btit cn unit - 1 - ip addressing

8/18/2019 BTIT CN Unit - 1 - IP Addressing

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Subject Name : Computer Networks Subject Code : BTIT 603

Faculty:

Dr. Siddhartha Sankar BiswasAssistant ProfessorDepartment of Computer Science & EngineeringJamia Hamdard University, New Delhi

Branch : B.Tech. (Information Technology) Semester : 6th

IP Addressing (IPv4 Addressing)

T o p i c s

1

•

Introduction to IP Addressing

• Classful Addressing Concepts• Case Study of Classful Addressing

• CIDR Addressing Concepts• Case Study of CIDR Addressing

• Subnet addressing

• Network Address Translation (NAT)

18 March 2016

An Internet Protocol address (IP address) is a numericallabelassigned to each device (e.g., computer, printer)

participating in a computer network

that uses the Internet Protocol for communication.

An IP address serves two principal functions:

(i) Host or network interface identification and(ii) Location addressing.

The designers of the Internet Protocol defined an IPaddress as a 32-bit number

and this system is known as Internet Protocol Version 4 (IPv4).

IP Address

18 March 2016 2Dr. Siddhartha Sankar B iswas


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Due to the enormous growth of the Internet and thepredicted depletion of available addresses,

a new version of IP known as IPv6, using 128 bits for the

address, was developed in 1995

and its deployment has been ongoing since the mid-2000s.

IP addresses are binary numbers,but they are usually stored in text files

and displayed in human-readable notations,

Example: 172.16.254.1 (for IPv4),

and 2001:db8:0:1234:0:567:8:1 (for IPv6).18 March 2016 3Dr. Siddhartha Sankar B iswas

IP AddressesIPv4 Addresses

Classfull Addressing

IPv6 Addresses

Classless Inter-Domain Routing (CIDR)informally known as Classless Addressing

Class-A Addressing

Class-B Addressing

Class-C Addressing

Class-D AddressingClass-E Addressing



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IANA

RIR RIR RIRRIR RIR

ISP

USER

ISP ISP ISP ISP

USER USER USER USER

Only 5 RIRs around the world

N

n u m b e r s

N

n u m b e r s

N

n u m b e r s

N

n u m b e r s

N

n u m b e r s

M numbers M numbers M numbers M numbers M numbers

Internet Assigned Numbers Authority (IANA)manages the IP address space allocations globallyand delegates five Regional Internet Registries (RIRs)which further allocate IP address blocks to Local Internet Registries (LIR)

also known as Internet Service Providers (ISP) and other entities.




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Classfull AddressingA classful addressing architecture was used in the Internetsince 1981 until the introduction of Classless Inter-DomainRouting (CIDR) in 1993.Classful Addressing method divides the address space forIPv4 into five address classes.

Each class defines a different network size, with eachnetwork having different number of hosts.The Classes are classes A, B, C used for uncasting andmulticast network class D.The fifth class (E) address range is reserved for future orexperimental purposes.

Even after its discontinuation, Classfull Addressing termsare often still used erroneously by people working in IT.

Classful network concepts remain in practice only in limitedscope in the default configuration parameters of somenetwork software and hardware components.18 March 2016 7Dr. Siddhartha Sankar B iswas

Notations

01110010111000111100101010101010

114

Binary Notations

Dotted-Decimal Notations

227 202 170

Rough Note:Remember its Dotted Decimal and not simple Decimal.In simple decimal this binary number would be = 1927531178

For representing the binary into dotted decimal,we divide the 32 Bits into 4 octets of 8 Bits each.



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8 Bits 8 Bits 8 Bits 8 Bits

32 Bits

1st Octet 2nd Octet 3rd Octet 4th Octet

So, value of each octet can be between 0 – 255By 8 Bits we can represent upto 28 = 256 numbers

