Chapter 1: TCP/IP Architecture BY: Maliheh Bahekmat

Transcription

1 Chapter 1: TCP/IP Architecture BY: Maliheh Bahekmat 1

2 OSI Model 2

3 TCP/IP Protocol Stack 3

4 TCP/IP 4

5 TCP/IP vs OSI 5

6 Packet Encapsulation The data is sent down the protocol stack Each layer adds to the data by prepending headers 22Bytes 20Bytes 20Bytes 4Bytes 64 to 1500 Bytes 6

7 Addresses in TCP/IP 7

8 IP Addresses IP addresses are logical addresses (not physical) 32 bits. Includes a network ID and a host ID. Every host must have a unique IP address. IP addresses are assigned by a central authority (American Registry for Internet Numbers) 8

9 In classful addressing, the address space is divided into five classes: A, B, C, D, and E. 9

10 Occupation of the address space 10

11 Finding the class in binary notation 11

12 Finding the address class 12

13 Example 1 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 2 31 or 2,147,483,648 addresses. 13

14 Example 2 Find the class of the address: Solution The first bit is 0. This is a class A address. 14

15 Occupation of the address space 15

16 Finding the class in decimal notation 16

17 Example 3 Find the class of the address: Solution The first byte is 227 (between 224 and 239); the class is D. 17

18 Example 4 In Example 1 we showed that class A has 2 31 (2,147,483,648) addresses. How can we prove this same fact using dotted-decimal notation? Solution The addresses in class A range from to We notice that we are dealing with base 256 numbers here. 18

19 Each byte in the notation has a weight. The weights are as follows: 256 3, 256 2, 256 1, Last address: = 2,147,483,647 First address: = 0 If we subtract the first from the last and add 1, we get : 2,147,483,

20 IP Addresses IP 20

21 The four formats of IP Addresses Class A 0 NetID HostID B 10 NetID HostID C 110 NetID HostID D 1110 Multicast Address 8 bits 8 bits 8 bits 8 bits 21

22 IP Address Classes A B C D E 22

23 l l Class A 128 possible network IDs over 4 million host IDs per network ID l l Class B 16K possible network IDs 64K host IDs per network ID l Class C over 2 million possible network IDs l about 256 host IDs per network ID 23

24 Host and Network Addresses A single network interface is assigned a single IP address called the host address. A host may have multiple interfaces, and therefore multiple host addresses. Hosts that share a network all have the same IP network address (the network ID). 24

25 IP Broadcasting An IP broadcast addresses has a host ID of all 1s. IP broadcasting is not necessarily a true broadcast, it relies on the underlying hardware technology. Two types of broadcasting: Limited broadcasting Direct Broadcasting 25

26 Direct Broadcasting

27 Subnet Addresses An organization can subdivide it s host address space into groups called subnets. The subnet ID is generally used to group hosts based on the physical network topology. 10 NetID SubnetID HostID 27

28 Subnetting 28 IP

29 Subnetting router Subnet x Subnet x Subnet x 29

30 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 30

31 In classful addressing, the network address (the first address in the block) is the one that is assigned to the organization. 31

32 Two-level addressing in classful addressing 32

33 Two-level addressing in classful addressing 33

34 Example 5 Given the network address , 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 73. The addresses range from to

35 Sample internet 35

36 Network address 36

37 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. 37

38 Masking concept 38

39 AND operation 39

40 Network mask 40

41 Finding a network address 41

42 Example 6 Given the address and the default class A mask, find the beginning address (network address). Solution The default mask is , which means that only the first byte is preserved and the other 3 bytes are set to 0s. The network address is

43 IP addresses are designed with two levels of hierarchy. 43

44 A network with two levels of hierarchy (not subnetted) 44

45 A network with two levels of hierarchy (not subnetted) 45

46 Subnetting Subnets can simplify routing. IP subnet broadcasts have a hostid of all 1s. It is possible to have a single wire network with multiple subnets. 46

47 Hierarchy concept in a telephone number 47

48 Addresses in a network with and without subnetting 48

49 A network with three levels of hierarchy (subnetted) 49

50 Network mask and subnet mask 50

51 Default mask and subnet mask 51

52 Finding the Subnet Address Given an IP address, we can find the subnet address the same way we found the network address in the previous chapter. We apply the mask to the address. We can do this in two ways: straight or short-cut. 52

53 Straight Method In the straight method, we use binary notation for both the address and the mask and then apply the AND operation to find the subnet address. 53

54 Example 7 What is the subnetwork address if the destination address is and the subnet mask is ? Solution The subnetwork address is

