Δίκτυα και Επικοινωνίες

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Summarization & Subnets

VLSM

Working with VLSM Networks

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10.1.1.1

written in
dotted
decimal

format.

Four sections are
separated by dots.

Each section contains
a number between 0
and 255.

Dots separate the sections

Each section
contains a number
between 0 and 255

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10.1.1.1

Why is each section a
number between 0 and 255?

Computers operate in
binary, humans operate in
decimal.

Computers treat IP
large 32 digit binary
number, but this is hard
for people to do.

So, we split them up into
four smaller sections so
we can remember and
work with them better!

Dots separate the sections

Each section
contains a number
between 0 and 255

Why????

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10.1.1.1

32/4 == 8.

2
8

= 256.

But, computers
number starting at 0,
so to make a space of
256 numbers, we
number from 0 to 255.

00001010 00000001 00000001 00000001

8

8

8

8

32

Each 8 digit group
represents a number
between 0 and 255

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10.1.1.1

Each device on a network
is assigned an IP

fundamental parts:

The
network

portion,
which describes the
physical wire the device
is attached to.

The
host

portion, which
identifies the host on
that wire.

How can we tell the
difference between the
two sections?

00001010 00000001 00000001 00000001

Network

Host

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10.1.1.1

us where to split the
network and host sections.

Each place there is a 1 in
binary digit belongs to the
network portion of the

Each place there is a 0 in
binary digit belongs to the
host portion of the

00001010 00000001 00000001 00000001

Network

Host

255.255.255.0

11111111 11111111 11111111 00000000

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10.1.1.1

An alternative set of
terminology is:

The network portion of
prefix.

The host portion of the
host.

expressed as a
prefix
length,

which is a count of
the number of 1’s in the

00001010 00000001 00000001 00000001

Prefix

Host

11111111 11111111 11111111 00000000

8 + 8 + 8 = 24

10.1.1.1/24

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0’s in the host bits.

all 1’s in the host bits.

Packets sent to either
delivered to all the
hosts connected to the
wire.

10 1 1 0/24

00001010 000000011 00000001 00000000

prefix

host

these bits are 0, so this is the network address

10 1 1 255/24

00001010 000000011 00000001 11111111

prefix

host

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Two of the most
common questions
you are going to face
when dealing with IP

What’s the network?

What’s the host?

How dow we figure
this out?

192.168.100.80/26

????

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Working with Addresses (The Hard Way)

First, convert the IP
is easier than it looks.

Work with one octet at a
time.

Divide by two, farm out the
remainder on the side.

The bottom is the binary
MSD, the top the binary
LSD.

192

96

0

divide by 2

remainder

48

0

divide by 2

remainder

24

0

divide by 2

remainder

12

0

divide by 2

remainder

6

0

divide by 2

remainder

3

0

divide by 2

remainder

1

1

divide by 2

remainder

0

1

divide by 2

remainder

Left

Right

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Working with Addresses (The Hard Way)

Write down the IP

11000000 10101000 01100100 01010000

192 168 100 80

If you have a prefix
length, just wrote
down the number of
1’s. If you have a
computer the binary
as with the IP

11111111 11111111 11111111 11000000

8 +8 +8 +2 == 26

AND these two.

11000000 10101000 01100100 01000000

Convert back to
dotted decimal. This
is the network

192 168 100 64

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Working with Addresses (The Hard Way)

Write down the IP

11000000 10101000 01100100 01010000

192 168 100 80

If you have a prefix
length, just wrote
down the number of
1’s. If you have a
computer the binary
as with the IP

11111111 11111111 11111111 11000000

8 +8 +8 +2 == 26

NOR these two.

00000000 00000000 00000000 00010000

Convert back to
dotted decimal. This

0 0 0 16

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Working with Addresses (The Hard Way)

To convert from
binary to decimal, use
a simple chart.

indicated for each 1
set in the binary
number.

128

1

128

64

0

0

32

1

32

16

0

0

8

1

8

4

0

0

2

0

0

1

0

0

168

16

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Working with Addresses (The Easy Way)

First, if you are using a
to a prefix length.

For each octet in the
add 8 to the prefix length.

For the one octet that isn’t
255, convert to binary and
bits
--
or use a chart!

255.255.255.192

8 +8 +8 +2 == 26

192 == 11000000

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Working with Addresses (The Easy Way)

Take the prefix length and
divide by 8.

Take the resulting number,
and ignore those octets
--
these are all part of the

We’re going to use the
remainder to find the
fourth octet of the network

26/8 == 3
(remainder 2)

192.168.100.80/26

These three
octets are
part of the
network

The remainder tells us
in the fourth octet is

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Working with Addresses (The Easy Way)

Take the remainder, and
find the corresponding
“multiple” on the chart; in
this case, 64.

