One network. Four departments. Different sizes. One solution.
Your company just acquired a smaller firm. You've been given one /24 network to split across
4 departments of different sizes. How do you carve it up without wasting addresses? Subnetting.
An IP address is 32 bits split into 4 octets. The subnet mask defines where "network" ends and "host" begins.
IP Address — 192.168.10.0
1
1
0
0
0
0
0
0
.
1
0
1
0
1
0
0
0
.
0
0
0
0
1
0
1
0
.
0
0
0
0
0
0
0
0
Subnet Mask — /24 = 255.255.255.0
1
1
1
1
1
1
1
1
.
1
1
1
1
1
1
1
1
.
1
1
1
1
1
1
1
1
.
0
0
0
0
0
0
0
0
Network Bits (green)
The 1s in the mask identify the network portion. All devices on the same subnet share identical network bits. They can communicate directly without a router.
Host Bits (gold)
The 0s in the mask are your host address space. More host bits = more usable IPs per subnet but fewer subnets available.
CIDR notation counts the 1s: /24 means 24 network bits, 8 host bits.
192.168.10.0/24 is identical to 192.168.10.0 / 255.255.255.0
Slide 3 of 10
The Formula — 2^n - 2
One formula governs every subnetting question on the exam. Know it cold.
Usable Hosts = 2host bits- 2
Subtract 2 because the first address is the network address and the last is the broadcast address. Neither can be assigned to a host.
/26 — 6 host bits
2^6 - 2 = 62 hosts
4 subnets per /24. Good for medium departments.
/27 — 5 host bits
2^5 - 2 = 30 hosts
8 subnets per /24. Good for smaller groups.
/30 — 2 host bits
2^2 - 2 = 2 hosts
Point-to-point links only. Just two router interfaces.
Subnets per parent
2^(bits borrowed)
Borrowing 2 bits from /24 gives 2^2 = 4 subnets. Each one shrinks but you get more of them.
Common Mistake
The -2 is for hosts only. When counting how many subnets you get from borrowing bits, do NOT subtract 2. All-zeros and all-ones subnets are valid in modern networks (RFC 1878).
Your /24 has 256 total addresses (2^8). Carve it into /26 subnets: each has 64 addresses — 62 usable.
You get 4 subnets. 4 x 64 = 256. Every bit is accounted for.
Slide 4 of 10
Cheat Sheet — /24 Through /30
The range tested most frequently on the exam. Memorize the usable hosts column.
CIDR
Subnet Mask
Block Size
Usable Hosts
Subnets from /24
Common Use
/24
255.255.255.0
256
254
1
Full /24 — large LAN or floor segment
/25
255.255.255.128
128
126
2
Split a /24 in half
/26
255.255.255.192
64
62
4
Medium department, 4 per /24
/27
255.255.255.224
32
30
8
Small department or VLAN
/28
255.255.255.240
16
14
16
Server room, small team
/29
255.255.255.248
8
6
32
Very small cluster
/30
255.255.255.252
4
2
64
Point-to-point WAN link
Pattern: Block sizes are powers of 2 — 256, 128, 64, 32, 16, 8, 4.
Each additional bit on the CIDR prefix halves the block size. Every bit you gain for networks you lose from hosts.
Slide 5 of 10
Step-by-Step — /26 from /24
Given 192.168.10.0/24, create four /26 subnets. Work every step the same way every time.
Step 1
Find the block size
/26 means 6 host bits. Block size = 2^6 = 64.
Subnets increment in the fourth octet by 64.
Block = 64
Step 2
List the network addresses
Start at .0, add block size until you exceed 255:
.0, .64, .128, .192
4 subnets total
Step 3
Find the broadcast addresses
Broadcast = next subnet's network address minus 1.
Subnet .0 broadcasts at .63. Subnet .64 at .127.
.63 / .127 / .191 / .255
Step 4
Identify usable host range
First usable = network address + 1. Last usable = broadcast - 1.
Subnet 192.168.10.0/26: first = .1, last = .62.
62 usable hosts per subnet
192.168.10.0/26
Range: .1 - .62 Broadcast: .63
192.168.10.64/26
Range: .65 - .126 Broadcast: .127
192.168.10.128/26
Range: .129 - .190 Broadcast: .191
192.168.10.192/26
Range: .193 - .254 Broadcast: .255
Slide 6 of 10
VLSM — Allocate Largest First
Variable Length Subnet Masking: each department gets exactly what it needs. No waste.
4 departments: Engineering (60 hosts), HR (28 hosts), Management (12 hosts), Network team (2 hosts).
Giving everyone a /26 wastes most of the address space on smaller departments. VLSM fixes this.
