Understanding IP Addressing

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W H I T E PA P E R

Understanding

IP Addressing:

Everything You

Ever Wanted To

Know

Page 2

Understanding IP Addressing: Everything

You Ever Wanted To Know

CONTENTS

Internet Scaling Problems

Classful IP Addressing

Subnetting 7

Variable Length Subnet Masks (VLSM)

Classless Inter-Domain Routing (CIDR)

New Solutions for Scaling the Internet Address Space

IPv6 Resolves IPv4 Issues

Additional IPv6 Features

Keeping Current on Internet Addressing Issues

Appendix A - References

Appendix B - Classful IP Addressing

Appendix C - Subnetting Exercises

Appendix D - VLSM Exercise

Appendix E - CIDR Exercises

I I I

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Understanding IP Addressing:

Everything You Ever Wanted To Know

The Internet continues to grow at a phenomenal rate. This is reflected in

the tremendous popularity of the World Wide Web (WWW), the opportu-

nities that businesses see in reaching customers from virtual storefronts,

and the emergence of new ways of doing business. It is clear that expanding

business and public awareness will continue to increase demand for access

to resources on the Internet.

Internet Scaling Problems

Over the past few years, the Internet has experienced two major scaling

issues as it has struggled to provide continuous and uninterrupted

growth:

•

The eventual exhaustion of IP version 4 (IPv4) address space

•

The need to route traffic between the ever increasing number of net-

works that comprise the Internet

The first problem is concerned with the eventual depletion of the IP

address space. IPv4 defines a 32-bit address which means that there are

only 232 (4,294,967,296) IPv4 addresses available. As the Internet con-

tinues to grow, this finite number of IP addresses will eventually be

exhausted.

The address shortage problem is aggravated by the fact that portions of

the IP address space have not been efficiently allocated. Also, the tradi-

tional model of classful addressing does not allow the address space to

be used to its maximum potential. The Address Lifetime Expectancy

(ALE) Working Group of the Internet Engineering Task Force (IETF) has

expressed concerns that if the current address allocation policies are not

modified, the Internet will experience a near to medium term exhaus-

tion of its unallocated address pool. If the Internet’s address supply

problem is not solved, new users may be unable to connect to the global

Internet. More than half of all possible IPv4 addresses have been

assigned to ISPs, corporations, and government agencies, but only an

estimated 69 million addresses are actually in use.

F I G U R E 1 . N e t w o r k N u m b e r G ro w t h

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The second problem is caused by the rapid growth in the size of the

Internet routing tables. Internet backbone routers are required to main-

tain complete routing information for the Internet. Over recent years,

routing tables have experienced exponential growth as increasing num-

bers of organizations connect to the Internet. In December 1990 there

were 2,190 routes, in December 1995 there were more than 30,000

routes, and in December 2000 more than 100,000 routes.

Unfortunately, the routing problem cannot be solved by simply

installing more router memory and increasing the size of the routing

tables. Other factors related to the capacity problem include the grow-

ing demand for CPU horsepower to compute routing table/topology

changes, the increasingly dynamic nature of WWW connections and

their effect on router forwarding caches, and the sheer volume of infor-

mation that needs to be managed by people and machines. If the num-

ber of entries in the global routing table is allowed to increase without

bounds, core routers will be forced to drop routes and portions of the

Internet will become unreachable.

The long-term solution to these problems can be found in the wide-

spread deployment of IP Next Generation (IPng or IPv6). Currently,

IPv6 is being tested and implemented on the 6Bone network, which is

an informal collaborative project covering North America, Europe, and

Japan. 6Bone supports the routing of IPv6 packets, since that function

has not yet been integrated into many production routers. Until IPv6

can be deployed worldwide, IPv4 patches will need to be used and

modified to continue to provide the universal connectivity users have

come to expect.

U N D E R S TA N D I N G I P A D D R E S S I N G

F I G U R E 2 . G ro w t h o f I n t e r n e t R o u t i n g Ta b l e s

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Classful IP Addressing

When IP was first standardized in September 1981, the specification

required that each system attached to an IP-based Internet be assigned

a unique, 32-bit Internet address value. Systems that have interfaces to

more than one network require a unique IP address for each network

interface. The first part of an Internet address identifies the network on

which the host resides, while the second part identifies the particular

host on the given network. This creates the two-level addressing hierar-

chy that is illustrated in Figure 3.

In recent years, the network number field has been referred to as the

network prefix because the leading portion of each IP address identifies

the network number. All hosts on a given network share the same net-

work prefix but must have a unique host number. Similarly, any two

hosts on different networks must have different network prefixes but

may have the same host number.

Primary Address Classes

To provide the flexibility required to support networks of varying sizes,

the Internet designers decided that the IP address space should be

divided into three address classes-Class A, Class B, and Class C. This is

often referred to as classful addressing. Each class fixes the boundary

between the network prefix and the host number at a different point

within the 32-bit address. The formats of the fundamental address

classes are illustrated in Figure 4.

F I G U R E 3 . Tw o - L e v e l I n t e r n e t A d d re s s S t r u c t u re

F I G U R E 4 . P r i n c i p l e C l a s s f u l I P A d d re s s F o r m a t s

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One of the fundamental features of classful IP addressing is that each

address contains a self-encoding key that identifies the dividing point

between the network prefix and the host number. For example, if the

first two bits of an IP address are 1-0, the dividing point falls between

the 15th and 16th bits. This simplified the routing system during the

early years of the Internet because the original routing protocols did not

supply a deciphering key or mask with each route to identify the length

of the network prefix.

Class A Networks (/8 Prefixes)

Each Class A network address has an 8-bit network prefix, with the

highest order bit set to 0 (zero) and a 7-bit network number, followed

by a 24-bit host number. Today, Class A networks are referred to as

“/8s” (pronounced “slash eight” or just “eights”) since they have an 8-

bit network prefix.

A maximum of 126 (27 -2) /8 networks can be defined. The calculation

subtracts two because the /8 network 0.0.0.0 is reserved for use as the

default route and the /8 network 127.0.0.0 (also written 127/8 or

127.0.0.0/8) is reserved for the “loopback” function. Each /8 supports a

maximum of 224 -2

(16,777,214) hosts per network. The host calculation subtracts two

because the all-0s (all zeros or “this network”) and all-1s (all ones or

“broadcast”) host numbers may not be assigned to individual hosts.

Since the /8 address block contains 231 (2,147,483,648 ) individual

addresses and the IPv4 address space contains a maximum of 232

(4,294,967,296) addresses, the /8 address space is 50 percent of the total

IPv4 unicast address space.

Class B Networks (/16 Prefixes)

Each Class B network address has a 16-bit network prefix, with the two

highest order bits set to 1-0 and a 14-bit network number, followed by a

16-bit host number. Class B networks are now referred to as “/16s” since

they have a 16-bit network prefix.

A maximum of 16,384 (214 ) /16 networks can be defined with up to

65,534 (216-2) hosts per network. Since the entire /16 address block

contains 230 (1,073,741,824) addresses, it represents 25 percent of the

total IPv4 unicast address space.

Class C Networks (/24 Prefixes)

Each Class C network address has a 24-bit network prefix, with the

three highest order bits set to 1-1-0 and a 21-bit network number, fol-

lowed by an 8-bit host number. Class C networks are now referred to as

“/24s” since they have a 24-bit network prefix.

A maximum of 2,097,152 (221 ) /24 networks can be defined with up to

254 (28-2) hosts per network. Since the entire /24 address block con-

tains 229 (536,870,912) addresses, it represents 12.5 percent (or one-

eighth) of the total IPv4 unicast address space.

U N D E R S TA N D I N G I P A D D R E S S I N G

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Other Classes

In addition to the three most popular classes, there are two additional

classes. Class D addresses have their leading four bits set to 1-1-1-0 and

are used to support IP Multicasting. Class E addresses have their leading

four bits set to 1-1-1-1 and are reserved for experimental use.

Dotted-Decimal Notation

To make Internet addresses easier for people to read and write, IP

addresses are often expressed as four decimal numbers, each separated

by a dot. This format is called “dotted-decimal notation.”

Dotted-decimal notation divides the 32-bit Internet address into four 8-

bit fields and specifies the value of each field independently as a deci-

mal number with the fields separated by dots. Figure 5 shows how a

typical /16 (Class B) Internet address can be expressed in dotted-decimal

notation.

Table 1 displays the range of dotted-decimal values that can be assigned

to each of the three principle address classes. The “xxx” represents the

host number field of the address that is assigned by the local network

administrator.

F I G U R E 5 . D o t t e d D e c i m a l N o t a t i o n

TA B L E 1 . D o t t e d D e c i m a l R a n g e s f o r E a c h A d d re s s C l a s s

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Unforeseen Limitations to Classful Addressing

The original Internet designers never envisioned that the Internet

would grow into what it has become today. Many of the problems that

the Internet is facing today can be traced back to the early decisions

that were made during its formative years.

•

During the early days of the Internet, the seemingly unlimited

address space allowed IP addresses to be allocated to an organization

based on its request rather than its actual need. As a result, addresses

were freely assigned to those who asked for them without concerns

about the eventual depletion of the IP address space.

•

The decision to standardize on a 32-bit address space meant that there

were only 232 (4,294,967,296) IPv4 addresses available. A decision to

support a slightly larger address space would have exponentially

increased the number of addresses thus eliminating the current

address shortage problem.

•

The classful A, B, and C octet boundaries were easy to understand

and implement, but they did not foster the efficient allocation of a

finite address space. Problems resulted from the lack of a network

class that was designed to support medium-sized organizations. For

example, a /24, which supports 254 hosts, is too small while a /16,

which supports 65,534 hosts, is too large. In the past, sites with sev-

eral hundred hosts were assigned a single /16 address instead of two

/24 addresses. This resulted in a premature depletion of the /16 net-

work address space. Now the only readily available addresses for

medium-sized organizations are /24s, which have the potentially nega-

tive impact of increasing the size of the global Internet’s routing table.

Figure 6 shows basic class A, B, and C networks.

U N D E R S TA N D I N G I P A D D R E S S I N G

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The subsequent history of Internet addressing involved a series of steps

that overcame these addressing issues and supported the growth of the

global Internet.

Additional Practice with Classful Addressing

Appendix B provides exercises using Classful IP Addressing.

F I G U R E 6 . B a s i c C l a s s A , B , a n d C N e t w o r k s

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U N D E R S TA N D I N G I P A D D R E S S I N G

Subnetting

In 1985, RFC 950 defined a standard procedure to support the subnet-

ting, or division, of a single Class A, B, or C network number into

smaller pieces. Subnetting was introduced to overcome some of the

problems that parts of the Internet were beginning to experience with

the classful two-level addressing hierarchy, such as:

•

Internet routing tables were beginning to grow.

•

Local administrators had to request another network number from the

Internet before a new network could be installed at their site.

Both of these problems were attacked by adding another level of hierar-

chy to the IP addressing structure. Instead of the classful two-level hier-

archy, subnetting supports a three-level hierarchy. Figure 7 illustrates

the basic idea of subnetting, which is to divide the standard classful

host number field into two parts-the subnet number and the host num-

ber on that subnet.

Subnetting attacked the expanding routing table problem by ensuring

that the subnet structure of a network is never visible outside of the

organization’s private network. The route from the Internet to any sub-

net of a given IP address is the same, no matter which subnet the desti-

nation host is on. This is because all subnets of a given network number

use the same network prefix but different subnet numbers. The routers

within the private organization need to differentiate between the indi-

vidual subnets, but as far as the Internet routers are concerned, all of

the subnets in the organization are collected into a single routing table

entry. This allows the local administrator to introduce arbitrary com-

plexity into the private network without affecting the size of the Inter-

net’s routing tables.

Subnetting overcame the registered number issue by assigning each

organization one (or at most a few) network numbers from the IPv4

address space. The organization was then free to assign a distinct sub-

network number for each of its internal networks. This allowed the

organization to deploy additional subnets without obtaining a new net-

work number from the Internet.

F I G U R E 7 . S u b n e t A d d re s s H i e r a rc h y

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In Figure 8, a site with several logical networks uses subnet addressing

with a single /16 (Class B) network address. The router accepts all traffic

from the Internet addressed to network 130.5.0.0, and forwards traffic

to the interior subnetworks based on the third octet of the classful

address. The deployment of subnetting within the private network pro-

vides several benefits:

•

The size of the global Internet routing table does not grow because

the site administrator does not need to obtain additional address space

and the routing advertisements for all of the subnets are combined

into a single routing table entry.

•

The local administrator has the flexibility to deploy additional sub-

nets without obtaining a new network number from the Internet.

•

Route flapping (that is, the rapid changing of routes) within the pri-

vate network does not affect the Internet routing table since Internet

routers do not know about the reachability of the individual subnets-

they just know about the reachability of the parent network number.

Extended Network Prefix

Internet routers use only the network prefix of the destination address

to route traffic to a subnetted environment. Routers within the subnet-

ted environment use the extended network prefix to route traffic

between the individual subnets. The extended network prefix is com-

posed of the classful network prefix and the subnet number.

F I G U R E 9 . E x t e n d e d N e t w o r k P re f i x

F I G U R E 8 . S u b n e t t i n g t h e R o u t i n g R e q u i re m e n t s o f t h e

I n t e r n e t

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U N D E R S TA N D I N G I P A D D R E S S I N G

1 0

The extended network prefix has traditionally been identified by the

subnet mask. For example, if an administrator has the /16 address of

130.5.0.0 and wants to use the entire third octet to represent the subnet

number, the administrator must specify a subnet mask of 255.255.255.0.

The bits in the subnet mask and the Internet address have a one to one

correspondence. The bits of the subnet mask are set to 1 (one) if the sys-

tem examining the address should treat the corresponding bit in the IP

address as part of the extended network prefix. The bits in the mask are

set to 0 (zero) if the system should treat the bit as part of the host num-

ber. This numbering is illustrated in Figure 10.

The standards describing modern routing protocols often refer to the

extended network prefix length rather than the subnet mask. The pre-

fix length is equal to the number of contiguous one-bits in the tradi-

tional subnet mask. This means that specifying the network address

130.5.5.25 with a subnet mask of 255.255.255.0 can also be expressed as

130.5.5.25/24. The /<prefix length> notation is more compact and eas-

ier to understand than writing out the mask in its traditional dotted-

decimal format. This is illustrated in Figure 11.

Note that modern routing protocols still carry the subnet mask. None of

the Internet standard routing protocols have a 1-byte field in the header

that contains the number of bits in the extended network prefix. Each

routing protocol is still required to carry the complete four-octet subnet

mask.

