Multiple Access (1) Required reading: Garcia 6.1, 6.2.1, CSE 3213, Fall 2010 Instructor: N. Vlajic

Transcription

1 1 Multiple Access (1) Required reading: Garcia 6.1, 6.2.1, CSE 3213, Fall 2010 Instructor: N. Vlajic

2 Multiple Access Communications 2 Broadcast Networks aka multiple access networks multiple sending & receiving stations share the 1 M 2 3 Shared multiple access medium 4 5 same transmission medium advantages: 1) low cost infrastructure 2) all stations attached to the medium hear transmission from any other station routing not necessary disadvantages: access of multiple sending and receiving nodes to the shared medium must be coordinated 1) stations should not be transmitting simultaneously or interrupting each other 2) stations should not be able to monopolize the transmission/shared medium examples: LAN, cellular and satellite networks

3 Multiple Access Communications (cont.) 3 Approaches to Medium Sharing Medium Sharing Techniques Static Dynamic Medium Channelization Access Control Scheduling Random Access (1) Static Channelization static & collision free sharing partition medium into separate channels, which are then dedicated to particular users advantage: no collisions, perfect fairness each node gets a dedicated transmission rate R/N during each time interval (suitable for streaming data, e.g. voice streams) disadvantage: each user gets only a fraction of the full channel capacity, even when no other station is transmitting

4 Multiple Access Communications (cont.) 4 Example [ systems employing static channelization ] satellite = repeater uplink f 1 downlink f 2 uplink f3 downlink f 4 uplink f in downlink f out satellite network cellular network two frequency bands: one for uplink and one for downlink each station is allocated a channel in the uplink and in the downlink frequency band different approaches can be employed to create uplink/downlink channels (FDMA, TDMA, CDMA) although each station can theoretically transmit to and listen to any channel, stations remain within their pre-allocated channels to avoid interference

5 Multiple Access Communications (cont.) 5 (2) Dynamic Medium Access MAC Schemes the medium is shared on a per frame basis advantage: transmitting node transmits at the full rate of the channel (suitable for bursty data, e.g. short messages) disadvantage: simultaneous attempts by two or more stations to access the channel result in collision collision can be minimized through scheduling or random access control Ring Network Bus Network

6 Multiple Access Communications (cont.) 6 Medium Sharing Techniques Static Channelization FDMA TDMA Scheduling Dynamic Medium Access Control Reservation Polling Token Passing Random Access ALOHA CSMA continuous load heavier bursty load light bursty load

7 Random Access Techniques: ALOHA 7 ALOHA the earliest random-access method (1970s) still used in wireless cellular systems for its simplicity a station transmits whenever it has data to transmit, producing smallest possible delay receiver ACKs data if more than one frames are transmitted at the same time, they interfere with each other (collide) and are lost if ACK not received within timeout (2*propagation delay), the station picks random backoff time (to reduce likelihood of subseq. collisions) station retransmits frame after backoff time data ACK

8 Random Access Techniques: ALOHA (cont.) 8 Example [ Aloha throughput ] Station Transmission Time (F) R 1.2' Station Station ' R Complete Collision Partial Collision Broadcast channel

9 Random Access Techniques: ALOHA (cont.) 9 Vulnerable Period assume frames of constant length (L) & transmission time (X=L/R) consider a frame with starting transmission time t o the frame will be successfully transmitted if no other frame collides with it any transmission that begins in interval [t 0, t 0 +X], or in the prior X seconds leads to collision vulnerable period = [ t 0 X, t 0 + X ] First transmission Backoff period B Retransmission if necessary t 0 -X t 0 t 0 +X t 0 +X+2t prop t 0 +X+2t prop +B t Vulnerable period Time-out What is the probability of no other transmission, i.e. no collision, in the vulnerable period?!

10 Random Access Techniques: ALOHA (cont.) 10 Throughput definitions and assumptions: S(S out ) throughput: average # of successful frame transmiss. per X sec (if network operates under stable conditions S out = S in, where S in - arrival rate of new frames to the system) G load average # of overall transmission attempts! per X sec P succ probability of a successful frame transmission S = P succ G How to find P succ? suppose a frame is ready for transmission at time t 0 frame will be transmitted successfully if no other frame attempts transmission X sec before and after t 0 random backoff spreads retransmissions so that frame transmission (arrivals) are equally likely at any instant in time Poisson process!!! stations new packets old packets S in G channel Collision? No, with P succ Yes, with (1-P succ ) S out t 0 -X 2X t 0 t 0 + X ready for transmission

11 Random Access Techniques: ALOHA (cont.) 11 if general, if frame arrivals are equally likely at any instant in time, and arrivals occur at an average rate of λ [arrivals per sec] Poisson process P[k arrivals in T seconds] = ( T) k! k λ λt e in our case, λ=g/x [arrivals per second] and T=2X, hence P[k transmissions in 2X seconds] = (2G) k! k e 2G accordingly, probability of successful transmission (no collision) is: P succ = P[0 transmissions in 2X seconds] = e 2G and throughput S = G P succ = G e 2G t 0 -X t 0 t 0 + X

