Subnet Cheat Sheet
Networking · CIDR, netmasks, host ranges
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The Core Pattern
| CIDR | Netmask | Network Part | Host Bits | Usable Hosts |
|---|---|---|---|---|
| /8 | 255.0.0.0 | X . _ . _ . _ | 24 bits | ~16,000,000 |
| /16 | 255.255.0.0 | X.X . _ . _ | 16 bits | ~65,000 |
| /24 | 255.255.255.0 | X.X.X . _ | 8 bits | 254 |
| /25 | 255.255.255.128 | X.X.X . _ (half) | 7 bits | 126split octet |
Mental Shortcut
The slash ÷ 8 = how many octets are locked.
/8 → 1 locked octet → 255.0.0.0
/16 → 2 locked octets → 255.255.0.0
/24 → 3 locked octets → 255.255.255.0
Broadcast & Host Range — Always the Same Trick
| Address Type | Rule |
|---|---|
| Network address | Last free octet = 0 |
| Broadcast | Last free octet = 255 |
| Host range | Everything in between — .1 to .254 across the free octets |
The 5-Question Flow — Same Every Time
- Netmask → how many octets are locked? (slash ÷ 8)
- Host bits → 32 minus the slash number
- Hosts → 2^(host bits) − 2
- Network addr → given address, last free octet = 0
- Broadcast → last free octet = 255
- Range → .1 to .254 across the free octets
Split Octets — What Your Mind Should Do
Octet bit values, left to right (always this order):
| Bit 1 | Bit 2 | Bit 3 | Bit 4 | Bit 5 | Bit 6 | Bit 7 | Bit 8 |
|---|---|---|---|---|---|---|---|
| 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
| stolen (network) ↑ | ↑ free (host) | ||||||
Example above shows /28 — 4 bits stolen (28 − 24 = 4). Gold = locked, grey = free.
The mental path — step by step for any split CIDR:
- 1Remainderslash ÷ 8 → quotient = full locked octets, remainder = bits stolen/25: 25÷8 = 3 R1 → 1 bit stolen
- 2Stolen valueread the bit value from the table above (start at 128, count right)1 bit → 128 | 2 bits → 128+64=192 | 3 bits → 224
- 3Netmasklast octet = sum of stolen bit values; full octets before it = 255/25 → 255.255.255.128
- 4Block size256 − last-octet netmask (or: value of the last stolen bit)256 − 128 = 128 → subnets start at .0 and .128
- 5Network addrround the 4th octet of your address down to the nearest multiple of block size192.168.1.50/25 → 50 rounds to 0 → network = 192.168.1.0
- 6Broadcastnetwork 4th-octet + block − 10 + 128 − 1 = 127 → broadcast = 192.168.1.127
- 7Host rangenetwork+1 to broadcast−1192.168.1.1 → 192.168.1.126
Alternate Trick — Network Address by Division
Skip the rounding-down guesswork — divide it out instead.
- 1. Block size (B) = 256 − mask value in that octet
- 2. Divide the address octet by B, ignore remainder (just count how many full block sizes fit)
- 3. Multiply that count back by B → network octet
- 4. Zero out everything after it
One-liner: (octet ÷ B, round down) × B
Split CIDR Quick-Reference
| CIDR | Bits stolen | Netmask | Block size | Usable hosts | Subnet starts at… |
|---|---|---|---|---|---|
| /25 | 1 | 255.255.255.128 | 128 | 126 | .0 and .128 |
| /26 | 2 | 255.255.255.192 | 64 | 62 | .0, .64, .128, .192 |
| /27 | 3 | 255.255.255.224 | 32 | 30 | .0, .32, .64, .96 … |
| /28 | 4 | 255.255.255.240 | 16 | 14 | .0, .16, .32, .48 … |
| /29 | 5 | 255.255.255.248 | 8 | 6 | .0, .8, .16, .24 … |
| /30 | 6 | 255.255.255.252 | 4 | 2 | .0, .4, .8, .12 … (point-to-point links) |
Alternate Trick — The Block-Size Ladder (Skip-Counting)
No division at all — list the multiples of the block size and see which two the address falls between.
- 1. Block size (B) = 256 − mask value in the split octet
- 2. List multiples of B: 0, B, 2B, 3B … up through 256
- 3. Find the two consecutive rungs the address octet falls between — the lower rung is the network address in that octet
- 4. Next rung up − 1 = broadcast in that octet. Everything after the split octet: 0 for network, 255 for broadcast
Block size reference — mask value in the split octet → block size:
| 128 | 192 | 224 | 240 | 248 | 252 | 254 | 255 |
|---|---|---|---|---|---|---|---|
| 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |
Block size = 256 − mask value. E.g. mask 240 → 256 − 240 = 16.
