1. 6/9/2015 Unit-4 : Transport Layer 1 CS 1302 Computer Networks — Unit - 4 — — Transport Layer — Text Book Behrouz.A. Forouzan, “Data communication and Networking”, Tata McGrawHill, 2004

2. Transport Layer 6/9/20152Unit-4 : Transport Layer

3. Position of transport layer 6/9/20153Unit-4 : Transport Layer

4. Transport layer duties 6/9/20154Unit-4 : Transport Layer

5. Chapters Chapter 22 Process-to-Process Delivery Chapter 23 Congestion Control and QoS 6/9/20155Unit-4 : Transport Layer

6. Chapter 22 Process-to-Process Delivery: UDP and TCP 6/9/20156Unit-4 : Transport Layer

7. 22.1 Process-to-Process Delivery Client-Server Paradigm Addressing Multiplexing and Demultiplexing Connectionless/Connection-Oriented Reliable/Unreliable 6/9/20157Unit-4 : Transport Layer

8. The transport layer is responsible for process-to-process delivery. Note: 6/9/20158Unit-4 : Transport Layer

9. Figure 22.1 Types of data deliveries 6/9/20159Unit-4 : Transport Layer

10. Figure 22.2 Port numbers 6/9/201510Unit-4 : Transport Layer

11. Figure 22.3 IP addresses versus port numbers 6/9/201511Unit-4 : Transport Layer

12. Figure 22.4 IANA ranges 6/9/201512Unit-4 : Transport Layer

13. Figure 22.5 Socket address 6/9/201513Unit-4 : Transport Layer

14. Figure 22.6 Multiplexing and demultiplexing 6/9/201514Unit-4 : Transport Layer

15. Figure 22.7 Connection establishment 6/9/201515Unit-4 : Transport Layer

16. Figure 22.8 Connection termination 6/9/201516Unit-4 : Transport Layer

17. Figure 22.9 Error control 6/9/201517Unit-4 : Transport Layer

18. 22.2 UDP Port Numbers User Datagram Applications 6/9/201518Unit-4 : Transport Layer

19. UDP is a connectionless, unreliable protocol that has no flow and error control. It uses port numbers to multiplex data from the application layer. Note: 6/9/201519Unit-4 : Transport Layer

20. Table 22.1 Well-known ports used by UDP PortProtocolDescription 7EchoEchoes a received datagram back to the sender 9DiscardDiscards any datagram that is received 11UsersActive users 13DaytimeReturns the date and the time 17QuoteReturns a quote of the day 19ChargenReturns a string of characters 53NameserverDomain Name Service 67BootpsServer port to download bootstrap information 68BootpcClient port to download bootstrap information 69TFTPTrivial File Transfer Protocol 111RPCRemote Procedure Call 123NTPNetwork Time Protocol 161SNMPSimple Network Management Protocol 162SNMPSimple Network Management Protocol (trap) 6/9/201520Unit-4 : Transport Layer

21. Figure 22.10 User datagram format 6/9/201521Unit-4 : Transport Layer

22. The calculation of checksum and its inclusion in the user datagram are optional. Note: 6/9/201522Unit-4 : Transport Layer

23. UDP is a convenient transport-layer protocol for applications that provide flow and error control. It is also used by multimedia applications. Note: 6/9/201523Unit-4 : Transport Layer

24. 22.3 TCP Port Numbers Services Sequence Numbers Segments Connection Transition Diagram Flow and Error Control Silly Window Syndrome 6/9/201524Unit-4 : Transport Layer

25. Table 22.2 Well-known ports used by TCP PortProtocolDescription 7 EchoEchoes a received datagram back to the sender 9DiscardDiscards any datagram that is received 11UsersActive users 13DaytimeReturns the date and the time 17QuoteReturns a quote of the day 19ChargenReturns a string of characters 20FTP, DataFile Transfer Protocol (data connection) 21FTP, ControlFile Transfer Protocol (control connection) 23TELNETTerminal Network 25SMTPSimple Mail Transfer Protocol 53DNSDomain Name Server 67BOOTPBootstrap Protocol 79Finger 80HTTPHypertext Transfer Protocol 111RPCRemote Procedure Call 6/9/201525Unit-4 : Transport Layer

