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Multiple Access Chapter 12

Multiple Access Chapter 12. Application Presentation Session Transport Network Data Link Physical. Recall - OSI Reference Model. Node-to-node communications: Framing, Error Control, Flow Control, Physical Addressing, Access Control. Data Comm Layers.

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Multiple Access Chapter 12

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  1. Multiple AccessChapter 12 Fall 2009

  2. Application Presentation Session Transport Network Data Link Physical Recall - OSI Reference Model Node-to-node communications: Framing, Error Control, Flow Control, Physical Addressing, Access Control Data Comm Layers Fall 2009

  3. Major Data Link Tasks Node-to-node communication – design & procedures for communication between two adjacent nodes • Framing – the data link layer divides or organizes the raw bits provided by the physical layer into frames based on the particular technology – also adds a header to frame that specifies the address of the Tx and Rx • Flow Control – the data link layer make sure that the rate at which the Tx sends bits and frames to the Rx is a rate the Rx can handle. • Error Control - the data link layer adds some reliability to the physical layer by detecting and retransmitting damaged, duplicate or lost frames • Access Control – the data link layer determines which node has access to or control of the transport medium at any given time (sharing the transport). Fall 2009

  4. Two Functionality-oriented Sublayers Deals with the creating a reliable link Flow Control, Error Control Called Logical Link Control Sublayer Deals with resolving access to a shared medium Called media access control (MAC) sublayer Fall 2009

  5. Taxonomy of Multiple-access Protocols Multiple stations can access the link at the same time by the sharing of bandwidth (via time, frequency or code) No station has control over another for the link; random access; collision can occur Stations coordinate together in deciding which station has access to the link Fall 2009

  6. Frames in a Pure ALOHA Network ALOHA was developed at the University of Hawaii – developed for wireless radio LAN Stations can send anytime they want to – as a result, collision occur if 2 or more send at the same time If a collision occurs, frames are destroyed – needs to be re-transmitted – random wait time for each to re-send to avoid SECOND collision (called back-off time, TB) Fall 2009

  7. Procedure for pure ALOHA protocol If a collision occurs, frames are destroyed – needs to be re-transmitted – random wait time for each to re-send to avoid SECOND collision (called back-off time, TB) The time a station should wait before determining if there is a problem is the “worst case” wait time – the time it takes a frame to travel back-and-forth between the furthest separated stations A random “wait to resend time” or back-off time is generated using one of these formulas If the ACK comes within the wait time, great, if NOT, increment the “transmission attempt” count If the max # of retransmission attempts, Kmax, is reached, the station must give up and retry later Fall 2009

  8. ALOHA Vulnerable Time and Throughput Vulnerable Time – the time there is a possibility of a collision – given Tfr is the time to send a fixed-length frame Vulnerable Time = 2 × Tfr Throughput – average number of successful transmissions - where G equals the average number of frames generated by the system during one frame transmission time Throughput = G × e-2G Fall 2009

  9. Frames in a slotted ALOHA network With Pure ALOHA, the stations can send anytime, and as a result, collision can occur. With Slotted ALOHA, the time is divided into slots and the stations are forced to send ONLY at the beginning of the time slot. Fall 2009

  10. Vulnerable Time = Tfr Throughput = G x e-G Fall 2009

  11. Space/time Model of the Collision in CSMA In minimizing the chance of a collision and increasing performance, the Carrier Sense Multiple Access (CSMA) method was developed. For the CSMA method, the station senses or listens to the medium before sending – if it senses a signal already on the medium it waits before sensing again – if it doesn’t sense a signal, it sends. At t2 (very short time right after t1), C senses transport and senses no signal, therefore sends signal in opposite direction At t1, B senses transport and senses no signal, therefore sends signal Collision first occurs here Fall 2009

  12. Vulnerable Time in CSMA At t1, A sends frame The first bit of the frame reaches B, C and D at these points in time Therefore, if B, C or D send a frame in the opposite direction with in this time (gray area), there would be a collision –vulnerable time This is also the time it takes the first to propagate the entire transport Fall 2009

