ECE 4110 – Internetwork Programming Subnetting, Supernetting, and Classless Addressing.

ECE 4110 – Internetwork Programming

Subnetting, Supernetting, and

Classless Addressing

Spring 2003Tuna Tugcu ECE 4110 – Internetwork Programming 2

IP Addresses IP address is 32 bits long. Conceptually the address is the pair

(NETID, HOSTID). IP domain name and addresses

managed: Formerly, by IANA (Internet Assigned

Numbers Authority) Now by ICANN (Internet Corporation for

Assigned Names and Numbers)

3

Classful Addressing (revisited)


Classful Addressing (cont’d)

Historically, a class A address was assigned to networks with > 216 (65,536) hosts.

Class B to networks with 28 (256) to 216 hosts.

Class C to networks with < 218 hosts. Class E was reserved for future use.



Scaling issues: Eventual exhaustion of the IPv4 address

space. Ability to route traffic between ever

increasing number of networks that comprise the internet.

IPv4 uses 32 bit address 232 4.3 billion addresses, and six billion lives on earth at present.

CIDR (Classless Inter-Domain Routing) slowed down address exhaustion.



Class A all 0s network number is used to represent the “default” route (0.0.0.0).

A routing table entry that means any destination not matching any other table entry should be sent to the default route.

Class A all 1s network number is loopback. Packet never leaves the machine. Used for testing the protocol stack or accesing

processes on the same machine. Another block (netid 10) in Class A address

space is reserved for private use. 27 - 3 = 125 class A networks.



16 class B blocks are reserved for private addresses.

214 - 16 = 16,368 class B blocks. 256 Class C blocks are used for

private addresses. 221-256 = 2,096,896 class C

networks.



Most of class E is wasted. Nobody wants class C. In the beginning:

IP addresses were assigned based on request, not need.

32-bit address was thought to be plenty. Classful addressing is easy to understand

and implement, but not efficient. Class C blocks are too small, class B blocks are too large.


Subnetting Makes the subnet structure of a

network invisible outside the organization’s private network.

External Internet does not need to know internal subnet structure.

To reach any host, external routers only need to know the path to the “gateway” router for the entire subnetwork.

Subnetting reduces the size of routing tables.


Subnetting (cont’d)

At boot time, a machine gets its own IP address (E.g., stored on disk).

Host also needs to know how many bits are used for subnet ID and how many for host ID. This is the subnet mask.

Subnet mask is 32 bit value containing “one bits” for the network ID and subnet ID, “zero valued bits” for host ID.



When a host is given its own IP address and its subnetwork mask it can then figure out: Am I class A, B, or C address?

Look at higher order bits. Where is the boundary between the

network ID and the subnet ID? Defined by class definition.

Where is the boundary between the subnet ID and the host ID?

Host ID is 0s in mask.



All 0’s host number is used to identify the base network (or subnetwork).

All 1’s host number represents broadcast address for the network (or subnetwork).

It is also possible to deploy network numbers from the private address space for internal connectivity (RFC 1918). A Network Address Translator (NAT) must be

used to provide external internet access.


A Network with Two Levels ofHierarchy (not subnetted)

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A Network with Three Levels ofHierarchy (subnetted)


Addresses in a Network withand without Subnetting

“Site+Subnet id” is called extended network prefix.


Analogy from the Telephone Network

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Default Mask vs Subnet Mask


Finding the Subnet Address: Straight Method 11001000 00101101 00100010 00111000 11111111 11111111 11110000 00000000 11001000 00101101 00100000 00000000

The subnetwork address is 200.45.32.0.


Finding the Subnet Address: Short-Cut Method


Comparison of a Default Mask and a Subnet Mask

The number of subnets must be a power of 2.


Example

A company is granted the site address 201.70.64.0 (class C). The company needs six subnets. Design the subnets.


Example (Sol’n) The number of 1s in the default

mask is 24 (class C). 6 is not a power of 2. The next number that is a power of 2 is 8. We need 3 more 1s in the subnet mask. The total number of 1s in the subnet mask is

27 (24 + 3). Subnet mask:

11111111 11111111 11111111 11100000, or 255.255.255.224

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Example (Sol’n)


Variable Length Subnet Masks (VLSM) IP network is subdivided in

unequal pieces. Each subnetwork has its own

mask thus “extended-network-prefixes” have different lengths.

Allows size of subnets to reflect the number of required host addresses in each subdivision.


VLSM (cont’d)

We want to use: Smaller subnets (longer masks) where we have

fewer nodes. Larger subnets (shorter masks) where we have

more nodes. Routing protocols must carry extended-

network-prefix information with each route advertisement.

OK to use: OSPF, RIPV2, IS-IS, CISCO’s E–IGRP they are all classless

Routers must use a forwarding algorithm based on longest match.