F I G U R E 1 0 . S u b n e t M a s k

F I G U R E 1 1 . E x t e n d e d N e t w o r k P re f i x L e n g t h

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Subnet Design Considerations

The deployment of an addressing plan requires careful thought. Four

key questions that must be answered before any design should be

undertaken are:

1 How many total subnets does the organization need today?

2 How many total subnets will the organization need in the future?

3 How many hosts are on the organization’s largest subnet today?

4 How many hosts will there be on the organization’s largest subnet in

the future?

The first step in the planning process is to take the maximum number of

subnets required and round up to the nearest power of two. For exam-

ple, if an organization needs nine subnets, 23 (or 8) will not provide

enough subnet addressing space, so the network administrator will

need to round up to 24 (or 16).

The network administrator must always allow adequate room for

growth. For example, although 14 subnets are required today, 16 sub-

nets might not be enough in two years when the 17th subnet needs to

be deployed. In this case, it would be wise to select 25 (or 32) as the

maximum number of subnets.

The second step is to ensure that there are enough host addresses for

the organization’s largest subnet. If the largest subnet needs to support

50 host addresses today, 25 (or 32) will not provide enough host address

space so the network administrator will need to round up to 26 (or 64).

The final step is to make sure that the organization’s address allocation

provides enough bits to deploy the required subnet addressing plan.

For example, if the organization has a single /16, it could easily deploy 4

bits for the subnet number and 6 bits for the host number. However, if

the organization has several /24s and it needs to deploy nine subnets, it

may have to subnet each of its /24s into four subnets (using 2 bits) and

then build the network by combining the subnets of three /24 network

numbers.

An alternative solution would be to deploy network numbers from the

private address space (RFC 1918) for internal connectivity and use a

Network Address Translator (NAT) to provide external Internet access.

Subnet Example #1

Given

An organization is assigned the network number 193.1.1.0/24 and it

needs to define six subnets. The largest subnet is required to support 25

hosts.

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U N D E R S TA N D I N G I P A D D R E S S I N G

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Defining the Subnet Mask / Extended Prefix Length

The first step in defining the subnet mask is to determine the number of

bits required to define the six subnets. Since a network address can

only be subnetted along binary boundaries, subnets must be created in

blocks of powers of two [2 (21), 4 (22), 8 (23), 16 (24), and so on]. Thus,

it is impossible to define an IP address block such that it contains

exactly six subnets. For this example, the network administrator must

define a block of 8 (23) and have two unused subnets that can be

reserved for future growth.

Since 8 = 23, three bits are required to enumerate the eight subnets in

the block. In this example, the organization is subnetting a /24 so it will

need three more bits, or a /27, as the extended network prefix. A 27-bit

extended network prefix can be expressed in dotted-decimal notation

as 255.255.255.224. This notation is illustrated in Figure 12.

A 27-bit extended network prefix leaves 5 bits to define host addresses

on each subnet. This means that each subnetwork with a 27-bit prefix

represents a contiguous block of 25 (32) individual IP addresses. How-

ever, since the all-0s and all-1s host addresses cannot be allocated, there

are 30 (25-2) assignable host addresses on each subnet.

Defining the Subnet Numbers

The eight subnets will be numbered 0 through 7. Throughout the

remainder of this paper, the XXX notation indicates the binary repre-

sentation of the number. The 3-bit binary representation of the decimal

values 0 through 7 are: 0 (000 ), 1 (001 ), 2 (010 ), 3 (011 ), 4 (100 ), 5

(101 ), 6 (110 ), and 7 (111 ).

In general, to define Subnet #N, the network administrator places the

binary representation of N into the bits of the subnet number field. For

example, to define Subnet #6, the network administrator simply places

the binary representation of 6 (110 ) into the 3 bits of the subnet num-

ber field.

F I G U R E 1 2 . E x a m p l e # 1 - D e f i n i n g t h e S u b n e t

M a s k / E x t e n d e d P re f i x L e n g t h

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1 3

The eight subnet numbers for this example are listed in the following

code sample. The underlined portion of each address identifies the

extended network prefix, while the bold digits identify the 3 bits repre-

senting the subnet number field:

Base Net: 11000001.00000001.00000001 .00000000 = 193.1.1.0/24

Subnet #0: 11000001.00000001.00000001.000 00000 = 193.1.1.0/27

Subnet #1: 11000001.00000001.00000001.001 00000 = 193.1.1.32/27

Subnet #2: 11000001.00000001.00000001.010 00000 = 193.1.1.64/27

Subnet #3: 11000001.00000001.00000001.011 00000 = 193.1.1.96/27

Subnet #4: 11000001.00000001.00000001.100 00000 = 193.1.1.128/27

Subnet #5: 11000001.00000001.00000001.101 00000 = 193.1.1.160/27

Subnet #6: 11000001.00000001.00000001.110 00000 = 193.1.1.192/27

Subnet #7: 11000001.00000001.00000001.111 00000 = 193.1.1.224/27

An easy way to verify that the subnets are correct is to ensure that they

are all multiples of the Subnet #1 address. In this example, all subnets

are multiples of 32: 0, 32, 64, 96, and so on.

The All-0s Subnet and All-1s Subnet

When subnetting was first defined in RFC 950, it prohibited the use of

the all-0s and the all-1s subnets. The reason for this restriction was to

eliminate situations that could potentially confuse a classful router.

Today a router can be both classless and classful at the same time-it

could be running RIP-1 (classful protocol) and BGP-4 (Border Gateway

Protocol Version 4-a classless protocol) at the same time.

With respect to the all-0s subnet, a router requires that each routing

table update include the route/<prefix length> pair to differentiate

between a route to the all-0s subnet and a route to the entire network.

For example, when using RIP-1which does not supply a mask or prefix

length with each route, the routing advertisements for subnet

193.1.1.0/27 and for network 193.1.1.0/24 are identical-193.1.1.0. With-

out somehow knowing the prefix length or mask, a router cannot tell

the difference between a route to the all-0s subnet and the route to the

entire network. This example is illustrated in Figure 13.

F I G U R E 1 3 . D i ff e re n t i a t i n g B e t w e e n a R o u t e t o t h e A l l - 0 s

S u b n e t a n d t h e E n t i re N e t w o r k

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U N D E R S TA N D I N G I P A D D R E S S I N G

1 4

Regarding the all-1s subnet, a router requires that each routing table

entry include the prefix length so that it can determine whether a

broadcast (directed or all-subnets) should be sent only to the all-1s sub-

net or to the entire network. For example, when the routing table does

not contain a mask or prefix length for each route, confusion can occur

because the same broadcast address (193.1.1.255) is used for both the

entire network 193.1.1.0/24 and the all-1s subnet 193.1.1.224/27. This

issue is illustrated in Figure 14.

Defining Host Addresses for Each Subnet

According to Internet practices, the host number field of an IP address

cannot contain all 0-bits or all 1-bits. The all-0s host number identifies

the base network (or subnetwork) number, while the all-1s host number

represents the broadcast address for the network (or subnetwork).

In our current example, there are 5 bits in the host number field of each

subnet address. This means that each subnet represents a block of 30

host addresses (25 -2 = 30, note that the 2 is subtracted because the

all-0s and the all-1s host addresses cannot be used). The hosts on each

subnet are numbered 1 through 30.

In general, to define the address assigned to Host #N of a particular

subnet, the network administrator places the binary representation of N

into the subnet’s host number field. For example, to define the address

assigned to Host #15 on Subnet #2, the network administrator simply

places the binary representation of 15 (011112 ) into the 5-bits of Sub-

net #2’s host number field.

F I G U R E 1 4 . I d e n t i f y i n g a B ro a d c a s t t o t h e A l l 1 s S u b n e t

a n d t h e E n t i re N e t w o r k

Page 17

The valid host addresses for Subnet #2 in this example are listed in the

following sample code. The underlined portion of each address identi-

fies the extended network prefix, while the bold digits identify the 5-

bit host number field:

Subnet #2: 11000001.00000001.00000001.010 00000 = 193.1.1.64/27

Host #1: 11000001.00000001.00000001.010 00001 = 193.1.1.65/27

Host #2: 11000001.00000001.00000001.010 00010 = 193.1.1.66/27

Host #3: 11000001.00000001.00000001.010 00011 = 193.1.1.67/27

Host #4: 11000001.00000001.00000001.010 00100 = 193.1.1.68/27

Host #5: 11000001.00000001.00000001.010 00101 = 193.1.1.69/27

Host #15: 11000001.00000001.00000001.010 01111 = 193.1.1.79/27

Host #16: 11000001.00000001.00000001.010 10000 = 193.1.1.80/27

Host #27: 11000001.00000001.00000001.010 11011 = 193.1.1.91/27

Host #28: 11000001.00000001.00000001.010 11100 = 193.1.1.92/27

Host #29: 11000001.00000001.00000001.010 11101 = 193.1.1.93/27

Host #30: 11000001.00000001.00000001.010 11110 = 193.1.1.94/27

The valid host addresses for Subnet #6 are listed in the following sam-

ple code. The underlined portion of each address identifies the

extended network prefix, while the bold digits identify the 5-bit host

number field:

Subnet #6: 11000001.00000001.00000001.110 00000 = 193.1.1.192/27

Host #1: 11000001.00000001.00000001.110 00001 = 193.1.1.193/27

Host #2: 11000001.00000001.00000001.110 00010 = 193.1.1.194/27

Host #3: 11000001.00000001.00000001.110 00011 = 193.1.1.195/27

Host #4: 11000001.00000001.00000001.110 00100 = 193.1.1.196/27

Host #5: 11000001.00000001.00000001.110 00101 = 193.1.1.197/27

Host #15: 11000001.00000001.00000001.110 01111 = 193.1.1.207/27

Host #16: 11000001.00000001.00000001.110 10000 = 193.1.1.208/27

Host #27: 11000001.00000001.00000001.110 11011 = 193.1.1.219/27

Host #28: 11000001.00000001.00000001.110 11100 = 193.1.1.220/27

Host #29: 11000001.00000001.00000001.110 11101 = 193.1.1.221/27

Host #30: 11000001.00000001.00000001.110 11110 = 193.1.1.222/27

Defining the Broadcast Address for Each Subnet

The broadcast address for Subnet #2 is the all-1s host address or:

11000001.00000001.00000001.010 11111 = 193.1.1.95

Note that the broadcast address for Subnet #2 is exactly one less than

the base address for Subnet #3 (193.1.1.96). This is always the case-the

broadcast address for Subnet #n is one less than the base address for

Subnet #(n+1).

1 5

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The broadcast address for Subnet #6 is simply the all-1s host address

or:

11000001.00000001.00000001.110 11111 = 193.1.1.223

Again, the broadcast address for Subnet #6 is exactly one less than the

base address for Subnet #7 (193.1.1.224).

Subnet Example #2

Given

An organization is assigned the network number 140.25.0.0/16 and it

must create a set of subnets that supports up to 60 hosts on each subnet.

Defining the Subnet Mask / Extended Prefix Length

The first step is to determine the number of bits required to define 60

hosts on each subnet. Since a block of host addresses can only be

assigned along binary boundaries, host address blocks can only be cre-

ated in powers of two. This means that it is impossible to create a block

that contains exactly 60 host addresses.

To support 60 hosts, the network administrator must define a minimum

address block of 62 (26-2) host addresses. However, this choice would

only provide two unused host addresses on each subnet for future

growth, which is not likely to support additional growth. The network

administrator must define a block of 126 (27-2) host addresses with 66

addresses on each subnet for future growth. A block of 126 host

addresses requires 7 bits in the host number field.

The next step is to determine the subnet mask/extended prefix length.

Since 7 bits of the 32-bit IP address are required for the host number

field, the extended prefix must be a /25 (25 = 32-7). A 25-bit extended

network prefix can be expressed in dotted-decimal notation as

255.255.255.128. This notation is illustrated in Figure 15.

F I G U R E 1 5 . E x a m p l e # 2 - D e f i n i n g t h e S u b n e t

M a s k / E x t e n d e d P re f i x L e n g t h

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1 7

Figure 15 shows that the 25-bit extended prefix assigns 9 bits to the

subnet number field. Since 29 = 512, nine bits allow the definition of

512 subnets. Depending on the organization’s requirements, the net-

work administrator could have elected to assign additional bits to the

host number field (allowing more hosts on each subnet) and reduce the

number of bits in the subnet number field (decreasing the total number

of subnets that can be defined).

Although this example creates a rather large number of subnets, it illus-

trates what happens to the dotted- decimal representation of a subnet

address when the subnet number bits extend across an octet boundary.

Note that the same type of confusion can occur when the host number

bits extend across an octet boundary.

Defining Each of the Subnet Numbers

The 512 subnets will be numbered 0 through 511. The 9-bit binary rep-

resentation of the decimal values 0 through 511 are: 0 (0000000002 ), 1

(0000000012 ), 2 (0000000102 ), 3 (0000000112 ), ..., 511 (1111111112 ).

To define Subnet #3, the network administrator places the binary rep-

resentation of 3 (0000000112 ) into the 9 bits of the subnet number

field. The 512 subnet numbers for this example are listed in the follow-

ing sample code. The underlined portion of each address identifies the

extended network prefix, while the bold digits identify the 9 bits repre-

senting the subnet number field:

Base Net: 10001100.00011001 .00000000.00000000 = 140.25.0.0/16

Subnet #0: 10001100.00011001.00000000.0 0000000 = 140.25.0.0/25

Subnet #1: 10001100.00011001.00000000.1 0000000 = 140.25.0.128/25

Subnet #2: 10001100.00011001.00000001.0 0000000 = 140.25.1.0/25

Subnet #3: 10001100.00011001.00000001.1 0000000 = 140.25.1.128/25

Subnet #4: 10001100.00011001.00000010.0 0000000 = 140.25.2.0/25

Subnet #5: 10001100.00011001.00000010.1 0000000 = 140.25.2.128/25

Subnet #6: 10001100.00011001.00000011.0 0000000 = 140.25.3.0/25

Subnet #7: 10001100.00011001.00000011.1 0000000 = 140.25.3.128/25

Subnet #8: 10001100.00011001.00000100.0 0000000 = 140.25.4.0/25

Subnet #9: 10001100.00011001.00000100.1 0000000 = 140.25.4.128/25

Subnet #510: 10001100.00011001.11111111.0 0000000 = 140.25.255.0/25

Subnet #511: 10001100.00011001.11111111.1 0000000 = 140.25.255.128/25

Note that the sequential subnet numbers are not sequential when

expressed in dotted-decimal notation. This can be confusing to people

who expect dotted-decimal notation to make IP addressing easier. In

this example, the dotted-decimal notation obscures the subnet number-

ing scheme.

