# Chapter 5. IPv4 Addresses. TCP/IP Protocol Suite 1 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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1 Chapter 5 IPv4 Addresses TCP/IP Protocol Suite 1 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2 Chapter Outline TCP/IP Protocol Suite 2

3 5-1 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. TCP/IP Protocol Suite 3

5 Note An IPv4 address is 32 bits long. Note The IPv4 addresses are unique and universal. TCP/IP Protocol Suite 5

6 Note The address space of IPv4 is 2 32 or 4,294,967,296. Note Numbers in base 2, 16, and 256 are discussed in Appendix B. TCP/IP Protocol Suite 6

7 Figure 5.1 Dotted-decimal notation Hexadecimal: 75951DEA TCP/IP Protocol Suite 7

8 Example 5.1 Change the following IPv4 addresses from binary notation to dotted-decimal notation. a b c d Solution We replace each group of 8 bits with its equivalent decimal number (see Appendix B) and add dots for separation: a b c d TCP/IP Protocol Suite 8

9 Example 5.2 Change the following IPv4 addresses from dotted-decimal notation to binary notation. a b c d Solution We replace each decimal number with its binary equivalent: a b c d TCP/IP Protocol Suite 9

10 Example 5.3 Find the error, if any, in the following IPv4 addresses: a b c d 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. TCP/IP Protocol Suite 10

11 Example 5.4 Change the following IPv4 addresses from binary notation to hexadecimal notation. a b Solution We replace each group of 4 bits with its hexadecimal equivalent. Note that 0X (or 0x) is added at the beginning or the subscript 16 at the end. a. 0X810B0BEF or 810B0BEF 16 b. 0XC1831BFF or C1831BFF 16 TCP/IP Protocol Suite 11

12 Example 5.5 Find the number of addresses in a range if the first address is and the last address is Solution We can subtract the first address from the last address in base 256 (see Appendix B). The result is in this base. To find the number of addresses in the range (in decimal), we convert this number to base 10 and add 1 to the result.. TCP/IP Protocol Suite 12

13 Example 5.6 The first address in a range of addresses is If the number of addresses in the range is 32, what is the last address? Solution We convert the number of addresses minus 1 to base 256, which is We then add it to the first address to get the last address. Addition is in base 256. TCP/IP Protocol Suite 13

14 Figure 5.2 Bitwise NOT operation TCP/IP Protocol Suite 14

15 Example 5.7 TCP/IP Protocol Suite 15

16 Figure 5.3 Bitwise AND operation TCP/IP Protocol Suite 16

17 Example 5.8 TCP/IP Protocol Suite 17

18 Figure 5.4 Bitwise OR operation TCP/IP Protocol Suite 18

19 Example 5.9 TCP/IP Protocol Suite 19

20 5-2 CLASSFUL ADDRESSING IP addresses, when started a few decades ago, used the concept of classes. This architecture is called classful addressing. In the mid-1990s, a new architecture, called classless addressing, was introduced that supersedes the original architecture. In this section, we introduce classful addressing because it paves the way for understanding classless addressing and justifies the rationale for moving to the new architecture. Classless addressing is discussed in the next section. TCP/IP Protocol Suite 20

21 Figure 5.5 Occupation of address space When IP addressing was first started, it used a concept called classful addressing. A newer concept called classless addressing is slowly replacing it though. Regarding classful addressing, the address space is divided into five classes: A, B, C, D and E. Class # of addresses Percent of the Space TCP/IP Protocol Suite 21

22 Figure 5.6 Finding the class of address Start Class: A Class: B Class: C Class: D Class: E TCP/IP Protocol Suite 22

23 Example 5.10 Find the class of each address: a b c d 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. TCP/IP Protocol Suite 23

24 Example 5.11 Find the class of each address: a b c d 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. TCP/IP Protocol Suite 24

25 Figure 5.8 Netid and hostid TCP/IP Protocol Suite 25

26 Figure 5.9 Blocks in Class A Class A has 128 blocks or network ids First byte is the same (netid), the remaining 3 bytes can change (hostids) Network id 0 (first), Net id 127 (last) and Net id 10 are reserved leaving 125 ids to be assigned to organizations/companies Each block contains 16,777,216 addresses this block should be used by large organizations. The first address in the block is called the network address defines the network of the organization Example Netid 73 is assigned Last address is reserved Recall: routers have addressees TCP/IP Protocol Suite 26

