Find the network address, broadcast, usable host range and host count of any IPv4 or IPv6 CIDR block, with netmask, wildcard and equal subnets.
By AdminDrafted with AI writing toolsUpdated Checked examples: 7
192.168.10.77 sits inside 192.168.10.64/26, which has 64 addresses (62 usable, 192.168.10.65 to 192.168.10.126). Private (RFC 1918). Splitting it gives 4 × /28 of 16 addresses each.
| # | Subnet | First usable | Last usable | Broadcast |
|---|---|---|---|---|
| 1 | 192.168.10.64/28 | 192.168.10.65 | 192.168.10.78 | 192.168.10.79 |
| 2 | 192.168.10.80/28 | 192.168.10.81 | 192.168.10.94 | 192.168.10.95 |
| 3 | 192.168.10.96/28 | 192.168.10.97 | 192.168.10.110 | 192.168.10.111 |
| 4 | 192.168.10.112/28 | 192.168.10.113 | 192.168.10.126 | 192.168.10.127 |
The first 26 bits name the network; the other 6 number the addresses inside it.
The all-zeros host is the network itself and the all-ones host is the broadcast, so neither is assigned.
The new prefix adds the borrowed bits; the second line counts addresses in each subnet.
In the formula, A is the address, M the mask, N the network, L the last address and H the conventional IPv4 host count (prefixes /0–/30). A prefix length p keeps the first p bits of an address as the network and leaves the rest to number hosts. The calculator ANDs the address with the netmask to find the network address, ORs that with the wildcard mask (the inverted netmask) to find the last address, and counts 2^(32 − p) addresses for IPv4 or 2^(128 − p) for IPv6.
Engineers use it to plan VLANs, firewall rules and cloud VPC ranges. The default, 192.168.10.77/26, belongs to 192.168.10.64/26: 64 addresses, 62 of them usable from .65 to .126, with .127 as the broadcast. Splitting it 4 ways gives four /28 subnets of 16 addresses.
IPv4 counts leave out the network and broadcast addresses except on /31 point-to-point links (RFC 3021) and /32 host routes. IPv6 has no broadcast. This calculator leaves out the subnet-router anycast address except for /127 point-to-point links and /128 single addresses; actual assignment also depends on address scope, other reservations and platform rules. Splits are always into a power of two equal subnets; for a VLSM plan, split one of the results again.
An IPv4 block written /p has 2^(32 − p) addresses and 2^(32 − p) − 2 usable hosts, because the first address names the network and the last is the broadcast. A /26 has 2⁶ = 64 addresses and 62 hosts: 192.168.10.77/26 sits in 192.168.10.64/26, with hosts .65 to .126.
The rule covers /0 to /30. A /31 point-to-point link uses both of its addresses, and a /32 is a single host.
Work out the block size: find the first mask octet that isn't 255 and subtract it from 256. A /26 mask is 255.255.255.192, so blocks are 256 − 192 = 64 addresses long and start at .0, .64, .128 and .192.
The same method works in any octet. For 172.16.5.4 with mask 255.255.240.0 (/20), third-octet blocks are 256 − 240 = 16 wide. The network is 172.16.0.0 and the broadcast 172.16.15.255, leaving 4,094 hosts.
In binary, the network address is the address ANDed with the mask, and the broadcast is the network with every host bit set to 1.
Each extra prefix bit halves the block. A netmask must be a run of ones followed by zeros, so 255.0.255.0 is not a valid mask.
The wildcard mask used in Cisco access lists and OSPF is the inverse: subtract each octet from 255. A /26 has wildcard 0.0.0.63.
| Address and prefix | Network | Netmask | Usable | Usable range |
|---|---|---|---|---|
| 192.168.10.77/26 | 192.168.10.64/26 | 255.255.255.192 | 62 | 192.168.10.65 – 192.168.10.126 |
| 172.16.5.4/255.255.240.0 | 172.16.0.0/20 | 255.255.240.0 | 4094 | 172.16.0.1 – 172.16.15.254 |
| 10.0.0.0/31 | 10.0.0.0/31 | 255.255.255.254 | 2 | 10.0.0.0 – 10.0.0.1 |
| 203.0.113.5/32 | 203.0.113.5/32 | 255.255.255.255 | 1 | 203.0.113.5 – 203.0.113.5 |
| 0.0.0.0/0 | 0.0.0.0/0 | 0.0.0.0 | 4294967294 | 0.0.0.1 – 255.255.255.254 |
The /31 row follows RFC 3021 for point-to-point links, and the /32 row is a single host route. 0.0.0.0/0 covers the whole IPv4 space, which is how a default route is written.
