1 CHAPTER 12 : IPV4 1 Contents • Introduction • Function of Network Layers • IPv4 Notation – Binary Notation – Dotted Decimal Notation IP 4 Add i • IPv4 Addressing – Classful Addressing – Classless Addressing 2
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CHAPTER 12 : IPV41
Contents
• Introduction
• Function of Network Layers
• IPv4 Notation– Binary Notation
– Dotted Decimal Notation
IP 4 Add i• IPv4 Addressing– Classful Addressing
– Classless Addressing2
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Introduction
• Communication at network layeryEg. host to host, computercommunicate with Internet
• This level of communicationrequired a global address scheme
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Functions of network layer
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Network layer duties
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NoteNote::
An IP address is a 32-bit address.
The IP addresses are unique and universal defines the connection to the Internet
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NoteNote::
The address space of IPv4 is 232
or 4,294,967,296
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IPv4 Notations
• Binary Notation
di l d i 32 bi- displayed in 32 bits
- each octet is often referred to as a byte
- Eg.
01110101 10010101 00011101 00000010
• Dotted-Decimal NotationDotted Decimal Notation- usually written in decimal form with a decimalpoint (dot) separating the byte
- Eg. 117.149.29.2 8
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Figure. Dotted-decimal notation
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Example 1Example 1Change the following IP addresses from binary notation to dotted-decimal notation.
a. 10000001 00001011 00001011 11101111
b. 11111001 10011011 11111011 00001111
SolutionSolution
We replace each group of 8 bits with its equivalent decimal number and dd d t f tiadd dots for separation:
a. 129.11.11.239b. 249.155.251.15
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Example 2Example 2Change the following IP addresses from dotted-decimal notation to binary notation.
a. 111.56.45.78
b. 75.45.34.78
SolutionSolution
We replace each decimal number with its binary equivalent
a. 01101111 00111000 00101101 01001110b. 01001011 00101101 00100010 01001110
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IPv4 Addressing
– Classful Addressing– Classless Addressing
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NoteNote::
In classful addressing, the address space is divided into five
classes: A, B, C, D, and E.
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Finding the classes in binary and dotted-decimal notation
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Find the class of each address.a. 00000001 00001011 00001011 11101111b 11000001 10000011 00011011 11111111
Example Example 33
b. 11000001 10000011 00011011 11111111c. 14.23.120.8d. 252.5.15.111
a. The first bit is 0. This is a class A address.
SolutionSolution
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.
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Table. Number of blocks and block size in classful IPv4 addressing
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Note
In classful addressing, a large part of the available addresses were wasted.
Note
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- IP address is divided into netid and hostid
Netid and Hostid- IP address is divided into netid and hostid- varying lengths, depend on the class of theaddress.Mask (Default Mask)
- A 32-bit address contiguous 1s and contiguous 0s- it can be also written as /nit can be also written as /n- n leftmost bits are 1s and (32-n) rightmost bits are0s- Eg. 234.56.78.69/8
11111111 00000000 00000000 0000000018
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ClasslessInterdomain Routing
Table. Default masks for classful addressingg
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Note
Classful addressing, which is almost obsolete, is replaced with classless
addressing.
Note
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• To overcome the address depletion
Classless Addressing
• To overcome the address depletion• No classes• Addresses are granted in blocks (range)• Size of block varies based on the nature &
size of entity
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To simplify the handling addresses, the Internet
Restriction
authorities impose three restrictions:1. The addresses in a block must be contiguous2. The number of addresses in a block must be
a power of 2 (1, 2, 4, 8, 16 …)3. The first address must be evenly divisible by
the number of addressesthe number of addresses
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The next figure shows a block of addresses,in both binary and dotted-decimal notation,granted to a small business that needs 16
Example
gaddresses.
