Mastering Subnet Mask and Subnetting: A Comprehensive Guide

Subnet Mask

The subnet mask is a 32-bit number, which tells about which is the network portion and which is the host portion of an IP address. As we know from classful IP addresses, that is the network portion and the host portion. But in the case of subnetting, we don’t know about the network and host portion. For this reason, we are using a subnet mask. It will consist of either continuous ones or continuous zeros,

E.g. class a subnet mask is 11111111. 00000000. 00000000. 00000000

An IP address of class A is 10.0.0.0 255.0.0.0

10.0.0.0 is a Class A network and 255.0.0.0 is its subnet mask.

Subnetting means the subdivision of a network. Subnetting subdivides a larger network into smaller networks;

Why We Use Subnetting

The answer is the shortage of IP addresses. If a company reserves any classful IP network, then it will reserve the whole of that network. Similarly, if any other company wants only one IP address in the classful network, it will also reserve another full network which will be a waste of IP addresses.

For example, if a person got the IP 192.168.1.1 255.255.255.0, in 192.168.1.0 network, then there is a need for only IP and he is wasting 256-1=255, so he has wasted 255 IP addresses in one network.

When the use of the internet started becoming popular in 1990, IP addresses were close to running out. Similarly, At that time, it was seen that the growth of the internet would stop. To overcome this problem, there were some solutions in which subnetting was one of them. So there was a solution that divided a larger network into smaller ones and controlled the wastage of IP addresses.

The second one was NAT, while the third one was classless IP addressing. We divided iP addresses into classes, e.g. class A, class B, class C, class D, and class E. Dividing an IP address into classes was a waste of IP addresses. So they decided that the IP address would not be divided into classes and the IP address would not relate to any class. There will be no boundary of a class.

Advantages of Subnetting

Division of a local network

A greater number of networks

Simplified addressing

Better security

Smaller collision and broadcast domain

Enhance management

Good administration

How to do subnetting

Suppose there is an IP address, 192.168.1.0 255.255.255.0

Then write its subnet mask in binary form

1111 1111. 1111 1111. 1111 1111. 0000 0000

We divided the subnet into two parts, the network portion and the host portion. 1 represents the network portion while 0 represents the host portion. So, There are 24 bits of network and 8 bits of host.

The number of hosts in this IP is 2ⁿ, put n=8 because there are 8 zeros which represent the host portion, so 2⁸=256.

The 256 hosts are bigger, which will create greater collision and broadcast domain. So we will try to decrease this network into a smaller one by borrowing some host bits into network bits according to our needs. As there are Three usable classes in IP addresses, we are going to take the first Class network in the subnetting.

Class C

Suppose we take one bit from the host portion and add it to network bits as

11111111. 11111111. 11111111. 00000000

11111111. 11111111. 11111111. 10000000

We borrowed one bit in the last octet, so the number of networks and hosts are below according to the formula:

Finding the total number of networks = 2ⁿ

Finding the total number of hosts per network = 2ⁿ-2

First, find the total number of networks.

As we borrow only one bit, so the value of n=1, put it in the formula,

2¹=2*1=2, so the total number of networks will be Two.

While

The total number of hosts = 2ⁿ-2,

As the total number of zeros in the last octet is 7, so put n=7,

2⁷-2= 2*2*2*2*2*2*-2= 128-2 = 126, the total number of hosts per network = 126.

As 2⁷=128 in the host portion, 128 is called block size, so the block size of each network is 128 means 128 + 128 = 256, and each network will contain 128 host IDs.

As we have two networks,

So 1^st network = 192.168.1.0 255.255.255.128

2^nd network = 192.168.128.0 255.255.255.128

So finding the network ID, 1^st host IP, last host IP, and broadcast IP in 1^st network

Note:- How did 128 in the subnet come, what is the reason beyond 128,

128 64 32 16 8 4 2 1 2ⁿ where n= 0, 1,2,3,4,5,6,7

1 0 0 0 0 0 0 0

As only one bit is on and all the other bits are off (zero) only takes 128,

192.168.1.0 255.255.255.128 OR 192.168.1.0/25

Network ID = 192.168.1.0 255.255.255.128

1^st host IP = 192.168.1.1 255.255.255.128

Last host IP = 192.168.1.126 255.255.255.128

Broadcast IP = 192.168.1.127 255.255.255.128

Now take the 2nd network and find its network ID, 1st host IP, last host IP, and broadcast IP.

