COMMUNICATION NETWORKS ECE 422 DATA COMMUNICATION & COMPUTER - PowerPoint PPT Presentation

MULTIPLE ACCESS IN COMMUNICATION NETWORKS ECE 422 – DATA COMMUNICATION & COMPUTER NETWORKS Wednesday, 12 February 2020 1

WHERE ARE WE IN THE SYLLABUS? Course Content: Introduction: Overview of Data Communications and Networking. Physical Layer: Analog and Digital, Analog Signals, Digital Signals, Analog versus Digital, Data Rate Limits, Transmission Impairment, More about signals. Digital Transmission: Line coding, Block coding, Sampling, Transmission mode. Analog Transmission: Modulation of Digital Data; Telephone modems, modulation of Analog signals. Multiplexing: FDM, TDM, CDM. Transmission Media: Guided Media, Unguided media (wireless). Data Link Layer: Error Detection and correction - Types of Errors, Detection, Error Correction; Data Link Control and Protocols-Flow and Error Control, Stop-and-wait ARQ. Go-Back-N ARQ, Selective Repeat ARQ, HDLC. Point-to-Point Access- Point – to-Point Protocol (PPP), PPP Stack, Multiple Access Random Access, Controlled Access, Channelization. Network Layer: Host to Host Delivery: Internetworking, addressing and Routing Network Layer Protocols: ARP, IPV4, ICMP, IPV6 and ICMPV6 Transport Layer: Process to Process Delivery: UDP; TCP congestion control and Quality of service. Application Layer: Client Server Model, Socket Interface, Domain Name System (DNS): Electronic Mail (SMTP) and file transfer (FTP) HTTP and WWW. Local area Network: Ethernet - Traditional Ethernet, Fast Ethernet, Gigabit Ethernet; Token bus, token ring; Wireless LANs - IEEE 802.11, Bluetooth virtual circuits: Frame Relay and ATM. Industrial Communication and Control Networks: Transmission methods, Network topology, Contemporary networks – Profibus, Controller Area Network (CAN), DeviceNet, CANopen, Actuator Sensor Interface (AS-1),Industrial Ethernet. 2

DATA LINK The Data Link Layer 1. Makes the physical layer appear error- free to the upper layer. 2. Functions include: a) Physical addressing b) Flow-control c) Error Control d) Access Control Suppose the Physical Media is Multipoint (Ethernet) or Broadcast (Terrestrial Radio or Satellite)? 3

NEED FOR RULES FOR MEETINGS OR ASSEMBLIES & APPLICATION TO COMPUTER NETWORKS For effective Communication, we require that no two people: 1. Should speak at the same time, 2. Interrupt each other, 3. Monopolize the discussion. We can therefore introduce control measure as follows: a) Have a chairman selecting who speaks and limiting speech time, b) Use a microphone automated switch-off to place a limit on speech time, The alternative, is to have a free for all meeting Computer Networks Apply the Same Rules when many Computers/Devices share the (Shouting match!) same communication Media 4

CATEGORIES OF MULTIPLE ACCESS PROTOCOLS Multiple Access Protocols Random Access Controlled Access Channelization 1. Aloha 1. Reservation 1. FDMA 2. CSMA 2. Polling 2. TDMA 3. CSMA/CD 3. Token Passing 3. CDMA 4. CSMA/CA Covered in ECE 416 5

RANDOM ACCESS 1. Each station has access to the media without being controlled by another station. 2. If more than one station transmit at the same time, then there is an access conflict- collision and the frames are either discarded or modified. 3. A random access protocol must address the B C A following questions: a) When can the station access the medium? b) What can the station do if the medium is busy? c) How can the station determine the success or failure of the transmission? E F d) What can the station do if there is an access conflict? 6

HISTORY OF RANDOM ACCESS 1. ALOHA was the earliest random access method. 2. It was developed at the University of Hawaii in early 1970. 3. It is a system for coordinating and arbitrating access to a shared communication Networks channel. 4. It was designed for a radio (wireless) LAN, but it can be used on any shared medium. 5. There are two different types of ALOHA: a) Pure ALOHA b) Slotted ALOHA 7

ILLUSTRATION OF PURE ALOHA 1. Assume each station sends two frames. 2. In the diagram, there are a total of eight frames send on the shared medium. 3. Some of these frames collide because multiple frames are in contention for the shared channel. 4. The Receiver acknowledges each correctly received frame. 5. If the Transmitting station does not receive ACK before timeout, it retransmits the frame. 8

PURE ALOHA FLOWCHART 𝑶 𝒄𝒈 = 𝟏 Transmitter Sends Receiver Listens is is 𝑶 𝒄𝒈 ≥ 𝑶 𝑴𝒋𝒏? 𝑶 𝒄𝒈 = 𝑶 𝒄𝒈 + 𝟐 9

VULNERABLE TIME FOR PURE ALOHA Vulnerable time = 2 × 𝑢 If a station wants to transmit Once the station has started data, its transmission may transmitting data, its damage the transmission of transmission may be disrupted another packet which over the t seconds if another commenced transmission up to t station starts transmitting. 10 seconds before.

