How many usable hosts are in a /24?

254. A /24 leaves eight host bits, so it spans 2^8 = 256 addresses, and two of those are reserved: the all-zeros address names the network itself and the all-ones address is the broadcast. The same subtraction gives 126 hosts on a /25, 62 on a /26, 30 on a /27, 14 on a /28, 6 on a /29 and 2 on a /30. Going the other way, a /16 holds 65,534 hosts and a /8 holds 16,777,214.

What is a subnet mask and what does it actually do?

A subnet mask is a 32-bit number written as a solid run of ones followed by a solid run of zeros — 255.255.255.0 is twenty-four ones then eight zeros. The ones mark the bits of an address that identify the network and the zeros mark the bits that identify a host inside it. A host ANDs its own address with the mask, ANDs the destination address with the same mask, and compares: matching results mean the destination is on the local link, different results mean the packet goes to the default gateway. That single comparison is the mask's entire job.

What does the /24 in an IP address mean?

It is CIDR notation, and the number is simply how many leading bits of the 32-bit address belong to the network. /24 means the first 24 bits are the network and the remaining 8 identify a host, which is exactly the same thing as the mask 255.255.255.0. CIDR replaced the old class A/B/C system in 1993 precisely so a prefix could fall anywhere, not just on the 8-, 16- and 24-bit boundaries the classes allowed.

Why do you subtract 2 from the number of addresses?

Because two addresses in every ordinary subnet are reserved rather than assignable. The address with all host bits set to zero is the network address, the name of the subnet itself, and the address with all host bits set to one is the directed broadcast, which reaches every host on that link at once. Neither can be configured on an interface. The exceptions are a /31, where RFC 3021 removed the reservation entirely for point-to-point links, and a /32, which is a single host route.

What is a /31 subnet, and why does it have 2 usable hosts?

A /31 has exactly two addresses, and RFC 3021 says both are assignable. On a point-to-point link there is nowhere for a broadcast to go that a unicast would not already reach, so the network and broadcast reservations serve no purpose and are dropped. Using /31s instead of /30s on router-to-router links halves the address waste — a /30 spends four addresses to connect two interfaces. Most modern routers support /31 links; some older or embedded stacks still insist on a /30.

What is the difference between a subnet mask and a wildcard mask?

They are bitwise inverses of each other. A subnet mask puts ones on the network bits — 255.255.255.0 — while the matching wildcard mask puts ones on the bits that are allowed to vary — 0.0.0.255. Subtract each octet of the mask from 255 and you have the wildcard. Cisco access lists and OSPF network statements take the wildcard form, which is why the two get confused; the arithmetic underneath is identical.

Which IP address ranges are private?

RFC 1918 reserves three blocks for private use: 10.0.0.0/8, 172.16.0.0/12 and 192.168.0.0/16. They are never routed on the public internet, so anyone can use them behind NAT. Three more ranges are commonly mistaken for private: 127.0.0.0/8 is loopback, 169.254.0.0/16 is link-local and appears when DHCP fails, and 100.64.0.0/10 is carrier-grade NAT space an ISP uses between its own equipment and your router.

Tech

Subnet Calculator

A prefix splits 32 bits in two: network = IP AND mask, and usable hosts = 2^(32 − prefix) − 2. Enter an address and a prefix for the network address, broadcast, first and last host, wildcard and mask — plus the bits themselves with the network/host boundary drawn on, a block splitter, and the full /0 to /32 chart.

Your network

IPv4 address

Paste a whole CIDR block and it will split itself

/24

Prefix length/24

How many of the 32 bits identify the network

/0/8/16/24/32

Subnet mask

Type a dotted-decimal mask and the prefix follows

Network

192.168.1.0/24

254 usable hostsPrivate (RFC 1918)Class C

Address in binary

11000000101010000000000100001010

24 network bits · 8 host bits

Mask in binary

11111111111111111111111100000000

Network address: 192.168.1.0
Broadcast address: 192.168.1.255
First usable host: 192.168.1.1
Last usable host: 192.168.1.254
Subnet mask: 255.255.255.0
Wildcard mask: 0.0.0.255
Total addresses: 256
Usable hosts: 254

Host range

192.168.1.1 → 192.168.1.254

192.168.1.0 names the network and 192.168.1.255 is the broadcast, so neither can be given to a machine.

