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Chapter 8

Chapter 8. Internet Protocol. Objectives . Upon completion you will be able to:. Understand the format and fields of a datagram Understand the need for fragmentation and the fields involved Understand the options available in an IP datagram Be able to perform a checksum calculation

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Chapter 8

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  1. Chapter 8 Internet Protocol Objectives Upon completion you will be able to: • Understand the format and fields of a datagram • Understand the need for fragmentation and the fields involved • Understand the options available in an IP datagram • Be able to perform a checksum calculation • Understand the components and interactions of an IP package TCP/IP Protocol Suite

  2. Figure 8.1Position of IP in TCP/IP protocol suite IP: connectionless, unreliable, packet switching w/ datagrams TCP/IP Protocol Suite

  3. 8.1 DATAGRAM A packet in the IP layer is called a datagram, a variable-length packet consisting of two parts: header and data. The header is 20 to 60 bytes in length and contains information essential to routing and delivery. Some of the fields: VER - version numbers, 4 and 6 HLEN - header length in 4-byte words. Value of 5 means 20 byte header TCP/IP Protocol Suite

  4. Figure 8.2IP datagram TCP/IP Protocol Suite

  5. Figure 8.3Service type or Differentiated Services - DS field This field was previously called Service type. Now called Differentiated Services. Precedence bits - never used. Similar to PRI bits in IPv6. TOS (Type of Service) bits - see next slide. TCP/IP Protocol Suite

  6. Table 8.1 Types of service If you want to send a packet with a special type of service, use one of the above 5-bit sets. TCP/IP Protocol Suite

  7. Table 8.2 Default types of service Some apps have default service types. TCP/IP Protocol Suite

  8. If we call these 8 bits Differentiated Services (and not the older Service Type), then the first six bits are called code- points. Table 8.3 Values for codepoints When the 3 right-most bits are 0, the 3 left-most bits are the same as the precedence bits from the previous slides. When the 3 right-most bits are not all 0s, the 6 bits define 64 Services based on the priority assignment by the Internet or local authorities. Assignments have not yet been finalized. TCP/IP Protocol Suite

  9. Figure 8.4Encapsulation of a small datagram in an Ethernet frame The total length field defines the total length of the datagram including the header. Total length field is 16 bits, or 65,535 bytes. Of which 20 to 60 bytes are the header. If an IP datagram is short, and is packaged into an Ethernet frame, remember that the minimum payload size of an Ethernet frame is 46 bytes (to avoid being mistaken for a runt). TCP/IP Protocol Suite

  10. Identification, Flags, and Fragmentation offset are all used to perform fragmentation, which we will cover shortly. Time to Live - 8-bit field, so Time to Live can be set to 255. As it passes thru a router, the router decrements the counter. When counter hits 0, the datagram is deleted (and ICMP sends an error message back to the source). Why might a host set the Time to Live field to 1? TCP/IP Protocol Suite

  11. Figure 8.5Multiplexing The Protocol field (8 bits) identifies the upper layer protocol that is using IP for transmission of its data. TCP/IP Protocol Suite

  12. Table 8.4 Protocols These are some common values that are used in the Protocol field. TCP/IP Protocol Suite

  13. Example 1 An IP packet has arrived with the first 8 bits as shown: 01000010 The receiver discards the packet. Why? TCP/IP Protocol Suite

  14. Example 1 An IP packet has arrived with the first 8 bits as shown: 01000010 The receiver discards the packet. Why? SolutionThere is an error in this packet. The 4 left-most bits (0100) show the version, which is correct. The next 4 bits (0010) show the header length; which means (2 × 4 = 8), which is wrong. The minimum number of bytes in the header must be 20. The packet has been corrupted in transmission. TCP/IP Protocol Suite

  15. Example 2 In an IP packet, the value of HLEN is 1000 in binary. How many bytes of options are being carried by this packet? TCP/IP Protocol Suite

