Data analysis and stochastic modeling

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1 Data analysis and stochastic modeling Lecture 7 An introduction to queueing theory Guillaume Gravier guillaume.gravier@irisa.fr with a lot of help from Paul Jensen s course jensen/ormm/instruction/powerpoint/or_models_09/14_queuing.ppt 1. A gentle introduction to probability 2. Data analysis 3. Cluster analysis 4. Estimation theory and practice 5. Mixture models 6. Random processes Where are we? 7. Queing systems discrete time Markov chains revisited continuous time processes queueing systems 8. Hypothesis testing What are we gonna talk about today? Typology of random processes Another look at discrete time Markov chains Continuous time processes continuous time Markov chains Birth and death processes Queueing theory what is a queueing system? M/M* queues other types of queues and queue networks A process is a collection of random variables {X(t) t T } over an index set T where X(t) take values in a state space I. T is discrete continuous discrete discrete-time continuous-time chain chain I is continuous discrete-time continuous-time continuous-state continuous-state process process

2 Markov property Discrete time Markov chains (DTMC) Markov property [slight change of notations from previous lecture] A stochastic process {X(t) t T } is a Markov process if, for any t 0 < t 1 <... < t n < t, the conditional distribution of X(t) given X(t 0 ),...,X(t n ) depends only on X(t n ), i.e. P[X(t) x X(t n ) x n,...,x(t 0 ) t 0 ] = P[X(t) x X(t n ) x n ] A Markov chain is said to be homogeneous if P[X(t) x X(t n ) x n ] = P[X(t t n ) x X(0) x n ]. Andreï A. Markov P[X n = x n X 0 = x 0,...,X n 1 = x n 1 ] = P[X n = x n X n 1 = x n 1 ] We want to study the following quantities (for homogeneous chains only) state of the process at step n: p i (n) = P[X n = i] n-step transition probability: p ij (n) = P[X m+n = j X m = i] The process is entirely defined by the initial distribution: p(0) = [p 0 (0),p 1 (0),...] the transition matrix: P = [p ij = p ij (1) = P[X n = j X n 1 = i]] n-step transitions Asymptotic behavior From the Markovian assumption, we can note that P[X t+m+n = j X t = i] = p ij (m + n) = k p ik (m)p kj (n), v i = lim n P[X n = i]? which gives us the following property (m 1, n n 1) P(n) = P P(n 1) = P n. The marginal distribution of X n is given by P[X n = i] = p i (n) = P[X 0 = k]p[x n = i X 0 = k] = k k and the state of the system at step n is given by [p 0 (n), p 1 (n),...] = p(n) = p(0) P n. p k (0)p ki (n), v i = long run proportion of time spent in state i when n, p ij (n) becomes independent of i and n all rows of P n converge toward a common limit v the limit satisfies v = v P under the constraint that v i 0, i v i = 1. conditions for the limit to be defined aperiodic the limiting state probabilities exists irreductible aperiodic the limit exists and is independent of the initial probabilities p(0) irreductible aperiodic with recurrent states (or finite state irreductible aperiodic) v is the unique stationary probability vector

3 Continuous time Markov chains (CTMC) P[X t = x X t0 = x 0,...,X tn 1 = x n 1 ] = P[X t = x X tn 1 = x n 1 ] We want to study the following quantities (for homogeneous chains only) state of the process at time t: p i (t) = P[X t = i] n-step transition probability: p ij (t) = P[X t = j X 0 = i] The process is entirely defined by the initial distribution: p(0) = [p 0 (0),p 1 (0),...] the transition probabilities: p ij (t) Properties of the transition probabilities p ij (t) 0 t j p ij(t) = 1 i, t p ij (t + s) = k p ik(t)p ki (s) t, s Transition probabilities and transition rates dealing with transition probabilities that are a function of time is a mess! use transition rates rather than probabilities p ij (t) q ij = lim t 0 t Q = P (0) and properties of the transition rate matrix d P(t) = QP(t) = P(t)Q dt P[X t+h = j X t = i] = q ij h + o(h) p ii (t, t + h) = 1 q ii h + o(h) q i = q ii = lim t 0 p ii (t) 1 t i j Jump process and embedded Markov chain T i is the time of the i-th jump the time to the next jump follows an exponential model P[T 1 > t X 0 = i] = e q it Y n = X Tn is a DTMC with transition probabilities p ij = q ij q i if i j and q i 0 0 if i j and q i = 0 0 if i = j and q i 0 1 if i = j and q i = 0 Limiting probabilities Under certain conditions, we have for a CTMC X lim p ij(t) = v j = t π j/q j π k /q k where π is the unique invariant measure associated with Y. The limiting probabilities, if they exist, are obtained by solving the equation system vq = 0 v1 = 1 k