26. Figure 22.11 Stream delivery 6/9/201526Unit-4 : Transport Layer

27. Figure 22.12 Sending and receiving buffers 6/9/201527Unit-4 : Transport Layer

28. Figure 22.13 TCP segments 6/9/201528Unit-4 : Transport Layer

29. Example 1 Imagine a TCP connection is transferring a file of 6000 bytes. The first byte is numbered 10010. What are the sequence numbers for each segment if data are sent in five segments with the first four segments carrying 1000 bytes and the last segment carrying 2000 bytes? Solution The following shows the sequence number for each segment: Segment 1 ==> sequence number: 10,010 (range: 10,010 to 11,009) Segment 2 ==> sequence number: 11,010 (range: 11,010 to 12,009) Segment 3 ==> sequence number: 12,010 (range: 12,010 to 13,009) Segment 4 ==> sequence number: 13,010 (range: 13,010 to 14,009) Segment 5 ==> sequence number: 14,010 (range: 14,010 to 16,009) 6/9/201529Unit-4 : Transport Layer

30. The bytes of data being transferred in each connection are numbered by TCP. The numbering starts with a randomly generated number. Note: 6/9/201530Unit-4 : Transport Layer

31. The value of the sequence number field in a segment defines the number of the first data byte contained in that segment. Note: 6/9/201531Unit-4 : Transport Layer

32. The value of the acknowledgment field in a segment defines the number of the next byte a party expects to receive. The acknowledgment number is cumulative. Note: 6/9/201532Unit-4 : Transport Layer

33. Figure 22.14 TCP segment format 6/9/201533Unit-4 : Transport Layer

34. Figure 22.15 Control field 6/9/201534Unit-4 : Transport Layer

35. Table 22.3 Description of flags in the control field FlagDescription URGThe value of the urgent pointer field is valid. ACKThe value of the acknowledgment field is valid. PSHPush the data. RSTThe connection must be reset. SYNSynchronize sequence numbers during connection. FINTerminate the connection. 6/9/201535Unit-4 : Transport Layer

36. Figure 22.16 Three-step connection establishment 6/9/201536Unit-4 : Transport Layer

37. Figure 22.17 Four-step connection termination 6/9/201537Unit-4 : Transport Layer

38. Table 22.4 States for TCP StateDescription CLOSEDThere is no connection. LISTENThe server is waiting for calls from the client. SYN-SENTA connection request is sent; waiting for acknowledgment. SYN-RCVDA connection request is received. ESTABLISHEDConnection is established. FIN-WAIT-1 The application has requested the closing of the connection. FIN-WAIT-2The other side has accepted the closing of the connection. TIME-WAITWaiting for retransmitted segments to die. CLOSE-WAITThe server is waiting for the application to close. LAST-ACKThe server is waiting for the last acknowledgment. 6/9/201538Unit-4 : Transport Layer

39. Figure 22.18 State transition diagram 6/9/201539Unit-4 : Transport Layer

40. A sliding window is used to make transmission more efficient as well as to control the flow of data so that the destination does not become overwhelmed with data. TCP’s sliding windows are byte-oriented. Note: 6/9/201540Unit-4 : Transport Layer

41. Figure 22.19 Sender buffer 6/9/201541Unit-4 : Transport Layer

42. Figure 22.20 Receiver window 6/9/201542Unit-4 : Transport Layer

43. Figure 22.21 Sender buffer and sender window 6/9/201543Unit-4 : Transport Layer

44. Figure 22.22 Sliding the sender window 6/9/201544Unit-4 : Transport Layer

45. Figure 22.23 Expanding the sender window 6/9/201545Unit-4 : Transport Layer

46. Figure 22.24 Shrinking the sender window 6/9/201546Unit-4 : Transport Layer

47. In TCP, the sender window size is totally controlled by the receiver window value (the number of empty locations in the receiver buffer). However, the actual window size can be smaller if there is congestion in the network. Note: 6/9/201547Unit-4 : Transport Layer

48. Some points about TCP’s sliding windows: Note: The source does not have to send a full window’s worth of data. The size of the window can be increased or decreased by the destination. The destination can send an acknowledgment at any time. 6/9/201548Unit-4 : Transport Layer

49. Figure 22.25 Lost segment 6/9/201549Unit-4 : Transport Layer

50. Figure 22.26 Lost acknowledgment 6/9/201550Unit-4 : Transport Layer

51. Figure 22.27 TCP timers 6/9/201551Unit-4 : Transport Layer

52. Multiplexing 6/9/201552Unit-4 : Transport Layer

53. Figure 6.1 Dividing a link into channels 6/9/201553Unit-4 : Transport Layer

54. Figure 6.2 Categories of multiplexing 6/9/201554Unit-4 : Transport Layer

55. 6.1 FDM Multiplexing Process Demultiplexing Process The Analog Hierarchy Other Applications of FDM Implementation 6/9/201555Unit-4 : Transport Layer

