(Subnet) Addresses - JNNCE ECE Manjunath

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Transcript (Subnet) Addresses - JNNCE ECE Manjunath

Unit –6

Network Layer: Logical Addressing

04\08\2010 Unit-VI Network Layer 1

• •

Overview Ipv4 addresses Ipv6 addresses

04\08\2010 Unit-VI Network Layer 2

ADDRESSING

Four levels of addresses are used in an internet employing the TCP/IP protocols: physical, logical, port, and specific .

04\08\2010 Unit-VI Network Layer 3

Physical Addressing

• A network adapter has a unique and permanent physical address. • A Physical address

is also called MAC address

is a 48 bit flat address burned into the ROM of the NIC (Network Interface Card) card at the factory which is a Layer1 device of the OSI model. • On a local area network , low-lying hardware-conscious protocols deliver data across the physical network using the adapter's physical address . • On a basic ethernet network, for example, a computer sends messages directly onto the transmission medium. • The network adapter of each computer listens to every transmission on the local network to determine whether a message is addressed to its own physical address.

04\08\2010 Unit-VI Network Layer 4

Physical Addressing

04\08\2010 Unit-VI Network Layer 5

Logical Addressing

• • • A Logical address also called IP address is a 32 bit address assigned to each system in a network. This works in Layer-3 of OSI Model . This would be generally the IP address .

04\08\2010 Unit-VI Network Layer 6

Logical Addressing

04\08\2010 Unit-VI Network Layer 7

Logical Addressing

04\08\2010 Unit-VI Network Layer 8

Logical Addressing

04\08\2010 Unit-VI Network Layer 9

Logical Addressing

04\08\2010 Unit-VI Network Layer 10

IP Addresses

04\08\2010 Unit-VI Network Layer 11

04\08\2010 The physical addresses will change from hop to hop, but the logical addresses usually remain the same.

Unit-VI Network Layer 12

Port Address

A single wire connects the network to the distant computer, but there may be many applications on that machine-a web server, an ftp server, a telnet server, etc.-waiting for somebody to connect. So the question arises: How do you use one wire and one IP address to connect to the right application? The answer: Ports.

Port address is transport layer ID (similar to IP in Network Layer) which identify the application on the host.

A port address is a 16-bit address represented by one decimal number as shown.

Telnet Mail (smtp, or send mail) World Wide Web Port 80 Post Office (pop, or get mail) Port 110 News (nntp) Port 23 Port 25 Port 119 04\08\2010 Unit-VI Network Layer 13

04\08\2010

IPv4 ADDRESSES

Unit-VI Network Layer 14

IPv4 ADDRESSES

An IPv4 address is a 32-bit address that uniquely and universally defines the connection of a device (for example, a computer or a router) to the Internet.

• • • • Address Space Notations Classful Addressing Classless Addressing Network Address Translation (NAT) 04\08\2010 Unit-VI Network Layer 15

• • • IPv4 ADDRESSES IPv4 protocol address has an address space An address is the total number of addresses used by the protocol. If a protocol uses N bits to define an address the address space is 2 N value. • • • • Notations Binary Notation and Dotted Decimal Notation Binary Notation: 32 bits are used each octet is referred as byte, 4 byte address Dotted Decimal Notation: Written in Decimal point and each byte is separated by dots.

04\08\2010 Unit-VI Network Layer 16

IPv4 ADDRESSES An IPv4 address is 32 bits long.

The IPv4 addresses are unique and universal.

• • • An IP address is a 32-bit sequence of 1s and 0s.

To make the IP address easier to use, the address is usually written as four decimal numbers separated by periods. This way of writing the address is called the dotted decimal format.

The address space of IPv4 is 2 32 or 4,294,967,296.

