Logical Addressing - IPv4 Addresses

Logical Addressing - IPv4 Addresses

An IPv4 address is a 32-bit address that uniquely and universally defines the connection of a device (for example, a computer or a router) to the Internet. IPv4 addresses are unique. They are unique in the sense that each address defines one, and only one, connection to the Internet. Two devices on the Internet can never have the same address at the same time. The IPv4 addresses are universal in the sense that the addressing system must be accepted by any host that wants to be connected to the Internet.

1. Address Space

A protocol such as IPv4 that defines addresses has an address space. An address space is the total number of addresses used by the protocol. If a protocol uses N bits to define an address, the address space is 2N because each bit can have two different values (0 or 1) and N bits can have 2N values.

IPv4 uses 32-bit addresses, which means that the address space is 232 or 4,294,967,296 (more than 4 billion). This means that, theoretically, if there were no restrictions, more than 4 billion devices could be connected to the Internet.

2. Notations

There are two prevalent notations to show an IPv4 address: binary notation and dotted decimal notation.

a. Binary Notation

In binary notation, the IPv4 address is displayed as 32 bits. Each octet is often referred to as a byte. So it is common to hear an IPv4 address referred to as a 32-bit address or a 4-byte address. The following is an example of an IPv4 address in binary notation.

01110101 10010101 00011101 00000010

b. Dotted-Decimal Notation

To make the IPv4 address more compact and easier to read, Internet addresses are usually written in decimal form with a decimal point (dot) separating the bytes. The following is the dotted-decimal notation of the above address:

117.149.29.2

Figure 3.1 shows an IPv4 address in both binary and dotted-decimal notation. Note that because each byte (octet) is 8 bits, each number in dotted-decimal notation is a value ranging from 0 to 255

Example 3.1

Change the following IPv4 addresses from binary notation to dotted-decimal notation.

a)10000001 00001011 00001011 11101111

b)11000001 10000011 00011011 11111111

Solution

We replace each group of 8 bits with its equivalent decimal number and add dots for separation.

a)129.11.11.239

b)193.131.27.255

Example 3.2

Change the following IPv4 addresses from dotted-decimal notation to binary notation.

a)111.56.45.78

b)221.34.7.82

Solution

We replace each decimal number with its binary equivalent.

a)01101111 00111000 00101101 01001110

b)11011101 00100010 00000111 01010010

Example 3.3

Find the error, if any, in the following IPv4 addresses.

a)111.56.045.78

b)221.34.7.8.20

c) 75.45.301.14

d)11100010.23.14.67

Solution

a)There must be no leading zero (045).

b)There can be no more than four numbers in an IPv4 address.

c) Each number needs to be less than or equal to 255 (301 is outside this range).

d)A mixture of binary notation and dotted-decimal notation is not allowed.

3. Classful Addressing

IPv4 addressing, at its inception, used the concept of classes. This architecture is called classful addressing. In classful addressing, the address space is divided into five classes: A, B, C, D, and E.

We can find the class of an address when given the address in binary notation or dotted-decimal notation. If the address is given in binary notation, the first few bits can immediately tell us the class of the address. If the address is given in decimal-dotted notation, the first byte defines the class. Both methods are shown in Figure 3.2.

Example 3.4

Find the class of each address.

a)00000001 00001011 00001011 11101111

b)11000001 10000011 00011011 11111111

c) 14.23.120.8

d)252.5.15.111

Solution

a)The first bit is O. This is a class A address.

b)The first 2 bits are 1; the third bit is O. This is a class C address.

c) The first byte is 14 (between 0 and 127); the class is A.

d)The first byte is 252 (between 240 and 255); the class is E.

3.1 Classes and Blocks

One problem with classful addressing is that each class is divided into a fixed number of blocks with each block having a fixed size as shown in Table 3.1.

Class A addresses were designed for large organizations with a large number of attached hosts or routers. Class B addresses was designed for midsize organizations with tens of thousands of attached hosts or routers. Class C addresses were designed for small organizations with a small number of attached hosts or routers. Class D addresses were designed for multicasting. The class E addresses were reserved for future use. In c1assfnl addressing, a large part of the available addresses were wasted

3.2 Netid and Hostid

In classful addressing, an IP address in class A, B, or C is divided into netid and hostid.

These parts are of varying lengths, depending on the class of the address. The netid is in color, the hostid is in white. Note that the concept does not apply to classes D and E. In class A, one byte defines the netid and three bytes define the hostid. In class B, two bytes define the netid and two bytes define the hostid. In class C, three bytes define the netid and one byte defines the hostid.