The first three rows are private addresses. RFC 1918 reserves 10.0.0.0/8, 172.16.0.0/12 and 192.168.0.0/16 for internal networks. The /32 example, 203.0.113.5, comes from a separate block that RFC 5737 sets aside for documentation.
Borrowing k host bits makes 2ᵏ equal subnets, each with a prefix k bits longer. Splitting 192.168.10.64/26 four ways borrows 2 bits and gives four /28s of 16 addresses and 14 hosts each, starting at .64, .80, .96 and .112.
A request for 3 subnets rounds up to 4, because binary splits come in powers of two. For subnets of different sizes (VLSM), split one of the results again, for example one /28 into two /29s.
IPv6 has no broadcast address. A /64, the standard size for one LAN, holds 2⁶⁴ = 18,446,744,073,709,551,616 addresses. A /48 contains 65,536 of those /64s, and a /56 contains 256.
The first address of an IPv6 subnet is the subnet-router anycast address and isn't given to a host, which leaves 2⁶⁴ − 1 in a /64. A /127 point-to-point link (RFC 6164) uses both of its addresses, and a /128 is a single address.
AWS and Azure reserve 5 addresses in every ordinary IPv4 subnet: the network address, the next three and the last. A /24 there has 251 assignable addresses, not 254.
The smallest AWS subnet, a /28, leaves 11 addresses, and Azure's smallest, a /29, leaves 3. Plan cloud capacity with 2^(32 − p) − 5.
To see the same addresses as bits or hex, use the binary and hex converter.
Checked against: Independent calculation
Checked against: RFC 3021 §2
Checked against: List = [203.0.113.5]; RFC 5737 §3
Checked against: Independent calculation
A /24 has 256 addresses and 254 usable hosts, because the all-zeros address names the network and the all-ones address is the broadcast. The IPv4 rule is 2^(32 − prefix) − 2: a /25 gives 126, a /26 62, a /27 30, a /28 14, a /29 6 and a /30 2. The exceptions are /31, where RFC 3021 lets a point-to-point link use both addresses, and /32, a single host.
They say the same thing in two ways. CIDR notation writes the number of network bits after a slash, as in /26; a subnet mask writes those bits as a dotted address with that many leading ones, so /26 is 255.255.255.192 and /20 is 255.255.240.0. CIDR, introduced in 1993 and now defined in RFC 4632, replaced the fixed class A, B and C boundaries, so any prefix from /0 to /32 is valid.
A wildcard mask is the subnet mask with every bit inverted, so 255.255.255.192 becomes 0.0.0.63. Cisco IOS access lists and OSPF network statements use it: a 0 bit must match and a 1 bit is ignored. To get it, subtract each octet of the netmask from 255. For a /26 the last octet is 63, one less than the block's 64 addresses.
For ordinary IPv4 subnets, both clouds reserve 5 addresses: the network address, the next three (for the router and DNS, or held for future use) and the last address. A /24 in an AWS VPC or Azure virtual network therefore has 251 assignable addresses instead of the 254 shown here. The smallest subnet AWS allows is a /28, which leaves 11; Azure's smallest is a /29, which leaves 3. AWS BYOIP ranges are an exception: all addresses in the range can be used. This calculator reports conventional IP host counts, not provider-specific assignable capacity.
A /64 holds 2^64 = 18,446,744,073,709,551,616 addresses. It is the standard size for one IPv6 LAN because RFC 4291 requires 64-bit interface identifiers for most unicast addresses, and stateless autoconfiguration (SLAAC) builds addresses from them. A /48 contains 65,536 of those /64 subnets, and a /56 contains 256.
This calculator includes 7 worked examples with independently sourced answers. They run in the test suite, and you can run them here too.