How to split a /24 network into four /26 subnets

Short answer

To split an IPv4 /24 into /26 subnets, borrow two bits from the host part. That produces 2² = 4 equal subnets. Each /26 contains 64 addresses: one network address, 62 traditionally usable host addresses, and one broadcast address.

Complete example using 192.168.10.0/24

A /26 mask is 255.255.255.192. Its block size in the last octet is:

256 - 192 = 64

The network boundaries are therefore 0, 64, 128, and 192:

Enter one address from each row with /26 in the IPv4 subnet calculator to verify its network, broadcast, and usable range.

Where the four networks come from

A /24 leaves eight host bits. Moving to /26 assigns two of them to the subnet number:

00xxxxxx → 0–63
01xxxxxx → 64–127
10xxxxxx → 128–191
11xxxxxx → 192–255

Six host bits remain, giving 2⁶ = 64 addresses per block. The four blocks are contiguous, do not overlap, and cover the original /24 exactly.

A reusable procedure

1. Confirm that the starting network is aligned to its /24 boundary.
2. Determine the borrowed bits: four subnets require two.
3. Add them to the prefix: /24 + 2 = /26.
4. Convert the prefix to a mask and calculate the increment, 64.
5. List networks by adding that increment within the original block.
6. Reserve each block’s first address as network and last as broadcast.

Common mistakes

• Counting one subnet’s broadcast as the next subnet’s network.
• Starting from an address that is not the actual /24 boundary.
• Confusing four subnets with four hosts.
• Subtracting two addresses before calculating total block size.
• Choosing /25, which produces only two subnets inside a /24.

RFC 4632 defines CIDR notation through a prefix length. This example uses fixed-size subnets; if the required blocks have different sizes, the task becomes a VLSM allocation and needs a different plan.