EXAMPLE - PURE ALOHA VUNERABILITY A pure ALOHA network transmits 500-bit frames on a shared channel of 1 Mbps. Under what circumstances will this frame be collision-free? SOLUTION 𝐺𝑠𝑏𝑛𝑓 𝑇𝑗𝑨𝑓 500 • Frame Transmission time, 𝑢 = 𝑈𝑠𝑏𝑜𝑡𝑛𝑗𝑡𝑡𝑗𝑝𝑜 𝑇𝑞𝑓𝑓𝑒 = 1,000,000 = 500𝜈𝑡𝑓𝑑 • The Vulnerable time = 2𝑢 = 2 × 500 = 1,000𝜈 sec = 1 𝑛𝑡𝑓𝑑 • Therefore there will be no collision if no station transmits in a period of 1 msec, i.e 0.5 msec before or 0.5 msec after the station starts transmitting. B C A t t 11 0.5 ms 0.5 ms

PURE ALOHA THROUGHPUT 1. Let us call G the average number of frames generated by a pure ALOHA system during a frame transmission time i.e attempts per packet. 2. Then the average number of successful transmissions, S is given by: 𝑇 = 𝐻𝑓 −2𝐻 3. The maximum throughput occurs when G = 0.5, i.e 𝑇 𝑛𝑏𝑦 = 0.5𝑓 −1 = 0.184 12

EXAMPLE QUESTION: A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the throughput if the system -all stations together- produces the average of: a) 1000 frames per second b) 500 frames per second c) 250 frames per second 13

SOLUTION (a) 200 The frame transmission time is 200,000 or 1 ms. (a) If the system creates 1000 frames per second, this is 1 frame per second. The load, G = 1 Or S = Ge -2G = 1xe -2 = 0.135 or 13.5% Throughput = 1000 x 0.135 = 135 frames per second 14

SOLUTION (b) 200 The frame transmission time is 200,000 or 1 ms. 1 (a) If the system creates 500 frames per second, this is 2 frame per second. 1 The load, G = 2 Or 1 2 e -1 = 0.184 or 18.4% S = Throughput = 500 x 0.184 = 93 frames per second 15

SOLUTION (c) 200 The frame transmission time is 200,000 or 1 ms. 1 (a) If the system creates 250 frames per second, this is 4 frame per second. 1 The load, G = 4 Or 1 4 e -0.5 = 0.152 or 15.2% S = Throughput = 250 x 0.152 = 36 frames per second 16

SLOTTED ALOHA 1. The efficiency of the pure ALOHA system can be improved by “slotting” time into segments whose duration is exactly equal to the transmission time of a single packet (assuming constant-length packets). 2. In slotted ALOHA, each user starts transmitting only at the beginning of a slot. 3. As a result, when two packets conflict, they will only overlap completely rather than partially, thus increasing channel efficiency. Slot-3 Slot-2 Slot-1 B C A t t 17

COLLISSIONS IN SLOTTED ALOHA • xxx 18

VULNERABILITY OF SLOTTED ALOHA Vulnerable time = 𝑈 19

THROUGHPUT IN SLOTTED ALOHA 1. Let us call G the average number of frames generated by a pure ALOHA system during one frame transmission time. 2. Then the average number of successful transmissions, S is given by: 𝑇 = 𝐻𝑓 −𝐻 3. The maximum throughput occurs when G = 1, i.e 𝑇 𝑛𝑏𝑦 = 1𝑓 −2 = 0.368 20

PURE ALOHA Vs SLOTTED ALOHA THROUGHPUT 21

EXAMPLE A slotted ALOHA network transmits 200-bit frames using a shared channel with a 200-kbps bandwidth. Find the throughput if the system (all stations together) produces a) 1000 frames per second b) 500 frames per second c) 250 frames per second 22

ANSWER (a) 200 The frame transmission time is 200,000 or 1 ms (a) System produces 1000 frames per second Let us call G the average number of frames generated In this case G = 0.001 x 1,000 = 1 S =G x e -G or S =0.368 (or 36.8 percent) (was 13.5% for pure Aloha!) Throughput is therefore 1000 x 0.0368 =368 frames. Therefore only 368 out of 1000 frames will probably survive. 23

ANSWER (b) 200 The frame transmission time is 200,000 or 1 ms (b) System produces 500 frames per second 1 In this case G = 500 x 0.001 = 2 2 𝑓 − 1 1 S =G x e -G = 2 =0.303 ( or 30.3 percent) (was 18.4 % for pure Aloha!) Throughput is 500 x 0.0303 =151. Only 151 frames out of 500 will probably survive. 24

CARRIER SENSE MULTIPLE ACCESS 1. Carrier sense multiple access (CSMA) requires that each station first listen to the medium (or check the state of the medium) before sending. 2. It is based on the principle "sense before transmit" or "listen before talk.“ 3. Compared with previous cases of ALOHA, it reduces the possibility of collision, but it does not eliminate it. 25