04\08\2010 Unit-VI Network Layer 17

04\08\2010

Classful Addressing

Unit-VI Network Layer 18

Internet Addresses (IP Addresses) Defined when IP was standardized in 1981 IP addresses are 32-bit long and consist of: • • a network address part – network identifier a host address part – host number within that network IP addresses are grouped into classes (A,B,C) depending on the size of the network identifier and the host part of the address A fourth class (Class D) was defined later (1988) for Multicast addresses 04\08\2010 Unit-VI Network Layer 19

Internet Address Classes Class A • 126 networks (0 and 127 reserved) (1 byte starts from but MSB bit is always 0) • Assigned to very large size networks where number of hosts 65K to16M Class B • • 16384 networks Assigned to Intermediate size networks where number of hosts 256 to 65K Class C • • 2097152 networks Assigned to smaller networks where #hosts < 256 04\08\2010 Unit-VI Network Layer 20

Finding the classes in binary and dotted-decimal notation

Number of blocks and block size in classful IPv4 addressing 04\08\2010 Unit-VI Network Layer 21

Every IP address has two parts: 1.

Network 2.

Host IP addresses are divided into classes A,B and C to define large, medium, and small networks. The Class D address was created to enable multicasting. IETF reserves Class E addresses for its own research.

04\08\2010 Unit-VI Network Layer 22

Reserved IP ADDRESSES Certain host addresses are reserved and cannot be assigned to devices on a network. An IP address that has binary 0s in all host bit positions is reserved for the

network address

. An IP address that has binary 1s in all host bit positions is reserved for the

broadcast address

. 04\08\2010 Unit-VI Network Layer 23

Example

Change the following IPv4 addresses from binary notation to dotted-decimal notation.

Solution

04\08\2010 Unit-VI Network Layer 24

Example

Change the following IPv4 addresses from dotted-decimal notation to binary notation.

Solution

04\08\2010 Unit-VI Network Layer 25

Example

Find the error, if any, in the following IPv4 addresses.

Solution a. There must be no leading zero (045).

b. There can be no more than four numbers.

c. Each number needs to be less than or equal to 255.

d. A mixture of binary notation and dotted-decimal notation is not allowed.

04\08\2010 Unit-VI Network Layer 26

Example

Find the class of each address.

0 0000001 00001011 00001011 11101111 110 00001 10000011 00011011 11111111

14 .23.120.8

252 .5.15.111

Solution a. The first bit is 0. This is a class A address.

b. The first 2 bits are 1; the third bit is 0. This is a class C address.

c. The first byte is 14; the class is A.

d. The first byte is 252; the class is E.

04\08\2010 Unit-VI Network Layer 27

Netid and Hostid

In classful addressing an IP address in class A,B, C is divided into netid and hostid In class A one byte defines the netid and 3 bytes defines the host ID In class B 2 byte defines the netid and 2 bytes defines the host ID In class C 3 byte defines the netid and 1 bytes defines the host ID Unit-VI Network Layer 28 04\08\2010

Mask

The mask helps to find the netid and hostid In class A first 8 bits defines the netid; the next 24 bits hostid , hence in this first 8 are 1s . /n i.e 8 or 16 or 24 shows the mask for each class.

This /n notation is called Classless Interdomain Routing (CIDR)

Default masks for classful addressing

Unit-VI Network Layer 29 04\08\2010

04\08\2010

Subnets

Unit-VI Network Layer 30

Problems with Classes Class A usually too big Class C often too small Not enough Class Bs Inefficient utilisation of address space Solution: Extending the network part of the address:

Subnetting

In classful addressing, a large part of the available addresses were wasted.

04\08\2010 Unit-VI Network Layer 31

Subnetting

Subnets

A campus network consisting of LANs for various departments 04\08\2010 Unit-VI Network Layer 32

Subnetting

Subnet Mask

Subnet masks are applied to an IP address to identify the

Network

portion and the

Host

portion of the address.

bitwise logical AND operation

between the address and the subnet mask s performed in order to find the

Network Address

or number.

04\08\2010 Default Subnet Masks Class A - 255.0.0.0

• 11111111.

