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Medium Access Control. Tanenbaum (Chapter 4) Others References: Walrand J., Communication Networks: A first Course Bertesekas and Gallager, Data Networks . Where in the OSI Reference Model ?. Application. Presentation. Session. Transport. Medium Access Control. Network. Link Layer.

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## Medium Access Control

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**Medium Access Control**Tanenbaum (Chapter 4) Others References: Walrand J., Communication Networks: A first Course Bertesekas and Gallager, Data Networks**Where in the OSI Reference Model ?**Application Presentation Session Transport Medium Access Control Network Link Layer Data Link Layer Physical Medium Access Control**Why Do We Need a MAC Layer ?**• Let us consider different topologies • Point-to-point channel • full duplex • half duplex • Broadcast channel Medium Access Control**Management of Broadcast Channels**• Contention free: allocating statically shares of the channel to all stations (by TDM, FDM, or other) • Contention: let stations compete for the ALL channel Servers One server Medium Access Control**Objectives of This Chapter:Establish A Key Result**• 1) Efficiency: with heavy load, contention free systems are better than contention systems. • 2) For average delay: contention systems are better than contention free (Whatever is the protocol! ) Introduction to Queueing Theory Medium Access Control**Objectives of This Chapter (cont’d):Establish some**Properties • Relationship between a MAC protocol and physical properties of network: • Upperbound on efficiency • Maximum length of medium • Bandwidth • Minimum packet size Medium Access Control**Objectives of This Chapter (cont’d):Techniques of Analysis**• Initiate students to the techniques used to analyze the performance of MAC protocols: • Aloha protocol • Slotted Aloha • CSMA • CSMA/CD Medium Access Control**Objective 1: Establish A Key Result on Average Delay**Queueing Theory**Introduction to Analytical Modeling**L.Kleinrock, “Queueing Systems” R.Jain, “The Art of Computer Systems Performance Analysis” E. Modiano, MIT course on Communications**Packet Switched Networks**PS PS PS PS PS PS PS Buffer Packet Switch PS Medium Access Control**Queues Everywhere**• Processes waiting for CPU • I/O Disk requests • Network interface card • IP input (output) queues • Events (mouse click, keyboard…) • ……………………… Medium Access Control**Queueing Theory**• What creates queues? • Randomness!!! • Used for analyzing network performance • In packet networks, randomness comes from: • random packet arrival • Random packet length • Information of interest: • Delay in buffer (queueing delay) • Buffer size Medium Access Control**Queueing Theory**Servers jobs Job Sources Or customers Medium Access Control**A Queueing System A/S/m/B/K/SD**• A : Interarrival time of jobs distribution • S : The service time distribution • m : Number of servers • B : Capacity of the system (max # of jobs allowed) • K : population size • SD : Service discipline • Default values B = ∞; K=∞; and Z=FCFS • For A and B • G : General distribution • M : Exponential Distribution (memoryless property) • D : Deterministic Medium Access Control**Example**• M/D/2/15/20000/FCFS • Time between arrivals exponentially distributed • Service time constant (no variation) • 2 servers • System capacity is 15 (2 places for currently served + 13 waiting) • Population is 20000 (20000 customers will ever come to the system) • Service discipline is first come first served Medium Access Control**Queueing Models**• Model good for • Requests received by Google • Customers waiting in line • Packets waiting to be transmitted over a line • Information of interest • Average number of customers in the system • Average delay experienced by a customer Medium Access Control**Queueing System: Variables of Interest**• Mean arrival rate: l • Mean service rate: m (service time per job) • Number of customers in system: n • Waiting time (in queue+service): w Servers jobs Job Sources Or customers Medium Access Control**A Key Result : Little’s Law**Arrivals Departure Black Box Mean # in system = Mean arrival rate X mean time spent in system E(n) = l.E(w) Remarks : - Result independent of A/B distributions or SD - Can be applied to all system or part of it - Crowded system long delays Medium Access Control**A Key Result : Little’s LawExample**A monitor on an HTTP server showed that the average time to satisfy a request was about 50 milliseconds. The requests arrival rate is 200 requests per second. What