Transcript Digital Communication Network TCS 3164

IPv4 Addresses
Data Communication
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INTRODUCTION
 The identifier used in the IP layer of the TCP/IP
protocol suite to identify each device connected
to the Internet is called the Internet address or
IP address.
 An IPv4 address is a 32-bit address that
uniquely and universally defines the connection
of a host or a router to the Internet;
 an IP address is the address of the interface.
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Note
An IPv4 address is 32 bits long.
The IPv4 addresses are unique
and universal.
The address space of IPv4 is
232 or 4,294,967,296.
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Figure Dotted-decimal notation
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Example
Change the following IPv4 addresses from binary notation to
dotted-decimal notation.
a. 10000001 00001011 00001011 11101111
b. 11000001 10000011 00011011 11111111
c. 11100111 11011011 10001011 01101111
d. 11111001 10011011 11111011 00001111
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
c. 231.219.139.111
d. 249.155.251.15
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Example
Change the following IPv4 addresses from dotted-decimal
notation to binary notation.
a. 111.56.45.78
b. 221.34.7.82
c. 241.8.56.12
d. 75.45.34.78
Solution
We replace each decimal number with its binary equivalent:
a. 01101111 00111000 00101101 01001110
b. 11011101 00100010 00000111 01010010
c. 11110001 00001000 00111000 00001100
d. 01001011 00101101 00100010 01001110
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Example
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 should be no leading zeroes (045).
b. We may not have more than 4 bytes in an IPv4 address.
c. Each byte should be less than or equal to 255.
d.A mixture of binary notation and dotted-decimal notation.
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Finding the class of address
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Finding the class of an address
using continuous checking
0
Class: A
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1
1
Start
0
0
Class: B
Class: C
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1
0
Class: D
Class: E
Example
Find the class of each address:
a. 00000001 00001011 00001011 11101111
b. 11000001 10000011 00011011 11111111
c. 10100111 11011011 10001011 01101111
d. 11110011 10011011 11111011 00001111
Solution
See the procedure in Figure 5.7.
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 bit is 1; the second bit is 0. This is a class B address.
d. The first 4 bits are 1s. This is a class E address.
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Example
Find the class of each address:
a. 227.12.14.87
b. 193.14.56.22
c. 14.23.120.8
d. 252.5.15.111
Solution
a.
The first byte is 227 (between 224 and 239); the class is D.
b.
The first byte is 193 (between 192 and 223); the class is C.
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.
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Blocks in Class A
Millions of class A addresses
are wasted.
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Blocks in Class B
Many class B addresses are wasted.
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Blocks in Class C
Not so many organizations are so small to
have a class C block.
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The single block in Class D
Class D addresses are made of one block,
used for multicasting.
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The single block in Class E
The only block of class E addresses was
reserved for future purposes.
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Sample Internet
The network address is the identifier of a
network.
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Network mask
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Restrictions

The number of addresses in a block must be a
power of 2

The first address must be evenly divisible by the
number of addresses.

For example, if a block contains 4 addresses, the first
address must be divisible by 4.
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Example 4 - Find first address
What is the first address in the block if one of the
addresses is 167.199.170.82/27?
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Example 4 - Find first address
What is the first address in the block 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 following shows the process:
Address in binary:
10100111 11000111 10101010
Keep the left 27 bits: 10100111 11000111 10101010
Result in CIDR notation: 167.199.170.64/27
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01010010
01000000
Example 5 – Find first address
Using binary
What is the first address in the block if one of the
addresses is 140.120.84.24/20?
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Example 5 – Find first address
Using binary
What is the first address in the block if one of the
addresses is 140.120.84.24/20?
Solution
Figure 5.3 shows the solution. The first, second, and
fourth bytes are easy; for the third byte we keep the bits
corresponding to the number of 1s in that group. The
first address is 140.120.80.0/20.
See Next Slide
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Figure 5.3
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Example 5
Example 6 – Find first address Another
solution method
Find the first address in the block if one of the
addresses is 140.120.84.24/20.
Solution
The first, second, and fourth bytes are as defined in the
previous example. To find the third byte, we write 84 as
the sum of powers of 2 and select only the leftmost 4
(m is 4) as shown in Figure 5.4. The first address is
140.120.80.0/20.
See Next Slide
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Example 7 - Number of addresses
Find the number of addresses in the block if one of
the addresses is 140.120.84.24/20.
Solution
The prefix length is 20. The number of addresses in the
block is 232−20 or 212 or 4096. Note that
this is a large block with 4096 addresses.
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Example 8 - Find last address
Find the last address in the block if one of the
addresses is 140.120.84.24/20.
Solution
We found in the previous examples that the first
address is 140.120.80.0/20 and the number of addresses
is 4096. To find the last address, we need to add 4095
(4096 − 1) to the first address.
See Next Slide
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Example 8 (Continued)
To keep the format in dotted-decimal notation, we need
to represent 4095 in base 256 (see Appendix B) and do
the calculation in base 256. We write 4095 as 15.255.
(256 divides into 4095 15 times with a remainder of 255.) We then add
the first address to this number (in base 255) to obtain
the last address as shown below:
140 . 120 . 80 . 0
15 . 255
------------------------140 . 120 . 95 . 255
The last address is 140.120.95.255/20.
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Alternate Method

Not crazy about base 256 arithmetic? Do it in
binary.
 .80.0 in binary is 01010000.00000000
 4095 in binary is 1111 11111111
 Add the two values:


01011111.11111111
This is .95.255
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Example 9 - find last address
Another method
Find the last address in the block if one of the
addresses is 140.120.84.24/20.
Solution
The mask has twenty 1s and twelve 0s. The
complement of the mask has twenty 0s and twelve 1s.
In other words, the mask complement is
00000000 00000000 00001111 11111111
or 0.0.15.255. We add the mask complement to the
beginning address to find the last address.
See Next Slide
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Example 9
(Continued)
We add the mask complement to the beginning address
to find the last address.
140 . 120 . 80 . 0
0 . 0 . 15 . 255
---------------------------140 . 120 . 95 . 255
The last address is 140.120.95.255/20.
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Example 10 - find the block
Find the block if one of the addresses is
190.87.140.202/29.
Solution
We follow the procedure in the previous examples to
find the first address, the number of addresses, and the
last address. To find the first address, we notice that the
mask (/29) has five 1s in the last byte. So we write the
last byte as powers of 2 and retain only the leftmost
five as shown below:
See Next Slide
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Example 10 (Continued)
202
➡ 128 + 64 + 0 + 0 + 8 + 0 + 2 + 0
The leftmost 5 numbers are ➡ 128 + 64 + 0 + 0 + 8
The first address is 190.87.140.200/29
The number of addresses is 232−29 or 8. To find the last address,
we use the complement of the mask. The mask has twenty-nine
1s; the complement has three 1s. The complement is 0.0.0.7. If
we add this to the first address, we get 190.87.140.207/29. In
other words, the first address is 190.87.140.200/29, the last
address is 190.87.140.207/20. There are only 8 addresses in this
block.
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Example 11
Show a network configuration for the block in the previous
example.
Solution
The organization that is granted the block in the previous
example can assign the addresses in the block to the hosts in its
network. However, the first address needs to be used as the
network address and the last address is kept as a special address
(limited broadcast address). Figure 5.5 shows how the block can
be used by an organization. Note that the last address ends with
207, which is different from the 255 seen in classful addressing.
See Next Slide
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Note:
In classless addressing, the last address
in the block does not necessarily end in
255.
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