00000000.00000000.00000000

Class B - 255.255.0.0

• 11111111.11111111.

00000000.00000000

Class C - 255.255.255.0

• 11111111.11111111.11111111.

00000000 Unit-VI Network Layer 33

Subnetting

Logical Bitwise AND Operation

Example •

140.179.240.200

It’s a Class B, so the subnet mask is:

•

255.255.0.0

In Binary: 10001100.10110011.11110000.11001000

11111111.11111111.00000000.00000000

10001100.10110011.00000000.00000000

By doing this, the computer has found that Network Address is

140.179.0.0

04\08\2010 Unit-VI Network Layer 34

Subnetting

Another Example:

Suppose we have the address of: 206.15.143.89?

What class is it?

Class C

What is the subnet mask?

255.255.255.0

What is the Network Address ?

206.15.143.

0

What is the host portion of the address?

0.0.0 .89

Unit-VI Network Layer 35 04\08\2010

Subnetting

You can manipulate your subnet mask in order to create more network addresses. 04\08\2010 If you have a Class C network, how many individual host addresses can you have?

• • 1 to 254 Remember, you can’t have all “0”s and all “1”s in the host portion of the address ( Reserved address ). • So we cannot use 206.25.143.0 (all “0”s) or 206.25.143.255 (all “1”s) as a host address.\ • Remember, an address of all “0”s or all “1”s cannot be used in the last octet (or host portion). All “0”s signify the Network Address and all “1”s signify the broadcast address Unit-VI Network Layer 36

Subnetting

Example We have 1 Class C Network (206.15.143.0) And we have 254 host address (1 to 254) But what if our LAN has 5 networks has no more than 30 hosts on it?

in it and each network Do we apply for 4 more Class C licenses, so we have one for each network?

We would be wasting 224 addresses on each network, a total of 1120 addresses 04\08\2010 Unit-VI Network Layer 37

Subnetting

Subnetting is a way of taking an existing class license and breaking it down to create more Network Addresses.

This will always reduce the number of host addresses for a given network.

Subnetting makes more efficient use of the address.

04\08\2010 Unit-VI Network Layer 38

Subnetting

How Does Subnetting Work?

Additional bits can be added (changed from 0 to 1) to the subnet mask to further subnet, or breakdown, a network.

When the

logical AND

is done by the computer, the result will give it a new Network (or Subnet) Address.

04\08\2010 Unit-VI Network Layer 39

Subnetting

We ask our ISP for a Class C license.

They give us the Class C bank of 206.15.143.0

This gives us 1 Network (206.15.143.0) with the potential for 254 host addresses (206.15.143.

1 to 206.15.143.

254 ).

But we have a LAN made up of 5 Networks with the largest one serving 25 hosts .

So we need to

Subnet

our 1 IP address...

04\08\2010 Unit-VI Network Layer 40

Subnetting

So How Does This Work?

To calculate the number of subnets (networks ) and/or hosts , we need to do some math: Use the formula 2 n -2 where the n can represent either how many subnets ( networks ) needed OR how many hosts per subnet needed (where -2 is 000000000 and 11111111 addresses are not used).

04\08\2010 Unit-VI Network Layer 41

Subnetting

So How Does This Work?

We know we need at least 5 subnets . So 2 3 -2 will give us 6 subnet addresses (Network Addresses).

We know we need at least 25 hosts 2 5 -2 will give us 30 hosts per network. per subnet (network).

This will work, because we can steal the first 3 bits from the host’s portion of the address to give to the network portion and still have 5 (8-3) left for the host portion : 04\08\2010 Unit-VI Network Layer 42

Subnetting

Break it down:

Let’s go back to what portion is what: We have a Class C address:

NNNNNNNN.NNNNNNNN.NNNNNNNN.

HHHHHHHH

With a Subnet mask of: 11111111.11111111.11111111.

