Chapter_04_05_06

TCP/IP Protocol Suite 1Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Chapter 4

Introduction to Network

Layer

TCP/IP Protocol Suite 2

Chapter Outline

4.1 Introduction

4.2 Switching

4.3 Packet Switching

4.4 Network Layer Services

4.5 Other Network Layer Issues


4-1 INTRODUCTION

At the conceptual level, we can think of the global Internet as a black box network that connects millions (if not billions) of computers in the world together. At this level, we are only concerned that a message from the application layer in one computer reaches the application layer in another computer.


A B

Physical

Data link

Network

Transport

Application

Physical

Data link

Network

Transport

Application

Internet

Figure 4.1 Internet as a block box

Message Message


Figure 4.2 Internet as a combination of LANs and WANs connected together


4-2 SWITCHING

From the previous discussion, it is clear that the passage of a message from a source to a destination involves many decisions. When a message reaches a connecting device, a decision needs to be made to select one of the output ports through which the packet needs to be send out. In other words, the connecting device acts as a switch that connects one port to another port.


Topics Discussed in the Section

Circuit Switching Packet Switching


In circuit switching, the whole message is sent from the source to the

destination without being divided into packets.

Note


A good example of a circuit-switched network is the early telephone systems in which the path was established between a caller and a callee when the telephone number of the callee was dialed by the caller. When the callee responded to the call, the circuit was established. The voice message could now flow between the two parties, in both directions, while all of the connecting devices maintained the circuit. When the caller or callee hung up, the circuit was disconnected. The telephone network is not totally a circuit-switched network today.

Example 4.1


In packet switching, the message is first divided into manageable packets at the

source before being transmitted. The packets are assembled at the

destination.

Note


4-3 PACKET SWITHING

The network layer is designed as a packet-switched network. This means that the packet at the source is divided into manageable packets, normally called datagrams. Individual datagrams are then transferred from the source to the destination. The received datagrams are assembled at the destination before recreating the original message. The packet-switched network layer of the Internet was originally designed as a connectionless service, but recently there is a tendency to change this to a connection-oriented service.


Topics Discussed in the Section

Connectionless Service Connection-Oriented Service


Sender Network

Network

ReceiverOut of orderR3

R4

R5

R1 R2

A connectionlesspacket-swtiched network

Figure 4.3 A connectionless packet-switched network

4 3 2 1

1

2

3

42

3 31

4 43 21


Figure 4.4 Forwarding process in a connectionless network

1 2

43

Outputinterface

Destinationaddress

Routing table

12

3

AB

HDestinationaddress

SA DA Data

Send the packetout of interface 2

SA DA Data

LegendSA: Source addressDA: Destination address


In a connectionless packet-switched network, the forwarding decision

is based on the destination address of the packet.

Note


Time Time Time Time

SouceDestination

Figure 4.5 Delay in a connectionless network

1

2

3

Totaldelay


Sender

Network

Network

ReceiverR4

R5

R1 R2

R3

A connection-orientedpacket-switched network

Packets

Virtual circuit

Legend

4 3 2 1

Figure 4.6 A connection-oriented packet switched network

4 3 2 1

4

3

2

1

4 3 2 1 4 3 2 1


In a connection-oriented packet switched network, the forwarding

decision is based on the label of the packet.

Note


1

1

1

4

4

2

2 2

3

3

3

A

B

Network

Network

R3

R5

R4

R1 R2

Request packet

Virtual circuit

Legend

A to B

Figure 4.8 Sending request packet in a virtual-circuit network

A to B

1

A to B2

A to B3

A to B

4

LabelPort Port3141

Label

OutgoingIncoming

A to B

Port Port3661

Label Label

OutgoingIncoming

A to B

Label

A to B

Port Port4221

Label

OutgoingIncoming


Time Time Time Time

Source Destination

Figure 4.11 Delay in a connection-oriented network

1

2

Transmissiontime

3

4

5

Setup

Teardow

n

Total

delay

6

7


Chapter 4

Objectives Upon completion you will be able to:

IP Addresses:Classful Addressing

• Understand IPv4 addresses and classes• Identify the class of an IP address• Find the network address given an IP address• Understand masks and how to use them• Understand subnets and supernets


4.1 INTRODUCTION4.1 INTRODUCTION

The identifier used in the IP layer of the TCP/IP protocol suite to identify each device connected to the Internet is called the Internet address or IP address. An IP address is a 32-bit address that uniquely and universally defines the connection of a host or a router to the Internet. IP addresses are unique. They are unique in the sense that each address defines one, and only one, connection to the Internet. Two devices on the Internet can never have the same address.

The topics discussed in this section include:

Address SpaceNotation


An IP address is a 32-bit address.