0to

255

0to

255

0to

255

0to

255

1 1 1 1 1 1 1 1

= 0

= 255

0 0 0 0 0 0 0 0


Class A Addressing

00

Leading

Bit

24 Bits7 Bits

Network ID(net ID) Host ID


0 0 0 0 0 0 0 0

0 1 1 1 1 1 1 1

= 0

= 127

to

0to127

0 - 255 0 - 2550 - 255

Range of Class A IP Addresses

0.0.0.0to 127.255.255.255

Max.No. of possible Networks= 27

= 128

Max. No. of possible Hosts per Network= 224

= 16777216

Range of 1st Octet



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Class B Addressing

1

LeadingBits

16 Bits14 Bits



1 0 0 0 0 0 0 0

1 0 1 1 1 1 1 1

= 128

= 191

to

128 - 191 0 - 255 0 - 2550 - 255

Range of Class B IP Addresses128.0.0.0

to 191.255.255.255

Max. No. of possible Networks= 214

= 16384

Max No. of possible Hosts per Network= 216

= 65536

Range of 1st Octet

1 01 0


Class C AddressingLeadingBits

8 Bits21 Bits



1 1 0 0 0 0 0 0

1 1 0 1 1 1 1 1

= 192

= 223

to

192 - 223 0 - 255 0 - 2550 - 255

Range of Class C IP Addresses192.0.0.0

to 223.255.255.255


= 2097152

Max No. of possible Hosts per Network= 28

= 256

Range of 1st Octet

11 11 1 01 1 0



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Class D Addressing(used for Multicasting)

LeadingBits

28 Bits

Multicast


1 1 1 0 0 0 0 0

1 1 1 0 1 1 1 1

= 224

= 239

to

224 - 239 0 - 255 0 - 2550 - 255Range of Class D IP Addresses

224.0.0.0to

239.255.255.255


= 1Max No. of possible Hosts

(for multicasting) per Network= 228

= 268435456

Range of 1st Octet

11 11 1 11 1 1 01 1 1 0


Class E Addressing

(Reserved for Future Use)Leading

Bits

28 Bits

(To be defined in future)


1 1 1 1 0 0 0 0

1 1 1 1 1 1 1 1

= 240

= 255

to

240 - 255 0 - 255 0 - 2550 - 255

Range of Class E IP Addresses240.0.0.0

to 255.255.255.255

Max. No. of possible Networksand

Max No. of possible Hosts per Network

Range of 1st Octet

11 11 1 11 1 1 11 1 1 1

Not yet defined18 March 2016 14Dr. Siddhartha Sankar B iswas


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Class LeadingBits

N/WIDSize(Bits)

HostIDSize(Bits)

Max. No.of

N/W

Max. No.of Hostsper N/W

IPRange

27 =128

214 =16384

221 =2097152

0224 =

16777216

216 =65536

28 =256

A

B

C

D

E

7 24

10

0.0.0.0

to 127.255.255.255

14 16128.0.0.0

to 191.255.255.255

110 21 8192.0.0.0

to 223.255.255.255

1110 0 28

20 = 1 228 =

268435456224.0.0.0

to 239.255.255.255

1111224.0.0.0

to 239.255.255.255

FutureUse

Reserved for Future Use.Parameters Still Not Defined.

Data is Multicastto all Hosts


A

50%

B

25%

C

13%

D6%

E

6% Total IPv4 Addresses

(232 = 4294967296)

Class A = 128 x 16777216 = 2147483648

Class B = 16384 x 65536 = 1073741824

Class C = 2097152 x 256 = 536870912

Class D = 1 x 268435456 = 268435456

Class E = 268435456

Max. No.of hostsper N/W

Max. No.of IP

addressespossible

% ofaddressesin eachclass

Max. No.of N/W

Let us find the number ofIP addresses available ineach class

18 March 2016 16Dr. Siddhartha Sankar Biswas


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Case 1: If suppose there needs to be 110 sub-networks and each of themwould require to 50,000 to 60,000 hosts?

Ans: ClassA or ClassB

What do you think , which Class of network would a network engineerimplement in a organization


Ans: Class ACase 3: If suppose there needs to be 200 sub-networks and each of themwould require to 50,000 to 60,000 hosts?

Ans: Class B


?

Lets do some case study


Classless Inter-Domain Routing (CIDR) Addressing(Classless Addressing)

The flaws in Classful Addressing scheme combined with thefact of fast growth of Internet lead to a situation whereIP addresses were nearly depleted.

Though the number of devices on Internet in mid 80’s orearly 90’s were certainly less than 232

but with due to technical short coming in Classfuladdressing mechanism, there came a situation where

Internet Assigned Numbers Authority (IANA) was facingproblem in allocating IP address,specially Class A and Class B addresses.

Therefore Internet Engineering Task Force (IETF)developed a new IPv4 addressing scheme known asClassless Addressing, technically known asClassless Inter-Domain Routing(CIDR) addressing scheme.