VLSM (cont’d)


VLSM: Example 1 Destination: 10.1.2.5 =

00001010.00000001.00000010.00000101 Routing table has entries for:

Route #1: 10.1.0.0/ 2410.1.0.0/24 =

[00001010.00000001.00000000].00000000 Route #2: 10.1.2.0/ 24

10.1.2.0/24 = [00001010.00000001.00000010].00000000

Route #3: 10.1.0.0/ 1610.1.0.0/16 =

[00001010.00000001].00000000.00000000 Route #2 has the longest matching prefix.


Sidenote: Private Networks IANA has reserved the following addresses as

private networks (RFC – 1918). 10.0.0.0 - 10.255.255.255

(10.0.0.0/8 prefix) 172.16.0.0 - 172.31.255.255

(172.16.0.0/12 prefix) 192.168.0.0 - 192.168.255.255

(192.168.0.0/16 prefix) Private networks are not routable on the

Internet. They can be used simultaneously by many

organizations.


VLSM: Example 2 An organization has the “private” network

prefix 192.168.0.0/16 and plans to deploy VLSM.

Here is their plan:

16 Equal size blocks => 4 bits

192.168.0.0/16

0 1 2 3 … 12 13 14 15

0 1 … 30 31

0 1 … 6 7

0 1 … 14 15


VLSM: Example 2 (cont’d)

To have 16 equal size blocks 24 =16 =>4 bits are needed beyond /16 => 20 extended-network-prefix.

Each of these blocks has 212=4096 addresses inside.



Base Prefix: 11000000.10101000.00000000.00000000 = 192.168.0.0/16

Subnet # 0: 11000000.10101000.00000000.00000000 = 192.168.0.0/20

Subnet # 1: 11000000.10101000.00010000.00000000 = 192.168.16.0/20

Subnet # 2: 11000000.10101000.00100000.00000000 = 192.168.32.0/20

Subnet # 3: 11000000.10101000.00110000.00000000 = 192.168.48.0/20

Subnet # 4: 11000000.10101000.01000000.00000000 = 192.168.64.0/20

…… Subnet # 13: 11000000.10101000.11010000.00000000 =

192.168.208.0/20 Subnet # 14: 11000000.10101000.11100000.00000000 =

192.168.224.0/20 Subnet # 15: 11000000.10101000.11110000.00000000 =

192.168.240.0/20


VLSM: Example 3 While at this level in the hierarchical

addressing plan, let’s look at the resulting host addresses available for Subnet #3.

Define the host addresses for Subnet #34 (192.168.48.0/20).

192.168.0.0/16

0 1 2 3 … 12 13 14 15

0 1 … 30 31

0 1 … 6 7

0 1 … 14 15



Subnet #3: 11000000.10101000.00110000.00000000 = 192.168.48.0 /20

Host #1: 11000000.10101000.00110000.00000001 = 192.168.48.1 Host #2: 11000000.10101000.00110000.00000010 = 192.168.48.2 Host #3: 11000000.10101000.00110000.00000011 = 192.168.48.3 …… …… Host #4093: 11000000.10101000.00111111.11111101 =

192.168.63.253 Host #4094: 11000000.10101000.00111111.11111110 =

192.168.63.254

The broadcast address for Subnet #34 is found by setting the host-number field to all 1s.

Broadcast : 11000000.1010100.00111111.11111111 = 192.168.63.255


VLSM: Example (Notation) We are working with a /16 base address

Subnet #34 is the subnet with value 3 20-bit extended network prefix (a 4-bit field has been added to the /16 base

address) Subnet #144-#144 is Sub-subnet #14

under Subnet #14. Both subnet and sub-subnet fields are 4 bits.

Thus, the extended network prefix is now 24 bits long.


VLSM: Example 4 While still at the same level, let’s look

at the 16 subnets we have under Subnet #144 (192.168.224.0/20).

192.168.0.0/16

0 1 2 3 … 12 13 14 15

0 1 … 30 31

0 1 … 6 7

0 1 … 14 15



Subnet # 14: 11000000.10101000.11100000.00000000 = 192.168.224.0/20

Subnet # 14-0: 11000000.10101000.11100000.00000000 = 192.168.224.0/24

Subnet # 14-1: 11000000.10101000.11100001.00000000 = 192.168.225.0/24

Subnet # 14-2: 11000000.10101000.11100010.00000000 = 192.168.226.0/24

Subnet # 14-3: 11000000.10101000.11100011.00000000 = 192.168.227.0/24

Subnet # 14-4: 11000000.10101000.11100100.00000000 = 192.168.228.0/24

…… …… Subnet # 14-14: 11000000.10101000.11101110.00000000 =

192.168.238.0/24 Subnet # 14-15: 11000000.10101000.11101111.00000000 =

192.168.239.0/24


VLSM: Example 5 Still at the same level, let’s look at the host

addresses on one of these new sub – subnets, Sub-subnet #144-#34 (192.168.227.0/24).