Find the split octet from the CIDR prefix — don't guess it from the address:
| Prefix range | Split octet |
|---|---|
| /1 – /8 | 1st octet |
| /9 – /16 | 2nd octet |
| /17 – /24 | 3rd octet |
| /25 – /31 | 4th octet |
e.g. /23 → 255.255.254.0 → the mask value isn't 255 until the 3rd octet, so the split octet is the 3rd — read it straight off the prefix, not off which number in the address "looks splittable."
Ladder Worked Examples
192.168.1.77/264th octet
Mask 255.255.255.192 · split-octet value 192 · block = 256 − 192 = 64
Ladder
77 falls between 64 and 128 → round down to 64
| Network addr | 192.168.1.64 |
| Broadcast | 192.168.1.127 |
10.0.0.130/254th octet
Mask 255.255.255.128 · split-octet value 128 · block = 256 − 128 = 128
Ladder
130 falls between 128 and 256 → round down to 128
| Network addr | 10.0.0.128 |
| Broadcast | 10.0.0.255 |
172.16.45.200/274th octet
Mask 255.255.255.224 · split-octet value 224 · block = 256 − 224 = 32
Ladder
200 falls between 192 and 224 → round down to 192
| Network addr | 172.16.45.192 |
| Broadcast | 172.16.45.223 |
172.30.9.240/284th octet
Mask 255.255.255.240 · split-octet value 240 · block = 256 − 240 = 16
Ladder
240 falls between 240 and 256 → round down to 240
| Network addr | 172.30.9.240 |
| Broadcast | 172.30.9.255 |
192.168.200.14/294th octet
Mask 255.255.255.248 · split-octet value 248 · block = 256 − 248 = 8
Ladder
14 falls between 8 and 16 → round down to 8
| Network addr | 192.168.200.8 |
| Broadcast | 192.168.200.15 |
192.168.5.201/304th octet
Mask 255.255.255.252 · split-octet value 252 · block = 256 − 252 = 4
Ladder
201 falls between 200 and 204 → round down to 200
| Network addr | 192.168.5.200 |
| Broadcast | 192.168.5.203 |
10.20.131.55/233rd octet
Mask 255.255.254.0 · split-octet value 254 · block = 256 − 254 = 2
Ladder
131 falls between 130 and 132 → round down to 130
| Network addr | 10.20.130.0 |
| Broadcast | 10.20.131.255 |
Common mistakes
- Don't confuse the mask value (e.g. 224) with the computed network/broadcast octet (e.g. 192) — the mask value is never the answer by itself.
- Watch for off-by-one errors converting CIDR → mask: /23 = 254, not 248 — count the stolen bits from the prefix, don't eyeball it.
- Always sanity-check a candidate network address by confirming it's actually divisible by the block size (e.g. 192 ÷ 32 = 6 exactly — no remainder means it's a valid rung on the ladder).
Worked Example — 192.168.10.0/24
| Question | Answer |
|---|---|
| Netmask | 255.255.255.0 |
| Host bits | 32 − 24 = 8 |
| Usable hosts | 2⁸ − 2 = 254 |
| Network address | 192.168.10.0 |
| Broadcast | 192.168.10.255 |
| Host range | 192.168.10.1 → 192.168.10.254 |
Practice Problems
10.50.0.0/8
| Netmask | 255.0.0.0 |
| Usable hosts | ~16,000,000 |
| Network addr | 10.0.0.0 |
| Broadcast | 10.255.255.255 |
| Host range | 10.0.0.1 → 10.255.255.254 |
172.20.0.0/16
| Netmask | 255.255.0.0 |
| Usable hosts | ~65,000 |
| Network addr | 172.20.0.0 |
| Broadcast | 172.20.255.255 |
| Host range | 172.20.0.1 → 172.20.255.254 |
192.168.100.0/24
| Netmask | 255.255.255.0 |
| Usable hosts | 254 |
| Network addr | 192.168.100.0 |
| Broadcast | 192.168.100.255 |
| Host range | 192.168.100.1 → 192.168.100.254 |
192.168.1.0/25split octet
| Netmask | 255.255.255.128 |
| Usable hosts | 126 |
| Network addr | 192.168.1.0 |
| Broadcast | 192.168.1.127 |
| Host range | 192.168.1.1 → 192.168.1.126 |