How to solve 192.168.12.77/26
A /26 leaves 6 host bits: 2⁶ = 64 addresses in a block. The boundaries in the last octet are 0, 64, 128 and 192. The host 77 falls between 64 and 127.
| Value | Answer |
|---|---|
| Network | 192.168.12.64 |
| Broadcast | 192.168.12.127 |
| Subnet mask | 255.255.255.192 |
| First usable host | 192.168.12.65 |
| Last usable host | 192.168.12.126 |
| Usable hosts | 62 |
What changes at /23?
The block boundary moves into the third octet. For 10.20.5.77/23, the network is 10.20.4.0 and the broadcast is 10.20.5.255. There are 512 addresses, or 510 usable hosts. Do not treat every prefix as a fourth-octet problem.
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