CS 1302 Computer Networks — Unit - 4 — — Transport Layer —

CS 1302 Computer Networks — Unit - 4 — — Transport Layer —
Text Book Behrouz .A. Forouzan, “Data communication and Networking”, Tata McGrawHill, 2004 4/12/2019 Unit-4 : Transport Layer

Unit-4 : Transport Layer
4/12/2019 Unit-4 : Transport Layer

Position of transport layer 4/12/2019 Unit-4 : Transport Layer

Transport layer duties 4/12/2019 Unit-4 : Transport Layer

Chapters Chapter 22 Process-to-Process Delivery Chapter 23 Congestion Control and QoS 4/12/2019 Unit-4 : Transport Layer

Process-to-Process Delivery: UDP and TCP
Chapter 22 Process-to-Process Delivery: UDP and TCP 4/12/2019 Unit-4 : Transport Layer

Process-to-Process Delivery Client-Server Paradigm Addressing Multiplexing and Demultiplexing Connectionless/Connection-Oriented Reliable/Unreliable 4/12/2019 Unit-4 : Transport Layer

The transport layer is responsible for process-to-process delivery.
Note: The transport layer is responsible for process-to-process delivery. 4/12/2019 Unit-4 : Transport Layer

Figure Types of data deliveries 4/12/2019 Unit-4 : Transport Layer

Figure Port numbers 4/12/2019 Unit-4 : Transport Layer

Figure IP addresses versus port numbers 4/12/2019 Unit-4 : Transport Layer

Figure IANA ranges 4/12/2019 Unit-4 : Transport Layer

Figure Socket address 4/12/2019 Unit-4 : Transport Layer

Figure Multiplexing and demultiplexing 4/12/2019 Unit-4 : Transport Layer

Figure Connection establishment 4/12/2019 Unit-4 : Transport Layer

Figure Connection termination 4/12/2019 Unit-4 : Transport Layer

Figure Error control 4/12/2019 Unit-4 : Transport Layer

UDP Port Numbers User Datagram Applications 4/12/2019 Unit-4 : Transport Layer

Note: UDP is a connectionless, unreliable protocol that has no flow and error control. It uses port numbers to multiplex data from the application layer. 4/12/2019 Unit-4 : Transport Layer

Table Well-known ports used by UDP Port Protocol Description 7 Echo Echoes a received datagram back to the sender 9 Discard Discards any datagram that is received 11 Users Active users 13 Daytime Returns the date and the time 17 Quote Returns a quote of the day 19 Chargen Returns a string of characters 53 Nameserver Domain Name Service 67 Bootps Server port to download bootstrap information 68 Bootpc Client port to download bootstrap information 69 TFTP Trivial File Transfer Protocol 111 RPC Remote Procedure Call 123 NTP Network Time Protocol 161 SNMP Simple Network Management Protocol 162 Simple Network Management Protocol (trap) 4/12/2019 Unit-4 : Transport Layer

Figure User datagram format 4/12/2019 Unit-4 : Transport Layer

Note: The calculation of checksum and its inclusion in the user datagram are optional. 4/12/2019 Unit-4 : Transport Layer

Note: UDP is a convenient transport-layer protocol for applications that provide flow and error control. It is also used by multimedia applications. 4/12/2019 Unit-4 : Transport Layer

TCP Port Numbers Services Sequence Numbers Segments Connection Transition Diagram Flow and Error Control Silly Window Syndrome 4/12/2019 Unit-4 : Transport Layer

Table Well-known ports used by TCP Port Protocol Description 7 Echo Echoes a received datagram back to the sender 9 Discard Discards any datagram that is received 11 Users Active users 13 Daytime Returns the date and the time 17 Quote Returns a quote of the day 19 Chargen Returns a string of characters 20 FTP, Data File Transfer Protocol (data connection) 21 FTP, Control File Transfer Protocol (control connection) 23 TELNET Terminal Network 25 SMTP Simple Mail Transfer Protocol 53 DNS Domain Name Server 67 BOOTP Bootstrap Protocol 79 Finger 80 HTTP Hypertext Transfer Protocol 111 RPC Remote Procedure Call 4/12/2019 Unit-4 : Transport Layer