56. Figure 6.3 FDM 6/9/201556Unit-4 : Transport Layer

57. FDM is an analog multiplexing technique that combines signals. Note: 6/9/201557Unit-4 : Transport Layer

58. Figure 6.4 FDM process 6/9/201558Unit-4 : Transport Layer

59. Figure 6.5 FDM demultiplexing example 6/9/201559Unit-4 : Transport Layer

60. Example 1 Assume that a voice channel occupies a bandwidth of 4 KHz. We need to combine three voice channels into a link with a bandwidth of 12 KHz, from 20 to 32 KHz. Show the configuration using the frequency domain without the use of guard bands. Solution Shift (modulate) each of the three voice channels to a different bandwidth, as shown in Figure 6.6. 6/9/201560Unit-4 : Transport Layer

61. Figure 6.6 Example 1 6/9/201561Unit-4 : Transport Layer

62. Example 2 Five channels, each with a 100-KHz bandwidth, are to be multiplexed together. What is the minimum bandwidth of the link if there is a need for a guard band of 10 KHz between the channels to prevent interference? Solution For five channels, we need at least four guard bands. This means that the required bandwidth is at least 5 x 100 + 4 x 10 = 540 KHz, as shown in Figure 6.7. 6/9/201562Unit-4 : Transport Layer

63. Figure 6.7 Example 2 6/9/201563Unit-4 : Transport Layer

64. Example 3 Four data channels (digital), each transmitting at 1 Mbps, use a satellite channel of 1 MHz. Design an appropriate configuration using FDM Solution The satellite channel is analog. We divide it into four channels, each channel having a 250-KHz bandwidth. Each digital channel of 1 Mbps is modulated such that each 4 bits are modulated to 1 Hz. One solution is 16- QAM modulation. Figure 6.8 shows one possible configuration. 6/9/201564Unit-4 : Transport Layer

65. Figure 6.8 Example 3 6/9/201565Unit-4 : Transport Layer

66. Figure 6.9 Analog hierarchy 6/9/201566Unit-4 : Transport Layer

67. Example 4 The Advanced Mobile Phone System (AMPS) uses two bands. The first band, 824 to 849 MHz, is used for sending; and 869 to 894 MHz is used for receiving. Each user has a bandwidth of 30 KHz in each direction. The 3- KHz voice is modulated using FM, creating 30 KHz of modulated signal. How many people can use their cellular phones simultaneously? Solution Each band is 25 MHz. If we divide 25 MHz into 30 KHz, we get 833.33. In reality, the band is divided into 832 channels. 6/9/201567Unit-4 : Transport Layer

68. 6.2 WDM Wave Division Multiplexing 6/9/201568Unit-4 : Transport Layer

69. Figure 6.10 WDM 6/9/201569Unit-4 : Transport Layer

70. WDM is an analog multiplexing technique to combine optical signals. Note: 6/9/201570Unit-4 : Transport Layer

71. Figure 6.11 Prisms in WDM multiplexing and demultiplexing 6/9/201571Unit-4 : Transport Layer

72. 6.3 TDM Time Slots and Frames Interleaving Synchronizing Bit Padding Digital Signal (DS) Service T Lines Inverse TDM More TDM Applications 6/9/201572Unit-4 : Transport Layer

73. Figure 6.12 TDM 6/9/201573Unit-4 : Transport Layer

74. TDM is a digital multiplexing technique to combine data. Note: 6/9/201574Unit-4 : Transport Layer

75. Figure 6.13 TDM frames 6/9/201575Unit-4 : Transport Layer

76. Example 5 Four 1-Kbps connections are multiplexed together. A unit is 1 bit. Find (1) the duration of 1 bit before multiplexing, (2) the transmission rate of the link, (3) the duration of a time slot, and (4) the duration of a frame? Solution We can answer the questions as follows: 1. The duration of 1 bit is 1/1 Kbps, or 0.001 s (1 ms). 2. The rate of the link is 4 Kbps. 3. The duration of each time slot 1/4 ms or 250  s. 4. The duration of a frame 1 ms. 6/9/201576Unit-4 : Transport Layer

