Cisco ip-addressing

2© 2003, Cisco Systems, Inc. All rights reserved.RST-2002

IP Addressing


IP Addressing

• Basic Addressing

• Working with Addresses

• Summarization & Subnets

• VLSM

• Working with VLSM Networks

• Classful Addressing

• Working with 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

Each section contains a number between 0 and 255


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

Why????


Basic Addressing

10.1.1.1

• 32/4 == 8.

• 28 = 256.

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

00001010 00000001 00000001 000000018 8 8 8

32

Each 8 digit group represents a number between 0 and 255


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

Ne

two

rk

Ho

st


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

Ne

two

rk

Ho

st

255.255.255.0

11111111 11111111 11111111 00000000


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

Pre

fix

Ho

st

11111111 11111111 11111111 00000000

8 + 8 + 8 = 24

10.1.1.1/24


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 0/2400001010 000000011 00000001 00000000

prefix host

these bits are 0, so this is the network address

10 1 1 255/2400001010 000000011 00000001 11111111

prefix host

these bits are 1, so this is the broadcast address


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?

192.168.100.80/26????


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

ftR

igh

t



Write down the IP address.

11000000 10101000 01100100 01010000192 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 110000008 +8 +8 +2 == 26

AND these two. 11000000 10101000 01100100 01000000

Convert back to dotted decimal. This is the network address.

192 168 100 64



Write down the IP address.

11000000 10101000 01100100 01010000192 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 110000008 +8 +8 +2 == 26

NOR these two. 00000000 00000000 00000000 00010000

Convert back to dotted decimal. This is the host address.

0 0 0 16



• 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

168


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!

255.255.255.1928 +8 +8 +2 == 26

192 == 11000000



• 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

The remainder tells us what the network address

in the fourth octet is



• 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 == 6464 x 2 == 128

Remainder == 2

Network is 64!

80 - 64 == 16

192.168.100.64/26

16 Hosts!



• 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

192.168.100.127 is the broadcast address



• 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

The remainder tells us what the network address

in the third octet is



• 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 == 1004 x 26 == 104

Remainder == 6

Third octet is 100!Set the fourth octet to 0.

192.168.100.0/22



• 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 == 10241024 – 2 == 1022 hosts

Remainder == 6

“0” octets == 11 x 256 == 256



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


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 network.

10.1

.1.2

10.1

.1.4

10.1

.1.7

10.1

.1.8

10.1.1.0/26



• 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.

10.1.1.0/2610.1.1.64/26

10.1.1.2/3210.1.1.4/3210.1.1.7/3210.1.1.8/32

These host addresses are described by this network

10.1.1.0/25

These networks are described by this network



10.1.1.0 through 10.1.1.31.

00001010 00000001 00000001 0000000010 1 1 0

11111111 11111111 11111111 11000000

10.1.1.32 through 10.1.1.63.

00001010 00000001 00000001 0100000010 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 0000000010 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!



• 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.

10.1.1.0/2610.1.1.64/26

10.1.1.2/3210.1.1.4/3210.1.1.7/3210.1.1.8/32

These host addresses are subnets of this network

10.1.1.0/25

These networks are subnets of this network


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.

10.1.1.0/2410.1.2.0/2510.1.2.128/2610.1.2.192/27

All of these are valid in the same network!


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).

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


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.

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



• 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

Start the next block at 10.1.1.64



• Next block is 49 hosts.



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


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









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





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




8 7 6 5 4 3 2 1

1 2 4 8 16 32 64 128

(14 + 2) == 16

24 + 4 == 28


next 14 hosts




• Last block is 3 hosts.




• 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


next 14 hosts



• 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

sub

net

s

sub

net

ssu

bn

et

sup

ern

etsu

per

net

sup

ern

et


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 within a major net.


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

11.0.0.0/8100.0.0.0/8120.0.0.0/8

Class B 110X 16 128.0.0.0/16 through 191.0.0.0/16

130.1.0.0/16148.45.0.0/16190.100.0.0/16

Class C 1110 24 192.0.0.0/24 through 223.0.0.0/24

193.1.3.0/24193.1.4.0/24192.2.5.0/24


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.

10.1.1.0/2410.1.2.0/2410.1.3.0/2510.1.3.128/25

11.1.1.0/2511.1.1.128/26

two different prefix lengths in the same major network


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 subtracting the network mask from the natural mask, and then using the chart.



• 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.

10.1.1.0/2510.0.0.0/8 is class A

25 – 8 == 1717/8 == 2, 1 remaining



• 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 == 1717/8 == 2, 1 remaining

65536 – 2 == 65534 networks



• 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 0/2400001010 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



• 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 0/2400001010 11111111 11111111 00000000

natural network

natural host

configured network these bits are 1, so this is

the broadcast network

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



• 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.

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



• 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


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 (Experimental)

11111x 240.0.0.0 through ....


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 masksS 144.2.2.0/24 [1/0] via 208.0.12.11S 144.2.3.0/29 [1/0] via 208.0.12.11C 208.0.7.0/24 is directly connected, Serial0/0C 208.0.6.0/24 is directly connected, FastEthernet0/0C 208.0.0.0/24 is directly connected, FastEthernet0/1S 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 mask in the same network address space

Date post:	01-Jul-2015
Category:	Technology
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