IP addressing 2

Dariusz Dwornikowski,

Andrzej Stroiński

Problems

• equal subnetting (RFC 950, 1985)
• unequal subnetting (RFC 1009, 1987)
• CIDR (RFC 1517-1520, 1993)

Subnetting

A company has been given one network address.

The company has 4 buildings.

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How to allocate addresses ?

How to separate networks ?

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The solution is to divide one network into few smaller ones.

Equal length subnetting

• "borrow" bits from computer part to extend the network part
• enumerate subnets in the borrowed part
• calculate network address, range and broadcast for every new subnet
• calculate new masks for new subnets

Network part

Computer part

New network part

Borrowed bits

Example

Divide 150.10.0.0/16 into 4 subnets:

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• Decide how many bits we need to borrow to enumerate 4 subnets - 2 bits
• Enumerate subnets with combinations on the borrowed bits

00 - the first one

01 - the second one

10 - the third one

11 - the fourth one

Example

Network part

Computer part 14 bits

- -

Two bits size

Before

After

Example: First subnet

• Calculate network address
• first 16 bits is untouched
• the borrowed part 00
• computer part : 14 zeroes
• decimal: 150.10.0.0
• Calculate new network mask
• mask needs to be extended by two borrowed bits
• new mask is 18 ones long
• decimal: 255.255.192.0 or /18

Network part

Computer part (14 bits)

0 0

Example: First subnet

• Calculate range
• smallest computer number 00000000000001
• decimal 150.10.0.1
• largest comp. number 11111111111110
• decimal 150.10.63.254
• Calculate broadcast address
• computer part filled with ones
• decimal 150.10.63.255

Example

Example: solution

Unequal subnetting (VLSM)

Variable Length Subnet Mask (VLSM)

• We want some subnets smaller, some bigger

Solution: multiple subnetting, divide to smaller subnets, then divide these subnets again.

VLSM Example

Let's divide 150.10.64.0/18 into 4 more subnets.

VLSM Example

150.10.0.0/18

150.10.128.0/18

150.10.192.0/18

150.10.64.0/20

150.10.80.0/20

150.10.96.0/20

150.10.112.0/20

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This way routers outside 150.10.0.0/16 can have only one entry in their routing tables

Routes to subnets need to be handled inside the network (supernet).

VLSM - applicability condition

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lets assume that routers R1 and R2 are using RIP algorithm in order to designate routing paths

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used VSLM network is 100.0.0.0/8

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Problem:

• router R2 is sending message to router R1 with network address 100.192.0.0
• what network mask address should router R1 use in order to handle this request?

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Solution:

• (applicability condition) - with each protocol message sent in the network, both addresses should be present: network address and mask address

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VLSM usability condition

VLSM can only be used in routing protocols which announce IP address with its mask.

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RIP v2

OSPF

EIGRP

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CIDR (Classless Inter-Domain Routing)

• classless (mask can be shorter than the min. mask in classful)
• allows for aggregating network addresses into on supernet address
• saving space in routing tables
• used with VLSM
• address ranges given according to demand
• widely used

CIDR allocation

In classful allocation - small flexibility:

A class - /8 mask - over 16 million addresses

B class - /16 mask - 65536 addresses

C class - /24 mask - 256 addresses

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CIDR allocates according to the need:

/27 - 32 addresses

/26 - 64 addresses

...

/13 - 524288 addresses

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CIDR aggregation

• given IP address range, find the longest matching sequence of bits
• cut the remaining part and shorten the mask

Example:

200.1.0.0/24 11001000.00000001.000000 00.00000000

200.1.1.0/24 11001000.00000001.000000 01.00000000

200.1.2.0/24 11001000.00000001.000000 10.00000000

200.1.3.0/24 11001000.00000001.000000 11.00000000

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Supernet address: 200.1.0.0/22

CIDR aggregation

dest

mask

gateway

dest

mask

gateway

CIDR aggregation

• sometimes aggregation to only one address is not possible (it would include non-existing address)

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Example:

200.1.48.0/24 - 200.1.79.0/24

Aggregates to:

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200.1.48.0/20 for (200.1.48.0 - 200.1.63.0)

and

200.1.64.0/20 for (200.1.64.0 - 200.1.79.0)

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first aggregation

second aggregation

Exercises:

• Divide 200.10.20.0/24 to 8 equal subnets
• Aggregate networks
• 202.1.0.0/24 - 202.1.63.0/24
• 202.1.24.0/24 - 202.1.39.0/24
• For the diagram below, propose a network addressing, given the demands and a network address 200.200.50.0/23.

Diagram for ex. 3

R1

R2

R3

R4

200.200.50.0/23

100

120

70

50

30

30