27 Note Millions of class A addresses are wasted. TCP/IP Protocol Suite 27

28 Figure 5.10 Blocks in Class B Class B is divided into 16,384 blocks (65,536 addresses each) 16 blocks are reserved First 2 bytes are the same (netid), the remaining 2 bytes can change (hostids) For example, Network id covers addresses to Network id is the last netid for this block Example Netid is assigned Last address is reserved Recall: routers have addresses TCP/IP Protocol Suite 28

29 Note Many class B addresses are wasted. TCP/IP Protocol Suite 29

30 Figure 5.11 Blocks in Class C Class C is divided into 2,097,152 blocks (256 addresses each) 256 blocks are reserved First 3 bytes are the same (netid), the remaining 1 byte can change (hostids) For example, Network id covers addresses to TCP/IP Protocol Suite 30

31 Note Not so many organizations are so small to have a class C block. TCP/IP Protocol Suite 31

32 Figure 5.12 The single block in Class D Note Class D addresses are made of one block, used for multicasting. TCP/IP Protocol Suite 32

33 Figure 5.13 The single block in Class E Note The only block of class E addresses was reserved for future purposes. TCP/IP Protocol Suite 33

34 Note The range of addresses allocated to an organization in classful addressing was a block of addresses in Class A, B, or C. TCP/IP Protocol Suite 34

35 Figure 5.14 Two-level addressing in classful addressing TCP/IP Protocol Suite 35

36 Figure 5.15 Information extraction in classful addressing netid First address 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 TCP/IP Protocol Suite 36

37 Example 5.13 An address in a block is given as Find the number of addresses in the block, the first address, and the last address. Solution Figure 5.16 shows a possible configuration of the network that uses this block. 1. The number of addresses in this block is N = 2 32 n = 16,777, To find the first address, we keep the leftmost 8 bits and set the rightmost 24 bits all to 0s. The first address is /8, in which 8 is the value of n. 3. To find the last address, we keep the leftmost 8 bits and set the rightmost 24 bits all to 1s. The last address is TCP/IP Protocol Suite 37

38 Figure 5.16 Solution to Example 5.13 TCP/IP Protocol Suite 38

39 Example 5.14 An address in a block is given as Find the number of addresses in the block, the first address, and the last address. Solution Figure 5.17 shows a possible configuration of the network that uses this block. 1. The number of addresses in this block is N = 2 32 n = 65, To find the first address, we keep the leftmost 16 bits and set the rightmost 16 bits all to 0s. The first address is /16, in which 16 is the value of n. 3. To find the last address, we keep the leftmost 16 bits and set the rightmost 16 bits all to 1s. The last address is TCP/IP Protocol Suite 39

40 Figure 5.17 Solution to Example 5.14 TCP/IP Protocol Suite 40

41 Example 5.15 An address in a block is given as Find the number of addresses in the block, the first address, and the last address. Solution Figure 5.17 shows a possible configuration of the network that uses this block. 1. The number of addresses in this block is N = 2 32 n = To find the first address, we keep the leftmost 24 bits and set the rightmost 8 bits all to 0s. The first address is /16, in which 24 is the value of n. 3. To find the last address, we keep the leftmost 24 bits and set the rightmost 8 bits all to 1s. The last address is /16. TCP/IP Protocol Suite 41

42 Figure 5.18 Solution to Example 5.15 TCP/IP Protocol Suite 42

43 Figure 5.19 Sample Internet TCP/IP Protocol Suite 43

44 Note The network address is the identifier of a network. TCP/IP Protocol Suite 44

45 Figure 5.20 Network addresses TCP/IP Protocol Suite 45

47 Figure 5.21 Network mask TCP/IP Protocol Suite 47

48 Figure 5.22 Finding a network address using the default mask If bit is ANDed with 1, it s preserved If bit is ANDed with 0, it s changed to a 0. A simple way to determine the netid for un-subnetted cases: (1) if mask byte is 255, retain corresponding byte of the address, (2) if mask byte is 0, set corresponding address byte to 0. TCP/IP Protocol Suite 48

