Step 5 — Network & Broadcast Addresses
Every subnet has two reserved addresses: the network address (first) and the broadcast address (last). Understanding how to calculate these is essential for subnetting.
Network Address
The network address is the first address in a subnet. It identifies the network itself and cannot be assigned to a host.
How to find it:
Set all host bits to 0
Example: 192.168.1.150/24
Broadcast Address
The broadcast address is the last address in a subnet. Packets sent to this address are delivered to all hosts on the subnet.
How to find it:
Set all host bits to 1
Example: 192.168.1.150/24
Step-by-Step Method
For more complex subnets, here's a reliable method:
Example: Find network and broadcast for 172.16.45.200/21
Step 1: Identify the "interesting" octet
/21 means 21 network bits = 8 + 8 + 5 = third octet is "interesting"
Step 2: Find the block size
Block size = 256 - subnet mask value in interesting octet
/21 → mask is 255.255.248.0 → Block size = 256 - 248 = 8
Step 3: Find the network address
Third octet is 45. Find largest multiple of 8 that's ≤ 45
8 × 5 = 40 ✓ (8 × 6 = 48 is too big)
Network: 172.16.40.0
Step 4: Find the broadcast address
Next network starts at 40 + 8 = 48
Broadcast = next network - 1 = 172.16.48.0 - 1
Broadcast: 172.16.47.255
Block Size Reference
The block size tells you where subnet boundaries fall. Memorize this:
| Mask Value | CIDR (in octet) | Block Size | Subnet Boundaries |
|---|---|---|---|
| 128 | /25, /17, /9 | 128 | 0, 128 |
| 192 | /26, /18, /10 | 64 | 0, 64, 128, 192 |
| 224 | /27, /19, /11 | 32 | 0, 32, 64, 96, 128... |
| 240 | /28, /20, /12 | 16 | 0, 16, 32, 48, 64... |
| 248 | /29, /21, /13 | 8 | 0, 8, 16, 24, 32... |
| 252 | /30, /22, /14 | 4 | 0, 4, 8, 12, 16... |
More Worked Examples
10.50.100.75/26
- Block size: 256 - 192 = 64
- Fourth octet: 75. Largest multiple of 64 ≤ 75 is 64
- Network:
10.50.100.64 - Broadcast: 64 + 64 - 1 =
10.50.100.127
192.168.50.200/27
- Block size: 256 - 224 = 32
- Fourth octet: 200. Largest multiple of 32 ≤ 200 is 192 (32×6)
- Network:
192.168.50.192 - Broadcast: 192 + 32 - 1 =
192.168.50.223
172.20.130.50/19
- Interesting octet: 3rd (19 bits = 8+8+3)
- Block size: 256 - 224 = 32
- Third octet: 130. Largest multiple of 32 ≤ 130 is 128 (32×4)
- Network:
172.20.128.0 - Broadcast:
172.20.159.255(next network at 160)
Practice Exercises
Find the network and broadcast addresses:
Store your work
These have two outputs (network + broadcast). Treat this as a saved scratchpad.
Saved locally in your browser for this device.
- 192.168.1.100/24
- 10.20.30.40/28
- 172.16.100.200/22
- 192.168.50.130/25
- 10.0.45.67/20
Show Answers
- Network:
192.168.1.0, Broadcast:192.168.1.255 - Block size 16. 40 is in 32-47 block.
Network:10.20.30.32, Broadcast:10.20.30.47 - Block size 4 in 3rd octet. 100 is in 100-103 block.
Network:172.16.100.0, Broadcast:172.16.103.255 - Block size 128. 130 is in 128-255 block.
Network:192.168.50.128, Broadcast:192.168.50.255 - Block size 16 in 3rd octet. 45 is in 32-47 block.
Network:10.0.32.0, Broadcast:10.0.47.255
Checkpoint
Before moving on, make sure you can:
- Calculate the network address for any IP/CIDR combination
- Calculate the broadcast address for any IP/CIDR combination
- Use the block size method for quick calculations
- Identify which octet is the "interesting" octet