Step 1: Bits to borrow
To get 8 subnets, we need 2ⁿ ≥ 8, so n = 3 bits are borrowed from the host portion.
New prefix = /16 + 3 = /19
Step 2: Subnet mask
The 3 borrowed bits fall in the third octet: 11100000 = 224
- Dotted decimal: 255.255.224.0
- CIDR: /19
Step 3: Block size and subnets
Block size = 256 − 224 = 32, so the third octet increases by 32 for each subnet. Each subnet has 13 host bits, giving
2¹³ − 2 = 8,190 usable hosts.
| Subnet | Network Address | First Usable | Last Usable | Broadcast |
| 1 | 10.10.0.0 | 10.10.0.1 | 10.10.31.254 | 10.10.31.255 |
| 2 | 10.10.32.0 | 10.10.32.1 | 10.10.63.254 | 10.10.63.255 |
| 3 | 10.10.64.0 | 10.10.64.1 | 10.10.95.254 | 10.10.95.255 |
| 4 | 10.10.96.0 | 10.10.96.1 | 10.10.127.254 | 10.10.127.255 |
| 5 | 10.10.128.0 | 10.10.128.1 | 10.10.159.254 | 10.10.159.255 |
| 6 | 10.10.160.0 | 10.10.160.1 | 10.10.191.254 | 10.10.191.255 |
| 7 | 10.10.192.0 | 10.10.192.1 | 10.10.223.254 | 10.10.223.255 |
| 8 | 10.10.224.0 | 10.10.224.1 | 10.10.255.254 | 10.10.255.255 |
Answer
- Subnet mask: 255.255.224.0 (/19)
- Third subnet: 10.10.64.0/19
- First usable IP: 10.10.64.1
- Last usable IP: 10.10.95.254
— rhs