00000000 We need to steal 3 bits from the host it to the Network portion: portion to give

NNNNNNNN.NNNNNNNN.NNNNNNNN.NNN

HHHHH

04\08\2010 Unit-VI Network Layer 43

Subnetting

Break it down: NNNNNNNN.NNNNNNNN.NNNNNNNN.NNN

HHHHH

This will change our subnet mask to the following: 11111111.11111111.11111111.111

00000 Above is how the computer will see our new subnet mask, but we need to express it in decimal form as well: 255.255.255.224

128+64+32=224 04\08\2010 Unit-VI Network Layer 44

Subnetting

What address is what?

Which of our 254 addresses will be a Subnet (or Network) address and which will be our host addresses?

Because we are using the first 3 bits for our subnet mask, we can configure them into eight different ways (binary form): 04\08\2010 Unit-VI Network Layer 45

Subnetting

What address is what?

Which of our 254 addresses will be a Subnet (or Network) address addresses?

and which will be our host 04\08\2010 Because we are using the first 3 bits for our subnet mask, we can configure them into eight different ways (binary form): 000 001 010 100 110 011 101 111 Unit-VI Network Layer 46

Subnetting

What address is what?

We cannot use all “0”s or all “1”s 000 001 010 100 011 101 110 111 •

We are left with 6 useable network numbers.

Unit-VI Network Layer 04\08\2010 47

Subnetting

Network (Subnet) Addresses

Remember our values: 128 64 32 16 Now our 3 bit configurations: 8 0 0 0 1 1 1 0 1 1 0 0 1 1 0 1 0 1 0 H H H H H H H H H H H H 4 2 1 Equals H H H H H H H H H H H H H H H H H H 32 64 96 128 160 192 04\08\2010 Unit-VI Network Layer 48

Subnetting

Network (Subnet) Addresses

0 0 0 1 1 1 0 1 1 0 0 1 1 0 1 0 1 0 h h h h h h h h h h h h h h h h h h h h h h h h h h h h h h 32 64 96 128 160 192 04\08\2010

Each of these numbers becomes the Network Address of their subnet...

Unit-VI Network Layer 49

Subnetting

Network (Subnet) Addresses

206.15.143.32

206.15.143.64

206.15.143.96

206.15.143.128

206.15.143.160

206.15.143.192

04\08\2010 Unit-VI Network Layer 50

Subnetting

host Addresses

The device assigned the first address will receive the first number AFTER the network address shown before.

0 0 206.15.143.33 or 32+1 1 0 0 0 0 1 And the last address in the Network will look like this: 0 0 1 206.15.143.62

1 1 1 1 0 04\08\2010 *Remember, we cannot use all “1”s, that is the broadcast address (206.15.143.63) Unit-VI Network Layer 51

Subnetting

Host Addresses

The next network will start at 206.15.143.64

The first IP address on this subnet network will receive: 206.15.143.65

0 1 0 0 0 0 0 1 And the last address in the Network will receive: 206.15.143.94

0 1 0 1 1 1 1 0 04\08\2010 * Remember, the broadcast address (206.15.143.95) Unit-VI Network Layer 52

Subnetting

Can you figure out the rest?

Network: Host Range

206.15.143.32

206.15.143.64

206.15.143.96

206.15.143.128

206.15.143.160

206.15.143.192

206.15.143.

32 to 206.15.143.

62 206.15.143.

65 to 206.15.143.

94 206.15.143.

97 to 206.15.143.

126 206.15.143.

129 to 206.15.143.

158 206.15.143.

161 to 206.15.143.

190 206.15.143.

193 to 206.15.143.

