Sujoy's Tech Blog: How to Subnet? Subnetting in a Nutshell for Dummies! Learn Subnetting in 10 Minutes!! [Everything Explained]

Hi, I'm Sujoy from India and today I'll tell you how to solve any subnetting practice questions easily! You may be preparing for your CCNA exam or want to learn everyday subnetting. I'll tell you how to subnet in your head easily.

IP Address Basics- An IP address is a 32 bit binary number, represented in 4 blocks of 8bits each. Each 8bit block is called an Octet. For human comprehension, each octet is written in Decimal form, separated by dots. This representation is called 'Dotted-Decimal Representation'. IP addresses are of 5 'classes', which is generally identified by 1st octet of an IP address.
Class A (0-127), Class B (128-191), Class C (192-223), Class D (224-239), Class E (240-254).
Example- 192.168.1.115, which is Class C address, since first octet 192 falls within range (192-223).

Subnet Mask Basics- Subnet Mask is a 32 bit binary number. Just like IP address, the subnet mask is also represented in 4 blocks of 8bits each. Subnet mask, when written with an IP address, represents which subnet the given IP address belongs. Example- 255.255.255.224 or /27. Here /27 means out of 32 bit subnet mask, first 27 bits are reserved from left.
Subnet Mask Categorization- /8-/15 only for Class A, /16-/23 Class A & B, /24-/30 Class A, B & C.

Let's take a question- 192.168.1.115/27. I'll tell you how to do Eleven subnetting calculations easily.

Weighted Binary Chart (represents an octet)

2⁷ 2⁶ 2⁵ 2⁴ 2³ 2² 2¹ 2⁰

128 64 32 16 8 4 2 1

Let's Begin the Calculation!

1. Subnet Mask- The given ip address is 192.168.1.115, which is Class C IP address, means first 3 octets or 24bits of subnet mask is already reserved (which is 255.255.255.0). Now there is /27, means 27-24 = 3 more bits are reserved from 4th octet. Total 27 bits are reserved.
Now from Weighted Binary Chart, if we reserve 3bits from left, the mask value will be 128+64+32 = 224. So complete subnet mask for this subnet = 255.255.255.224.

2. Block Size- Block size is 2^unreserved bits. Here in 4th octet, 3 bits are reserved. Unreserved bits 8-3 = 5. From Weighted Binary Chart, 2^5 = 32, so block size is 32. Block size is very important.

3. Network ID/Network Address- Now we will count in block of 32, which is our block size.
0-32-64-96-128-160-192-224-256. 4th octet of our IP address is 115, which falls in range 96-128. So our network ID is 192.168.1.96 (by starting value of range).

4. Broadcast Address- It's the last IP address of the subnet. Our range is 96-128. Last IP address is 127, because from 128 next subnet starts (128-160). Broadcast address = 192.168.1.127.

5. Start IP Address- It's the first IP address after Network ID. In our subnet, Network ID is 96, so Start IP address = 192.168.1.97.

6. End IP Address- It's the last IP address before Broadcast Address. In our subnet, Broadcast Address is 127, so End IP address = 192.168.1.126.

7. Range of Valid Assignable IP Addresses for This Subnet- 192.168.1.97 - 192.168.1.126.
These IP addresses can be assigned to host devices (such as PCs in that subnet).

8. Number of Subnets- It's given by 2^reserved bits. Reserved bits are 3, so number of subnets = 2^3 = 8. Let's look at this, 0-32-64-96-128-160-192-224-256. Our first subnet is 0-31, next is 32-63, and so on, total 8 subnets.

9. Number of Hosts Per Subnet- It's given by (2^unreserved bits - 2). Or Block Size - 2. Unreserved bits are 5, and 2^5 is 32 (which is our block size). And 32-2 = 30 hosts per subnet. Let's verify!
Our valid IP address range is 192.168.1.97 - 192.168.1.126. If you count to 126, starting from 97, you'll count 30 addresses.

10. IP 192.168.1.115 is in Which Subnet? It's is fourth subnet. Our first subnet is 0-31, second is 32-63, third is 64-95, fourth is 96-127. IP 192.168.1.115 falls in 4th subnet.

11. Represent 192.168.1.115 in Binary- As I mentioned earlier, an IP address is actually a binary number. Let's write each octet in binary format. We'll use Weighted Binary Chart for this calculation.

128 64 32 16 8 4 2 1
1 1 0 0 0 0 0 0 = 192
1 0 1 0 1 0 0 0 = 168
0 0 0 0 0 0 0 1 = 1
0 1 1 1 0 0 1 1 = 115

So the binary representation is- 192.168.1.115 = 11000000.10101000.00000001.01110011
Here's the logic- to make 192 (1st octet), we need 128+64=192. We've taken those two bits, and placed '1' below 128 and 64 in Weighted Binary Chart. We didn't take any other bits so we placed 0 below them.
Similarly for 168, we've taken 128+32+8 = 168. So we've placed '1' below 128, 32 and 8, and so on.

You can watch video version of this article here- https://youtu.be/91ZZ7xVLjmU