IPv4 Addressing: Interfaces, Subnets, Masks and CIDR

The picture: a street name and a house number#

Think of a postal address like “Maple Street, number 14”. The street name gets the letter to the right street. The house number gets it to the right door on that street. Every house on Maple Street shares the street name, and no two share a number.

An IPv4 address works the same way. The first part says which network you are on, like the street. The rest says which device on that network, like the house number. Far-away routers only care about the street. Only the router on your street cares about the house number.

The one twist: an IPv4 address does not have a fixed spot where the street name ends. A small number written after a slash, like /24, tells you where the line falls.

Vocabulary, before anything uses it#

Addresses belong to interfaces#

A laptop on Wi-Fi usually has one interface in use, so one IPv4 address. A router joins several links, so it has several interfaces, and each one needs its own address. A router with four links has four IP addresses.

An IPv4 address is 32 bits long. That allows 2³² = 4,294,967,296 different addresses. To make them readable, we split the 32 bits into four bytes and write each byte as a decimal number from 0 to 255.

The bit values in a byte, left to right, are 128, 64, 32, 16, 8, 4, 2, 1. Every conversion is adding up the columns that hold a 1.

Finding the subnets in a network#

Exams often show a drawing of routers and links and ask how many subnets there are. There is a reliable recipe:

1. Imagine unplugging every interface from its router, so each router stands alone.
2. What is left is a set of separate “islands” of connected interfaces.
3. Each island is one subnet. Count them.

Two things trip people up. A LAN (local area network, such as one office’s wired network) with many hosts behind one switch is one island, because a switch is not a router. And a cable joining two routers directly is an island too: a subnet with just two interfaces in it.

Reading a.b.c.d/x: prefix, mask and host count#

The x after the slash is the prefix length: how many bits from the left are the network part. The other 32 − x bits are the host part. Write h for that number of host bits.

addresses in the block = 2^h     usable host addresses = 2^h − 2

You subtract 2 because the all-zeros host part is the network address and the all-ones host part is the broadcast address. Neither goes on an interface.

One exception: on a cable joining exactly two routers, RFC 3021 allows a /31 (2 addresses) with no network or broadcast address. Use the “minus 2” rule unless a question raises /31 on purpose.

Designing subnets from one block#

An organization splits its one block among its departments. Every subnet must start on a multiple of its own size.

Equal-size subnets

Borrowing n bits from the host part gives 2ⁿ subnets of equal size.

Variable-size subnets

Real departments are not the same size. Giving each one a block that fits is called VLSM, variable-length subnet masking. The method:

1. For each group, add 2 to the hosts needed and round up to a power of two. That is its block size.
2. Sort the groups from largest block to smallest.
3. Hand out blocks from the start of the range in that order. Big blocks first keeps every block lined up on a multiple of its size.

That leaves 198.51.100.228 to 198.51.100.255, 28 addresses, spare for later.

Classful addressing, as history#

Before CIDR, the prefix length was fixed by the first few bits of the address. This was classful addressing, and you may still see it in older material.

The sizes jumped too far: a class C was too small for most organizations and a class B far too big. CIDR fixed this by letting the prefix be any length. People still say “class C” loosely to mean “a /24”, but routers ignore classes completely.

Practice#

Work in binary for the first few, even where you could guess. Write 128 64 32 16 8 4 2 1 across the top of your page.

1. Write 172.29.6.130 in binary.
2. Write 11000110 00110011 01100100 00001010 in dotted-decimal.
3. Router R1 connects three LANs and has a direct cable to R2 and another to R3. R2 and R3 each connect one LAN and are not cabled to each other. How many subnets are there?
4. A host has address 10.14.77.200/20. Give the subnet mask, network address, broadcast address, usable range and number of usable hosts.
5. A department needs 500 host addresses. What is the longest prefix (smallest block) that fits, and how many usable hosts does it give?
6. 192.168.40.0/24 is split into equal /27 subnets. How many subnets are there, how many usable hosts does each have, and what is the usable range of the third one?
7. Are 192.0.2.61 and 192.0.2.66 on the same subnet if the prefix is /25? What if it is /26?
8. An organization needs 2,000 host addresses. Compare what classful addressing could give it with what CIDR can give it.

Solutions#

1. 10101100 00011101 00000110 10000010. 172 = 128 + 32 + 8 + 4; 29 = 16 + 8 + 4 + 1; 6 = 4 + 2; 130 = 128 + 2.
2. 198.51.100.10. 11000110 = 128 + 64 + 4 + 2 = 198; 00110011 = 32 + 16 + 2 + 1 = 51; 01100100 = 64 + 32 + 4 = 100; 00001010 = 8 + 2 = 10.
3. 7 subnets. Unplug every router interface: the LAN islands are 3 (on R1) + 1 (on R2) + 1 (on R3) = 5, and the two router-to-router cables are 2 more.
4. Mask 255.255.240.0; network 10.14.64.0; broadcast 10.14.79.255; usable 10.14.64.1 to 10.14.79.254; 4,094 hosts. A /20 keeps 4 bits of the third byte. 77 = 0100|1101; zeroing the low 4 bits gives 01000000 = 64, setting them to ones gives 01001111 = 79. Host bits: 32 − 20 = 12, so 2¹² − 2 = 4,094.
5. A /23, with 510 usable hosts. 500 + 2 = 502 addresses needed. The next power of two is 512 = 2⁹, so 9 host bits and a prefix of 32 − 9 = 23. 512 − 2 = 510.
6. 8 subnets, 30 usable hosts each; the third is 192.168.40.65 to 192.168.40.94. A /27 borrows 3 bits from the /24, so 2³ = 8 subnets of 32 addresses. They start at .0, .32, .64, ... The third is 192.168.40.64/27: network .64, broadcast .95, usable .65 to .94.
7. Same subnet under /25; different subnets under /26. Under /25 the blocks are .0 to .127 and .128 to .255, and both addresses are in the first. Under /26 the blocks are .0 to .63 and .64 to .127, so 61 and 66 fall on either side of the boundary at 64.
8. Classful: a class B, wasting 63,534 addresses. CIDR: a /21 with 2,046 usable hosts. (The alternative was eight separate class C networks, 2,032 usable hosts, but that means eight routing-table entries instead of one.) A class C has only 254 usable hosts, too few. A class B has 65,534, and 65,534 − 2,000 = 63,534 go unused. With CIDR, 2,002 rounds up to 2,048 = 2¹¹, a /21, wasting only 46.

Test-readiness checklist#

• ☐  I can explain why a router has one IP address per interface.
• ☐  I can convert between dotted-decimal and binary without a calculator.
• ☐  I can count the subnets in a drawing by unplugging the router interfaces.
• ☐  I can turn a /x into a subnet mask, and a mask back into a /x.
• ☐  I can find the network address, broadcast address and usable range for any a.b.c.d/x.
• ☐  I can split a block into equal subnets, and into variable-size subnets, largest first.
• ☐  I can say what classful addressing was and why CIDR replaced it.

Frequently asked questions#

Something here not clear? A topic you wish we covered? Tell us. We read every message, and a request is the fastest way to get a guide written.