Subnet Masking in Computer Network--CST 2nd year by Tanushree Bhadra

Network Layer:
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.

An IPv4 address is 32 bits long.
Note

The address space of IPv4 is
232 or 4,294,967,296.
Note

19.5
Figure 19.1 Dotted-decimal notation and binary notation for an IPv4 address

Change the following IPv4 addresses from binary notation
to dotted-decimal notation.
Example 19.1
Solution
We replace each group of 8 bits with its equivalent decimal
number (see Appendix B) and add dots for separation.

Change the following IPv4 addresses from dotted-decimal
notation to binary notation.
Example 19.2
Solution
We replace each decimal number with its binary equivalent
(see Appendix B).

Find the error, if any, in the following IPv4 addresses.
Example 19.3
Solution
a. There must be no leading zero (045).
b. There can be no more than four numbers.
c. Each number needs to be less than or equal to 255.
d. A mixture of binary notation and dotted-decimal
notation is not allowed.

In classful addressing, the address space is divided into five
classes:
A, B, C, D, and E.
Note

Figure 19.2 Finding the classes in binary and dotted-decimal notation

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
Example 19.4
Solution
a. The first bit is 0. This is a class A address.
b. The first 2 bits are 1; the third bit is 0. This is a class C
address.
c. The first byte is 14; the class is A.
d. The first byte is 252; the class is E.

Table 19.1 Number of blocks and block size in classful IPv4 addressing

In classful addressing, a large part of the available addresses were wasted.
Note

Table 19.2 Default masks for classful addressing

Classful addressing, which is almost obsolete, is replaced with classless addressing.
Note

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.
Note

The first address in the block can be found by setting the rightmost
32 − n bits to 0s.
Note

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 00100111
If we set 32−28 rightmost bits to 0, we get
11001101 00010000 00100101 0010000
or
205.16.37.32
Example 19.6

The last address in the block can be found by setting the rightmost
32 − n bits to 1s.
Note

Find the last address for the block in Example 19.6.
Solution
The binary representation of the given address is
11001101 00010000 00100101 00100111
If we set 32 − 28 rightmost bits to 1, we get
11001101 00010000 00100101 00101111
or
205.16.37.47
Example 19.7

The number of addresses in the block can be found by using the formula
232−n.
Note

Find the number of addresses in Example 19.6.
Example 19.8
Solution
The value of n is 28, which means that number
of addresses is 2 32−28 or 16.

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.
Example 19.9

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 1s;
the result is 0 otherwise.
Example 19.9 (continued)

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.

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

Figure 19.4 A network configuration for the block 205.16.37.32/28

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.
Note

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.
Example 19.10

Solution
Figure 19.9 shows the situation.
Group 1
For this group, each customer needs 256 addresses. This
means that 8 (log2 256) bits are needed to define each host.
The prefix length is then 32 − 8 = 24. The addresses are

Group 2
means that 7 (log2 128) bits are needed to define each host.
The prefix length is then 32 − 7 = 25. The addresses are

Group 3
means that 6 (log264) bits are needed to each host. The
prefix length is then 32 − 6 = 26. The addresses are
Number of granted addresses to the ISP: 65,536
Number of allocated addresses by the ISP: 40,960
Number of available addresses: 24,576

19.33
Figure 19.9 An example of address allocation and distribution by an ISP

An IPv6 address is 128 bits long.
Note

Reference
Data Communications and Networking By Behrouz
A.Forouzan

Subnet Masking in Computer Network--CST 2nd year by Tanushree Bhadra

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