Given host 172.16.45.7/22, find the subnet mask, network address, broadcast address, and the usable host range.

Mask 255.255.252.0; network 172.16.44.0, broadcast 172.16.47.255, usable hosts 172.16.44.1 – 172.16.47.254 (1022 hosts).

* Magic number 4 in the 3rd octet: 45 falls in the 44–47 block. *

This one crosses an octet boundary because the /22 boundary sits inside the 3rd octet.

Step-by-step:

1. /22 mask = 255.255.252.0 (3rd octet 11111100). The interesting octet is the 3rd.
2. Magic number = 256 − 252 = 4 → 3rd-octet networks step by 4: 0, 4, 8, …, 44, 48, …
3. The host's 3rd octet is 45, which falls in the 44–47 block → network 3rd octet = 44.
4. Network address: 172.16.44.0/22
5. Broadcast: 172.16.47.255 (last address before the next network 172.16.48.0).
6. Usable range: 172.16.44.1 – 172.16.47.254
7. Host bits = 32 − 22 = 10 → usable hosts = 2^10 − 2 = 1022.

Gotcha: with a /22 the host portion spans the 4th octet and part of the 3rd, so the network number is rarely the same as the 3rd octet you started with.

Go deeper:

• ipcalc (jodies.de) — enter 172.16.45.7/22 and check the network, broadcast and 1022-host range.
• Subnetting Mastery (PracticalNetworking) — worked examples where the boundary sits inside the 3rd octet.