192.168.1.128 255.255.255.128 OR 192.168.128.0/25

Network ID = 192.168.1.128 255.255.255.128

1^st host IP = 192.168.1.129 255.255.255.128

Last host IP = 192.168.1.254 255.255.255.128

Broadcast IP = 192.168.1.255 255.255.255.128

Two questions arise,

1^st question, why put minus two (-2) in finding hosts

2^nd question is why we put 128 in the subnet mask.

1^st question is, why do we put minus two (-2) in finding the host formula,

The answer is, that in every network, we waste two host IPs.

One IP is network ID which we can’t assign to any host and the 2^nd is broadcast ID which we also do not give this ID to any host.

The 2^nd question is why we put 128 in the subnet mask.

The answer is, that if we borrow one bit from the host part, then the subnet mask will be 128 and its CIDR notation will be /25 in class C.

Similarly, if we borrow two bits from the host portion, then the subnet mask will be 128+64=192, and its CIDR notation will be/26 in Class C, according to the below table.

If we borrow two bits from the host portion, then 2ⁿ = 2² = 2*2 = 4 networks,

We will make 4 networks while remaining hosts, 8-2 = 6 bits for hosts, put n=6 in the hosts’ formula, To find the number of hosts in a network = 2ⁿ-2 = 2⁶-2= 62 hosts.

Total number of networks = 4, total number of hosts = 62 hosts.

The block size is 2⁶= 64, the block size is 64, while the total network is 4, so 64*4 = 256, and each network will contain 64 hosts.

By taking 1^st network of 192.168.1.0/26

192.168.1.0 255.255.255.192 OR 192.168.1.0/26

Note:-

How 192 in the subnet mask come here we also know that we took 2 bits from the host.

Ans:-

128 64 32 16 8 4 2 1 2n where n= 0, 1,2,3,4,5,6,7

1 1 0 0 0 0 0 0

[So 128+64=192]

Network ID = 192.168.1.0 255.255.255.192

1^st host IP = 192.168.1.1 255.255.255.192

Last host IP = 192.168.1.62 255.255.255.192

Broadcast IP = 192.168.1.63 255.255.255.192

1^st Network 1^st network = 192.168.1.0/26 192.168.1.0 255.255.255.192 or 192.168.1.0/26 Network ID = 192.168.1.0 255.255.255.192 1^st host IP = 192.168.1.1 255.255.255.192 Last host IP = 192.168.1.62 255.255.255.192 Broadcast IP= 192.168.1.63 255.255.255.192	2^nd Network 2^nd network = 192.168.1.64/26 192.168.1.64 255.255.255.192 or 192.168.1.64/26 Network ID = 192.168.1.64 255.255.255.192 1^st host IP = 192.168.1.65 255.255.255.192 Last host IP = 192.168.1.126 255.255.255.192 Broadcast IP= 192.168.1.127 255.255.255.192
3^rd Network 3^rd network = 192.168.1.0/26 192.168.1.128 255.255.255.192 or 192.168.1.128/26 Network ID = 192.168.1.128 255.255.255.192 1^st host IP = 192.168.1.129 255.255.255.192 Last host IP = 192.168.1.190 255.255.255.192 Broadcast IP= 192.168.1.191 255.255.255.192	4^th Network 4^th network = 192.168.1.192/26 192.168.1.192 255.255.255.192 or 192.168.1.192/26 Network ID = 192.168.1.192 255.255.255.192 1^st host IP = 192.168.1.193 255.255.255.192 Last host IP = 192.168.1.254 255.255.255.192 Broadcast IP= 192.168.1.255 255.255.255.192

Class C 2-bits networks

If we take 3 bits from the host portion, then what will be the total networks, and how many total host IDs?