Note:


The IP addresses are unique.

Note:


The address space of IPv4 is232 or 4,294,967,296.

Note:


Figure 4.1 Dotted-decimal notation


Change the following IP addresses from dotted decimal notation to binary notation.

a. 114.34.2.8

b. 129.14.6.8

c. 208.34.54.12

d. 238.34.2.1

Example

Solutiona. 01110010 00100010 00000010 00001000b. 10000001 00001110 00000110 00001000c. 11010000 00100010 00110110 00001100d. 11101110 00100010 00000010 00000001


The binary, decimal, and hexadecimal number systems are reviewed in

Appendix B.

Note:


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

a. 10000001 00001011 00001011 11101111b. 11000001 10000011 00011011 11111111c. 11100111 11011011 10001011 01101111d. 11111001 10011011 11111011 00001111

Example 1

SolutionWe replace each group of 8 bits with its equivalent decimal number (see Appendix B) and add dots for separation:

a. 129.11.11.239 b. 193.131.27.255c. 231.219.139.111 d. 249.155.251.15


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

a. 111.56.45.78 b. 221.34.7.82c. 241.8.56.12 d. 75.45.34.78

Example 2

SolutionWe replace each decimal number with its binary equivalent:

a. 01101111 00111000 00101101 01001110b. 11011101 00100010 00000111 01010010c. 11110001 00001000 00111000 00001100d. 01001011 00101101 00100010 01001110


Find the error, if any, in the following IP addresses:

a. 111.56.045.78 b. 221.34.7.8.20

c. 75.45.301.14 d. 11100010.23.14.67

Example 3

Solution

a. There are no leading zeroes in dotted-decimal notation (045).

b. We may not have more than four numbers in an IP address.

c. In dotted-decimal notation, each number is less than or equal to 255; 301 is outside this range.

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


Change the following IP addresses from binary notation to hexadecimal notation.

a. 10000001 00001011 00001011 11101111

b. 11000001 10000011 00011011 11111111

Example 4

SolutionWe replace each group of 4 bits with its hexadecimal equivalent (see Appendix B). Note that hexadecimal notation normally has no added spaces or dots; however, 0X (or 0x) is added at the beginning or the subscript 16 at the end to show that the number is in hexadecimal.

a. 0X810B0BEF or 810B0BEF16

b. 0XC1831BFF or C1831BFF16


4.2 CLASSFUL ADDRESSING

IP addresses, when started a few decades ago, used the concept of classes. This architecture is called classful addressing. In the mid-1990s, a new architecture, called classless addressing, was introduced and will eventually supersede the original architecture. However, part of the Internet is still using classful addressing, but the migration is very fast.


Recognizing ClassesNetid and HostidClasses and BlocksNetwork AddressesSufficient InformationMaskCIDR NotationAddress Depletion


Figure 4.2 Occupation of the address space


Figure 4.3 Finding the class in binary notation


How can we prove that we have 2,147,483,648 addresses in class A?

Example 5

SolutionIn class A, only 1 bit defines the class. The remaining 31 bits are available for the address. With 31 bits, we can have 231

or 2,147,483,648 addresses.


Find the class of each address:

a. 00000001 00001011 00001011 11101111b. 11000001 10000011 00011011 11111111c. 10100111 11011011 10001011 01101111d. 11110011 10011011 11111011 00001111

Example 6

SolutionSee the procedure in Figure 4.4.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 bit is 0; the second bit is 1. This is a class B address.d. The first 4 bits are 1s. This is a class E address..


Figure 4.5 Finding the class in decimal notation


Find the class of each address:

a. 227.12.14.87 b.193.14.56.22 c.14.23.120.8d. 252.5.15.111 e.134.11.78.56

Example 7

Solutiona. The first byte is 227 (between 224 and 239); the class is D.b. The first byte is 193 (between 192 and 223); the class is C.c. The first byte is 14 (between 0 and 127); the class is A.d. The first byte is 252 (between 240 and 255); the class is E.e. The first byte is 134 (between 128 and 191); the class is B.


Figure 4.6 Netid and hostid


Figure 4.7 Blocks in class A


Figure 4.8 Blocks in class B


Figure 4.9 Blocks in class C


Given the network address 17.0.0.0, find the class, the block, and the range of the addresses.

Example 9

SolutionThe class is A because the first byte is between 0 and 127. The block has a netid of 17. The addresses range from 17.0.0.0 to 17.255.255.255.



Example 10

SolutionThe class is B because the first byte is between 128 and 191. The block has a netid of 132.21. The addresses range from 132.21.0.0 to 132.21.255.255.