Try a block

No sign-up · Free

The basics

What a subnet actually is

An IPv4 address is not really four numbers. It is one 32-bit integer that we write in four chunks of eight bits because that is easier to say out loud. Subnetting is the act of drawing a line through those 32 bits: everything on the left names a network, everything on the right names a host inside it. Move the line and you trade network count against host count — that is the whole subject.

32 bits, four octets

One number wearing a disguise

Each dotted number is one octet, eight bits, a value from zero to two hundred and fifty-five. Nothing about the dots is meaningful to the maths; they are punctuation. Every operation in this page happens on the bits underneath.

192.168.1.10 = 11000000.10101000.00000001.00001010

prefix = network bits

The line through the bits

The prefix length n says how many leading bits belong to the network. The remaining 32 − n bits count the addresses inside it. A longer prefix means more subnets, each smaller; a shorter one means fewer subnets, each larger.

A /26 keeps 26 network bits and 6 host bits → 62 usable hosts

192.168.1.130 as a /26

11000000101010000000000110000010

26 network bits, then 6 host bits — the accent bar is the subnet boundary

The mask

How a subnet mask really works

The mask is how a machine finds that boundary without being told where it is. It is 32 bits of solid ones followed by solid zeros, and it is used with one operation: bitwise AND. AND returns one only when both inputs are one, so ANDing an address with a mask keeps the network bits and wipes the host bits to zero. What is left is the network address.

Address bit	Mask bit	AND
0	0	0
0	1	0
1	0	0
1	1	1

Address 192.168.1.130

11000000101010000000000110000010

AND mask 255.255.255.192

11111111111111111111111111000000

= network 192.168.1.128

11000000101010000000000110000000

The host bits are gone; the network bits survived untouched.

ones then zeros, never mixed

Why 255.255.0.255 is not a mask

A legal mask is a single unbroken run of ones. If the ones were allowed to be scattered the network bits would not be contiguous, routing prefixes could not be compared by length, and the whole idea of a prefix would collapse. Every stack rejects a discontiguous mask.

255.255.240.0 is legal · 255.255.0.255 is rejected

wildcard = NOT mask

The mask turned inside out

Invert every bit and you get the wildcard mask, where ones mark the bits allowed to vary. Access lists and routing protocol statements use this form. Subtract each octet from two hundred and fifty-five to convert between them in your head.

Mask 255.255.255.192 → wildcard 0.0.0.63

Method

How to subnet an address by hand

Four steps take you from an address and a prefix to every number a router cares about. None of them need a calculator once the powers of two are familiar.

1
Turn the prefix into a mask
Write n ones, then 32 − n zeros, and read the result back in octets. A prefix that is not a multiple of eight leaves one octet part-full — that is the only one worth thinking about.
/26 → 255.255.255.192
2
AND the address with the mask
Every host bit becomes zero and every network bit survives. The result is the network address — the name of the subnet, and the address you write in a routing table.
192.168.1.130 AND 255.255.255.192 = 192.168.1.128
3
Set every host bit to one for the broadcast
The top of the block is the directed broadcast. Equivalently, add 2^(32 − n) − 1 to the network address, or OR the wildcard mask onto it. Both give the same answer.
192.168.1.128 OR 0.0.0.63 = 192.168.1.191
4
Count what is left in the middle
The block holds 2^(32 − n) addresses. Take out the network and broadcast and the rest are assignable, running from one above the network to one below the broadcast.
2^6 − 2 = 62 hosts · 192.168.1.129 – 192.168.1.190

In binary

Splitting a network, worked bit by bit

The exam question, done the long way once so the shortcut makes sense: take a /24, borrow bits from the host side, and enumerate the subnets that fall out. Every figure below is computed by the same code the calculator above runs on.