Figure Stream delivery 4/12/2019 Unit-4 : Transport Layer

Figure Sending and receiving buffers 4/12/2019 Unit-4 : Transport Layer

Figure TCP segments 4/12/2019 Unit-4 : Transport Layer

Example 1 Imagine a TCP connection is transferring a file of 6000 bytes. The first byte is numbered 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) 4/12/2019 Unit-4 : Transport Layer

Note: The bytes of data being transferred in each connection are numbered by TCP. The numbering starts with a randomly generated number. 4/12/2019 Unit-4 : Transport Layer

Note: The value of the sequence number field in a segment defines the number of the first data byte contained in that segment. 4/12/2019 Unit-4 : Transport Layer

Note: 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. 4/12/2019 Unit-4 : Transport Layer

Figure TCP segment format 4/12/2019 Unit-4 : Transport Layer

Figure Control field 4/12/2019 Unit-4 : Transport Layer

Table Description of flags in the control field Flag Description URG The value of the urgent pointer field is valid. ACK The value of the acknowledgment field is valid. PSH Push the data. RST The connection must be reset. SYN Synchronize sequence numbers during connection. FIN Terminate the connection. 4/12/2019 Unit-4 : Transport Layer

Figure Three-step connection establishment 4/12/2019 Unit-4 : Transport Layer

Figure Four-step connection termination 4/12/2019 Unit-4 : Transport Layer

Table States for TCP State Description CLOSED There is no connection. LISTEN The server is waiting for calls from the client. SYN-SENT A connection request is sent; waiting for acknowledgment. SYN-RCVD A connection request is received. ESTABLISHED Connection is established. FIN-WAIT-1 The application has requested the closing of the connection. FIN-WAIT-2 The other side has accepted the closing of the connection. TIME-WAIT Waiting for retransmitted segments to die. CLOSE-WAIT The server is waiting for the application to close. LAST-ACK The server is waiting for the last acknowledgment. 4/12/2019 Unit-4 : Transport Layer

Figure State transition diagram 4/12/2019 Unit-4 : Transport Layer

Note: 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. 4/12/2019 Unit-4 : Transport Layer

Figure Sender buffer 4/12/2019 Unit-4 : Transport Layer

Figure Receiver window 4/12/2019 Unit-4 : Transport Layer

Figure Sender buffer and sender window 4/12/2019 Unit-4 : Transport Layer

Figure Sliding the sender window 4/12/2019 Unit-4 : Transport Layer

Figure Expanding the sender window 4/12/2019 Unit-4 : Transport Layer

Figure Shrinking the sender window 4/12/2019 Unit-4 : Transport Layer

Note: 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. 4/12/2019 Unit-4 : Transport Layer

Note: Some points about TCP’s sliding windows: 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. 4/12/2019 Unit-4 : Transport Layer

Figure Lost segment 4/12/2019 Unit-4 : Transport Layer

Figure Lost acknowledgment 4/12/2019 Unit-4 : Transport Layer

Figure TCP timers 4/12/2019 Unit-4 : Transport Layer

Multiplexing 4/12/2019 Unit-4 : Transport Layer

Figure Dividing a link into channels 4/12/2019 Unit-4 : Transport Layer

Figure Categories of multiplexing 4/12/2019 Unit-4 : Transport Layer

6.1 FDM Multiplexing Process Demultiplexing Process The Analog Hierarchy Other Applications of FDM Implementation 4/12/2019 Unit-4 : Transport Layer

Figure FDM 4/12/2019 Unit-4 : Transport Layer

FDM is an analog multiplexing technique that combines signals.
Note: FDM is an analog multiplexing technique that combines signals. 4/12/2019 Unit-4 : Transport Layer

