420 likes | 740 Vues
CHAPTER 6. MEDIUM ACCESS CONTROL PROTOCOL. Figure 6.1 Taxonomy of multiple-access protocols discussed in this chapter. 6-1 RANDOM ACCESS.
E N D
CHAPTER 6 MEDIUM ACCESS CONTROL PROTOCOL
Figure 6.1 Taxonomy of multiple-access protocols discussed in this chapter
6-1 RANDOM ACCESS In random access or contention methods, no station is superior to another station and none is assigned the control over another. No station permits, or does not permit, another station to send. At each instance, a station that has data to send uses a procedure defined by the protocol to make a decision on whether or not to send. Topics discussed in this section: ALOHACarrier Sense Multiple Access Carrier Sense Multiple Access with Collision Detection Carrier Sense Multiple Access with Collision Avoidance
ALOHA • Wireless link to provide data transfer between main campus & remote campuses of University of Hawaii • Simplest solution: just do it • A station transmits whenever it has data to transmit • If more than one frames are transmitted, they interfere with each other (collide) and are lost • If ACK not received within timeout, then a station picks random backoff time (to avoid repeated collision) • Station retransmits frame after backoff time First transmission Retransmission Backoff period B t t0 t0+X t0-X t0+X+2tprop + B t0+X+2tprop Vulnerable period Time-out
Throughput of ALOHA • Collisions are means for coordinating access • Max throughput is rmax=1/2e (18.4%) • Bimodal behavior: • Small G, S≈G • Large G, S↓0 • Collisions can snowball and drop throughput to zero e-2 = 0.184
Example 6.1 The stations on a wireless ALOHA network are a maximum of 600 km apart. If we assume that signals propagate at 3 × 108 m/s, we find Tp = (600 × 105 ) / (3 × 108 ) = 2 ms. Now we can find the value of TB for different values of K . a. For K = 1, the range is {0, 1}. The station needs to| generate a random number with a value of 0 or 1. This means that TB is either 0 ms (0 × 2) or 2 ms (1 × 2), based on the outcome of the random variable.
Example 6.1 (continued) b. For K = 2, the range is {0, 1, 2, 3}. This means that TBcan be 0, 2, 4, or 6 ms, based on the outcome of the random variable. c. For K = 3, the range is {0, 1, 2, 3, 4, 5, 6, 7}. This means that TB can be 0, 2, 4, . . . , 14 ms, based on the outcome of the random variable. d. We need to mention that if K > 10, it is normally set to 10.
Example 6.2 A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the requirement to make this frame collision-free? Solution Average frame transmission time Tfr is 200 bits/200 kbps or 1 ms. The vulnerable time is 2 × 1 ms = 2 ms. This means no station should send later than 1 ms before this station starts transmission and no station should start sending during the one 1-ms period that this station is sending.
Note The throughput for pure ALOHA is S = G × e −2G . The maximum throughput Smax = 0.184 when G= (1/2).
Example 6.3 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 a. 1000 frames per second b. 500 frames per second c. 250 frames per second. Solution The frame transmission time is 200/200 kbps or 1 ms. a. If the system creates 1000 frames per second, this is 1 frame per millisecond. The load is 1. In this case S = G× e−2 G or S = 0.135 (13.5 percent). This means that the throughput is 1000 × 0.135 = 135 frames. Only 135 frames out of 1000 will probably survive.
Example 6.3 (continued) b. If the system creates 500 frames per second, this is (1/2) frame per millisecond. The load is (1/2). In this case S = G × e −2G or S = 0.184 (18.4 percent). This means that the throughput is 500 × 0.184 = 92 and that only 92 frames out of 500 will probably survive. Note that this is the maximum throughput case, percentagewise. c. If the system creates 250 frames per second, this is (1/4) frame per millisecond. The load is (1/4). In this case S = G × e −2G or S = 0.152 (15.2 percent). This means that the throughput is 250 × 0.152 = 38. Only 38 frames out of 250 will probably survive.
Slotted ALOHA • Time is slotted in X seconds slots • Stations synchronized to frame times • Stations transmit frames in first slot after frame arrival • Backoff intervals in multiples of slots Backoff period B t (k+1)X t0+X+2tprop kX t0+X+2tprop+ B Time-out Vulnerableperiod Only frames that arrive during prior X seconds collide
Note The throughput for slotted ALOHA is S = G × e−G . The maximum throughput Smax = 0.368 when G = 1.
Example 6.4 A slotted ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the throughput if the system (all stations together) produces a. 1000 frames per second b. 500 frames per second c. 250 frames per second. Solution The frame transmission time is 200/200 kbps or 1 ms. a. If the system creates 1000 frames per second, this is 1 frame per millisecond. The load is 1. In this case S = G× e−G or S = 0.368 (36.8 percent). This means that the throughput is 1000 × 0.0368 = 368 frames. Only 386 frames out of 1000 will probably survive.