3.3 Mask

Although  the  length  of  the  netid  and  hostid  (in  bits)  is  predetermined  in  classful addressing, we can also use a mask (also called the default mask), a 32-bit number made of contiguous Is followed by contiguous as. The masks for classes A, B, and C are shown in Table 3.2. The concept does not apply to classes D and E

The mask can help us to find the netid and the hostid. For example, the mask for a class A address has eight 1s, which means the first 8 bits of any address in class A define the netid; the next 24 bits define the hostid. The last column of Table 3.2 shows the mask in the form /n where n can be 8, 16, or 24 in classful addressing. This notation is also called slash notation orClassless Interdomain Routing (CIDR) notation. The notation is used in classless addressing,

3.4 Subnetting

If an organization was granted a large block in class A or B, it could divide the addresses

into several contiguous groups and assign each group to smaller networks (called subnets). Subnetting increases the number of Is in the mask.

3.5 Supernetting

In supernetting, an organization can combine several class C blocks to create a larger range of addresses. In other words, several networks are combined to create a supernetwork or a supemet. An organization can apply for a set of class C blocks instead of just one.

3.6 Address Depletion

The flaws in classful addressing scheme combined with the fast growth of the Internet led to the near depletion of the available addresses. Yet the number of devices on the Internet is much less than the 232 address space. We have run out of class A and B addresses, and a class C block is too small for most midsize organizations. One solution that has alleviated the problem is the idea of classless addressing.

4. Classless Addressing

To overcome address depletion and give more organizations access to the Internet, classless addressing was designed and implemented. In this scheme, there are no classes, but the addresses are still granted in blocks.

4.1 Address Blocks

In classless addressing, when an entity, small or large, needs to be connected to the Internet, it is granted a block (range) of addresses. The size of the block varies based on the nature and size of the entity. For example, a household may be given only two addresses; a large organization may be given thousands of addresses. An ISP, as the Internet service provider, may be given thousands or hundreds of thousands based on the number of customers it may serve.

Restrictions to simplify the handling of addresses, the Internet authorities impose three restrictions on classless address blocks:

1. The addresses in a block must be contiguous, one after another.

2. The number of addresses in a block must be a power of 2 (I, 2, 4, 8 ...).

3. The first address must be evenly divisible by the number of addresses.

Example 3.5

Figure 3.3 shows a block of addresses, in both binary and dotted-decimal notation, granted to a small business that needs 16 addresses.

We can see that the restrictions are applied to this block. The addresses are contiguous. The number of addresses is a power of 2 (16 = 24), and the first address is divisible by 16. The first address, when converted to a decimal number, is 3,440,387,360, which when divided by 16 results in 215,024,210.

4.2 Mask

A better way to define a block of addresses is to select any address in the block and the mask. As we discussed before, a mask is a 32-bit number in which the n leftmost bits are 1s and the 32 - n rightmost bits are 0s. However, in classless addressing the mask for a block can take any value from 0 to 32. It is very convenient to give just the value of n preceded by a slash.

First Address: The first address in the block can be found by setting the 32 - n rightmost bits inthe binary notation of the address to 0s. The first address in the block can be found by setting the rightmost 32 - n bits to 0s.

Example 3.6

A block of addresses is granted to a small organization. We know that one of the addresses is 205.16.37.39/28. What is the first address in the block?

Solution

The binary representation of the given address is 11001101 00010000 00100101 00100 I 11. If we set 32 - 28 rightmost bits to 0, we get 11001101 000100000100101 0010000 or 205.16.37.32.

Last Address: The last address in the block can be found by setting the 32 - n rightmost bits inthe binary notation of the address to Is. The last address in the block can be found by setting the rightmost 32 - n bits to Is.

Example 3.7

Find the last address for the block in Example 3.6.

Solution

The binary representation of the given address is 11001101 000100000010010100100111. If we set 32 - 28 rightmost bits to 1, we get 11001101 00010000 001001010010 1111 or 205.16.37.47.

Number of Addresses: The number of addresses in the block is the difference between the lastand first address. It can easily be found using the formula 232- n.

Example 3.8

Find the number of addresses in Example 3.6.

Solution

The value of n is 28, which means that number of addresses is 232- 28 or 16.

Example 3.9

Another way to find the first address, the last address, and the number of addresses is to represent the mask as a 32-bit binary (or 8-digit hexadecimal) number. This is particularly useful when we are writing a program to find these pieces of information. In Example 19.5 the /28 can be represented as 11111111 11111111 11111111 11110000 (twenty-eight 1s and four 0s).

Find

a. The first address

b. The last address

c. The number of addresses

Solution

a. The first address can be found by ANDing the given addresses with the mask. ANDing here is done bit by bit. The result of ANDing 2 bits is 1 if both bits are Is; the result is 0 otherwise.

Address: 11001101 00010000 00100101 00100111

Mask: 11111111 11111111 11111111 11110000

First address: 11001101 00010000 00100101 00100000

b. The last address can be found by ORing the given addresses with the complement of the mask. ORing here is done bit by bit. The result of ORing 2 bits is 0 if both bits are 0s; the result is 1 otherwise. The complement of a number is found by changing each 1 to 0 and each 0 to 1.

Address: 11001101 00010000 00100101 00100111

Mask complement: 00000000 00000000 00000000 00001111

Last address: 11001101 00010000 00100101 00101111

c. The number of addresses can be found by complementing the mask, interpreting it as a decimal number, and adding 1 to it.