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Defining Host Addresses for Each Subnet

In this example there are 7 bits in the host number field of each subnet

address, which means that each subnet represents a block of 126 host

addresses. The hosts on each subnet are numbered 1 through 126.

The valid host addresses for Subnet #3 are listed in the following sam-

ple code. The underlined portion of each address identifies the

extended network prefix, while the bold digits identify the 7-bit host

number field:

Subnet #3: 10001100.00011001.00000001.1 0000000 = 140.25.1.128/25

Host #1: 10001100.00011001.00000001.1 0000001 = 140.25.1.129/25

Host #2: 10001100.00011001.00000001.1 0000010 = 140.25.1.130/25

Host #3: 10001100.00011001.00000001.1 0000011 = 140.25.1.131/25

Host #4: 10001100.00011001.00000001.1 0000100 = 140.25.1.132/25

Host #5: 10001100.00011001.00000001.1 0000101 = 140.25.1.133/25

Host #6: 10001100.00011001.00000001.1 0000110 = 140.25.1.134/25

Host #62: 10001100.00011001.00000001.1 0111110 = 140.25.1.190/25

Host #63: 10001100.00011001.00000001.1 0111111 = 140.25.1.191/25

Host #64: 10001100.00011001.00000001.1 1000000 = 140.25.1.192/25

Host #65: 10001100.00011001.00000001.1 1000001 = 140.25.1.193/25

Host #123: 10001100.00011001.00000001.1 1111011 = 140.25.1.251/25

Host #124: 10001100.00011001.00000001.1 1111100 = 140.25.1.252/25

Host #125: 10001100.00011001.00000001.1 1111101 = 140.25.1.253/25

Host #126: 10001100.00011001.00000001.1 1111110 = 140.25.1.254/25

Defining the Broadcast Address for Each Subnet

The broadcast address for Subnet #3 is the all-1s host address or:

10001100.00011001.00000001.1 1111111 = 140.25.1.255

The broadcast address for Subnet #3 is exactly one less than the base

address for Subnet #4 (140.25.2.0).

Additional Practice with Subnetworks

Appendix C provides exercises using subnetting.

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Variable Length Subnet Masks (VLSM)

In 1987, RFC 1009 specified how a subnetted network could use more

than one subnet mask. When an IP network is assigned more than one

subnet mask, it is considered a network with (VLSM) since the

extended network prefixes have different lengths.

RIP-1 Permits Only a Single Subnet Mask

When using RIP-1, subnet masks have to be uniform across the entire

network prefix. RIP-1 allows only a single subnet mask to be used

within each network number because it does not provide subnet mask

information as part of its routing table update messages. In the absence

of this information, RIP-1 is forced to make assumptions about the mask

that should be applied to any of its learned routes.

How does a RIP-1 based router know what mask to apply to a route

when it learns a new route from a neighbor? If the router has a subnet

of the same network number assigned to a local interface, it assumes

that the learned subnetwork was defined using the same mask as the

locally configured interface. However, if the router does not have a sub-

net of the learned network number assigned to a local interface, the

router has to assume that the network is not subnetted and applies the

route’s natural classful mask.

For example, assume that Port 1 of a router has been assigned the IP

address 130.24.13.1/24 and that Port 2 has been assigned the IP address

200.14.13.2/24. If the router learns about network 130.24.36.0 from a

neighbor, it applies a /24 mask since Port 1 is configured with another

subnet of the 130.24.0.0 network. However, when the router learns

about network 131.25.0.0 from a neighbor, it assumes a “natural” /16

mask since no other masking information is available. How does a RIP-1

based router know whether it should include the subnet number bits in

a routing table update to a RIP-1 neighbor? A router executing RIP-1

will only advertise the subnet number bits on another port if the update

port is configured with a subnet of the same network number. If the

update port is configured with a different subnet or network number,

the router will only advertise the network portion of the subnet route

and zero-out the subnet number field.

For example, assume that Port 1 of a router has been assigned the IP

address 130.24.13.1/24 and that Port 2 has been assigned the IP address

200.14.13.2/24. Also, assume that the router has learned about network

130.24.36.0 from a neighbor. Since Port 1 is configured with another

subnet of the 130.24.0.0 network, the router assumes that network

130.24.36.0 has a /24 subnet mask. When it comes to advertise this

route, the router advertises 130.24.36.0 on Port 1, but it only advertises

130.24.0.0 on Port 2.

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For these reasons, RIP-1 is limited to a single subnet mask for each net-

work number. However, there are several advantages to be gained if

more than one subnet mask can be assigned to a given IP network num-

ber:

•

Multiple subnet masks permit more efficient use of an organization’s

assigned IP address space.

•

Multiple subnet masks permit route aggregation which can signifi-

cantly reduce the amount of routing information at the backbone

level within an organization’s routing domain.

Efficient Use of Assigned IP Address Space

VLSM supports more efficient use of an organization’s assigned IP

address space. The earlier limitation of supporting only a single subnet

mask across a given network prefix locked the organization into a fixed

number of fixed sized subnets.

For example, assume that a network administrator configured the

130.5.0.0/16 network with a /22 extended network prefix, as shown in

Figure 16. A /16 network with a /22 extended network prefix would

permit 64 subnets (26), each of which could support a maximum of

1,022 hosts (210-2).

Please refer to Figure 16. This configuration would be suitable if the

organization wanted to deploy a number of large subnets, but what

about the occasional small subnet containing only 20 or 30 hosts? Since

a subnetted network could have only a single mask, the network admin-

istrator would still be required to assign the 20 or 30 hosts to a subnet

with a 22-bit prefix. This assignment would waste approximately 1,000

IP host addresses for each small subnet deployed. Limiting the associa-

tion of a network number with a single mask did not encourage the

flexible and efficient use of an organization’s address space. One solu-

tion to this problem was to allow a subnetted network to be assigned

more than one subnet mask.

F I G U R E 1 6 . 1 3 0 . 5 . 0 / 1 6 w i t h a / 2 2 E x t e n d e d N e t w o r k P re f i x

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For example, assume that the network administrator was also allowed to

configure the 130.5.0.0/16 network with a /26 extended network prefix,

as shown in Figure 17. A /16 network address with a /26 extended net-

work prefix would permit 1,024 subnets (210), each of which would

support a maximum of 62 hosts (26 -2). The /26 prefix would be ideal

for small subnets with less than 60 hosts, while the /22 prefix would be

well suited for larger subnets containing up to 1,000 hosts.

Route Aggregation

VLSM also allows the recursive division of an organization’s address

space so that it can be reassembled and aggregated to reduce the

amount of routing information at the top level. Conceptually, a network

is first divided into subnets, then some of the subnets are divided into

sub-subnets, and some of the sub subnets are divided into sub-subnets.

This allows the detailed structure of routing information for one subnet

group to be hidden from routers in another subnet group.

11.0.0.0./8 11.1.0.0/16

11.2.0.0/16

11.3.0.0/16

11.252.0.0/16

11.253.0.0/16

11.254.0.0/16 11.1.1.0/24

11.1.2.0/24

11.1.253.0/24

11.1.254.0/24

11.253.32.0/19

11.253.64.0/19

11.253.160.0/19

11.253.192.0/19 11.1.253.32/27

11.1.253.64/27

11.1.253.160/27

11.1.253.192/27

F I G U R E 1 7 . 1 3 0 . 5 . 0 / 1 6 w i t h a / 2 6 E x t e n d e d N e t w o r k P re f i x

Page 24

In Figure 18, the 11.0.0.0/8 network is first configured with a /16

extended network prefix. The 11.1.0.0/16 subnet is then configured

with a /24 extended network prefix and the 11.253.0.0/16 subnet is

configured with a /19 extended network prefix. Note that the recursive

process does not require that the same extended network prefix be

assigned at each level of the recursion. Also, the recursive subdivision

of the organization’s address space can be carried out as far as the net-

work administrator needs to take it.

U N D E R S TA N D I N G I P A D D R E S S I N G

2 2

F I G U R E 1 9 . R o u t e A g g re g a t i o n , R e d u c i n g R o u t i n g Ta b l e

S i z e

F I G U R E 1 8 . R e c u r s i v e D i v i s i o n o f a N e t w o r k P re f i x

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Figure 19 illustrates how a planned and thoughtful allocation of VLSM

can reduce the size of an organization’s routing tables. Notice how

Router D can summarize the six subnets behind it into a single adver-

tisement (11.1.253.0/24) and how Router B can aggregate all subnets

behind it into a single advertisement (11.1.0.0/16). Likewise, Router C

can summarize the six subnets behind it into a single advertisement

(11.253.0.0/16). Finally, since the subnet structure is not visible outside

of the organization, Router A injects a single route into the global Inter-

net’s routing table-11.0.0.0/8 (or 11/8).

VLSM Design Considerations

When developing a VLSM design, the network designer must recur-

sively ask the same set of questions as for a traditional subnet design.

The same set of design decisions must be made at each level of the hier-

archy:

1 How many total subnets does this level need today?

2 How many total subnets will this level need in the future?

3 How many hosts are on this level’s largest subnet today?

4 How many hosts will be on this level’s largest subnet be in the future?

At each level, the design team must ensure that they have enough extra

bits to support the required number of subentities in the next levels of

recursion.

Assume that a network is spread out over a number of sites. For exam-

ple, if an organization currently has three campuses, it probably needs 3

bits of subnetting (23 = 8) to allow the addition of more campuses in the

future. Now, within each campus, there is likely to be a secondary level

of subnetting to identify each building. Finally, within each building, a

third level of subnetting might identify each of the individual work-

groups. Following this hierarchical model, the top level is determined

by the number of campuses, the middle level is based on the number of

buildings at each site, and the lowest level is determined by the maxi-

mum number of subnets and maximum number of users per subnet in

each building.

The deployment of a hierarchical subnetting scheme requires careful

planning. It is essential that the network designers recursively work

their way down through their addressing plan until they get to the bot-

tom level. At the bottom level, they must make sure that the leaf sub-

nets are large enough to support the required number of hosts. When

the addressing plan is deployed, the addresses from each site must be

aggregated into a single address block that keeps the backbone routing

tables from becoming too large.

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Requirements for Deploying VLSM

The successful deployment of VLSM has three prerequisites:

•

The routing protocols must carry extended network prefix informa-

tion with each route advertisement.

•

All routers must implement a consistent forwarding algorithm based

on the “longest match.”

•

For route aggregation to occur, addresses must be assigned so that

they have topological significance.

Routing Protocols Must Carry Extended Network Prefix Lengths

Routing protocols, such as OSPF and I-IS-IS, enable the deployment of

VLSM by providing the extended network prefix length or mask value

along with each route advertisement. This permits each subnetwork to

be advertised with its corresponding prefix length or mask. If the rout-

ing protocols did not carry prefix information, a router would have to

either assume that the locally configured prefix length should be

applied, or perform a look-up in a statically configured prefix table that

contains all of the required masking information. The first alternative

cannot guarantee that the correct prefix is applied, and static tables do

not scale since they are difficult to maintain and subject to human error.

To deploy VLSM in a complex topology, the administrator must select

OSPF or I-IS-IS as the Interior Gateway Protocol (IGP) rather than RIP-1.

Note that RIP-2, defined in RFC 1388, improves the RIP protocol by

allowing it to carry extended network prefix information. Therefore,

RIP-2 supports the deployment of VLSM.

Forwarding Algorithm Based on the Longest Match

All routers must implement a consistent forwarding algorithm based on

the longest match algorithm. The deployment of VLSM means that the

set of networks associated with extended network prefixes may mani-

fest a subset relationship. A route with a longer extended network pre-

fix describes a smaller set of destinations than the same route with a

shorter extended network prefix. As a result, a route with a longer

extended network prefix is more specific while a route with a shorter

extended network prefix is less specific. Routers must use the route

with the longest matching extended network prefix (most specific

matching route) when forwarding traffic.

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For example, if a packet’s destination IP address was 11.1.2.5 and there

were three network prefixes in the routing table (11.1.2.0/24,

11.1.0.0/16, and 11.0.0.0/8), the router would select the route to

11.1.2.0/24. The 11.1.2.0/24 route would be selected because its prefix

has the greatest number of corresponding bits in the Destination IP

address of the packet. This concept is illustrated in Figure 20.

A very subtle but extremely important issue is that since the destina-

tion address matches all three routes, it must be assigned to a host that

is attached to the 11.1.2.0/24 subnet. If the 11.1.2.5 address is assigned

to a host that is attached to the 11.1.0.0/16 or 11.0.0.0/8 subnet, the

routing system will never route traffic to the host since the “longest

match algorithm” assumes that the host is part of the 11.1.2.0/24 sub-

net. Great care must be taken when assigning host addresses to ensure

that every host is reachable.

Topologically Significant Address Assignment

Since OSPF and I-IS-IS convey the extended network prefix information

with each route, the VLSM subnets can be scattered throughout an

organization’s topology. However, to support hierarchical routing and

reduce the size of an organization’s routing tables, addresses should be

assigned so that they are topologically significant.

Hierarchical routing requires that addresses be assigned to reflect the

actual network topology. This reduces the amount of routing informa-

tion by aggregating the set of addresses assigned to a particular region

of the topology into a single routing advertisement for the entire set.

Hierarchical routing allows this to be done recursively at various points

within the hierarchy of the routing topology. If addresses do not have a

topological significance, they cannot be aggregated and the size of the

routing tables cannot be reduced.

F I G U R E 2 0 . B e s t M a t c h R o u t e w i t h L o n g e s t P re f i x ( M o s t

S p e c i f i c )

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VLSM Example

Given

An organization has been assigned the network number 140.25.0.0/16

and it plans to deploy VLSM. Figure 21 provides a graphic display of

the VLSM design for the organization.

The first step of the subnetting process divides the base network

address into 16 equally sized address blocks. Then Subnet #1 is divided

into 32 equally sized address blocks and Subnet #14 is divided into 16

equally sized address blocks. Finally, Subnet #14-14 is divided into

eight equally sized address blocks.

Define the 16 Subnets of 140.25.0.0/16

The first step in the subnetting process divides the base network

address into 16 equally sized address blocks, as illustrated in Figure 22.

Since 16 = 24, four bits are required to identify each of the 16 subnets.

This means that the organization needs four more bits, or a /20, in the

extended network prefix to define the 16 subnets of 140.25.0.0/16.

Each of these subnets represents a contiguous block of 212 (or 4,096)

network addresses.