  13. Behavior of three persistence methods After a station senses, what happens if the channel is busy or idle ? The station continuously senses while the channel is busy, as soon as the channel is idle, it sends. Higher probability collision can occur because some other station could have sent at the same time. If the channel is busy, the station waits a random amount of time and retry – if idle, it sends. Lower probability of collision because no 2 stations will have the same random wait time – however, less efficient because the station could be waiting when it could be sending The station continuously senses while the channel is busy, as soon as the channel is idle, (1) with probability p, it sends its frameOR (2) with probability q = 1 – p, the station waits for the beginning of the next time slot and checks the line again – if the line is idle, it goes to step (1), if busy, it uses a back-off procedure Fall 2009

  14. Collision of the first bit in CSMA/CD The Carrier Sense Multiple Access (CSMA) method does NOT specify a procedure when a collision occurs - Carrier Sense Multiple Access with Collision Detection (CSMA/CD) does At t1, A sends a signal At t2, C sends a signal a collision occurs and a jam signal is sent to inform the collision participants At t4, A receives jam signal and aborts At t3, C receives jam signal and aborts Fall 2009

  15. Timing in CSMA/CA In a wireless case (versus wired case), stations would need to avoid collisions. So the Carrier Sense Multiple Access with collision detection (CSMA/CD) method was adapted to avoid collisions Carrier Sense Multiple Access with Collision Avoidance (CSMA/CA). The station continuously senses while the channel is busy, as soon as the channel is idle, it waits an IFS (interframe space) amount of time – enough time for the first bit to reach depending on the distance of the other station After the IFS, it waits an amount of time equal to various increments of “contention time” – contention time is broken down into a randomly generated number of time slots Fall 2009

  16. The shorter the IFS, the higher priority the station will have • if channel is idle after IFS time, station sends • if channel is busy after IFS time, the additional contention time wait is 1 time slot and a retry • if the channel is still busy, it then waits additional 2 time slots for the contention time, • if the channel is still busy, it then waits additional 4 time slots for the contention time, Fall 2009

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  18. Controlled Access • In controlled access, the stations consult one another to find which station has the right to send. • A station cannot send unless it has been authorized by other stations. • Three popular controlled-access methods. Reservation Polling Token Passing Fall 2009

  19. Reservation Access Method In the reservation method, a station needs to make a reservation before sending data. Time is divided into intervals and in each interval, a reservation frame precedes the data In first time interval, stations 1, 3 and 4 made reservations in that order In second time interval, only station 1 made a reservation #slots in the reservation frame equals the # of stations Fall 2009

  20. Select and Poll Functions in Polling Access Method One device is the primary and the other devices are secondary All data exchanges made through the primary Primary dictates which device can use the transport If primary wants to send data, it select (notify) the secondary it will send data If primary wants to receive data, it polls (ask) the secondary if it wants to send data or not Fall 2009

  21. Token Ring LAN • Token Ring is a protocol defined by IEEE 802.5 • Use a token passing ACCESS method • Token Passing Method • During idle times (network not being used), a token circulates • The token is passed station to station until a station needs to send data • When the station sends it’s data, it holds the token • The data (or frame) circulates and get re-generated by each station • The Rx COPIES the frame (based on destination address) • The RX sends data back to the original Tx • Token is then released to circulate Fall 2009

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  24. Logical Ring and Physical Topology in token-passing access method • Doesn’t have to be physically laid out as a ring Fall 2009

  25. CHANNELIZATION Channelizationis a multiple-access method in which the available bandwidth of a link is shared in time, frequency, or through code, between different stations. Three channelization protocols. Frequency-Division Multiple Access (FDMA) Time-Division Multiple Access (TDMA) Code-Division Multiple Access (CDMA) Fall 2009

  26. Frequency-division multiple access (FDMA) In FDMA, the available bandwidth of the common channel is divided into bands that are separated by guard bands. Fall 2009

  27. Time-division multiple access (TDMA) In TDMA, the bandwidth is just one channel that is timeshared between different stations. Fall 2009