1
Write the block you were given in binary
The network is 192.168.10.0/24. Twenty-four ones in the mask means the first three octets are fixed and the last eight bits are yours to hand out.
Network 192.168.10.0
11000000101010000000101000000000
Mask 255.255.255.0
11111111111111111111111100000000
24 network bits · 8 host bits · 254 usable hosts
2
Decide how many bits to borrow
Each bit you move from the host side to the network side doubles the number of subnets. Four subnets needs two bits, because two bits count to four. Always round up to the next power of two — you cannot borrow half a bit.
2^2 = 4 subnets ≥ 4 wanted
3
Move the boundary and read off the new mask
The prefix grows by the number of bits borrowed. Those two extra ones sit at the top of the fourth octet, which in decimal is 128 + 64 = 192.
New mask 255.255.255.192
11111111111111111111111111000000
/24 + 2 borrowed bits = /26 · 6 host bits left
/24 + 2 = /26 → 255.255.255.192
4
Work out the block size from the mask
Subtract the interesting mask octet from 256 and you have the step between consecutive subnets. That number — the block size, or magic number — is the only arithmetic you need to enumerate them.
256 − 192 = 64 addresses per subnet

Count in block-size steps and fill in the ranges

Start at the parent network and add the block size repeatedly. Each result is a network address; the address one below the next network is that subnet's broadcast, and everything strictly between them is assignable.

Subnet	Network	Host range	Broadcast
1	192.168.10.0/26	192.168.10.1 – 192.168.10.62	192.168.10.63
2	192.168.10.64/26	192.168.10.65 – 192.168.10.126	192.168.10.127
3	192.168.10.128/26	192.168.10.129 – 192.168.10.190	192.168.10.191
4	192.168.10.192/26	192.168.10.193 – 192.168.10.254	192.168.10.255

4 subnets × 62 usable hosts = 248 addressable machines, down from 254

Worked examples

Five real allocations

The same arithmetic across the block sizes people actually deploy — an office LAN, a router link, a campus split, a small DMZ, and four networks merged back into one. Switch tabs to watch the working move.

One office, one flat network

A single office needs addresses for laptops, printers and phones. The default private /24 is the obvious allocation and nobody has to think about masks again.

Given 192.168.10.0/24 mask 255.255.255.0

Host bits: 32 − 24 = 8

Addresses: 2^8 = 256

Usable: 256 − 2 = 254

Range: 192.168.10.1 → 192.168.10.254

Broadcast: 192.168.10.255

192.168.10.0/24 → 254 usable hosts

The whole last octet is host space, so the mask lines up with the dots and every address in the office shares the first three numbers.

Subnet splitter

Cut a block into equal subnets

Give it a parent block and the size you want the pieces to be. It borrows the bits, works out how many subnets that makes, and lists each one with its network, host range and broadcast address. Long lists are capped at the first sixty-four rows; the rest continue the same fixed step.

Split a block

Parent block

The network you have been allocated

/24

New subnet size/26

Every piece comes out the same size

Subnets

Addresses each

Usable each

New mask

/26

192.168.1.0/24 borrows 2 bits → 2^2 = 4 subnets of 64 addresses

#	Network	First host	Last host	Broadcast
1	192.168.1.0/26	192.168.1.1	192.168.1.62	192.168.1.63
2	192.168.1.64/26	192.168.1.65	192.168.1.126	192.168.1.127
3	192.168.1.128/26	192.168.1.129	192.168.1.190	192.168.1.191
4	192.168.1.192/26	192.168.1.193	192.168.1.254	192.168.1.255

Vlsm

Variable-length masks: stop giving everyone the same size

Equal-size subnetting wastes space the moment your departments are different sizes. VLSM fixes it by sizing each subnet to its own requirement, largest first, packing them from the bottom of the block. The rule that makes it work: every subnet must begin on a multiple of its own size, so the big blocks have to be placed before the small ones.