222

04\08\2010 Unit-VI Network Layer 53

Subnetting

How the computer finds the Network Address:

200.15.143.89 An address on the subnet 225.225.225.224 The new subnet mask When the computer does the

Logical Bitwise AND Operation

it will come up with the following Network Address ( or Subnet Address ): 11001000.00001111.10001111.01011001= 200.15.143.89

11111111.11111111.11111111.111

00000 = 255.255.255.224

11001000.00001111.10001111.01000000 = 200.15.143.64

04\08\2010 This address falls on our 2nd Subnet (Network) Unit-VI Network Layer 54

04\08\2010

Classless Addressing

Unit-VI Network Layer 55

Classless Addressing

Classfull Addressing: drawbacks

Classful Addressing + Subnetting • at least one route per class is advertised in routing updates

Number of networks is doubling faster than once per year Memory

is not growing that fast Only a few

routers

can keep the current number of routes Unit-VI Network Layer 56 04\08\2010

Classless Addressing

Overview: (Classful) IPv4 Addressing Limits

Provides IP scheme with limitations: • • • Class A – 126 networks: 16,777,214 hosts each Class B – 65,000 networks: 65,534 hosts each Class C – 2 million networks: 254 hosts each While available addresses were running out, only 3% of assigned addresses were actually being used!

• Subnet zero, broadcast addresses, pool of unused addresses at Class A and B sites, etc.

04\08\2010 Unit-VI Network Layer 57

Classless Addressing Introduced by CIDR - Classless Inter Domain Routing Networks are grouped (aggregated) into blocks Blocks of networks are advertised New way of thinking: • There are no network numbers , but just prefixes • address space There are no subnet masks , just prefix lengths Classless addresses notation 10.181.215.32 /27 10.181.215.32 with mask 255.255.255.224

Binary representation of mask: 11111111.11111111.11111111.11100000

04\08\2010 Unit-VI Network Layer 58

Classless Address Notation Hosts Prefix Classful 04\08\2010 . . .

8 16 32 64 128 256 . . .

4096 8192 16384 32768 65535 . . .

. . .

/29 /28 /27 /26 /25 /24 . . .

/20 /19 /18 /17 /16 . . .

. . .

1 C . . .

16 C’s 32 C’s 64 C’s 128 C’s 1 B Subnet Mask . . .

255.255.255.248

255.255.255.240

255.255.255.224

255.255.255.192

255.255.255.128

255.255.255.0

. . .

255.255.240.0

255.255.224.0

255.255.192.0

255.255.128.0

255.255.0.0

. . .

Classless Addressing

Rules:

The address in a block must be contiguous.

The number of address in a block must be a power of 2 (1, 2, 4, 8, . . .) 3.

The first address must be evenly divisible by the number of address .

04\08\2010 Unit-VI Network Layer 60

Example

Figure 19.3 shows a block of addresses, in both binary and dotted decimal notation, granted to a small business that needs 16 addresses.

The addresses are contiguous. The number of addresses is a power of 2 (16 = 2 4 ), and the first address is divisible by 16. The first address, when converted to a decimal number, is 3,440,387,360, which when divided by 16 results in 215,024,210.

04\08\2010 Unit-VI Network Layer 61

Classless Addressing Mask: In 32 bit in which n leftmost bits are 1s and the 23-n rightmost bits are 0s In IPv4 addressing, a block of addresses can be defined as x.y.z.t /

in which x.y.z.t defines one of the addresses and the /

defines the mask.

The first address in the block can be found by setting the rightmost 32 −

bits to 0s.

The last address in the block can be found by setting the rightmost 32 − n bits to 1s.

The number of addresses in the block can be found by using the formula 2 32−n .

04\08\2010 Unit-VI Network Layer 62

Example

A block of addresses is granted to a small organization. We know that one of the addresses is 205.16.37.39/28. What is the first address in the block?

Solution: The binary representation of the given address is 11001101 00010000 00100101 00100111 If we set 32−28 rightmost bits to 0, we get 11001101 00010000 00100101 0010000 or 205.16.37.32.

04\08\2010 Unit-VI Network Layer 63

Example

Find the last address for the block 205.16.37.39/28.