We can see that the restrictions are appliedto this block. The addresses are contiguous.The number of addresses is a power of 2 (16
24) d th fi t dd i di i ibl b 16= 24), and the first address is divisible by 16.The first address, when converted to adecimal number, is 3,440,387,360, whichwhen divided by 16 results in 215,024,210. 23
Figure. A block of 16 addresses granted to a small organization
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Note
In IPv4 addressing, a block of addresses can be defined as
x.y.z.t /nin which x y z t defines one of the
Note
in which x.y.z.t defines one of the addresses and the /n defines the mask.
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Note
The first address in the block can be found by setting the rightmost
32 − n bits to 0s.
Note
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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?
Example 4Example 4
The binary representation of the given address is11001101 00010000 00100101 00100111
If we set 32−28 rightmost bits to 0, we get 11001101 00010000 00100101 0010000
SolutionSolution
or 205.16.37.32.
This is actually the block shown in Figure slide 24
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Note
The last address in the block can be found by setting the rightmost
32 − n bits to 1s.
Note
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Find the last address for the block in Example 4
Example Example 55
SolutionSolutionThe binary representation of the given address is
11001101 00010000 00100101 00100111If we set 32 − 28 rightmost bits to 1, we get
11001101 00010000 00100101 00101111 or
205.16.37.47Thi i t ll th bl k h i Fi lid 24This is actually the block shown in Figure slide 24
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Note
The number of addresses in the block can be found by using the formula
232−n.
Note
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Find the number of addresses in Example 4
Example Example 66
Sol tionSol tion
The value of n is 28, which means thatNumber of addresses is 2 32−28 or 16.
SolutionSolution
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Another way to find the first address, the last address,and the number of addresses is to represent the maskas a 32-bit binary (or 8-digit hexadecimal) number.This is particularly useful when we are writing a
ExampleExample
This is particularly useful when we are writing aprogram to find these pieces of information. InExample 5 the /28 can be represented as
11111111 11111111 11111111 11110000 (twenty-eight 1s and four 0s).
Finda. The first addressb. The last addressc. The number of addresses.
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a Th fi t dd b f d b ANDi th i
Example Example ... continued... continued
SolutionSolution
a. The first address can be found by ANDing the givenaddresses with the mask. ANDing here is done bit bybit. The result of ANDing 2 bits is 1 if both bits are 1s; the result is 0 otherwise.
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b. The last address can be found by ORing the
SolutionSolution
Example Example ... continued... continued
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.
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SolutionSolution
Example Example ... continued... continued
c. The number of addresses can be found bycomplementing the mask, interpreting it as a decimal number, and adding 1 to it.
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Note
The first address in a block is normally not assigned to any device; it is used as the network address that
represents the organization
Note
represents the organization to the rest of the world.
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1) In a block of addresses, we know the IP address of onehost is 25.34.12.56/16. What are the first address(network address) and the last address (limited
ExerciseExercise
broadcast address) in this block?
1) Write the following masks in slash notation (/n)a. 255.255.255.0b. 255.0.0.0c. 255.255.224.0d 255 255 240 0
Idawaty A
hmad : S
em: 2008/2009
d. 255.255.240.0
3) Find the range of addresses in the following blocksa. 123.56.77.32/29 b. 17.34.16.0/23
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mester 2
NoteNote::
IP addresses are designed with two (no subnetting) or three
(subnetting) levels of hierarchy.
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Note
Each address in the block can be considered as a two-level
hierarchical structure: the leftmost n bits (prefix) define
the network;the network;the rightmost 32 − n bits define
the host.39
Figure .A frame in a character-oriented protocol
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Figure. A network with two levels of hierarchy
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Three –Level of Hierarchy:Subnetting
Eg. Suppose an organization is given the block17.12.40.0/26, which contains 64 addresses.The organization has three offices and needsto divide the addresses into three subblocks of32, 16 and 16 addresses.,
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Figure. Configuration and addresses in a subnetted network
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Figure. Three-level hierarchy in an IPv4 address
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Table Addresses for private networksTable. Addresses for private networks
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Figure. A NAT implementation
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