Segment	Hosts needed	Prefix	Network	Host range	Broadcast	Spare
Sales floor	100	/25	192.168.4.0	192.168.4.1 – 192.168.4.126	192.168.4.127	26
Engineering	50	/26	192.168.4.128	192.168.4.129 – 192.168.4.190	192.168.4.191	12
Operations	25	/27	192.168.4.192	192.168.4.193 – 192.168.4.222	192.168.4.223	5
Guest Wi-Fi	10	/28	192.168.4.224	192.168.4.225 – 192.168.4.238	192.168.4.239	4
Link to branch A	2	/31	192.168.4.240	192.168.4.240 – 192.168.4.241	—	0
Link to branch B	2	/31	192.168.4.242	192.168.4.242 – 192.168.4.243	—	0

192.168.4.0/24 · 244 of 256 addresses allocated, 12 left for growth. An equal-size split able to hold the largest segment would need /25 blocks of 126 hosts and would run out after two.

Sizing

The prefix each host count needs

Round the requirement up to the next power of two, then remember the two reserved addresses. Always size for the machines you will have, not the ones you have today.

Hosts needed	Smallest prefix	Mask	Usable
2	/31	255.255.255.254	2
5	/29	255.255.255.248	6
10	/28	255.255.255.240	14
25	/27	255.255.255.224	30
50	/26	255.255.255.192	62
100	/25	255.255.255.128	126
200	/24	255.255.255.0	254
500	/23	255.255.254.0	510
1,000	/22	255.255.252.0	1,022
5,000	/19	255.255.224.0	8,190
60,000	/16	255.255.0.0	65,534

Shortcut

The block size, or “magic number”

Nobody subnets in binary at a whiteboard. They use the block size: find the one octet the mask is neither all ones nor all zeros in, subtract that octet from two hundred and fifty-six, and you have the step between consecutive networks. Count in that step and the subnet boundaries fall out.

256 − mask octet

Find the step

The interesting octet is the one the prefix lands inside. Subtracting it from two hundred and fifty-six gives the number of addresses in each subnet, which is also the gap between one network address and the next.

A /26 masks to 255.255.255.192 → 256 − 192 = 64

count in steps

Walk the boundaries

Start at zero in that octet and keep adding the block size until you pass two hundred and fifty-five. Every number you land on is a network address, and every address just below the next one is a broadcast.

/26 networks: .0 · .64 · .128 · .192

works in any octet

Not just the last one

A prefix between eight and sixteen steps through the second octet, one between sixteen and twenty-four through the third. The method is the same; only the column you are counting in changes.

A /20 masks to 255.255.240.0 → 256 − 240 = 16 in the third octet

Reference

Every CIDR prefix from /0 to /32

The complete table: the dotted-decimal mask, the wildcard mask, how many addresses the block holds, how many of those can be assigned to machines, and how it compares to a /24. All thirty-three rows, computed rather than typed.