Solution: The binary representation of the given address is 11001101 00010000 00100101 00100111 If we set 32 − 28 rightmost bits to 1, we get 11001101 00010000 00100101 00101111 or 205.16.37.47

Find the number of addresses in Example 19.6.

The value of n is 28, of addresses is 2 32−28 or 16.

which means that number

04\08\2010 Unit-VI Network Layer 64

Example

Another way to find the first address, the last address, and the number of addresses is to represent the mask as a 32-bit binary (or 8 digit hexadecimal) number. This is particularly useful when we are writing a program to find these pieces of information. In Example 19.5 the /28 can be represented as 11111111 11111111 11111111 11110000 (twenty-eight 1s and four 0s).

Find a.

The first address b.

The last address c.

The number of addresses.

04\08\2010 Unit-VI Network Layer 65

Example

Solution a.

The first address can be found by ANDing the given addresses with the mask. ANDing here is done bit by bit. The result of ANDing 2 bits is 1 if both bits are 1s; the result is 0 otherwise.

04\08\2010 Unit-VI Network Layer 66

Example

The last address can be found by ORing the given addresses with the complement of the mask. ORing here is done bit by bit. The result of ORing 2 bits is 0 if both bits are 0s; the result is 1 otherwise. The complement of a number is found by changing each 1 to 0 and each 0 to 1.

04\08\2010 Unit-VI Network Layer 67

Example

The number of addresses can be found by complementing the mask, interpreting it as a decimal number, and adding 1 to it.

04\08\2010 Unit-VI Network Layer 68

Network Addresses

The first address in a block is normally not assigned to any device; it is used as the network address that represents the organization to the rest of the world.

The router has 2 addresses one belongs to the granted block the other belongs to the network that is at other side of the router.

69 04\08\2010 Unit-VI Network Layer

Hierarchy

Hierarchy in a telephone network in North America

IP addresses have levels of hierarchy.

In North America telephone network has 3 levels of hierarchy.

1 st level defines the area code,2 nd level exchange and the last level defines the connection of the local loop.

04\08\2010 Unit-VI Network Layer 70

Hierarchy

Two levels of hierarchy in an IPv4 address

Each address in the block can be considered as a two-level hierarchical structure: The leftmost

bits ( prefix ) define the network ; The rightmost 32 − n bits define the host , and is called as suffix.

04\08\2010 Unit-VI Network Layer 71

Hierarchy

Three-level hierarchy in an IPv4 address

An organization that is granted a block of addresses may create clusters of networks called subnets and divide the addresses between the different networks.

The rest of the world considers the organization as one entity; however internally has several subnets .

All messages are sent to the router , router routes to subnets .

04\08\2010 Unit-VI Network Layer 72

Example

Suppose an organization is given the block 17.12.14.0/26, which contains 64 addressees. The organization has three offices and needs to divide the addresses into three subblocks of 32, 16, and16 addresses. Find the new masks.

Soln: 1.

Mask for the first subnet is n1, then2 32-n1 must be 32 i.e n1=27 2.

Mask for the second subnet is n2, then2 32-n2 must be 16 i.e n2=28 Mask for the third subnet is n3, then2 32-n3 must be 16 i.e n3=28 We can find the subnet addresses from one of addresses in the subnet In subnet 1 the addresses 17.12.14.29/27 can give us the subnet address if the mask is of /27 Host: 00010001

00001100 00001110 00011101 Mask: 27 Subnet: 00010001

04\08\2010

00001100 00001110

Unit-VI Network Layer

0000000 => 17.12.14.0

Example

In subnet 2 the addresses 17.12.14.45/28 can give us the subnet address if the mask is of /28 Host: 00010001

00001100 00001110 00101101 Mask: 28 Subnet: 00010001 00001100 00001110 0010000 => 17.12.14.32

In subnet 3 the addresses 17.12.14.50/28 can give us the subnet address if the mask is of /28 Host: 00010001

00001100 00001110 00110010 Mask: 28 Subnet: 00010001 00001100 00001110 0011000 => 17.12.14.48

04\08\2010 Unit-VI Network Layer 74

Configuration and addresses in a subnetted network

04\08\2010 Unit-VI Network Layer 75

Addresses Allocation

• • Global Authority called Internet Corporation for Assigned Names and Addresses(ICANN).

