Subnet for 500 hosts
A network that needs 500 hosts fits in a /23, mask 255.255.254.0, with 510 usable addresses.
A network built for 500 hosts needs a /23. The mask is 255.255.254.0, and its 510 usable addresses leave 10 spare for growth.
The next smaller block, a /24, only allows 254 usable addresses, which is not enough for 500 hosts. That makes /23 the smallest block that fits.
Enter any address below to see the exact network, broadcast, and host range for a /23 block.
Calculate a /23 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).
11000000.00000000.00000010.0000000011111111.11111111.11111110.00000000Example: 192.0.2.0/23
Taking 192.0.2.0/23 as a worked example, a /23 block breaks down like this:
| Network address | 192.0.2.0 |
|---|---|
| Usable range | 192.0.2.1 - 192.0.3.254 |
| Broadcast address | 192.0.3.255 |
| Total addresses | 512 |
| Usable hosts | 510 |
Frequently asked questions
What subnet mask do I need for 500 hosts?
A network with 500 hosts fits in a /23, which uses the mask 255.255.254.0 and provides 510 usable addresses.
How many addresses does a /23 block have in total?
A /23 has 512 total addresses, 510 of them usable (the total minus the network and broadcast addresses, except for /31 and /32).
Does a /23 waste any addresses for 500 hosts?
Yes, 10 addresses of headroom. A /23 provides 510 usable addresses for the 500 you need, and the next smaller block, a /24, does not hold enough.
How do I work this out myself?
Find the smallest power of two, minus two, that is at least 500. That takes 9 host bits, which leaves a /23 prefix. See how to calculate a subnet mask for the full method.