Prefix	Subnet mask	Wildcard mask	Addresses	Usable hosts	Relative to a /24	Note
/0	0.0.0.0	255.255.255.255	4,294,967,296	4,294,967,294	16,777,216 × /24	The whole IPv4 space / default route
/1	128.0.0.0	127.255.255.255	2,147,483,648	2,147,483,646	8,388,608 × /24
/2	192.0.0.0	63.255.255.255	1,073,741,824	1,073,741,822	4,194,304 × /24
/3	224.0.0.0	31.255.255.255	536,870,912	536,870,910	2,097,152 × /24
/4	240.0.0.0	15.255.255.255	268,435,456	268,435,454	1,048,576 × /24
/5	248.0.0.0	7.255.255.255	134,217,728	134,217,726	524,288 × /24
/6	252.0.0.0	3.255.255.255	67,108,864	67,108,862	262,144 × /24
/7	254.0.0.0	1.255.255.255	33,554,432	33,554,430	131,072 × /24
/8	255.0.0.0	0.255.255.255	16,777,216	16,777,214	65,536 × /24	One classful class A
/9	255.128.0.0	0.127.255.255	8,388,608	8,388,606	32,768 × /24
/10	255.192.0.0	0.63.255.255	4,194,304	4,194,302	16,384 × /24
/11	255.224.0.0	0.31.255.255	2,097,152	2,097,150	8,192 × /24
/12	255.240.0.0	0.15.255.255	1,048,576	1,048,574	4,096 × /24
/13	255.248.0.0	0.7.255.255	524,288	524,286	2,048 × /24
/14	255.252.0.0	0.3.255.255	262,144	262,142	1,024 × /24
/15	255.254.0.0	0.1.255.255	131,072	131,070	512 × /24
/16	255.255.0.0	0.0.255.255	65,536	65,534	256 × /24	One classful class B
/17	255.255.128.0	0.0.127.255	32,768	32,766	128 × /24
/18	255.255.192.0	0.0.63.255	16,384	16,382	64 × /24
/19	255.255.224.0	0.0.31.255	8,192	8,190	32 × /24
/20	255.255.240.0	0.0.15.255	4,096	4,094	16 × /24
/21	255.255.248.0	0.0.7.255	2,048	2,046	8 × /24
/22	255.255.252.0	0.0.3.255	1,024	1,022	4 × /24
/23	255.255.254.0	0.0.1.255	512	510	2 × /24
/24	255.255.255.0	0.0.0.255	256	254	1 × /24	One classful class C
/25	255.255.255.128	0.0.0.127	128	126	1/2 of a /24
/26	255.255.255.192	0.0.0.63	64	62	1/4 of a /24
/27	255.255.255.224	0.0.0.31	32	30	1/8 of a /24
/28	255.255.255.240	0.0.0.15	16	14	1/16 of a /24
/29	255.255.255.248	0.0.0.7	8	6	1/32 of a /24
/30	255.255.255.252	0.0.0.3	4	2	1/64 of a /24	Classic point-to-point link
/31	255.255.255.254	0.0.0.1	2	2	1/128 of a /24	Point-to-point, RFC 3021
/32	255.255.255.255	0.0.0.0	1	1	1/256 of a /24	Single host route

/31 and /32 break the minus-two rule on purpose: a /31 is a two-address point-to-point link under RFC 3021 and a /32 is a single host route.

Special ranges

Private, loopback, link-local and the rest

Not every IPv4 block is routable on the public internet. These are the reservations you will actually meet — the three RFC 1918 private ranges everyone builds on, plus the blocks that look private but are not.

Block	Range	Addresses	Name	RFC	What it is for
255.255.255.255/32	255.255.255.255 – 255.255.255.255	1	Limited broadcast	RFC 919	Reaches every host on the local link and is never routed.
192.0.2.0/24	192.0.2.0 – 192.0.2.255	256	Documentation (TEST-NET-1)	RFC 5737	Reserved for examples and manuals so no real network is disrupted.
198.51.100.0/24	198.51.100.0 – 198.51.100.255	256	Documentation (TEST-NET-2)	RFC 5737	A second documentation block, used when an example needs two networks.
203.0.113.0/24	203.0.113.0 – 203.0.113.255	256	Documentation (TEST-NET-3)	RFC 5737	A third documentation block, common in RFC examples.
169.254.0.0/16	169.254.0.0 – 169.254.255.255	65,536	Link-local (APIPA)	RFC 3927	Self-assigned when DHCP fails. Reachable only on the local link.
198.18.0.0/15	198.18.0.0 – 198.19.255.255	131,072	Benchmarking	RFC 2544	Set aside for testing network devices, not for production traffic.
172.16.0.0/12	172.16.0.0 – 172.31.255.255	1,048,576	Private (RFC 1918)	RFC 1918	16 contiguous /16s. The usual choice for medium campus networks.
192.168.0.0/16	192.168.0.0 – 192.168.255.255	65,536	Private (RFC 1918)	RFC 1918	256 /24s. What almost every home router hands out.
100.64.0.0/10	100.64.0.0 – 100.127.255.255	4,194,304	Carrier-grade NAT	RFC 6598	Shared address space an ISP uses between its NAT and your router.
10.0.0.0/8	10.0.0.0 – 10.255.255.255	16,777,216	Private (RFC 1918)	RFC 1918	16.7 million addresses. The default for large enterprise networks.
127.0.0.0/8	127.0.0.0 – 127.255.255.255	16,777,216	Loopback	RFC 1122	Traffic never leaves the host. 127.0.0.1 is localhost.
0.0.0.0/8	0.0.0.0 – 0.255.255.255	16,777,216	This network	RFC 1122	Only valid as a source address while a host is still learning its own.
224.0.0.0/4	224.0.0.0 – 239.255.255.255	268,435,456	Multicast	RFC 5771	One-to-many group addresses. Not assigned to an interface.
240.0.0.0/4	240.0.0.0 – 255.255.255.255	268,435,456	Reserved (class E)	RFC 1112	Held back for future use and dropped by most stacks.