ICANN allocates addresses to ISP, ISP grants addresses to its customers.

Unit-VI Network Layer 04\08\2010 76

Example

An ISP is granted a block of addresses starting with 190.100.0.0/16 (65,536 addresses). The ISP needs to distribute these addresses to three groups of customers as follows: 1.

The first group has 64 customers; each needs 256 addresses.

The second group has 128 customers; each needs 128 addresses.

The third group has 128 customers; each needs 64 addresses.

Design the subblocks and find out how many addresses are still available after these allocations.

04\08\2010 Unit-VI Network Layer 77

Example

Group 1: In this group, each customer needs 256 addresses. That is 8 (log2 256) bits are needed to define each host. The prefix length is then 32 − 8 = 24. The addresses are Group 2: In this group, each customer needs 128 addresses. This means that 7 (log2 128) bits are needed to define each host. The prefix length is then 32 − 7 = 25. The addresses are

04\08\2010 Unit-VI Network Layer 78

Example

Group 3 For this group, each customer needs 64 addresses. This means that 6 (log 2 64) bits are needed to each host. The prefix length is then 32 − 6 = 26. The addresses are Number of granted addresses to the ISP: 65,536 Number of allocated addresses by the ISP: 40,960 Number of available addresses: 24,576

04\08\2010 Unit-VI Network Layer 79

Example

An example of address allocation and distribution by an ISP

04\08\2010 Unit-VI Network Layer 80

04\08\2010

Network Addresses Translation (NAT)

Unit-VI Network Layer 81

Network Addresses Translation (NAT) • • • • •

Private vs Public IP Addresses

Whatever connects directly into Internet must have public (globally unique) IP address There is a shortage of public IPv4 address So Private IP addresses can be used within a private network Three address ranges are reserved for private usage

•

10.0.0.0/8

• •

172.16.0.0/16 to 172.31.0.0/16 192.168.0.0/24 to 192.168.255.0/24 A private IP is mapped to a Public IP, when the machine has to access the Internet

Unit-VI Network Layer 82 04\08\2010

NAT

NAT (Network Address Translation) Maps Private IPs to Public IPs It is required because of shortage of IPv4 Address

04\08\2010 Unit-VI Network Layer 83

Static NAT : Maps unique Private IP to unique Public IP Dynamic NAT : Maps Multiple Private IP to a Pool of Public IPs (Port Address Translation : Maps a Public IP and Port Number to a service in Private IP)

04\08\2010 Unit-VI Network Layer 84

Network Addresses Translation (NAT)

• The Internet authorities have reserved three sets of addresses as private addresses • Any organization can use an address out of this set without permission from the Internet authorities .

• Therese addresses are unique inside the organization , but they are not unique globally .

• The router will not forward a packet that has theses addresses as destination addresses.

• The site have one single connection to the global Internet through Router that runs the NAT software.

Addresses for private networks

04\08\2010 Unit-VI Network Layer 85

Network Addresses Translation (NAT)

A NAT implementation

04\08\2010 Unit-VI Network Layer 86

Network Addresses Translation (NAT)

Addresses in Translation

• • Outgoing packets go through the NAT router replaces the source address in the packet with the global NAT address.

All incoming packet destination address are replaced by private address. 04\08\2010 Unit-VI Network Layer 87

Network Addresses Translation (NAT) • • When the router translates the source address of the outgoing packet it also makes note of the destination address.

When response comes back from destination address it checks for its source address from translation table 04\08\2010 Unit-VI Network Layer 88

Network Addresses Translation (NAT)

Five-column translation table

04\08\2010 Unit-VI Network Layer 89

• • NAT and ISP An ISP that serves dial up customers can use NAT to conserve addresses.

