Subnetting Made Simple:

An Algorithmic Approach

By

Munmi Kakoty

21st century has been marked by rapid rise in packet based technologies. IP has been the frontrunner in this field. We are gradually moving towards an all-packet based platform. This paper aims at bringing to fore certain curious mathematical properties of the 32-bit IPv4 structure. These properties can be put to good use in calculating:

a) Subnet Maskb) IPs of subnets and hosts (esp. deeply embedded subnet and hosts of an arbitrarily

large network)

Before we delve into the technical details let’s get an overview of the IPv4 32-bit structure (not header structure!!).

What is an IPv4 address and why do we need it in first place?

IPv4 or IP version-4 address is a 32-bit dotted decimal number that is used for Layer-3 addressing i.e. to identify networks and hosts within the network at Layer-3 level. It is due to IP addresses that we can route packets from one network to another.

Properties of IPv4 structure:

a) It is 32-bit long. This calculates to 232 addresses at our disposal.b) The 32-bits are grouped into a group of 8-bits called an Octet. IPv4 has 4*Octets.

To understand subnetting we need to understand two approaches to classifying the IPv4address space:

a) Classful addressing approachb) Classless addressing approach or CIDR (Classless Inter Domain Routing)

Let us understand each of them step by step.

Classful Approach: This approach of addressing scheme divides the IPv4 address space into classes. In other words IPv4 addresses are grouped into different groups otherwise known as classes. This approach defines 5 classes:

a) Class Ab) Class Bc) Class Cd) Class De) Class E

2-level hierarchy model:

In this approach the 32-bit address space is divided into two-level hierarchy: Network Bits and Host Bits

Network Bits Host Bits

Each class has got certain properties and these are illustrated in a very brief manner below.

Class A: The properties of the Class A IP range is highlighted below:

0Network Bits Host Bits

a) Network Bits, n = 8-bitsb) Host Bits, h = 24-bitsc) First bit of the network part is 0.d) Number of networks= 28 or 256e) Number of hosts= 224 or 16777216f) IP address range= 0.x.x.x to 127.x.x.xg) First network of Class A is 0.0.0.0 or All-zero network. This is a reserved network

hence can’t be used.h) The network 127.x.x.x is used for loopback testing and hence can’t be used for

normal addressing.i) Usable IP address range= 1.x.x.x to 126.x.x.x

Class B: The properties of the Class B IP range is highlighted below:

10Network Bits Host Bits

a) Network Bits, n = 16-bitsb) Host Bits, h = 16-bits c) First two bits of the network part are 1 0.d) Number of networks= 216 or 65536e) Number of hosts= 216 or 65536f) IP address range= 128.0.x.x to 191.0.x.x

Class C: The properties of the Class C IP range is highlighted below:

110Network Bits Host

Bits

a) Network Bits, n = 24-bitsb) Host Bits, h = 8-bits c) First three bits of the network part are 110.d) Number of networks= 224 or 16777216e) Number of hosts= 28 or 256f) IP address range= 128.0.x.x to 191.0.x.x

Class D is used for multicasting and Class E is reserved. These classes will not be discussedin details.

Properties of Class D:

a) First 4-bits of the 32-bits is 1110b) No network or Host part defined

Properties of Class E:

a) First 4-bits of the 32-bits is 1111b) No network or Host part defined

Limitations of classful addressing approach:

The classful addressing approach has the following major limitations:

a) Class A contains small number of network with a relatively large number of hosts while Class C contains large number of network with relatively small number of hosts. This sort of size extremities has led to lot of wastage of IPs and rapid depletion of the intermediate sized Class B network.

b) Addressing system is quite rigid. Say if you have 100 hosts one has to buy a Class C network even leading to wastage of 156 IPs if there is no utilization of the same in near future.

3-level hierarchy model:

In this model another level of hierarchy namely subnet bits (borrowed from host bits) are added. This level leads to:

a) Addressing becomes more flexible as we can now break a large network into smaller network of our requirement.

b) Wastage of IP is minimized.c) Introduction of subnet mask.

Network Bits Subnetwork Bits Host Bits

What is a subnet?

A subnetwork, or subnet, is a logically visible, distinctly addressed part of a single IP network.

What is a subnet mask?

