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VLSM — cut one /24 into four right-sized subnets
Enterprise Intermediate 5 devices 6 checked tasks 100 points Plan lab
One address block, 192.168.10.0/24, and four needs: 50 hosts, 25 hosts, a router-to-router link and 10 hosts. Give each the smallest subnet that fits — /26, /27, /30, /28 — address the routers from the plan, route between them, and prove all three LANs can reach each other. You build it in a browser, type real commands at each device, and every task below is
checked against the running configuration — so you find out it is wrong before anybody else does.
Open this lab in the simulator
The simulator runs in your browser with nothing to install. This lab is part of the full simulator: open it free for 7 days with a new account, then with a plan. It needs an account, which keeps your progress.
The scenario
One address block, 192.168.10.0/24, and four needs: 50 hosts, 25 hosts, a router-to-router link and 10 hosts. Give each the smallest subnet that fits — /26, /27, /30, /28 — address the routers from the plan, route between them, and prove all three LANs can reach each other.
Your role: a network engineer on the enterprise team. Level: Intermediate Done when: all 6 goals show as passed (100 points).
Network status
Devices: 2 routers, 3 PCs, cabled as drawn. No Layer 1 or Layer 2 faults: every link is connected and up. LAN-A (50 hosts) is a host with 192.168.10.10/26, gateway 192.168.10.1. LAN-B (25 hosts) is a host with 192.168.10.70/27, gateway 192.168.10.65. LAN-C (10 hosts) is a host with 192.168.10.120/28, gateway 192.168.10.113. R1, R2 start with an empty configuration. Known problems
None: nothing is misconfigured. The network is simply not finished: what the goals list is what you have to build or secure. Goals
R1 Gi0/0 = 192.168.10.1/26 (LAN A: 62 usable hosts) (15 pts) R1 Gi0/2 = 192.168.10.65/27 (LAN B: 30 usable hosts) (15 pts) The link R1–R2 is a /30: R1 192.168.10.97, R2 192.168.10.98 (20 pts) R2 Gi0/0 = 192.168.10.113/28 (LAN C: 14 usable hosts) (15 pts) LAN A reaches LAN C (20 pts) LAN B reaches LAN C and LAN C reaches LAN B (15 pts) The activity, in order
R1 Gi0/0 = 192.168.10.1/26 (LAN A: 62 usable hosts) Why: A router interface routes nothing until it has an address and is up. Every other task builds on this one. Verify with: show ip interface brief R1 Gi0/2 = 192.168.10.65/27 (LAN B: 30 usable hosts) Why: A router interface routes nothing until it has an address and is up. Every other task builds on this one. Verify with: show ip interface brief The link R1–R2 is a /30: R1 192.168.10.97, R2 192.168.10.98 Why: A router interface routes nothing until it has an address and is up. Every other task builds on this one. Verify with: show ip interface brief R2 Gi0/0 = 192.168.10.113/28 (LAN C: 14 usable hosts) Why: A router interface routes nothing until it has an address and is up. Every other task builds on this one. Verify with: show ip interface brief LAN A reaches LAN C Why: A ping needs a route in BOTH directions. This is the end-to-end proof that everything before it worked. Verify with: ping 192.168.10.120 LAN B reaches LAN C and LAN C reaches LAN B Why: A ping needs a route in BOTH directions. This is the end-to-end proof that everything before it worked. Verify with: ping 192.168.10.120 Tips
Work in the order given: later goals depend on the earlier ones. After every change, run the verify command before moving on; do not guess. Stuck? Read the verify output first. It names the interface, neighbour or route that is wrong. Common mistake: Placing a /27 on a boundary that is not a multiple of 32 Common mistake: Forgetting the network and broadcast addresses when counting hosts Common mistake: Overlapping the point-to-point /30 Why this matters
VLSM lets one block be cut into subnets of different sizes so no address space is wasted. The method is always the same: allocate the largest need first, on its natural boundary, then the next.
What you will be able to do
Choose the smallest prefix that fits each requirement
Lay subnets on correct boundaries without overlap
Address and route a small network from the plan
The network
ISR 2911, PC — 5 devices in all, already cabled.