Example 11

SolutionThe class is C because the first byte is between 192 and 223. The block has a netid of 220.34.76. The addresses range from 220.34.76.0 to 220.34.76.255.


If the network address is given, we can find the block and the range of addresses in the block.

What about the reverse??? If the address is given, can we find

the network address??? If there is no subnets. Then we can

use the concept of “MASK”


Figure 4.10 Masking concept


Figure 4.11 AND operation


Table 4.2 Default masks


Given the address 23.56.7.91, find the beginning address (network address).

Example 12

SolutionThe default mask is 255.0.0.0, which means that only the first byte is preserved and the other 3 bytes are set to 0s. The network address is 23.0.0.0.



Example 13

SolutionThe default mask is 255.255.0.0, which means that the first 2 bytes are preserved and the other 2 bytes are set to 0s. The network address is 132.6.0.0.



Example 14

SolutionThe default mask is 255.255.255.0, which means that the first 3 bytes are preserved and the last byte is set to 0. The network address is 201.180.56.0.


4.3 OTHER ISSUES

In this section, we discuss some other issues that are related to addressing in general and classful addressing in particular.


Multihomed DevicesLocation, Not NamesSpecial AddressesPrivate AddressesUnicast, Multicast, and Broadcast Addresses


Table 4.3 Special addresses


Figure 4.13 Network address


Figure 4.14 Example of direct broadcast address


Figure 4.18 Example of loopback address

• Is used to test the IP software on the machine.

• When this address is used, packet will never leave the machine. It simply return to the protocol software


Table 4.5 Addresses for private networks

• A number of blocks in each class are assigned for private use.

• These addresses are used either in isolation or in connection with network address translation techniques.


A router has an IP address of 108.5.18.22. It sends a direct broadcast packet to all the host in the network. What are the source and destination IP addresses used in this packet

Example

Solution

Source address: 108.5.18.22 Destination address: 108.255.255.255


A router has an IP address of 185.67.89.34 needs loopback testing. What are the source and destination IP addresses

Example

Solution

Source address: 185.67.89.34 Destination address: 127.X.Y.Z


4.4 SUBNETTING AND SUPERNETTINGIn the previous sections we discussed the problems associated with classful addressing. Specifically, the network addresses available for assignment to organizations are close to depletion. This is coupled with the ever-increasing demand for addresses from organizations that want connection to the Internet. In this section we briefly discuss a solution: subnetting


SubnettingSupernettingSupernet MaskObsolescence


IP addresses are designed with two levels of hierarchy.

Note:


Figure 4.20 A network with two levels of hierarchy (not subnetted)


Figure 4.21 A network with three levels of hierarchy (subnetted)

• Network (Site)• Subnet• Host


Figure 4.22 Addresses in a network with and without subnetting


Figure 4.23 Hierarchy concept in a telephone number


Introduction to Subnetting

To create a subnet address, a network administrator borrows bits from the host field

and designates them as the subnet field.


Subnetting To allocate a part of the host address space to network addresses and leave the remaining part to other physical networks It can get most out of the limited 32-bit IP address space and reduce the size of the routing table in a large internetwork A netmask to determine which bits in the IP address space represent the sub-network addresses


Introduction to Subnetting

Host bits must be reassigned (or “borrowed”) as network bits.

The starting point is always the leftmost host bit.

3 bits borrowed allows 23-2 or 6 subnets




Figure 4.24 Default mask and subnet mask


What is the subnetwork address if the destination address is 200.45.34.56 and the subnet mask is 255.255.240.0?

Example 15

SolutionWe apply the AND operation on the address and the subnet mask.

Address ➡ 11001000 00101101 00100010 00111000

Subnet Mask ➡ 11111111 11111111 11110000 00000000

Subnetwork Address ➡ 11001000 00101101 00100000 00000000.

Subnetwork address 200.45.32.0


Figure 4.25 Comparison of a default mask and a subnet mask


Figure 4.27 Comparison of subnet, default, and supernet masks


Find the contiguous mask in each case

a) 1024 subnets in Class A

b) 256 subnets in Class B

c) 32 subnets in Class C

d) 4 subnets in Class C

Example

Solutiona. 2x = 1024 x = log21024 = 10 mask is 255.255.192.0b. 2x = 256 x = log2256 = 8 mask is 255.255.255.0c. 2x = 32 x = log232 = 5 mask is 255.255.255.248d. 2x = 4 x = log24 = 2 mask is 255.255.255.192


What is the maximum number of subnets in each area

a) Class A: mask 255.255.192.0

b) Class B: mask 255.255.192.0

c) Class C: mask 255.255.255.192

d) Class C: mask 255.255.255.240

Example

Solutiona. 10 extra 1s 210 = 1024 subnetsb. 2 extra 1s 22 = 4 subnetsc. 2 extra 1s 22 = 4 subnetsd. 4 extra 1s 24 = 16 subnets