77. In a TDM, the data rate of the link is n times faster, and the unit duration is n times shorter. Note: 6/9/201577Unit-4 : Transport Layer

78. Figure 6.14 Interleaving 6/9/201578Unit-4 : Transport Layer

79. Example 6 Four channels are multiplexed using TDM. If each channel sends 100 bytes/s and we multiplex 1 byte per channel, show the frame traveling on the link, the size of the frame, the duration of a frame, the frame rate, and the bit rate for the link. Solution The multiplexer is shown in Figure 6.15. 6/9/201579Unit-4 : Transport Layer

80. Figure 6.15 Example 6 6/9/201580Unit-4 : Transport Layer

81. Example 7 A multiplexer combines four 100-Kbps channels using a time slot of 2 bits. Show the output with four arbitrary inputs. What is the frame rate? What is the frame duration? What is the bit rate? What is the bit duration? Solution Figure 6.16 shows the output for four arbitrary inputs. 6/9/201581Unit-4 : Transport Layer

82. Figure 6.16 Example 7 6/9/201582Unit-4 : Transport Layer

83. Figure 6.17 Framing bits 6/9/201583Unit-4 : Transport Layer

84. Example 8 We have four sources, each creating 250 characters per second. If the interleaved unit is a character and 1 synchronizing bit is added to each frame, find (1) the data rate of each source, (2) the duration of each character in each source, (3) the frame rate, (4) the duration of each frame, (5) the number of bits in each frame, and (6) the data rate of the link. Solution See next slide. 6/9/201584Unit-4 : Transport Layer

85. Solution (continued) We can answer the questions as follows: 1. The data rate of each source is 2000 bps = 2 Kbps. 2. The duration of a character is 1/250 s, or 4 ms. 3. The link needs to send 250 frames per second. 4. The duration of each frame is 1/250 s, or 4 ms. 5. Each frame is 4 x 8 + 1 = 33 bits. 6. The data rate of the link is 250 x 33, or 8250 bps. 6/9/201585Unit-4 : Transport Layer

86. Example 9 Two channels, one with a bit rate of 100 Kbps and another with a bit rate of 200 Kbps, are to be multiplexed. How this can be achieved? What is the frame rate? What is the frame duration? What is the bit rate of the link? Solution We can allocate one slot to the first channel and two slots to the second channel. Each frame carries 3 bits. The frame rate is 100,000 frames per second because it carries 1 bit from the first channel. The frame duration is 1/100,000 s, or 10 ms. The bit rate is 100,000 frames/s x 3 bits/frame, or 300 Kbps. 6/9/201586Unit-4 : Transport Layer

87. Figure 6.18 DS hierarchy 6/9/201587Unit-4 : Transport Layer

88. Table 6.1 DS and T lines rates ServiceLine Rate (Mbps) Voice Channels DS-1T-11.54424 DS-2 DS-2T-2 6.312 6.31296 DS-3 DS-3T-3 44.736 44.736 672 672 DS-4T-4274.1764032 6/9/201588Unit-4 : Transport Layer

89. Figure 6.19 T-1 line for multiplexing telephone lines 6/9/201589Unit-4 : Transport Layer

90. Figure 6.20 T-1 frame structure 6/9/201590Unit-4 : Transport Layer

91. Table 6.2 E line rates E Line Rate (Mbps) Voice Channels E-1 2.048 2.048 30 30 E-2 8.448 8.448 120 120 E-3 34.368 34.368 480 480 E-4139.2641920 6/9/201591Unit-4 : Transport Layer

92. Figure 6.21 Multiplexing and inverse multiplexing 6/9/201592Unit-4 : Transport Layer

93. Client-Server Model: Socket Interface 6/9/201593Unit-4 : Transport Layer

94. 24.1 Client-Server Model Relationship Concurrency Processes 6/9/201594Unit-4 : Transport Layer

95. Figure 24.1 Client-server model 6/9/201595Unit-4 : Transport Layer

96. Figure 24.2 Client-server relationship 6/9/201596Unit-4 : Transport Layer

97. Figure 24.3 Connectionless iterative server 6/9/201597Unit-4 : Transport Layer

98. Figure 24.4 Connection-oriented concurrent server 6/9/201598Unit-4 : Transport Layer

99. 24.2 Socket Interface Sockets Connectionless Iterative Server Connection-Oriented Server 6/9/201599Unit-4 : Transport Layer