49 Example 5.16 A router receives a packet with the destination address Show how the router finds the network address of the packet. Solution Since the class of the address is B, we assume that the router applies the default mask for class B, to find the network address. TCP/IP Protocol Suite 49

50 Recall IP Addresses: Classful Addressing Class # of addresses Percent of the Space A 2 31 = % B 2 30 = % C 2 29 = % D 2 28 = % E 2 28 = % TCP/IP Protocol Suite 50

52 5-bit Address Space Illustration 1-bit Netid case 16 addresses/block Number of blocks: 2 Address range per block: 0 to 15 Netids: 0, 1 Network Addresses : 00000, Broadcast Addresses: 01111, TCP/IP Protocol Suite 52

53 5-bit Address Space Illustration 2-bit Netid Case 8 addresses/block Number of blocks: 4 Address range per block: 0 to 7 Netids: 00, 01, 10, 11 Network Addresses : 00000, 01000, 10000, Broadcast Addresses: 00111, 01111, 10111, TCP/IP Protocol Suite 53

54 5-bit Address Space Illustration 3-bit Netid Case 4 addresses/block Number of blocks: 8 Address range per block: 0 to 3 Netids: 000, 001, 010, 011, 100, 101, 110, 111 Network Addresses : 00000, 00100, 01000, , 10100, 11000, Broadcast Addresses: 00011, 00111, 01011, , 10111, 11011, TCP/IP Protocol Suite 54

55 Mixing 3-bit & 2-bit Cases 4 addresses/block and 8 addresses/block Number of blocks: 6 Address range per block: 0 to 3 and 0 to 7 Netids: 000, 001, 010, 011, 10, 11 Network Addresses : 00000, 00100, 01000, , Broadcast Addresses: 00011, 00111, 01011, , TCP/IP Protocol Suite 55

56 Subnetting When we talked about CLASSFUL addressing we realized the problem of wasted host addresses and depleting available network addresses. In subnetting, a network is divided into several smaller networks called subnetworks or subnets each subnet will have it s own address Typically, there are 2 steps in reaching a destination: first we must reach the network (netid) and then we reach the destination (hostid) TCP/IP Protocol Suite 56

57 A network with two levels of hierarchy (not subnetting) The 2 level approach is not enough some times you can only have 1 physical network in example, all host are at the same level no grouping TCP/IP Protocol Suite 57

58 A network with three levels of hierarchy (subnetted) TCP/IP Protocol Suite 58

59 Addresses in a network with and without subnetting With subnetting, there are 3 levels (versus 2 levels). Partition the hostid space into subnetid and hostid. (1 st ) network, (2 nd ) subnetwork and (3 rd ) host TCP/IP Protocol Suite 59

60 Figure 5.25 Network mask and subnetwork mask TCP/IP Protocol Suite 60

61 Similar to Hierarchy concept in a telephone number TCP/IP Protocol Suite 61

63 Example 5.20 In Example 5.19, we divided a class B network into four subnetworks. The value of n = 16 and the value of n 1 = n 2 = n 3 = n 4 = 16 + log 2 4 = 18. This means that the subnet mask has eighteen 1s and fourteen 0s. In other words, the subnet mask is which is different from the network mask for class B ( ). TCP/IP Protocol Suite 63

64 Example 5.21 In Example 5.19, we show that a network is divided into four subnets. Since one of the addresses in subnet 2 is , we can find the subnet address as: The values of the first, second, and fourth bytes are calculated using the first short cut for AND operation. The value of the third byte is calculated using the second short cut for the AND operation. TCP/IP Protocol Suite 64

65 Example What is the subnetwork address if the destination address is and the subnet mask is ? Solution The subnetwork address is TCP/IP Protocol Suite 65

66 Recall - 5-bit Address Space Illustration 1-bit Netid case (no subnets) 16 addresses/block Number of blocks: 2 Address range per block: 0 to 15 Netids: 0, 1 Network Addresses : 00000, Broadcast Addresses: 01111, TCP/IP Protocol Suite 66