Also called netmask. A customer-configurable value for increasing the number of bits within an IPv4 address that are used for network identification. Use of a subnet mask allows one IP address to be subdivided into multiple networks (called subnets). Within a subnet mask, the bits set to 1 specify the portion of the address used to identify networks, while the bits set to 0 identify the hosts.

Default subnet masks: There are three default subnet masks corresponding to the three classes. They are:

a) Class A Subnet Mask represented by 255.0.0.0b) Class B Subnet Mask represented by 255.255.0.0c) Class C Subnet Mask represented by 255.255.255.0

The standard subnet mask is only for reference purpose. Two approaches for exploiting the advantage 3-level hierarchical addressing scheme are used:

a) Fixed Length Subnet Masking or FLSMb) Variable Length Subnet Masking or VLSM

FLSM is based on classful addressing approach while VLSM is based on CIDR or classless addressing approach.

Brief description of FLSM and VLSM is presented for understanding.

FLSM or Fixed Length Subnet Masking:

FLSM is based on the philosophy “one-size-fits-all” design. FLSM breaks a network into equal sized subnet. FLSM is applicable to classful networks i.e. Class A, B & C.

To illustrate let’s take a Class C network 192.9.1.0 255.255.255.0. Using FLSM method the number of subnet bits, S= 2, 3, 4, 5, & 6.

Subnet Bits, S FLSM Subnet Mask

No. Ofsubnets, 2N

Host Bits, H No. of usable hosts, 2H-2

2 255.255.255.192 4 6 62

3 255.255.255.224 8 5 30

4 255.255.255.240 16 4 14

5 255.255.255.248 32 3 6

6 255.255.255.252 64 2 2

Say if we want 31 hosts/subnet, from the table we find that we have to use FLSM value 255.255.255.192 and break the network into 4 subnet containing 62 hosts each. But we require only 31 hosts/subnet wastage of 31 host addresses/subnet. Thus FLSM introduces flexibility in addressing but doesn’t help much to avoid wastage of IP addresses.

The solution- VLSM or Variable Length Subnet Masking was introduced.

VLSM or Variable Length Subnet Masking:

A Variable Length Subnet Mask (VLSM) is a means of allocating IP addressing resources to

subnets according to their individual need rather than some general network-wide rule.

VLSM is based on CIDR or Classless Inter Domain Routing. CIDR eliminates the concept of

class. The CIDR notation table is given below.

CIDR Notation Table:

Subnet Mask CIDR Prefix

Total IP's Usable IP's

255.255.255.255 /32 1 1

255.255.255.254 /31 2 0

255.255.255.252 /30 4 2

255.255.255.248 /29 8 6

255.255.255.240 /28 16 14

255.255.255.224 /27 32 30

255.255.255.192 /26 64 62

255.255.255.128 /25 128 126

255.255.255.0 /24 256 254

255.255.254.0 /23 512 510

255.255.252.0 /22 1024 1022

255.255.248.0 /21 2048 2046

255.255.240.0 /20 4096 4094

255.255.224.0 /19 8192 8190

255.255.192.0 /18 16,384 16,382

255.255.128.0 /17 32,768 32,766

255.255.0.0 /16 65,536 65,534

255.254.0.0 /15 131,072 131,070

255.252.0.0 /14 262,144 262,142

255.248.0.0 /13 524,288 524,286

255.240.0.0 /12 1,048,576 1,048,574

255.224.0.0 /11 2,097,152 2,097,150

255.192.0.0 /10 4,194,304 4,194,302

255.128.0.0 /9 8,388,608 8,388,606

255.0.0.0 /8 16,777,216 16,777,214

254.0.0.0 /7 33,554,432 33,554,430

252.0.0.0 /6 67,108,864 67,108,862

248.0.0.0 /5 134,217,728 134,217,726

240.0.0.0 /4 268,435,456 268,435,454

224.0.0.0 /3 536,870,912 536,870,910

192.0.0.0 /2 1,073,741,824 1,073,741,822

128.0.0.0 /1 2,147,483,648 2,147,483,646

0.0.0.0 /0 4,294,967,296 4,294,967,294

Getting hands dirty……..

Let’s solve a problem to get up close and personal with VLSM and CIDR.

*Remember the network bits do not change.

Base IP: 140.25.0.0/16 (first 16-bits belong to network)

a) Problem: Break the base network into 16 equally sized sub-networks.

