Subnet for 150 hosts
A network that needs 150 hosts fits in a /24, mask 255.255.255.0, with 254 usable addresses.
A network built for 150 hosts needs a /24. The mask is 255.255.255.0, and its 254 usable addresses leave 104 spare for growth.
The next smaller block, a /25, only allows 126 usable addresses, which is not enough for 150 hosts. That makes /24 the smallest block that fits.
Enter any address below to see the exact network, broadcast, and host range for a /24 block.
Calculate a /24 network
Accepts CIDR (192.0.2.0/24), IP + mask (192.0.2.0 255.255.255.0), or a bare IP (defaults to /24).
11000000.00000000.00000010.0000000011111111.11111111.11111111.00000000Example: 192.0.2.0/24
Taking 192.0.2.0/24 as a worked example, a /24 block breaks down like this:
| Network address | 192.0.2.0 |
|---|---|
| Usable range | 192.0.2.1 - 192.0.2.254 |
| Broadcast address | 192.0.2.255 |
| Total addresses | 256 |
| Usable hosts | 254 |
Frequently asked questions
What subnet mask do I need for 150 hosts?
A network with 150 hosts fits in a /24, which uses the mask 255.255.255.0 and provides 254 usable addresses.
How many addresses does a /24 block have in total?
A /24 has 256 total addresses, 254 of them usable (the total minus the network and broadcast addresses, except for /31 and /32).
Does a /24 waste any addresses for 150 hosts?
Yes, 104 addresses of headroom. A /24 provides 254 usable addresses for the 150 you need, and the next smaller block, a /25, does not hold enough.
How do I work this out myself?
Find the smallest power of two, minus two, that is at least 150. That takes 8 host bits, which leaves a /24 prefix. See how to calculate a subnet mask for the full method.