Subnetting ends more CCNA attempts than any other topic. Not OSPF. Not VLANs. Subnetting — the thing candidates swear they'll "figure out on exam day."
They don't. The Cisco CCNA (200-301) exam runs about $300 USD per attempt (verify current vendor pricing before you budget — Cisco adjusts regional pricing), and walking in shaky on subnetting is the most expensive way to fail it. Subnetting underpins routing, access lists, DHCP, and NAT. Miss it and the exam finds you everywhere.
The good news: subnetting is a closed system. Same table, same three steps, every single time. Learn it until it's boring and it stops costing you points.
The cheat sheet: CIDR to subnet mask
Memorize this table and you've done half the work. These are the masks the exam actually uses:
CIDRSubnet maskTotal addressesUsable hosts/24255.255.255.0256254/25255.255.255.128128126/26255.255.255.1926462/27255.255.255.2243230/28255.255.255.2401614/29255.255.255.24886/30255.255.255.25242
The pattern behind the last octet is worth memorizing too. Each bit you add doubles the mask value in a fixed sequence: 128, 192, 224, 240, 248, 252, 254, 255. A /26 means 2 bits on in the last octet (192). A /29 means 5 bits on (248). Once you see the sequence, you can rebuild any mask from the CIDR in seconds.
Two rules that never change: total addresses = 2host bits, and usable hosts = total minus 2 (one for the network address, one for broadcast). A /26 has 6 host bits → 26 = 64 addresses → 62 usable.
The only math you need: the block size
Forget converting to binary on exam day. You need exactly one formula:
Block size = 256 − the interesting octet value
The "interesting octet" is the last octet in the mask that isn't 255 or 0. For 255.255.255.224, the interesting octet is 224, so the block size is 256 − 224 = 32. That means every subnet in this scheme is 32 addresses wide: .0–.31, .32–.63, .64–.95, and so on.
Network addresses always fall on multiples of the block size. That's the whole trick. If the block size is 32 and the IP is 192.168.10.77, the network is 192.168.10.64 — because 64 is the multiple of 32 at or below 77.
The 3-step method (worked example)
Take 192.168.10.77/27. Find the network address, broadcast address, and first usable host.
Step 1: Find the block size. /27 = 255.255.255.224. Interesting octet is 224. Block size = 256 − 224 = 32.
Step 2: Find the network address. Multiples of 32 in the last octet: 0, 32, 64, 96… The IP's last octet is 77, which sits between 64 and 96. Network address = 192.168.10.64.
Step 3: Fill in the rest. Broadcast is one less than the next network: 96 − 1 = 192.168.10.95. First usable host is network + 1 = 192.168.10.65. Last usable is broadcast − 1 = 192.168.10.94.
That's it. Same three steps whether the interesting octet is the second, third, or fourth. When it's the third octet (like a /22), the block-size math lands there instead — the method doesn't change.
Practice questions
Try these before scrolling to the answers. Aim for under 60 seconds each — that's exam pace.
- What is the subnet mask for /27 in dotted-decimal notation?
- How many usable host addresses does a /26 network provide?
- A host is configured as 10.45.7.200/22. What is the broadcast address of its subnet?
- You need to divide 172.16.0.0/16 into at least 14 subnets. What subnet mask should you use?
- Is 192.168.1.200/26 a valid host address? If so, what is its subnet's network address?
Answers (worked)
- 255.255.255.224. A /27 is 8 + 8 + 8 + 3 bits. Three bits on in the last octet = 128 + 64 + 32 = 224. (Or read it straight off the cheat-sheet table.)
- 62. A /26 leaves 6 host bits (32 − 26). 26 = 64 total addresses, minus network and broadcast = 62 usable.
- 10.45.7.255. /22 = 255.255.252.0 — the interesting octet is the third (252). Block size = 256 − 252 = 4. Multiples of 4 in the third octet: 0, 4, 8… The IP's third octet is 7, which falls in the 4–7 block. Network = 10.45.4.0, next network = 10.45.8.0, so broadcast = 10.45.7.255. (First usable would be 10.45.4.1.)
- 255.255.240.0 (/20). You need 2n ≥ 14 subnets. 23 = 8 is too few; 24 = 16 works. Borrow 4 bits from the /16 → /20. A /20 is 255.255.240.0 (third octet: 4 bits on = 240; block size 16). Each subnet gets 212 = 4,096 addresses (4,094 usable) — plenty of room.
- Yes — network 192.168.1.192. /26 = 255.255.255.192, block size 64. Multiples: 0, 64, 128, 192… The last octet 200 falls in the 192–255 block, so the network is 192.168.1.192 and .200 is a valid host (usable range .193–.254, broadcast .255).
How did you do? If any of these took more than a minute, that's your signal: drill the table and the block-size math until a /27 takes 30 seconds flat. Speed is the skill — the exam gives you roughly 72 seconds per question, and a four-minute subnetting problem steals time from everything after it.
Make it boring, then move on
The fastest way to get your reps in: grab the free Subnetting Practice Worksheet and work it with the cheat sheet above until you stop reaching for the table. Then, when subnetting is boring, you're ready for the rest of the exam — because the CCNA te sts whether you can configure and troubleshoot, not just calculat
The 30-Day Exam Sprint Planner (free) puts the study guide and practice pack in one checkout: every exam domain weighted like the real test, plus exam-style questions with full explanations so test day feels like a rerun. At roughly 2% of the $300 exam fee, it's the cheapest insurance against a retake.
Subnetting isn't hard. It's just unfamiliar — right up until it isn't. Get your reps in now and it becomes the easiest points on the exam.