The largest multiple of 64
that will fit into 80 is 64, so
the network is 64.

“set aside” earlier, and the
network
(prefix!)

is
192.168.100.64/26.

80
-

64 == 16, so the host

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6

5

4

3

2

1

1

2

4

8

16

32

64

128

64 x 1 == 64

64 x 2 == 128

Remainder == 2

Network is 64!

80
-

64 == 16

192.168.100.64/26

16 Hosts!

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Working with Addresses (The Easy Way)

How many hosts are in this
network? The remainder tells
minus the network and
hosts.

that number to the network

The key is to work in octets,
rather than trying to work
with the entire IP address at
once!

8

7

6

5

4

3

2

1

1

2

4

8

16

32

64

128

Remainder == 2

64
-

2 == 62 hosts

64 + (64
-

1) == 127

192.168.100.127 is the

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Working with Addresses (The Easy Way)

What if the prefix length is
less than 24?

Take the prefix length and
divide by 8.

Take the resulting number,
and ignore those octets
--
these are all part of the

We’re going to use the
remainder to find the
third

octet of the network

22/8 == 2
(remainder 6)

192.168.100.80/22

These three
octets are
part of the
network

The remainder tells us
in the third octet is

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Working with Addresses (The Easy Way)

Take the remainder, and
find the corresponding
“multiple” on the chart; in
this case, 4.

The largest multiple of 64
that will fit into 80 is 64, so
the network is 64.

Add the two octets we “set
aside” earlier, and make
any octets after the
network 0’s (the fourth
octet).

The network
(prefix!)

is
192.168.100.0/22.

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7

6

5

4

3

2

1

1

2

4

8

16

32

64

128

4 x 25 == 100

4 x 26 == 104

Remainder == 6

Third octet is 100!

Set the fourth octet to 0.

192.168.100.0/22

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Working with Addresses (The Easy Way)

To find the number of
hosts, take the number of
octets set to 0, which is 1
in this case (the fourth
octet), and multiply by
256.

Next, take the number
relating to the remainder
from the chart, and
multiple this by the
number we just found
above.

Subtract two.

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6

5

4

3

2

1

1

2

4

8

16

32

64

128

4 x 256 == 1024

1024

2 == 1022 hosts

Remainder == 6

“0” octets == 1

1 x 256 == 256

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Working with Addresses (The Easy Way)

The key is to work in octets, rather than
trying to work with the entire IP address at
once!

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Summarization & Subnets

(prefix!) represents a set
of hosts attached to a
wire.

We can abstract this, and
simply say that a prefix
represents a set of

We can say that we’ve
“summarized” information
to the physical wire by
referring to the entire
group as a single network.

10.1.1.2

10.1.1.4

10.1.1.7

10.1.1.8

10.1.1.0/26

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Summarization & Subnets

In effect, we’ve shortened the
(prefix!),

and lengthened the
in effect describing more
hosts (destinations) in a

If we can shorten the prefix
length to describe multiple
hosts with a single network
shorten the prefix length so a
describes two networks?

We can! It’s called
summarization,

or just
summarization.

10.1.1.0/26

10.1.1.64/26

10.1.1.2/32

10.1.1.4/32

10.1.1.7/32

10.1.1.8/32

These
host
are
described
by this
network

10.1.1.0/25

These
networks
are
described
by this
network

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Summarization & Subnets

10.1.1.0 through
10.1.1.31.

00001010 00000001 00000001 00000000

10 1 1 0

11111111 11111111 11111111 11000000

10.1.1.32 through
10.1.1.63.

00001010 00000001 00000001 01000000

10 1 1 64

11111111 11111111 11111111 11000000

10.1.1.0 through
10.1.1.63, so it’s the
same space!

00001010 00000001 00000001 00000000

10 1 1 0

11111111 11111111 11111111 10000000

Changing the mask bit from 1 to 0, which shortens the
prefix length, means the bit in the two networks that
distinguish them from one another are now considered
host bits!

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Summarization & Subnets

A network which is a part
of another network is
called a
subnet.

There is another term, the
supernet, but it’s
definition depends on
whether you are using
VLSM subnetting, or
calssful subnetting, so it
will be defined in the next
two sections.

10.1.1.0/26

10.1.1.64/26

10.1.1.2/32

10.1.1.4/32

10.1.1.7/32

10.1.1.8/32

These
host
are
subnets of
this
network

10.1.1.0/25

These
networks
are
subnets of
this
network

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VLSM

VLSM: Variable Length

It simply means that the
space.