Solution:

Network bits, N=16

No. of subnets required=16

Nearest power of 2 greater than or equal to 16 is 24>=16

Therefore no. of subnet bits, S=4

Length of subnet mask, (N+S) = (16+4) =20

CIDR Notation= /20

No. of host bits, H= [32-(N+S)] = [32-20] = 12

Network Bits, (N) Subnet Bits, (S) Length of subnet mask, (N+S)

Host Bits, [32-(N+S)]

16 4 20 12

Base Network: 10001100.00011001 .00000000.00000000 = 140.25.0.0/16Subnet #0: 10001100.00011001.0000 0000.00000000 = 140.25.0.0/20Subnet #1: 10001100.00011001.0001 0000.00000000 = 140.25.16.0/20Subnet #2: 10001100.00011001.0010 0000.00000000 = 140.25.32.0/20Subnet #3: 10001100.00011001.0011 0000.00000000 = 140.25.48.0/20Subnet #4: 10001100.00011001.0100 0000.00000000 = 140.25.64.0/20Subnet #5: 10001100.00011001.0101 0000.00000000 = 140.25.80.0/20Subnet #6: 10001100.00011001.0110 0000.00000000 = 140.25.96.0/20Subnet #7: 10001100.00011001.0111 0000.00000000 = 140.25.112.0/20Subnet #8: 10001100.00011001.1000 0000.00000000 = 140.25.128.0/20Subnet #9: 10001100.00011001.1001 0000.00000000 = 140.25.144.0/20Subnet #10: 10001100.00011001.1010 0000.00000000 = 140.25.160.0/20Subnet #11: 10001100.00011001.1011 0000.00000000 = 140.25.176.0/20Subnet #12: 10001100.00011001.1100 0000.00000000 = 140.25.192.0/20Subnet #13: 10001100.00011001.1101 0000.00000000 = 140.25.208.0/20Subnet #14: 10001100.00011001.1110 0000.00000000 = 140.25.224.0/20Subnet #15: 10001100.00011001.1111 0000.00000000 = 140.25.240.0/20

b) Problem: Break the base network in such a way that there are 16 hosts/subnet.

Solution:

Network bits, N=16

No. of hosts per subnet=16

Nearest power of 2 greater than or equal to 16 is 24>=16

Therefore no. of host bits, H=4

No. of subnet bits, S= [32-(N+H)] = [32-20] = 12

Length of subnet mask, (N+S) = (16+12) =28

CIDR Notation= /28

Network Bits, (N) Subnet Bits, [32-(N+H)]

Length of subnet mask, (N+S)

Host Bits, (H)

16 12 28 4

Base Network: 10001100.00011001 .00000000.00000000 = 140.25.0.0/16Subnet #0: 10001100.00011001.0000 0000.00000000 = 140.25.0.0/28Subnet #1: 10001100.00011001.0000 0000.00010000 = 140.25.0.16/28Subnet #2: 10001100.00011001.0000 0000.00100000 = 140.25.0.32/28Subnet #3: 10001100.00011001.0000 0000.00110000 = 140.25.0.48/28

:::Subnet#268435454: 10001100.00011001.1111 1111.11100000 = 140.25.0.48/28Subnet # 268435455: 10001100.00011001.1111 1111.11110000 = 140.25.255.240/28

So we have got the formulae:

1) If number of subnets are specified

Network Bit = N

Subnet Bits = S such that 2S>= (No. of required subnets)

Length of subnet = (N+S)

CIDR notation = / (N+S)

Host Bits, H = [32-(N+S)] and No. of usable host IP = (2H-2)

2) If number of hosts are specified

Network Bit = N

Host Bits = H such that 2H>= (No. of required hosts)

Subnet Bits, S = [32-(N+H)]

No. of subnet = 2S

Length of subnet = (N+S)

CIDR notation = / (N+S)

The method described above to find the subnets is known as Binary Approach.

c) Problem: Define the hosts for Subnet#1 obtained in problem (b).

