What is CLASSLESS
ADDRESSING in networking?
Number of Addresses in a Block
There is only one requirement on the number
of blocks of addresses; it must be a power. of 2 (2, 4, 8, . . .). A family can be given a two-address block. A small company could receive 16 addresses. A large
family may be given 1024 addresses.
Beginning Address
The starting address must be evenly divisible
by the number of addresses. For instance, if a block contains 4 addresses, the beginning
address must be divisible by 4. If the block has
less than 256 addresses, we need to check only
the rightmost byte. If it has less than 65,536
addresses, we need to check only the two
rightmost bytes, and so on.
Example 9
Which of the following can be the starting address of a
block that contains 16 addresses?
205.16.37.32
190.16.42.44
17.17.33.80
123.45.24.52
Solution
The stated address 205.16.37.32 is eligible as 32 is
divisible by 16. The address 17.17.33.80 is eligible
because 80 is divisible by 16.
Example 10
Which one following can be the starting address of a
block that contains 1024 addresses?
205.16.37.32
190.16.42.0
17.17.32.0
123.45.24.52
Solution
The number who is divisible by 1024, the rightmost byte of an
address should be 0 and the second rightmost byte
must be divisible by 4. Only the address 17.17.32.0
meets this condition.
Slash notation is also called
CIDR
notation.
Example 11
A small group is given a block with the starting
address and the prefix length 205.16.37.24/29 (in slash
notation). What is the range of the block?
Solution
The beginning address is 205.16.37.24. If we are to locate the
last address we keep the first 29 bits same and change the
last 3 bits to 1s.
Beginning:11001111 00010000 00100101 00011000
Ending : 11001111 00010000 00100101 00011111
There are only 8 addresses in this block.
Example 12
We can find the range of many addresses in Example 11 by
a different method. We can argue that the length of the suffix
is 32 - 29 or 3. So there are 2
3 to 8 addresses in this block.
If the first address is 205.16.37.24, the last address is
205.16.37.31 (24 + 7 = 31).
Note
A block in classes A, B, and C
can easily be represented in slash
notation as
A.B.C.D/ n
where n is
8 (class A), 16 (class B), or
24 (class C) eighter .
Example 13
What is the network address if one of the addresses is
167.199.170.82/27?
Solution
The prefix length is 27, which means that we must
keep the first 27 bits as is and change the remaining
bits (5) to 0s. The 5 bits can target only the last byte.
The last byte is 01010010. Changing the last 5 bits
to 0s, we get 01000000 or 64. The network address
is 167.199.170.64/27.
Example 14
A group is granted the
block 130.34.12.64/26.
The organization needs to have four subnets. What are the
subnet addresses and the range of addresses for each
subnet?
Solution
The suffix length is 6. This indicates the total of addresses in the block is 64 (26
). If we create
four segmented piece of a larger IP network (subnets), each segmented piece of a larger IP network. will have 16 addresses.
Let us first find the subnet prefix (subnet mask).
We need four subnets, which means we need to add
two more 1s to the site prefix. The subnet prefix is
then /28.
Subnet 1: 130.34.12.64/28 to 130.34.12.79/28.
Subnet 2 : 130.34.12.80/28 to 130.34.12.95/28.
Subnet 3: 130.34.12.96/28 to 130.34.12.111/28.
Subnet 4: 130.34.12.112/28 to 130.34.12.127/28.
See Figure 5.15
Example 15
An ISP is granted a block of addresses starting with
190.100.0.0/16. The ISP needs to distribute these
addresses to three groups of customers as follows:
1. The first group has 64 customers; each requires 256 addresses.
2. The second group has 128 customers; each needs 128 addresses.
3. The third group also has 128 customers; each requires 64 addresses.
Rearrange the subblocks and give the slash notation for each
subblock. If we find out that how many addresses are still available
after these allocations.
Solution
Group 1
For this organization, each customer required 256 addresses.
So we knew that the suffix length is 8 (28 = 256). The
prefix length is then 32 - 8 = 24.
01: 190.100.0.0/24 190.100.0.255/24
02: 190.100.1.0/24 190.100.1.255/24
…………………………………..
64: 190.100.63.0/24190.100.63.255/24
Total = 64 256 = 16,384
Group 2
For this organization, each customer required 128 addresses.
This show that the suffix length is 7 (27 = 128). Thus, the prefix length is 25 times 32.
The addresses
are:
001: 190.100.64.0/25 190.100.64.127/25
002: 190.100.64.128/25 190.100.64.255/25
003: 190.100.127.128/25 190.100.127.255/25
Total = 128 128 = 16,384
Group 3
For this organization, each customer required 64 addresses.
We knew that the suffix length is 6 (26 = 64). The
prefix length is then 32 - 6 = 26.
001:190.100.128.0/26 190.100.128.63/26
002:190.100.128.64/26 190.100.128.127/26
…………………………
128:190.100.159.192/26 190.100.159.255/26
Total = 128 64 = 8,192
Number of granted addresses: 65,536
Number of allocated addresses: 40,960
Number of available addresses: 24,576
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