Subnetting and VLSM Design Lab

Key Takeaways

  • Performance-based questions reward a repeatable method - size networks, allocate largest first, then verify - more than mental shortcuts.
  • Variable Length Subnet Masking (VLSM) allocates larger subnets before smaller ones so aligned address space is not wasted or trapped.
  • Each subnet design must identify network ID, usable host range, broadcast address, prefix length, and default gateway.
  • Plan for growth, point-to-point links, management networks, and reserved space before assigning any addresses.
  • Fast verification checks block size, overlap, gateway placement inside the subnet, and host capacity.
Last updated: June 2026

Why Subnetting PBQs Matter on N10-009

The CompTIA Network+ N10-009 exam contains a maximum of 90 questions in 90 minutes, a mix of multiple-choice and performance-based questions (PBQs). PBQs usually appear in the first few minutes of the exam, are weighted more heavily than single multiple-choice items, and consume time quickly. The passing score is 720 on a 100-900 scale. A common mistake is to burn fifteen minutes on the first subnetting PBQ - flag it, move on, and return after answering the faster multiple-choice items.

Subnetting PBQs typically ask you to complete a table, drag subnets into the correct slot, or choose the best mask for several departments. The safest approach is always the same: size each network, allocate the largest network first, then verify that no ranges overlap.

Scenario

You are given the private block 10.40.8.0/22. Design subnets for a branch office:

NetworkRequired usable hostsNotes
Staff220User workstations and printers
Voice110IP phones
Wireless corporate70Managed laptops
Servers26Local services and appliances
Network management12Switch, AP, and firewall management
WAN link A2Point-to-point link
WAN link B2Point-to-point link

The /22 spans 10.40.8.0 through 10.40.11.255 - that is 1024 total addresses. A /22 borrows two host bits from a /24, so it covers four contiguous /24 blocks (8, 9, 10, 11).

Step 1: Choose Prefix Lengths

Usable hosts = 2^(host bits) - 2. The minus two accounts for the reserved network and broadcast addresses.

Required usable hostsSmallest prefixUsable hostsBlock size
220/24254256
110/25126128
70/25126128
26/273032
12/281416
2/3024
2/3024

Note that 70 hosts will not fit in a /26 (62 usable), so it rounds up to a /25. This rounding catches many test takers.

Step 2: Allocate Largest to Smallest

NetworkSubnetUsable rangeBroadcastGateway
Staff10.40.8.0/24.8.1 - .8.25410.40.8.25510.40.8.1
Voice10.40.9.0/25.9.1 - .9.12610.40.9.12710.40.9.1
Wireless10.40.9.128/25.9.129 - .9.25410.40.9.25510.40.9.129
Servers10.40.10.0/27.10.1 - .10.3010.40.10.3110.40.10.1
Management10.40.10.32/28.10.33 - .10.4610.40.10.4710.40.10.33
WAN A10.40.10.48/30.10.49 - .10.5010.40.10.5110.40.10.49
WAN B10.40.10.52/30.10.53 - .10.5410.40.10.5510.40.10.53

Unused space remains from 10.40.10.56 through 10.40.11.255 - roughly 456 addresses for growth, another VLAN, loopbacks, or future WAN circuits.

PBQ Method (Repeatable Six Steps)

  1. Write host requirements from largest to smallest.
  2. Convert each requirement to the smallest prefix that fits (round up).
  3. Start at the first address in the assigned block.
  4. Move forward by the block size of each subnet (block size = 256 minus the mask value in the interesting octet).
  5. Place the default gateway at the first usable address consistently.
  6. Verify each broadcast address is exactly one less than the next network ID - no gaps, no overlap.

Fast Mask Reference

Memorize this table cold; it eliminates arithmetic under pressure.

PrefixMaskUsable hostsBlock size
/24255.255.255.0254256
/25255.255.255.128126128
/26255.255.255.1926264
/27255.255.255.2243032
/28255.255.255.2401416
/29255.255.255.24868
/30255.255.255.25224

A fast trick: the interesting octet is wherever the mask is neither 255 nor 0. The block size is 256 minus that octet value. For /27 (mask .224) the block size is 256 - 224 = 32, so subnets start at .0, .32, .64, .96, and so on.

Worked Mental Check

Given 10.40.10.32/28, which subnet does host 10.40.10.40 belong to? Block size for /28 is 16, so subnets begin at .0, .16, .32, .48. Address .40 falls in the .32 block; usable range .33 - .46; broadcast .47. The gateway .33 is valid; .47 would be the broadcast and must never be a gateway.

Common PBQ Traps

  • Allocating small subnets first, leaving no aligned space for a larger network.
  • Using the broadcast or network address as a host or gateway.
  • Forgetting a /30 has only two usable addresses for a point-to-point link.
  • Overlapping two subnets in the same range because block size was miscalculated.
  • Choosing a prefix that meets today's count but ignores a stated growth requirement.
  • Mixing CIDR notation with the wrong dotted-decimal mask (for example pairing /26 with 255.255.255.128).
Test Your Knowledge

A subnet must support 70 usable IPv4 hosts. What is the smallest common prefix length that satisfies the requirement?

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B
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D
Test Your Knowledge

Using block size, what is the broadcast address of 10.40.10.32/28?

A
B
C
D
Test Your Knowledge

Which VLSM allocation method most reduces wasted space and overlap risk?

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B
C
D
Test Your KnowledgeMulti-Select

Which values should you verify in a completed subnetting PBQ? Select three.

Select all that apply

The usable range excludes the network and broadcast addresses
No subnet overlaps another assigned subnet
The prefix length supports the required host count
Every subnet uses the same default gateway IP address
The subnet with the most hosts is always assigned last