Given a Host IP address 192.168.45.32 and Subnet mask

of 255.255.255.128. Calculate the following

• Number of Subnet bits

• Maximum number of Subnets

• Number of Host bits

• Maximum number of hosts

• Subnetwork Address

• Broadcast Address

• Show subnets

(Subnetwork address, valid host, broadcast address)

Example


Answer• Number of Subnet bits - 1

• Maximum number of Subnets - 2

• Number of Host bits - 7

• Maximum number of hosts - 126

• Subnetwork Address – 195.168.45.0

• Broadcast Address – 195.168.45.127

• Show subnets Subnet Valid Hosts Broadcast 192.168.45.0 192.168.45.1 to 192.168.45.126 192.168.45.127 192.168.45.128 192.168.45.129 to 192.168.45.254 192.168.45.255



Given a Host IP address 221.131.72.32 and Subnet mask

of 255.255.255.224. Calculate the following

• Number of Subnet bits

• Maximum number of Subnets

• Number of Host bits

• Maximum number of hosts

• Subnetwork Address

• Broadcast Address

• Show first 12 subnets


Example


Answer Number of Subnet bits -3

• Maximum number of Subnets - 8

• Number of Host bits -5

• Maximum number of hosts - 30

• Subnetwork Address – 221.131.72.32

•

• Broadcast Address – 221.131.72.63

Show first 12 subnets Subnet , Valid Hosts , Broadcast 221.131.72.32 , 221.131.72.33 to 221.131.72.62 , 221.131.72.63 221.131.72.64 , 221.131.72.65 to 221.131.72.94 , 221.131.72.95 221.131.72.96 , 221.131.72.97 to 221.131.72.126 , 221.131.72.127 221.131.72.128 , 221.131.72.129 to 221.131.72.158 , 221.131.72.159 221.131.72.160 , 221.131.72.161 to 221.131.72.190 , 221.131.72.191 221.131.72.192 , 221.131.72.193 to 221.131.72.222 , 221.131.72.223 221.131.72.224 , 221.131.72.225 to 221.131.72.254 , 221.131.72.255


Chapter 5

Objectives Upon completion you will be able to:

IP Addresses:Classless Addressing

• Understand the concept of classless addressing• Be able to find the first and last address given an IP address• Be able to find the network address given a classless IP address• Be able to create subnets from a block of classless IP addresses• Understand address allocation and address aggregation


5.1 VARIABLE-LENGTH BLOCKS

In classless addressing variable-length blocks are assigned that belong to no class. In this architecture, the entire address space (232 addresses) is divided into blocks of different sizes.


RestrictionsFinding the BlockGranted Block


Figure 5.2 Format of classless addressing address

In classless addressing the address must always be accompanied by the mask

Written in CIDR (Classless InterDomain Routing) notation


Prefix and Suffix The slash (/n) defines the number of bits (prefix length) that defines the network (~netid). n is the number of ones (1) in the mask The remaining bits (32 – n) are the bits left for the suffix (~hostid)


Example

Write the following mask in slash notationa. 255.255.255.0b. 255.0.0.0c. 255.255.224.0d. 255.255.240.0

Answer


Example

Find the range of addresses in the following block

a. 123.56.77.32/29

Answer


Example

Find the range of addresses in the following block200.17.21.128/27

Answer


Table 5.1 Prefix lengths


Classful addressing is a special case of classless addressing.

Note:


Network, First and Last Addresses

First address in the block can be found by ANDing the mask with the address or by setting the suffix bits (32 – n) all to 0s Number of addresses in the block can be found using 232-n

Last address in the block is found by setting the suffix bits all to 1s


What is the first address in the block if one of the addresses is 167.199.170.82/27?

Example 4

Address in binary: 10100111 11000111 10101010 01010010

Keep the left 27 bits: 10100111 11000111 10101010 01000000

Result in CIDR notation: 167.199.170.64/27

SolutionThe prefix length is 27, which means that we must keep the first 27 bits as is and change the remaining bits (5) to 0s. The following shows the process:


What is the first address in the block if one of the addresses is 140.120.84.24/20?

Example 5

See Next Slide

SolutionFigure 5.3 shows the solution. The first, second, and fourth bytes are easy; for the third byte we keep the bits corresponding to the number of 1s in that group. The first address is 140.120.80.0/20.


Figure 5.3 Example 5


Find the first address in the block if one of the addresses is 140.120.84.24/20.

Example 6

See Next Slide

SolutionThe first, second, and fourth bytes are as defined in the previous example. To find the third byte, we write 84 as the sum of powers of 2 and select only the leftmost 4 (m is 4) as shown in Figure 5.4. The first address is 140.120.80.0/20.




