Chapter 2 1

IP Addressing

“If we all did the things we are capable of doing, we would literally astound ourselves”

- Thomas Alva Edison, 1847-1931

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Objectives• Recognize and describe the various IP address classes

from A to E, and explain how they’re composed and used

• Describe the IPv4 address limitations, and how techniques like Classless Inter-Domain Routing (CIDR) and use of private IP addresses with Network Address Translation (NAT) ease those limitations

• Define the terms subnet and supernet, and apply subnetting and supernetting concepts in solving specific network design problems

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IP Addressing Basics

• Different addressing schemes:

– Symbolic (eg: www.bcit.ca)

– Logical numeric (eg: 172.16.1.10)

– Physical numeric (eg: 6 byte MAC addresses)

• Symbolic addresses are easier to remember than a numeric address such as 199.95.72.8

• Physical numeric addresses are MAC layer addresses associated with the Data Link layer (of the OSI Reference model)

• Logical numeric addresses are IP addresses associated with the Network layer

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IP Addressing IPv4 uses 32-bit addresses, commonly represented in

dotted decimal notation. Eg: 11000000 00001100 00001010 00000101 (in binary)

192 12 10 5 (each octet in decimal) Written as: 192.12.10.5 (in dotted decimal notation)

Classful Addresses Address range is divided into 5 classes (A to E) Each address has two parts:

• Network address (Net id) and Host address (Host id)• A two-level hierarchy

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0 Net id Host id (24 bits)Class A

1 Net id Host id (16 bits)Class B 0

1 Net id Host id (8 bits)Class C 01

1 Multicast group idClass D 01 1

1 Reserved for future useClass E 01 1 1

Classful Addresses

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Address ranges for different classes

Class Range

A 1.0.0.0 to 126.255.255.255

B 128.0.0.0 to 191.255.255.255

C 192.0.0.0 to 223.255.255.255

D 224.0.0.0 to 239.255.255.255

E 240.0.0.0 to 255.255.255.255

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Classful Addresses

• Class A - only ~125 networks possible

– Each network can support 16,777,214 hosts (2^24 - 2)

– 0.0.0.0 is not assigned to a specific network

– The address range 10.x.x.x (x: 0-255) is reserved for private network use (as per RFC 1918)

– 127.x.x.x (x: 0-255) is reserved for loopback testing

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Classful Addresses

• Class B - for moderate to large networks– Each network can support 65,534 hosts (2^16 -

2)– The address range 172.16.0.0 to 172.31.255.255

is reserved for private use

• Class C - for small networks– Each network can support 254 hosts (2^8 - 2)– The address range 192.168.0.0 -192.168.255.255

is reserved for private use

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Types of Addresses• Unicast: data sent to a single host (or, an interface on a

machine)• Broadcast: sent to all hosts on a network

– Directed broadcast - host id with all 1’s• Eg: A packet sent to 190.10.255.255 is received by

all hosts on the network 190.10.0.0• Routers may forward these broadcast packets

– Limited broadcast - 255.255.255.255• Never forwarded by a router

• Multicast: sent to a set of hosts that belong to a “multicast” group

• Host id with all 0’s is not assigned as a host address, but identifies the network.

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Subnetting

• A network can be divided into sub-networks internally, by dividing the host portion of an IP address into a subnet id and a host id within the subnetwork (a three-level hierarchy)

• This activity of stealing bits from the host portion to further subdivide the network portion of an address is called subnetting a network address, or subnetting

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Subnet Mask

• A 32-bit subnet mask identifies the network and subnet bits in an IP address

• If a bit value is 1 in the subnet mask, the corresponding bit in the IP address is considered part of the network address

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Subnet Masks

• The simplest form of subnet masking uses a technique called constant-length subnet masking (CLSM), in which each subnet includes the same number of hosts and represents a simple division of the address space made available by subnetting into multiple equal segments

• Another form of subnet masking uses a technique called variable-length subnet masking (VLSM) and permits a single network address to be subdivided into multiple subnets, in which subnets need not all be the same size

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Subnetting Example 1:

• An large organization is assigned with the network address 190.10.0.0/16. It needs to support about 150 subnets for different locations. In each subnet, it needs to support about 200 hosts.

• As the first step, decide the number of bits needed from host bits to represent the subnet ID.

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Subnetting Example 1:

• Subnetting the network 190.10.0.0 by using 8 bits of the 16 host id bits– Subnet mask: 255.255.255.0– Possible subnets: 2^8 => 256– Possible hosts per subnet: 2^8 - 2 => 254– Addresses of subnetworks:

• 190.10.0.0 (Subnet #0)

• 190.10.1.0 (Subnet #1)

• ….