67 5-bit Address Space Illustration 1-bit Subnet case Number of blocks/networks: 2 Number subnets per block: 2 8 addresses/subnet Address range per subnet: 0 to 7 Subnet ids: 0, 1 Network Addresses : 00000, 01000, 10000, Broadcast Addresses: 00111, 01111, 10111, TCP/IP Protocol Suite 67

68 5-bit Address Space Illustration 2-bit Subnet case Number of blocks/networks: 2 Number subnets per block: 4 4 addresses/subnet Address range per subnet: 0 to 3 Subnet ids: 00, 01, 10, 11 Network Addresses : 00000, 00100, 01000, , 10100, 11000, Broadcast Addresses: 00011, 00111, 01011, , 10111, 11011, TCP/IP Protocol Suite 68

69 Illustrating the mask concept (1 of 3) What is the mask? TCP/IP Protocol Suite 69

70 Illustrating the mask concept (2 of 3) What is the mask (subnet mask)? TCP/IP Protocol Suite 70

71 Illustrating the mask concept (3 of 3) What is the mask (subnet mask)? TCP/IP Protocol Suite 71

72 The number of subnets must be a power of 2. Determine the number of subnets added by looking at the number of 1s added to the default mask and performing 2 raised to that number For example, 2 3 = 8 subnets TCP/IP Protocol Suite 72

73 Example A company is granted the site address (class C). The company needs six subnets. Design the subnets. Solution The number of 1s in the default mask is 24 (class C). The company needs six subnets. This number 6 is not a power of 2. The next number that is a power of 2 is 8 (2 3 ). We need 3 more 1s in the subnet mask. The total number of 1s in the subnet mask is 27 (24 + 3). The total number of 0s is 5 (32-27). The mask would be TCP/IP Protocol Suite 73

74 Solution (Continued) or The number of subnets is 8. The number of addresses in each subnet is 2 5 (5 is the number of 0s) or 32. TCP/IP Protocol Suite 74

75 Solution (Continued) TCP/IP Protocol Suite 75

76 Example Solution A company is granted the site address (class B). The company needs 1000 subnets. Design the subnets. The number of 1s in the default mask is 16 (class B). The company needs 1000 subnets. This number is not a power of 2. The next number that is a power of 2 is 1024 (2 10 ). We need 10 more 1s in the subnet mask. The total number of 1s in the subnet mask is 26 ( ). The total number of 0s is 6 (32-26). The mask is or The number of subnets is The number of addresses in each subnet is 2 6 (6 is the number of 0s) or 64. TCP/IP Protocol Suite 76

77 Solution (Continued) Subtract 63 from 255 to get 192 TCP/IP Protocol Suite 77

78 Supernetting Although class A and B addresses are dwindling there are plenty of class C addresses The problem with C addresses is, they only have 256 hostids not enough for any midsize to large size organization especially if you plan to give every computer, printer, scanner, etc. multiple IP addresses Supernetting allows an organization the ability to combine several class C blocks in creating a larger range of addresses Note: breaking up a network = subnetting Note: combining Class-C networks = supernetting TCP/IP Protocol Suite 78

79 Assigning or Choosing Class C Blocks When assigning class C block, the choices of blocks need to follow a set of rules: #1 the # of blocks must be a power of 2 #2 blocks must be contiguous (no gaps between blocks) #3 the 3 rd byte of the first address in the superblock must be evenly divisible by the number of blocks ie. if the # of blocks is N, the 3 rd byte must be divisible by N TCP/IP Protocol Suite 79

80 Example A company needs 600 addresses. Which of the following set of class C blocks can be used to form a supernet for this company? a b c d Solution a: No, there are only three blocks. Must be a power of 2 b: No, the blocks are not contiguous. c: No, 31 in the first block is not divisible by 4. d: Yes, all three requirements are fulfilled. (1. Power of 2, 2. Contiguous and 3. 3 rd byte of 1 st address is divisible by 4: 32/4=8) TCP/IP Protocol Suite 80