12 Random Access Techniques: ALOHA (cont.) 12 S vs. G in Pure ALOHA NOTE: our analysis assumed that many nodes share a common channel & have comparable transmiss. rates (if only 1 node uses the medium, S=1) initially, as G increases S increases until it reaches S max after that point the network enters unstable operating conditions in which collisions become more likely and the number of backlogged stations increases (consequently, S in > S out ) max throughput of ALOHA (S max = 0.184) occurs at G=0.5, which corresponds to a total arrival rate of one frame per vulnerable period S max = max ALOHA throughput = 18% of channel capacity S - throughput per frame time stable operating area (S in ~S out ) unstable operating area (S in S ou ) G - attempts per frame time 4 18% of channel utilization, with Aloha, is not encouraging. But, with everyone transmitting at will we could hardly expected a 100% success rate.

13 Random Access Techniques: ALOHA (cont.) 13 Example [ Aloha ] a) What is the vulnerable period (in milliseconds) of a pure ALOHA broadcast system with R=50-kbps wireless channel, assuming 1000-byte frames. b) What is the maximum throughput S of such a channel (system), in kbps? a) (frame transmission time =) X = 1000 bytes / 50 kbps = 8000 bits / 50 kbps X = 160 milliseconds vulnerable period = 2*X = 320 milliseconds b) From theory, max throughput = 0.18 * R = 0.18 * 50 kbps = kbps

14 Random Access Techniques: Slotted ALOHA 14 Slotted ALOHA improved ALOHA, with reduced probability of collision assumptions: time is divided into slots of size X=L/R (one frame time) nodes start to transmit only at the beginning of a slot nodes are synchronized so that each node knows when the slots begin operation: 1) when node has a fresh frame to send, it waits until next frame slot and transmits 3) if there is a collision, node retransmits the frame after a backoff-time (backoff-time = multiples of time-frames)

15 Random Access Techniques: Slotted ALOHA (cont.) 15 Example [ Aloha vs. Slotted Aloha ] A B C D E Pure ALOHA tic tic tic tic tic tic tic tic tic tic tic tic tic A B C D E Slotted ALOHA

16 Random Access Techniques: Slotted ALOHA (cont.) 16 Vulnerable Period of Slotted ALOHA consider one arbitrary packet P that becomes ready for transmission at some time t during the time slot [k, k+1] packet P will be transmitted successfully if no other packet becomes available for transmission during the same time slot vulnerable period = [ t 0 X, t 0 ] P succ = P[0 arrivals in X seconds] = e G S = G P succ = G e G kx packet P (k+1)x Vulnerable period period in which packet P becomes ready for transmission Backoff period B t 0 +X+2t prop t 0 +X+2t prop + B Time-out t

17 Random Access Techniques: Slotted ALOHA (cont.) 17 S vs. G in Slotted ALOHA max throughput of Slotted ALOHA (S max = 0.36) occurs at G=1, which corresponds to a total arrival rate of one frame per vulnerable period S max = 0.36 max Slotted ALOHA throughput = 36% of actual channel capacity S Throughput (ALOHA) ideal no collision Ideal (no collisions): R Slotted ALOHA: S=G*e -G Slotted ALOHA: Re -R Slotted ALOHA: S=G*e -2G Pure ALOHA: Re -2R R G Slotted ALOHA vs. Pure ALOHA slotted ALOHA reduces vulnerability to collision, but also adds a waiting period for transmission if contention is low, it will prevent very few collisions, & delay many of the (few) packets that are sent

18 Random Access Techniques: Slotted ALOHA (cont.) 18 Example [ slotted Aloha ] Measurements of slotted ALOHA channel with an infinite number of users show that 10% of the slots are idle. a) What is the channel load, G? b) What is the throughput, S? c) Is the channel underloaded or overloaded? a) 10% of slots idle frame will be successfully transmitted if sent in those 10% of slots P succ = 0.1 According to theory, P succ = e -G G = -ln(p succ ) = - ln(0.1) = 2.3 b) According to theory, S = P succ *G = G*e -G as G=2.3 and e -G =0.1 S = 0.23 c) Whenever G>1, the channel is overloaded, so it is overloaded in this case.

19 Random Access Techniques: Slotted ALOHA (cont.) 19 Example [ slotted Aloha in cellular (GSM) systems ]

20 Arrival: passengers arrive randomly and independently a Poisson process. Passenger arrivals are equally likely at any instant in time. average arrival rate = λ [passenger / sec] 20 average # of arrived passengers in T [sec]: average # of passengers in T [sec] = λt probability of having exactly k passengers in line after T [sec]: P[k arrivals in T seconds] = k λt) λt ( k! e Departure: passengers What is arrive the probability in batches that exactly NOT a 1 Poisson passenger process. arrives to the station, off the buss, in T sec? λ [passenger / sec]

21 21 P[k ( arrivals in T seconds] = T) k! k λ λt e T=1 k