Guides2026-11-08•9 min read

FLSM vs. VLSM Subnetting: How to Design Multi-Subnet CCNA Network Topologies

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FLSM vs. VLSM Subnetting: How to Design Multi-Subnet CCNA Network Topologies

Why does dividing a /24 network into four equal /26 subnets waste 60 IPv4 addresses when one of those links is just a 2-router WAN connection? Here is how Variable-Length Subnet Masking (VLSM) solves address exhaustion.

When designing an enterprise or branch network from a single assigned IPv4 block (such as 192.168.10.0/24), network engineers choose between two subnetting methodologies:

FLSM (Fixed-Length Subnet Masking): Every subnet uses the exact same subnet mask (and therefore has the exact same number of usable host addresses).
VLSM (Variable-Length Subnet Masking, documented in IETF RFC 1878): Subnets can use different prefix lengths (/25, /26, /27, /30, etc.) tailored to the exact number of hosts required on each LAN or WAN segment.

On the Cisco CCNA (200-301) exam, you will frequently be asked to design a VLSM addressing plan or identify why two router interfaces fail to route because of overlapping VLSM subnets.

You can validate every subnet block in your VLSM plan using the IPv4 Subnet Calculator and deploy the topology live in the Web Packet Tracer & Network Simulator.

1. The Problem with Fixed-Length Subnet Masking (FLSM)

Suppose your organization is allocated 192.168.10.0/24 (256 total IPv4 addresses) and needs to address 4 network segments:

LAN A (Engineering): 100 hosts
LAN B (Sales): 50 hosts
LAN C (IT Operations): 25 hosts
WAN Link (R1 $\leftrightarrow$ R2): 2 router interfaces

What Happens If You Try FLSM?

To create 4 equal-sized subnets from a /24, you borrow 2 bits to create four /26 subnets (64 total IPs, 62 usable hosts each):

Failure 1 (Too Small for LAN A): LAN A needs 100 hosts, which cannot fit inside a /26 (max 62 usable hosts)! And if you use a /25 (126 usable hosts), you only get 2 subnets total—leaving no subnets for LAN C or the WAN link.
Failure 2 (Massive Address Waste on WAN Link): Even if a /26 were used on the point-to-point WAN link between R1 and R2, assigning a 64-address /26 block to a link that only ever has 2 routers wastes $62 - 2 = \mathbf{60}$ usable IPv4 addresses!

2. How Variable-Length Subnet Masking (VLSM) Solves This

VLSM ( conceptually "subnetting a subnet") lets you slice 192.168.10.0/24 into right-sized blocks—provided you follow one ironclad rule:

The Golden Rule of VLSM Allocation:
Always sort your subnets from largest host requirement to smallest host requirement, and always allocate each block starting at a valid boundary multiple of its own block size ($2^h$).

Step-by-Step VLSM Allocation Walkthrough (`192.168.10.0/24`)

Let us allocate our 4 segments in descending order of size:

192.168.10.0/24 (256 Total Addresses: .0 to .255)
├── [.0   – .127]  LAN A (100 hosts) ──► 192.168.10.0/25   (126 usable)
└── [.128 – .255]  Remaining /25 split further:
    ├── [.128 – .191]  LAN B (50 hosts)  ──► 192.168.10.128/26 (62 usable)
    └── [.192 – .255]  Remaining /26 split further:
        ├── [.192 – .223]  LAN C (25 hosts)  ──► 192.168.10.192/27 (30 usable)
        └── [.224 – .255]  Remaining /27 split further:
            ├── [.224 – .227]  WAN R1↔R2 (2 IPs) ──► 192.168.10.224/30 (2 usable)
            └── [.228 – .255]  Unallocated spare space (28 IPs free for future growth!)

Here is the completed, non-overlapping VLSM Addressing Table:

Segment Name	Required Hosts	Allocated CIDR Prefix	Subnet Mask	Usable Host Range	Broadcast IP	Unused Capacity
1. LAN A (Engineering)	`100`	`192.168.10.0/25`	`255.255.255.128`	`.1` – `.126`	`.127`	26 hosts
2. LAN B (Sales)	`50`	`192.168.10.128/26`	`255.255.255.192`	`.129` – `.190`	`.191`	12 hosts
3. LAN C (IT Ops)	`25`	`192.168.10.192/27`	`255.255.255.224`	`.193` – `.222`	`.223`	5 hosts
4. WAN (`R1` $\leftrightarrow$ `R2`)	`2`	`192.168.10.224/30`	`255.255.255.252`	`.225` – `.226`	`.227`	0 hosts

3. Why Allocating Smallest-First Causes Overlapping Subnets

What happens if a junior administrator allocates the smallest subnet first (192.168.10.0/30, covering .0–.3) and then tries to allocate a /25 starting at .4 (192.168.10.4/25)?

Because a /25 has a block size of 128, its valid network boundaries are strictly .0 and .128.
If you configure 192.168.10.4 255.255.255.128 on a router interface, the router performs a bitwise AND (4 AND 128 = 0), meaning 192.168.10.4/25 is merely a host IP inside 192.168.10.0/25 (0–127)—which completely overlaps with 192.168.10.0/30 (0–3)!

Key Rule: A block of size $B = 2^h$ can only start at an address that is an exact integer multiple of $B$ ($0, B, 2B, 3B\dots$). Allocating from largest block size to smallest block size automatically guarantees every subnet lands on a valid boundary.

4. Hands-On Lab: Building a VLSM Topology in Web Packet Tracer

Test your VLSM addressing plan live in the Web Packet Tracer & Network Simulator:

Open Web Packet Tracer and click Load Sample Topology.