F I G U R E 2 2 . S i x t e e n S u b n e t s f o r 1 4 0 . 2 5 . 0 . 0 / 1 6

F I G U R E 2 1 . A d d re s s S t r a t e g y f o r V L S M E x a m p l e

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The 16 subnets of the 140.25.0.0/16 address block are listed in the fol-

lowing code sample. The subnets are numbered 0 through 15. The

underlined portion of each address identifies the extended network pre-

fix, while the bold digits identify the 4 bits representing the subnet

number field:

Base Network: 10001100.00011001 .00000000.00000000 = 140.25.0.0/16

Subnet #0: 10001100.00011001.0000 0000.00000000 = 140.25.0.0/20

Subnet #1: 10001100.00011001.0001 0000.00000000 = 140.25.16.0/20

Subnet #2: 10001100.00011001.0010 0000.00000000 = 140.25.32.0/20

Subnet #3: 10001100.00011001.0011 0000.00000000 = 140.25.48.0/20

Subnet #4: 10001100.00011001.0100 0000.00000000 = 140.25.64.0/20

Subnet #13: 10001100.00011001.1101 0000.00000000 = 140.25.208.0/20

Subnet #14: 10001100.00011001.1110 0000.00000000 = 140.25.224.0/20

Subnet #15: 10001100.00011001.1111 0000.00000000 = 140.25.240.0/20

Define the Host Addresses for Subnet #3 (140.25.48.0/20)

Figure 23 shows the host addresses that can be assigned to Subnet #3

(140.25.48.0/20).

Since the host number field of Subnet #3 contains 12 bits, there are

4,094 valid host addresses (212 -2) in the address block. The hosts are

numbered 1 through 4,094. The valid host addresses for Subnet #3 are

listed in the following sample code. The underlined portion of each

address identifies the extended network prefix, while the bold digits

identify the 12-bit host number field:

Subnet #3: 10001100.00011001.0011 0000.00000000 = 140.25.48.0/20

Host #1: 10001100.00011001.0011 0000.00000001 = 140.25.48.1/20

Host #2: 10001100.00011001.0011 0000.00000010 = 140.25.48.2/20

Host #3: 10001100.00011001.0011 0000.00000011 = 140.25.48.3/20

Host #4093: 10001100.00011001.0011 1111.11111101 = 140.25.63.253/20

Host #4094: 10001100.00011001.0011 1111.11111110 = 140.25.63.254/20

F I G U R E 2 3 . H o s t A d d re s s f o r S u b n e t # 3 ( 1 4 0 . 2 5 . 4 8 . 0 / 2 0 )

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The broadcast address for Subnet #3 is the all-1s host address or:

10001100.00011001.0011 1111.11111111 = 140.25.63.255

The broadcast address for Subnet #3 is exactly one less than the base

address for Subnet #4 (140.25.64.0).

Define the Sub-Subnets for Subnet #14 (140.25.224.0/20)

After the base network address is divided into 16 subnets, Subnet #14

is subdivided into 16 equally sized address blocks. This division is illus-

trated in Figure 24.

Since 16 = 24, four more bits are required to identify each of the 16

subnets. This means that the organization will need to use a /24 as the

extended network prefix length. The 16 subnets of the 140.25.224.0/20

address block are listed in the following sample code. The subnets are

numbered 0 through 15. The underlined portion of each sub-subnet

address identifies the extended network prefix, while the bold digits

identify the 4 bits representing the sub-subnet number field:

Subnet #14: 10001100.00011001.1110 0000.00000000 = 140.25.224.0/20

Subnet #14-0: 10001100.00011001.1110 0000 .00000000 = 140.25.224.0/24

Subnet #14-1: 10001100.00011001.1110 0001 .00000000 = 140.25.225.0/24

Subnet #14-2: 10001100.00011001.1110 0010 .00000000 = 140.25.226.0/24

Subnet #14-3: 10001100.00011001.1110 0011 .00000000 = 140.25.227.0/24

Subnet #14-4: 10001100.00011001.1110 0100 .00000000 = 140.25.228.0/24

Subnet #14-14: 10001100.00011001.1110 1110 .00000000 = 140.25.238.0/24

Subnet #14-15: 10001100.00011001.1110 1111 .00000000 = 140.25.239.0/24

F I G U R E 2 4 . S u b - S u b n e t s f o r S u b n e t # 1 4 ( 1 4 0 . 2 5 . 2 2 4 . 0 / 2 0 )

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Define Host Addresses for Subnet #14-3 (140.25.227.0/24)

Figure 25 shows the host addresses that can be assigned to Subnet #14-

3 (140.25.227.0/24).

Each of the subnets of Subnet #14-3 has 8 bits in the host number field.

This means that each subnet represents a block of 254 valid host

addresses (28 -2). The hosts are numbered 1 through 254.

The valid host addresses for Subnet #14-3 are listed in the following

sample code. The underlined portion of each address identifies the

extended network prefix, while the bold digits identify the 8-bit host

number field:

Subnet #14 3: 10001100.00011001.11100011 .00000000 = 140.25.227.0/24

Host #1 10001100.00011001.11100011 .00000001 = 140.25.227.1/24

Host #2 10001100.00011001.11100011 .00000010 = 140.25.227.2/24

Host #3 10001100.00011001.11100011 .00000011 = 140.25.227.3/24

Host #4 10001100.00011001.11100011 .00000100 = 140.25.227.4/24

Host #5 10001100.00011001.11100011 .00000101 = 140.25.227.5/24

Host #253 10001100.00011001.11100011 .11111101 = 140.25.227.253/24

Host #254 10001100.00011001.11100011 .11111110 = 140.25.227.254/24

The broadcast address for Subnet #14-3 is the all-1s host address or:

10001100.00011001.11100011. 11111111 = 140.25.227.255

The broadcast address for Subnet #14-3 is exactly one less than the base

address for Subnet #14-4 (140.25.228.0).

F I G U R E 2 5 . H o s t A d d re s s e s f o r S u b n e t # 1 4 - 3

( 1 4 0 . 2 5 . 2 2 7 . 0 / 2 4 )

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Define the Sub-Subnets for Subnet #14-14 (140.25.238.0/24)

After Subnet #14 is divided into 16 subnets, Subnet #14-14 is subdi-

vided into eight equally sized address blocks, as shown in Figure 26.

Since 8 = 23, three more bits are required to identify each of the eight

subnets. This means that the organization will need to use a /27 as the

extended network prefix length.

The eight subnets of the 140.25.238.0/24 address block are listed in the

following sample code. The subnets are numbered 0 through 7. The

underlined portion of each sub-subnet address identifies the extended

network prefix, while the bold digits identify the 3 bits representing

the subnet-number field:

Subnet #14-14: 10001100.00011001.11101110 .00000000 = 140.25.238.0/24

Subnet#14-14-0: 10001100.00011001.11101110.000 00000 = 140.25.238.0/27

Subnet#14-14-1: 10001100.00011001.11101110.001 00000 = 140.25.238.32/27

Subnet#14-14-2: 10001100.00011001.11101110.010 00000 = 140.25.238.64/27

Subnet#14-14-3: 10001100.00011001.11101110.011 00000 = 140.25.238.96/27

Subnet#14-14-4: 10001100.00011001.11101110.100 00000 = 140.25.238.128/27

Subnet#14-14-5: 10001100.00011001.11101110.101 00000 = 140.25.238.160/27

Subnet#14-14-6: 10001100.00011001.11101110.110 00000 = 140.25.238.192/27

Subnet#14-14-7: 10001100.00011001.11101110.111 00000 = 140.25.238.224/27

F I G U R E 2 6 . S u b - S u b n e t s f o r S u b n e t # 1 4 - 1 4

( 1 4 0 . 2 5 . 2 3 8 . 0 / 2 4 )

Page 33

3 1

Define Host Addresses for Subnet #14-14-2 (140.25.238.64/27)

Figure 27 shows the host addresses that can be assigned to Subnet #14-

14-2 (140.25.238.64/27).

Each of the subnets of Subnet #14-14 has 5 bits in the host number

field. This means that each subnet represents a block of 30 valid host

addresses (25 -2). The hosts will be numbered 1 through 30.

The valid host addresses for Subnet #14-14-2 are listed in the following

sample code. The underlined portion of each address identifies the

extended network prefix, while the bold digits identify the 5-bit host

number field:

Subnet#14-14-2: 10001100.00011001.11101110.010 00000 = 140.25.238.64/27

Host #1 10001100.00011001.11101110.010 00001 = 140.25.238.65/27

Host #2 10001100.00011001.11101110.010 00010 = 140.25.238.66/27

Host #3 10001100.00011001.11101110.010 00011 = 140.25.238.67/27

Host #4 10001100.00011001.11101110.010 00100 = 140.25.238.68/27

Host #5 10001100.00011001.11101110.010 00101 = 140.25.238.69/27

Host #29 10001100.00011001.11101110.010 11101 = 140.25.238.93/27

Host #30 10001100.00011001.11101110.010 11110 = 140.25.238.94/27

The broadcast address for Subnet #14-14-2 is the all-1s host address or:

10001100.00011001.11011100.010 11111 = 140.25.238.95

The broadcast address for Subnet #6-14-2 is exactly one less than the

base address for Subnet #14-14-3 (140.25.238.96).

Additional Practice with VLSM

Appendix D provides exercises for using VLSM.

F I G U R E 2 7 . H o s t A d d re s s e s f o r S u b n e t # 1 4 - 1 4 - 2

( 1 4 0 . 2 5 . 2 3 8 . 6 4 / 2 7 )

Page 34

U N D E R S TA N D I N G I P A D D R E S S I N G

3 2

Classless Inter-Domain Routing (CIDR)

By 1992, the exponential growth of the Internet was raising serious con-

cerns among members of the IETF about the ability of the Internet’s

routing system to scale and support future growth. These problems

were related to:

•

The near-term exhaustion of the Class B network address space

•

The rapid growth in the size of the global Internet’s routing tables

•

The eventual exhaustion of the 32-bit IPv4 address space

Throughout the Internet’s growth, the first two problems listed became

critical and the response to these immediate challenges was the develop-

ment of Classless Inter-Domain Routing (CIDR). The third problem,

which is of a more long-term nature, is currently being explored by the

IP Next Generation (IPng or IPv6) working group of the IETF.

CIDR was officially documented in September 1993 in RFC 1517, 1518,

1519, and 1520. CIDR supports two important features that benefit the

global Internet routing system:

•

CIDR eliminates the traditional concept of Class A, Class B, and Class

C network addresses.

•

CIDR supports route aggregation where a single routing table entry

can represent the address space of thousands of traditional classful

routes. This allows a single routing table entry to specify how to route

traffic to many individual network addresses. Route aggregation helps

control the amount of routing information in the Internet’s backbone

routers, reduces route flapping (rapid changes in route availability),

and eases the local administrative burden of updating external rout-

ing information.

Without the rapid deployment of CIDR in 1994 and 1995, the Internet

routing tables would have in excess of 70,000 classful routes and the

Internet would probably not be functioning today.

CIDR Promotes the Efficient Allocation of the IPv4 Address Space

CIDR eliminates the traditional concept of Class A, Class B, and Class C

network addresses and replaces them with the generalized concept of a

network prefix. Routers use the network prefix, rather than the first 3

bits of the IP address, to determine the dividing point between the net-

work number and the host number. As a result, CIDR supports the

deployment of arbitrarily sized networks rather than the standard 8-bit,

16-bit, or 24-bit network numbers associated with classful addressing.

In the CIDR model, each piece of routing information is advertised with

a bit mask (or prefix length). The prefix length is a way of specifying

the number of leftmost contiguous bits in the network portion of each

routing table entry. For example, a network with 20 bits of network

number and 12 bits of host number would be advertised with a 20-bit

prefix length (/20). The IP address advertised with the /20 prefix could

Page 35

be a former Class A, Class B, or Class C address. Routers that support

CIDR do not make assumptions based on the first three bits of the

address, they rely on the prefix length information provided with the

route.

In a classless environment, prefixes are viewed as bitwise contiguous

blocks of the IP address space. For example, all prefixes with a /20 pre-

fix represent the same amount of address space (212 or 4,096 host

addresses). Furthermore, a /20 prefix can be assigned to a traditional

Class A, Class B, or Class C network number. Figure 28 shows how each

of the following /20 blocks represent 4,096 host addresses-

10.23.64.0/20, 130.5.0.0/20, and 200.7.128.0/20.

Table 3 provides information about the most commonly deployed CIDR

address blocks. The table shows that a /15 allocation can also be speci-

fied using the traditional dotted-decimal mask notation of 255.254.0.0.

Also, a /15 allocation contains a bitwise contiguous block of 128K

(131,072) IP addresses that can be classfully interpreted as two Class B

networks or 512 Class C networks.

3 3

F I G U R E 2 8 . B i t w i s e C o n t i g u o u s A d d re s s B l o c k s

TA B L E 3 . C I D R A d d re s s B l o c k s

Page 36

Host Implications for CIDR Deployment

There may be severe host implications when CIDR-based networks are

deployed. Since many hosts are classful, their user interface will not

permit them to be configured with a mask that is shorter than the nat-

ural mask for a traditional classful address.

For example, to deploy 200.25.16.0 as a /20 to define a network capable

of supporting 4,094 (212 -2) hosts, ensure that the software executing

on each end station will allow a traditional Class C (200.25.16.0) to be

configured with a 20-bit mask since the natural mask for a Class C net-

work is a 24-bit mask. If the host software supports CIDR, shorter

masks can be configured.

There will be no host problems by deploying the 200.25.16.0/20 (a tra-

ditional Class C) allocation as a block of 16 /24s since non-CIDR hosts

will interpret their local /24 as a Class C. Likewise, 130.14.0.0/16 (a tra-

ditional Class B) could be deployed as a block of 255 /24s since the hosts

will interpret the /24s as subnets of a /16. If host software supports the

configuration of shorter than expected masks, the network manager has

tremendous flexibility in network design and address allocation.

Efficient Address Allocation

How does CIDR lead to the efficient allocation of the IPv4 address

space? In a classful environment, an Internet Service Provider (ISP) can

only allocate /8, /16, or /24 addresses. In a CIDR environment, the ISP

can carve out a block of its registered address space that specifically

meets the needs of each client, provides additional room for growth,

and does not waste a scarce resource.

Assume that an ISP has been assigned the address block 206.0.64.0/18.

