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Lecture 13 – Continuous-Time Markov Chains

Lecture 13 – Continuous-Time Markov Chains. Topics Markovian property Exponential distribution Rate matrix ATM Example Birth and death processes Queuing systems. Markovian Property for CTMCs. Stochastic process : { Y t : t  0 }, where Y t is a nonnegative integer

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Lecture 13 – Continuous-Time Markov Chains

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  1. Lecture 13 – Continuous-Time Markov Chains • Topics • Markovian property • Exponential distribution • Rate matrix • ATM Example • Birth and death processes • Queuing systems

  2. Markovian Property for CTMCs Stochastic process: {Yt : t 0}, where Yt is a nonnegative integer CTMC is similar to that of a DTMC except “one-step” has no meaning in continuous time so the Markovian property must hold for all future times instead of just for one step. Definition 1: The process Y = {Yt : t ≥ 0} with state space S is a CTMC if the following condition holds for all jÎS, and t, s ≥ 0 Pr{Yt+s = j | Yu, u ≤ s} = Pr{Yt+s = j | Ys }. In addition, the chain is said to have stationary transitions if Pr{Yt+s = j | Ys = i} = Pr{Yt = j | Y0 = i}. Interpretation: First equation says that the conditional distribution of the future Yt+s given the present Ys and the past Yu, 0 ≤ u ≤ s, depends only on the present and is independent of the past. Second equation says that Pr{Yt+s = j | Ys = i} is independent of s.

  3. Example of Definition 1 • Problem: Suppose that a CTMC enters state i at, say, time 0 and does not leave during the next 15 minutes; i.e., a transition does not occur. • Question: What is the probability that a transition will not occur in the next 5 minutes? • Approach: Markovian property tells us that the probability that the process will remain in state i during the interval [15, 20] is just the unconditional probability that it stays in state i for at least 5 minutes. • Solution: Let Ti denote the amount of time that the process stays in state i before making a transition into a different state. Then • Pr{Ti > 20 | Ti > 15} = Pr{Ti > 5} • or, in general, • Pr{Ti > s + t | Ti > s} = Pr{Ti > t} • for all s, t ≥ 0. Hence, the random variable Ti is memoryless and so is exponentially distributed.

  4. Generalization of Example • The Markovian property gives • Pr{Ti > s + t | Ti > s} = Pr{ Ti > t} • for all s, t ≥ 0. • Implication: The random variable Ti is memoryless and thus is exponentially distributed. • Alternative definition of CTMC: A stochastic process having the properties that each time it enters state i, • (i) the amount of time it spends in that state before making a transition into a different state is exponentially distributed with mean, say, 1/li , and • (ii) when the process leaves state i it next enters state j with some probability, say, pij , where pij must satisfy • pii = 0, for all iÎS • Sjpij = 1, for all i ÎS

  5. ATM Example (max of 5 in system) • Statistics • Average time between arrivals = 30 sec (0.5 min): l = 2/min • Average service time = 24 sec (0.4 min): m = 2.5/min • Design questions • How many ATMs should there be? • Should the foyer be expanded? • State-transition network

  6. Exponential Distribution pdf: f(t) = le–lt for t ≥ 0 CDF: F(t) = 1 – e–lt for t ≥ 0 Parameters: Mean = 1/l Var = (1/l)2 Exponential distribution with Mean = 0.5 (l = 2)

  7. Poisson Process When the duration of the time between events is exponentially distributed, the number of occurrences of the event in a given time interval has a Poisson distribution. • Pr{k arrivals in time t} = for k = 0, 1,… • For the ATM example with l = 2, the expected number of arrivals in the interval [0, t] is 2t.

  8. Rate Diagram • For CTMCs, activities are better represented by their rate of occurrence, so rather than using a state-transition network we use a ratediagram or rate network. • This network is easily constructed from the state-transition network by replacing the activity designation by the activity rate. ATM Network Transient analysis

  9. Rate Matrix A computationally more convenient alternative to the rate diagram is the rate matrixR whose element rij is the transition rate from state i to j. In general, rij = lpij or rij = mpij General rate matrix Rate matrix for ATM example

  10. Transient Analysis • Determine the probability that the system will be in a particular state at time t. • The transient probabilities are a function of the initial state. • Unconditional probability vector: • q(t) = (q0(t), q1(t), q2(t),…,qm-1(t)) • Requirement:

  11. Transient Analysis (cont’d) • For some small interval of time ∆, let n = t/∆ be the number of steps or increments required to represent t. • The transient solution of the process can be approximated at time t = n∆ with a DTMC by solving the following equation: • q(n∆) = q(0)P(n) or q(n∆+∆) = q(n∆)P • where P is a state-transition matrix determined from the rate matrix R.

