Internet Technology

Chapter 5

IPv4
Addresses

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

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

Chapter 5.1 Introduction
Outline
5.2 Classful Addressing

5.3 Classless Addressing

5.4 Special Addresses

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
Notation
Range of Addresses
Operations


Note

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.


Figure 5.1 Dotted-decimal notation


Example 5.1
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 5.2
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 5.3
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 5.4
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 810B0BEF 16
b. 0XC1831BFF or C1831BFF 16

Example 5.5
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 5.6
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.


Figure 5.2 Bitwise NOT operation


Example 5.7


Figure 5.3 Bitwise AND operation


Example 5.8


Figure 5.4 Bitwise OR operation


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.

Classes
Classes and Blocks
Two-Level Addressing
 Three-Level Addressing: Subnetting
 Supernetting


Figure 5.5 Occupation of address space


Figure 5.6 Finding the class of address


Figure 5.7 Finding the class of an address using continuous checking

1 1 1 1
Start
0 0 0 0

Class: A Class: B Class: C Class: D Class: E


Example 5.10
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 5.11
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
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.


Figure 5.8 Netid and hostid


Figure 5.9 Blocks in Class A


Note

Millions of class A addresses
are wasted.


Figure 5.10 Blocks in Class B


Note

Many class B addresses are wasted.


Figure 5.11 Blocks in Class C


Note

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


Figure 5.12 The single block in Class D


Note

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


Figure 5.13 The single block in Class E


Note

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.


Figure 5.14 Two-level addressing in classful addressing


Example 5.12


Figure 5.15 Information extraction in classful addressing

netid 000 ... 0
First address


Example 5.13
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 = 2 32−n
= 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.

Figure 5.16 Solution to Example 5.13


Example 5.14
An address in a block is given as 180.8.17.9. Find the
number of addresses in the block, the first address,
and the last address.

Solution
= 65,536.
bits and set the rightmost 16 bits all to 0s. The
first address is 18.8.0.0/16, in which 16 is the value
of n .
address
TCP/IP Protocol Suite 45 is

Example 5.15
An address in a block is given as 200.11.8.45. Find
the number of addresses in the block, the first
address, and the last address.

Solution
= 256.
bits and set the rightmost 8 bits all to 0s. The first
address is
200.11.8.0/16, in which 24 is the value of n .
address
TCP/IP Protocol Suite 47 is

Figure 5.19 Sample Internet


Note

The network address is the identifier of
a network.


Figure 5.20 Network addresses


Figure 5.21 Network mask


Figure 5.22 Finding a network address using the default mask


Example 5.16
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 5.17
Three-level addressing can be found in the telephone
system if we think about the local part of a telephone
number as an exchange and a subscriber connection:

in which 626 is the area code, 358 is the exchange,
and 1301 is the subscriber connection.


Example 5.18
Figure 5.23 shows a network using class B addresses
before subnetting. We have just one network with
almost 216 hosts. The whole network is connected,
through one single connection, to one of the routers
in the Internet. Note that we have shown /16 to show
the length of the netid (class B).


Figure 5.23 Example 5.18


Example 5.19
Figure 5.24 shows the same network in Figure 5.23
after subnetting. The whole network is still connected
to the Internet through the same router. However, the
network has used a private router to divide the
network into four subnetworks. The rest of the
Internet still sees only one network; internally the
network is made of four subnetworks. Each
subnetwork can now have almost 214 hosts. The
network can belong to a university campus with four
different schools (buildings). After subnetting, each
school has its own subnetworks, but still the whole
campus is one network for the rest of the Internet.
Note that /16 and /18 show the length of the netid and
subnetids.


Figure 5.25 Network mask and subnetwork mask


Example 5.20
In Example 5.19, we divided a class B network into
four subnetworks. The value of n = 16 and the value
of
n 1 = n 2 = n 3 = n 4 = 16 + log 2 4 = 18.
This means that the subnet mask has eighteen 1s
and fourteen 0s. In other words, the subnet mask is
255.255.192.0 which is different from the network
mask for class B (255.255.0.0).


Example 5.21
In Example 5.19, we show that a network is divided
into four subnets. Since one of the addresses in
subnet 2 is 141.14.120.77, we can find the subnet
address as:

The values of the first, second, and fourth bytes are
calculated using the first short cut for AND operation.
The value of the third byte is calculated using the
second short cut for the AND operation.


Figure 5.26 Comparison of subnet, default, and supernet mask


5-3 CLASSLESS ADDRESSING

Subnetting and supernetting in classful
addressing did not really solve the address
depletion problem. With the growth of the
Internet, it was clear that a larger address
space was needed as a long-term solution.
Although the long-range solution has already
been devised and is called IPv6, a short-term
solution was also devised to use the same
address space but to change the distribution
of addresses to provide a fair share to each
organization. The short-term solution still uses
IPv4 addresses, but it is called classless

Variable –Length Blocks
Two-Level Addressing
 Block Allocation
 Subnetting


Figure 5.27 Variable-length blocks in classless addressing


Note

In classless addressing, the prefix
defines the network and the suffix
defines the host.