This block represents 16,384 (214) IP addresses, which can be inter-

preted as 64 /24s. If a client requires 800 host addresses, rather than

assigning a Class B address (and wasting approximately 64,700

addresses) or four individual Class C addresses (and introducing four

new routes into the global Internet routing tables), the ISP could assign

the client the address block 206.0.68.0/22, which is a block of 1,024

(210) IP addresses (four contiguous /24s). The efficiency of this alloca-

tion is illustrated in Figure 29.

U N D E R S TA N D I N G I P A D D R E S S I N G

3 4

F I G U R E 2 9 . C I D R E ff i c i e n t A d d re s s A l l o c a t i o n

Page 37

3 5

CIDR Address Allocation Example

For this example, assume that an ISP owns the address block

200.25.0.0/16. This block represents 65,536 (216) IP addresses (or 256

/24s).

The ISP wants to allocate the smaller 200.25.16.0/20 address block,

which represents 4,096 (212) IP addresses (or 16 /24s).

Address Block 11001000.00011001.00010000.00000000 200.25.16.0/20

In a classful environment, the ISP is forced to use the /20 as 16 individ-

ual /24s.

However, in a classless environment, the ISP is free to cut up the pie

any way it wants. It could slice the original pie into pieces (each one-

half of the address space) and assign one portion to Organization A,

then cut the other half into two pieces (each one-fourth of the address

space) and assign one piece to Organization B, and then slice the

remaining fourth into two pieces (each one-eighth of the address space)

and assign them to Organization C and Organization D. Each of the orga-

nizations is free to allocate the address space within its “Intranetwork”

as desired. This example is illustrated in Figure 31.

F I G U R E 3 1 . S l i c i n g t h e P i e - C l a s s l e s s E n v i o r n m e n t

F I G U R E 3 0 . S l i c i n g t h e P i e - C l a s s f u l E n v i o r n m e n t

Page 38

The following steps explain how to assign addresses with classless inter-

domain routing.

Step #1: Divide the address block 200.25.16.0/20 into two equally sized

slices. Each block represents one-half of the address space, or 2,048

(211) IP addresses.

ISP’s Block 11001000.00011001.00010000.00000000 200.25.16.0/20

Org A: 11001000.00011001.00010000.00000000 200.25.16.0/21

Reserved: 11001000.00011001.00011000.00000000 200.25.24.0/21

Step #2: Divide the reserved block (200.25.24.0/21) into two equally

sized slices. Each block represents one-fourth of the address space, or

1,024 (210) IP addresses.

Reserved 11001000.00011001.00011000.00000000 200.25.24.0/21

Org B: 11001000.00011001.00011000.00000000 200.25.24.0/22

Reserved 11001000.00011001.00011100.00000000 200.25.28.0/22

Step #3: Divide the reserved address block (200.25.28.0/22) into two

equally sized blocks. Each block represents one-eighth of the address

space, or 512 (29) IP addresses.

Reserved 11001000.00011001.00011100.00000000 200.25.28.0/22

Org C: 11001000.00011001.00011100.00000000 200.25.28.0/23

Org D: 11001000.00011001.00011110.00000000 200.25.30.0/23

Comparing CIDR to VLSM

CIDR and VLSM both allow a portion of the IP address space to be

recursively divided into subsequently smaller pieces. The difference is

that with VLSM, the recursion is performed on the address space previ-

ously assigned to an organization and is invisible to the global Internet.

CIDR, on the other hand, permits the recursive allocation of an address

block by an Internet Registry to a high-level ISP, a mid-level ISP, a low-

level ISP, and a private organization’s network.

Like VLSM, the successful deployment of CIDR has three prerequisites:

•

The routing protocols must carry network prefix information with

each route advertisement.

•

All routers must implement a consistent forwarding algorithm based

on the longest match.

•

For route aggregation to occur, addresses must be assigned so that

they are topologically significant.

Controlling the Growth of Internet’s Routing Tables

CIDR helps control the growth of the Internet’s routing tables by reduc-

ing the amount of routing information. This process requires that the

Internet be divided into addressing domains. Within a domain, detailed

information is available about all of the networks that reside in the

domain. Outside of an addressing domain, only the common network

prefix is advertised. This allows a single routing table entry to specify a

route to many individual network addresses.

U N D E R S TA N D I N G I P A D D R E S S I N G

3 6

Page 39

3 7

Figure 32 illustrates how the allocation described in the previous CIDR

example helps reduce the size of the Internet routing tables. Assume

that a portion of the ISP’s address block (200.25.16.0/20) has been allo-

cated as described in the previous example:

•

Organization A aggregates eight /24s into a single advertisement

(200.25.16.0/21)

•

Organization B aggregates four /24s into a single advertisement

(200.25.24.0/22)

•

Organization C aggregates two /24s into a single advertisement

(200.25.28.0/23)

•

Organization D aggregates two /24s into a single advertisement

(200.25.30.0/23)

Then the ISP can inject the 256 /24s in its allocation into the Internet

with a single advertisement-200.25.0.0/16.

Note that route aggregation by means of BGP-4 (the protocol that allows

CIDR aggregation) is not automatic. The network engineers must config-

ure each router to perform the required aggregation. The successful

deployment of CIDR allows the number of individual networks on the

Internet to expand while minimizing the number of routes in the Inter-

net routing tables.

F I G U R E 3 2 . R e d u c e d S i z e o f I n t e r n e t R o u t i n g Ta b l e s

Page 40

Routing in a Classless Environment

Figure 33 illustrates the routing advertisements for Organization A,

which was discussed in the previous CIDR example.

Since all of Organization A’s routes are part of the ISP #1’s address

block, the routes to Organization A are implicitly aggregated by means

of ISP #1’s aggregated announcement to the Internet. In other words,

the eight networks assigned to Organization A are hidden behind a sin-

gle routing advertisement. Using the longest match forwarding algo-

rithm, Internet routers will route traffic to host 200.25.17.25 to ISP #1,

which will in turn route the traffic to Organization A. Now assume that

Organization A changes its network provider to a different ISP (ISP #2),

as illustrated in Figure 34.

To manage the size of the Internet routing tables, Organization A can

obtain a block of ISP #2’s address space and renumber the address. This

would allow the eight networks assigned to Organization A to be hid-

den behind the aggregate routing advertisement of ISP #2. Unfortu-

nately, renumbering is a labor-intensive task that could be very

difficult, if not impossible, for Organization A.

U N D E R S TA N D I N G I P A D D R E S S I N G

3 8

F I G U R E 3 4 . O r g a n i z a t i o n A C h a n g e s N e t w o r k P ro v i d e r s t o

I S P # 2

F I G U R E 3 3 . R o u t i n g A d v e r t i s e m e n t s f o r O r g a n i z a t i o n A

Page 41

3 9

The best strategy is for Organization A to retain ownership of its

address space and have ISP #2 advertise an “exception” (more specific)

route into the Internet. The exception route allows all traffic for

200.25.0.0/16 to be sent to ISP #1, with the exception of the traffic to

200.25.16.0/21. This routing is accomplished by having ISP #2 adver-

tise, in addition to its own 199.30.0.0/16 block, a route for

200.25.16.0/21. Refer to Figure 35.

Using the longest match forwarding algorithm, Internet routers will

route traffic addressed to host 200.25.17.25 to ISP #2, which will in

turn route the traffic to Organization A. Clearly, the introduction of a

large number of exception routes can reduce the effectiveness of the

CIDR deployment and eventually cause Internet routing tables to begin

exploding again.

Additional Practice with CIDR

Appendix E provides exercises using CIDR.

F I G U R E 3 5 . I S P # 2 ’s M o re S p e c i f i c R o u t e i n t o t h e I n t e r n e t

Page 42

U N D E R S TA N D I N G I P A D D R E S S I N G

4 0

New Solutions for Scaling the Internet Address Space

As we enter the 21st century, the problems of IPv4 address shortages

and expanding Internet routing tables are still with us. The good news

is that CIDR is working. The bad news is that recent growth trends indi-

cate that the number of Internet routes is increasing at an exponential

rate. The Internet must find a way to keep the routing table growth lin-

ear. The IETF is continuing its efforts to develop solutions that will

overcome these problems, enabling the continued growth and scalability

of the Internet.

Appeal to Return Unused IP Network Prefixes

RFC 1917 requests that the Internet community return unused address

blocks to the Internet Assigned Numbers Authority (IANA) for redistri-

bution. This includes unused network numbers, addresses for networks

that will never be connected to the global Internet for security reasons,

and sites that are using a small percentage of their address space. RFC

1917 also petitions ISPs to return unused network prefixes that are out-

side of their assigned address blocks.

Address Allocation for Private Internets

RFC 1918 requests that organizations use the private Internet address

space for hosts that require IP connectivity within their enterprise net-

work, but do not require external connections to the global Internet.

The IANA has reserved the following three address blocks for private

Internets:

•

10.0.0.0 - 10.255.255.255 (10/8 prefix)

•

172.16.0.0 - 172.31.255.255 (172.16/12 prefix)

•

192.168.0.0 - 192.168.255.255 (192.168/16 prefix)

Any organization that elects to use addresses from these reserved blocks

can do so without contacting the IANA or an Internet registry. Since

these addresses are never injected into the global Internet routing sys-

tem, the address space can simultaneously be used by many different

organizations.

The disadvantage to this addressing scheme is that it requires an organi-

zation to use a Network Address Translator (NAT) for global Internet

access. However, the use of the private address space and a NAT make it

much easier for clients to change their ISP without renumbering or

“punching holes” in a previously aggregated advertisement. A benefit

of this addressing scheme to the Internet is that it reduces the demand

for IP addresses so large organizations may require only a small block of

the globally unique IPv4 address space.

Page 43

4 1

Address Allocation from the Reserved Class A Address Space

An Internet-Draft, “Observations on the Use of Components of the Class

A Address Space within the Internet,” explores the allocation of the

upper-half of the currently reserved Class A address space through dele-

gated registries. As the demand for IP addresses continues to grow, it

may be necessary to allocate the 64.0.0.0/2 address space. The 64.0.0.0/2

address block is huge and represents 25 percent of the IPv4 unicast

address space.

Implications of Address Allocation Policies

An Internet-Draft, “Implications of Various Address Allocation Policies

for Internet Routing,” discusses the fundamental issues that must be

considered as the Internet develops new unicast address allocation and

management policies. The draft compares the benefits and limitations of

an “address ownership” policy with an “address lending” policy.

Address ownership means that when an address block is assigned to an

organization, it remains allocated to that organization for as long as the

organization wants to keep it. This means that the address block is

portable and that the organization could use it to gain access to the

Internet no matter where the organization connects to the Internet.

Address lending means that an organization obtains its address block on

a loan basis. If the loan ends, the organization can no longer use the

borrowed address block. It must obtain new addresses and must renum-

ber those addresses before using them.

Hierarchical routing requires that addresses reflect the network topol-

ogy in order to permit route aggregation. The draft argues that two fun-

damental problems break the hierarchical addressing and routing model

supported by CIDR:

•

The continued existence of routes prior to CIDR that cannot be aggre-

gated.

•

Organizations that switch ISPs and continue to use addresses from

their previous ISP’s address block.

The new ISP cannot aggregate the old address block as part of its aggre-

gation, so it must inject an exception route into the Internet. If the

number of exception routes continues to increase, it will erode the ben-

efits of CIDR and prevent the scalability of the Internet’s routing sys-

tem.

The draft recommends that large providers, which can express their

destinations with a single prefix, be assigned address blocks following

the address ownership model. However, all allocations from these

providers to a downstream client should follow the address lending

model. This means that if an organization changes its provider, the loan

is canceled and the client is required to renumber addresses.

Page 44

U N D E R S TA N D I N G I P A D D R E S S I N G

4 2

This draft has generated a tremendous amount of discussion within the

Internet community about the concept of address ownership and what

it means in the context of global routing.

Administrators of smaller organizations that want to own their

addresses have concerns about the difficulty of renumbering and their

lack of self-determination if their provider or their provider’s upstream

provider changes its provider.

Finally, ISPs have concerns because the term “large provider” has not

been defined. At this time, the discussion continues since any criteria

recommended by the IETF is bound to be perceived as unfair by some.

Procedures for Internet/Enterprise Renumbering (PIER)

In the face of the address ownership versus address lending debate, it is

clear that renumbering has become an issue. Procedures for

Internet/Enterprise Renumbering (PIER) is a working group of the IETF

charged with the task of developing a renumbering strategy.

RFC 1916 is a request by PIER for the Internet community to provide

assistance in the development of a series of documents describing how

an organization might proceed to renumber its network. The ultimate

goal of these documents is to provide education and practical experi-

ence to the Internet community.

Market-Based Allocation of IP Address Blocks

An Internet-Draft, “Suggestions for Market-Based Allocation of IP

Address Blocks,” is a proposal to make IPv4 address assignments trans-

ferable and condones the exchange of money as part of the transfer pro-

cedure. The draft suggests that the Internet community embrace the

profit motive as an incentive to motivate organizations to act in ways

that will improve resource use. This proposal is similar to another pro-

posal to introduce financial incentives for route aggregation (that is,

have ISPs levy a charge for each route advertised). The idea is to move

the decisions regarding scarce resources from a political atmosphere to a

financial environment that is better suited to deal with scarcity.

Page 45

IPv6 Resolves IPv4 Issues

With the growth of the Internet and its possible extension to additional

devices, such as TVs, toasters, and coffee makers, all IPv4 solutions pro-

posed for scaling the Internet address space will only delay the

inevitable. There are just not enough IPv4 addresses. The IETF has pro-

duced a comprehensive set of specifications to define what is commonly

known as the next-generation IP protocol (“IPng” or “IPv6”). IPv6

eliminates the need for VLSM, CIDR, and much more.

Introduction to IPv6 Addressing

IPv6 increases the IP address size from 32 bits to 128 bits to support

more levels of the addressing hierarchy, a much greater number of

addressable nodes, and simpler auto-configuration. IPv6 supports

approximately 340,282,366,920,938,463,463,374,607,431,768,211,456

possible IP addresses. IPv6 text representation is very different from

IPv4. The address form can be written three ways (preferred, com-

pressed, and mixed) and it offers three different types of addresses (uni-

cast, anycast, and multicast).

The preferred form is the full IPv6 address in hexadecimal values which

is X:X:X:X:X:X:X:X, where each X refers to a four-digit hexadecimal inte-

ger (16 bits). Each digit consists of four bits, each integer consists of

four digits, and each address consists of eight integers which totals 128

bits (4 x 4 x 8 = 128). A colon must be included to separate each inte-

ger. Note that the integers are hexadecimal integers and the letters A

through F represent the numbers 10 through 15. Figure 36 depicts a full

hexadecimal to binary IPv6 address.