82 Explain Supernetting Conceptually Back out this bit from netid into host id Causes these 2 blocks to combine as a single block TCP/IP Protocol Suite 82

83 Figure 5.26 Comparison of subnet, default, and supernet mask TCP/IP Protocol Suite 83

84 5-3 CLASSLESS ADDRESSING Subnetting and supernetting in classful addressing did not really solve the address depletion problem. With the growth of the Internet, it was clear that a larger address space was needed as a long-term solution. Although the long-range solution has already been devised and is called IPv6, a short-term solution was also devised to use the same address space but to change the distribution of addresses to provide a fair share to each organization. The short-term solution still uses IPv4 addresses, but it is called classless addressing. TCP/IP Protocol Suite 84

85 Classless addressing Variable-length blocks are used that belong to no classes. TCP/IP Protocol Suite 85

86 Figure 5.27 Variable-length blocks in classless addressing the number of addresses in a block; it must be a power of 2 (2 0, 2 1, 2 2, 2 3,...). TCP/IP Protocol Suite 86

87 Figure 5.28 Prefix and suffix Note In classless addressing, the prefix defines the network and the suffix defines the host. TCP/IP Protocol Suite 87

88 Note The prefix length in classless addressing can be 1 to 32. TCP/IP Protocol Suite 88

89 Example 5.22 What is the prefix length and suffix length if the whole Internet is considered as one single block with 4,294,967,296 addresses? Solution In this case, the prefix length is 0 and the suffix length is 32. All 32 bits vary to define 2 32 = 4,294,967,296 hosts in this single block. TCP/IP Protocol Suite 89

90 Example 5.23 What is the prefix length and suffix length if the Internet is divided into 4,294,967,296 blocks and each block has one single address? Solution In this case, the prefix length for each block is 32 and the suffix length is 0. All 32 bits are needed to define 2 32 = 4,294,967,296 blocks. The only address in each block is defined by the block itself. TCP/IP Protocol Suite 90

91 Figure 5.29 Slash notation The number of addresses in a block is inversely related to the value of the prefix length, n. A small n means a larger block; a large n means a small block. TCP/IP Protocol Suite 91

92 Note In classless addressing, we need to know one of the addresses in the block and the prefix length to define the block. TCP/IP Protocol Suite 92

93 Example 5.26 The following addresses are defined using slash notations. a. In the address /8, the network mask is The mask has eight 1s and twenty-four 0s. The prefix length is 8; the suffix length is 24. b. In the address /16, the network mask is The mask has sixteen 1s and sixteen 0s.The prefix length is 16; the suffix length is 16. c. In the address /27, the network mask is The mask has twenty-seven 1s and five 0s. The prefix length is 27; the suffix length is 5. TCP/IP Protocol Suite 93

95 Example 5.27 One of the addresses in a block is /27. Find the number of addresses in the network, the first address, and the last address. Solution The value of n is 27. The network mask has twenty-seven 1s and five 0s. It is a. The number of addresses in the network is 2 32 n = 32. b. We use the AND operation to find the first address (network address). The first address is /27. TCP/IP Protocol Suite 95

96 Example 5.27 Continued c. To find the last address, we first find the complement of the network mask and then OR it with the given address: The last address is /27. TCP/IP Protocol Suite 96

97 Example 5.28 One of the addresses in a block is /24. Find the number of addresses, the first address, and the last address in the block. Solution The network mask is a. The number of addresses in the network is = 256. b. To find the first address, we use the short cut methods discussed early in the chapter. The first address is /24. TCP/IP Protocol Suite 97

98 Example 5.28 Continued c. To find the last address, we use the complement of the network mask and the first short cut method we discussed before. The last address is /24. TCP/IP Protocol Suite 98

99 Example 5.29 One of the addresses in a block is /20. Find the number of addresses, the first address, and the last address in the block. Solution The network mask is a. The number of addresses in the network is = b. To find the first address, we apply the first short cut to bytes 1, 2, and 4 and the second short cut to byte 3. The first address is /20. TCP/IP Protocol Suite 99