4 Homogeneous Poisson process Classical discrete distributions: Poisson N(t) = number of occurences of an event (with rate ) in [0,t] P[N(t + τ) N(t) = k] = e τ (τ) k k! Poisson law a counting process with the following property is a Poisson process lim P[N(t + τ) N(t) > 1 N(t + τ) N(t) 1] = 0 τ memoryless process interarrival times are independent and exponentially distributed (mean ) A Poisson P(α) is defined as P[X = k] = k e Probability of k events occuring at a rate c over an interval of duration t ( = ct) Approximation of the binomial for small p and large n E[X] = V [X] = k! Siméon D. Poisson [Illustration: Skbkekas (from wikipedia.org)] Poisson process and CTMC A Poisson process is a continuous time Markov chain with rate transition matrix Q = Birth and death process N(t) = size of a population at time t Birth distribution n = birth rate for a population of size n remaining time until the next birth exp( n ) Death distribution n = death rate for a population of size n remaining time until the next death exp( n ) 0 1 k 1 k k+1 N(t) is a CTMC with transition rate diagram

5 Birth and death process equilibrium Expected rate in = Expected rate out The equilibrium v is given by solving the system whose solution is given by vq = 0 v1 = 1 with S = i v 0 = S 0... i i. v i = i 1 S 1... i flow into 0 1 v 1 = 0 v 0 flow out of 0 flow into 1 0 v v 2 = ( )v 1 flow out of 1 flow into 2 1 v v 3 = ( )v 2 flow out of 2. flow into i i 1 v i 1 + i+1 v i+1 = ( i + i )v i flow out of i. Particular case: if i = and i =, then v i = ρ i (1 ρ) with ρ = /. All this is true only if S < or, in the particular case, if ρ < 1. v 1 = 0 1 v 0, v 2 = 1 2 v 1 = v 0,... v k = k k... 1 v 0 What is a queueing system? What are queueing systems good for? A queueing system is usefull to answer the following questions input population entry queue service mechanism exit What is the average number of customers in the queue? in the system? What is the average time a customer spends in the queue? in the system? What is the probability of a customer to be rejected? Arrival process size of the population bulk vs. individual arrivals interarrival distribution customer balking Service process number of servers batch or single service service time distribution What is the fraction of time a server is idle? What is the probability distribution of a customer s waiting time in the queue? in the system? In turn, answering those questions supports decision making Should we have one or several queues? Queue What is the best queueing discipline for my application? finite or infinite discipline (FIFO, LIFO, priorities, etc.)

6 Kendall s notation A/B/N/m/s A and B = distribution of resp. the interarrival and the service time M exponential model E k Erlang model D deterministic time G arbitrary iid model Erlang distribution f(x;k,) = k x k 1 e x (k 1)! N = number of parallel servers m = max. number of customers in the system s = population size Examples: M/M/1/ /, M/M/S/ /, M/M/1/C/, M/M/1/K/K M/G/1, M/D/1, etc. Quantities of interest Quantities of interest (cont d) n = n = N(t) = P n (t) = mean arrival rate of entering customers when n customers are in the system mean service of the overall system when n customers are in the system number of customers in the system (queue and service) at time t the probability that there are exactly n customers in the system at time t R i = time spent in the system by the ith customer We are primarily interested in the following steady state quantities N : number of clients in the system 1 t L = E[N] = lim N(s)ds t t 0 R: time spent in the system by a client 1 W = E[R] = lim n n nr i i=1 N(t) and P n (t) are difficult to compute for an arbitrary time t study steady-state regime when t We can define in a similar way the two quantities excluding service time, L Q and W Q, where L = L S + L Q.