  28. Code-Division Multiple Access Conceptually – if any pair of devices “speak” a different language – then there could be multiple pairs of devices sharing the transport at the same time if the language of each pair is different In CDMA, one channel carries all transmissions simultaneously. Fall 2009

  29. Communication with Code • Four stations, 1-4, are connected to the same channel • Let data for channel x is dx and the code for channel x is cx • Also, if we multiply the code, cx by another code cy, the result will be 0 • And if we multiply the code, cx, by itself, the result will be 4. With these properties, the four stations can send data on a common channel by the station multiplying its data by its code and summing them together Fall 2009

  30. With this approach, if Station 2 wanted to hear what Station 1 is saying, Station 2 would multiply this by Station 1’s code, c1, then divide by 4 Fall 2009

  31. (d1*c1+d2*c2+d3*c3+d4*c4)*c1= (d1*c1*c1+d2*c2*c1+d3*c3*c1+d4*c4 *c1)*= 4*d1+0+0+0 • Then [4*d1]/4 = d1 Fall 2009

  32. Chip Sequences • Actually the code is a sequence of numbers called Chips • Each station is assigned a unique chip • The number of elements in each chip is equal to the number of stations. Example of chip codes for 4 stations Fall 2009

  33. You can multiply a chip sequence by a number (ie. 3 * [+1 +1 -1 -1] = [+3 +3 -3 -3] ) • You can multiply 2 equal chip sequences (ie. [+1 +1 -1 -1]* [+1 +1 -1 -1] = • 1+1+ 1+1 = 4 ) • You can multiply 2 different chip sequences (ie. [+1 +1 -1 -1]* [+1 +1 +1 +1] = • 1+1- 1- 1 = 0 ) • You can add 2 different chip sequences (ie. [+1 +1 -1 -1]+ [+1 +1 +1 +1] = • [+2 +2 0 0] ) Fall 2009

  34. Data representation in CDMA • A 0 bit is represented by -1 • A 1 bit is represented by +1 Fall 2009

  35. Sharing Channel in CDMA Given the code concept and properties, and data representation explained in the previous ppt slide, THE ACTUAL WAY CDMA SHARES THE CHANNEL IS ILLUSTRATED BELOW: Fall 2009

  36. General rule and Examples of Creating Walsh Tables The chip sequence generated for each station is NOT a random choice A Walsh Table is used to generate the appropriate chip sequence for each station. The rules are below: Each ROW, is a sequence of chips If we know the table for N sequences, we can create the table for 2N sequences Complement of WN W2 Complement of W1 Complement of W2 W1 Table for 2 is and therefore, the Table for 2x2 is Table for 1 is and therefore, the Table for 2x1 is The number of sequences in a Walsh table needs to be a power of 2 Fall 2009

  37. Example 1 Given W1 is Find the chips for a network with a. Two stations b. Four stations Solution We can use the rows of W2 and W4: a. For a two-station network, we have [+1 +1] and [+1 −1]. b. For a four-station network we have [+1 +1 +1 +1], [+1 −1 +1 −1], [+1 +1 −1 −1], and [+1 −1 −1 +1]. Fall 2009

  38. Example 2 What is the number of sequences if we have 90 stations in our network? Solution The number of sequences needs to be 2m (a power of 2). We need to choose m = 7 and N = 27 or 128. We can then use 90 of the sequences as the chips. Fall 2009

  39. Example 3 Prove that a receiving station can get the data sent by a specific sender if it multiplies the entire data on the channel by the sender’s chip code and then divides it by the number of stations. Solution Let us prove this for the first station, using our previous four-station example. We can say that the data on the channelD = (d1 ⋅ c1 + d2 ⋅ c2 + d3 ⋅ c3 + d4 ⋅ c4). The receiver which wants to get the data sent by station 1 multiplies these data by c1. When we divide the result by N, we get d1 . Fall 2009

  40. Exam 2 Outline • The exam will cover chapters 7, 8, 10, 11 and 12. • The exam will be 75 minutes • The exam will have fill-in questions, short problems and problems • The exam will be open book (closed notes, closed lecture notes) • Bring a calculator (no PDA or Laptop) Fall 2009

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