Markov Reliability and Availability Analysis. Markov Processes

Transcription

1 Markov Reliability and Availability Analysis Firma convenzione Politecnico Part II: Continuous di Milano e Time Veneranda Discrete Fabbrica State del Duomo di Milano Markov Processes Aula Magna Rettorato Mercoledì27 maggio 2015

2 Continuous Time Discrete State Markov Processes The stochastic process may be observed at: Discrete times DISCRETE-TIME DISCRETE-STATE MARKOV PROCESSES 0 t 1 t 2 t 3 t n T m t Continuously CONTINUOUS-TIME DISCRETE-STATE MARKOV PROCESS 0 T m t

3 The conceptual model: Continuous-Time 3 The stochastic process is observed continuously and transitions are assumed to occur continuously in time state j = 0 state j = 5 state j = 2 t 0 state j = N T m

4 The conceptual model: Finite State Space 4 The random process of system transition between states in time is described by a stochastic process {X(t); t 0} X t system state at time t X 3.6 = 5: the system is in state number 5 at time t = 3.6 OBJECTIVE: Computing the probability that the system is in a given state as a function of time, for all possible states Xt j, t 0, T, j 0,1, N P m...,

5 Needed Information 5 Objective: P X t = j, t 0, T m, j = 0,1,, N What do we need? Transition Probabilities!

6 The conceptual model: Transition Probabilities 6 Transition probability that the system moves to state j at time t + ν given that it is in state i at current time t and given the previous system history P X t j X t i, X u xu, 0 u t (i = 0, 1,, N, j = 0, 1,, N) i j t 0 u t t + ν T m

7 The conceptual model: Markov Assumption 7 IN GENERAL STOCHASTIC PROCESSES: the probability of a future state of the system usually depends on its entire life history PX t j X t i, X u xu, 0 u t IN MARKOV PROCESSES: the probability of a future state of the system only depends on its present state P X (i = 0, 1,, N, j = 0, 1,, N) t j X t i, X u xu, 0 u P X = t j X t i (i = 0, 1,, N, j = 0, 1,, N) THE PROCESS HAS NO MEMORY t

8 The conceptual model: homogeneous Markov process 8 If the transition probability depends on the interval ν and not on the individual times t and t + ν the transition probabilities are stationary the Markov process is homogeneous in time p ij t, t PX t j X t i p ij i j 0 T m t t 1 t 1 + ν i j 0 T m t t 2 t 2 + ν

9 The conceptual model: Transition Rates 9 HYPOTHESIS: The time interval ν = dt is small such that only one event (i.e., one stochastic transition) can occur within it p ij dt = P X t + dt = j X t = i = 1 e α ij dt ij dt = dt, (Taylor 1 st order expansion) dt0 dt lim 0 dt ij = transition rate from state i to state j

10 The conceptual model: The Transition Probability Matrix (1) 10 p ij dt dt dt, ij dt0 In analogy with the discrete-time formulation: dt lim 0 dt Discrete-time transition probability matrix Continuous-time transition probability matrix i j N 0 p p... p A 1 p p... p N p p... p N N N 0 N1 NN N 1 dt 0 j 01 dt... 0N dt j1 N A 10 dt 1 dt 1 j... 1N dt j0 j p ii dt = 1 p ij dt = 1 dt α ij + θ(dt) j i j i

11 The conceptual model: the fundamental matrix equation (1) 11 In analogy with the discrete-time formulation: P( t dt) P( t) A P 0 t Pt... P t 1 N t dtpt dt... P t dt P N 0 1 = N 1 dt 0 j 01 dt... 0N dt j1 N A 10 dt 1 dt 1 j... 1N dt j0 j First-equation: N P0 t dt 1 dt0 j P0 t 10P1 t dt... N 0PN tdt j1

12 The conceptual model: the fundamental matrix equation (2) 12 N P0 t dt 1 dt0 j P0 t 10P1 t dt... N 0PN tdt j1 subtract P 0 (t) on both sides t dt P t P t P t P t dt P t dt P t dt P t dt P t 0 N j1 0 j N 0 divide by dt N P0 0 0 P N dt j1 t P t P t j 0 let dt 0 N N lim dt0 P 0 t dt P0 t dt dp dt 0 N j1... P t P t P t 0 j N 0 N