The compressed form substitutes zero strings with double colons (::) to

compress the zeros. This method replaces zeros only when they fill a

complete 16-bit group, and the double colon can be used only once in

any given address. The double colon can also be used to compress the

leading or trailing zeros in an address. For example address

1080:0:0:0:8:800:200C:417A could be represented as

1080::8:800:200C:417A. In addition to replacing the zeros that complete

a 16-bit group, all zeroes that are to the left of a given 16-bit expression

may be left out. Table 4 shows the compressed form of some IPv6

addresses using the double colon.

4 3

F I G U R E 3 6 . H e x i d e c i m a l t o B i n a r y C o n v e r s i o n

Page 46

The third form will be useful in mixed IPv4/IPv6 environments. This

form is represented as X:X:X:X:X:X:X:X:D.D.D.D. Where the Xs repre-

sent the hexadecimal values of the six high-order 16-bit pieces of the

address. The Ds represent the standard IPv4 decimal value representa-

tion of the four low-order 8-bit pieces of the address. Table 5 displays

the mixed IPv4 and IPv6 address forms and the corresponding com-

pressed form.

In each address type, IPv6 identifies interfaces, not nodes. A node is

identified by a unicast address assigned to one of its interfaces.

Unicast Addressing

There are six types of unicast IPv6 addresses:

•

Aggregatable global unicast addresses

•

Link-local addresses

•

Site-local addresses

•

Special addresses

•

NSAP addresses

•

IPX addresses

U N D E R S TA N D I N G I P A D D R E S S I N G

4 4

TA B L E 5 . M i x e d I P v 4 a n d I P v 6 A d d re s s e s a n d T h e i r

C o r re s p o n d i n g C o m p re s s e d F o r m

TA B L E 4 . C o m p re s s e d F o r m o f A d d re s s i n g

Page 47

Aggregatable global unicast addresses can be routed globally on the

IPv6 portion of the Internet known as the 6bone (IPv6 backbone).

These types of addresses are equivalent to public IPv4 addresses. As the

name implies, aggregatable global unicast addresses are designed to be

aggregated or summarized to produce an efficient routing infrastruc-

ture.

Anycast Addressing

An IPv6 anycast address is assigned to more than one interface, typi-

cally belonging to different nodes, with the property that a packet sent

to an anycast address is routed to the “nearest” interface having that

address, according to the routing protocol’s measure of distance.

Anycast addresses, when used as part of a route sequence, permits a

node to select which of several Internet service providers it wants to

carry its traffic. This capability is sometimes called “source selected

policies.” Anycast addressing is implemented by configuring anycast

addresses to identify the set of routers belonging to internet service

providers (for example, one anycast address per Internet service

provider). The anycast addresses can be used as intermediate addresses

in an IPv6 routing header to cause a packet to be delivered by means of

a particular provider or sequence of providers. Other possible uses of

anycast addresses are to identify the set of routers attached to a particu-

lar subnet, or the set of routers providing entry into a particular rout-

ing domain.

Multicast Addressing

A multicast address is an identifier for a set of interfaces that typically

belong to different nodes. Each multicast group identifies two or more

packet recipients. A particular multicast address can be confined to a

single system, restricted within a specific site, associated with a partic-

ular network link, or distributed worldwide. When a packet is sent to a

multicast address, the protocol delivers the packet to all interfaces iden-

tified by that address.

IPv6 Efficient Hierarchy Addressing and Routing

IPv6 routing is almost exactly like IPv4 routing except for the length of

the address and the subnet mask. Although IPv4 uses a dotted-decimal

representation of the network prefix known as the subnet mask, a sub-

net mask is not used for IPv6. Only the prefix length notation is sup-

ported.

As in the CIDR notation, the prefix length consists of the bits that have

the fixed values or are the bits of the network identifier. For example,

FE80::2AA:FF:FE9A:4CA2/64 indicates that the first 64 bits of the

address are the network prefix. Prefix notation is also used to express

network or subnet identifiers. For example, 21DA:D3::/48 is a subnet.

The routing in IPv6 mirrors the classless concept, but IPv6 has a well-

defined set of boundaries from which to define sets of address space to

delegate downstream to other people who get Internet connectivity.

At the level of the Internet backbone where major enterprises and ISP

networks come together, it is necessary to maintain a hierarchical

addressing system similar to the national and internal telephone sys-

4 5

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tems. Large central-office phone switches, for example, need only a

three-digit national area code prefix compared to the current IPv4 sys-

tem, which uses a somewhat haphazard form of address hierarchy to

move traffic between networks attached to the Internet backbone. Fig-

ure 37 shows a globally routable unicast prefix divided into six sec-

tions.

The fields in the aggregatable global unicast address are:

•

FP - Indicates the Format Prefix (FP). The Format Prefix for globally

routable unicast prefixes will always have the same three bits in the

initial deployment of IPv6. These first three bits indicate to any rout-

ing entity on the Internet that this address is a globally routable uni-

cast address. The FP will be unique to each type of IPv6 address,

making it easier for routing entities to discern packet types and

process them according to the rules that apply to the respective

packet type. The FP serves as a delimiter, enabling a routing device to

make a quick decision as to how to handle the incoming packet. This

addressing scheme allows the rest of the packet to be used for other

valuable information.

•

TLA ID - Indicates the Top Level Aggregator (TLA). TLAs are admin-

istered by IANA and allocated to local Internet registries that, in turn,

allocate individual TLAs to large Internet service providers (ISPs). A

13-bit field allows up to 8,192 TLAs. Routers in the highest level of

the IPv6 Internet routing hierarchy (called default-free routers) do not

have a default route-only routes with 16-bit prefixes corresponding to

the allocated TLAs.

•

Res - Indicates bits reserved for future use in expanding the size of

either the TLA ID or the NLA ID. The size of this field is 8 bits.

•

NLA ID - Indicates the Next-Level Aggregator (NLA). The NLA ID is

used to identify a specific customer site. The size of this field is 24

bits. The NLA ID allows an ISP to create multiple levels of addressing

hierarchy within a network to organize addressing and routing for

downstream ISPs and to identify sites. The structure of the ISP’s net-

work is transparent to the default-free routers.

U N D E R S TA N D I N G I P A D D R E S S I N G

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F I G U R E 3 7 . G l o b a l l y R o u t a b l e I P v 6 U n i c a s t A d d re s s

A rc h i t e c t u re

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•

SLA ID - Indicates the Site-Level Aggregator (SLA) for the address.

The SLA ID is used by an individual organization to identify subnets

within its site. The size of this field is 16 bits. The organization can

use these 16-bits to create 65,536 subnets or multiple levels of

addressing hierarchy and an efficient routing infrastructure. With 16

bits of subnetting flexibility, an aggregatable global unicast prefix

assigned to an organization is equivalent to that organization being

allocated an IPv4 Class A network ID (assuming that the last octet is

used for identifying nodes on subnets). The structure of the cus-

tomer’s network is transparent to the ISP.

•

Interface ID - Indicates the interface on a specific subnet. The size of

this field is 64 bits.

With IPv6, small to regional network service providers and end users

will no longer have the ability to obtain address space directly from reg-

istries. Instead, TLAs will be assigned address blocks, which they will

in turn be in charge of managing and delegating to their downstream

connections (NLAs and SLAs). This shift in address management is

thought to be much more efficient than the current address management

policies. This policy places the burden on backbone TLA core providers

to carry the routes as transit. With this new hierarchical architecture,

the number of Internet Core routing entries that need to be carried is

decreased, thus limiting the scope of future Internet routing complex-

ity. Figure 38 depicts a generic addressed IPv6 Internet.

Figure 38 shows two TLAs, and a variety of NLAs, and SLAs in various

configurations. TLA I owns a Top Level Aggregator block. TLA I is

assigned 3FFE:2900::/24 as its TLA delegation, and TLA II is assigned

3FFE:4200::/24 as its TLA delegation. TLA I and TLA II must supply

each other with these routes for routing to operate properly between

TLA I and TLA II backbones.

TLA I must subdelegate blocks of address space to its NLA and SLA

customers. In this case, NLA I is assigned 3FFE:2900:1::/48, and NLA II

is assigned 3FFE:2900:2::/48. These NLAs then delegate blocks to their

customers out of this block.

4 7

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U N D E R S TA N D I N G I P A D D R E S S I N G

4 8

Assume SLA I is assigned 3FFE:2900:D:E::/63. Starting at the bottom

aggregators, SLA I must announce its block 3FFE:2900:1:10::/63 to NLA

I. Because this is a subset of NLA I’s space, NLA I is not required to

announce this SLA (from SLA I) to TLA I. A similar situation exists with

NLA II. TLA I only needs to hear the NLA aggregations that it dele-

gated to the two NLAs, regardless of how the NLAs have subdelegated

their space.

At this point, TLA I has to carry only three announcements for non-

backbone space. Note that the first two of these announcements are

simply subsets of the block assigned to TLA I. Therefore, in the bilat-

eral peering between TLA I and TLA II, only one route needs to be

exchanged between these peers. Although this is a limited example, it

shows the routing simplicity that has come to pass as a result of this

aggregation.

The IPv6 routing will greatly increase the long-term efficiency of the

Internet backbone routing tables for two main reasons. The current

IPv4 addresses are portable and can be taken with an individual who

leaves a certain service provider’s jurisdiction and goes to another

provider. This leads to many extraneous announcements in the core of

the Internet backbone as Network Service Providers lose the ability to

aggregate announcements properly.

The second reason is that only TLAs will be assigned address space

from the Numbering Authorities. Today, IANA is the responsible party

for numbering, which in turn delegates numbers to regional registries

such as ARIN, RIPE, and APNIC. These regional numbers authorities in

turn assign IPv4 address space to Internet providers, businesses, and

organizations that can demonstrate sufficient need for their own IP

blocks. Notice how this process leads to more small blocks being carried

in the core Internet table. If renumbering were simple, then getting IP

space directly from upstream providers of connectivity would not be

such an issue. Users who are dissatisfied with service could simply get

another provider and then renumber. By ensuring that only TLAs get

address space, only big blocks of space are delegated, which ensures

that aggregation can always occur.

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IPv6 Address Examples

As explained earlier, IPv6 subnetting can be compared to classless

addressing. The following list of host IDs shows how IPv6 addresses

could be written. Note that the 255 address limit that existed with IPv4

is no longer a limitation in IPv6, so there could be many more possibili-

ties in addition to what is listed. The underlined portion of each address

identifies the network prefix which is calculated through the prefix

length notation. In other words, since /48 is the prefix length notation,

then the first four integers will be the prefix length and the rest will be

the interface ID. Because each integer equals 16, the calculation is 48

divided by 16 equals 4.

Subnet #1: ABF2:45AF:2574:9980:7654:FCD4:FF26:0078/48

Host #1: ABF2:45AF:2574:9980:7654:FCD4:FF26:0080/48

Host #2: ABF2:45AF:2574:9980:7654:FCD4:FF26:0081/48

Host #3: ABF2:45AF:2574:9980:7654:FCD4:FF26:0082/48

Host #4: ABF2:45AF:2574:9980:7654:FCD4:FF26:0083/48

Host #5: ABF2:45AF:2574:9980:7654:FCD4:FF26:0084/48

Host #6: ABF2:45AF:2574:9980:7654:FCD4:FF26:0085/48

Host #62: ABF2:45AF:2574:9980:7654:FCD4:FF26:0086/48

Host #63: ABF2:45AF:2574:9980:7654:FCD4:FF26:0087/48

Host #64: ABF2:45AF:2574:9980:7654:FCD4:FF26:0088/48

Host #65: ABF2:45AF:2574:9980 :7654:FCD4:FF26:0089/48

Host #123: ABF2:45AF:2574:9980:7654:FCD4:FF26:0090/48

Host #124: ABF2:45AF:2574:9980:7654:FCD4:FF26:0091/48

Host #125: ABF2:45AF:2574:9980:7654:FCD4:FF26:0092/48

Host #126: ABF2:45AF:2574:9980 :7654:FCD4:FF26:0093/48

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Unlike IPv4, there are numerous addressing possibilities, which means

IPv6 is a much more robust protocol. Although, IPv6 will mainly be

used on the global Internet backbone figure 39 demonstrates the versa-

tility of IPv6 addressing.

F I G U R E 3 9 . I S P # 2 ’s M o re S p e c i f i c R o u t e i n t o t h e I n t e r n e t

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5 1

Additional IPv6 Features

Although IPv6 appears to have functionality similar to CIDRs, there are

many additional features that were not included in IPv4. These features

make IPv6 much more robust and convenient. These changes include a

streamlined IPv6 header, stateless configuration, built in security, bet-

ter QoS, and increased real-time performance.

Streamlined IPv6 Header

The IPv6 header has a new format that is designed to keep header over-

head to a minimum. This format is achieved by moving both nonessen-

tial fields and option fields to extension headers that are placed after

the IPv6 header. The streamlined IPv6 header provides more efficient

processing at intermediate routers.

Stateless and Stateful Configuration

IPv6 supports both stateful and stateless address configurations. IPv6

will work with or without a DHCP server. With stateless address config-

uration, hosts on a link automatically configure themselves with IPv6

addresses for the link (called link-local addresses) and with addresses

derived from prefixes advertised by local routers. Even in the absence of

a router, hosts on the same link can automatically configure themselves

with link-local addresses and communicate without manual configura-

tion.

Built-in Security

Support for IPSec is an IPv6 protocol suite requirement. This require-

ment provides a standards-based solution for network security and pro-

motes interoperability between different IPv6 implementations.

Better Support for QoS

New fields in the IPv6 header define how traffic is handled and identi-

fied. Traffic identification using a flow label field in the IPv6 header

allows routers to identify and provide special handling for packets

belonging to a flow, which is a series of packets between a source and

destination. Because the traffic is identified in the IPv6 header, support

for QoS can be achieved even when the packet payload is encrypted

through IPSec.

Real-Time Performance

IPv6 offers a packet prioritization feature that provides the real-time

and near real-time applications an improved response time. Conse-

quently, IPv6 will become the protocol of choice for those applications.