192.168.1.0 255.255.255.0

Taking its masks

11111111. 11111111. 11111111. 00000000

11111111. 11111111. 11111111.11100000

192.168.1.0/27, /27 means that we took 3 bits from the host portion, so a formula is 2ⁿ, put n=3, 2³=2*2*2 = 8, the total number of networks will 8 while finding the total number of hosts,

Number of zeros in host portion = 5

Formula for finding a host in a network = 2ⁿ-2, put n=5, 2⁵-2=30,

There will be a total number of hosts in a network = 30 hosts.

Finding 1^st network IDs

192.168.1.0 255.255.255.224

Network ID = 192.168.1.0 255.255.255.224

1^st host ID = 192.168.1.1 255.255.255.224

Last host ID= 192.168.1.30 255.255.255.224

Broadcast ID= 192.168.1.31 255.255.255.224

For simplification, take all the 8 networks in a single table

1^st Network 1^st network = 192.168.1.0/27 192.168.1.0 255.255.255.224 or 192.168.1.0/27 Network ID = 192.168.1.0 255.255.255.224 1^st host IP = 192.168.1.1 255.255.255.224 Last host IP = 192.168.1.30 255.255.255.224 Broadcast IP= 192.168.1.31 255.255.255.224	2^nd Network 2^nd network = 192.168.1.32/27 192.168.1.32 255.255.255.224 or 192.168.1.32/27 Network ID = 192.168.1.32 255.255.255.224 1^st host IP = 192.168.1.33 255.255.255.224 Last host IP = 192.168.1.62 255.255.255.224 Broadcast IP= 192.168.1.63 255.255.255.224
3^rd Network 3^rd network = 192.168.1.64/27 192.168.1.64 255.255.255.224 or 192.168.1.64/27 Network ID = 192.168.1.64 255.255.255.224 1^st host IP = 192.168.1.65 255.255.255.224 Last host IP = 192.168.1.94 255.255.255.224 Broadcast IP= 192.168.1.95 255.255.255.224	^4th Network 3^th network = 192.168.1.96/27 192.168.1.0 255.255.255.224 or 192.168.1.96/27 Network ID = 192.168.1.96 255.255.255.224 1^st host IP = 192.168.1.97 255.255.255.224 Last host IP = 192.168.1.126 255.255.255.224 Broadcast IP= 192.168.1.127 255.255.255.224
5^th Network 5^th network = 192.168.1.128/27 192.168.1.128 255.255.255.224 or 192.168.1.128/27 Network ID = 192.168.1.128 255.255.255.224 1^st host IP = 192.168.1.129 255.255.255.224 Last host IP = 192.168.1.158 255.255.255.224 Broadcast IP= 192.168.1.159 255.255.255.224	6^th Network 6^th network = 192.168.1.160/27 192.168.1.160 255.255.255.224 or 192.168.1.160/27 Network ID = 192.168.1.160 255.255.255.224 1^st host IP = 192.168.1.161 255.255.255.224 Last host IP = 192.168.1.190 255.255.255.224 Broadcast IP= 192.168.1.191 255.255.255.224
7^th Network 7^th network = 192.168.1.192/27 192.168.1.192 255.255.255.224 or 192.168.1.192/27 Network ID = 192.168.1.192 255.255.255.224 1^st host IP = 192.168.1.193 255.255.255.224 Last host IP = 192.168.1.222 255.255.255.224 Broadcast IP= 192.168.1.223 255.255.255.224	8^th Network 8^th network = 192.168.1.224/27 192.168.1.224 255.255.255.224 or 192.168.1.224/27 Network ID = 192.168.1.224 255.255.255.224 1^st host IP = 192.168.1.225 255.255.255.224 Last host IP = 192.168.1.254 255.255.255.224 Broadcast IP= 192.168.1.255 255.255.255.224

Class C 3 bits subnetting

Class B Subnetting

The classful network of class B consists of 16 bits of network portion and 16 bits of host portion. But if we want to subdivide the classful class B network, then 1^st take one IP address of class B as 172.16.0.0 255.255.0.0,

The subnet mask of the class B network

11111111. 11111111. 00000000. 00000000

Network portion = 16 1’s

Host portion = 16 zeros

So borrow one bit from the host portion as

11111111. 11111111. 00000000. 00000000

11111111. 11111111. 10000000. 00000000

We took one bit from the host portion and On it, so the formula for finding the network IDs, 2n, put n=1, 21= 2*1= 2, so the total number of network = 2 networks

