In many situations, we require a large number of subnets. For example, the class B network address 130.20.0.0 has a default mask of 255.255.0.0 or /16 Prefix. So, this address has 16 network bits in the network portion and 16 host bits in the host portion. The 16 bits in the host portion are available to borrow for creating subnets.
The table in the figure below highlights all the possible scenarios for subnetting a /16 prefix. The total number of hosts in a network with a/16 prefix is ( 216-2 =65534 ). This is an extensive network. We can subnet this network according to our requirements for better management and performance.
| Prefix Length | Subnet Mask | Subnet in Binary | Available Networks | Usable Hosts per Network |
|---|---|---|---|---|
| /17 | 255.255.128.0 | 11111111.11111111.10000000.00000000 (1 borrowed bit) | 2¹ = 2 | 2¹⁵ − 2 = 32,766 |
| /18 | 255.255.192.0 | 11111111.11111111.11000000.00000000 (2 borrowed bits) | 2² = 4 | 2¹⁴ − 2 = 16,382 |
| /19 | 255.255.224.0 | 11111111.11111111.11100000.00000000 (3 borrowed bits) | 2³ = 8 | 2¹³ − 2 = 8,190 |
| /20 | 255.255.240.0 | 11111111.11111111.11110000.00000000 (4 borrowed bits) | 2⁴ = 16 | 2¹² − 2 = 4,094 |
| /21 | 255.255.248.0 | 11111111.11111111.11111000.00000000 (5 borrowed bits) | 2⁵ = 32 | 2¹¹ − 2 = 2,046 |
| /22 | 255.255.252.0 | 11111111.11111111.11111100.00000000 (6 borrowed bits) | 2⁶ = 64 | 2¹⁰ − 2 = 1,022 |
| /23 | 255.255.254.0 | 11111111.11111111.11111110.00000000 (7 borrowed bits) | 2⁷ = 128 | 2⁹ − 2 = 510 |
| /24 | 255.255.255.0 | 11111111.11111111.11111111.00000000 (8 borrowed bits) | 2⁸ = 256 | 2⁸ − 2 = 254 |
| /25 | 255.255.255.128 | 11111111.11111111.11111111.10000000 (9 borrowed bits) | 2⁹ = 512 | 2⁷ − 2 = 126 |
| /26 | 255.255.255.192 | 11111111.11111111.11111111.11000000 (10 borrowed bits) | 2¹⁰ = 1,024 | 2⁶ − 2 = 62 |
| /27 | 255.255.255.224 | 11111111.11111111.11111111.11100000 (11 borrowed bits) | 2¹¹ = 2,048 | 2⁵ − 2 = 30 |
| /28 | 255.255.255.240 | 11111111.11111111.11111111.11110000 (12 borrowed bits) | 2¹² = 4,096 | 2⁴ − 2 = 14 |
| /29 | 255.255.255.248 | 11111111.11111111.11111111.11111000 (13 borrowed bits) | 2¹³ = 8,192 | 2³ − 2 = 6 |
Example: You are a network administrator for a large enterprise that requires 60 sub-networks. Your Public address is 130.20.0.0/16.
Borrowing bits from those mentioned above, the /16 prefix network should start in the third octet, going from left to right. Borrow a single bit one by one until the calculation reaches 60. You can also consult the table in the above figure, which displays the number of subnets and the number of hosts per subnet. We can also create a custom table for 60 subnets. The table below displays the number of subnets that can be made when borrowing bits from the third octet. Notice that the Class B network can borrow up to 14 host bits.
IP Address – 130.20.0.0
Subnet Mask – 255.255.0.0 of /16
Network Bits (N) – 16
Host Bits (H) – 16
Required Sub-networks – 60
For 60 sub-networks, we borrow 6 bits from the third octet for a /22 prefix. VLSM can further adjust sizes (e.g., /24 for smaller sites). Each network’s prefix will change
from /16 + 6 =/22. The subnet mask will be 255.255.252.0 for each Subnetwork, and with 6 borrowed bits, we can make 64 subnets. For Network ID, we will follow the following procedure.
There are 6 borrowed bits. These borrowed bits will be arranged according to the network number, as in the following table.
| Network Number | Borrowed bits arrangement in the third octet | Remarks |
| 0 | 00000000 | The first six digits are the binary of 0 |
| 1 | 00000100 | The first six digits are the binary of 1 |
| 2 | 00001000 | The first six digits are the binary of the 2 |
| 3 | . | The first six digits are the binary of 3 |
| . | . | . |
| . | . | . |
| 50 | 11001000 | The first six digits are the binary of the 50 |
| . | . | . |
| . | . | . |
| 62 | 11111000 | The first six digits are the binary of the 62 |
| 63 | 11111100 | The first six digits are the binary of the 63 |
These binary patterns determine the address ranges, calculating network IDs (e.g., 130.20.80.0), broadcast IPs (e.g., 130.20.83.255), and usable IP ranges (e.g., 130.20.80.1 – 130.20.83.254) for each subnet. For example, we have required the abovementioned parameters for subnet numbers 20, 40, and 55.
Here’s the extracted subnetting calculation from the image showing multiple subnets based on custom binary patterns:
📌 Subnetting Calculation Summary
🔹 Net-20 (Borrowed Bits Binary: 010100)
- Network Address:130.20.01010000.00000000 = 130.20.80.0
- First Usable IP: 130.20.01010000.00000001 = 130.20.80.1
- Last Usable IP: 130.20.01010011.11111110 = 130.20.83.254
- Broadcast Address: 130.20.01010011.11111111 = 130.20.83.255
- Network Prefix: /22
- (20 → Binary
010100
🔹 Net-30 (Borrowed Bits Binary: 101000)
- Network Address: 130.20.01111000.00000000 = 130.20.120.0
- First Usable IP: 130.20.01111000.00000001 = 130.20.120.1
- Last Usable IP: 130.20.01111011.11111110 = 130.20.123.254
- Broadcast Address: 130.20.01111011.11111111 = 130.20.123.255
- Network Prefix: /22
- (30 → Binary
011110, positioned as the first 6 bits in the third octet)
🔹 Net-55 (Borrowed Bits Binary: 110111)
- Network Address: 130.20.11011100.00000000 = 130.20.220.0
- First Usable IP: 130.20.11011100.00000001 = 130.20.220.1
- Last Usable IP: 130.20.11011111.11111110 = 130.20.223.254
- Broadcast Address: 130.20.11011111.11111111 = 130.20.223.255
- Network Prefix: /22
- (55 → Binary
110111, positioned as the first 6 bits in the third octet)
We can do the same process for all 64 sub-networks. So now we can use 50 sub-networks from the 64 sub-networks
Calculating the Hosts for Subnets
To calculate hosts for each subnet, look at the third and fourth octets. After borrowing 6 bits for the subnet, two host bits remain in the third octet and 8 host bits in the fourth octet, for 10 bits in the host portion. So, apply the host calculation formula. There are only 1022 usable host addresses available for each /22 subnet.
Conclusion
Classless subnetting with extensive borrowing ensures scalability for growing networks and aligns with IPv6 /64 prefixes.