Any prefix length is
allowed in the network at
any point.

10.1.1.0/24

10.1.2.0/25

10.1.2.128/26

10.1.2.192/27

All of these are valid in
the same network!

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VLSM

At this point, you pretty much already know
VLSM! You already know how to find the network
hosts in a network.

Two other common problems in working with
VLSM networks remain:

Building summary addresses from groups of
networks. We won’t cover this here (maybe later in
routing).

Building network addressing schemes from a given
number of hosts and networks.

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Working with VLSM Networks

You have 5 subnets with the following numbers of hosts
on them: 58, 14, 29, 49, 3

You are given the address space 10.1.1.0/24.

Determine what subnets you could use to fit these hosts
into it.

How to solve this:

Order the networks from the largest to the smallest.

Find the smallest number in the chart that will fit the number
of the largest number of hosts + 2.

Continue through each space needed until you either run out
of space, or you finish.

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Working with VLSM Networks

58, 14, 29, 49, 3: reorder to

Smallest number larger than
(58 + 2) is 64. 64 is 2 bits.

24 bits of prefix length in the
for 26.

First network is 10.1.1.0/26.

The next network is 10.1.1.0 +
64, so we start the next
“round” at 10.1.1.64.

8

7

6

5

4

3

2

1

1

2

4

8

16

32

64

128

32 < (58 + 2) < 64

24 + 2 == 26

10.1.1.0/26 takes
care of the first 58
hosts

Start the next block
at 10.1.1.64

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Working with VLSM Networks

Next block is 49 hosts.

Smallest number larger than
(49 + 2) is 64. 64 is 2 bits.

24 bits of prefix length in the
for 26.

We start this block at
10.1.1.64, so network is
10.1.1.64/26.

The next network is 10.1.1.64
+ 64, so we start the next
“round” at 10.1.1.128.

8

7

6

5

4

3

2

1

1

2

4

8

16

32

64

128

32 < (49 + 2) < 64

24 + 2 == 26

10.1.1.64/26 takes
care of the next 49
hosts

Start the next block
at 10.1.1.128

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Working with VLSM Networks

Next block is 29 hosts.

Smallest number larger than
(29 + 2) is 32. 32 is 3 bits.

24 bits of prefix length in the
for 27.

We start this block at
10.1.1.128, so network is
10.1.1.128/27.

The next network is
10.1.1.128 + 32, so we start
the next “round” at
10.1.1.160.

8

7

6

5

4

3

2

1

1

2

4

8

16

32

64

128

16 < (29 + 2) < 32

24 + 3 == 27

10.1.1.128/27
takes care of the
next 29 hosts

Start the next block
at 10.1.1.160

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Working with VLSM Networks

Next block is 14 hosts.

Smallest number larger than
(14 + 2) is 16. 16 is 4 bits
(actually equal, but it still
works!).

24 bits of prefix length in the
for 28.

We start this block at
10.1.1.160, so network is
10.1.1.160/27.

The next network is
10.1.1.160 + 16, so we start
the next “round” at
10.1.1.176.

8

7

6

5

4

3

2

1

1

2

4

8

16

32

64

128

(14 + 2) == 16

24 + 4 == 28

10.1.1.160/28
takes care of the
next 14 hosts

Start the next block
at 10.1.1.176

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Working with VLSM Networks

Last block is 3 hosts.

Smallest number larger than
(3 + 2) is 8. 8 is 5 bits.

24 bits of prefix length in the
for 29.

We start this block at
10.1.1.176, so network is
10.1.1.176/29.

This is the last block of
hosts, so we’re done!

8

7

6

5

4

3

2

1

1

2

4

8

16

32

64

128

4 < (5 + 2) < 8

24 + 5 == 29

10.1.1.176/29
takes care of the
next 14 hosts

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Working with VLSM Networks

A subnet is any network
which is “part of” a larger
network space.

A supernet is any network
which covers a larger
space than a given
network, including the
space covered by the
network.

10.1.2.0/24

10.1.1.0/24

10.1.0.0/23

10.1.2.0/25

10.1.2.128/25

10.1.2.128/26

subnets

subnets

subnet

supernet

supernet

supernet

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Classful subnetting is similar to VLSM,
with two more rules:

The IP address space is divided into
“classes,” with each class having a specific
“natural” prefix length. Each block of address
space is called a “major net.”

You cannot have more than one prefix length
within a major net.