  12. Transition Matrix for Transient Analysis • Let ai be the sum of all transition rates out of state i; that is, • and let pijrij. Then

  13. Transition Matrix for ATM Example Rate network

  14. Transient Analysis for ATM Example • Assume system is empty at t = 0. • We wish to approximate the transient probabilities at t = 1 min. • Initial probability vector: q(0) = (1, 0, 0, 0, 0, 0) • Use equation q(n∆) = q(0)P(n) • Number of steps: n = t/∆ = 1/∆ • Case 1: ∆ = 0.05 n = 20 steps (1 min) • q(20∆) = q(1) = (0.433, 0.291, 0.162, 0.075, 0.029, 0.011) • Case 2: ∆ = 0.025 n = 40 steps (1 min) • q(40∆) = q(1) = (0.435, 0.291, 0.160, 0.073, 0.029, 0.011) • (almost identical)

  15. Transient Solution for ATM Example (∆ = 0.05, 0 t 1)

  16. Steady-State Solutions • Definition: The probability that the system is in state i is constant (independent of initial conditions). • Steady-state probability for state i: iP = limtq(t) • Vector: • Calculations in Chapter 15: must solve m simultaneous linear equations in m unknowns. • ATM example: • After 1 min with ∆ = 0.25 • q(1) = (0.435, 0.291, 0.160, 0.073, 0.029, 0.011) • In the limit • P = (0.271, 0.217, 0.173, 0.139, 0.111, 0.089)

  17. Transient Computations for ATM Examplewith ∆ = 0.025

  18. System Statistics • Provide managerial insights • Evaluate system performance and quality of service • Evaluate design options • ATM Example: • Proportion of time ATM is idle: • Efficiency (proportion of time busy): • Proportion of customers rejected: • Proportion of customers who wait: • Expected number in system:

  19. ATM Example (cont’d) • Expected number in queue: • Throughput rate (average number passing through the system): • Balking rate (average number of customers lost): • Average time in system (given by Little’s law):

  20. ATM Design Alternatives • Performance summary (contradictory?) • Busy 73% of time • Space in foyer less than 40% utilized; that is, • (average no. in systems / 5)  100% = 37.36% • 9% of customers lost • Average wait in queue = 60(1.139/1.822) = 37 sec • Options • Add machines • Expand size of foyer • Add human teller

  21. Add More ATMs Rate diagram for 3 ATMs:  = 2,  = 2.5 Comparative analysis

  22. Add Human Teller • Performance • Average service rate for teller: m1 = 1/min • Average service rate for ATM: m2 = 2.5/min • Arrival rate: l = 2/min • Two-server queuing system • Indices: teller = 1; ATM = 2 • State variables: s = (s1, s2, s3)

  23. Explanation: s = (110); teller and ATM are busy, no customers are waiting. Add Human Teller (cont’d) • Events • Arrival = a • Service completion for teller = d1 • Service completion for ATM = d2 • State-transition network

  24. Add Human Teller (cont’d) • Event rates • Arrival: l = 2/min • Service completion for teller: m1 = 1 • Service completion for ATM: m2 = 2.5 • Rate diagram

  25. Add Human Teller (cont’d) • Rate matrixR = (rij) • where rij = transition rate from state i to state j Explanation: r43 =m1 + m2 = 1 + 2.5 = 3.5 where state 4 = (111) and state 3 = (110)

  26. Comparisons For ATM Example Steady-state solution for human teller: s = (000) (010) (100) (110) (111) (112) (113) πP=(0.214, 0.046, 0.313, 0.205, 0.117, 0.067, 0.038)

  27. Properties • Markov process if time between arrivals has exponential distribution • No steady state [transient probabilities are governed by Poisson distribution: • pk(t) = (lt)ke-lt/k!, k = 0,1,2,… ] • Probability of N(t) arrivals in time t is n: Pr{N(t) n} = Pure Birth Processes Example: Hurricanes Rate matrix (let li be arrival rate for state i)

  28. Rate matrix • Let mi be completion rate for state i • State space S = (0,1,…,10} Pure Death Processes • Examples • Delivery of packages • Completion of 10 course study units Steady state probability vector: πP = (1,0,…,0) State 0 is an absorbing state