  16. Example 2 In an IP packet, the value of HLEN is 1000 in binary. How many bytes of options are being carried by this packet? SolutionThe HLEN value is 8, which means the total number of bytes in the header is 8 × 4 or 32 bytes. The first 20 bytes are the base header, the next 12 bytes are the options. TCP/IP Protocol Suite

  17. Example 3 In an IP packet, the value of HLEN is 516 and the value of the total length field is 002816 . Howmany bytes of data are being carried by this packet? TCP/IP Protocol Suite

  18. Example 3 In an IP packet, the value of HLEN is 516 and the value of the total length field is 002816 . Howmany bytes of data are being carried by this packet? SolutionThe HLEN value is 5, which means the total number of bytes in the header is 5 × 4 or 20 bytes (no options). The total length is 40 bytes, which means the packet is carrying 20 bytes of data (40 − 20). TCP/IP Protocol Suite

  19. Example 4 An IP packet has arrived with the first few hexadecimal digits as shown below: 45000028000100000102 . . . How many hops can this packet travel before being dropped? The data belong to what upper layer protocol? TCP/IP Protocol Suite

  20. Example 4 An IP packet has arrived with the first few hexadecimal digits as shown below: 45000028000100000102 . . . How many hops can this packet travel before being dropped? The data belong to what upper layer protocol? SolutionTo find the time-to-live field, we skip 8 bytes (16 hexadecimal digits). The time-to-live field is the ninth byte, which is 01. This means the packet can travel only one hop. The protocol field is the next byte (02), which means that the upper layer protocol is IGMP (see Table 8.4). TCP/IP Protocol Suite

  21. 8.2 FRAGMENTATION The format and size of a frame depend on the protocol used by the physical network. A datagram may have to be fragmented to fit the protocol regulations. The topics discussed in this section include: Maximum Transfer Unit (MTU) Fields Related to Fragmentation TCP/IP Protocol Suite

  22. Figure 8.6MTU MTU - Maximum Transfer Unit TCP/IP Protocol Suite

  23. Table 8.5 MTUs for some networks Max datagram size for IP is 65535 bytes. So if we have a max sized datagram to send over Ethernet, what do we do? TCP/IP Protocol Suite

  24. Figure 8.7Flags field A datagram can be fragmented by the source host or any router in the path. Reassembly is done only by the destination host. Most fields are copied from one fragment to the next. The 3 fields that are not copied are the flags, fragmentation offset, and the total length. (And the checksum of course is recalculated.) The Identification field is copied from one fragment to the next. The Do Not Fragment bit is set to 1 if the network is not supposed to fragment this datagram. (If it has to be fragmented, it is tossed.) The More Fragments bit is set to 1 if there are more fragments following this one. TCP/IP Protocol Suite

  25. Figure 8.8Fragmentation example The Fragmentation Offset tells what position this fragment is in the whole stream. The offset counts by 8. So if a fragment is supposed to start at byte 400, the offset equals 50. TCP/IP Protocol Suite

  26. Figure 8.9Detailed fragmentation example TCP/IP Protocol Suite

  27. Example 5 A packet has arrived with an M bit value of 0. Is this the first fragment, the last fragment, or a middle fragment? Do we know if the packet was fragmented? TCP/IP Protocol Suite

  28. Example 5 A packet has arrived with an M bit value of 0. Is this the first fragment, the last fragment, or a middle fragment? Do we know if the packet was fragmented? SolutionIf the M bit is 0, it means that there are no more fragments; the fragment is the last one. However, we cannot say if the original packet was fragmented or not. A non-fragmented packet is considered the last fragment. TCP/IP Protocol Suite

  29. Example 7 A packet has arrived with an M bit value of 1 and a fragmentation offset value of zero. Is this the first fragment, the last fragment, or a middle fragment?. TCP/IP Protocol Suite

  30. Example 7 A packet has arrived with an M bit value of 1 and a fragmentation offset value of zero. Is this the first fragment, the last fragment, or a middle fragment?. SolutionBecause the M bit is 1, it is either the first fragment or a middle one. Because the offset value is 0, it is the first fragment. TCP/IP Protocol Suite

