Subnet for 5 hosts
A network that needs 5 hosts fits in a /29, mask 255.255.255.248, with 6 usable addresses.
A network built for 5 hosts needs a /29. The mask is 255.255.255.248, and its 6 usable addresses leave 1 spare for growth.
The next smaller block, a /30, only allows 2 usable addresses, which is not enough for 5 hosts. That makes /29 the smallest block that fits.
Enter any address below to see the exact network, broadcast, and host range for a /29 block.
Calculate a /29 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).
11000000.00000000.00000010.0000000011111111.11111111.11111111.11111000Example: 192.0.2.0/29
Taking 192.0.2.0/29 as a worked example, a /29 block breaks down like this:
| Network address | 192.0.2.0 |
|---|---|
| Usable range | 192.0.2.1 - 192.0.2.6 |
| Broadcast address | 192.0.2.7 |
| Total addresses | 8 |
| Usable hosts | 6 |
Frequently asked questions
What subnet mask do I need for 5 hosts?
A network with 5 hosts fits in a /29, which uses the mask 255.255.255.248 and provides 6 usable addresses.
How many addresses does a /29 block have in total?
A /29 has 8 total addresses, 6 of them usable (the total minus the network and broadcast addresses, except for /31 and /32).
Does a /29 waste any addresses for 5 hosts?
Yes, 1 address of headroom. A /29 provides 6 usable addresses for the 5 you need, and the next smaller block, a /30, does not hold enough.
How do I work this out myself?
Find the smallest power of two, minus two, that is at least 5. That takes 3 host bits, which leaves a /29 prefix. See how to calculate a subnet mask for the full method.