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Problem Statement

What do you think , which Class of network would a network engineerimplement in a organization

If suppose there needs to be 500 sub-networks and each of themwould require to 50,000 to 70,000 hosts?

Solution : Not possible

Rough Note: Though by some indirect methods this can be implemented in someunconventional fashion,

but no straight forward, simple and conventional solution is possible.


0

LeadingBit

24 Bits7 Bits


LeadingBits

16 Bits14 Bits

Network ID(net ID)

Host ID

Class A

Class B

Reason for the Problem

1 0



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23 Bits9 Bits


Solution to the Problem

Max. No. of possible Networks

= 29= 512

Max. No. of possible Hosts per Network= 223

= 8388608

Let us remove the class system where Net ID and Host ID have fixedno. of bits assigned.

Lets make the no. of bits required for defining Net ID and Host IDmore flexible and as per requirement.

And since we are eliminating class system, we would be able to freethe leading bits which were used as identification purpose.


23 Bits9 Bits


Solution to the Problem


= 512Max. No. of possible Hosts per Network

= 223

= 8388608

Let us remove the class system where Net ID and Host ID have fixedno. of bits assigned.

Lets make the no. of bits required for defining Net ID and Host IDmore flexible and as per requirement.

And since we are eliminating class system, we would be able to freethe leading bits which were used as identification purpose.



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23 Bits9 Bits

Network ID(net ID)

Host ID

0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0

In Binary Notation : 0.0.0.0

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1

IP Address RangeStarting IP

Last IP

In Binary Notation : 255.255.255.255


23 Bits9 Bits

Network ID

(net ID)

Host ID

0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0

In Binary Notation : 0.0.0.0 / 9

1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 1 1 1 1 1 1 1

Network Address RangeStarting Network Address

Last Network Address

In Binary Notation : 255.128.0.0 / 918 March 2016 24Dr. Siddhartha Sankar B iswas


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Address Blocks

In classless addressing, when an entity, small or

large, needs to be connected to the Internet,it is always granted a block (range) of addresses.

The size of the block (the number of addresses)varies based on the nature and size of the entity.

For example,a household may be given only one or twoaddresseswhereas a large organisation may be giventhousand of addresses.


RestrictionsTo simplify the handling of addresses,the Internet authorities impose three restrictions onclassless address blocks:

1.The addresses in a block must be contagiousi.e. one after another.

2.The number of addresses in a block must be a

power of 2 (1,2,4,8,.,.,.,).

3.The first address must be evenly divisible bythe number of addresses.



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Example: A block of address granted to a business house which requires 14 IP addresses

205.16.37.32

205.16.37.47

… …

First

Last

We can see that the restrictions are applied to this block1.We can see that the addresses provided are contiguous.2.The number of addresses provided are 16 , which is a power of 2.

(24 = 16).3.The first address when converted to decimal number is

3440387360, which is evenly divisible by total no. of addresses =16. Step 1: Dotted Decimal to Binary Value

205.16.37.32 = 11001101 00010000 00100101 00100000

Step 2: Binary Value to Decimal Value

11001101 00010000 00100101 00100000 = 3440387360

Step 3: Divide the Decimal Value by number of addresses

3440387360 / 16 = 2150124210 (evenly divisible)18 March 2016 27Dr. Siddhartha Sankar B iswas

Mask

As we know seen that in classless IPv4 addressing,addresses are given in contiguous block.But writing the whole block can be time consuming thing,so there needs a better way of defining the blocks insteadof writing all the numbers.

In IPv4 addressing, a block of addresses can be defined as

x.y.z.t / n , where

In which x.y.z.t defines one of the addresses from

and /n defines the mask (this is known as CIDR notation)

Which means mask of the block is an IP addresses with n leftmost bits are 1’s and rest of the (32-n) bits are 0’s.

therefore a better way of defining a block of IPv4classless IP addresses is developed known as masking.



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First IP Address of the blockThe first IP address of the block can be found by doingAND operation between the given address and the mask.

(This is by default subnet address of the given Network)

Last IP Address of the blockThe last IP address of the block can be found by doing ORoperation between the given address and the compliment ofthe mask.

Number of Address in the block

Number of addresses in the block can be found byconverting the compliment of the mask from its binaryvalue to decimal valueand then adding it with 1 (decimal).


Example

A block of IP address is granted to an organization.

We know that one of the address is 205.16.37.39/28.