While the number of host ID,

As we took 1 bit from the host portion, so total number of host bits = 16-1=15 bits

Formula for finding number of hosts = 2ⁿ– 2, put n=15, 2¹⁵-2 = 32768-2=32766

Total number of host IDs = 32766

172.16.0.0 255.255.128.0 or 172.16.0.0/17

Take 1^st network

Network ID =172.16.0.0 255.255.128.0

1^st Host ID = 172.16.0.1 255.255.128.0

Last Host ID = 172.16.127.254 255.255.128.0

Broadcast ID = 172.16.127.255 255.255.128.0

While the 2^nd network 172.16.128.0 255.255.128.0

Network ID = 172.16.128.0 255.255.128.0

1^st Host ID = 172.16.128.1 255.255.128.0

Last Host ID = 172.16.128.254 255.255.128.0

Broadcast ID = 172.16.255.255 255.255.128.0

This time if we borrow two bits from the host portion. The classful B network subnet mask is below.

11111111.11111111. 00000000.00000000

11111111. 11111111. 11000000. 00000000

Number of network = 2ⁿ, put n=2, 2²= 2*2 =4,

Total number of hosts = 16-2= 14, put n=14 in formula = 2¹⁴–2= 16384,

1^st network

172.16.0.0 255.255.192.0 or 172.16.0.0/26

172.16.0.0 /26

Network ID = 172.16.0.0 255.255.192.0

First Host ID = 172.16.0.1 255.255.192.0

Last Host ID = 172.16.63.254 255.255.192.0

Broadcast ID = 172.16.63.255 255.255.192.0

Taking 2^nd Network

Network ID = 172.16.128.0 255.255.192.0

First Host ID= 172.16.128.129 255.255.192.0

Last Host ID = 172.16.91.254 255.255.192.0

Broadcast ID = 172.16.255.255 255.255.192.0

172.16.0.0255.255.0.0

The subnet is in binary form 11111111.11111111.00000000. 00000000

There are 16 bits for the network and 16 bits for the host portion.

Taking three bits from the host portion

Take three bits from the host portion of Class B. The classful B network subnet mask is below 11111111. 11111111. 00000000.00000000

11111111. 11111111. 11100000. 00000000

Number of network = 2ⁿ, put n=3, 2³= 2*2*2 =8,

Total number of host bits = 16-3= 13, put n=13 in formula = 2¹³–2= 8192-2,

So the total number of hosts per network is 8190

1st Network 1^st Network = 172.16.0.0/19 172.16.0.0 255.255.224.0 or 172.16.0.0/27 Network ID = 172.16.0.0 255.255.224.0 First Host ID = 172.16.0.1 255.255.224.0 Last Host ID = 172.16.31.254 255.255.224.0 Broadcast ID = 172.16.31.255 255.255.224.0	2^nd Network 2^nd Network = 172.16.32.0/19 172.16.32.0 255.255. or 172.16.32.0/27 Network ID = 172.16.32.0 255.255.224.0 First Host ID = 172.16.32.33 255.255.224.0 Last Host ID = 172.16..62 255.255.224.0 Broadcast ID = 172.16.32.63 255.255.224.0
3^rd Network 3^nd Network = 172.16.64.0/19 172.16.64.0 255.255.224.0 or 172.16.64.0/27 Network ID = 172.16.64.0 255.255.224.0 First Host ID = 172.16.64.1 255.255.224.0 Last Host ID = 172.16.95.254 255.255.224.0 Broadcast ID = 172.16.95.255 255.255.224.0	^4th Network 4^nd Network = 172.16.96.0/19 172.16.96.0 255.255.224.0 or 172.16.96.0/27 Network ID = 172.16.96.0 255.255.224.0 First Host ID = 172.16.96.1 255.255.224.0 Last Host ID = 172.16.127.254 255.255.224.0 Broadcast ID = 172.16.127.255 255.255.224.0
5^th Network 5^th Network = 172.16.128.0/19 172.16.128.0 255.255.224.0 or 172.16.128.0/27 Network ID = 172.16.128.0 255.255.224.0 First Host ID = 172.16.128.1 255.255.224.0 Last Host ID = 172.16.159.254 255.255.224.0 Broadcast ID = 172.16.159.255 255.255.224.0	6^th Network 6^th Network = 172.160.0/19 172.16.160.0 255.255.224.0 or 172.16.160.0/27 Network ID = 172.16.160.0 255.255.224.0 First Host ID = 172.16.160.1 255.255.224.0 Last Host ID = 172.16.191.254 255.255.224.0 Broadcast ID = 172.16.191.255 255.255.224.0
7^th Network 7^th Network = 172.16.192.0/19 172.16.192.0 255.255.224.0 or 172.16.192.0/27 Network ID = 172.16.192.0 255.255.224.0 First Host ID = 172.16.192.1 255.255.224.0 Last Host ID = 172.16.223.254 255.255.224.0 Broadcast ID = 172.16.223.255 255.255.224.0	8^th Network 8^th Network = 172.16.224.0/19 172.16.224.0 255.255.224.0 or 172.16.224.0/27 Network ID = 172.16.224.0 255.255.224.0 First Host ID = 172.16.224.1 255.255.224.0 Last Host ID = 172.16.255.254 255.255.224.0 Broadcast ID = 172.16.255.255 255.255.224.0