Fa0 Gi0/0 Fa0 Gi0/2 Gi0/1 Gi0/1 Fa0 Gi0/0 R1 R2 LAN-A (50 hosts) 192.168.10.10/26 · gw 192.168.10.1 LAN-B (25 hosts) 192.168.10.70/27 · gw 192.168.10.65 LAN-C (10 hosts) 192.168.10.120/28 · gw 192.168.10.113 Drawn from this lab: devices, cables, ports and the addresses already set. Device Model Address already set
R1 ISR 2911 configure it R2 ISR 2911 configure it LAN-A (50 hosts) PC 192.168.10.10 / 255.255.255.192 gw 192.168.10.1 LAN-B (25 hosts) PC 192.168.10.70 / 255.255.255.224 gw 192.168.10.65 LAN-C (10 hosts) PC 192.168.10.120 / 255.255.255.240 gw 192.168.10.113
Cabling
LAN-A (50 hosts) Fa0 ↔ R1 Gi0/0 LAN-B (25 hosts) Fa0 ↔ R1 Gi0/2 R1 Gi0/1 ↔ R2 Gi0/1 LAN-C (10 hosts) Fa0 ↔ R2 Gi0/0
What is configured at the start
Nothing is configured to start with beyond the addresses above: you build the configuration from a clean start.
Tasks, and how each one is checked
Every task is decided by reading the live network — a neighbour in FULL, a ping that succeeds, a line in the running configuration — not by looking for a keyword you typed. The same command shows you what the lab sees.
# Task Pts How to verify it yourself
1 R1 Gi0/0 = 192.168.10.1/26 (LAN A: 62 usable hosts) 15 On R1: show ip interface brief — Gi0/0 has 192.168.10.1 and is up/up. 2 R1 Gi0/2 = 192.168.10.65/27 (LAN B: 30 usable hosts) 15 On R1: show ip interface brief — Gi0/2 has 192.168.10.65 and is up/up. 3 The link R1–R2 is a /30: R1 192.168.10.97, R2 192.168.10.98 20 On R1: show ip interface brief — Gi0/1 has 192.168.10.97 and is up/up. and On R2: show ip interface brief — Gi0/1 has 192.168.10.98 and is up/up. 4 R2 Gi0/0 = 192.168.10.113/28 (LAN C: 14 usable hosts) 15 On R2: show ip interface brief — Gi0/0 has 192.168.10.113 and is up/up. 5 LAN A reaches LAN C 20 On LAN-A (50 hosts): ping 192.168.10.120 must succeed. 6 LAN B reaches LAN C and LAN C reaches LAN B 15 On LAN-B (25 hosts): ping 192.168.10.120 must succeed. and On LAN-C (10 hosts): ping 192.168.10.70 must succeed.
Stuck? Ask the coach
Tell the coach what you see or what you tried. It gives you a hint, not the answer, the way a senior engineer would. Included with a free trial or any plan.
Ask the coach
Common mistakes
Placing a /27 on a boundary that is not a multiple of 32
Forgetting the network and broadcast addresses when counting hosts
Overlapping the point-to-point /30
Reference solution
Try it yourself first — the lab checks your work either way. This is one configuration that makes every task pass, device by device:
R1 interface g0/0
ip address 192.168.10.1 255.255.255.192
no shutdown
exit
interface g0/2
ip address 192.168.10.65 255.255.255.224
no shutdown
exit
interface g0/1
ip address 192.168.10.97 255.255.255.252
no shutdown
exit
ip route 192.168.10.112 255.255.255.240 192.168.10.98 R2 interface g0/0
ip address 192.168.10.113 255.255.255.240
no shutdown
exit
interface g0/1
ip address 192.168.10.98 255.255.255.252
no shutdown
exit
ip route 192.168.10.0 255.255.255.192 192.168.10.97
ip route 192.168.10.64 255.255.255.224 192.168.10.97 Where this is examined
CCNA (IPv4 addressing), Network+ (subnetting, VLSM).
This lab practises these topics in the zNetLaB certification courses:
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