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
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
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
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
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
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?
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?
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.
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.