Solution:

Subnet #1: 10001100.00011001.0000 0000.00010000 = 140.25.0.16/28Host#1: 10001100.00011001.0000 0000.00010001 = 140.25.0.17/28Host#2: 10001100.00011001.0000 0000.00010010 = 140.25.0.18/28:Host#32: 10001100.00011001.0000 0000.00011111 = 140.25.0.31/28

The method applied in solving the above problems can be termed as Binary approach. This approach is highly recommendable if you are new to subnetting. This method though has a disadvantage: It’s a bit slow in case of locating a subnet or host that is deeply embedded. For this purpose a new faster algorithm has been developed based on the shifting of octets. These will be introduced later. Let’s have a brief tryst with binary numbers.

Binary numbers:

Binary or Base-2 numbers are denoted by 1 and 0. Any number can be represented by a string of 1s and 0s. The place values from left to right are in the range 2N-1*[0, 1], where Nbelongs to [0, 1, 2, 3…n] i.e. to represent say the string 11001 in decimal we have: (20*1+21*0+22*0+23*1+24*1) [Bits from left to right]

= 25

Powers N-1 …… 6 5 4 3 2 1 0

Place Value

2N-1 …… 26 25 24 23 22 21 20

Converting a decimal number into binary number:

Say we want to convert (25)10 to (25)2 so how do we do it???

*(N)b the subscript, b denotes the base and N denotes the number

Divide (25)10 by 2 repeatedly till the dividend is 1 or 0 and record the remainders obtained in each division stage and then arrange them in reverse order to give (25)2.

1. 25/2= 2*12+1; Remainder Stage-1: 1

2. 12/2= 2*6+0; Remainder Stage-2: 0

3. 6/2= 2*3+0; Remainder Stage-3: 0

4. 3/2= 2*1+1; Remainder Stage-4: 1

5. Dividend= 1

Thus the (25)2 is 11001 the bits arranged in reverse order.

Generalizing the process of arranging the bits:

0 or 1 Remainder at Stage-(N-1)

Remainder at Stage-(N-2)

……... Remainder at Stage-2

Remainder at Stage-1

In the above example of converting (25)10 to (25)2:

0 or 1 Remainder

Stage-4

Remainder

Stage-3

Remainder

Stage-2

Remainder

Stage-1

1 1 0 0 1

For elaborate reference please refer any books on binary mathematics or just Google…!!!!!!

Examples:

1) Address: 192.9.1.0/24

N = 24, S = 4 & H = 4

CIDR notation = / (N+S) = / (24+4) = /28

Subnet-0#192.9.1.0000 0000 = 192.9.1.0/28

Subnet-1#192.9.1.0001 0000 = 192.9.1.16/28

:

Subnet-15#192.9.1.1111 0000 = 192.9.1.240/28

Host-2 of Subnet-1#192.9.1.0001 0010 = 192.9.1.18/28

2) Address: 172.1.0.0/16

N = 16, S = 7 & H = 9

CIDR notation = / (N+S) = / (16+7) = /23

Subnet-0#172.1.0000 0000.0000 0000 = 172.1.0.0/23

Subnet-1#172.1.0000 0010.0000 0000 = 172.1.2.0/23

:

Subnet-127#172.1.1111 1110.0000 0000 = 172.1.254.0/23

Host-2 of Subnet-1#172.1.0000 0010.0001 0010 = 172.1.2.2/23

3) Address: 10.0.0.0/8

N = 8, S = 7 & H = 17

CIDR notation = / (N+S) = / (8+7) = /15

Subnet-0#10.0000 0000.0000 0000.0000 0000 = 10.0.0.0/15

Subnet-1#10.0000 0010. 0000 0000.0000 0000 = 10.2.0.0/15

:

Subnet-127#10.1111 1110. 0000 0000.0000 0000 = 10.254.0.0/15

Host-32769 of Subnet-1#10.0000 0010. 1000 0000.0000 0001 = 10.2.128.1/15

Address: 10.0.0.0/8

N = 8, S = 3 & H = 21

CIDR notation = / (N+S) = / (8+3) = /11

Subnet-0#10.0000 0000.0000 0000.0000 0000 = 10.0.0.0/15

Subnet-1#10.0010 0000. 0000 0000.0000 0000 = 10.32.0.0/15

:

Subnet-7#10.1110 0000. 0000 0000.0000 0000 = 10.224.0.0/15

Host-524289 of Subnet-1#10.0010 1000. 000 0000.0000 0001 = 10.40.0.1/15

Readers can try out few more problems on their own applying the method. So happy subnetting.......!!!!!!!!!!!!!!!!!!!!!!!!!!