Figure 5.28 Prefix and suffix


Note

The prefix length in classless
addressing can be 1 to 32.


Example 5.22
What is the prefix length and suffix length if the whole
Internet is considered as one single block with
4,294,967,296 addresses?

Solution
In this case, the prefix length is 0 and the suffix
length is 32. All 32 bits vary to define 232 =
4,294,967,296 hosts in this single block.


Example 5.23
What is the prefix length and suffix length if the
Internet is divided into 4,294,967,296 blocks and
each block has one single address?

Solution
In this case, the prefix length for each block is 32 and
the suffix length is 0. All 32 bits are needed to define
232 = 4,294,967,296 blocks. The only address in
each block is defined by the block itself.


Example 5.24
The number of addresses in a block is inversely
related to the value of the prefix length, n . A small n
means a larger block; a large n means a small block.


Figure 5.29 Slash notation


Note

In classless addressing, we need to
know one of the addresses in the block
and the prefix length to define the block.


Example 5.25
In classless addressing, an address cannot per se
define the block the address belongs to. For example,
the address 230.8.24.56 can belong to many blocks
some of them are shown below with the value of the
prefix associated with that block:


Example 5.26
The following addresses are defined using slash
notations.

a. In the address 12.23.24.78/8, the network mask is
255.0.0.0. The mask has eight 1s and twenty-four
0s. The prefix length is 8; the suffix length is 24.

b. In the address 130.11.232.156/16, the network
mask is 255.255.0.0. The mask has sixteen 1s and
sixteen 0s.The prefix length is 16; the suffix length
is 16.

c. In the address 167.199.170.82/27, the network
mask is 255.255.255.224. The mask has twenty-
seven 1s and five 0s. The prefix length is 27; the
TCP/IP Protocol Suite 5.
suffix length is 76

Example 5.27
One of the addresses in a block is
167.199.170.82/27. Find the number of addresses in
the network, the first address, and the last address.

Solution
The value of n is 27. The network mask has twenty-
seven 1s and five 0s. It is 255.255.255.240.
a. The number of addresses in the network is 2 32 − n =
32.
b. We use the AND operation to find the first address
(network address). The first address is
167.199.170.64/27.


Example 5.27 Continued
c. To find the last address, we first find the
complement of the network mask and then OR it
with the given address: The last address is
167.199.170.95/27.


Example 5.28
One of the addresses in a block is 17.63.110.114/24.
Find the number of addresses, the first address, and
the last address in the block.

Solution
The network mask is 255.255.255.0.
a. The number of addresses in the network is 2 32 − 24 =
256.
b. To find the first address, we use the short cut
methods
discussed early in the chapter. The first address
is
17.63.110.0/24.


c. To find the last address, we use the complement of
the network mask and the first short cut method we
discussed before. The last address is
17.63.110.255/24.


Example 5.29
One of the addresses in a block is 110.23.120.14/20.
Find the number of addresses, the first address, and
the last address in the block.

Solution
The network mask is 255.255.240.0.
a. The number of addresses in the network is 2 32 − 20 =
4096.
b. To find the first address, we apply the first short
cut to bytes 1, 2, and 4 and the second short cut
to byte 3. The first address is 110.23.112.0/20.


c. To find the last address, we apply the first short
cut to bytes 1, 2, and 4 and the second short cut
to byte 3. The OR operation is applied to the
complement of the mask. The last address is
110.23.127.255/20.


Example 5.30
An ISP has requested a block of 1000 addresses. The
following block is granted.

a. Since 1000 is not a power of 2, 1024 addresses
are granted (1024 = 210).
b. The prefix length for the block is calculated as n =
32 −
log21024 = 22.
c. The beginning address is chosen as 18.14.12.0
(which is
divisible by 1024).

The granted block is 18.14.12.0/22. The first address
is 18.14.12.0/22 and the last address is
18.14.15.255/22.

Example 5.31
Assume an organization has given a class A block as
73.0.0.0 in the past. If the block is not revoked by the
authority, the classless architecture assumes that the
organization has a block 73.0.0.0/8 in classless
addressing.


Note

The restrictions applied in allocating
addresses for a subnetwork are
parallel to the ones used to allocate
addresses for a network.


