📘 08-08: Subnet Calculation Using Binary¶
💡 Why Use Binary in Subnetting?¶
Subnetting is fundamentally a binary operation.
- IP addresses are stored in binary
- Subnet masks operate at the bit level
Understanding binary makes subnetting:
- more intuitive
- less error-prone
🧮 Example 1: Basic Subnetting (/24 → /26)¶
🔹 Given:¶
- IP Address: 192.168.1.0
- Default Subnet Mask: 255.255.255.0 (/24)
- New Subnet Mask: 255.255.255.192 (/26)
🔹 Step 1: Identify Network and Host Bits¶
- 1s → Network portion
- 0s → Host portion
Before subnetting:
- 255.255.255.0
- 11111111.11111111.11111111.00000000
After subnetting:
- 255.255.255.192
- 11111111.11111111.11111111.11000000
Result:
- First 26 bits → Network
- Last 6 bits → Host
🧮 Step 2: Number of Subnets¶
Borrowed bits = 2
Number of subnets = 2² = 4
🧮 Step 3: Number of Hosts per Subnet¶
Remaining host bits = 6
- Total addresses = 2⁶ = 64
- Usable hosts = 64 − 2 = 62
🧮 Step 4: Address/Subnet Ranges (Binary Understanding)¶
IP Address (binary):
11000000.10101000.00000001.00000000
Keep first 26 bits fixed (network part).
-
Lowest address:
11000000.10101000.00000001.00000000 → 192.168.1.0 -
Highest address:
11000000.10101000.00000001.00111111 → 192.168.1.63
Subnet ranges (increment = 64):
| Subnet | Network Address | Broadcast Address |
|---|---|---|
| 1 | 192.168.1.0 | 192.168.1.63 |
| 2 | 192.168.1.64 | 192.168.1.127 |
| 3 | 192.168.1.128 | 192.168.1.191 |
| 4 | 192.168.1.192 | 192.168.1.255 |
🧮 Step 5: Host Range (Example: Subnet 1)¶
- Network ID: 192.168.1.0
- Broadcast ID: 192.168.1.63
- Usable Hosts: 192.168.1.1 → 192.168.1.62
🧮 Example 2: Subnetting a Class B Network (/16 → /18)¶
🔹 Given:¶
- IP Address: 172.16.0.0
- Default Subnet Mask: /16
- New Subnet Mask: /18
🔹 Step 1: Identify Network and Host Bits¶
Before subnetting:
- 255.255.0.0
- 11111111.11111111.00000000.00000000
After subnetting:
- 255.255.192.0
- 11111111.11111111.11000000.00000000
Result:
- First 18 bits → Network
- Last 14 bits → Host
🧮 Step 2: Number of Subnets¶
Borrowed bits = 2
Number of subnets = 2² = 4
🧮 Step 3: Number of Hosts per Subnet¶
Remaining host bits = 14
- Total addresses = 2¹⁴ = 16384
- Usable hosts = 16384 − 2 = 16382
🧮 Step 4: Address/Subnet Ranges (Binary Understanding)¶
IP Address (binary):
10101100.00010000.00000000.00000000
Keep first 18 bits fixed.
-
Lowest address:
10101100.00010000.00000000.00000000 → 172.16.0.0 -
Highest address:
10101100.00010000.00111111.11111111 → 172.16.63.255
🔹 Understanding Multi-Octet Range¶
Total addresses per subnet = 16384
- Each 4th octet cycle = 256 addresses
- 16384 / 256 = 64 values in the 3rd octet
So: - 3rd octet ranges from 0 → 63
🧮 Subnet Ranges¶
| Subnet | Network Address | Broadcast Address |
|---|---|---|
| 1 | 172.16.0.0 | 172.16.63.255 |
| 2 | 172.16.64.0 | 172.16.127.255 |
| 3 | 172.16.128.0 | 172.16.191.255 |
| 4 | 172.16.192.0 | 172.16.255.255 |
🧮 Step 5: Host Range (Example: Subnet 1)¶
- Network ID: 172.16.0.0
- Broadcast ID: 172.16.63.255
- Usable Hosts: 172.16.0.1 → 172.16.63.254
📌 Note¶
The block size method (a faster approach for determining subnet ranges) is covered in:
➡️ 08-06: Subnetting Examples
💡 Key Idea¶
Subnetting using binary helps you:
- clearly visualize network vs host bits
- understand how address ranges are formed
- avoid calculation mistakes
Previous: Subnetting Examples | All Notes | Module 08 Exercise | Next: Implementing Subnets on a Network