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U N D E R S TA N D I N G I P A D D R E S S I N G

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Keeping Current on Internet Addressing Issues

General Internet Information

Internet Monthly Reports discuss the accomplishments, milestones, and

problems discovered on the Internet. They are available from:

http://info.internet.isi.edu/1/in-notes/imr

Minutes of the most recent IETF Proceedings are available from:

http://www.ietf.cnri.reston.va.us/proceedings/directory.html

Information about the size and content of the Internet routing table is

available on the Merit Web pages:

http://www.ra.net/~ra/statistics/routes.html

CIDR Deployment (CIDRD)

To obtain current general information about the CIDRD working group

of the IETF and its charter:

http://www.ietf.cnri.reston.va.us/html.charters/cidrd-charter.html

To subscribe to the CIDRD mailing list: cidrd-request@iepg.org

Internet-Drafts published by the CIDRD working group are available

from:

http://www.ietf.cnri.reston.va.us/ids.by.wg/cidrd.html

Procedures for Internet/Enterprise Renumbering (PIER)

General information about the PIER working group of the IETF and its

charter is available from:

http://www.ietf.cnri.reston.va.us/html.charters/pier-charter.html

To subscribe to the PIER mailing list:

pier-request@isi.edu

Papers developed by PIER are available from:

http://www.isi.edu:80/div7/pier/

Dynamic Host Configuration (DHCP)

To obtain information about the DHCP working group, current Internet-

Drafts, and Requests for Comments:

http://www.ietf.cnri.reston.va.us/html.charters/dhc-charter.html

To access the DHCP Home Page:

http://charlotte.acns.nwu.edu/internet/tech/dhcp/

To subscribe to the DHCP mailing list:

host-conf-request@sol.eg.bucknell.edu

To review the DHCP mail list archive:

ftp://ftp.bucknell.edu/pub/dhcp

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IPv6 (IPNGWG)

To obtain information about the IPv6 working group, current Internet-

Drafts, and Requests for Comments:

http://www.ietf.cnri.reston.va.us/html.charters/ipngwg-charter.html

To access the IPng Home Page:

http://playground.sun.com/pub/ipng/html/ipng-main.html

To subscribe to the IPng mailing list:

majordomo@sunroof.eng.sun.com

To review the IPng mail list archive:

ftp://parcftp.xerox.com/pub/ipng

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U N D E R S TA N D I N G I P A D D R E S S I N G

5 4

Appendix A - References

Requests for Comments

Requests for Comments are available from:

http://ds.internic.net/ ds/dspg2intdoc.html

950 J. Mogul, J. Postel, “Internet Standard Subnetting Procedure,”

08/01/1985. (Pages=18) (STD 5)

985 National Science Foundation, Network Technical Advisory Group,

“Requirements for Internet Gateways-draft,” 05/01/1986. (Pages=23)

(Obsoleted by RFC1009)

1009 R. Braden, J. Postel, “Requirements for Internet Gateways,”

06/01/1987. (Pages=55) (Obsoletes RFC985) (STD 4) (Obsoleted by

RFC1716)

1245 J. Moy, “OSPF Protocol Analysis,” 08/08/1991. (Pages=12)

1246 J. Moy, “Experience with the OSPF Protocol,” 08/08/1991.

(Pages=31)

1247 J. Moy, “OSPF Version 2,” 08/08/1991. (Pages=189) (Format=.txt,

.ps) (Obsoletes RFC1131) (Obsoleted by RFC1583)

1338 V. Fuller, T. Li, K. Varadhan, J. Yu, “Supernetting: An Address

Assignment and Aggregation Strategy,” 06/26/1992. (Pages=20) (Obso-

leted by RFC1519)

1366 E. Gerich, “Guidelines for Management of IP Address Space,”

10/22/1992. (Pages=8) (Obsoleted by RFC1466)

1466 E. Gerich, “Guidelines for Management of IP Address Space,”

05/26/1993. (Pages=10) (Obsoletes RFC1366)

1517 R. Hinden, “Applicability Statement for the Implementation of

Classless Inter-Domain Routing (CIDR) ,” 09/24/1993. (Pages=4)

1518 Y. Rekhter, T. Li, “An Architecture for IP Address Allocation with

CIDR,” 09/24/1993. (Pages=27)

1519 V. Fuller, T. Li, J. Yu, K. Varadhan, “Classless Inter-Domain Routing

(CIDR): An Address Assignment and Aggregation Strategy,” 09/24/1993.

(Pages=24) (Obsoletes RFC1338)

1520 Y. Rekhter, C. Topolcic, “Exchanging Routing Information Across

Provider Boundaries in the CIDR Environment,” 09/24/1993. (Pages=9)

1583 J. Moy, “OSPF Version 2,” 03/23/1994. (Pages=212) (Obsoletes

RFC1247)

1716 P. Almquist, F. Kastenholz, “Toward Requirements for IP Routers,”

11/04/1994. (Pages=186) (Obsoletes RFC1009) (Obsoleted by RFC1812)

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1721 G. Malkin, “RIP Version 2 Protocol Analysis,” 11/15/1994.

(Pages=4) (Obsoletes RFC1387)

1722 G. Malkin, “RIP Version 2 Protocol Applicability Statement,”

11/15/1994. (Pages=5)

1723 G. Malkin, “RIP Version 2 Carrying Additional Information,”

11/15/1994. (Pages=9) (Updates RFC1058) (Obsoletes RFC1388)

1724 G. Malkin, F. Baker, “RIP Version 2 MIB Extension,” 11/15/1994.

(Pages=18) (Obsoletes RFC1389)

1812 F. Baker, “Requirements for IP Version 4 Routers,” 06/22/1995.

(Pages=175) (Obsoletes RFC1716)

1900 B. Carpenter, Y. Rekhter, “Renumbering Needs Work,” 02/28/1996.

(Pages=4)

1916 H. Berkowitz, P. Ferguson, W. Leland, P. Nesser, “Enterprise

Renumbering: Experience and Information Solicitation,” 02/28/1996.

(Pages=8)

1917 P. Nesser, “An Appeal to the Internet Community to Return

Unused IP Network (Prefixes) to the IANA,” 02/29/1996. (Pages=10)

1918 Y. Rekhter, R. Moskowitz, D. Karrenberg, G. de Groot, E. Lear,

“Address Allocation for Private Internets,” 02/29/1996. (Pages=9) (Obso-

letes RFC1627)

Internet-Drafts

Internet-Drafts are available from:

http://www.ietf.cnri.reston.va.us/1id-abstracts. html

“Suggestions for Market-Based Allocation of IP Address Blocks,”

<draft-ietf-cidrd-blocks- 00.txt>, P. Resnick, 02/23/1996. (24,590 bytes)

“Observations on the use of Components of the Class A Address Space

within the Internet,” <draft-ietf-cidrd-classa-01.txt>, G.Huston,

12/22/1995. (21,347 bytes)

Classless in-addr.arpa Delegation,” <draft-ietf-cidrd-classless-inaddr-

00.txt>, H. Eidnes, G. de Groot, 01/18/1996. (13,224 bytes)

“Implications of Various Address Allocation Policies for Internet Rout-

ing,” <draft-ietf-cidrd- addr-ownership-07.txt>, Y. Rekhter, T. Li,

01/15/1996. (34,866 bytes)

“Suggestions for Market-Based Allocation of IP Address Blocks,”

<draft-ietf-cidrd-blocks- 00.txt>, P. Resnick, 02/23/1996. (24,590 bytes)

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Textbooks

Comer, Douglas E. Internetworking with TCP/IP Volume 1 Principles,

Protocols, and Architecture Second Edition, Prentice Hall, Inc. Engle-

wood Cliffs, New Jersey, 1991

Huitema, Christian. Routing in the Internet, Prentice Hall, Inc. Engle-

wood Cliffs, New Jersey, 1995

Stevens, W. Richard. TCP/IP Illustrated: Volume 1 The Protocols, Addi-

son Wesley Publishing Company, Reading MA, 1994

Wright, Gary and W. Richard Stevens. TCP/IP Illustrated: Volume 2 The

Implementation, Addison Wesley Publishing Company, Reading MA,

1995

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Appendix B - Classful IP Addressing

Practice Exercises

1 Complete the following table which provides practice in converting a

number from binary notation to decimal format.

2 Complete the following table which provides practice in converting a

number from decimal notation to binary format.

3 Express 145.32.59.24 in binary format and identify the address class:

4 Express 200.42.129.16 in binary format and identify the address class:

5 Express 14.82.19.54 in binary format and identify the address class:

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Solutions to Classful IP Addressing Practice Exercises

1 Complete the following table which provides practice in converting a

number from binary notation to decimal format.

Complete the following table which provides practice in converting a number

from decimal notation to binary format.

Express 145.32.59.24 in binary format and identify the classful prefix length:

10010001.00100000.00111011.00011000 /16 or Class B

Express 200.42.129.16 in binary format and identify the classful prefix

length:

11001000.00101010.10000001.00010000 /24 or Class C

Express 14.82.19.54 in binary format and identify the classful prefix length:

00001110.01010010. 00010011.00110110 /8 or Class A

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Appendix C - Subnetting Exercises

Subnetting Exercise #1

Assume that you have been assigned the 132.45.0.0/16 network block.

You need to establish eight subnets.

1 __________ binary digits are required to define eight subnets.

2 Specify the extended network prefix that allows the creation of eight

subnets.

_______________________________________________________

3 Express the subnets in binary format and dotted-decimal notation:

#0 ____________________________________________________

#1 ____________________________________________________

#2 ____________________________________________________

#3 ____________________________________________________

#4 ____________________________________________________

#5 ____________________________________________________

#6 ____________________________________________________

#7 ____________________________________________________

4 List the range of host addresses that can be assigned to Subnet #3

(132.45.96.0/19).

_______________________________________________________

5 What is the broadcast address for Subnet #3 (132.45.96.0/19)?

_______________________________________________________

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Subnetting Exercise #2

1 Assume that you have been assigned the 200.35.1.0/24 network

block. Define an extended network prefix that allows the creation of

20 hosts on each subnet.

________________________________________________________

2 What is the maximum number of hosts that can be assigned to each

subnet?

________________________________________________________

3 What is the maximum number of subnets that can be defined?

________________________________________________________

4 Specify the subnets of 200.35.1.0/24 in binary format and dotted-dec-

imal notation.

________________________________________________________

5 List the range of host addresses that can be assigned to Subnet #6

(200.35.1.192/27).

________________________________________________________

6 What is the broadcast address for subnet 200.35.1.192/27?

_______________________________________________________

U N D E R S TA N D I N G I P A D D R E S S I N G

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Solution for Subnetting Exercise #1

Assume that you have been assigned the 132.45.0.0/16 network block.

You need to establish eight subnets.

1 Three binary digits are required to define the eight subnets.

2 Specify the extended network prefix that allows the creation of eight

subnets.

/19 or 255.255.224.0

3 Express the subnets in binary format and dotted-decimal notation:

Subnet #0: 10000100.00101101. 000 00000.00000000 = 132.45.0.0/19

Subnet #1: 10000100.00101101. 001 00000.00000000 = 132.45.32.0/19

Subnet #2: 10000100.00101101. 010 00000.00000000 = 132.45.64.0/19

Subnet #3: 10000100.00101101. 011 00000.00000000 = 132.45.96.0/19

Subnet #4: 10000100.00101101. 100 00000.00000000 = 132.45.128.0/19

Subnet #5: 10000100.00101101. 101 00000.00000000 = 132.45.160.0/19

Subnet #6: 10000100.00101101. 110 00000.00000000 = 132.45.192.0/19

Subnet #7: 10000100.00101101. 111 00000.00000000 = 132.45.224.0/19

4 List the range of host addresses that can be assigned to Subnet #3

(132.45.96.0/19).

Subnet #3: 10000100.00101101.011 00000.00000000 = 132.45.96.0/19

Host #1: 10000100.00101101.011 00000.00000001 = 132.45.96.1/19

Host #2: 10000100.00101101.011 00000.00000010 = 132.45.96.2/19

Host #3: 10000100.00101101.011 00000.00000011 = 132.45.96.3/19

Host #8190: 10000100.00101101.011 11111.11111110 = 132.45.127.254/19

5 What is the broadcast address for Subnet #3 (132.45.96.0/19)?

10000100.00101101.011 11111.11111111 = 132.45.127.255/19

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Solution for Subnetting Exercise #2

1 Assume that you have been assigned the 200.35.1.0/24 network

block. Define an extended network prefix that allows the creation of

20 hosts on each subnet.

A minimum of 5 bits are required to define 20 hosts so the extended

network prefix is /27 (27 = 32-5).

2 What is the maximum number of hosts that can be assigned to each

subnet?

The maximum number of hosts on each subnet is 25 -2, or 30.

3 What is the maximum number of subnets that can be defined?

The maximum number of subnets is 23, or 8.

4 Specify the subnets of 200.35.1.0/24 in binary format and dotted-dec-

imal notation.

Subnet #0: 11001000.00100011.00000001. 000 00000 = 200.35.1.0/27

Subnet #1: 11001000.00100011.00000001. 001 00000 = 200.35.1.32/27

Subnet #2: 11001000.00100011.00000001. 010 00000 = 200.35.1.64/27

Subnet #3: 11001000.00100011.00000001. 011 00000 = 200.35.1.96/27

Subnet #4: 11001000.00100011.00000001. 100 00000 = 200.35.1.128/27

Subnet #5: 11001000.00100011.00000001. 101 00000 = 200.35.1.160/27

Subnet #6: 11001000.00100011.00000001. 110 00000 = 200.35.1.192/27

Subnet #7: 11001000.00100011.00000001. 111 00000 = 200.35.1.224/27

5 List the range of host addresses that can be assigned to Subnet #6

(200.35.1.192/27).

Subnet #6: 11001000.00100011.00000001. 110 00000 = 200.35.1.192/27

Host #1: 11001000.00100011.00000001.110 00001 = 200.35.1.193/27

Host #2: 11001000.00100011.00000001.110 00010 = 200.35.1.194/27

Host #3: 11001000.00100011.00000001.110 00011 = 200.35.1.195/27

Host #29: 11001000.00100011.00000001.110 11101 = 200.35.1.221/27

Host #30: 11001000.00100011.00000001.110 11110 = 200.35.1.222/27

6 What is the broadcast address for subnet 200.35.1.192/27?

11001000.00100011.00000001.110 11111 = 200.35.1.223

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Appendix D - VLSM Exercise

Given

An organization has been assigned the network number 140.25.0.0/16

and it plans to deploy VLSM. Figure D-1 provides a graphic display of

the VLSM design for the organization.

To arrive at this design, the first step of the subnetting process divides

the base network address into eight equally sized address blocks. Then

Subnet #1 is divided into 32 equally sized address blocks and Subnet

#6 is divided into 16 equally sized address blocks. Finally, Subnet #6-

14 is divided into eight equally sized address blocks.