  31. Example 8 A packet has arrived in which the offset value is 100. What is the number of the first byte? Do we know the number of the last byte? TCP/IP Protocol Suite

  32. Example 8 A packet has arrived in which the offset value is 100. What is the number of the first byte? Do we know the number of the last byte? SolutionTo find the number of the first byte, we multiply the offset value by 8. This means that the first byte number is 800. We cannot determine the number of the last byte unless we know the length of the data. TCP/IP Protocol Suite

  33. Example 9 A packet has arrived in which the offset value is 100, the value of HLEN is 5 and the value of the total length field is 100. What is the number of the first byte and the last byte? TCP/IP Protocol Suite

  34. Example 9 A packet has arrived in which the offset value is 100, the value of HLEN is 5 and the value of the total length field is 100. What is the number of the first byte and the last byte? SolutionThe first byte number is 100 × 8 = 800. The total length is 100 bytes and the header length is 20 bytes (5 × 4), which means that there are 80 bytes in this datagram. If the first byte number is 800, the last byte number must be 879. TCP/IP Protocol Suite

  35. 8.3 OPTIONS The header of the IP datagram is made of two parts: a fixed part and a variable part. The variable part comprises the options that can be a maximum of 40 bytes. The topics discussed in this section include: Format Option Types TCP/IP Protocol Suite

  36. Figure 8.10Option format Not all routers/hosts use these options, but they must be ready to do so if they are present in the datagram. Copy - tells whether to copy this option into a fragment Class - defines the general purpose of the option TCP/IP Protocol Suite

  37. Figure 8.11Categories of options As we just saw, only 6 options in use currently. The single-byte options are only 1 byte in length and do not require length or data fields. TCP/IP Protocol Suite

  38. Figure 8.12No operation option Used as a filler between options. For example, can be used to align the next option on a 16- or 32-bit boundary. TCP/IP Protocol Suite

  39. Figure 8.13End of option option Denotes the end of the options and that the data is next. TCP/IP Protocol Suite

  40. Figure 8.14Record route option Records the route a datagram takes thru routers. Can only record 9 routers, since max size of the header is 60 bytes, 20 bytes for base header, leaving only 40 bytes for options. TCP/IP Protocol Suite

  41. Figure 8.15Record route concept The Pointer field (4, then 8, then 12, then 16) is the byte number of the first available space. TCP/IP Protocol Suite

  42. Figure 8.16Strict source route option For when a datagram has to follow a given, fixed route. TCP/IP Protocol Suite

  43. Figure 8.17Strict source route concept First hop address is here Second hop address is here Note that as hops are made, next hop is replaced with address of router you just went thru TCP/IP Protocol Suite

  44. Figure 8.18Loose source route option Similar to fixed route - each router in the list must be visited, but other routers can be visited too. TCP/IP Protocol Suite

  45. Figure 8.19Timestamp option Can be used if you want to record the time the datagram visits each router. Time in milleseconds, Universal Time. O-Flow bits (overflow bits) record the number of routers that could not add their timestamp because no more fields were available. TCP/IP Protocol Suite

  46. Figure 8.20Use of flag in timestamp The Flag bits tell the router whether to do one of the following operations: TCP/IP Protocol Suite

  47. Figure 8.21Timestamp concept TCP/IP Protocol Suite

  48. Example 10 Which of the six options must be copied to each fragment? TCP/IP Protocol Suite

  49. Example 10 Which of the six options must be copied to each fragment? SolutionWe look at the first (left-most) bit of the code for each option. a. No operation: Code is 00000001; not copied.b. End of option: Code is 00000000; not copied.c. Record route: Code is 00000111; not copied.d. Strict source route: Code is 10001001; copied.e. Loose source route: Code is 10000011; copied.f. Timestamp: Code is 01000100; not copied. TCP/IP Protocol Suite

  50. Example 11 Which of the six options are used for datagram control and which are used for debugging and management? TCP/IP Protocol Suite

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