History

CIDR versus the old class A, B and C

Before 1993 the prefix was not written down at all — it was inferred from the first few bits of the address. That system, classful addressing, allowed exactly three network sizes, and the middle one was so much bigger than most organisations needed that the address space was being burned through at an alarming rate. CIDR replaced it by making the prefix explicit and free to sit anywhere.

Class	Leading bits	First octet	Implied prefix	Notes
A	0	0.0.0.0 – 127.255.255.255	/8	128 networks of 16,777,214 hosts each.
B	10	128.0.0.0 – 191.255.255.255	/16	16,384 networks of 65,534 hosts each.
C	110	192.0.0.0 – 223.255.255.255	/24	2,097,152 networks of 254 hosts each.
D	1110	224.0.0.0 – 239.255.255.255	—	Multicast — not assigned to interfaces.
E	1111	240.0.0.0 – 255.255.255.255	—	Reserved, never allocated for general use.

the prefix was implied

Classful: three sizes, take it or leave it

The first bits of the address decided the size of the network, so there was no way to ask for something in between. Companies that needed more than a class C took a class B and left almost all of it idle.

A class B held 65,534 hosts — handed to organisations needing a few hundred

the prefix is written down

CIDR: any boundary, and routes that merge

Making the prefix explicit did two things at once: allocations could be sized properly, and adjacent blocks could be advertised as one shorter prefix. That second trick, supernetting, is what kept the global routing table from exploding.

Four consecutive /24s advertise as one /22 — 1,024 addresses in a single route

Worth knowing

Rules worth remembering

Six facts that decide whether a subnetting answer is right, most of which are the edge cases a quick formula gets wrong.

two addresses are never assignable

The network and the broadcast

The all-zeros host address names the subnet and the all-ones host address is the directed broadcast. Neither goes on an interface, which is why small blocks feel so wasteful.

A /30 has 4 addresses but only 2 usable

RFC 3021

A /31 has two usable addresses

On a point-to-point link a broadcast reaches nothing a unicast would not, so the reservation was dropped. Both addresses are assignable, which halves the waste of the traditional four-address link.

10.0.0.0/31 → 10.0.0.0 and 10.0.0.1, no broadcast

prefix /32

A /32 is one address

A host route. It appears on loopback interfaces, in routing tables pointing at a single server, and in firewall rules that mean one machine and nothing else.

8.8.8.8/32 → exactly 1 host, no network or broadcast pair

the mask is local

Two ends can disagree, and will break

Nothing carries the mask across the wire. Each host uses its own, so a mismatch shows up as traffic that works in one direction, or works to some addresses on the link and not others.

One end on a /24 and the other on a /25 will each reach the other only half the time

subnet zero is fine now

The first and last subnets are usable

Old equipment refused the all-zeros and all-ones subnets of a block. That restriction was lifted long ago and every current platform allows both, though the habit lives on in old study material.

A /24 split into /26s gives four usable subnets, not two

longest prefix wins

How a router picks between overlapping routes

Routes are not tried in order. The most specific match — the longest prefix — always wins, which is what lets a specific block be pulled out of a larger aggregate.

A packet matching both a /16 and a /24 follows the /24

Common mistakes

Where subnetting answers go wrong

Almost every wrong answer in a subnetting exercise is one of these six, and none of them are arithmetic slips.

