Ip Address Ppt

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By: mahio (36 month(s) ago) Private Comment

simple and elegant ... I liked it.

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Presentation Transcript

Slide 1:

Chapter 4 IP Addresses: Classful Addressing

Slide 2:

CONTENTS INTRODUCTION CLASSFUL ADDRESSING OTHER ISSUES A SAMPLE INTERNET

Slide 3:

INTRODUCTION 4.1

Slide 4:

An IP address is a 32-bit address.

Slide 5:

The IP addresses are unique.

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Address Space addr15 addr1 addr2 addr41 addr31 addr226 ………….. ………….. ………….. ………….. ………….. ………….. …………..

Slide 7:

RULE: addr15 addr1 addr2 addr41 addr31 addr226 ………….. ………….. ………….. ………….. ………….. ………….. ………….. If a protocol uses N bits to define an address, the address space is 2N because each bit can have two different values (0 and 1) and N bits can have 2N values.

Slide 8:

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

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01110101 10010101 00011101 11101010 Binary Notation

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Figure 4-1 Dotted-decimal notation

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0111 0101 1001 0101 0001 1101 1110 1010 Hexadecimal Notation 75 95 1D EA 0x75951DEA

Slide 12:

The binary, decimal, and hexadecimal number systems are reviewed in Appendix B.

Slide 13:

Example 1 Change the following IP address from binary notation to dotted-decimal notation. 10000001 00001011 00001011 11101111 Solution 129.11.11.239

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Example 2 Change the following IP address from dotted-decimal notation to binary notation. 111.56.45.78 Solution 01101111 00111000 00101101 01001110

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Example 3 Find the error, if any, in the following IP address: 111.56.045.78 Solution There are no leading zeroes in dotted-decimal notation (045).

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Example 3 (continued) Find the error, if any, in the following IP address: 75.45.301.14 Solution In dotted-decimal notation, each number is less than or equal to 255; 301 is outside this range.

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Example 4 Change the following IP addresses from binary notation to hexadecimal notation. 10000001 00001011 00001011 11101111 Solution 0X810B0BEF or 810B0BEF16

Slide 18:

CLASSFUL ADDRESSING 4.2

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Figure 4-2 Occupation of the address space

Slide 20:

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

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Figure 4-3 Finding the class in binary notation

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Figure 4-4 Finding the address class

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Example 5 How can we prove that we have 2,147,483,648 addresses in class A? Solution In class A, only 1 bit defines the class. The remaining 31 bits are available for the address. With 31 bits, we can have 231 or 2,147,483,648 addresses.

Slide 24:

Example 6 Find the class of the address: 00000001 00001011 00001011 11101111 Solution The first bit is 0. This is a class A address.

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Example 6 (Continued) Find the class of the address: 11000001 10000011 00011011 11111111 Solution The first 2 bits are 1; the third bit is 0. This is a class C address.

Slide 26:

Figure 4-5 Finding the class in decimal notation

Slide 27:

Example 7 Find the class of the address: 227.12.14.87 Solution The first byte is 227 (between 224 and 239); the class is D.

Slide 28:

Example 7 (Continued) Find the class of the address: 193.14.56.22 Solution The first byte is 193 (between 192 and 223); the class is C.

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Example 8 In Example 4 we showed that class A has 231 (2,147,483,648) addresses. How can we prove this same fact using dotted-decimal notation? Solution The addresses in class A range from 0.0.0.0 to 127.255.255.255. We notice that we are dealing with base 256 numbers here.

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Solution (Continued) Each byte in the notation has a weight. The weights are as follows: 2563 , 2562, 2561, 2560 Last address: 127 ? 2563 + 255 ? 2562 + 255 ? 2561 + 255 ? 2560 = 2,147,483,647 First address: = 0 If we subtract the first from the last and add 1, we get 2,147,483,648.

Slide 31:

Figure 4-6 Netid and hostid

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Figure 4-7 Blocks in class A

Slide 33:

Millions of class A addresses are wasted.

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Figure 4-8 Blocks in class B

Slide 35:

Many class B addresses are wasted.

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Figure 4-9 Blocks in class C

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The number of addresses in a class C block is smaller than the needs of most organizations.

Slide 38:

Class D addresses are used for multicasting; there is only one block in this class.

Slide 39:

Class E addresses are reservedfor special purposes; most of the block is wasted.

Slide 40:

Network Addresses The network address is the first address. The network address defines the network to the rest of the Internet. Given the network address, we can find the class of the address, the block, and the range of the addresses in the block

Slide 41:

In classful addressing, the network address (the first address in the block) is the one that is assigned to the organization.

Slide 42:

Example 9 Given the network address 17.0.0.0, find the class, the block, and the range of the addresses. Solution The class is A because the first byte is between 0 and 127. The block has a netid of 17. The addresses range from 17.0.0.0 to 17.255.255.255.

Slide 43:

Example 10 Given the network address 132.21.0.0, find the class, the block, and the range of the addresses. Solution The class is B because the first byte is between 128 and 191. The block has a netid of 132.21. The addresses range from 132.21.0.0 to 132.21.255.255.

Slide 44:

Example 11 Given the network address 220.34.76.0, find the class, the block, and the range of the addresses. Solution The class is C because the first byte is between 192 and 223. The block has a netid of 220.34.76. The addresses range from 220.34.76.0 to 220.34.76.255.

Slide 45:

Mask A mask is a 32-bit binary number that gives the first address in the block (the network address) when bitwise ANDed with an address in the block.

Slide 46:

Figure 4-10 Masking concept

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Figure 4-11 AND operation

Slide 48:

The network address is the beginning address of each block. It can be found by applying the default mask toany of the addresses in the block (including itself). It retains the netid of the block and sets the hostid to zero.

Slide 49:

Example 12 Given the address 23.56.7.91 and the default class A mask, find the beginning address (network address). Solution The default mask is 255.0.0.0, which means that only the first byte is preserved and the other 3 bytes are set to 0s. The network address is 23.0.0.0.

Slide 50:

Example 13 Given the address 132.6.17.85 and the default class B mask, find the beginning address (network address). Solution The default mask is 255.255.0.0, which means that the first 2 bytes are preserved and the other 2 bytes are set to 0s. The network address is 132.6.0.0.

Slide 51:

Example 14 Given the address 201.180.56.5 and the class C default mask, find the beginning address (network address). Solution The default mask is 255.255.255.0, which means that the first 3 bytes are preserved and the last byte is set to 0. The network address is 201.180.56.0.

Slide 52:

We must not apply the default mask of one class to an address belonging to another class.

Slide 53:

OTHER ISSUES 4.13

Slide 54:

Figure 4-12 Multihomed devices

Slide 55:

Figure 4-13 Network addresses

Slide 56:

Figure 4-14 Example of direct broadcast address

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Figure 4-15 Example of limited broadcast address

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Figure 4-16 Example of this host on this address

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Figure 4-17 Example of specific host on this network

Slide 60:

Figure 4-18 Example of loopback address

Slide 61:

Private Addresses A number of blocks in each class are assigned for private use. They are not recognized globally. These blocks are depicted in Table 4.4

Slide 62:

Unicast, Multicast, and Broadcast Addresses Unicast communication is one-to-one. Multicast communication is one-to-many. Broadcast communication is one-to-all.