1. An address is 32 bits

An IPv4 address is 32 bits, written as four groups of 8 bits (octets) in decimal. Each bit in an octet is worth a power of two, from 128 on the left to 1 on the right.

37 = 32 + 4 + 1, so its bits are 00100101. To convert, go from left to right and write 1 whenever the bit value fits into what is left.

2. The mask splits the address in two

The prefix length, like /26, says how many bits from the left are the network part. The rest are host bits. The mask writes the same thing as 1s for network bits and 0s for host bits.

3. Five numbers every subnet has

• Network address: all host bits 0. It names the subnet and is not given to a host.
• Broadcast address: all host bits 1. It reaches every host in the subnet.
• First usable host: network + 1.
• Last usable host: broadcast − 1.
• Usable hosts: 2^{host bits} − 2.

4. The block-size shortcut

You do not need binary for most questions. Find the octet the mask cuts through, then:

1. Block size = 256 − the mask value in that octet.
2. Network addresses in that octet are multiples of the block size.
3. The network address is the biggest multiple that is not above the host's octet.
4. The broadcast address is one below the next multiple.

Worked example

5. Choosing a prefix for N hosts

Add 2 (network and broadcast), then find the smallest power of two that is at least that big. The exponent is the number of host bits. The prefix is 32 minus that.

Example: 100 hosts need 102 addresses. 2⁶ = 64 is too small and 2⁷ = 128 fits, so 7 host bits and a /25.

6. Splitting a network

To make more subnets, borrow host bits and add them to the prefix. Borrowing b bits gives 2^b subnets, each with 2^{32 − new prefix} addresses.

7. VLSM step by step

Variable-length subnet masks let each network get the size it needs instead of one size for all. The trick is to always work with blocks: sizes that are powers of two.

1. List what each network needs. Write the number of hosts for every LAN and add 2 for the network and broadcast addresses. A point-to-point link needs 2 hosts, so 4 addresses.
2. Round each up to a power of two. 100 + 2 = 102 becomes 128. 50 + 2 = 52 becomes 64. The block size gives the prefix: 128 addresses means 7 host bits, so /25.
3. Sort largest first. Big blocks need big alignment, so they go first.
4. Allocate in order. Start at the first address of the parent network. Each next subnet starts where the previous one ended (previous network + its block size).
5. Check alignment and fit. Every network address must be a multiple of its own block size, and the last block must end inside the parent network. What is left over is free for growth.

8. Special address ranges

Classes still show up in exams: A starts 0 to 127, B 128 to 191, C 192 to 223, D (multicast) 224 to 239, E (reserved) 240 to 255. Real networks use CIDR, so the class does not decide the mask.

9. Prefix cheat sheet

Block size is the step between network addresses in the octet the mask cuts through (or 1 for a whole-octet mask in the last octet).

10. Common mistakes

• Forgetting to subtract 2 for usable hosts, or subtracting 2 on a /31 or /32.
• Using the host's own octet as the network address instead of rounding down to a multiple of the block size.
• Mixing up the wildcard mask with the subnet mask. The wildcard is 255 minus each mask octet.
• Starting VLSM allocations at an address that is not a multiple of the block size.
• Counting from 0: the first subnet is number 1, so subnet j starts at (j − 1) × block size.

Ready to try it? Open the exercises.