Class B 3 bits subnetting

Class A Subnetting

There are 8 bits for the network portion and 24 bits for the host portion in the Class A network. If we divide the Class A network into small classes then from the previous method, we will borrow bits from the host portion.

Suppose the IP address of class A is 10.0.0.0 255.0.0.0,

The subnet mask of the class B network

11111111. 00000000. 00000000. 00000000

Network portion = 8 1’s

Host portion = 24 zeros

So borrow one bit from the host portion as

11111

111. 00000000. 00000000. 00000000

11111111. 10000000. 10000000. 00000000

Class A One-Bit Subnetting

We took one bit from the host portion and On it, so the formula for finding the network IDs, 2n, put n=1, 21= 2*1= 2, so the total number of network = 2 networks

While the number of host IDs,

As we took 1 bit from the host portion, so total number of host bits = 24-1=23 bits

The formula for finding number of hosts per hosts=2ⁿ–2, put n=23,

2²³-2=8388608-2=8388606

Total number of host IDs per network 8388606

10.0.0.0 255.128.0.0 or 10.0.0.0/9

1^st network	2^nd network
Network ID =10.0.0.0 255.128.0.0 1^st Host ID = 10.0.0.1 255.128.0.0 Last Host ID = 10.127.255.254 255.128.0.0 Broadcast ID = 10.127.255.255 255.128.0.0	Network ID =10.128.0.0 255.128.0.0 1^st Host ID = 10.128.0.1 255.128.0.0 Last Host ID = 10.127.255.254 255.128.0.0 Broadcast ID = 10.127.255.255 255.128.0.0

Cass A One bit subnetting

Class Two Bits Subnetting

Borrow two bits from the host portion to the network portion.

11111111.00000000. 00000000. 00000000

11111111.11000000. 00000000. 00000000

Bits borrowed= 2 bits

Total number of networks= 2n,

Put n=2, 2²=2*2 =4,

Total number of networks 4

While the total number of host bits is 24-2= 22, we did minus 2 because we borrowed two bits from the host portion.

The formula for finding the total number of hosts per network= 2ⁿ-2,

Put n=22,

2²²-2=4194304-2= 4194302

So the total number of hosts per network = 4194302

Suppose we have a class A IP= 10.0.0.0, then its subnet mask will 255.192.0.0 or 10.0.0.0/10

1^st network 1^st Network= 10.0.0.0/10 Network ID =10.0.0.0 255.192.0.0 1^st Host ID = 10.0.0.1 255.192.0.0 Last Host ID = 10.63.255.254 255.192.0.0 Broadcast ID = 10.63.255.255 255.192.0.0	2^nd network 2^nd Network= 10.64.0.0/10 Network ID =10.64.0.0 255.192.0.0 1^st Host ID = 10.64.0.1 255.192.0.0 Last Host ID = 10.127.255.254 255.192.0.0 Broadcast ID = 10.127.255.255 255.192.0.0
3^rd network 3^rd Network= 10.128.0.0/10 Network ID =10.128.0.0 255.192.0.0 1^st Host ID = 10.128.0.1 255.192.0.0 Last Host ID = 10.191.255.254 255.192.0.0 Broadcast ID = 10.191.255.255 255.192.0.0	4^th network 4^th Network= 10.192.0.0/10 Network ID =10.192.0.0 255.192.0.0 1^st Host ID = 10.192.0.1 255.192.0.0 Last Host ID = 10.192.255.254 255.192.0.0 Broadcast ID = 10.192.255.255 255.192.0.0

