Step 8 — Subnetting Practice: Intermediate
These problems involve non-/24 networks, third-octet subnetting, and more complex calculations. This is closer to what you'll see in real enterprise environments.
Problem 1: Third Octet Subnetting
172.16.45.200/21
Find: Network, Broadcast, First Host, Last Host, Usable Hosts
Your answer (saved)
Write: network, broadcast, first host, last host, usable hosts.
Saved locally in your browser for this device.
Show Solution
/21 = 255.255.248.0 → Block size = 8 in third octet
Third octet is 45. Blocks: 0, 8, 16, 24, 32, 40, 48...
45 falls in the 40-47 block
| Network | 172.16.40.0 |
| First Host | 172.16.40.1 |
| Last Host | 172.16.47.254 |
| Broadcast | 172.16.47.255 |
| Usable Hosts | 2,046 (2¹¹ - 2) |
Problem 2: /22 Network
10.100.130.50/22
Find: Network, Broadcast, First Host, Last Host, Usable Hosts
Your answer (saved)
Write: network, broadcast, first host, last host, usable hosts.
Saved locally in your browser for this device.
Show Solution
/22 = 255.255.252.0 → Block size = 4 in third octet
Third octet is 130. Blocks: 128, 132, 136...
130 falls in the 128-131 block
| Network | 10.100.128.0 |
| First Host | 10.100.128.1 |
| Last Host | 10.100.131.254 |
| Broadcast | 10.100.131.255 |
| Usable Hosts | 1,022 (2¹⁰ - 2) |
Problem 3: Small /28 Subnet
192.168.50.177/28
Find: Network, Broadcast, First Host, Last Host, Usable Hosts
Your answer (saved)
Write: network, broadcast, first host, last host, usable hosts.
Saved locally in your browser for this device.
Show Solution
/28 = 255.255.255.240 → Block size = 16
Blocks: 0, 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192...
177 falls in the 176-191 block
| Network | 192.168.50.176 |
| First Host | 192.168.50.177 |
| Last Host | 192.168.50.190 |
| Broadcast | 192.168.50.191 |
| Usable Hosts | 14 |
Problem 4: Point-to-Point Link
10.255.255.253/30
Find: Network, Broadcast, First Host, Last Host, Usable Hosts
Your answer (saved)
Write: network, broadcast, first host, last host, usable hosts.
Saved locally in your browser for this device.
Show Solution
/30 = 255.255.255.252 → Block size = 4
253 ÷ 4 = 63.25, so block starts at 63 × 4 = 252
| Network | 10.255.255.252 |
| First Host | 10.255.255.253 |
| Last Host | 10.255.255.254 |
| Broadcast | 10.255.255.255 |
| Usable Hosts | 2 |
/30 subnets are commonly used for point-to-point router links.
Problem 5: Large /19 Network
172.20.130.50/19
Find: Network, Broadcast, First Host, Last Host, Usable Hosts
Your answer (saved)
Write: network, broadcast, first host, last host, usable hosts.
Saved locally in your browser for this device.
Show Solution
/19 = 255.255.224.0 → Block size = 32 in third octet
Third octet is 130. Blocks: 0, 32, 64, 96, 128, 160...
130 falls in the 128-159 block
| Network | 172.20.128.0 |
| First Host | 172.20.128.1 |
| Last Host | 172.20.159.254 |
| Broadcast | 172.20.159.255 |
| Usable Hosts | 8,190 (2¹³ - 2) |
Problem 6: Subnet Membership
Three servers need to communicate without routing:
- Server A: 10.10.100.50/21
- Server B: 10.10.105.200/21
- Server C: 10.10.108.10/21
Which servers can communicate directly?
Your answer (saved)
Write your reasoning and which servers share a subnet.
Saved locally in your browser for this device.
Show Solution
/21 = Block size 8 in third octet
Server A (100) → Block 96-103 → Network: 10.10.96.0
Server B (105) → Block 104-111 → Network: 10.10.104.0
Server C (108) → Block 104-111 → Network: 10.10.104.0
Server B and C can communicate directly (same subnet)
Server A needs a router to reach B and C
Problem 7: Design - Multiple Departments
You're assigned 10.50.0.0/16 for your organization. You need to create subnets for 4 departments, each requiring at least 4,000 hosts. What CIDR should you use?
Your answer (saved)
Write the CIDR you chose and how you calculated host capacity.
Saved locally in your browser for this device.
Show Solution
Need 4,000 hosts per subnet:
2¹² - 2 = 4,094 hosts ✓ (2¹¹ - 2 = 2,046 not enough)
So we need 12 host bits → 32 - 12 = /20
Answer: Use /20 subnets
- 16 possible subnets (borrowing 4 bits from /16)
- 4,094 usable hosts per subnet
- Subnets: 10.50.0.0/20, 10.50.16.0/20, 10.50.32.0/20, 10.50.48.0/20
Problem 8: AWS VPC Style
10.0.37.128/20
This is a common AWS VPC subnet range. Find the complete subnet info.
Your answer (saved)
Write: mask, network, host range, broadcast, usable hosts.
Saved locally in your browser for this device.
Show Solution
/20 = 255.255.240.0 → Block size = 16 in third octet
Third octet is 37. Blocks: 0, 16, 32, 48...
37 falls in the 32-47 block
| Network | 10.0.32.0 |
| First Host | 10.0.32.1 |
| Last Host | 10.0.47.254 |
| Broadcast | 10.0.47.255 |
| Usable Hosts | 4,094 |
Note: AWS reserves 5 IPs per subnet, so actual usable is 4,089.
Checkpoint
If you can solve these problems confidently, you have solid subnetting fundamentals. The next steps cover advanced topics: VLSM and supernetting.