Station A begins transmission at t = 0 A Station A captures channel at t = tprop A Carrier Sensing Multiple Access (CSMA) • A station senses the channel before it starts transmission • If busy, either wait or schedule backoff (different options) • If idle, start transmission • Vulnerable period is reduced to tprop(due to channel capture effect) • When collisions occur they involve entire frame transmission times • If tprop >X (or if a>1), no gain compared to ALOHA or slotted ALOHA
CSMA Options • Transmitter behavior when busy channel is sensed • 1-persistent CSMA (most greedy) • Start transmission as soon as the channel becomes idle • Low delay and low efficiency • Non-persistent CSMA (least greedy) • Wait a backoff period, then sense carrier again • High delay and high efficiency • p-persistent CSMA (adjustable greedy) • Wait till channel becomes idle, transmit with prob. p; or wait one mini-slot time & re-sense with probability 1-p • Delay and efficiency can be balanced Sensing
Example 6.5 A network using CSMA/CD has a bandwidth of 10 Mbps. If the maximum propagation time (including the delays in the devices and ignoring the time needed to send a jamming signal, as we see later) is 25.6 μs, what is the minimum size of the frame? Solution The frame transmission time is Tfr = 2 × Tp = 51.2 μs. This means, in the worst case, a station needs to transmit for a period of 51.2 μs to detect the collision. The minimum size of the frame is 10 Mbps × 51.2 μs = 512 bits or 64 bytes. This is actually the minimum size of the frame for Standard Ethernet.
Figure 6.11 Energy level during transmission, idleness, or collision
6-2 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. We discuss three popular controlled-access methods. Topics discussed in this section: ReservationPollingToken Passing
Reservations Systems • Centralized systems: A central controller accepts requests from stations and issues grants to transmit • Frequency Division Duplex (FDD): Separate frequency bands for uplink & downlink • Time-Division Duplex (TDD): Uplink & downlink time-share the same channel • Distributed systems: Stations implement a decentralized algorithm to determine transmission order Central Controller
M 1 2 3 Reservation Systems Reservation interval Frame transmissions d r d d r d d d Time Cycle n Cycle (n + 1) r = • Transmissions organized into cycles • Cycle: reservation interval + frame transmissions • Reservation interval has a minislot for each station to request reservations for frame transmissions
Reservation System Options • Centralized or distributed system • Centralized systems: A central controller listens to reservation information, decides order of transmission, issues grants • Distributed systems: Each station determines its slot for transmission from the reservation information • Single or Multiple Frames • Single frame reservation: Only one frame transmission can be reserved within a reservation cycle • Multiple frame reservation: More than one frame transmission can be reserved within a frame • Channelized or Random Access Reservations • Channelized (typically TDMA) reservation: Reservation messages from different stations are multiplexed without any risk of collision • Random access reservation: Each station transmits its reservation message randomly until the message goes through
(a) 5 r 3 5 3 3 r 3 5 r 3 r 5 r 8 8 t (b) 8 5 r 3 5 3 3 r 3 5 r 3 r 5 r 8 8 t Example • Initially stations 3 & 5 have reservations to transmit frames • Station 8 becomes active and makes reservation • Cycle now also includes frame transmissions from station 8
Efficiency of Reservation Systems • Assume minislot duration = vX • TDM single frame reservation scheme • If propagation delay is negligible, a single frame transmission requires (1+v)X seconds • Link is fully loaded when all stations transmit, maximum efficiency is: • TDM k frame reservation scheme • If k frame transmissions can be reserved with a reservation message and if there are M stations, as many as Mk frames can be transmitted in XM(k+v) seconds • Maximum efficiency is:
ρmax = = 1 1 + 2.71v XX(1+ev) Random Access Reservation Systems • Large number of light traffic stations • Dedicating a minislot to each station is inefficient • Slotted ALOHA reservation scheme • Stations use slotted Aloha on reservation minislots • On average, each reservation takes at least e minislot attempts • Effective time required for the reservation is 2.71vX
Polling Systems • Centralized polling systems: A central controller transmits polling messages to stations according to a certain order • Distributed polling systems: A permit for frame transmission is passed from station to station according to a certain order • A signaling procedure exists for setting up order Central Controller
Polling System Options • Service Limits: How much is a station allowed to transmit per poll? • Exhaustive: until station’s data buffer is empty (including new frame arrivals) • Gated: all data in buffer when poll arrives • Frame-Limited: one frame per poll • Time-Limited: up to some maximum time • Priority mechanisms • More bandwidth & lower delay for stations that appear multiple times in the polling list • Issue polls for stations with message of priority k or higher
Figure 6.14 Select and poll functions in polling access method
Comparison of MAC approaches • Aloha & Slotted Aloha • Simple & quick transfer at very low load • Accommodates large number of low-traffic bursty users • Highly variable delay at moderate loads • Efficiency does not depend on a • CSMA-CD • Quick transfer and high efficiency for low delay-bandwidth product • Can accommodate large number of bursty users • Variable and unpredictable delay
Comparison of MAC approaches • Reservation • On-demand transmission of bursty or steady streams • Accommodates large number of low-traffic users with slotted Aloha reservations • Can incorporate QoS • Handles large delay-bandwidth product via delayed grants • Polling • Generalization of time-division multiplexing • Provides fairness through regular access opportunities • Can provide bounds on access delay • Performance deteriorates with large delay-bandwidth product