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Probability Review Thinh Nguyen Probability Theory Review Sample space Bayes’ Rule Independence Expectation Distributions Sample Space - Events Sample Point Sample Space S The outcome of a random experiment The set of all possible outcomes Discrete and Continuous Events A set of outcomes, thus a subset of S Certain, Impossible and Elementary Set Operations Union A B Intersection A B Complement AC Properties S A B AC Commutation A B B A Associativity A B C A B C Distribution A B C A B A C De Morgan’s Rule C A B AC BC A B Axioms and Corollaries Axioms 0 P A PS 1 If A B P A B P A P B If A1, A2, … are pairwise exclusive P Ak P Ak k 1 k 1 Corollaries P AC 1 P A P A 1 P 0 P A B P A P B P A B Conditional Probability Conditional Probability of event A given that event B has occurred P A | B A B S P A B P B If B1, B2,…,Bn a partition of S, then B1 B2 P A P A | B1 P B1 ... P A | B j P B j (Law of Total Probability) A B AC A B3 Bayes’ Rule If B1, …, Bn a partition of S then P B j | A P A B j P A P A | B j P B j n P A | B PB k 1 posterior k k likelihood prior evidence Event Independence Events A and B are independent if P A B P A P B If two events have non-zero probability and are mutually exclusive, then they cannot be independent Random Variables Random Variables The Notion of a Random Variable The outcome is not always a number Assign a numerical value to the outcome of the experiment Definition A function X which assigns a real number X(ζ) to each outcome ζ in the sample space of a random experiment S ζ X(ζ) = x x Sx Cumulative Distribution Function Defined as the probability of the event {X≤x} FX x P X x Fx(x) 1 Properties 0 FX x 1 x lim FX x 1 Fx(x) lim FX x 0 ¾ x x if a b then FX a FX a P a X b FX b FX a P X x 1 FX x 1 ½ ¼ 0 1 2 3 x Types of Random Variables Continuous Probability Density Function fX x FX x dFX x dx x f X t dt Discrete Probability Mass Function PX xk P X xk FX x PX xk u x xk k Probability Density Function The pdf is computed from fX(x) dFX x fX x dx Properties P a X b f X x dx b fX(x) a FX x x f X t dt 1 f X t dt For discrete r.v. f X x PX xk x xk k dx P x X x dx f X x dx x Expected Value and Variance The expected value or mean of X is E X tf X t dt E X xk PX xk 2 Var X 2 E X E X k The standard deviation of X is Std X Var X Properties E c c The variance of X is Properties E cX cE X Var c 0 E X c E X c Var cX c2Var X Var X c Var X Queuing Theory Example nSend a file over the internet packet Modem card buffer link (fixed rate) place C Computation (Queuing) transmission propagation Delay Models B A time Queue Model Practical Example Multiserver queue Multiple Single-server queues Standard Deviation impact Queueing Time Queuing Theory The theoretical study of waiting lines, expressed in mathematical terms input server queue Delay= queue time +service time output The Problem Given One or more servers that render the service A (possibly infinite) pool of customers Some description of the arrival and service processes. Describe the dynamics of the system Evaluate its Performance If there is more than one queue for the server(s), there may also be some policy regarding queue changes for the customers. Common Assumptions The queue is FCFS (FIFO). We look at steady state : after the system has started up and things have settled down. State=a vector indicating the total # of customers in each queue at a particular time instant (all the information necessary to completely describe the system) Notation for queuing systems A/B/c/d A = the interarrival time distribution B = the service time distribution c = the number of servers d = the queue size limit nomitted :Where A and B can be D for Deterministic distribution M for Markovian (exponential) distribution G for General (arbitrary) distribution if infinite The M/M/1 System Poisson Process Exponential server queue output Arrivals follow a Poisson process nReadily amenable for analysis nReasonable for a wide variety of situations a(t) = # of arrivals in time interval [0,t] = mean arrival rate