should be the buffer size (unit is requests) at the http server For the requests ? Little’s law states E(n) = l. E(w) Here l =200 req/s and E(w) = 0.050 s The expected number of customers in the system is E(n) E(n) = 200 x 0.050 = 10 It would be safe to have a buffer size of 20 Medium Access Control**Arrival Process**• Packets arrive according to some random process • There are many stochastic processes. • A nice stochastic process is the Poisson process • Mean arrival rate of l packets per second • Prob(n arrivals during T) = [(l.T)n e-lT]/n! Medium Access Control**Example**• The number of phone calls arriving to a switch can be closely modeled as a Poisson process. Suppose that the mean arrival rate is 100 per hour. • What is the probability to receive 10 calls in 6 minutes? T = 0.1 hour, l = 100/h Prob(n arrivals during T) = [(l.T)n e-lT]/n! Prob(10 arrivals during 0.1 h)= (100x0.1)10xe-(100x0.1)/10!=0.12 Prob(10 arrivals during 15 mn) = 0.003 Medium Access Control**InterArrival Times of Poisson Process**• This is the time between consecutive arrivals. IA is a continuous random variable • What is its probability distribution function? • Prob(IA <= t) = 1 – Prob(IA > t) • = 1 - Prob(0 arrivals within t) • Prob(0 arrivals during t) = [(l.t)0 e-lt]/0!= e-lt • So, Prob(IA <= t) = 1 - e-lt Medium Access Control**InterArrival Times of Poisson Process (2)**• The cumulative distribution function (CDF) is: • Prob(IA <= t) = 1 - e-lt • The probability distribution function is the derivative of CDF, i.e., PDF = l.e-lt • This what is called the exponential distribution • This distribution is largely used to model the service times, time between error losses, .. Medium Access Control**Memoryless Property**• Def: A random variable X is said to be without memory, or memoryless, • P(X>s+t|X>t) = P(X>s) for all s, t ≥ 0 • In words: “When I get to the bus station, I am told that the probability that the bus comes within the next 10 minutes is 0.90. After one hour waiting, I am told that the probability that the bus comes within the next 10 minutes is still 0.90 ” • Important result : X is memoryless iff it is exponentially distributed Medium Access Control**More Examples**• Suppose that the time to graduate from AU is exponentially distributed with mean 4 years. • Given that a student already spent 3 years at AU, what is the expected remaining time before he graduates? Medium Access Control**Properties of Poisson Process (PP)**• Merging of K P.P with mean rate li results in a P.P with mean rate the sum of the li’s. • Splitting (randomly) a P.P with mean rate l with probabilities pi results in P.Ps with mean rates pi. l. l1 Sli l2 li ln p1.l1 p2.l2 l pi.li pn.ln Medium Access Control**Analysis of an M/M/1**Queue • Interarrival exponentially distributed (Memoryless) • Service time exponentially distributed (Memoryless) • One server • Infinite capacity of system • Infinite population • First come, first served Jobs Source Server Medium Access Control**Analysis of an M/M/1 (2)**• We are interested in the average number of jobs in the system, the average waiting time in the system, or in the probability to have a given number of customers in the system. • Notations • N(t): number of jobs in system at time t • Pn(t) = Prob{N(t)=n} • Pn = lim t-->∞ Pn(t) • l= arrival rate • m = service rate (service time = 1/ m) Medium Access Control**Markov Chain for M/M/1 System**l.d l.d l.d l.d l.d l.d 0 1 2 3 4 5 • Circle i = state i means there are i customers in the system • What is the probability Prob(i,j), i.e, the probability of transition from state i and j? • Prob(j,j+1) = l.d, and Proj(j,j-1) = m.d 1-l.d m.d m.d m.d m.d m.d m.d Medium Access Control**Remarks about M/M/1**• l > m, otherwise the system will be instable • After some time in operation, an M/M/1 gets into some equilibrium. • When in equilibrium l.Pi = m.Pi+1 l l l l l l 0 1 2 3 4 5 m m m m m m l i m Medium Access Control**Determining Pn**l l l l l l 0 1 2 3 4 5 m m m m m m l.Po = m.P1 l.P1 = m.P2 ………….. l.Pn = m.Pn+1 P1 = (l\m).P0 P2 = (l\m).P1 ………….. Pn+1 = (l\m).Pn Pn = (l\m)n.P0 Using the equations above and the sum SPi = 1, we can derive Pi’s SPi = S (l\m)i.P0 = P0.S (l\m)i = P0/(1-(l\m)) = P0/(1-r) = 1 P0 =(1-r) Pi = ri.(1-r) Medium Access Control**Mean Queue Length E(N) = r/(1-r)r =l/m is the traffic**intensity • The average number of customers in the system is E(N) • E(N) = Si.ri.(1-r) = r/(1-r) • E(N) = (1-r).Si.ri • E(N) = r/(1-r) • E(N) = l/(m-l) Medium Access Control**Mean Waiting Time E(W) = 1/(1-r)m**• What is the average time in the system? (queueing delay + service time) • We use Little’s formula: E(N) = l.E(W) Medium Access Control**Prob{Server is busy} ?