000000

000001

111110

111111


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.0

00

26 2 =

62hosts

.64

01

26 2 =

62hosts

.128

10

26 2 =

62hosts

.192

11

26 2 =

62hosts





192.168.32.64/26 > 01

192.168.32.65 /26 > 01

192.168.32.126 /26 > 01

192.168.32.127 /26 > 01

000000

000001

111110

111111


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.0

00

26 2 =

62hosts

.64

01

26 2 =

62hosts

.128

10

26 2 =

62hosts

.192

11

26 2 =

62hosts





192.168.32.128/26 > 10

192.168.32.129 /26 > 10

192.168.32.190 /26 > 10

192.168.32.191 /26 > 10

000000

000001

111110

111111


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.0

00

26 2 =

62hosts

.64

01

26 2 =

62hosts

.128

10

26 2 =

62hosts

.192

11

26 2 =

62hosts





192.168.32.192/26 > 11

192.168.32.193 /26 > 11

192.168.32.254 /26 > 11

192.168.32.255 /26 > 11

000000

000001

111110

111111


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Three Types ofThree Types of

Subnetting QuestionsSubnetting Questions

Those that deal with Those that deal with

The number of HOSTSThe number of HOSTS

The number of SUBNETSThe number of SUBNETS

The MULTIPLIERThe MULTIPLIER


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The Number of HOSTSThe Number of HOSTS

Given (Class C) : 192.168.32.66 /27

How many HOSTS can be supported per subnet?

/27 = 000

Subnet bits

00000

Host bits

2255 22= 3030

You must subtract 2 because of the 32 possible combinations,

the following combinations are not useable for host IDs :

all host bits =0, this is a network and/or subnetwork ID

all host bits =1, this is a broadcast ID.

This leaves 30 out of the 32 combinations for host IDs


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The Number of SubnetsThe Number of Subnets

Given (Class C) : 192.168.32.66 /27

How many Subnets can be supported?

/27 = 000

Subnet bits

00000

Host bits

2233 = 88

You can support up to 8 subnets with 3 subnet bits


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The MULTIPLIERThe MULTIPLIER

Given : 192.168.32.66 /27

What subnet contains this IP address ?

Step 1 : Find the multiplier : M =32 (see slide 15)

Step 2 : Divide the subnetted octet by M and drop the fractional portion

of the answer :

66 / 32 = 2.0625

Step 3 : Multiply M by the integer result from Step 2 :

32 * 2 =64

The subnet that contains this IP address is Subnet 64


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Class B Network Example

172.16.0.0/16

10101100.00010000.00000000.00000000

Required : 45 subnets each having at least 500 hosts

How manybits are required to represent 45 subnets ?

6bits are required : 26 = 64

10101100.00010000.00000000.00000000Network bits Subnet bits Host bits

Are 10 host bits enough to represent 500 hosts ?Yes

Why ? 210 2 = 1022 - more than enough for 500 hosts!


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172.16.0.0/22

10101100.00010000.00000000.00000000

10101100.00010000.00000100.00000001

10101100.00010000.00000111.11111110

Minimum host with subnet ID 4

Maximum host with subnet ID 4

172.16.4.1 /22

172.16.7.254 /22


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172.16.0.0/22

10101100.00010000.00000000.00000000

10101100.00010000.00000100.00000001

10101100.00010000.00000111.11111110

Minimum host with subnet ID 4

Maximum host with subnet ID 4

172.16.44.1 /22

172.16.77.254 /22

Notice the change in the 33rdrd octetoctet even though we are still in

the same subnet


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Internet ResourcesInternet Resources

www.semsim.com/ccna/tutorial/subnetting/subnetting.html

www.learntosubnet.com

http://www.cisco.com/en/US/tech/tk365/technologies_tech_note09186a00800a67f5.shtml

Documents

Basic Sub Netting