Find the number of addresses in the block if one of the addresses is 140.120.84.24/20.

Example 7

SolutionThe prefix length is 20. The number of addresses in the block is 232−20 or 212 or 4096. Note thatthis is a large block with 4096 addresses.


Using the first method, find the last address in the block if one of the addresses is 140.120.84.24/20.

Example 8

See Next Slide

SolutionWe found in the previous examples that the first address is 140.120.80.0/20 and the number of addresses is 4096. To find the last address, we need to add 4095 (4096 − 1) to the first address.


To keep the format in dotted-decimal notation, we need to represent 4095 in base 256 (see Appendix B) and do the calculation in base 256. We write 4095 as 15.255. We then add the first address to this number (in base 255) to obtain the last address as shown below:

Example 8 (Continued)

140 . 120 . 80 . 0 15 . 255

-------------------------140 . 120 . 95 . 255

The last address is 140.120.95.255/20.


Using the second method, find the last address in the block if one of the addresses is 140.120.84.24/20.

Example 9

See Next Slide

SolutionThe mask has twenty 1s and twelve 0s. The complement of the mask has twenty 0s and twelve 1s. In other words, the mask complement is

00000000 00000000 00001111 11111111

or 0.0.15.255. We add the mask complement to the beginning address to find the last address.


140 . 120 . 80 . 0 0 . 0 . 15 . 255----------------------------140 . 120 . 95 . 255


We add the mask complement to the beginning address to find the last address.

The last address is 140.120.95.255/20.


Find the block if one of the addresses is 190.87.140.202/29.

Example 10

See Next Slide

SolutionWe follow the procedure in the previous examples to find the first address, the number of addresses, and the last address. To find the first address, we notice that the mask (/29) has five 1s in the last byte. So we write the last byte as powers of 2 and retain only the leftmost five as shown below:


202 ➡ 128 + 64 + 0 + 0 + 8 + 0 + 2 + 0

The leftmost 5 numbers are ➡ 128 + 64 + 0 + 0 + 8

The first address is 190.87.140.200/29


The number of addresses is 232−29 or 8. To find the last address, we use the complement of the mask. The mask has twenty-nine 1s; the complement has three 1s. The complement is 0.0.0.7. If we add this to the first address, we get 190.87.140.207/29. In other words, the first address is 190.87.140.200/29, the last address is 190.87.140.207/29. There are only 8 addresses in this block.


Show a network configuration for the block in the previous example.

Example 11

See Next Slide

SolutionThe organization that is granted the block in the previous example can assign the addresses in the block to the hosts in its network. However, the first address needs to be used as the network address and the last address is kept as a special address (limited broadcast address). Figure 5.5 shows how the block can be used by an organization. Note that the last address ends with 207, which is different from the 255 seen in classful addressing.




In classless addressing, the last address in the block does not

necessarily end in 255.

Note:


In CIDR notation, the block granted is defined by the first address and the

prefix length.

Note:


5.2 SUBNETTING

When an organization is granted a block of addresses, it can create subnets to meet its needs. The prefix length increases to define the subnet prefix length.


Finding the Subnet MaskFinding the Subnet AddressesVariable-Length Subnets


In fixed-length subnetting, the number of subnets is a power of 2.

Note:


An organization is granted the block 130.34.12.64/26. The organization needs 4 subnets. What is the subnet prefix length?

Example 12

SolutionWe need 4 subnets, which means we need to add two more 1s (log2 4 = 2) to the site prefix. The subnet prefix is then /28.


In Fixed length subnetting find the number of 1s that must be added to the mask if the number of desired subnets is

a. 2 b. 62 c. 122 d. 250

Example 12-1

Solution


What is the maximum number of subnets if the prefix length of a block is

a. 18 b. 10 c. 27 d. 31

Example 12-2

Solution


What are the subnet addresses and the range of addresses for each subnet in the previous example-12?

Example 13

See Next Slide

SolutionFigure 5.6 shows one configuration.


The site has 232−26 = 64 addresses. Each subnet has 232–28 = 16 addresses. Now let us find the first and last address in each subnet.


See Next Slide

1. The first address in the first subnet is 130.34.12.64/28, using the procedure we showed in the previous examples. Note that the first address of the first subnet is the first address of the block. The last address of the subnet can be found by adding 15 (16 −1) to the first address. The last address is 130.34.12.79/28.



2.The first address in the second subnet is 130.34.12.80/28; it is found by adding 1 to the last address of the previous subnet. Again adding 15 to the first address, we obtain the last address, 130.34.12.95/28.

