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Fault Tree Analysis Part 11 – Markov Model. State Space Method Example: parallel structure of two components Possible System States: 0 (both components.

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Presentation on theme: "Fault Tree Analysis Part 11 – Markov Model. State Space Method Example: parallel structure of two components Possible System States: 0 (both components."— Presentation transcript:

1 Fault Tree Analysis Part 11 – Markov Model

2 State Space Method Example: parallel structure of two components Possible System States: 0 (both components in failed state); 1 (component 1 functioning, component 2 in failed state); 2 (component 2 functioning, component 1 in failed state); 3 (both components functioning).

3 State Space Diagram 23 10

4 Markov Processes The event means that the system at time t is in state j and j = 1, 2, …,r. The probability of this event is denoted by The transitions between the states may be described by a stochastic process A stochastic process satisfying the Markov property is called the Markov process.

5 Markov Property Given that a system is in state i at time t, i.e. X(t)=i, the future states X(t+v) do not depends on the previous states X(u), u<t. For all possible x(u) and 0 ≦ u<t.

6 Stationary Transition Probability A Markov process with stationary transition properties is often called a process with no memory.

7 Properties of Transition Probabilities Chapman-Kolmogorov equation

8 Transition Rate

9 Derivation of State Equation (1) From Chapman-Kolmogorov equation Substitute

10 Derivation of State Equation (2) After dividing by Δt, letting Δt→0, we get the state equations.

11 State Equations

12 Simplified State Equations Since the initial state is known, the state equations can be simplified by omitting the first index i

13 State Equations in Matrix Notation Let Then where

14 Additional Properties Notice that the sums of the columns of the transition rate matrix add up to zero. The following constraint must be imposed The mean staying time in state j

15 Example Consider a single component with two states: 1 (the component is working) and 0 (the component is in a failed state). Thus, The state equations:

16 Example Since It can be derived that

17 Frequency of Departure from State j to State k The unconditional probability of a departure from state j to state k in the time interval (t, t+Δt] is The frequency of departure

18 Frequency of Departure from State j at Steady State At steady state The total frequency

19 Frequency of Arrival to State j at Steady State The frequency of arrival from state k to state j at the steady state The total frequency of arrivals to state j (from state equations at steady state)

20 Visit Frequency The visit frequency to state j is defined as the expected number of visits to state j per unit time.

21 Mean Duration of a Visit The total departure rate from state j Since the departure rate is constant, the duration of a stay in state j should be exponentially distributed with parameter Thus, the mean duration of stay is

22 A Useful Relation The mean proportion of time the system is spending in state j ( ) A special case is the formula for unavailability under corrective maintenance policy

23 System Availability Let S={1, 2, …, r} be the set of all possible states of a system. Let B denote the subset of states in which the system is functioning. Let F=S-B denote the states in which the system is failed. Then, the average (or long-term) system availability and unavailability are


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