55 Short-Cut Method ** If the byte in the mask is 255, copy byte in the address. ** If the byte in the mask is 0, replace the byte in the address with 0. ** If the byte in the mask is neither 255 nor 0, we write the mask and the address in binary and apply the AND operation. 55

56 Example 8 What is the subnetwork address if the destination address is and the mask is ? Solution See Figure next slide 56

57 Example 57

58 Comparison of a default mask and a subnet mask 58

59 Example 9 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). 59

60 Solution (Continued) 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 is 60

61 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. See next slide 61

62 Example : 62

63 supernetwork 63

64 supernetwork In subnetting, we need the first address of the subnet and the subnet mask to define the range of addresses 64

65 Rules: The number of blocks must be a power of 2 (1, 2, 4, 8, 16,...). The blocks must be contiguous in the address space (no gaps between the blocks). The third byte of the first address in the superblock must be evenly divisible by the number of blocks. In other words, if the number of blocks is N, the third byte must be divisible by N. 65

66 Example 10: A company needs 600 addresses. Which of the following set of class C blocks can be used to form a supernet for this company?

67 Solution 8 1: No, there are only three blocks. 2: No, the blocks are not contiguous. 3: No, 31 in the first block is not divisible by 4. 4: Yes, all three requirements are fulfilled. 67

68 CLASSLESS ADDRESSING 68

69 Variable-length blocks 69

70 Number of Addresses in a Block There is only one condition on the number of addresses in a block; it must be a power of 2 (2, 4, 8,...). A household may be given a block of 2 addresses. A small business may be given 16 addresses. A large organization may be given 1024 addresses. 70

71 Beginning Address The beginning address must be evenly divisible by the number of addresses. 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. 71

72 Example 11 Which of the following can be the beginning address of a block that contains 16 addresses? Solution The address is eligible because 32 is divisible by 16. The address is eligible because 80 is divisible by

73 Example 12 Which of the following can be the beginning address of a block that contains 1024 addresses? Solution To be divisible by 1024, the rightmost byte of an address should be 0 and the second rightmost byte must be divisible by 4. Only the address meets this condition. 73

74 Figure 5-14 Slash notation 74

75 Slash notation is also called CIDR notation. 75

76 Slash notation 76

77 Example 13 A small organization is given a block with the beginning address and the prefix length /29 (in slash notation). What is the range of the block? Solution The beginning address is To find the last address we keep the first 29 bits and change the last 3 bits to 1s. Beginning: Ending : There are only 8 addresses in this block. 77

78 A block in classes A, B, and C can easily be represented in slash notation as A.B.C.D/ n where n is either 8 (class A), 16 (class B), or 24 (class C). 78

79 Example 14 What is the network address if one of the addresses is /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 5 bits affect only the last byte. The last byte is Changing the last 5 bits to 0s, we get or 64. The network address is /27. 79

80 Example 15 An organization is granted the block /26. The organization needs to have four subnets. What are the subnet addresses and the range of addresses for each subnet? Solution The suffix length is 6. This means the total number of addresses in the block is 64 (2 6 ). If we create four subnets, each subnet will have 16 addresses. 80

81 Solution (Continued) Let us first find the subnet prefix (subnet mask). We need four subnets, which means we need to add two more 1s to the site prefix. The subnet prefix is then /28. Subnet 1: /28 to /28. Subnet 2 : /28 to /28. Subnet 3: /28 to /28. Subnet 4: /28 to /28. See next slide 81

82 Example 15 82

83 Example 16 An ISP is granted a block of addresses starting with /16. The ISP needs to distribute these addresses to three groups of customers as follows: 1. The first group has 64 customers; each needs 256 addresses. 2. The second group has 128 customers; each needs 128 addresses. 3. The third group has 128 customers; each needs 64 addresses. Design the subblocks and give the slash notation for each subblock. Find out how many addresses are still available after these allocations. 83

84 Mapping IP Addresses to Hardware Addresses IP Addresses are not recognized by hardware. If we know the IP address of a host, how do we find out the hardware address? The process of finding the hardware address of a host given the IP address is called Address Resolution 84

85 Address Resolution Protocol ARP is a broadcast protocol is used by a sending host when it knows the IP address of the destination but needs the Ethernet address. ARP provides mapping 32bit IP address <-> 48bit MAC address <-> 00-C0-4F

86 Cont.. ARP cache maintains the recent mappings from IP addresses to MAC addresses every host on the network receives the request Each host checks the request against it s IP address - the right one responds. ARP request broadcast on Ethernet Destination host ARP layer responds 86

