Cisco-IP-Addressing

IP Addressing

•

Basic Addressing

•

Working with Addresses

•

Summarization & Subnets

•

VLSM

•

Working with VLSM Networks

•

Classful Addressing

Basic Addressing

10.1.1.1

•

IP addresses are

written in

dotted

decimal

format.

•

Four sections are

separated by dots.

•

Each section contains

a number between 0

and 255.

Dots separate the sections

Basic Addressing

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

addresses as a single 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

Basic Addressing

10.1.1.1

•

32/4 == 8.

•

2

= 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

Basic Addressing

10.1.1.1

• Each device on a network is assigned an IP address.

• Each IP address has two 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

N

et

w

o

rk

H

o

s

Basic Addressing

10.1.1.1

• The network mask shows us where to split the

network and host sections.

• Each place there is a 1 in the network mask, that binary digit belongs to the network portion of the

address.

• Each place there is a 0 in the network mask, that binary digit belongs to the host portion of the address.

00001010 00000001 00000001 00000001

N

et

w

o

rk

H

o

s

t

255.255.255.0

Basic Addressing

10.1.1.1

• An alternative set of terminology is:

• The network portion of the address is called the

prefix.

• The host portion of the address is called the host. • The network mask is

expressed as a prefix length, which is a count of the number of 1’s in the subnet mask.

00001010 00000001 00000001 00000001

P

re

fix

H

o

s

t

11111111 11111111 11111111 00000000

8 + 8 + 8 = 24

Basic Addressing

• The network address is the IP address with all 0’s in the host bits.

• The broadcast address is the IP address with all 1’s in the host bits.

• Packets sent to either address will be

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

Working with Addresses

•

Two of the most

common questions

you are going to face

when dealing with IP

addresses are:

• What’s the network?

• What’s the host?

•

How dow we figure

this out?

Working with Addresses (The Hard Way)

• First, convert the IP

address into binary. This 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 Le

ft

R

ig

h

Working with Addresses (The Hard Way)

Write down the IP

address. 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 network mask,

computer the binary as with the IP

address.

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

address.

Working with Addresses (The Hard Way)

Write down the IP

address. 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 network mask,

computer the binary as with the IP

address.

11111111 11111111 11111111 11000000

8 +8 +8 +2 == 26

NOR these two. _{00000000 00000000 00000000 00010000}

Convert back to

dotted decimal. This is the host address.

Working with Addresses (The Hard Way)

•

To convert from

binary to decimal, use

a simple chart.

•

Add the number

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

Working with Addresses (The Easy Way)

• First, if you are using a network mask, convert it to a prefix length.

• For each octet in the

network mask that is 255, add 8 to the prefix length.

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

255.255.255.192

8 +8 +8 +2 == 26

Working with Addresses (The Easy Way)

• Take the prefix length and divide by 8.

• Take the resulting number, and ignore those octets out of the IP address--these are all part of the network address!

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

fourth octet of the network address.

26/8 == 3

(remainder 2)

192.168.100.80/26

These three octets are part of the network

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.

• Add the three octets we “set aside” earlier, and the network (prefix!) is 192.168.100.64/26. • 80 - 64 == 16, so the host

address is 16.

8 7 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

Working with Addresses (The Easy Way)

• How many hosts are in this network? The remainder tells us there are 64 addresses, minus the network and

broadcast addresses, so 62 hosts.

• To find the broadcast address, subtract 1 from the number of hosts, and add that number to the network address.

• 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 addresses

64 + (64 - 1) == 127

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 out of the IP address--these are all part of the network address!

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

octet of the network address.

22/8 == 2

(remainder 6)

192.168.100.80/22

These three octets are part of the network

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.

8 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.

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.

8 7 6 5 4 3 2 1 1 2 4 8 16 32 64 128

4 x 256 == 1024 1024 – 2 == 1022 hosts

Remainder == 6

Working with Addresses (The Easy Way)

•

The key is to work in octets, rather than

Summarization & Subnets

• A single network address (prefix!) represents a set of hosts attached to a wire. • We can abstract this, and

simply say that a prefix

represents a set of reachable addresses.

• We can say that we’ve

“summarized” information

about the hosts attached to the physical wire by referring to the entire group as a single

Summarization & Subnets

• In effect, we’ve shortened the network part of the address

(prefix!), and lengthened the host portion of the address, in effect describing more hosts

(destinations) in a single address.

• If we can shorten the prefix length to describe multiple hosts with a single network address, why can’t we shorten the prefix length so a single network address describes two networks?

• We can! It’s called address summarization, or just

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

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

VLSM

• VLSM: Variable Length

Subnet Masking

• It simply means that the entire IP address space is treated as one flat address 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

VLSM

•

At this point, you pretty much already know VLSM!

You already know how to find the network address,

broadcast address, and number of 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).

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:

• Start with the chart!

• 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.

Working with VLSM Networks

• 58, 14, 29, 49, 3: reorder to 58, 49, 29, 14, 3. Start with 58. • Smallest number larger than

(58 + 2) is 64. 64 is 2 bits.

• 24 bits of prefix length in the address space given, add 2 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

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 address space given, add 2 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

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 address space given, add 3 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

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 address space given, add 4 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

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 address space given, add 5 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

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

Classful Addressing

•

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

Classful Addressing

Network Class Beginning Digits in Binary

Natural Prefix Length

Range of

Addresses 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

Classful Addressing

• It’s illegal to have multiple network masks within a 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

Working with Classful Addressing

•

You can find the network address,

broadcast address, and number of hosts

as we described earlier.

•

You can find the number of networks by

Working with Classful Addressing

• 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

Working with Classful Addressing

•

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

Working with Classful Addressing

•

Subnet 0

• The network with all the between the host and the

natural major net set to 0.

• This only exists in classful

addressing 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

Working with Classful Addressing

•

Broadcast Subnet

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

• This only exists in calssful address schemes.

10 255 255 0/24 00001010 11111111 11111111 00000000

natural network

natural host

configured

network these bits are 1, so this is

the broadcast network

Working with Classful Addressing

• 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:

• Start with the chart!

• 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.

Working with Classful Addressing

• 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

Private & Special Address Space

Address Space Range of Addresses

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 Range of Addresses

Class D (Multicast)

11110x 224.0.0.0 through 239.255.255.255 Class E

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