7 Little s law The M/G/1 model and the PASTA property For any queueing system with a steady state, with an average arrival rate of, L = W. Example: if the average waiting time is 2h and customers arrive at an average rate of 3 per hour, then the average number of customers in the queue is 6. The same equality holds for L Q and W Q. Note: for a constant death rate, W = W Q + 1/. Poisson arrival process with an average arrival rate of FIFO discipline no assumption on service time (apart from the iid one) memoryless property is not verified N(t) is not Markovian study a snapshot of the system immediately after each arrival embedded Markov chain X n = N(t n ) Poisson Arrivals See Temporal Average (PASTA) π k = lim t P[N(t) = k] = lim n P[X n = k] [R. Wolf. Poisson arrivals see time averages. Op. Res., 20: , 1982] Death and birth queues Steady state results memoryless arrival process interarrival law is exp( n ) given n customers in the system probability of an arrival in [t, t + dt] = n dt + o(t) memoryless service process service law is exp( n ) when n customers are in the system probability of a departure in [t, t + dt] = n dt + o(t) Durations and transitions to/from state i stays in i exp( i + i ) transits to i 1 with probability i /( i + i ) transits to i + 1 with probability i /( i + i ) π 1 = 0 π 0, π 2 = 1 π 1 = 0 1 π 0,... π k = k π k... 1 expected number in the system L = E[N] = kπ k expected number in the queue L Q = E[N Q ] = average (effective) arrival rate = efficienty of the system k π k k=0 k=1 E = (L L Q )/s (k s)π k k=s

8 The three server example The three server example (cont d) Consider a birth-and-death queueing system with s = 3 servers and the following characteristics: average arrival rate = 5 per hour average service rate = 2 per hour customers balk when 6 are already in the system k = π k = What is the probability that... all servers are idle? P[N(t) 0] = a customer will not have to wait? P[N(t) 2] = π 0 + π 1 + π 2 = 0.45 a customer will have to wait? P[N(t) 3] = 0.55 a customer balks? P[N(t) = 6] = The three server example (cont d) The methodology Expected number in queue: L Q = π 4 + 2π 5 + 3π 6 = 0.7 Expected number in service: L S = π 1 + 2π 2 + 3(1 π 0 π 1 π 2 ) = Efficiency of the system: E = L S /s = 74.8 % Average arrival rate: = (π π 5 ) = per hour 1. model the system and construct the rate diagram 2. develop the balance equation and solve for π i i = 0, 1,... use general birth-and-death process results whenever they apply 3. use steady-state distribution to L and L Q 4. use Little s law to get W and W Q Expected waiting times using Little s law with W S = L S / = 0.5 hours W Q = L Q / = hours W = L/ = W S + W Q = hours 3 3

9 The M/M/1 model exponential interarrival time with constant parameter k = k exponential service time with constant parameter k = k π n = ρ n (1 ρ) 5 6 The repairman example A maintenance worker must maintain 2 machines, where the 2 machines operate simultaneously when both are up. We make the following hypotheses: time until a machine breaks down exp with mean 10 hours time until a machine is fixed exp with mean 8 hours the repairman can only fix one machine at a time M/M/1/2/2 L = nπ n = n=1 2 L Q = L (1 π 0 ) = ( ) W = L/ = 1 W Q = L Q / = ( ) The repairman example (cont d) flow into 0 1 π 1 = 2π 0 flow out of 0 flow into 1 2π 0 + π 2 = ( + )π 1 flow out of 1 normalize π 1 = π 2 π 0 = π 1 = π 2 = = L = π 1 + 2π 2 = L Q = π 2 = 0.33 W = L = 11.55hour W Q = L Q = 3.56hour Proportion of time the repairman is busy = π 1 + π 2 = Proportion of time machine #1 is working = π π 1 = The telephone answering example A utility company wants to determine a staffing plan for its customer representatives. Calls arrive at an average rate of 10 per minute and the average service time is 1 minute. Determine the number of operators that would provide a satisfactory service to the calling population. Parameters: = 10, = 1,ρ = /s < 1 s > 10 M/M/11 M/M/12 M/M/13 L Q W Q P[T Q = 0] P[T q > 1]

10 One last for the road Networks of queues Packets arrive to a router at a rate of 1.5 min. The mean routing time is 30 seconds and if more than 3 packets are in the queue, an alternative routeur is seeked. Does this setting meet the following criteria: no more than 5% of the packets goes to another routeur and no more than 10% of the packets stay more than 1 min in the queue? Using the M/M/1/4 model, you should be able to demonstrate that the balking probability = π 4 =.104 does not meet the 5 % criteria the probability P[T Q > 1] = does not meet the 10 % objective However, a M/M/2/5 configuration does and so you need to buy one more routeur! 3 3