  29. Pure Death Process Example • Assume all units have the same completion rate: • rk,k–1 = µk = µ, k = 1,…,10 • Then transient probabilities are: • p10–k(t ) = (mt )ke-mt/k !, 0  k < 10, and • p0(t ) = 1 – p1(t ) – · · · – p10(t ) • Let m = 1 completions per week • Probability of completing k units in t = 14 weeks:

  30. Pure Death Process Example (cont’d) Let m = 1 completions per week Probability of k units remaining in t = 14 weeks: • Transient probabilities for k units remaining: • pk(t ) = (mt )10–ke-mt/(10–k) !, 0  k < 10, • and • p0(t ) = 1 – p1(t ) – · · · – p10(t )

  31. General Birth and Death Processes • Examples • Repair shop for a taxi company • Intensive care unit in hospital (turnover of nurses) • Rate matrix • Assume 7 states • Typically, m and l depend on state • Steady state probabilities, pP, will exist

  32. Queuing Systems Input source Customers Departures Service mechanism Queue Queue Discipline: Order in which customers are served; FIFO, LIFO, Random, Priority Five Field Notation: Arrival distribution / Service distribution / Number of servers / Maximum number in the system / Number in the calling population

  33. Queuing Notation Distributions (interarrival and service times) M = Exponential D = Constant time Ek = Erlang GI = General independent (arrivals only) G = General Parameters s = number of servers K = Maximum number in system N = Size of calling population

  34. Finite queue: e.g., Airline reservation system (M/M/s/K) a. Customer arrives but then leaves b. No more arrivals after K Characteristics of Queues Infinite queue: e.g., Mail order company (GI/G/s)

  35. Characteristics of Queues (continued) Finite input source: e.g., Repair shop for taxi company (N vehicles) with s service bays and limited capacity parking lot (K – s spaces). Each repair takes 1 day (GI/D/s/K/N). In this diagram N = K so we have GI/D/s/K/K system.

  36. State-transition network Single Channel Queue – Two Kinds of Service Bank teller: normal service (d), travelers checks (c), idle (i) Let p = portion of customers who buy travelers checks after normal service s1 = number in system, where s1 Î {0,1,2,... } s2 = status of teller, where s2Î {i,d,c} s = (s1, s2)

  37. Single Channel Queue for Bank (cont’d) • State transitions w.r.t. customer departures from teller • Current state: s = ( j,d), j = 1,2,… (teller busy) • Next state: either • s' = ( j–1,d), departure with probability 1 – p, or • s' = ( j,c), get checks with probability p • State transitions w.r.t. customer departures after purchasing travelers checks • Current state: s = ( j,c), j = 1,2,… (customer buying checks) • Next state: s' = ( j–1,d), departure with probability 1 • State transitions w.r.t. customer arrivals • Current state: s = ( j,dor c), j = 1,2,… (teller or checks busy) • Next state: s' = ( j+1,dor c), arrival with probability 1

  38. Single Channel Queue for Bank (cont’d) • Rate of transitions • Event: x (arrival or departure) • Rate of event x : gx (where ga = l, gd = m1, gc = m2) • Conditional probability: p(s,s' | x ) or p(i,j|x ) • Computations: rij = gxp(i,j |x ) • States -- assume limited no. customers at teller: K = 2 • s0 = (0,i ), s1 = (1,d ), s2 = (1,c ), s3 = (2,d ), s4 = (2,c ) • Rate matrix

  39. State-transition network Part Processing with Rework • Consider a machining operation in which there is a 0.4 probability that upon completion, a processed part will not be within tolerance. • Machine is in one of 3 states [s = { (0), (1), (2)}]: • 0 = idle • 1 = working on part for first time • 2 = reworking part Events a = arrival d1 = service completion from state 1 d2 = service completion from state 2

  40. Classification of States Accessible: Possible to go from state i to state j (path exists in the network from i to j). Two states communicate if both are accessible from each other. A system is irreducible if all states communicate. State i is recurrent if the system will return to it some time in the future after leaving it. If a state is not recurrent, it is transient.

  41. What You Should Know About Markov Chains • Definition of a CTMC. • What the difference is between a DTMC and a CTMC. • What the rate matrix and rate diagram are. • What is meant by a transient solution • What is meant by a steady-state solution. • What a birth-death process is. • Classification of the various types of queuing systems.

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