• 190.10.255.0 (Subnet #255)

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Subnetting Example 1 ...

• For Subnet #0:– A typical host address is 190.10.0.x where x = 1 to

254 (eg: 190.10.0.5), with a subnet mask of 255.255.255.0

– Also written as: 190.10.0.5/24 (without having to write the subnet mask) - Binary Count notation

– “24” identifies the number of contiguous 1 bits in the subnet mask and is called the “length of the Extended-Network-Prefix”

– Directed broadcast addresses of subnet #0:• 190.10.0.255

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Subnetting Example 2:

• An organization is assigned with network address 193.1.1.0/24. It needs to define 6 subnets for internal departments. The largest subnet need to support 25 hosts.

• Step 1: Determine the no. of bits needed from the host id bits (8 in this case) to define 6 subnets – 3 bits => 8 subnets (2 extra for future expansion)

• Step 2: Determine whether the remaining host id bits (5 in this case) is sufficient for max. hosts needed per subnet

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Subnetting Example 2 ...

• Step 2 continued …– 5 bits => 2^5 - 2 => 30 hosts per subnet

• Subnet mask for each subnet:– 11111111 11111111 11111111 11100000– 255.255.255.224

• Extended network prefix for each subnet: /27

• Network addresses:– Base network: 193.1.1.0/24– Subnet #0: 193.1.1.0/27 – Subnet #7: 193.1.1.224/27

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Subnetting Example 2 ...

• Valid host addresses for Subnet #2:– 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#16: 11000001.00000001.00000001.010 10000 = 193.1.1.80/27– ….– Host#30: 11000001.00000001.00000001.010 11110 = 193.1.1.94/27

• Broadcast address for each subnet:– Host id with all 1’s

– For Subnet #2 above: • 11000001.00000001.00000001.010 11111 = 193.1.1.95/27

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More Examples ...

• A host IP address is 193.27.100.110/26. Determine:

– the subnet address

– directed broadcast address for the subnet

– maximum number of possible hosts on the subnet

– maximum number of possible subnets (assuming constant length subnet masking)

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To find the subnet address ...

• When a host IP address is given, to find the subnet address: – convert the dotted decimal address to binary notation

(not necessary to convert decimal digits containing solely network bits to binary)

– identify the host bits in the IP address, using the subnet mask or the extended network prefix

– set all these host bits to zero– convert the resulting binary number back to dotted

decimal notation

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To find the subnet address ...

• In 193.27.100.110/26, there are 26 network bits (26 most significant bits) and 6 (32-26) host bits

• This means, the decimal digit 110 contains 2 network bits (2 most significant bits) and 6 host bits (6 least significant bits)

• decimal 110 => binary 01 101110• Host bits are: 101110• Setting host bits to 0 => 01 000000 => 64 (decimal)• Therefore, subnet address = 193.27.100.64/26

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To find the broadcast address ...

• When a host IP address is given, to find the broadcast address: – convert the dotted decimal address to binary notation

(not necessary to convert decimal digits containing solely network bits to binary)

– identify the host bits in the IP address, using the subnet mask or the extended network prefix

– set all these host bits to 1– convert the resulting binary number back to dotted

decimal notation

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To find the broadcast address ...

• As discussed previously, host bits are: 101110

• Setting host bits to 1 => 01 111111 => 127 (decimal)

• Therefore, broadcast address = 193.27.100.127/26

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To find the maximum number of possible hosts in a subnet ...

• Number of host bits = 6 (32-26)

• Max. possible addresses per subnet = 2^6 = 64

• As host bits with all 0’s and all 1’s are not valid host addresses, max. number of hosts possible

= 64-2 => 62

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To find the maximum number of subnets ...

• Number of subnet bits = 26 - 24 => 2

(where: 26 = total number of network bits

24 = default network bits in the given Class C address)

• Max. possible subnets = 2^2 = 4

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Variable Length Subnet Masks (VLSM)

• A limitation of having only a single subnet mask across a given network-prefix is that once the mask is selected, it locks the organization into a fixed number of fixed-sized subnets.

• In Subnetting Example 1 (subnetting 190.10.0.0 using 8 bits of the host id), there are 256 possible subnets with 254 hosts each. – If a small subnet needs only a max. of 10 hosts, this

wastes IP addresses • A solution is to allow a subnetted network to use more

than one subnet mask (RFC 1009)

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VLSM Example:• An organization is assigned the network number

140.25.0.0/16. It plans to divide the address space into 16 equal sized blocks (subnets 0-15), and then to sub-divide subnet #14 into 16 equal-sized blocks.