13 The conceptual model: The Transition Probability Matrix 13 Extending to the other equations: N... dp P t A *, A *... dt j 01 0N j1 α 00 N 10 1 j 1N j0 j1 α 11 TRANSITION RATE MATRIX It will be indicated as A System of linear, first-order differential equations in the unknown state probabilities P t, j 0,1,2,..., N, t 0 j

14 Exercise 1 Consider a system made by one component which can be in two states: working or failed. Assume constant failure rate λ and constant repair rate μ. You are required to: Draw the Markov diagram Find the transition rate matrix, A 14 MARKOV DIAGRAM A TRANSITION RATE MATRIX

15 Exercise 1 (Solution) 15 Discrete states = 0 component working 1 component failed Transition rates = λ rate of failure (i.e., from 0 to 1) = μ rate of repair (i.e., from 1 to 0) MARKOV DIAGRAM A TRANSITION RATE MATRIX

16 Exercise 2 16 Consider a system made by N identical components which can be in two states: working or failed. Assume constant failure rate λ, that N repairman are available and that the single component repair rate is constant and equal to μ. You are required to: Draw the Markov diagram Find the transition rate matrix, A

17 Exercise 2 - Solution 17 Consider a system made by N identical components which can be in two states: working or failed. Assume constant failure rate λ, that N repairmen are available and that the component repair rate is constant and equal to μ. You are required to: Draw the Markov diagram Find the transition rate matrix -System discrete states: State 0: none failed, all components function State 1: one component failed, N-1 function State 2: two components failed, N-2 function State N: all components failed, none function

18 Exercise 2 - Solution 18 HYPOTHESES: - one event (failure or repair of one component) can occur in the small t - the events are mutually exclusive all functioning all failed Explanation: Probability of transition 0 1 = = probability { anyone of the N components fails in Δt } = = probability { component 1 fails in Δt or component 2 fails in Δt or } = = probability { component 1 fails in Δt } + probability { component 2 fails in Δt } + = = λδt + λδt + = NλΔt

19 Exercise 3: system with N identical components and 1 repairman available Consider a system made by N identical components which can be in two states: working or failed. Assume constant failure rate λ, that 1 repairman is available and that the component repair rate is constant and equal to μ. You are required to: Draw the Markov diagram Find the transition rate matrix

20 Exercise 3: Solution SYSTEM CHARACTERISTICS: - The system is made of N identical components - Each component can be in two states: working or failed - The transition rates are constant = λ rate of failure = μ rate of repair - One repairman is available all functioning all failed

21 21 Solution to the Fundamental Equation

22 Solution to the fundamental equation of the Markov process continuous in time 22 dp P t A dt P0 C where N... j1 N A System of linear, first-order differential equations in the unknown state probabilities 0 j 01 0N 10 1 j 1N j0 j1 P t, j 0,1,2,..., N, t 0 j USE LAPLACE TRANSFORM

23 Solution to the fundamental equation of of the Markov process continuous in time: the Lapace Transform Method 23 st Laplace Transform: P % j s L Pj t e P 0 j t dt, j 0,1,..., N dpj t First derivative: L s P% js Pj0, j 0,1,..., N dt Apply the Laplace operator to dp P t A dt d P L dt t L P t A First derivative sp% s C P% s A Linearity P % s C s I A 1 P t = inverse transform of P ~ s

24 Solution to the fundamental equation of of the Markov process continuous in time: steady state probabilities 24 At steady state d P t dt Solve the (linear) system: 0 d P t dt A N j0 0 j 1 A A 0 P t It can be shown that j N D i0 j D i j 0,1,2,..., N D = determinant of the square matrix obtained from j by deleting the j-th row and column A

25 Exercise 4: one component/one repairman Solution to the fundamental equation (1) 25 Consider a system made by one component which can be in two states: working ( 0 ) or Failed ( 1 ). Assume constant failure rate λ and constant repair rate μ an that the component is working at t = 0: C = [1 0] - You are required to find the component steady state and instantaneous availability

26 26 Exercise 4: one component/one repairman Solution to the fundamental equation (2) Solve Compute 1 P s C si A % 1 si A s s s s s s s A si s s A si det 1

27 Exercise 4: one component/one repairman Solution to the fundamental equation (3) 27 Anti-Transform P ~ s s 1 ax It is known that 1 L e s a s s ss and L 1 s 1 s a 1 a 1 e ax STATE PROBABILITY VECTOR P() t e e t t P 0 t system instantaneous availability (probability of being in operational state 0 at time t) P 1 t system instantaneous unavailability (probability of being in failed state 1 at time t)

