IPv4 addressing, subnets and CIDR

What takes a packet to the right place on the Internet is the IP address. Routers look not at the whole address but at its prefix. That is why the midterm almost always has questions like “what is the network address of this address, what is the broadcast, how many hosts are there, and which interface does it go to according to this table?” They all rest on the same few bit operations. Someone who knows these rules solves these questions in minutes:

1. Network address = IP AND mask. The host bits are zeroed.
2. Broadcast = all host bits set to 1. That is the network address + block size - 1.
3. Number of hosts = $2^{h} - 2$, where $h$ is the number of host bits.
4. Block size = 256 - the value of the mask octet. Boundaries are always multiples of this number.
5. In routing, the longest prefix wins (longest prefix match).

The IPv4 address and the mask

An IPv4 address is a 32-bit number. To make it readable, it is split into four 8-bit parts (octets) and written in decimal: 192.168.10.77. An address consists of two parts: the network part and the host part. Interfaces with the same network part are on the same subnet and reach each other without a router.

The subnet mask tells where the split is. In the mask, the network bits are 1 and the host bits are 0. The short notation is the prefix length: /26 means “the first 26 bits are the network”.

In every subnet two addresses cannot be given to hosts: the network address, whose host bits are all 0, and the broadcast address, whose host bits are all 1.

CIDR: classless addressing

In the past, addresses were divided into classes (A: /8, B: /16, C: /24). An organization that wanted 2000 hosts did not fit in a class C, so it got a class B, and 63 thousand addresses were wasted. CIDR (Classless Inter-Domain Routing) removed these fixed boundaries: the prefix can be any length, written as a.b.c.d/x. Today the classes are only a historical term. If “classful” appears on an exam, the old system is meant.

Network address, broadcast, host range

Quick method: call the octet whose mask is neither 255 nor 0 the interesting octet. The block size is $256 - \text{mask}$. Find the multiple of the block size that contains that octet of the address.

Example 1: 192.168.10.77/26

mask        255.255.255.192      block size 256-192 = 64 (4th octet)
multiples of 64: 0, 64, 128, 192   77 → between 64 and 127
network     192.168.10.64
broadcast   192.168.10.127
hosts       192.168.10.65  ...  192.168.10.126
number of hosts 2^6 - 2 = 62

The same thing at the bit level: 77 = 01001101. The mask 192 = 11000000. AND: 01000000 = 64.

Example 2: 172.16.45.200/20

The interesting octet is the third: the /20 mask is 255.255.240.0, with block size 16.

multiples of 16: 0, 16, 32, 48     45 → between 32 and 47
network     172.16.32.0
broadcast   172.16.47.255         (3rd octet 47, 4th octet always 255)
hosts       172.16.32.1  ...  172.16.47.254
number of hosts 2^12 - 2 = 4094

PITFALL · the fourth octet. If the interesting octet is the third, the fourth octet of the network address is 0 and that of the broadcast is 255. Someone who writes 172.16.45.200 and says “network 172.16.32.200” forgot to zero the host bits.

Dividing into subnets (subnetting)

If you borrow $n$ of the host bits for the network part, you get $2^n$ equal subnets.

Example 3: split 200.10.20.0/24 into 4 equal subnets. $2^2 = 4$, so borrow 2 bits: /26. The block size is 64.

Example 3b: split 172.16.0.0/16 into /22 subnets. The question usually asks for three things: how many subnets, how many hosts in each, and which one is the n-th subnet.

borrowed bits  22 - 16 = 6       → 2^6 = 64 subnets
host bits      32 - 22 = 10      → 2^10 - 2 = 1022 hosts
block size     256 - 252 = 4     (in the 3rd octet)
subnet 1       172.16.0.0/22
subnet 2       172.16.4.0/22
subnet 3       172.16.8.0/22
subnet 64      172.16.252.0/22

The third octet of the n-th subnet is $(n - 1) \times 4$. Don’t forget to count from zero: subnet 3 starts at 8, not at 12.

VLSM: subnets of different sizes

In a real network, departments have different needs. VLSM (Variable Length Subnet Mask) gives each department the prefix that fits its need. Rule: start with the largest need, and make each subnet start at a multiple of its own block size.

Example 4: with 10.0.0.0/24, three LANs of 100, 50 and 20 hosts and a router link of 2 hosts.

100 hosts → 2^7-2=126 is enough → /25 → 10.0.0.0/25    (.1 - .126,   bc .127)
 50 hosts → 2^6-2=62  is enough → /26 → 10.0.0.128/26  (.129 - .190, bc .191)
 20 hosts → 2^5-2=30  is enough → /27 → 10.0.0.192/27  (.193 - .222, bc .223)
  2 hosts → 2^2-2=2   is enough → /30 → 10.0.0.224/30  (.225 - .226, bc .227)
first free address: 10.0.0.228

TIP · why do we start from the largest? If you start with a small one, a large block has to leave a gap to align itself to a multiple of its own size. Going from large to small, the blocks line up one after another with no gaps.

Route aggregation

The second gain of CIDR: advertising neighbouring networks with a single route. This is called route aggregation or supernetting. Router tables shrink.

Example 5: 198.51.96.0/24, .97.0/24, .98.0/24, .99.0/24 in one route.

 96 = 01100000
 97 = 01100001
 98 = 01100010
 99 = 01100011
      ^^^^^^      common first 6 bits
prefix = 16 + 6 = 22   →   198.51.96.0/22

Aggregation works cleanly only when the blocks are aligned and adjacent. You can’t aggregate 97 and 100 into a single /22.

Longest prefix match

Each row of a router’s forwarding table contains a prefix and an output interface. A destination address can match more than one row. The router chooses the longest prefix, because it is the more specific information.

11.20.0.0/16    → 0
11.20.64.0/18   → 1      (3rd octet 64 to 127)
11.20.72.0/21   → 2      (3rd octet 72 to 79)
0.0.0.0/0       → 3      (default)

11.20.75.9   → /16, /18, /21 match → longest /21 → 2
11.20.100.1  → /16, /18 match      → /18       → 1
11.20.70.3   → /16, /18 (not 72-79)→ /18       → 1
11.21.0.1    → only default        → 3

PITFALL · not the first match. It is wrong to read the table from top to bottom and pick the first row that matches. Find all the matches and look at the prefix length.

Private addresses and special-purpose blocks

RFC 1918 private addresses are not routed on the Internet. They are used in home and company networks, and are translated to a public address by NAT when leaving (next page).

Where do addresses come from? A host usually gets its address through DHCP (Dynamic Host Configuration Protocol): the four steps discover, offer, request, ACK. An organization gets its address block from its ISP, and ISPs get theirs from regional registries (under ICANN).

Common mistakes

135.46.56.0/22   eth0
135.46.60.0/22   eth1
135.46.56.0/21   eth2
0.0.0.0/0        eth3