**Medium Access Control**Prob{Server is idle} ?**Medium Access Control**Analysis of M/M/m (m=3)**l l l l l l 0 1 2 3 4 5 m 2m 3m 3m 3m 3m lPo = mP1 lP1 = 2mP2 ………….. lPn = 3mPn+1 Using the equations above and the sum Spi = 1, we can derive Pi’s Medium Access Control**Example**• Comparison source partition vs global FCFS • Let the system have n sources and m servers.Jobs generate at each source as P.P with rate l • Job’s computation time is exponentially distributed 1/m • Source partition is m M/M/1 while global FCFS may be viewed as M/M/m with Arrival rate n.l. Medium Access Control**A Fundamental Result In Queueing Theory**• One powerful server for all is better than one weak server for each one ! • Why ? Better utilization, as a dedicated channel may stay IDLE. One server Medium Access Control**Analysis of MAC Protocols**Read Tanenbaum Chapter 4 Key sections (Intro, 4.1, 4.2.1,4.2.2, 4.3)**Multiple Access Protocols**• Competing stations (possible collisions) • Aloha • Slotted Aloha • CSMA • CSMA/CD • Collision-free • Bit-map protocol • Binary countdown • Limited contention • Adaptive tree walk protocol Medium Access Control**Pure Aloha**• Designed by Abramson (wireless) • A station emits whenever it has something to send • If other station emits, a collision happens • If collision, frame must be resent • Best possible utilization at high load 18% Medium Access Control**Analysis of Pure Aloha**• Assume infinite population of users • Let Tr=time to trasmit a frame (“frame time”) • The population generates a traffic that is Poisson with mean N per “frame time” (new frames) • Since there are also retransmissions, the total traffic generated is Poisson with mean G • What is the throughput S? • S = G.Po where Po is the probability that a frame does not suffer a collision Medium Access Control**Analysis of Pure Aloha (2) (Example)**• Suppose that the bandwidth is 10 Mbps, and packet size is 1500 bytes • Tr= 1500.8/10 Mbps = 1.2 ms • Possible values for N (mean number of frames generated per time frame): between 0 and 1 • Values for generated traffic (G > N) • No retransmissions at all G = N • Low load G~N • High load G > N • Po will be higher at low load Medium Access Control**Analysis of Pure Aloha (3)**• What is the probability to generate k frames during a “frame time”? • Prob(k arrivals during 1) = [(G.1)k e-G.1]/k! (page 20) • What is the probability Po that that a frame does not suffer a collision? Tr t0 t0+Tr t0+2Tr t0+3Tr Medium Access Control**Analysis of Pure Aloha (4)**• Po is the probability that ZERO frame is generated during the vulnerable period • Po = Prob(0 arrivals during 2) = • = [(G.2)0 e-G.2]/0! • Po = e-G.2 • S = G.Po = G.e-G.2 (When is S maximum?) • The maximum throughput S is 1/(2.e) = 18.4% Tr t0 t0+Tr t0+2Tr t0+3Tr Medium Access Control**Slotted Aloha**• Designed by Roberts (wireless) • Requires synchronization and division of time in discrete slots • A station emits whenever it has something to send AND must wait for beginning of slot • Best possible utilization at high load 37% t t0 t0+t t0+2t t0+3t Medium Access Control**Analysis of Slotted Aloha (2)**• The key: slotted time reduces the vulnerable period to Tr (instead of 2Tr). • Po = Prob(0 arrivals during 1) = • = [(G.1)0 e-G.1]/0! • Po = e-G.1 • S = G.Po = G.e-G (When is S maximum?) • The maximum throughput S is 1/e = 36% • Exercise: derive the average number of transmissions of one frame before being successful. t t0 t0+t t0+2t t0+3t Medium Access Control**Carrier Sense Multiple AccessCSMA**• Designed by Metcalfe, analyzed by Kleinrock and Tobaggi • A station listens to the channel before sending. • If channel busy, wait until it becomes idle • When channel free, send with probability 1 • Are collisions still possible? Medium Access Control**Collisions with CSMA**• Key: signal takes time tp to propagate • Problem 1: If two stations are listening to grab the channel… • Problem 2: If S1 starts transmitting, S2 may well send during tp. • Problem 3: S1 is not aware of the collision Sender S1 Sender S2 tp Medium Access Control**Solving Problem 1: p-persistent CSMA**• 1-persistent: after collision, waits a random time and starts over (slide 48) • Non persistent: if channel busy, the station does not keep listening, but rather waits for a random time before listening again. • p-persistent: for slotted, if idle, send with probability p, otherwise defers to next slot • Much better utilization than Aloha, may go beyond 95%. Medium Access Control

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