Subnetting 101

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

IP:192.168.1.10010110Mask:255.255.255.00000000
Network: 192.168.1.0(host bits → all 0s)

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

IP:192.168.1.10010110Mask:255.255.255.00000000
Broadcast: 192.168.1.255(host bits → all 1s)

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 ValueCIDR (in octet)Block SizeSubnet Boundaries
128/25, /17, /91280, 128
192/26, /18, /10640, 64, 128, 192
224/27, /19, /11320, 32, 64, 96, 128...
240/28, /20, /12160, 16, 32, 48, 64...
248/29, /21, /1380, 8, 16, 24, 32...
252/30, /22, /1440, 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:

  1. 192.168.1.100/24
  2. 10.20.30.40/28
  3. 172.16.100.200/22
  4. 192.168.50.130/25
  5. 10.0.45.67/20
Show Answers
  1. Network: 192.168.1.0, Broadcast: 192.168.1.255
  2. Block size 16. 40 is in 32-47 block.
    Network: 10.20.30.32, Broadcast: 10.20.30.47
  3. Block size 4 in 3rd octet. 100 is in 100-103 block.
    Network: 172.16.100.0, Broadcast: 172.16.103.255
  4. Block size 128. 130 is in 128-255 block.
    Network: 192.168.50.128, Broadcast: 192.168.50.255
  5. 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