Reconfigure the sample topology from default /24 masks to our VLSM design:

On R1, open the CLI console and update Gi0/0 (LAN A) and Gi0/1 (LAN B):

enable
configure terminal
interface Gi0/0
 ip address 192.168.10.1 255.255.255.128
 no shutdown
exit
interface Gi0/1
 ip address 192.168.10.129 255.255.255.192
 no shutdown
end
show ip route

Inspect the show ip route output on R1:

Codes: C - connected, S - static

C    192.168.10.0/25 is directly connected, GigabitEthernet0/0
C    192.168.10.128/26 is directly connected, GigabitEthernet0/1

Select PC1 and set its IPv4 to 192.168.10.10, Subnet Mask 255.255.255.128 (/25), and Default Gateway 192.168.10.1.
Select PC2 and set its IPv4 to 192.168.10.140, Subnet Mask 255.255.255.192 (/26), and Default Gateway 192.168.10.129.
Run ping 192.168.10.140 from PC1—even though both PCs begin with 192.168.10.x, PC1 recognizes via its /25 mask that .140 is in a different subnet and routes the packet cleanly through R1!

5. Legitimate Sources & Standards References

IETF RFC 1878: Variable Length Subnet Table For IPv4 (T. Pummill, B. Manning, December 1995).
IETF RFC 4632: Classless Inter-domain Routing (CIDR): The Internet Address Assignment and Aggregation Plan (August 2006).
Cisco CCNA 200-301 Exam Topics (v1.1): Domain 1.0 Network Fundamentals (1.6 Configure and verify IPv4 addressing and subnetting to meet network requirements).

6. CCNA 200-301 Exam Prep: Questions & Detailed Answers

Question 1

You are designing a VLSM addressing scheme from 10.10.10.0/24 for a branch office that needs a Sales VLAN (55 hosts), an Engineering VLAN (25 hosts), a Voice VLAN (12 hosts), and two point-to-point router links (2 hosts each). Which subnet mask should be assigned to the Sales VLAN (55 hosts)?

A. /25 (255.255.255.128)
B. /26 (255.255.255.192)
C. /27 (255.255.255.224)
D. /28 (255.255.255.240)

Correct Answer: B
Explanation: To accommodate 55 usable hosts, you need $2^h - 2 \ge 55$.

With $h = 5$ (/27), $2^5 - 2 = 30$ usable hosts (too small).
With $h = 6$ (/26, 255.255.255.192), $2^6 - 2 = \mathbf{62}$ usable hosts, which fits 55 hosts while leaving three /26-sized quarters of the /24 space for Engineering (/27 = 30 hosts), Voice (/28 = 14 hosts), and the two WAN links (/30 = 2 hosts each).

Question 2

In a VLSM design using 192.168.1.0/24, Subnet 1 is assigned 192.168.1.0/26 and Subnet 2 is assigned 192.168.1.64/27. What is the very first valid Network ID available for Subnet 3 if Subnet 3 requires a /28 mask?

A. 192.168.1.80/28
B. 192.168.1.96/28
C. 192.168.1.128/28
D. 192.168.1.72/28

Correct Answer: B
Explanation:

Subnet 1 (192.168.1.0/26, block size 64) uses .0 through .63.
Subnet 2 (192.168.1.64/27, block size 32) uses .64 through .95.
Therefore, the next available address in the /24 block is .96, which is a valid multiple of 16 (the block size of a /28), making 192.168.1.96/28 (.96–.111) the next contiguous subnet.

Question 3

Why must routing protocols such as OSPF, EIGRP, RIPv2, and BGP (classless routing protocols) be used in a network that implements VLSM, rather than legacy classful routing protocols like RIPv1?

A. Classful routing protocols do not include the subnet mask (prefix length) in their routing updates.
B. Classless routing protocols encrypt VLSM subnet masks.
C. Legacy classful protocols only support IPv6.
D. VLSM requires Layer-2 switches to run BGP.

Correct Answer: A
Explanation: Legacy classful routing protocols (such as RIPv1 and IGRP) did not transmit subnet mask information inside routing protocol updates—routers had to assume the classful /8, /16, or /24 mask or the local interface's fixed mask. Classless routing protocols (OSPF, EIGRP, RIPv2, IS-IS, BGP) explicitly include the CIDR prefix length / subnet mask alongside every network prefix in routing updates, enabling VLSM and CIDR route summarization.

Question 4

An administrator configures Gi0/0 with 192.168.5.33/27 and tries to configure Gi0/1 on the same router with 192.168.5.50/28. Why is this configuration invalid?

A. A router cannot have two interfaces with different subnet masks.
B. 192.168.5.50/28 (192.168.5.48 – 192.168.5.63) overlaps with 192.168.5.33/27 (192.168.5.32 – 192.168.5.63).
C. 192.168.5.33 is a network ID.
D. /28 masks are only allowed on serial interfaces.

Correct Answer: B
Explanation: 192.168.5.33/27 belongs to the subnet 192.168.5.32/27, which spans addresses .32 through .63. Meanwhile, 192.168.5.50/28 belongs to the subnet 192.168.5.48/28, which spans .48 through .63. Because .48–.63 lies completely inside .32–.63, the two subnets overlap, which violates Layer-3 routing requirements on a router.

Question 5

How many /30 point-to-point WAN subnets can be carved out of a single /27 address block?

A. 4
B. 6
C. 8
D. 16

Correct Answer: C
Explanation: Moving from /27 to /30 borrows $30 - 27 = 3$ bits, producing $2^3 = \mathbf{8}$ /30 subnets (since $32 \text{ total IPs} \div 4 \text{ IPs per } /30 = 8$).