192.168.0.0/16

0 1 2 3 … 12 13 14 15

0 1 … 30 31

0 1 … 6 7

0 1 2 3 … 14 15



Subnet #144-#34: 11000000.10101000.11100011.00000000 = 192.168.227.0/24

Host #1: 11000000.10101000.11100011.00000001 = 192.168.227.1 Host #2: 11000000.10101000.11100011.00000010 = 192.168.227.2 Host #3: 11000000.10101000.11100011.00000011 = 192.168.227.3 Host #4: 11000000.10101000.11100011.00000100 = 192.168.227.4 Host #5: 11000000.10101000.11100011.00000101 = 192.168.227.5 …… …… Host #253: 11000000.10101000.11100011.11111101 = 192.168.227.253 Host #254: 11000000.10101000.11100011.11111110 = 192.168.227.254

The broadcast address for Subnet # 144–#34 is determined by setting all the bits in the host number field to 1:

Broadcast: 11000000.10101000.11100011.11111111 = 192.168.227.255


VLSM: Example 6 Let’s go down one more level, and look at

sub-subnet #144–#144 (192.168.238.0/24).

192.168.0.0/16

0 1 2 3 … 12 13 14 15

0 1 … 30 31

0 1 … 6 7

0 1 … 14 15



Since 8 = 23 , 3 more bits are required to identify each of eight subnets. Extended-network-prefix length will be /27.

Subnet #14-14: 11000000.10101000.11101110.00000000 = 192.168.238.0/24

Subnet #14-14-0: 11000000.10101000.11101110.00000000 = 192.168.238.0/27 Subnet #14-14-1: 11000000.10101000.11101110.00100000 =

192.168.238.32/27 Subnet #14-14-2: 11000000.10101000.11101110.01000000 =

192.168.238.64/27 Subnet #14-14-3: 11000000.10101000.11101110.01100000 =

192.168.238.96/27 Subnet #14-14-4: 11000000.10101000.11101110.10000000 =

192.168.238.128/27 Subnet #14-14-5: 11000000.10101000.11101110.10100000 =

192.168.238.160/27 Subnet #14-14-6: 11000000.10101000.11101110.11000000 =

192.168.238.192/27 Subnet #14-14-7: 11000000.10101000.11101110.11100000 =

192.168.238.224/27


Going the Other Way: Supernetting Several class C blocks combined to

create larger blocks. Two methods:

Pick the blocks to be merged randomly. Routers outside the organization treat each

block separately. Pick them according to a set of rules:

Number of blocks must be a power of 2. Blocks must be contiguous. If number of blocks is N, third byte of the first

address must be divisible by N.


Supernetting (cont’d)


Comparison of Subnet, Sefault, and Supernet Masks


Supernetting: Example 1 Question:

We need to make a supernetwork out of 16 class C blocks. What is the supernet mask?

Solution: For 16 blocks, we need to change four 1s to

0s in the default mask. So, the mask is 11111111 11111111 11110000

00000000 or 255.255.240.0


Supernetting: Example 2 A supernet has a first address of

205.16.32.0 and a supernet mask of 255.255.248.0. A router receives three packets with the following destination addresses: 205.16.37.44 205.16.42.56 205.17.33.76

Which packet belongs to the supernet?


Supernetting: Example 2 (sol’n)

Apply the supernet mask to see if we can find the beginning address.

205.16.37.44 AND 255.255.248.0 205.16.32.0 205.16.42.56 AND 255.255.248.0 205.16.40.0 205.17.33.76 AND 255.255.248.0 205.17.32.0

Only the first address belongs to this supernet.


Supernetting: Example 3

A supernet has a first address of 205.16.32.0 and a supernet mask of 255.255.248.0.

How many blocks are in this supernet and what is the range of addresses?


Supernetting: Example 3 (sol’n)

Supernet mask has 21 1s. Default mask has 24 1s.

(difference of 3 bits) 23 or 8 blocks in this supernet.

Blocks are 205.16.32.0 to 205.16.39.0.

First address is 205.16.32.0. Last address is 205.16.39.255.


CIDR: Classless InterDomain Routing CIDR removes concept of class A, B, C

network addresses. Instead, uses concept of network prefix. CIDR supports arbitrary sized networks. Routers use network prefix instead of

first 3 bits of IP address to determine dividing point between network number and host number.


CIDR (cont’d)

Constraints: Block size must be a power of 2. Beginning address must be divisible by

block size.E.g.: If the block contains 4 addresses, the

starting address cannot be 193.140.196.2, because:11000001 10001100 11000100 0000001011000001 10001100 11000100 0000001111000001 10001100 11000100 0000010011000001 10001100 11000100 00000101

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ECE 4110 – Internetwork Programming Subnetting, Supernetting, and Classless Addressing.

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