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TCP/IP Protocol Suite 1
Copyright 漏 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Chapter 5
IPv4
Addresses

OBJECTIVES:
OBJECTIVES:
飦� To introduce the concept of an address space in general and the
address space of IPv4 in particular.
飦� To discuss the classful architecture and the blocks of addresses
available in each class.
飦� To discuss the idea of hierarchical addressing and how it has
been implemented in classful addressing.
飦� To explain subnetting and supernetting for classful architecture.
飦� To discuss classless addressing, that has been devised to solve the
problems in classful addressing.
飦� To discuss some special blocks and some special addresses in
each block.
飦� To discuss NAT technology and show how it can be used to
alleviate of address depletion.

Chapter
Chapter
Outline
Outline
5.1 Introduction
5.1 Introduction
5.2 Classful Addressing
5.2 Classful Addressing
5.3 Classless Addressing
5.3 Classless Addressing
5.4 Special Addresses
5.4 Special Addresses
5.5 NAT
5.5 NAT

5-1 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 IPv4 address is a 32-bit address that
uniquely and universally defines the connection of a
host or a router to the Internet; an IP address is the
address of the interface.

Topics Discussed in the Section
飪糔otation
飪糝ange of Addresses
飪糘perations

An IPv4 address is 32 bits long.
Note
The IPv4 addresses are unique
and universal.
Note

The address space of IPv4 is
232
or 4,294,967,296.
Note
Numbers in base 2, 16, and 256 are
discussed in Appendix B.
Note

Figure 5.1 Dotted-decimal notation

Change the following IPv4 addresses from binary notation to
dotted-decimal notation.
a. 10000001 00001011 00001011 11101111
b. 11000001 10000011 00011011 11111111
c. 11100111 11011011 10001011 01101111
d. 11111001 10011011 11111011 00001111
Solution
We 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. 193.131.27.255
c. 231.219.139.111
d. 249.155.251.15
Example
Example 5.1

Change the following IPv4 addresses from dotted-decimal
notation to binary notation.
a. 111.56.45.78
b. 221.34.7.82
c. 241.8.56.12
d. 75.45.34.78
Solution
We replace each decimal number with its binary equivalent:
a. 01101111 00111000 00101101 01001110
b. 11011101 00100010 00000111 01010010
c. 11110001 00001000 00111000 00001100
d. 01001011 00101101 00100010 01001110
Example
Example 5.2

Find the error, if any, in the following IPv4 addresses:
a. 111.56.045.78
b. 221.34.7.8.20
c. 75.45.301.14
d. 11100010.23.14.67
Solution
a. There should be no leading zeroes (045).
b. We may not have more than 4 bytes in an IPv4 address.
c. Each byte should be less than or equal to 255.
d. A mixture of binary notation and dotted-decimal notation.
Example
Example 5.3

Change the following IPv4 addresses from binary notation to
hexadecimal notation.
a. 10000001 00001011 00001011 11101111
b. 11000001 10000011 00011011 11111111
Solution
We replace each group of 4 bits with its hexadecimal
equivalent. Note that 0X (or 0x) is added at the beginning or the
subscript 16 at the end.
a. 0X810B0BEF or 810B0BEF16
b. 0XC1831BFF or C1831BFF16
Example
Example 5.4

Find the number of addresses in a range if the first address is
146.102.29.0 and the last address is 146.102.32.255.
Solution
We can subtract the first address from the last address in base
256 (see Appendix B). The result is 0.0.3.255 in this base. To
find the number of addresses in the range (in decimal), we
convert this number to base 10 and add 1 to the result..
Example
Example 5.5

The first address in a range of addresses is 14.11.45.96. If the
number of addresses in the range is 32, what is the last
address?
Solution
We convert the number of addresses minus 1 to base 256,
which is 0.0.0.31. We then add it to the first address to get the
last address. Addition is in base 256.
Example
Example 5.6

Figure 5.2 Bitwise NOT operation

Example
Example 5.7

Figure 5.3 Bitwise AND operation

Example
Example 5.8

Figure 5.4 Bitwise OR operation

Example
Example 5.9

5-2 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 that supersedes the original architecture.
In this section, we introduce classful addressing
because it paves the way for understanding
classless addressing and justifies the rationale for
moving to the new architecture. Classless
addressing is discussed in the next section.

飪糃lasses
飪糃lasses and Blocks
飪糡wo-Level Addressing
飪� Three-Level Addressing: Subnetting
飪� Supernetting

In classful addressing,
In classful addressing,
the address space is
the address space is
divided into five classes:
divided into five classes:
A
A,
, B
B,
, C
C,
, D
D, and
, and E
E.
.