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Network Class

Beginning
Digits in
Binary

Natural
Prefix
Length

Range of

Example Major
Networks

Class A

10XX

8

1.0.0.0/8
through
126.0.0.0/8

11.0.0.0/8

100.0.0.0/8

120.0.0.0/8

Class B

110X

16

128.0.0.0/16
through
191.0.0.0/16

130.1.0.0/16

148.45.0.0/16

190.100.0.0/16

Class C

1110

24

192.0.0.0/24
through
223.0.0.0/24

193.1.3.0/24

193.1.4.0/24

192.2.5.0/24

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It’s illegal to have multiple
single major network.

There cannot be a mix of
/24’s and /25’s in the
10.0.0.0/8 major network.

There cannot be a mix of
/25’s and /26’s in the
11.0.0.0/8 network.

10.1.1.0/24

10.1.2.0/24

10.1.3.0/25

10.1.3.128/25

11.1.1.0/25

11.1.1.128/26

two different
prefix lengths
in the same
major network

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You can find the network address,
as we described earlier.

You can find the number of networks by
subtracting the network mask from the
natural mask, and then using the chart.

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10.1.1.0/25 is in the
10.0.0.0 class A major
network.

The natural prefix length
for a class A network is /8.

Subtract the natural prefix
length from the actual
prefix length.

Divide by 8, holding the
remainder on the side.

10.1.1.0/25

10.0.0.0/8 is class A

25

8 == 17

17/8 == 2, 1 remaining

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Find the remainder in
the power of two’s
chart.

Multiply the result,
256, and the number
from the power of
two’s chart.

Subtract 2.

8

7

6

5

4

3

2

1

1

2

4

8

16

32

64

128

(256 x 2) x 128 == 65536

10.1.1.0/25

10.0.0.0/8 is class A

25

8 == 17

17/8 == 2, 1 remaining

65536

2 == 65534 networks

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Subnet 0

The network with
all the between the
host and the
natural major net
set to 0.

This only exists in
classful
schemes.

10 0 0 0/24

00001010 00000000 00000000 00000000

natural
network

natural
host

configured
network

these bits are 0, so this is subnet 0

10.0.0.0/16

10.0.1.0/16

172.31.0.0/24

172.31.1.0/24

192.168.100.0/25

Yes

No

Yes

No

Yes

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The network with
all the bits
between the host
and the natural
major network set
to 1.

This only exists in
schemes.

10 255 255 0/24

00001010 11111111 11111111 00000000

natural
network

natural
host

configured
network

these bits are 1, so this is the

10.255.0.0/16

10.255.0.0/24

172.31.255.0/24

172.31.255.0/25

192.168.100.128/25

Yes

No

Yes

No

Yes

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You have 5 subnets with the following numbers of hosts
on them: 58, 14, 29, 49, 3

You are given the address space 10.1.0.0/22.

Determine what subnets you could use to fit these hosts
into it.

How to solve this:

Find the largest set of hosts.

Find the smallest number in the chart that will fit the number
of the largest number of hosts + 2.

Use that prefix length for all the subnets
(remember you
cannot have different subnet masks within the same major
network).

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A subnet is any prefix with
a prefix length longer than
the natural prefix length of
the major network.

A supernet is any prefix
with a prefix length
shorter than the natural
prefix length of the major
network.

172.18.1.0/24

Subnet

10.2.0.0/9

Subnet

172.34.0.0/15

Supernet

192.168.44.64/25

Subnet

192.168.44.0/23

Supernet

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10.0.0.0/8

10.0.0.0 through 10.255.255.255

172.16.0.0/19

172.16.0.0 through 172.31.0.0

192.168.0.0/16

192.168.0.0 through 192.168.255.255

Network Class

Beginning Digits in Binary

Class D
(Multicast)

11110x

224.0.0.0 through
239.255.255.255

Class E
(Experimental)

11111x

240.0.0.0 through ....

48

48

48

© 2003, Cisco Sy stems, Inc. All rights reserv ed.

RST
-
2002

Cisco IOS Show IP Route

2651A#sho ip route

....

Gateway of last resort is not set

C 208.0.12.0/24 is directly connected, Serial0/2

....

S 208.1.10.0/24 [1/0] via 208.0.12.11

....

144.2.0.0/16 is variably subnetted, 2 subnets, 2 masks

S 144.2.2.0/24 [1/0] via 208.0.12.11

S 144.2.3.0/29 [1/0] via 208.0.12.11

C 208.0.7.0/24 is directly connected, Serial0/0

C 208.0.6.0/24 is directly connected, FastEthernet0/0

C 208.0.0.0/24 is directly connected, FastEthernet0/1

S 208.1.0.0/16 [1/0] via 208.0.12.11

two different prefix
lengths under the
same major
network

a supernet and natural