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Chap-05 IP Addresses Classfless

Chap-05 IP Addresses Classfless - Chapter5 IP Addresses...

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TCP/IP Protocol Suite 1 Chapter   5 Chapter   5 Objectives Upon completion you will be able to: IP Addresses: IP Addresses: Classless Addressing Classless Addressing Understand the concept of classless addressing Be able to find the first and last address given an IP address Be able to find the network address given a classless IP address Be able to create subnets from a block of classless IP addresses Understand address allocation and address aggregation
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TCP/IP Protocol Suite 2 5.1 VARIABLE-LENGTH BLOCKS In classless addressing variable-length blocks are assigned that belong In classless addressing variable-length blocks are assigned that belong to no class. In this architecture, the entire address space (232 addresses) to no class. In this architecture, the entire address space (232 addresses) is divided into blocks of different sizes. is divided into blocks of different sizes. The topics discussed in this section include: The topics discussed in this section include: Restrictions Restrictions Finding the Block Finding the Block Granted Block Granted Block
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TCP/IP Protocol Suite 3 Figure 5.1 Variable-length blocks
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TCP/IP Protocol Suite 4 Which of the following can be the beginning address of a block that contains 16 addresses? a. 205.16.37.32 b .190.16.42.44 c . 17.17.33.80 d .123.45.24.52 Example 1 Solution Only two are eligible (a and c). The address 205.16.37.32 is eligible because 32 is divisible by 16. The address 17.17.33.80 is eligible because 80 is divisible by 16.
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TCP/IP Protocol Suite 5 Which of the following can be the beginning address of a block that contains 256 addresses? a. 205.16.37.32 b. 190.16.42.0 c. 17.17.32.0 d. 123.45.24.52 Example 2 Solution In this case, the right-most byte must be 0. As we mentioned in Chapter 4, the IP addresses use base 256 arithmetic. When the right-most byte is 0, the total address is divisible by 256. Only two addresses are eligible (b and c).
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TCP/IP Protocol Suite 6 Which of the following can be the beginning address of a block that contains 1024 addresses? a . 205.16.37.32 b .190.16.42.0 c . 17.17.32.0 d .123.45.24.52 Example 3 Solution In this case, we need to check two bytes because 1024 = 4 × 256. The right-most byte must be divisible by 256. The second byte (from the right) must be divisible by 4. Only one address is eligible (c).
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TCP/IP Protocol Suite 7 Figure 5.2 Format of classless addressing address
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TCP/IP Protocol Suite 8 Table 5.1 Table 5.1 Prefix lengths Prefix lengths
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TCP/IP Protocol Suite 9 Classful addressing is a special case of classless addressing. Note: Note:
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TCP/IP Protocol Suite 10 What is the first address in the block if one of the addresses is 167.199.170.82/27 ? Example 4 Address in binary: 10100111 11000111 10101010 01010010 Keep the left 27 bits: 10100111 11000111 10101010 010 00000 Result in CIDR notation: 167.199.170.64/27 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:
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TCP/IP Protocol Suite 11 What is the first address in the block if one of the addresses is 140.120.84.24/20 ?
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