• Using 4 bits for subnet id, 16 subnets of the 140.25.0.0/16 address block are:Base net: 10001100.00011001.00000000.00000000 = 140.25.0.0/16

Subnet #0: 10001100.00011001.00000000.00000000 = 140.25.0.0/20

Subnet #1: 10001100.00011001.00010000.00000000 = 140.25.16.0/20

….

Subnet #14: 10001100.00011001.11100000.00000000 = 140.25.224.0/20

Subnet #15: 10001100.00011001.11110000.00000000 = 140.25.240.0/20

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VLSM Example ...• Using 4 more bits for sub-subnet id, 16 sub-subnets

of Subnet #14 are:Subnet #14: 10001100.00011001.11100000.00000000 = 140.25.224.0/20

Subnet #14-0: 10001100.00011001.11100000.00000000 = 140.25.224.0/24

Subnet #14-1: 10001100.00011001.11100001.00000000 = 140.25.225.0/24

….

Subnet #14-14: 10001100.00011001.11101110.00000000 = 140.25.238.0/24

Subnet #14-15: 10001100.00011001.11101111.00000000 = 140.25.239.0/24

• Host addresses for Subnet #14-1:Host #1: 10001100.00011001.11100001.00000001 = 140.25.225.1/24

Host #2: 10001100.00011001.11100001.00000010 = 140.25.225.2/24

….

Host #254: 10001100.00011001.11100001.11111110 = 140.25.225.254/24

• Broadcast address for Subnet #14-1= 140.25.225.255

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The Vanishing IP Address Space

• Interim solutions for IPv4 address depletion problem:– IETF introduced a new way to carve up the IP address

space—Classless Inter-Domain Routing (CIDR)

– RFC 1918 reserves three ranges of IP addresses for private use—a single Class A (10.0.0.0-10.255.255.255), 16 Class Bs (172.16.0.0-172.31.255.255), AND 256 Class

Cs (192.168.0.0-192.168.255.255). When used together with Network Address Translation (a.k.a NAT), private IP addresses can help lift the “cap” on public IP addresses

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Classless Inter-Domain Routing (CIDR)

• Abandons the rigid address classes to eliminate the inefficiency in classful addressing

• CIDR ignores the traditional A, B, and C class

designations for IP addresses, and can therefore set

the network-host ID boundary wherever it wants

to.

• To use a CIDR address on any network, all routers

in the routing domain must “understand” CIDR

notation

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Classless Inter-Domain Routing (CIDR)

• Allows more efficient aggregation of routing info– Route Aggregation: Use of a single entry in a routing

table to represent address space of several networks– Reduces the size of routing tables in routers

• Allows Supernetting– Using contiguous blocks of Class C addresses to

simulate a single, large address space• Documented in RFCs 1517 to 1520• Eg: 192.125.61.8/20 identifies a network with a 20-bit

network prefix

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Supernets • Supernetting takes the opposite approach to subnetting:

by combining contiguous network addresses, it steals bits from the network portion and uses them to create a single, larger contiguous address space for host addresses

• Example: An organization has the following contiguous Class C 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

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Supernets

• The common prefix for all the 4 addresses is:

11010100 00111000 100001

• They can be aggregated as: 212.56.132.0 / 22

• In the Supernet, the network ID has 22 bits and the host ID has 10 bits

• The network address of supernet: 212.56.132.0/22

• The broadcast address of supernet: 212.56.135.255/22

• Valid Host addresses:

212.56.132.1/22 - 212.56.135.254/22

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Summary

• IP addresses allow identifying individual network interfaces (and therefore computers or other devices as well) on TCP/IP networks

• With Classful addressing, 5 address classes (A to E) are defined

• Classes A through C are assigned to individual hosts and consists of network ID and host ID portions

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Summary

• To help ease address scarcity, the IETF created a form of classless addressing called Classless Inter-Domain Routing (CIDR) that permits the network-host boundary basically anywhere

• Subnetting divides an assigned address space into smaller groups (subnetworks) by using bits from the host portion to form a subnetwork ID

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Summary

• Within the Class A, B, and C IP address ranges, the IETF has reserved private IP address ranges

• With CIDR, Supernetting is possible. Supernetting allows borrowing bits from the network portion (opposite of subnetting) to be used as host addresses, to form a “Supernet” by combining contiguous Class C addresses

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References

• RFC 1878, Variable Length Subnet Table For IPv4, Dec.1995

• http://www.mcmcse.com/articles/subnetting.shtml (on Subnetting Confusion)