3. Similarly, we find the first address of the third subnet to be 130.34.12.96/28 and the last to be 130.34.12.111/28.

4. Similarly, we find the first address of the fourth subnet to be 130.34.12.112/28 and the last to be 130.34.12.127/28.




An organization is granted a block of addresses with the beginning address 14.24.74.0/24. There are 232−24= 256 addresses in this block. The organization needs to have 11 subnets as shown below:

a. two subnets, each with 64 addresses.

b. two subnets, each with 32 addresses.

c. three subnets, each with 16 addresses.

d. four subnets, each with 4 addresses.

Design the subnets.

Example 14

See Next Slide For One Solution


1. We use the first 128 addresses for the first two subnets, each with 64 addresses. Note that the mask for each network is /26.

1st subnet 14.24.74.00 000000/26 2nd 14.24.74.01 000000/26

14.24.74.0/26 2nd 14.24.74.64/26

2. We use the next 64 addresses for the next two subnets, each with 32 addresses. Note that the mask for each network is /27.

1st subnet 14.24.74.100 00000/27 2nd 14.24.74.10100000/27

14.24.74.128/27 2nd 14.24.74.160/27

Example 14 (Continuted)

See Next Slide


3. We use the next 48 addresses for the next three subnets, each with 16 addresses. Note that the mask for each network is /28. 1st subnet 14.24.74.1100 0000/28 2nd 1101 = 208 3rd 1110

14.24.74.192/27 14.24.74.208/27 14.24.74.224/27

4. We use the last 16 addresses for the last four subnets, each with 4 addresses. Note that the mask for each network is /30. The subnet address for each subnet is given in the figure.1st subnet 14.24.74.111100/30 = 240 2nd 111101/30 = 244 3rd 111110/30 = 248 4th = 111111/30 = 252

14.24.74.240/30 14.24.74.244/30 14.24.74.248/30 14.24.74.252/30

Example 14 (Continuted)




As another example, assume a company has three offices: Central, East, and West. The Central office is connected to the East and West offices via private, point-to-point WAN lines. The company is granted a block of 64 addresses with the beginning address 70.12.100.128/26. The management has decided to allocate 32 addresses for the Central office and divides the rest of addresses between the two offices. Figure 5.8 shows the configuration designed by the management.

Example 15

See Next Slide


The company will have three subnets, one at Central, one at East, and one at West. The following lists the subblocks allocated for each network:


See Next Slide

a. The Central office uses the network address 70.12.100.128/27. This is the first address, and the mask /27 shows that there are 32 addresses in this network. Note that three of these addresses are used for the routers and the company has reserved the last address in the sub-block. The addresses in this subnet are 70.12.100.128/27 to 70.12.100.159/27. Note that the interface of the router that connects the Central subnet to the WAN needs no address because it is a point-to-point connection.



See Next Slide

b. The West office uses the network address 70.12.100.160/28. The mask /28 shows that there are only 16 addresses in this network. Note that one of these addresses is used for the router and the company has reserved the last address in the sub-block. The addresses in this subnet are 70.12.100.160/28 to 70.12.100.175/28. Note also that the interface of the router that connects the West subnet to the WAN needs no address because it is a point-to- point connection.



c. The East office uses the network address 70.12.100.176/28. The mask /28 shows that there are only 16 addresses in this network. Note that one of these addresses is used for the router and the company has reserved the last address in the sub-block. The addresses in. this subnet are 70.12.100.176/28 to 70.12.100.191/28. Note also that the interface of the router that connects the East subnet to the WAN needs no address because it is a point-to-point connection.




5.3 ADDRESS ALLOCATION

Address allocation is the responsibility of a global authority called the Internet Corporation for Assigned Names and Addresses (ICANN). It usually assigns a large block of addresses to an ISP to be distributed to its Internet users.


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:

Example 16

See Next Slide

a. The first group has 64 customers; each needs 256 addresses.b. The second group has 128 customers; each needs 128 addressesc. 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.


See Next Slide

SolutionFigure 5.9 shows the situation.




