CIDR Notation and Subnetting

Classful addressing gave every network a fixed size: $2^{24}$, $2^{16}$ or $2^8$ addresses. An organisation needing 2000 addresses had to take a class B block of 65,536 or several class C blocks, and every class C block was a separate entry in the Internet's routing tables. Classless Inter-Domain Routing (CIDR) removes the classes: a block is written a.b.c.d/n, meaning its first $n$ bits are the network prefix and the remaining $32 - n$ bits number the addresses inside it. Blocks of any power-of-two size can be allocated, split into subnets, and merged again into larger aggregates for routing.

Routers then forward by longest-prefix match: among all table entries whose prefix matches the destination, use the most specific one. Subnetting and forwarding calculations reduce to bit operations on the 32-bit address: AND with the mask, OR with its complement, and counting bits.

A mask must be contiguous ones then zeros; 255.255.0.255 is not a mask. The byte where the prefix ends is the only one needing arithmetic: in /20 the third byte keeps its top 4 bits, so block boundaries in that byte fall at multiples of $2^{8 - 4} = 16$.

Splitting a /n block by taking $s$ more prefix bits gives $2^s$ subnets of size $2^{32 - n - s}$. For conventional broadcast subnets with at least two host bits, each has $2^{32 - n - s} - 2$ usable host addresses; /31 point-to-point links use both addresses as endpoints (RFC 3021), while /32 denotes one address/host route. With variable-length subnet masks (VLSM), the subnets may have different sizes: give each need the smallest block with enough usable addresses, and allocate the largest blocks first so that every block starts at a multiple of its own size.

$2^k$ blocks of the same size $/m$ can be written as one /(m - k) exactly when they are contiguous and the first is aligned to a multiple of the combined size. A provider that owns 200.30.8.0/22 advertises one route for all four /24 blocks inside it.

A destination $d$ matches entry $p/n$ when the first $n$ bits of $d$ and $p$ agree. Of all matching entries the router uses the one with the largest $n$; the default route 0.0.0.0/0 matches everything and is used only when nothing else does. This is what lets a more specific route (a customer that moved to another provider) override an aggregate.

An organisation owns 198.51.100.0/24 and needs subnets for LANs of 100, 50 and 25 hosts and one point-to-point link (2 hosts). This worked plan assumes conventional /30 addressing for that link, rather than RFC 3021 /31 support; "smallest block" is relative to that assumption. With /31 support, two addresses suffice for the link.

The addresses .228 to .255 (28 addresses) remain free, for example as 198.51.100.228/30, .232/29 and .240/28. A LAN of 126 hosts would still fit a /25; one of 127 would need a /24.

For address $A$ and prefix length $0\le n\le32$, with $B = 2^{32 - n}$, the network value is the prefix base and the last address is the upper end of the block. That last address is the directed broadcast only for a conventional broadcast subnet with $n\le30$. Under RFC 3021, a /31 point-to-point link uses both addresses as endpoints and has no directed broadcast address; /32 denotes a single host address, not a separate network/broadcast pair:

In the byte where the prefix ends, with $j$ prefix bits in that byte, the network byte is the address byte rounded down to a multiple of $2^{8 - j}$.

Number of subnets and hosts from a mask. If the original block is /n and the subnet mask is /m, there are $2^{m - n}$ subnets. Each conventional broadcast subnet with $m\le30$ has $2^{32 - m} - 2$ usable host addresses; /31 point-to-point links use both addresses as endpoints (RFC 3021), while /32 denotes one address/host route.

Aggregation test. Blocks $N_1 < N_2 < \cdots < N_{2^k}$, each /m, aggregate to /(m - k) if and only if $N_{i+1} = N_i + 2^{32 - m}$ for all $i$ and $N_1$ is a multiple of $2^{32 - m + k}$.

172.21.147.200/20. The prefix ends 4 bits into the third byte; $147$ rounded down to a multiple of 16 is 144.

The block 130.50.0.0/16 is subnetted with mask 255.255.252.0 (/22): $2^{22 - 16} = 64$ subnets, each with $2^{10} - 2 = 1022$ usable hosts.

200.30.8.0/24, 200.30.9.0/24, 200.30.10.0/24 and 200.30.11.0/24 are contiguous, and 8 is a multiple of 4, so they aggregate to 200.30.8.0/22. The four blocks starting at 9, 10, 11 and 12 are contiguous too, but 9 is not a multiple of 4; the shortest exact cover is 200.30.9.0/24, 200.30.10.0/23 and 200.30.12.0/24, three routes.

200.10.64.0/18 holds $2^{14} = 16{,}384$ addresses, from 200.10.64.0 to 200.10.127.255.

Host 10.1.37.18/22 is in 10.1.36.0/22 (37 rounded down to a multiple of 4 is 36), with broadcast address 10.1.39.255.

192.168.10.5 and 192.168.13.200 agree in their first two bytes; in the third byte, 10 = 00001010 and 13 = 00001101 share their first five bits. The common prefix is $16 + 5 = 21$ bits, so the smallest block containing both is 192.168.8.0/21, from 192.168.8.0 to 192.168.15.255 (2048 addresses).

After the plan in "How it works", 28 addresses (.228 to .255) are free, yet no /27 (30 hosts) can be created: the only /27 boundary in that range is .224, and .224 to .227 is in use. The largest new subnet is the /28 at .240 (14 usable hosts); the free /29 blocks are .232, .240 and .248 (the last two together are that /28).

Common misconceptions.