Learning Objectives

• Calculate subnets where the boundary crosses into the third octet
• Understand /16 through /23 subnets
• Apply the block size method to any octet

When the Interesting Octet Changes

So far, we've worked with /24 and higher, where all the subnetting happens in the fourth octet. But what happens with prefixes like /20, /22, or /23?

The interesting octet is the octet where the network and host bits are mixed. For:

• /24 to /32: fourth octet is interesting
• /16 to /23: third octet is interesting
• /8 to /15: second octet is interesting

/23 Subnets

Block size in the third octet: 2. Mask: 255.255.254.0.

Addresses: 512. Usable hosts: 510.

For 10.10.0.0/23:

| Field | Value | |-------|-------| | Network Address | 10.10.0.0 | | First Usable | 10.10.0.1 | | Last Usable | 10.10.1.254 | | Broadcast | 10.10.1.255 | | Total | 512 |

The key insight: the /23 spans two /24s (10.10.0.x and 10.10.1.x).

/22 Subnets

Block size in the third octet: 4. Mask: 255.255.252.0.

Addresses: 1024. Usable hosts: 1022.

For 10.10.0.0/22:

| Field | Value | |-------|-------| | Network Address | 10.10.0.0 | | First Usable | 10.10.0.1 | | Last Usable | 10.10.3.254 | | Broadcast | 10.10.3.255 |

This spans four /24s: 10.10.0.x through 10.10.3.x.

/20 Subnets

Block size in the third octet: 16. Mask: 255.255.240.0.

Addresses: 4096. Usable hosts: 4094.

10.10.0.0/20 spans 10.10.0.x through 10.10.15.x.

The Method Still Works

The same block size method applies — just watch which octet you're working in:

Given 172.16.8.1/22:

1. /22 = interesting octet is the third (because it's between /16 and /23)
2. Block size = 2^(22 − 16) = 2^6 = 64? No, that's wrong...

Let me correct this. The formula for block size in the interesting octet:

octet position = ceil(prefix / 8)  → 3rd octet (bits 17-24)
bits in this octet = prefix − 16 (for /17 to /24)
block size = 2^(8 − bits_in_octet) = 2^(24 − prefix)

Actually, there's a simpler way. Just use:

block size = 256 − octet_value_of_mask

For /22: mask = 255.255.252.0, so third octet mask = 252. Block size in third octet = 256 − 252 = 4.

So for 172.16.8.1/22:

• Third octet = 8
• Block size = 4
• Network boundary: largest multiple of 4 ≤ 8 = 8
• Network: 172.16.8.0
• Broadcast: 172.16.11.255 (third octet: 8 + 4 − 1 = 11)
• First host: 172.16.8.1
• Last host: 172.16.11.254

Key Takeaways

• Identify the interesting octet: the one where the network/host boundary lies
• Block size = 256 − mask value in the interesting octet
• Network boundaries are multiples of the block size
• The pattern is the same at every octet

Next lesson: mixed practice.