Class A Two bits subnetting

Class A 3 bits Subnetting

Class A network borrowing 3 bits from the host portion

11111111.00000000. 00000000. 00000000

11111111.11100000.00000000.00000000

3 bits borrowed

Total number of networks= 2ⁿ,

Put n=3, 2³=2*2*2=8,

Total number of networks 8

While total number of host bits is 24-3= 21 because 3 bits are taken from the host part to the network portion for subnetting. The formula for finding the total number of hosts per network= 2ⁿ-2,

Put n=21,

2²¹-2=2097152-2= 2097150

So the total number of hosts per network is 2097150

Suppose we have a class A IP= 10.0.0.0, then its subnet mask will be 255.224.0.0 or 10.0.0.0/11

1st Network 1^st Network= 10.0.0.0/11 10.0.0.0 255.224.0.0 or 10.0.0.0/11 Network ID = 10.0.0.0 255.224.0.0 First Host ID = 10.0.0.1 255.224.0.0 Last Host ID = 10.31.255.254 255.224.0.0 Broadcast ID = 10.31.255.255 255.224.0.0	2^nd Network 2^nd Network = 10.32.0.0/11 10.32.0.0 255.224.0.0 or 10.32.0.0/11 Network ID = 10.32.0.0 255.224.0.0 First Host ID = 10.32.0.1 255.224.0.0 Last Host ID = 10.63.255.254 255.224.0.0 Broadcast ID = 10.63.255.255 255.224.0.0
3^rd Network 3^rd Network= 10.64.0.0/11 10.64.0.0 255.224.0.0 or 10.64.0.0/11 Network ID = 10.64.0.0 255.224.0.0 First Host ID = 10.64.0.1 255.224.0.0 Last Host ID = 10.95.255.254 255.224.0.0 Broadcast ID = 10.95.255.255 255.224.0.0	4^th Network 4^th Network= 10.96.0.0/11 10.96.0.0 255.224.0.0 or 10.96.0.0/11 Network ID = 10.96.0.0 255.224.0.0 First Host ID = 10.96.0.1 255.224.0.0 Last Host ID = 10.127.255.254 255.224.0.0 Broadcast ID = 10.127.255.255 255.224.0.0
5^th Network 5^th Network= 10.128.0.0/11 10.128.0.0 255.224.0.0 or 10.128.0.0/11 Network ID = 10.128.0.0 255.224.0.0 First Host ID = 10.128.0.1 255.224.0.0 Last Host ID = 10.159.255.254 255.224.0.0 Broadcast ID = 10.159.255.255 255.224.0.0	6^th Network 6^th Network= 10.160.0.0/11 10.160.0.0 255.224.0.0 or 10.160.0.0/11 Network ID = 10.160.0.0 255.224.0.0 First Host ID = 10.160.0.1 255.224.0.0 Last Host ID = 10.191.255.254 255.224.0.0 Broadcast ID = 10.191.255.255 255.224.0.0
7^th Network 7^th Network= 10.192.0.0/11 10.192.0.0 255.224.0.0 or 10.192.0.0/11 Network ID = 10.192.0.0 255.224.0.0 First Host ID = 10.192.0.1 255.224.0.0 Last Host ID = 10.223.255.254 255.224.0.0 Broadcast ID = 10.223.255.255 255.224.0.0	8^th Network 8^th Network= 10.224.0.0/11 10.224.0.0 255.224.0.0 or 10.128.0.0/11 Network ID = 10.224.0.0 255.224.0.0 First Host ID = 10.224.0.1 255.224.0.0 Last Host ID = 10.255.255.254 255.224.0.0 Broadcast ID = 10.255.255.255 255.224.0.0

Class A 3 bits subnetting

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