100. Figure 24.5 Socket structure 6/9/2015100Unit-4 : Transport Layer

101. Figure 24.6 Socket types 6/9/2015101Unit-4 : Transport Layer

102. Figure 24.7 Socket interface for connectionless iterative server 6/9/2015102Unit-4 : Transport Layer

103. Figure 24.8 Socket interface for connection-oriented concurrent server 6/9/2015103Unit-4 : Transport Layer

104. Congestion Control and Quality of Service 6/9/2015104Unit-4 : Transport Layer

105. 23.1 Data Traffic Traffic Descriptor Traffic Profiles 6/9/2015105Unit-4 : Transport Layer

106. Figure 23.1 Traffic descriptors 6/9/2015106Unit-4 : Transport Layer

107. Figure 23.2 Constant-bit-rate traffic 6/9/2015107Unit-4 : Transport Layer

108. Figure 23.3 Variable-bit-rate traffic 6/9/2015108Unit-4 : Transport Layer

109. Figure 23.4 Bursty traffic 6/9/2015109Unit-4 : Transport Layer

110. 23.2 Congestion Network Performance 6/9/2015110Unit-4 : Transport Layer

111. Figure 23.5 Incoming packet 6/9/2015111Unit-4 : Transport Layer

112. Figure 23.6 Packet delay and network load 6/9/2015112Unit-4 : Transport Layer

113. Figure 23.7 Throughput versus network load 6/9/2015113Unit-4 : Transport Layer

114. 23.3 Congestion Control Open Loop Closed Loop 6/9/2015114Unit-4 : Transport Layer

115. 23.4 Two Examples Congestion Control in TCP Congestion Control in Frame Relay 6/9/2015115Unit-4 : Transport Layer

116. TCP assumes that the cause of a lost segment is due to congestion in the network. Note: 6/9/2015116Unit-4 : Transport Layer

117. If the cause of the lost segment is congestion, retransmission of the segment does not remove the cause—it aggravates it. Note: 6/9/2015117Unit-4 : Transport Layer

118. Figure 23.8 Multiplicative decrease 6/9/2015118Unit-4 : Transport Layer

119. Figure 23.9 BECN 6/9/2015119Unit-4 : Transport Layer

120. Figure 23.10 FECN 6/9/2015120Unit-4 : Transport Layer

121. Figure 23.11 Four cases of congestion 6/9/2015121Unit-4 : Transport Layer

122. 23.5 Quality of Service Flow Characteristics Flow Classes 6/9/2015122Unit-4 : Transport Layer

123. 23.6 Techniques to Improve QoS Scheduling Traffic Shaping Resource Reservation Admission Control 6/9/2015123Unit-4 : Transport Layer

124. Figure 23.12 Flow characteristics 6/9/2015124Unit-4 : Transport Layer

125. Figure 23.13 FIFO queue 6/9/2015125Unit-4 : Transport Layer

126. Figure 23.14 Priority queuing 6/9/2015126Unit-4 : Transport Layer

127. Figure 23.15 Weighted fair queuing 6/9/2015127Unit-4 : Transport Layer

128. Figure 23.16 Leaky bucket 6/9/2015128Unit-4 : Transport Layer

129. Figure 23.17 Leaky bucket implementation 6/9/2015129Unit-4 : Transport Layer

130. A leaky bucket algorithm shapes bursty traffic into fixed-rate traffic by averaging the data rate. It may drop the packets if the bucket is full. Note: 6/9/2015130Unit-4 : Transport Layer

131. Figure 23.18 Token bucket 6/9/2015131Unit-4 : Transport Layer

132. The token bucket allows bursty traffic at a regulated maximum rate. Note: 6/9/2015132Unit-4 : Transport Layer

133. 23.7 Integrated Services Signaling Flow Specification Admission Service Classes RSVP 6/9/2015133Unit-4 : Transport Layer

134. Integrated Services is a flow-based QoS model designed for IP. Note: 6/9/2015134Unit-4 : Transport Layer

135. Figure 23.19 Path messages 6/9/2015135Unit-4 : Transport Layer

136. Figure 23.20 Resv messages 6/9/2015136Unit-4 : Transport Layer

137. Figure 23.21 Reservation merging 6/9/2015137Unit-4 : Transport Layer

138. Figure 23.22 Reservation styles 6/9/2015138Unit-4 : Transport Layer