Figure FDM process 4/12/2019 Unit-4 : Transport Layer

Figure FDM demultiplexing example 4/12/2019 Unit-4 : Transport Layer

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. 4/12/2019 Unit-4 : Transport Layer

Figure Example 1 4/12/2019 Unit-4 : Transport Layer

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 x 10 = 540 KHz, as shown in Figure 6.7. 4/12/2019 Unit-4 : Transport Layer

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. 4/12/2019 Unit-4 : Transport Layer

Figure Analog hierarchy 4/12/2019 Unit-4 : Transport Layer

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 In reality, the band is divided into 832 channels. 4/12/2019 Unit-4 : Transport Layer

6.2 WDM Wave Division Multiplexing 4/12/2019 Unit-4 : Transport Layer

Figure WDM 4/12/2019 Unit-4 : Transport Layer

WDM is an analog multiplexing technique to combine optical signals.
Note: WDM is an analog multiplexing technique to combine optical signals. 4/12/2019 Unit-4 : Transport Layer

Figure Prisms in WDM multiplexing and demultiplexing 4/12/2019 Unit-4 : Transport Layer

6.3 TDM Time Slots and Frames Interleaving Synchronizing Bit Padding Digital Signal (DS) Service T Lines Inverse TDM More TDM Applications 4/12/2019 Unit-4 : Transport Layer

Figure TDM 4/12/2019 Unit-4 : Transport Layer

TDM is a digital multiplexing technique to combine data.
Note: TDM is a digital multiplexing technique to combine data. 4/12/2019 Unit-4 : Transport Layer

Figure TDM frames 4/12/2019 Unit-4 : Transport Layer

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 s (1 ms). 2. The rate of the link is 4 Kbps. 3. The duration of each time slot 1/4 ms or 250 ms. 4. The duration of a frame 1 ms. 4/12/2019 Unit-4 : Transport Layer

Note: In a TDM, the data rate of the link is n times faster, and the unit duration is n times shorter. 4/12/2019 Unit-4 : Transport Layer

Figure Interleaving 4/12/2019 Unit-4 : Transport Layer

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. 4/12/2019 Unit-4 : Transport Layer

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. 4/12/2019 Unit-4 : Transport Layer

Figure Framing bits 4/12/2019 Unit-4 : Transport Layer

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. 4/12/2019 Unit-4 : Transport Layer

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 = 33 bits. 6. The data rate of the link is 250 x 33, or 8250 bps. 4/12/2019 Unit-4 : Transport Layer

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. 4/12/2019 Unit-4 : Transport Layer

Figure DS hierarchy 4/12/2019 Unit-4 : Transport Layer

Table 6.1 DS and T lines rates Service Line Rate (Mbps) Voice Channels DS-1 T-1 1.544 24 DS-2 T-2 6.312 96 DS-3 T-3 44.736 672 DS-4 T-4 4032 4/12/2019 Unit-4 : Transport Layer

Figure T-1 line for multiplexing telephone lines 4/12/2019 Unit-4 : Transport Layer

Figure T-1 frame structure 4/12/2019 Unit-4 : Transport Layer

Table 6.2 E line rates E Line Rate (Mbps) Voice Channels E-1 2.048 30 E-2 8.448 120 E-3 34.368 480 E-4 1920 4/12/2019 Unit-4 : Transport Layer

Figure Multiplexing and inverse multiplexing 4/12/2019 Unit-4 : Transport Layer

Client-Server Model: Socket Interface

Client-Server Model Relationship Concurrency Processes 4/12/2019 Unit-4 : Transport Layer

Figure Client-server model 4/12/2019 Unit-4 : Transport Layer

Figure Client-server relationship 4/12/2019 Unit-4 : Transport Layer

Figure Connectionless iterative server 4/12/2019 Unit-4 : Transport Layer

Figure Connection-oriented concurrent server 4/12/2019 Unit-4 : Transport Layer

Socket Interface Sockets Connectionless Iterative Server Connection-Oriented Server 4/12/2019 Unit-4 : Transport Layer