Suppose ISP has 1000 addresses but has 100,000 customers. Each of the customer is assigned a private network address. The ISP translates each addresses in outgoing packet to one of the 1000 global address.

An ISP and NAT

04\08\2010 Unit-VI Network Layer 90

IPv6 ADDRESSES

04\08\2010 Unit-VI Network Layer 91

IPv6 ADDRESSES

Despite all short-term solutions, address depletion is still a long-term problem for the Internet. This and other problems in the IP protocol itself have been the motivation for IPv6.

Structure Address Space 04\08\2010 Unit-VI Network Layer 92

IPv6 Addresses

• • • Structure: IPv6 address consists of 16 bytes or 128 bits specified in hexadecimal colon notation.

long and 128 bits are divided into length.

8 sections , each 2 bytes in 2 bytes in hex notation requires 4 hex digits .

IPv6 address in binary and hexadecimal colon notation

04\08\2010 Unit-VI Network Layer 93

IPv6 Addresses

Abbreviation

•

IP address in hexadecimal format is and contains many digits are zero .

very long

•

The leading zeros of a section are omitted .

Abbreviated IPv6 addresses

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Example

Expand the address 0:15::1:12:1213 to its original.

Solution We first need to align the left side of the double colon to the left of the original pattern and the right side of the double colon to the right of the original pattern to find how many 0s we need to replace the double colon.

This means that the original address is.

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IPv6 Addresses

Type prefixes for IPv6 addresses

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IPv6 Addresses

Type prefixes for IPv6 addresses (continued)

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IPv6 Addresses

Prefixes for provider-based unicast address

• • • Type Identifier: 3 bit field , defines the address as a provider based address Registry Identifier: 5 bit field indicates the agency that has registered . INTERNIC center for North America: RIPNIC center for European registration APNIC Asian and Pacific countries Provider Identifier: Internet Provider (ISP) 16 bit 04\08\2010 Unit-VI Network Layer 98

IPv6 Addresses

Prefixes for provider-based unicast address

• • • Subscriber Identifier: 24 bit length is used to identify subscriber (Organization) Subnet Identifier: Each organization has many subnets and 32 bit is used for identification Node Identifier: 48 bit is used to identify node connected to a subnet.

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IPv6 Addresses

Multicast address in IPv6

• • • used to define a group of hosts instead of just one Flag i s used define group of address as either permanent or transient.

Scope:

Anycast Addresses

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IPv6 Addresses

Reserved addresses in IPv6

• • Unspecified is used when host does not know its own address and sends an inquiry to find its address.

Loopback is used by a host to test itself without going into the network.

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IPv6 Addresses

• Compatible is used during the transition from IPv4 to IPv6. Node using IPv6 want to send a message to another node using IPv6, but message needs to pass through a part of network that still operates in IPv4.

• 04\08\2010 Mapped address is used when node has migrated to Ipv6 wants to send a packet to a node still using IPv4 Unit-VI Network Layer 102

IPv6 Addresses

Local addresses in IPv6

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IPv6 Addresses

A large number of consecutive IP address are available starting at 198.16.0.0. Suppose that four organizations, A, B, C, and D, request 4000, 2000, 4000, and 8000 addresses, respectively, and in that order. For each of these, give the first IP address assigned, the last IP address assigned, and the mask in the w.x.y.z/s notation.

To start with, all the requests are rounded up to a power of two. The starting address, ending address, and mask are as follows: A: 198.16.0.0 – 198.16.15.255 written as 198.16.0.0/20 B: 198.16.16.0 – 198.16.23.255 written as 198.16.16.0/21 C: 198.16.32.0 – 198.16.47.255 written as 198.16.32.0/20 D: 198.16.64.0 – 198.16.95.255 written as 198.16.64.0/19 04\08\2010 Unit-VI Network Layer 104