Subtracting two from a /31 or a /32: The blanket 2^(32 − n) − 2 rule gives zero usable hosts on a /31 and minus one on a /32. Both are wrong: a /31 has two usable addresses under RFC 3021 and a /32 has one. Handle the last two prefixes as special cases.
Reading the host address as the network address: 192.168.1.130/26 is not the network 192.168.1.130. AND it with the mask first: the network is 192.168.1.128 and the broadcast is 192.168.1.191. Always normalise to the network address before comparing anything.
Assuming the mask stops at a dot: Prefixes that are not multiples of eight split an octet in half, so the boundary sits inside a number rather than between two of them. A /20 subnets in the third octet and a /26 in the fourth.
Sizing subnets smallest-first in VLSM: A subnet must start on a multiple of its own size. Allocate a small block first and the next large one cannot start where you left off, so the gap is stranded. Always place the largest requirement first.
Forgetting the gateway needs an address too: A segment with fifty machines needs fifty-one usable addresses once the router interface is counted, plus anything for printers, access points and management. Size for the count you will reach, not the one on the ticket.
Treating 169.254.x.x or 100.64.x.x as private: Only 10.0.0.0/8, 172.16.0.0/12 and 192.168.0.0/16 are RFC 1918 private space. A 169.254 address means DHCP failed, and 100.64.0.0/10 belongs to your ISP's carrier-grade NAT — using either as your own LAN range causes problems that are hard to diagnose.

Ipv6

The same idea with a hundred and twenty-eight bits

IPv6 keeps CIDR notation and throws away almost everything else that makes IPv4 subnetting fiddly. The address is 128 bits rather than 32, written as eight groups of four hex digits, and the prefix means exactly what it means here: how many leading bits identify the network.

a /64 for every link

One size, by convention

Almost every ordinary IPv6 network is a slash sixty-four, regardless of how many machines are on it. Stateless address autoconfiguration depends on that half-and-half split, so shrinking a link prefix below it breaks things rather than saving anything.

2001:db8:abcd:1::/64 — the first 64 bits are the network, the last 64 the interface

no broadcast at all

Nothing to subtract

IPv6 replaced broadcast with multicast, and there is no reserved top address in a prefix. The minus-two rule simply does not exist, which is a relief given the block sizes involved.

A /64 holds 18,446,744,073,709,551,616 addresses and effectively all of them are usable

the site prefix

Why an ISP hands out a /48 or /56

You subnet at the site level, not the link level: carve your allocation into /64s and assign one per VLAN. The arithmetic is the same borrowing of bits, just with far more of them to spend.

A /48 contains 65,536 slash sixty-fours; a /56 contains 256

scarcity is gone

The maths differs because the pressure did

IPv4 subnetting is an exercise in conservation — that is why VLSM, NAT and the minus-two rule matter. IPv6 has enough space that planning is about readable structure rather than squeezing the last address out of a block.

One /64 link holds more addresses than the entire IPv4 internet, squared

The calculator above is IPv4 only. The one habit worth carrying across: normalise to the network address before you compare anything, in either protocol.

Quick answers

The blocks people look up most

The handful of prefixes that cover most real questions, answered in one line each.

How many hosts in a /24?254 usable, 256 total

What mask is a /26?255.255.255.192, 62 hosts

How many hosts in a /8?16,777,214 usable

What is the broadcast of 192.168.1.0/24?192.168.1.255

How many /24s in a /16?256 of them

Usable hosts on a /30?2 — a point-to-point link

Usable hosts on a /31?2, and no broadcast address

What does /32 mean?A single host, mask 255.255.255.255

FAQ

Common questions

Write the address and the mask as 32 bits and AND them together — every bit where both are 1 stays 1, everything else becomes 0. The result is the network address. Flip the mask to get the wildcard and OR it onto the network for the broadcast address. Everything strictly between those two is assignable, which gives the shortcut people actually use: a /n has 2^(32 − n) addresses and 2^(32 − n) − 2 usable hosts. For 192.168.1.10/24 that is 192.168.1.0 as the network, 192.168.1.255 as the broadcast, 192.168.1.1 to 192.168.1.254 for hosts, and 254 of them.

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