Group 1For this group, each customer needs 256 addresses. This means the suffix length is 8 (28 =256). The prefix length is then 32 − 8 = 24. The addresses are:


See Next Slide

1st Customer 190.100.0.0/24 190.100.0.255/242nd Customer 190.100.1.0/24 190.100.1.255/24. . .64th Customer 190.100.63.0/24 190.100.63.255/24Total = 64 × 256 = 16,384


Group 2For this group, each customer needs 128 addresses. This means the suffix length is 7 (27 =128). The prefix length is then 32 − 7 = 25. The addresses are:


See Next Slide

1st Customer 190.100.64.0/25 190.100.64.127/252nd Customer 190.100.64.128/25 190.100.64.255/25· · ·128th Customer 190.100.127.128/25 190.100.127.255/25

Total = 128 × 128 = 16,384



Group 3 For this group, each customer needs 64 addresses. This means the suffix length is 6 (26 = 64). The prefix length is then 32 − 6 = 26. The addresses are:

Example 16 (continued)

See Next Slide

1st Customer 190.100.128.0/26 190.100.128.63/26

2nd Customer 190.100.128.64/26 190.100.128.127/26· · ·128th Customer 190.100.159.192/26 190.100.159.255/26

Total = 128 × 64 = 8,192


Number of granted addresses to the ISP: 65,536

Number of allocated addresses by the ISP: 40,960

Number of available addresses: 24,576

Example 16 (continued)


26.3 NETWORK ADDRESS TRANSLATION (NAT)

Network address translation (NAT) allows a site to use a set of private addresses for internal communication and a set of global Internet addresses for communication with another site. The site must have only one single connection to the global Internet through a router that runs NAT software.


Address Translation Translation Table NAT and ISP


Figure 26.6 NAT


Figure 26.7 Address translation


Figure 26.8 Translation


Chapter 6

Upon completion you will be able to:

Delivery, Forwarding, and Routing of IP Packets

• Understand the different types of delivery and the connection • Understand forwarding techniques in classful addressing• Understand forwarding techniques in classless addressing• Understand how a routing table works• Understand the structure of a router

Objectives


6.1 DELIVERY

The network layer supervises delivery, the handling of the packets by the underlying physical networks. Two important concepts are the type of connection and direct versus indirect delivery.


Connection TypesDirect Versus Indirect Delivery


Delivery The network layer supervises delivery, the handling of the packets by the underlying physical networks Two important concepts are the type of connection and direct versus indirect delivery


Delivery Direct Delivery: where the destination is on the current network (always one at the end of a route) Indirect Delivery: where the destination is on a remote network so the packet must be passed to another router to be delivered (can be zero or more indirect deliveries)


Figure 6.1 Direct delivery


Figure 6.2 Indirect delivery


Example

A host with IP address 137.23.56.23/16 sends a packet to a host with IP address 137.23.67.9/16. Is the delivery direct or indirect? Assume no subnetting.

AnswerDirect; both hosts are on the same network (same netid, 137.23).


Example

A host with IP address 137.23.56.23/16 sends a packet to a host with IP address 137.23.67.9/16. Is the delivery direct or indirect? Assume no subnetting.

AnswerDirect; both hosts are on the same network (same netid, 137.23).


Example A host with IP address

137.23.56.23/16 sends a packet to a host with IP address 142.3.6.9/24. Is the delivery direct or indirect? Assume no subnetting.

AnswerIndirect; the hosts are on different networks (different netid, 137.23 and 142.3.6).


6.2 FORWARDING

Forwarding means to place the packet in its route to its destination. Forwarding requires a host or a router to have a routing table. .


Forwarding TechniquesForwarding with Classful AddressingForwarding with Classless AddressingCombination


Forwarding Techniques next-hop network specific entries host-specific entries default entry


Figure 6.3 Next-hop method

Reduces the size of the routing table by only recording the next hop for forwarding the packet


Figure 6.4 Network-specific method

Only one entry that defines the address of thedestination network


Figure 6.5 Host-specific routing

specific destination host address is recorded, usuallyused for control over routing


Figure 6.6 Default routing

All other destinations are covered by a single entry


Classful Routing Minimum of 3 columns

network destination address next-hop addresses interface number (outgoing port)

Default masks are used (mainly) to route packets to final destination networks (Internet) Applying the default mask to a destination address gives you the nodes network address which can be compared to the entries in the table Order of the routing table is: direct-delivery, host-specific, network-specific, default


Figure 6.7 Simplified forwarding module in classful address without subnetting


Figure 6.8 shows an imaginary part of the Internet. Show the routing tables for router R1.

Example 1

See Next Slide


Figure 6.8 Configuration for routing, Example 1


Solution

Figure 6.9 shows the three tables used by router R1. Note that some entries in the next-hop address column are empty because in these cases, the destination is in the same network to which the router is connected (direct delivery). In these cases, the next-hop address used by ARP is simply the destination address of the packet as we will see in Chapter 7.


See Next Slide


Figure 6.9 Tables for Example 1



Example



Example


Figure 6.10 Simplified forwarding module in classful address with subnetting


Figure 6.11 shows a router connected to four subnets.