Find the following

(a)The first address of the block.

(b)The last address assigned.

(c)Total IPs that the organization was assigned.



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Solution

Therefore Mask address is:

11111111 11111111 11111111 11110000

Mask is represented here by /28.

One of the IP address of the block = 205.16.37.39

In binary the IP address will be11001101 00010000 00100101 00100111

Mask Compliment is:

00000000 00000000 00000000 0000111118 March 2016 31Dr. Siddhartha Sankar B iswas

(a) First IP Address of the block

The first IP address of the block can be found by doingAND operation between the given address and the mask.

First IP Addressof the block

11001101 00010000 00100101 00100000

i.e. 205.16.37.32

IP Address 11001101 00010000 00100101 00100111

Mask 11111111 11111111 11111111 11110000

AND

(This is by default subnet address of the given Network)18 March 2016 32Dr. Siddhartha Sankar B iswas


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(b) Last IP Address of the block

The last IP address of the block can be found by doing ORoperation between the given address and the compliment of

the mask.

MaskCompliment

00000000 00000000 00000000 00001111

Last IP Addressof the block

11001101 00010000 00100101 00101111

i.e. 205.16.37.47

IP Address 11001101 00010000 00100101 00100111OR


(c) Number of Address in the block


MaskCompliment

00000000 00000000 00000000 00001111

Decimal value of the Mask Compliment = 15

Therefore,

Number of addresses in the block = 15 +1= 16



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Subnet Addressing

To understand subnet addressing, we must understand thefollowing concepts:

Hierarchy Network Addresses


28 BitsNetwork prefix

4 BitsHost Address

Two-Level Hierarchy

(No Subnetting possible)



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Example

An organization is assigned the block 205.16.37.32/28

Design the network configuration for the organization.


Solution

In order to design the network configuration for theOrganization which was granted 205.16.37.32/28

One of the IP = 205.16.37.32Which in binary is: 11001101 00010000 00100101 00100000

And Mask = 28Which in binary is: 11111111 11111111 11111111 11110000

We need to Find the following

(a)The first address of the block.

(b)The last address assigned.

(c) Size of the block(Total IPs that the organization was assigned).



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(a) First IP Address of the block



11001101 00010000 00100101 00100000

i.e. 205.16.37.32

IP Address 11001101 00010000 00100101 00100000

Mask 11111111 11111111 11111111 11110000

AND



(b) Last IP Address of the block

The last IP address of the block can be found by doing ORoperation between the given address and the compliment ofthe mask.

MaskCompliment

00000000 00000000 00000000 00001111


11001101 00010000 00100101 00101111

i.e. 205.16.37.47

IP Address 11001101 00010000 00100101 00100000OR



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(c) Number of Address in the block

Number of addresses in the block can be found byconverting the compliment of the mask from its binary

value to decimal valueand then adding it with 1 (decimal).

MaskCompliment

00000000 00000000 00000000 00001111


Therefore,



Rest of theInternet

205.16.37.33/28 205.16.37.34/28 205.16.37.46/28 205.16.37.47/28

Routers Internal IP 205.16.37.40/28

Routers External IP x.y.z.t /n

OrganizationNetwork

Range:205.16.37.32

to205.16.37.47

205.16.37.32/28

A network configuration for the block 205.16.37.32/28



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Host Address

Host ID

6 Bits

Three-Level Hierarchy(Subnetting)

Given below is a Two-Level Hierarchy

Let us divide it into Three-Level Hierarchy



Host Address

Sub-Networkprefix

1Bit

5Bits

1 Bit has been borrowed from host ID

andadded to network prefix to create 27 bit length subnetwork



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Host Address

Sub-Networkprefix

2Bits

4Bits

2 Bits has been borrowed from host IDand

added to network prefix to create 28 bit length subnetwork


Example

An organization is assigned the block 17.12.14.10/26.

The organization has 3 offices.

The network engineer wants to create 3 independentsub-blocks (i.e sub nets)

of size 32,16 and 16 addresses, for the offices.

Design the network configuration for the organization.