100 Example 5.29 Continued c. To find the last address, we apply the first short cut to bytes 1, 2, and 4 and the second short cut to byte 3. The OR operation is applied to the complement of the mask. The last address is /20. TCP/IP Protocol Suite 100

101 Block Allocation Three restrictions needed to be applied to the allocated block: a. The number of requested addresses, N, needs to be a power of 2. b. The value of prefix length, n, can be found from the number of addresses in the block. n = 32 log 2 N c. The beginning address needs to be divisible by the number of addresses of the block. For example, if a block contains 4 addresses, the beginning address must be divisible by 4. If the block has less than 256 addresses, we need to check only the rightmost byte. If it has less than 65,536 addresses, we need to check only the two rightmost bytes, and so on TCP/IP Protocol Suite 101

102 Example Which of the following can be the beginning address of a block that contains 16 addresses? a b c d Solution The address is eligible because 32 is divisible by 16. The address is eligible because 80 is divisible by 16. TCP/IP Protocol Suite 102

103 Example Which of the following can be the beginning address of a block that contains 1024 addresses? a b c d Solution To be divisible by 1024, the rightmost byte of an address should be 0 because any value in that first byte will be a fraction of 1024 (ie. 0 to 255). To be divisible by 1024, the rightmost byte should be 0 and the second rightmost byte must be divisible by 4 because for every unique number in the second byte position, there exist 256 addresses in the first byte position that maps to it. To get 1024 addresses overall, you will need an increment of 4 in the 2 nd byte position. Therefore, the 2 nd byte needs to be divisible by 4. Only the address meets this condition. TCP/IP Protocol Suite 103

104 TCP/IP Protocol Suite 104

105 Example Subnetting An organization is granted the block /26. The organization needs four subnetworks, each with an equal number of hosts. Design the subnetworks and find the information about each network. Solution The number of addresses for the whole network can be found as N = = 64. The first address in the network is /26 and the last address is /26. We now design the subnetworks: 1. We grant 16 addresses for each subnetwork to meet the first requirement (64/16 is a power of 2). 2. The subnetwork mask for each subnetwork is: TCP/IP Protocol Suite 105

106 Example 5.32 Continued 3. We grant 16 addresses to each subnet starting from the first available address. Figure 5.30 shows the subblock each subnet. Note that the starting address in each subnetwork is divisible by the number of addresses in that subnetwork. for TCP/IP Protocol Suite 106

107 Figure 5.30 Solution to Example 5.32 TCP/IP Protocol Suite 107

110 Figure 5.31 Solution to Example 5.33 TCP/IP Protocol Suite 110

111 Example 5.34 Assume a company has three offices: Central, East, and West. The Central office is connected to the East and West offices via private, WAN lines. The company is granted a block of 64 addresses with the beginning address /26. The management has decided to allocate 32 addresses for the Central office and divides the rest of addresses between the two other offices. 1. The number of addresses are assigned as follows: 2. We can find the prefix length for each subnetwork: TCP/IP Protocol Suite 111

112 Example 5.34 Continued 3. Figure 5.32 shows the configuration designed by the management. The Central office uses addresses /27 to /27. The company has used three of these addresses for the routers and has reserved the last address in the subblock. The East office uses the addresses /28 to /28. One of these addresses is used for the router and the company has reserved the last address in the subblock. The West office uses the addresses /28 to /28. One of these addresses is used for the router and the company has reserved the last address in the subblock. The company uses no address for the point-to-point connections in WANs. TCP/IP Protocol Suite 112

113 Figure 5.32 Example 5.34 TCP/IP Protocol Suite 113

114 Example Supernetting An ISP is granted a block of addresses starting with /16 (65,536 addresses). The ISP needs to distribute these addresses to three groups of customers as follows: The first group has 64 customers; each needs approximately 256 addresses. The second group has 128 customers; each needs approximately 128 addresses. The third group has 128 customers; each needs approximately 64 addresses. We design the subblocks and find out how many addresses are still available after these allocations. TCP/IP Protocol Suite 114