1 Specify the eight subnets of 140.25.0.0/16:

#0 ______________________________________________________

#1 ______________________________________________________

#2 ______________________________________________________

#3 ______________________________________________________

#4 ______________________________________________________

#5 ______________________________________________________

#6 ______________________________________________________

#7 ______________________________________________________

F I G U R E D - 1 . A d d re s s S t r a t e g y f o r V L S M E x a m p l e

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2 List the host addresses that can be assigned to Subnet #3

(140.25.96.0):

________________________________________________________

3 Identify the broadcast address for Subnet #3 (140.25.96.0):

_______________________________________________________

4 Specify the 16 subnets of Subnet #6 (140.25.192.0/19):

#6-0___________________________________________________

#6-1___________________________________________________

#6-2___________________________________________________

#6-3___________________________________________________

#6-4___________________________________________________

#6-5___________________________________________________

#6-6___________________________________________________

#6-7___________________________________________________

#6-8___________________________________________________

#6-9___________________________________________________

#6-10___________________________________________________

#6-11___________________________________________________

#6-12___________________________________________________

#6-13___________________________________________________

#6-14___________________________________________________

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5 List the host addresses that can be assigned to Subnet #6-3

(140.25.198.0/23):

_______________________________________________________

6 Identify the broadcast address for Subnet #6-3 (140.25.198.0/23):

_______________________________________________________

7 Specify the eight subnets of Subnet #6-14 (140.25.220.0/23):

#6-14-0_________________________________________________

#6-14-1_________________________________________________

#6-14-2_________________________________________________

#6-14-3_________________________________________________

#6-14-4_________________________________________________

#6-14-5_________________________________________________

#6-14-6_________________________________________________

#6-14-7_________________________________________________

8 List the host addresses that can be assigned to Subnet #6-14-2

(140.25.220.128/26):

_______________________________________________________

9 Identify the broadcast address for Subnet #6-14-2

(140.25.220.128/26):

_______________________________________________________

6 5

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6 6

Solution for VLSM Exercise

1 Specify the eight subnets of 140.25.0.0/16:

Base Network: 10001100.00011001 .00000000.00000000 = 140.25.0.0/16

Subnet #0: 10001100.00011001. 000 00000.00000000 = 140.25.0.0/19

Subnet #1: 10001100.00011001. 001 00000.00000000 = 140.25.32.0/19

Subnet #2: 10001100.00011001. 010 00000.00000000 = 140.25.64.0/19

Subnet #3: 10001100.00011001. 011 00000.00000000 = 140.25.96.0/19

Subnet #4: 10001100.00011001. 100 00000.00000000 = 140.25.128.0/19

Subnet #5: 10001100.00011001. 101 00000.00000000 = 140.25.160.0/19

Subnet #6: 10001100.00011001. 110 00000.00000000 = 140.25.192.0/19

Subnet #7: 10001100.00011001. 111 00000.00000000 = 140.25.224.0/19

2 List the host addresses that can be assigned to Subnet #3

(140.25.96.0):

Subnet #3: 10001100.00011001.011 00000.00000000 = 140.25.96.0/19

Host #1: 10001100.00011001.011 00000.00000001 = 140.25.96.1/19

Host #2: 10001100.00011001.011 00000.00000010 = 140.25.96.2/19

Host #3: 10001100.00011001.011 00000.00000011 = 140.25.96.3/19

Host #8189: 10001100.00011001.011 11111.11111101 = 140.25.127.253/19

Host #8190: 10001100.00011001.011 11111.11111110 = 140.25.127.254/19

3 Identify the broadcast address for Subnet #3 (140.25.96.0):

10001100.00011001.011 11111.11111111 = 140.25.127.255

4 Specify the 16 subnets of Subnet #6 (140.25.192.0/19):

Subnet #6: 10001100.00011001.110 00000.00000000 = 140.25.192.0/19

Subnet #6-0: 10001100.00011001.110 00000.00000000 = 140.25.192.0/23

Subnet #6-1: 10001100.00011001.110 00010.00000000 = 140.25.194.0/23

Subnet #6-2: 10001100.00011001.110 00100.00000000 = 140.25.196.0/23

Subnet #6-3: 10001100.00011001.110 00110.00000000 = 140.25.198.0/23

Subnet #6-4: 10001100.00011001.110 01000.00000000 = 140.25.200.0/23

Subnet #6-14: 10001100.00011001.110 11100.00000000 = 140.25.220.0/23

Subnet #6-15: 10001100.00011001.110 11110.00000000 = 140.25.222.0/23

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6 7

5 List the host addresses that can be assigned to Subnet #6-3 (140.25.198.0/23):

Subnet #6-3: 10001100.00011001.1100011 0.00000000 = 140.25.198.0/23

Host #1 10001100.00011001.1100011 0.00000001 = 140.25.198.1/23

Host #2 10001100.00011001.1100011 0.00000010 = 140.25.198.2/23

Host #3 10001100.00011001.1100011 0.00000011 = 140.25.198.3/23

Host #4 10001100.00011001.1100011 0.00000100 = 140.25.198.4/23

Host #5 10001100.00011001.1100011 0.00000110 = 140.25.198.5/23

Host #509 10001100.00011001.1100011 1.11111101 = 140.25.199.253/23

Host #510 10001100.00011001.1100011 1.11111110 = 140.25.199.254/23

6 Identify the broadcast address for Subnet #6-3 (140.25.198.0/23):

10001100.00011001.1100011 1.11111111 = 140.25.199.255

7 Specify the eight subnets of Subnet #6-14 (140.25.220.0/23):

Subnet #6-14: 10001100.00011001.1101110 000000000 = 140.25.220.0/23

Subnet#6-14-0: 10001100.00011001.1101110 0.00000000 = 140.25.220.0/26

Subnet#6-14-1: 10001100.00011001.1101110 0.01000000 = 140.25.220.64/26

Subnet#6-14-2: 10001100.00011001.1101110 0.10000000 = 140.25.220.128/26

Subnet#6-14-3: 10001100.00011001.1101110 0.11000000 = 140.25.220.192/26

Subnet#6-14-4: 10001100.00011001.1101110 1.00000000 = 140.25.221.0/26

Subnet#6-14-5: 10001100.00011001.1101110 1.01000000 = 140.25.221.64/26

Subnet#6-14-6: 10001100.00011001.1101110 1.10000000 = 140.25.221.128/26

Subnet#6-14-7: 10001100.00011001.1101110 1.11000000 = 140.25.221.192/26

8 List the host addresses that can be assigned to Subnet #6-14-2

(140.25.220.128/26):

Subnet#6-14-2: 10001100.00011001.11011100.10000000 = 140.25.220.128/26

Host #1 10001100.00011001.11011100.10000001 = 140.25.220.129/26

Host #2 10001100.00011001.11011100.10000010 = 140.25.220.130/26

Host #3 10001100.00011001.11011100.10000011 = 140.25.220.131/26

Host #4 10001100.00011001.11011100.10000100 = 140.25.220.132/26

Host #5 10001100.00011001.11011100.10000101 = 140.25.220.133/26

Host #61 10001100.00011001.11011100.10111101 = 140.25.220.189/26

Host #62 10001100.00011001.11011100.10111110 = 140.25.220.190/26

9 Identify the broadcast address for Subnet #6-14-2

(140.25.220.128/26):

10001100.00011001.11011100.10111111 = 140.25.220.191

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Appendix E - CIDR Exercises

1 List the individual network numbers defined by the CIDR block

200.56.168.0/21.

_______________________________________________________

2 List the individual network numbers defined by the CIDR block

195.24/13.

_______________________________________________________

3 Aggregate the following set of four /24 IP network addresses to the

highest degree possible.

212.56.132.0/24

212.56.133.0/24

212.56.134.0/24

212.56.135.0/24

_______________________________________________________

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6 9

4 Aggregate the following set of four /24 IP network addresses to the

highest degree possible.

212.56.146.0/24

212.56.147.0/24

212.56.148.0/24

212.56.149.0/24

_______________________________________________________

5 Aggregate the following set of 64 /24 IP network addresses to the

highest degree possible.

202.1.96.0/24 202.1.97.0/24

202.1.98.0/24

202.1.126.0/24

202.1.127.0/24

202.1.128.0/24

202.1.129.0/24

202.1.158.0/24

202.1.159.0/24[answer has three parts, so use three lines]

_______________________________________________________

6 How would you express the entire Class A address space as a single

CIDR advertisement?

_______________________________________________________

7 How would you express the entire Class B address space as a single

CIDR advertisement?

_______________________________________________________

8 How would you express the entire Class C address space as a single

CIDR advertisement?

_______________________________________________________

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Solutions for CIDR Exercises

1 List the individual network numbers defined by the CIDR block

200.56.168.0/21.

a. Express the CIDR block in binary format:

200.56.168.0/21 11001000.00111000.10101 000.00000000

b. The /21 mask is 3 bits shorter than the natural mask for a tradi-

tional /24. This means that the CIDR block identifies a block of 8 (or

23) consecutive /24 network numbers.

c. The range of /24 network numbers defined by the CIDR block

200.56.168.0/21 includes:

Net #0: 11001000.00111000.10101000 .xxxxxxxx 200.56.168.0

Net #1: 11001000.00111000.10101001 .xxxxxxxx 200.56.169.0

Net #2: 11001000.00111000.10101010 .xxxxxxxx 200.56.170.0

Net #3: 11001000.00111000.10101011 .xxxxxxxx 200.56.171.0

Net #4: 11001000.00111000.10101100 .xxxxxxxx 200.56.172.0

Net #5: 11001000.00111000.10101101 .xxxxxxxx 200.56.173.0

Net #6: 11001000.00111000.10101110 .xxxxxxxx 200.56.174.0

Net #7: 11001000.00111000.10101111 .xxxxxxxx 200.56.175.0

2 List the individual network numbers defined by the CIDR block

195.24/13.

a. Express the CIDR block in binary format:

195.24.0.0/13 11000011.00011000.00000000.00000000

b. The /13 mask is 11 bits shorter than the natural mask for a tradi-

tional /24. This means that the CIDR block identifies a block of 2,048

(or 211) consecutive /24 network numbers.

c. The range of /24 network numbers defined by the CIDR block

195.24/13 include:

Net #0: 11000011.00011000.00000000 .xxxxxxxx 195.24.0.0

Net #1: 11000011.00011000.00000001 .xxxxxxxx 195.24.1.0

Net #2: 11000011.00011000.00000010 .xxxxxxxx 195.24.2.0

Net #2045: 11000011.00011111.11111101 .xxxxxxxx 195.31.253.0

Net #2046: 11000011.00011111.11111110 .xxxxxxxx 195.31.254.0

Net #2047: 11000011.00011111.11111111 .xxxxxxxx 195.31.255.0

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3 Aggregate the following set of four /24 IP network addresses to the highest

degree possible.

212.56.132.0/24

212.56.133.0/24

212.56.134.0/24

212.56.135.0/24

a. List each address in binary format and determine the common pre-

fix for all of the addresses:

212.56.132.0/24 11010100.00111000.10000100.00000000

212.56.133.0/24 11010100.00111000.10000101.00000000

212.56.134.0/24 11010100.00111000.10000110.00000000

212.56.135.0/24 11010100.00111000.10000111.00000000

Common Prefix:

11010100.00111000.10000100.00000000

b. The CIDR aggregation is:

212.56.132.0/22

4 Aggregate the following set of four /24 IP network addresses to the

highest degree possible.

212.56.146.0/24

212.56.147.0/24

212.56.148.0/24

212.56.149.0/24

a. List each address in binary format and determine the common pre-

fix for all of the addresses:

212.56.146.0/24 11010100.00111000.10010010.00000000

212.56.147.0/24 11010100.00111000.10010011.00000000

212.56.148.0/24 11010100.00111000.10010100.00000000

212.56.148.0/24 11010100.00111000.10010101.00000000

b. Note that this set of four /24s cannot be summarized as a single /23.

212.56.146.0/23 11010100.00111000.10010010.00000000

212.56.148.0/23 11010100.00111000.10010100.00000000

c. The CIDR aggregation is:

212.56.146.0/23

212.56.148.0/23

Note that if two /23s are to be aggregated into a /22, then both /23s

must fall within a single /22 block. Since each of the two /23s is a

member of a different /22 block, they cannot be aggregated into a sin-

gle /22 (even though they are consecutive). They could be aggregated

into 222.56.144/21, but this aggregation would include four network

numbers that were not part of the original allocation. Hence, the

smallest possible aggregate is two /23s.

7 1

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7 2

5 Aggregate the following set of 64 /24 IP network addresses to the

highest degree possible.

202.1.96.0/24

202.1.97.0/24

202.1.98.0/24

202.1.126.0/24

202.1.127.0/24

202.1.128.0/24

202.1.129.0/24

202.1.158.0/24

202.1.159.0/24

a. List each address in binary format and determine the common pre-

fix for all of the addresses:

202.1.96.0/24 11001010.00000001.01100000.00000000

202.1.97.0/24 11001010.00000001.01100001.00000000

202.1.98.0/24 11001010.00000001.01100010.00000000

202.1.126.0/24 11001010.00000001.01111110.00000000

202.1.127.0/24 11001010.00000001.01111111.00000000

202.1.128.0/24 11001010.00000001.10000000.00000000

202.1.129.0/24 11001010.00000001.10000001.00000000

202.1.158.0/24 11001010.00000001.10011110.00000000

202.1.159.0/24 11001010.00000001.10011111.00000000

b. Note that this set of 64 /24s cannot be summarized as a single /19.

202.1.96.0/19 11001010.00000001.01100000.00000000

202.1.128.0/19 11001010.00000001.10000000.00000000

c. The CIDR aggregation is:

202.1.96.0/19

202.1.128.0/19

Similar to the previous example, if two /19s are to be aggregated into

a /18, the /19s must fall within a single /18 block. Since each of these

two /19s is a member of a different /18 block, they cannot be aggre-

gated into a single /18. They could be aggregated into 202.1/16, but

this aggregation would include 192 network numbers that were not

part of the original allocation. Thus, the smallest possible aggregate is

two /19s.

6 How would you express the entire Class A address space as a single

CIDR advertisement?

Since the leading bit of all Class A addresses is “0”, the entire Class A

address space can be expressed as 0/1.

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7 How would you express the entire Class B address space as a single

CIDR advertisement?

Since the leading two bits of all Class B addresses are “10”, the entire

Class B address space can be expressed as 128/2.

8 How would you express the entire Class C address space as a single

CIDR advertisement?

Since the leading three bits of all Class C addresses are “110”, the

entire Class C address space can be expressed as 192/3.

7 3

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To learn more about 3Com solutions, visit www.3com.com. 3Com Corporation is publicly traded on Nasdaq under

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