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1 IPv4 Addresses Data Communication. TCP/IP Protocol Suite2 INTRODUCTION  The identifier used in the IP layer of the TCP/IP protocol suite to identify.

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Presentation on theme: "1 IPv4 Addresses Data Communication. TCP/IP Protocol Suite2 INTRODUCTION  The identifier used in the IP layer of the TCP/IP protocol suite to identify."— Presentation transcript:

1 1 IPv4 Addresses Data Communication

2 TCP/IP Protocol Suite2 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.

3 3 An IPv4 address is 32 bits long. Note The IPv4 addresses are unique and universal. The address space of IPv4 is 2 32 or 4,294,967,296.

4 TCP/IP Protocol Suite4 Figure Dotted-decimal notation

5 TCP/IP Protocol Suite5 Change the following IPv4 addresses from binary notation to dotted-decimal notation. a b c d Example Solution We replace each group of 8 bits with its equivalent decimal number and add dots for separation: a b c d

6 TCP/IP Protocol Suite6 Change the following IPv4 addresses from dotted-decimal notation to binary notation. a b c d Example Solution We replace each decimal number with its binary equivalent: a b c d

7 TCP/IP Protocol Suite7 Find the error, if any, in the following IPv4 addresses: a b c d Example 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.

8 TCP/IP Protocol Suite8 Finding the class of address

9 TCP/IP Protocol Suite9 Finding the class of an address using continuous checking

10 TCP/IP Protocol Suite10 Find the class of each address: a b c d Example 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.

11 TCP/IP Protocol Suite11 Find the class of each address: a b c d Example 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.

12 TCP/IP Protocol Suite12 Blocks in Class A Millions of class A addresses are wasted.

13 TCP/IP Protocol Suite13 Blocks in Class B Many class B addresses are wasted.

14 TCP/IP Protocol Suite14 Blocks in Class C Not so many organizations are so small to have a class C block.

15 TCP/IP Protocol Suite15 The single block in Class D Class D addresses are made of one block, used for multicasting.

16 TCP/IP Protocol Suite16 The single block in Class E The only block of class E addresses was reserved for future purposes.

17 TCP/IP Protocol Suite17 Sample Internet The network address is the identifier of a network.

18 TCP/IP Protocol Suite18 Network mask

19 TCP/IP Protocol Suite19 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.

20 TCP/IP Protocol Suite20 What is the first address in the block if one of the addresses is /27? Example 4 - Find first address

21 TCP/IP Protocol Suite21 What is the first address in the block if one of the addresses is /27? Example 4 - Find first address Address in binary: Keep the left 27 bits: Result in CIDR notation: /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:

22 TCP/IP Protocol Suite22 What is the first address in the block if one of the addresses is /20? Example 5 – Find first address Using binary

23 TCP/IP Protocol Suite23 What is the first address in the block if one of the addresses is /20? Example 5 – Find first address Using binary See Next Slide 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 /20.

24 TCP/IP Protocol Suite24 Figure 5.3 Example 5

25 TCP/IP Protocol Suite25 Find the first address in the block if one of the addresses is /20. Example 6 – Find first address Another solution method See Next Slide 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 /20.

26 TCP/IP Protocol Suite26

27 TCP/IP Protocol Suite27 Find the number of addresses in the block if one of the addresses is /20. Example 7 - Number of addresses Solution The prefix length is 20. The number of addresses in the block is 2 32−20 or 2 12 or Note that this is a large block with 4096 addresses.

28 TCP/IP Protocol Suite28 Find the last address in the block if one of the addresses is /20. Example 8 - Find last address See Next Slide Solution We found in the previous examples that the first address is /20 and the number of addresses is To find the last address, we need to add 4095 (4096 − 1) to the first address.

29 TCP/IP Protocol Suite29 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 (256 divides into 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: Example 8 (Continued) The last address is /20.

30 TCP/IP Protocol Suite30 Alternate Method Not crazy about base 256 arithmetic? Do it in binary in binary is in binary is Add the two values: This is

31 TCP/IP Protocol Suite31 Find the last address in the block if one of the addresses is /20. Example 9 - find last address Another method See Next Slide 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 or We add the mask complement to the beginning address to find the last address.

32 TCP/IP Protocol Suite Example 9 (Continued) We add the mask complement to the beginning address to find the last address. The last address is /20.

33 TCP/IP Protocol Suite33 Find the block if one of the addresses is /29. Example 10 - find the block See Next Slide 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:

34 TCP/IP Protocol Suite ➡ The leftmost 5 numbers are ➡ The first address is /29 Example 10 (Continued) The number of addresses is 2 32−29 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 If we add this to the first address, we get /29. In other words, the first address is /29, the last address is /20. There are only 8 addresses in this block.

35 TCP/IP Protocol Suite35 Show a network configuration for the block in the previous example. Example 11 See Next Slide 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.

36 TCP/IP Protocol Suite36 In classless addressing, the last address in the block does not necessarily end in 255. Note:


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