IPv4 addressing and subnetting (the fast way)

Subnetting drove every networking exam for decades. Now it's mechanical — once you have the technique, every question takes under 15 seconds. This lesson is the technique. Practice the calculator below until it's automatic.

The address itself

IPv4 is 32 bits, written as four 8-bit numbers (octets) separated by dots: 192.168.1.42. Every octet is 0–255. Behind the scenes:

192       . 168       . 1         . 42
11000000   10101000   00000001   00101010

That's it. Everything else about addressing — classes, masks, subnets — is just rules for how to interpret which bits identify the network and which identify the host.

Mask and prefix length

A mask is a 32-bit number aligned with the address. Every 1 in the mask means "this bit is the network portion"; every 0 means "this bit is the host portion."

IP:    192.168.1.42        11000000.10101000.00000001.00101010
Mask:  255.255.255.0       11111111.11111111.11111111.00000000

The number of 1s in the mask is the prefix length: here, 24. We write the whole thing as CIDR: 192.168.1.42/24.

| Prefix | Mask | Hosts (usable) | |---|---|---| | /24 | 255.255.255.0 | 254 | | /25 | 255.255.255.128 | 126 | | /26 | 255.255.255.192 | 62 | | /27 | 255.255.255.224 | 30 | | /28 | 255.255.255.240 | 14 | | /29 | 255.255.255.248 | 6 | | /30 | 255.255.255.252 | 2 | | /31 | 255.255.255.254 | 2 (point-to-point) |

The "minus 2" is because every subnet reserves the all-zeros host bits for the network address and the all-ones for the broadcast. /31 is the exception for point-to-point links.

The fast technique

For any X.Y.Z.W/N:

1. Find which octet contains the prefix boundary: divide N by 8. The integer part says how many octets are "all network"; the remainder says how many bits of the next octet are network.
2. Compute the block size in that octet: block = 256 - mask_octet_value. Equivalently 2^(8 - remainder).
3. The network address aligns at a multiple of the block size in that octet.
4. The broadcast is network + block - 1.
5. Usable hosts: network + 1 to broadcast - 1.

Example: 10.0.0.135/27.

• 27 / 8 = 3 remainder 3 → the 4th octet has 3 network bits.
• Mask octet = 11100000 = 224. Block = 256 − 224 = 32.
• Network: round 135 down to multiple of 32 → 128. So 10.0.0.128.
• Broadcast: 128 + 32 − 1 = 159. So 10.0.0.159.
• Usable: 10.0.0.129 – 10.0.0.158 (30 hosts).

That's the whole game. Practice with the SubnetCalculator widget above — type random CIDRs and verify your mental answer.

Private vs public address space

| Range | CIDR | Use | |---|---|---| | 10.0.0.0 – 10.255.255.255 | 10.0.0.0/8 | Private (RFC 1918) — enterprises, big LANs | | 172.16.0.0 – 172.31.255.255 | 172.16.0.0/12 | Private | | 192.168.0.0 – 192.168.255.255 | 192.168.0.0/16 | Private — home routers | | 169.254.0.0/16 | | Link-local / APIPA (when DHCP fails) | | 127.0.0.0/8 | | Loopback (mostly 127.0.0.1) | | 224.0.0.0/4 | | Multicast | | 0.0.0.0/8, 240.0.0.0/4 | | Reserved / experimental | | Everything else | | Public — assigned by IANA / RIRs |

Private ranges need NAT to reach the internet — covered in Track D.

Special-purpose addresses

| Address | Meaning | |---|---| | 0.0.0.0 | "this host, this network" (DHCP-discovering host before it has an IP) | | 0.0.0.0/0 | Default route — "any destination" | | 255.255.255.255 | Limited broadcast (does not cross routers) | | <network>.<all-zeros host> | Network address — never assign to a host | | <network>.<all-ones host> | Directed broadcast |

What to remember

• Address is 32 bits + mask. Prefix length = number of mask 1-bits.
• Number of hosts = 2^(32-prefix) − 2, except /31 (= 2).
• The fast subnetting technique: find the octet, compute block = 256 − mask, round down for network, broadcast = network + block − 1.
• Memorise the private ranges (10/8, 172.16/12, 192.168/16).