Figure 4.2 Occupation of the address space

Figure 5.5 Occupation of address space

Figure 4.3 Finding the class in binary notation

Figure 5.7 Finding the class of an address using continuous checking
1
Class: A
0
Start
1
0
Class: B
1
0
Class: C
1
0
Class: D Class: E

How can we prove that we have 2,147,483,648 addresses in
class A?
Example 5
Solution
In 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 11101111
b. 11000001 10000011 00011011 11111111
c. 10100111 11011011 10001011 01101111
d. 11110011 10011011 11111011 00001111
Solution
See the procedure in Figure 5.7.
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 1; the second bit is 0. This is a class B
address.
d. The first 4 bits are 1s. This is a class E address.
Example
Example 5.10

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. 14.23.120.8
d. 252.5.15.111
e.134.11.78.56
Solution
a. 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.
Example
Example 5.11

In Example 5 we showed that class A has 231
(2,147,483,648)
addresses. How can we prove this same fact using dotted-
decimal notation?
Example 8
Solution
The addresses in class A range from 0.0.0.0 to
127.255.255.255. We need to show that the difference between
these two numbers is 2,147,483,648. This is a good exercise
because it shows us how to define the range of addresses
between two addresses. We notice that we are dealing with base
256 numbers here. Each byte in the notation has a weight. The
weights are as follows (see Appendix B):
See Next 狠狠撸

2563
, 2562
, 2561
, 2560
Example 8 (continued)
Last address: 127 脳 2563
+ 255 脳 2562
+
255 脳 2561
+ 255 脳 2560
= 2,147,483,647
First address: = 0
Now to find the integer value of each number, we multiply
each byte by its weight:
If we subtract the first from the last and add 1 to the result
(remember we always add 1 to get the range), we get
2,147,483,648 or 231
.

Figure 5.8 Netid and hostid

Millions of class A addresses
are wasted.
Note

Many class B addresses are wasted.
Note

Figure 4.9 Blocks in class C

Not so many organizations are so small
to have a class C block.
Note

Figure 5.12 The single block in Class D

Class D addresses are made of one
block, used for multicasting.
Note

Figure 5.13 The single block in Class E

The only block of class E addresses was
reserved for future purposes.
Note

The range of addresses allocated to an
organization in classful addressing
was a block of addresses in
Class A, B, or C.
Note

Figure 5.14 Two-level addressing in classful addressing

Example
Example 5.12

Figure 5.15 Information extraction in classful addressing
netid
First address
000 ... 0

Network Addresses
Network Addresses
The network address is the first address.
The network address defines the network to
the rest of the Internet.
Given the network address, we can find the
class of the address, the block, and the range
of the addresses in the block

An address in a block is given as 73.22.17.25. Find the number
of addresses in the block, the first address, and the last
address.
Solution
Figure 5.16 shows a possible configuration of the network that
uses this block.
1. The number of addresses in this block is N = 232鈭抧
=
16,777,216.
2. To find the first address, we keep the leftmost 8 bits and set
the rightmost 24 bits all to 0s. The first address is
73.0.0.0/8, in which 8 is the value of n.
3. To find the last address, we keep the leftmost 8 bits and set
the rightmost 24 bits all to 1s. The last address is
73.255.255.255.
Example
Example 5.13

Figure 5.16 Solution to Example 5.13

address.
Solution
uses this block.
=
65,536.
18.8.255.255.
Example
Example 5.14

address.
Solution
uses this block.
= 256.
200.11.8.255/24.
Example
Example 5.15

Figure 5.19 Sample Internet

The network address is the identifier of
a network.
Note

Figure 5.20 Network addresses

Mask
Mask
A mask is a 32-bit binary number that gives
the first address in the block (the network
address) when bitwise ANDed with an
address in the block.

Figure 5.21 Network mask

Figure 5.22 Finding a network address using the default mask

A router receives a packet with the destination address
201.24.67.32. Show how the router finds the network address of
the packet.
Solution
Since the class of the address is B, we assume that the router
applies the default mask for class B, 255.255.0.0 to find the
network address.
Example
Example 5.16

Example 12
Example 12
Given the address 23.56.7.91 and the
default class A mask, find the beginning
address (network address).
Solution
Solution
The 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
Example 13
default class B mask, find the beginning
Solution
Solution
The 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.

Example 14
Example 14
class C default mask, find the beginning
Solution
Solution
The default mask is 255.255.255.0,
which means that the first 3 bytes are
preserved and the last byte is set to 0.

We must not
We must not
apply the default mask
apply the default mask
of one class to
of one class to
an address belonging
an address belonging
to another class.
to another class.

Figure 4-12
Multihomed devices

221.45.71.0/24
221.45.71.20/24 221.45.71.178/24
221.45.71.64/24 221.45.71.126/24
Network:
Figure 5.38 Example of a directed broadcast address
Packet
Destination IP address:
221.45.71.255
Specific
Prefix Suffix
All 1s

221.45.71.20/24 221.45.71.178/24
221.45.71.64/24 221.45.71.126/24
Network
Figure 5.36 Example of limited broadcast address
Destination IP address:
255.255.255.255
Packet
Router blocks
the packet

Figure 5.35 Example of this host on this address
Source: 0.0.0.0
Destination: 255.255.255.255
Packet
DHCP Server

Figure 4-17
Example of specific host on this network

Figure 4-18
Example of loopback address
Transport layer
Application layer
Network layer
Process 1 Process 2
Destination address:127.x.y.z
Packet

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