Figure Socket structure 4/12/2019 Unit-4 : Transport Layer

Figure Socket types 4/12/2019 Unit-4 : Transport Layer

Figure Socket interface for connectionless iterative server 4/12/2019 Unit-4 : Transport Layer

Figure Socket interface for connection-oriented concurrent server 4/12/2019 Unit-4 : Transport Layer

Congestion Control and Quality of Service

Data Traffic Traffic Descriptor Traffic Profiles 4/12/2019 Unit-4 : Transport Layer

Figure Traffic descriptors 4/12/2019 Unit-4 : Transport Layer

Figure Constant-bit-rate traffic 4/12/2019 Unit-4 : Transport Layer

Figure Variable-bit-rate traffic 4/12/2019 Unit-4 : Transport Layer

Figure Bursty traffic 4/12/2019 Unit-4 : Transport Layer

Congestion Network Performance 4/12/2019 Unit-4 : Transport Layer

Figure Incoming packet 4/12/2019 Unit-4 : Transport Layer

Figure Packet delay and network load 4/12/2019 Unit-4 : Transport Layer

Figure Throughput versus network load 4/12/2019 Unit-4 : Transport Layer

Congestion Control Open Loop Open Loop Closed Loop 4/12/2019 Unit-4 : Transport Layer

Two Examples Congestion Control in TCP Congestion Control in Frame Relay 4/12/2019 Unit-4 : Transport Layer

Note: TCP assumes that the cause of a lost segment is due to congestion in the network. 4/12/2019 Unit-4 : Transport Layer

Note: If the cause of the lost segment is congestion, retransmission of the segment does not remove the cause—it aggravates it. 4/12/2019 Unit-4 : Transport Layer

Figure Multiplicative decrease 4/12/2019 Unit-4 : Transport Layer

Figure BECN 4/12/2019 Unit-4 : Transport Layer

Figure FECN 4/12/2019 Unit-4 : Transport Layer

Figure Four cases of congestion 4/12/2019 Unit-4 : Transport Layer

Quality of Service Flow Characteristics Flow Classes 4/12/2019 Unit-4 : Transport Layer

Techniques to Improve QoS Scheduling Traffic Shaping Resource Reservation Admission Control 4/12/2019 Unit-4 : Transport Layer

Figure Flow characteristics 4/12/2019 Unit-4 : Transport Layer

Figure FIFO queue 4/12/2019 Unit-4 : Transport Layer

Figure Priority queuing 4/12/2019 Unit-4 : Transport Layer

Figure Weighted fair queuing 4/12/2019 Unit-4 : Transport Layer

Figure Leaky bucket 4/12/2019 Unit-4 : Transport Layer

Figure Leaky bucket implementation 4/12/2019 Unit-4 : Transport Layer

Note: 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. 4/12/2019 Unit-4 : Transport Layer

Figure Token bucket 4/12/2019 Unit-4 : Transport Layer

The token bucket allows bursty traffic at a regulated maximum rate.
Note: The token bucket allows bursty traffic at a regulated maximum rate. 4/12/2019 Unit-4 : Transport Layer

Integrated Services Signaling Flow Specification Admission Service Classes RSVP 4/12/2019 Unit-4 : Transport Layer

Integrated Services is a flow-based QoS model designed for IP.
Note: Integrated Services is a flow-based QoS model designed for IP. 4/12/2019 Unit-4 : Transport Layer

Figure Path messages 4/12/2019 Unit-4 : Transport Layer

Figure Resv messages 4/12/2019 Unit-4 : Transport Layer

Figure Reservation merging 4/12/2019 Unit-4 : Transport Layer

Figure Reservation styles 4/12/2019 Unit-4 : Transport Layer

CS 1302 Computer Networks — Unit - 4 — — Transport Layer —

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CS 1302 Computer Networks — Unit - 4 — — Transport Layer —

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