Eng /26
HR /27
Mgmt /28
/30
140 addresses remaining
Engineering
192.168.10.0/26
Needs 60 — gets 62 usable
Hosts: .1 - .62
HR
192.168.10.64/27
Needs 28 — gets 30 usable
Hosts: .65 - .94
Management
192.168.10.96/28
Needs 12 — gets 14 usable
Hosts: .97 - .110
Network Team
192.168.10.112/30
Needs 2 — gets 2 usable
Hosts: .113 - .114
VLSM Rule: Always allocate the largest subnet first, then place smaller ones in the remaining space.
This prevents fragmentation and keeps address ranges contiguous.
Total used: 64 + 32 + 16 + 4 = 116 of 256 addresses.
Slide 7 of 10
The Magic Number Shortcut
256 minus the interesting mask octet = block size. Use this to find subnet boundaries without binary math.
Given any IP + mask: Divide the host octet by the block size (round down) to find the subnet.
Is 192.168.1.100 in a /26? Block = 64. 100 / 64 = 1 (integer). Subnet = 192.168.1.64/26.
Broadcast = 64 + 64 - 1 = .127.
Exam question type: "Which subnet does 172.16.4.200 belong to with a /27 mask?"
Block = 32. 200 / 32 = 6 remainder 8. Subnet = 172.16.4.192/27. Broadcast = .223. Usable: .193 - .222.
Slide 8 of 10
Special Subnets — /30, /31, and Supernetting
Not all subnets hold workstations. Some exist for router links, others for routing efficiency.
/30 — Classic Point-to-Point
Block size 4. Two usable hosts. Standard for WAN links between routers — one IP per end of the cable.
64 of these fit inside a single /24, making them ideal for conserving space in routing infrastructure.
Example: Router A gets 10.0.0.1/30, Router B gets 10.0.0.2/30. Broadcast: 10.0.0.3.
/31 — RFC 3021 Point-to-Point
Block size 2. The old formula gives 0 usable hosts (2^1 - 2 = 0) — so RFC 3021 changed the rule for point-to-point links:
both addresses are usable and there is no broadcast.
10.0.0.0/31: both .0 and .1 are valid host addresses.
Modern routers support /31. Saves 2 addresses per link compared to /30.
Supernetting / Route Summarization
The reverse of subnetting. Combine multiple contiguous networks into one larger prefix to reduce routing table size.
192.168.0.0/24 + 192.168.1.0/24 + 192.168.2.0/24 + 192.168.3.0/24
summarize to 192.168.0.0/22
Summarization Rule
Networks must be contiguous and start on a boundary aligned to the block size.
Summarized prefix = original prefix length minus log2(N).
Exam context: Supernetting questions appear in routing scenarios, not subnet math questions.
A router advertising one /22 summary route is cleaner and faster than four separate /24 routes.
Slide 9 of 10
Practice Problems
Work each one before reading the answer. Subnetting is a skill built through repetition, not memorization.
Q1: How many usable hosts does a /27 subnet provide?
2^5 - 2 = 30 hosts
5 host bits remain. Subtract 2 for network and broadcast.
Q2: What subnet does 10.0.0.100 belong to with mask 255.255.255.192?
Q6: Summarize 10.1.0.0/24 through 10.1.7.0/24 into one prefix.
10.1.0.0/21
8 contiguous /24s. /24 - log2(8) = /24 - 3 = /21.
On the N10-009 exam, subnetting questions often give an IP + mask and ask for the broadcast address,
valid host range, or which subnet an IP belongs to. The magic number shortcut handles all three in under 20 seconds.
Slide 10 of 10 | N10-009 Obj 1.4
You Can Subnet Now
Your company acquired a new firm. You were handed 192.168.10.0/24. Engineering needs 60 hosts, HR needs 28,
Management needs 12, and the Network team needs 2. Using VLSM, you allocate
.0/26 (62 usable) to Engineering, .64/27 (30 usable) to HR,
.96/28 (14 usable) to Management, and .112/30 (2 usable) to the Network team.
140 addresses remain for future growth. No waste. Problem solved.
5 Facts to Carry Out of This Presentation
1Usable hosts = 2^host bits - 2. Always subtract 2 for network address and broadcast.
2Magic number: 256 - mask octet = block size. Use it to find subnet boundaries instantly.
3VLSM: always allocate the largest subnet first, then fit smaller ones into the remaining space.
4/30 = 2 usable hosts for point-to-point. /31 (RFC 3021) = both addresses usable, no broadcast.
5Supernetting: combine N contiguous subnets. New prefix = original length minus log2(N) bits.