Example 5.32
An organization is granted the block 130.34.12.64/26.
The organization needs four subnetworks, each with
an equal number of hosts. Design the subnetworks
and find the information about each network.
Solution
The number of addresses for the whole network can
be found as N = 2 32 − 26 = 64. The first address in the
network is 130.34.12.64/26 and the last address is
130.34.12.127/26. We now design the subnetworks:
1. We grant 16 addresses for each subnetwork to
meet the first requirement (64/16 is a power of 2).
2. The subnetwork mask for each subnetwork is:


3. We grant 16 addresses to each subnet starting
from the first available address. Figure 5.30 shows
the subblock for each subnet. Note that the
starting address in each subnetwork is divisible
by the number of addresses in that subnetwork.


Example 5.33
An organization is granted a block of addresses with
the beginning address 14.24.74.0/24. The
organization needs to have 3 subblocks of addresses
to use in its three subnets as shown below:
❑ One subblock of 120 addresses.
Solution
There are 2 32 − 24 = 256 addresses in this block. The
first address is 14.24.74.0/24; the last address is
14.24.74.255/24.
a. The number of addresses in the first subblock is
not a power of 2. We allocate 128 addresses.
The subnet
mask is 25. The first address is 14.24.74.0/25; the
last

b. The number of addresses in the second subblock is
not a power of 2 either. We allocate 64 addresses.
The subnet mask is 26. The first address in this
block is 14.24.74.128/26; the last address is
14.24.74.191/26.
c. The number of addresses in the third subblock is
not a power of 2 either. We allocate 16
addresses. The subnet mask is 28. The first
address in this block is 14.24.74.192/28; the last
address is 14.24.74.207/28.
d. If we add all addresses in the previous subblocks,
the result is 208 addresses, which means 48
addresses are left in reserve. The first address in
this range is 14.24.74.209. The last address is
14.24.74.255.
e. Figure 5.31 shows the configuration of blocks. We

Example 5.34
Assume a company has three offices: Central, East,
and West. The Central office is connected to the East
and West offices via private, 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
other offices.
1. The number of addresses are assigned as follows:

2. We can find the prefix length for each subnetwork:


3. Figure 5.32 shows the configuration designed by
the management. The Central office uses addresses
70.12.100.128/27 to 70.12.100.159/27. The company
has used three of these addresses for the routers and
has reserved the last address in the subblock. The
East office
uses the addresses 70.12.100.160/28 to
70.12.100.175/28. One of these addresses is used for
the router and the company has reserved the last
address in the subblock. The West office uses the
addresses 70.12.100.160/28 to 70.12.100.175/28.
One of these addresses is used for the router and the
company has reserved the last address in the
subblock. The company uses no address for the
point-to-point connections in WANs.

Example 5.35
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:
❑ The first group has 64 customers; each needs
approximately 256 addresses.
❑ The second group has 128 customers; each needs
❑ The third group has 128 customers; each needs
We design the subblocks and find out how many
addresses are still available after these allocations.


Solution
Let us solve the problem in two steps. In the first
step, we allocate a subblock of addresses to each
group. The total number of addresses allocated to
each group and the prefix length for each subblock
can found as

Figure 5.33 shows the design for the first hierarchical
level. Figure 5.34 shows the second level of the
hierarchy. Note that we have used the first address
for each customer as the subnet address and have
reserved the last address as a special address.

Figure 5.33 Solution to Example 5.35: first step


Figure 5.34 Solution to Example 5.35: second step


5-4 SPECIAL ADDRESSES

In classful addressing some addresses were
reserved for special purposes. The classless
addressing scheme inherits some of these
special addresses from classful addressing.


 Special Blocks
 Special Addresses in each Block


Figure 5.35 Example of using the all-zero address

Source: 0.0.0.0
Destination: 255.255.255.255

Packet


Figure 5.36 Example of limited broadcast address

Network
221.45.71.64/24 221.45.71.126/24

Packet Destination IP address:
255.255.255.255

221.45.71.20/24 221.45.71.178/24

Router blocks
the packet


Figure 5.37 Example of loopback address

Process 1 Process 2 Application layer

Transport layer

Packet Network layer

Destination address:127.x.y.z


Figure 5.38 Example of a directed broadcast address

Network: 221.45.71.0/24
221.45.71.64/24 221.45.71.126/24

221.45.71.20/24 221.45.71.178/24
Prefix Suffix
Packet Specific All 1s

Destination IP address:
221.45.71.255


5-5 NAT

The distribution of addresses through ISPs
has created a new problem. If the business
grows or the household needs a larger
range, the ISP may not be able to grant the
demand because the addresses before and
after the range may have already been
allocated to other networks. In most
situations, however, only a portion of
computers in a small network need access to
the Internet simultaneously. A technology
that can help in this cases is network
address translation (NAT) .

 Address Translation
 Translation Table


Figure 5.39 NAT


Figure 5.40 Address resolution

172.18.3.1 Source: 172.18.3.1 Source: 200.24.5.8

172.18.3.2
Internet

172.18.3.20
Destination: 172.18.3.1 Destination: 200.24.5.8
Site using private addresses


Figure 5.41 Translation


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