Example 4

See Next Slide



Note several points. First, the site address is 145.14.0.0/16 (a class B address). Every packet with destination address in the range 145.14.0.0 to 145.14.255.255 is delivered to the interface m4 and distributed to the final destination subnet by the router. Second, we have used the address x.y.z.t/n for the interface m4 because we do not know to which network this router is connected. Third, the table has a default entry for packets that are to be sent out of the site. The router is configured to apply the mask /18 to any destination address.


Figure 6.11 Configuration for Example 4


The router in Figure 6.11 receives a packet with destination address 145.14.32.78. Show how the packet is forwarded.

Example 5

SolutionThe mask is /18. After applying the mask, the subnet address is 145.14.0.0. The packet is delivered to ARP with the next-hop address 145.14.32.78 and the outgoing interface m0.


A host in network 145.14.0.0 in Figure 6.11 has a packet to send to the host with address 7.22.67.91. Show how the packet is routed.

Example 6

SolutionThe router receives the packet and applies the mask (/18). The network address is 7.22.64.0. The table is searched and the address is not found. The router uses the address of the default router (not shown in figure) and sends the packet to that router.


In classful addressing we can have a routing table with three columns;

in classless addressing, we need at least four columns.

Note:


Classless Routing Minimum of 4 columns

mask needs to be specified for each entry Mask in each line of the routing table is applied to the destination address Routing table sorted in longest mask to shortest mask order


Figure 6.12 Simplified forwarding module in classless address


Make a routing table for router R1 using the configuration in Figure 6.13.

Example 7

SolutionTable 6.1 shows the corresponding table.

See Next Slide

See the table after the figure.


Figure 6.13 Configuration for Example 7


Table 6.1 Routing table for router R1 in Figure 6.13


Show the forwarding process if a packet arrives at R1 in Figure 6.13 with the destination address 180.70.65.140.

Example 8

SolutionThe router performs the following steps:

1. The first mask (/26) is applied to the destination address. The result is 180.70.65.128, which does not match the corresponding network address.

See Next Slide



2. The second mask (/25) is applied to the destination address. The result is 180.70.65.128, which matches the corresponding network address. The next-hop address (the destination address of the packet in this case) and the interface number m0 are passed to ARP for further processing.



Example 9

SolutionThe router performs the following steps:

See Next Slide


1. The first mask (/26) is applied to the destination address. The result is 201.4.22.0, which does not match the corresponding network address (row 1).

2. The second mask (/25) is applied to the destination address. The result is 201.4.22.0, which does not match the corresponding network address (row 2).

3. The third mask (/24) is applied to the destination address. The result is 201.4.22.0, which matches the corresponding network address. The destination address of the package and the interface number m3 are passed to ARP.




Example 10

SolutionThis time all masks are applied to the destination address, but no matching network address is found. When it reaches the end of the table, the module gives the next-hop address 180.70.65.200 and interface number m2 to ARP. This is probably an outgoing package that needs to be sent, via the default router, to some place else in the Internet.



Example 11

Solution



Example 12

Solution


Example

Find the topology of the network if the following routing table is for R1.

Mask Network Address

Next- Hop Address

Interface

/27 202.14.17.224

- M1

/18 145.23.192.0 - M0

Default default 130.56.12.4 m2

Answer


Now let us give a different type of example. Can we find the configuration of a router, if we know only its routing table? The routing table for router R1 is given in Table 6.2. Can we draw its topology?

Example 11

See Next Slide


Table 6.2 Routing table for Example 11


Example 11

SolutionWe know some facts but we don’t have all for a definite topology. We know that router R1 has three interfaces: m0, m1, and m2. We know that there are three networks directly connected to router R1. We know that there are two networks indirectly connected to R1. There must be at least three other routers involved (see next-hop column). We know to which networks these routers are connected by looking at their IP addresses. So we can put them at their appropriate place.

See Next Slide

(Continued)



We know that one router, the default router, is connected to the rest of the Internet. But there is some missing information. We do not know if network 130.4.8.0 is directly connected to router R2 or through a point-to-point network (WAN) and another router. We do not know if network140.6.12.64 is connected to router R3 directly or through a point-to-point network (WAN) and another router. Point-to-point networks normally do not have an entry in the routing table because no hosts are connected to them. Figure 6.14 shows our guessed topology.

See Next Slide


Figure 6.14 Guessed topology for Example 6


Example Show a routing table for a host that is totally isolated

Answer A host that is totally isolated needs no routing table. The routing table has no

entries.


Example

Show a routing table for a host that is connected to a LAN without being connected to the Internet

AnswerA routing table for a LAN not connected to the Internet and with no subnets can have a routing table with host-specific addresses. There is no next-hop address since all packets remain within the network.


Homework

Go To the following website

http://highered.mcgraw-hill.com/sites/0072967722/

Complete the Chapter 4-6 quizzes.



Documents

Chapter_04_05_06