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Mark Yu
Mark Yu

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Network Address Calculation: The Subnet Math That Matters

#networking #beginners #devops #security

Subnet math feels annoying until you need to debug a firewall rule, Kubernetes network policy, VPN route, or cloud security group.

Then suddenly this question matters:

Given an IP address and subnet mask, what network is it actually in?

Let’s calculate it directly.

The Rule 

network address = IP address AND subnet mask
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That is it.

The AND operation compares bits:

IP bit Mask bit Result
0 0 0
0 1 0
1 0 0
1 1 1

Example: 192.168.10.37/24

/24 means the subnet mask is:

255.255.255.0
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IP:

192.168.10.37
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Mask:

255.255.255.0
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Network:

192.168.10.0
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With /24, the first three octets are the network part. The last octet is host space.

Example: 192.168.10.37/26

/26 is more interesting.

Mask:

255.255.255.192
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The last octet in binary:

IP:   37  = 00100101
Mask: 192 = 11000000
AND:       00000000
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So the network address is:

192.168.10.0/26
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The /26 networks in the last octet move in blocks of 64:

Network Host range
192.168.10.0/26 .1 - .62
192.168.10.64/26 .65 - .126
192.168.10.128/26 .129 - .190
192.168.10.192/26 .193 - .254

37 falls into the first block.

Quick Mental Shortcut

For masks in the last octet:

block size = 256 - mask_octet
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For /26:

256 - 192 = 64
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So networks start at:

0, 64, 128, 192
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Find the block that contains the IP.

Python Check 

import ipaddress

ip = ipaddress.ip_interface("192.168.10.37/26")
print(ip.network)
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Output:

192.168.10.0/26
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This is what I use when I do not want to trust my tired brain during network debugging.

Why Developers Should Care

Subnet mistakes show up in:

  • cloud VPC rules
  • Docker/Kubernetes networking
  • VPN routing
  • database allowlists
  • office network debugging
  • zero-trust access rules

One wrong CIDR can expose too much or block the service you need.

Final Thought

Subnetting is not glamorous, but it is one of those fundamentals that saves time when production networking gets weird.

What subnet or CIDR mistake has bitten you before?

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