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Solution

Assigned block is : 17.12.14.10/26

Therefore one of the IP of this block is : 17.12.14.10(i.e. Host IP)

And Mask is : 26

Host IPAddress 00010001 00001100 00001110 00001010

Subnetmask 11111111 11111111 11111111 11000000

SubnetMask

Compliment00000000 00000000 00000000 00111111


First IP Address of the block



00010001 00001100 00001110 00000000

i.e. 17.12.14.0

IP Address 00010001 00001100 00001110 00001010

Mask 11111111 11111111 11111111 11000000

AND




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Last IP Address of the block

The last IP address of the block can be found by doing ORoperation between the given address and the compliment of

the mask.

MaskCompliment

00000000 00000000 00000000 00111111


00010001 00001100 00001110 00111111

i.e. 17.12.14.63

IP Address 00010001 00001100 00001110 00001010OR


Number of Address in the block


MaskCompliment

00000000 00000000 00000000 00111111


Therefore,




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The range of the block is :

17.12.14.0

to 17.12.14.63

Since the organizations needs 3 sub-networks,so we need to assign 3 subnet address,

one for each of the sub-networks.

But for defining the 3 subnet address

we need to find the masks for each of the subnets.

Total = 64 Addresses

This block is needed to be divided among the3 sub-networks having 32,16 and 16 addresses each.

So, before proceeding further,lets find the subnet masks for each of subnets.


Subnet Mask for Subnet-1



Subnet-1 needs 32 addresses.

Therefore Host ID must have = 5 bits.

Therefore Subnet mask = 32 - 5 = 27

Subnet-2 needs 16 addresses.


Therefore Subnet mask = 32 - 4 = 28

Subnet-3 has 16 addresses.


Therefore Subnet mask = 32 - 4 = 2818 March 2016 52Dr. Siddhartha Sankar B iswas


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17.12.14.0 / 27

to

17.12.14.31 / 27

Mask = 27

Sub-Network 1

Lets assign the following address sub-block to subnet-1

Lets take any one of the IP from the block = 17.12.14.30

Let us find the subnet address of the sub-network 1

Subnet address ofsubnet 1

00010001 00001100 00001110 00000000

i.e. 17.12.14.0

IP Address 00010001 00001100 00001110 00011110

Mask 11111111 11111111 1111111 11100000

AND



17.12.14.32 / 28

to

17.12.14.47 / 28

Mask = 28

Sub-Network 2





00010001 00001100 00001110 00100000

i.e. 17.12.14.32

IP Address 00010001 00001100 00001110 00100111

Mask 11111111 11111111 1111111 11110000AND




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17.12.14.48 / 28

to

17.12.14.63 / 28

Mask = 28

Sub-Network 3





00010001 00001100 00001110 00110000

i.e. 17.12.14.48

IP Address 00010001 00001100 00001110 00110111

Mask 11111111 11111111 1111111 11110000

AND



Internet

17.12.14.1/27

17.12.14.30/27

17.12.14.31/27

17.12.14.2/27

17.12.14.0/27Subnet addressOf Subnet -1


17.12.14.33/28

17.12.14.47/28

17.12.14.49/28

17.12.14.63/28


17.12.14.50/28

Network : 17.12.14.10/27

x.y.z.t/n

17.12.14.34/28

Subnet - 1

Subnet - 2

Subnet - 3



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Internet

172.18.3.1 172.18.3.2 172.18.3.20

172.18.3.30

200.24.5.8NAT Router

25.8.2.10

25.8.3.101

25.40.3.79

25.40.76.56

25.8.49.110

And many more …..

Sender

Responder

Organization’sNetwork

Network Address Translation

(NAT)


Source: 172.18.3.1Destination: 25.8.2.10

Source: 200.24.5.8Destination: 25.8.2.10

Translation Table(maintained atNAT Router)

Private External

Destination: 172.18.3.1Source: 25.8.2.10

Destination: 200.24.5.8Source: 25.8.2.10

25.8.2.10172.18.3.1

Sender to Responder : RequestStep 1.1: Datagram is sent to router from the sender.Step 1.2:Router receives the datagram and UPDATES

the Translation Table.Step 1.3: Router changes source address field

by deleting the Private address in theDatagram and adding its own External Address. Andthen forwards the datagram to the internet.

Step 1.1

Step 1.2

UpdationStep 1.2

Step 1.3

Step 2.1Step 2.2

CheckingStep 2.2

Step 2.3

Responder to Sender : ReplyStep 2.1: Datagram is received by router from the

internet.Step 2.2 : Router CHECKS the Translation Table.Step 2.3 : Router changes the destination address

field from its external address to theconcerned Private address.


btit cn unit - 1 - ip addressing

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