87 ARP (cont.) ARP does not need to be done every time an IP datagram is sent - hosts remember the hardware addresses of each other. Part of the ARP protocol specifies that the receiving host should also remember the IP and hardware addresses of the sending host. 87

88 ARP conversation HEY - Everyone please listen! Will please send me his/her Ethernet address? not me Hi Green! I m , and my Ethernet address is 87:A2:15:35:02:C3 88

89 ARP operation 89

90 Reverse Address Resolution The process of finding out the IP address of a host given a hardware address is called Reverse Address Resolution Reverse address resolution is needed by diskless workstations when booting. 90

91 ARP Operation ARP ARP ARP 91

92 RARP conversation HEY - Everyone please listen! My Ethernet address is 22:BC:66:17:01:75. Does anyone know my IP address? not me Hi Green! Your IP address is

93 RARP operation 93

94 ARP/RARP Packet Format IP IP ARP IP IP ARP CRC 94

95 IP: Internet Protocol Connectionless datagram service Performs: Routing Fragmentation/Reassembly Does not perform: Flow Control Sequence Control Duplicate Avoidance Error Recovery High Performance and Low Overhead 95

96 IP Datagrams IP provides connectionless, unreliable delivery of IP datagrams. Connectionless: each datagram is independent of all others. Unreliable: there is no guarantee that datagrams are delivered correctly or at all. 96

97 IP Routing IP uses next hop routing Uses routing table: Host entries for routes to specific hosts Network routes for routes to a network Default route (special network route) for all unknown destinations Route may be direct or through a router Classless Inter-Domain Routing (CIDR) requires network routes to explicitly specify a mask 97

98 IP Routing Source Application Transport Network Link Router Network Link Destination Application Transport Network Link Routing Table Destination IP address IP address of a next-hop router Flags Network interface specification 98

99 IP Routing To find a route, use first route found: Host route with matching destination Network route with matching destination network (only network numbers compared) If classful routing, use class-based mask With CIDR, search network routes with longest mask first Default route Once route found, packet is sent to gateway or destination (if gateway bit not set) 99

100 IP Routing 100

101 IP Encapsulation IP IP (IP ) ( ) 101

102 IP datagram 102

103 IP Datagram Structure IP IP IP IP padding IP IP 103

104 Identification Field IP = = = = = = 104

105 Type of Service (TOS) 0 : : : : : : : : 105

106 IP Datagram Fragmentation Each fragment (packet) has the same structure as the IP datagram. IP specifies that datagram reassembly is done only at the destination (not on a hop-by-hop basis). If any of the fragments are lost - the entire datagram is discarded (and an ICMP message is sent to the sender). 106

107 Maximum Transfer Unit (MTU) 107

108 IP Fragmentation MTU MTU IP IP IP 108

109 IP Fragmentation: Example MTU MTU IP IP MTU MTU MTU B A IP IP 109

110 IP Fragmentation: Example IP IP IP ( = ) IP ( = ) IP = ( ) 110

111 Flag field 111

112 Fragmentation example 112

113 Detailed example 113

114 Option format 114

115 Categories of options 115

116 No operation option 116

117 End of option option 117

118 Record route option 118

119 Record route concept 119

120 Strict source route option 120

121 Strict source route concept 121

122 Loose source route option 122

123 Timestamp option 123

124 Use of flag in timestamp 124

125 Timestamp concept 125

126 ICMP Services: Echo/Reply ICMP Echo ICMP Echo 126

127 ICMP Services: Redirect ( : (. ICMP 127

128 ICMP Internet Control Message Protocol ICMP is a protocol used for exchanging control messages. ICMP uses IP to deliver messages. ICMP messages are usually generated and processed by the IP software, not the user process. 128

129 ICMP Message Types Echo Request Echo Response Destination Unreachable Redirect Time Exceeded Redirect (route change) there are more

130 ICMP Used to report problems with delivery of IP Datagrams within an IP network ICMP Message Used by Ping, Tracerout commands 20bytes 4bytes IP Header ICMP Header ICMP Data Types and Codes Echo Request (type=8, code=0) Echo Reply(type=0, code=0) Type Code Checksum 1byte 1byte 2bytes Destination Unreachable(type=3, code=0) Time Exceeded(type=11, code=0) : Time-to-Live =0 130

131 ICMP Location ICMP ICMP TCP/UDP IP IP ICMP IP 131

132 ICMP Message Format ( ) + IP IP 132

133 ICMP Error Detection ICMP 133