28 Exercise 4: one component/one repairman Solution to the fundamental equation (4) 28 TIME-DEPENDENT STATE PROBABILITIES P 0 P 1 STEADY STATE PROBABILITIES 0 t t e 1 lim P 0 t lim P 1 t t t 1 MTBF 1 1 MTTR MTBF = average fraction of time the system is functioning t t e (system instantaneous availability) (system instantaneous unavailability) 1 MTTR 1 1 MTTR MTBF = average fraction of time the system is down (i.e., under repair)

29 29 Quantity of Interest

30 Frequency of departure from a state 30 Unconditional probability of arriving in state j in the next dt departing from state i at time t: P[X t + dt = j, X t = i] P[X t + dt = j, X t = i]=p X t + dt = j X t = i P X t = i = p ij dt P i (t) Frequency of departure from state i to state j: ν ij dep t = lim dt 0 p ij dt P i t dt = α ij P i (t) = (at steady state) v dep ij ij i Total frequency of departure from state i to any other state j: dep N v t P t P t = vi ii i i ij i ii i j0 ji (at steady state) dep

31 Frequency of arrival to a state 31 In analogy, the total frequency of arrivals to state i from any other state k: N arr i ki k k 0 ki N arr i ki k k 0 ki v t P t (at steady state) AT STEADY STATE: frequency of departures from state i = frequency of arrivals to state i

32 System Failure Intensity 32 SYSTEM FAILURE INTENSITY W f : Rate at which system failures occur Expected number of system failures per unit of time Rate of exiting a success state to go into one of fault S = set of success states of the system F = set of failure states of the system P i t W t P t f i i F is = probability of the system being in the functioning state i at time t λ i F = conditional (transition) probability of leaving success state i towards failure states

33 System Repair Intensity 33 SYSTEM REPAIR INTENSIT W r : Rate at which system repairs occur Expected number of system repairs per unit of time Rate of exiting a failed state to go into one of success F = set of failure states of the system P j t S = set of success states of the system = probability of the system being in the failure state j at time t μ j S = conditional (transition) probability of leaving failure state j towards success states W t P t r j j S jf

34 Exercise 5: one component/one repairman 34 Consider a system made by one component which can be in two states: working ( 0 ) or Failed ( 1 ). Assume constant failure rate λ and constant repair rate μ an that the component is working at t = 0: C = [1 0] - You are required to find the failure and repair intensities

35 Exercise 5: one component/one repairman Failure and repair intensities 35 functioning failed S = set of success states of the system = {0} F = set of failure states of the system = {1} λ i F = λ μ j S = μ W W r f W t P t f i i F is 2 e t t P t e 0 W t P t t t P t 1 r j j S jf

36 Sojourn Time in a state (1) 36 Time of occupance of state i (sojourn time) = T i Markov property and time homogeneity imply that if at time t the process is in state i, the time remaining in state i is independent of time already spent in state i P T i > t + s T i > t = P X t + u = i, 0 u s X τ = i, 0 τ t) = = P(X t + u = i, 0 u s X t = i) (by Markov property) = P(X u = i, 0 u s X 0 = i) (by homogeneity) = P(T i > s) Memoryless Property The only distribution satisfying the memoryless property is the Exponential distribution T i ~Exp

37 Sojourn Time in a state (2) 37 System departure rate from state i (at steady state): α ii T i ~Exp( α ii ) Expected sojourn time l i : average time of occupancy of state i l i = E T i = 1 α ii

38 Sojourn Time in a state (3) 38 Total frequency of departure at steady state: Average time of occupancy of state: li 1 ii dep i ii i dep i ii i i i l v i dep = v i arr i arr l i i The mean proportion of time i that the system spends in state i is equal to the total frequency of arrivals to state i multiplied by the mean duration of one visit in state i

39 System Availability 39 System instantaneous availability at time t = sum of the probabilities of being in a success state at time t S = set of success states of the system F = set of failure states of the system 1 1 p t P t q t P t is i jf 1 p% s Pi s Pj s s % % is In the Laplace domain jf j

40 System Reliability 40 TWO CASES: 1) Non-Reparaible Systems No repairs allowed 2) Reparaible Systems Repairs allowed

41 System Reliability: Non-Reparaible Systems 41 No repairs allowed Reliability = Availability Rt pt 1 qt In the Laplace Domain: Mean Time to Failure (MTTF): 1 R % s P % i s P % j s s is jf MTTF 0 st dt Rt e dt R0 P0 P s R t 0 s0 ~ is ~ i 1 s jf ~ j s0