Mask complement: 000000000 00000000 00000000 00001111

Number of addresses: 15 + 1 =16

4.3 Network Addresses

A very important concept in IP addressing is the network address. When an organization

is given a block of addresses, the organization is free to allocate the addresses to the devices that need to be connected to the Internet. The first address in the class, however, is normally (not always) treated as a special address. The first address is called the network address and defines the organization network. It defines the organization itself to the rest of the world.

The organization network is connected to the Internet via a router. The router has two addresses. One belongs to the granted block; the other belongs to the network that is at the other side of the router. We call the second address x.y.z.t/n because we do not know anything about the network it is connected to at the other side. All messages destined for addresses in the organization block (205.16.37.32 to 205.16.37.47) are sent, directly or indirectly, to x.y.z.t/n. We say directly or indirectly because we do not know the structure of the network to which the other side of the router is connected. The first address in a block is normally not assigned to any device; it is used as the network address that represents the organization to the rest of the world.

4.4 Hierarchy

IP addresses, like other addresses or identifiers we encounter these days, have levels of hierarchy. For example, a telephone network in North America has three levels of hierarchy. The leftmost three digits define the area code, the next three digits define the exchange, the last four digits define the connection of the local loop to the central office. Figure 3.5 shows the structure of a hierarchical telephone number

Two-Level Hierarchy: No Subnetting

An IP address can define only two levels of hierarchy when not subnetted. The n leftmost bits of the address x.y.z.tJn define the network (organization network); the 32 – n rightmost bits define the particular host (computer or router) to the network. The two common terms are prefix and suffix. The part of the address that defines the network is called the prefix; the part that defines the host is called the suffix. Figure 3.6 shows the hierarchical structure of an IPv4 address.

The prefix is common to all addresses in the network; the suffix changes from one device to another. Each address in the block can be considered as a two-level hierarchical structure: the leftmost n bits (prefix) define the network; the rightmost 32 - n bits define the host.

Note that applying the mask of the network, /26 to any of the addresses gives us the network address 17.12.14.0/26. We leave this proof to the reader. We can say that through subnetting, we have three levels of hierarchy. Note that in our example, the subnet prefix length can differ for the subnets as shown in Figure 3.8.

More Levels of Hierarchy

The structure of classless addressing does not restrict the number of hierarchical levels. An organization can divide the granted block of addresses into subblocks. Each subblock can in turn be divided into smaller subblocks. And so on. One example of this is seen in the ISPs. A national ISP can divide a granted large block into smaller blocks and assign each of them to a regional ISP. A regional ISP can divide the block received from the national ISP into smaller blocks and assign each one to a local ISP. A local ISP can divide the block received from the

regional ISP into smaller blocks and assign each one to a different organization. Finally, an organization can divide the received block and make several subnets out of it.

5. Address Allocation

The next issue in classless addressing is address allocation. The ultimate responsibility of address allocation is given to a global authority called the Internet Corporation for Assigned Names and Addresses (ICANN). However, ICANN does not normally allocate addresses to individual organizations. It assigns a large block of addresses to an ISP. Each ISP, in turn, divides its assigned block into smaller subblocks and grants the subblocks to its customers. In other words, an ISP receives one large block to be distributed to its Internet users. This is called address aggregation: many blocks of addresses are aggregated in one block and granted to one ISP.

Example 3.10

An ISP is granted a block of addresses starting with 190.100.0.0/16 (65,536 addresses). The ISP needs to distribute these addresses to three groups of customers as follows:

a)        The first group has 64 customers; each needs 256 addresses.

b)       The second group has 128 customers; each needs 128 addresses.

c)        The third group has 128 customers; each needs 64 addresses.

Design the subblocks and find out how many addresses are still available after these allocations.

Solution

Figure 3.9 shows the situation.

1. Group 1

For this group, each customer needs 256 addresses. This means that 8 (log2256) bits are needed to define each host. The prefix length is then 32 - 8 =24. The addresses are

1st Customer: 190.100.0.0/24 100.0.255/24

2nd Customer: 190.100.1.0/24190   190.100.1.255/24

64th Customer: 190.100.63.0/24      190.100.63.255/24

Total =64 X 256 =16,384

2. Group2

For this group, each customer needs 128 addresses. This means that 7 (10g2 128) bits are needed to define each host. The prefix length is then 32 - 7 =25. The addresses are

3. Group3

For this group, each customer needs 64 addresses. This means that 6 (log2 64) bits are needed to each host. The prefix length is then 32 - 6 =26. The addresses are

1st Customer: 190.100.128.0/26      190.100.128.63/26

2nd Customer: 190.100.128.64/26   190.100.128.127/26

128th Customer: 190.100.159.192/26 190.100.159.255/26

Total =128 X 64 =8192

Number of granted addresses to the ISP: 65,536

Number of allocated addresses by the ISP: 40,960

Number of available addresses: 24,576