115 Example 5.35 Continued Solution Let us solve the problem in two steps. In the first step, we allocate a subblock of addresses to each group. The total number of addresses allocated to each group and the prefix length for each subblock can found as Figure 5.33 shows the design for the first hierarchical level. Figure 5.34 shows the second level of the hierarchy. Note that we have used the first address for each customer as the subnet address and have reserved the last address as a special address. TCP/IP Protocol Suite 115

116 Figure 5.33 Solution to Example 5.35: first step TCP/IP Protocol Suite 116

117 Figure 5.34 Solution to Example 5.35: second step TCP/IP Protocol Suite 117

120 Figure 5.35 Example of using the all-zero address Source: Destination: Packet TCP/IP Protocol Suite 120

121 Example of this host on this address IP-less Host sending message to bootstrap server An address of all 0 s is used during bootstrap time if the host doesn t know it s IP address. The un-named host sends an all 0 source address and limited broadcast (all 1 s) destination address to the bootstrap server. Example of specific host on this network Host sending to some other specific host on a network An address with a netid of all 0 s is used by a host or router to send another host with in the same network a message. TCP/IP Protocol Suite 121

122 Figure 5.36 Example of limited broadcast address Network / / / /24 TCP/IP Protocol Suite 122

123 Figure 5.37 Example of loopback address The IP address with the 1 st byte equal to 127 is used for the loop back address. Loopback address is used to test software on a machine the packet never leaves the machine it returns to the protocol software Example: a ping command can send a packet with a loopback address as the destination address to see if the IP software is capable of receiving and processing a packet. TCP/IP Protocol Suite 123

124 Multicast addresses: /4 is reserved for multicast communication. TCP/IP Protocol Suite 124

126 Figure 5.38 Example of a directed broadcast address Network: / / / / /24 Packet TCP/IP Protocol Suite 126

128 5-5 NAT The distribution of addresses through ISPs has created a new problem. If the business grows or the household needs a larger range, the ISP may not be able to grant the demand because the addresses before and after the range may have already been allocated to other networks. In most situations, however, only a portion of computers in a small network need access to the Internet simultaneously. A technology that can help in this cases is network address translation (NAT). TCP/IP Protocol Suite 128

129 Figure 5.39 NAT Network Address Translation (NAT) allows a site to use a set of private addresses for internal communication and a set of global Internet addresses for communication with another site. The site must have only one single connection to the global Internet through a router that runs NAT software. The routers only 2 address: (1) the global IP address and (2) one private address TCP/IP Protocol Suite 129

130 Figure 5.40 Address Translation All packets leaving the network get assigned the global address as the source address (straightforward process) All packets coming into the network get their global destination address replaced with the appropriate private address (process is more involved) (explain this in the next ppt slide) TCP/IP Protocol Suite 130

132 NAT Using Multiple Global Addresses NAT Router with One GLOBAL address can only allow One private host to access the same EXTERNAL host with more global addresses, more private hosts can access the SAME external host A NAT Router with 8 global addresses can allow up to 8 private addresses (hosts) to access the SAME external host (simultaneously) can create up to 8 separate connections To create a many-to-many relationship, a 5-column table (versus 2-column table) is needed in reducing uncertainty by specifying port address and transport layer protocol TCP/IP Protocol Suite 132

133 An ISP and NAT An ISP serving customers can conserve addresses by using NAT. NOTE: think of customers as being apart of the ISP s private network before gaining access to the Global Internet. The ISP could assign a private address to each customer and when the customer leaves the private network, a translation would occur. Let an ISP with 100,000 customers be granted only 1000 global addresses - the ISP could assign private addresses to each 100,000 customers and the ISP translate the 100,000 source addresses for the outgoing packets with the 1000 global addresses TCP/IP Protocol Suite

134 Summary: To introduce the concept of an address space in general and the address space of IPv4 in particular. To discuss the classful architecture and the blocks of addresses available in each class. To discuss the idea of hierarchical addressing and how it has been implemented in classful addressing. To explain subnetting and supernetting for classful architecture. To discuss classless addressing, that has been devised to solve the problems in classful addressing. To discuss some special blocks and some special addresses in each block. To discuss NAT technology and show how it can be used to alleviate of address depletion. TCP/IP Protocol Suite 134

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