42 System Reliability 42 TWO CASES: 1) Non-reparaible systems No repairs allowed 2) Reparaible systems Repairs allowed

43 System Reliability: Reparaible Systems (1) Transform the failed states j F into absorbing states (the system cannot be repaired it is not possible to escape from a failed state) operating failed S F 0,1 2 operating A A = 2λ 2λ 0 μ μ + λ λ A = 2λ μ 2λ μ + λ The new matrix A' contains the transition rates for transitions only among the success states i S (the reduced system is virtually functioning continuously with no interruptions)

44 System Reliability: Reparaible Systems (2) Solve the reduced problem of ' for the probabilities of being in these (transient) safe states d P * A t, i S * dt t P * Reliability is t i A' R t P t Mean Time To Failure (MTTF) 0 % 0 % 0 MTTF R t dt P R * * NOTICE: in the reduced problem we have only transient states P 0 is i P i i i

45 Exercise 6 45 Consider a system made by 2 identical components in parallel. Each component can be in two states: working or failed. Assume constant failure rate λ, that 2 repairman are available and that the single component repair rate is constant and equal to μ. You are required to: find the system reliability find the system MTTF

46 Exercise 6: system with two identical components and two repairmen available (1) 46 both functioning one failed both failed 2 A TWO CASES: a) Parallel logic (1 out of 2) b) Series logic (2 out of 2)

47 Example 8: system with two identical components and two repairmen available (2) 47 PARALLEL LOGIC system operating system operating system failed 2 A System discrete states: State 0: system is operating (both components functioning) State 1: system is operating (only one of the two components functioning) State 2: system is failed (both components failed)

48 Example 8: system with two identical components and two repairmen available (3) Exclude all the failed states j F from the transition rate matrix System reliability R(t) :=probability of the system being in safe states 0 or 1 continuously from t = 0 PARTITION OF THE TRANSITION MATRIX A A' 0 2 2

49 Example 8: system with two identical components and two repairmen available (4) Solve the reduced problem of ' for the probabilities of being in these (transient) safe states * A t, i S P i In the time domain: d P dt P * P * t A' dp dt 2 2 P t In the Laplace domain: ~ P * ~ * 1 s P 0 si A' 1 0 ' 1 P% s si A

50 Example 8: system with two identical components and two repairmen available (5) 1 0 ' 1 P% s si A 2 2 with A' 50 si A' s 2 2 s si A' 1 1 s 2 s 2s 2 s 2 si 1 1 s 2 A' s 2 s s 0 1 where 0,

51 51 It is known that ' ~ * * s s s s s s s A si C s P () t t e e Rt s P s P s R 1 0 ~ ~ ~ SYSTEM RELIABILITY In the Laplace domain In the time domain SYSTEM RELIABILITY Example 8: system with two identical components and two repairmen available (6) L 1 1 s a =

52 Example 8: system with two identical components and two repairmen available (7) 52 MEAN TIME TO FAILURE MTTF R% 0 P% 0 1 Starting from P% s C s I : A ' 1 T w MTTF C A w with i i T 1 i0 ~ * 0 P i MTTF

53 Exercise 7 53 Consider a system made by 2 identical components in series. Each component can be in two states: working or failed. Assume constant failure rate λ, that 2 repairman are available and that the single component repair rate is constant and equal to μ. You are required to: find the system reliability find the system MTTF

54 Example 8: system with two identical components and two repairmen available (8) 54 both functioning one failed both failed 2 A TWO CASES: a) Parallel logic (1 out of 2) b) Series logic (2 out of 2)

55 Example 8: system with two identical components and two repairmen available (9) 55 SERIES LOGIC system operating system failed system failed 2 A System discrete states: State 0: system is operating (both components functioning) State 1: system is failed (only one of the two components functioning) State 2: system is failed (both components failed)

56 Example 8: system with two identical components and two repairmen available (9) Exclude all the failed states j F from the transition rate matrix System reliability R(t) :=probability of the system being in safe states 0 or 1 continuously from t = 0 PARTITION OF THE TRANSITION MATRIX A A'

57 Example 8: system with two identical components and two repairmen available (10) Solve the reduced problem of ' for the probabilities of being in these (transient) safe states * A t, i S P i Easy to solve in the time domain: dp dt P 0 P C A' which simplifies to * dp0 2 P dt 0 1 * P 0 * 0 Rt P t () 2 t 0 e