t = k ; k = 0,1,…. ; 0 Pr(exactly 1 arrival in [t,t+]) = Pr(no arrivals in [t,t+]) = 1- Pr(more than 1 arrival in [t,t+]) = 0 Pr(a(t) = n) = e- t ( t)n/n! Model for Interarrivals and Service times Customers arrive at times t0 < t1 < .... - Poisson distributed The differences between consecutive arrivals are the interarrival times : n = tn - t n-1 n in Poisson process with mean arrival rate exponentially distributed, , are Pr(n t) = 1 - e- t Service times are exponentially distributed, with mean service rate : Pr(Sn s) = 1 - e-s System Features Service times are independent service times are independent of the arrivals Both inter-arrival and service times are memoryless Pr(Tn > t0+t | Tn> t0) = Pr(Tn t) future events depend only on the present state This is a Markovian System Exponential Distribution given an arrival at time x P (x (x t )) P (( x ) t| x ) P ( x ) ) (1 e P ( (x t )) P ( x ) (1 e 1 P ( x ) P ( x ) (x t ) (x t ) x x t e (1 e ) e e x x e 1 (1 e ) Same as probability starting at time = 0 x ) Markov Models • n+1 Buffer Occupancy departure •n • n-1 t •n arrival t Probability of being in state n Pn (t t ) Pn (t )[(1 t )(1 t ) tt ] Pn 1 (t )[(t )(1 t )] Pn 1 (t )[(t )(1 t )] as t 0, Taylor series dPn (t ) Pn (t t ) Pn (t ) t dt Steady State Analysis Substituting for Pn (t t ) ( ) Pn Pn 1 Pn 1 Steady state P0 P1 Markov Chains 0 1 ... n-1 n n+1 Rate leaving n = Pn ( ) Rate arriving n = Pn 1 Pn 1 Steady State Pn ( ) Pn 1 Pn 1 State 0 P0 P1 Substituting Utilization P1 P0 P0 P2 ( ) P1 P0 P2 P1 P1 P0 P1 ( 1) P0 Substituting P1 P2 P0 ( 1) P0 P0 P0 P0 P0 2 2 Pn P0 n • Higher states have decreasing probability • Higher utilization causes higher probability of higher states What about P0 P0 Pn 1 P0 P0 n 0 n 0 n 0 1 P0 1 P0 1 1 Pn (1 ) n n Queue determined by n E(n), Average Queue Size q E (n) nPn n(1 ) (1 ) n n n 0 = 1- n 0 n 0 n Selecting Buffers E(N) 1/3 1 3 9 .25 .5 .75 .9 For large utilization, buffers grow exponentially Throughput Throughput=utilization/service time = /Ts For =.5 and Ts=1ms Throughput is 500 packets/sec Intuition on Little’s Law If a typical customer spends T time units, on the overage, in the system, then the number of customers left behind by that typical customer is equal to q Tq Applying Little’s Law M/M/1 Average Delay E (n) E (T ) or w Tw or q Tq E ( n) 1 1 E (T ) (1 ) (1 ) Ts 1 / Ts so E (T ) (1 ) (1 ) (1 ) Probability of Overflow n N 1 n N 1 P (n N ) pn (1 ) n N 1 Buffer with N Packets N 1 1 n pn 1 p0 p0 n 0 n 0 1 n 1 (1 ) p0 and pn N 1 N 1 1 1 N N (1 ) N N +1 pN (1 ) with 1 N 1 1 N Example Given Arrival rate of 1000 packets/sec Service rate of 1100 packets/sec Find 1000 0.91 1100 Utilization Probability of having 4 packets in the queue P4 (1 ) .062 4 P1 .082, P2 .075, P3 .068, P5 .056 Example E ( n) 9.99 1 With infinite buffers P(n 12) 121 .28 With 12 fixed buffers cell loss probabilit y ( 1 ) P12 .04 121 1 Pn .11,.10,.09,.09,.08,.07,.07,.06,.05,.05,.04 12 Application to Statistcal Multiplexing Consider one transmission line with rate R. Time-division Multiplexing Divide the capacity of the transmitter into N channels, each with rate R/N. R/N R/N R/N 1 T Statistical Multiplexing Buffering the packets coming from N streams into a single buffer and transmitting them one at a time. R T' 1 T N N N Network of M/M/1 Queues 1 1 3 4 2 3 2 1 1 2 Li i i i 2 1 2 3 3 1 3 1 2 3 L L1 L2 L3 T 1 i J i 1 i i M/G/1 Queue The customers pat at rate C since each customer pays C on the average and customers go through the queue per unit time. Assume that every customer in the queue pays at rate R when his or her remaining service time is equal R. time t, the customers pay at a rate At atogiven equal to the sum of the remaining service times of all the customer in the queue. The queue begin first S 2come-first served, this sum is equal to SQqueueing Total cost paid by a customer: the time of a customer who would 2 2 enter the queue at time t. E [ Q ] E [ S ] Expected cost paid by each customer: C S : Service Time Q : Queuing Time S 0 Q 2 E[Q] E[ S 2 ] E[Q] C 2 E[S 2 ] E[Q] 2(1 ) 1 T E[Q] S