1. Dependability Evaluation through Markovian model

2. Markovian model The combinatorial methods are unable to: - take care easily of the coverage factor - model the maintenance The Markov model is an alternative to the combinatorial methods. Two main concepts: - state - state transition

3. State and state transitions State: the state of a system represents all that must be known to describe the system at any given instant of time For the reliability/availability models each state represents a distinct combination of faulty and fault-free components State transitions govern the changes if state that occur within a system For the reliability/availability models each transition take place when one or more components change state for the event of a fault or for a repair action

4. State and state transitions (cnt.) State transitions are characterized by probabilities, such probability of fault, fault coverage and the probability of repair The probability of being in any given state, s, at some time,t+  t depends both: – the probability that the system was in a state from which it could transit to state state s given that the transition occurs during  t – the probability that the system was in state s at instant t and there was no event in the interval time  t The initial state should be any state, normally it is that representing all fault-free components IMPORTANT: IN A MARKOV CHAIN THE PROBABILITY TRANSITION DEPEND ONLY BY THE ACTUAL STATE, i.e. THE STATE CAPTURES THE HISTORY OF THE OLD TRANSITIONS

5. TMR reliability evaluation IO C1C1 C2C2 C3C3 r/n There are 4 components (1 voter + computation module), therefore each state is represented by 4 bit: if the component is fault-free then the bit value is 1 otherwise the bit value is 0. For example (1,1,1,1) represents the faut-free state For example (0,0,0,0) represents all components faulty

6. TMR reliability evaluation: states diagram

7. Markov chain reliability evaluation methodology State transition probability evaluation: If the fault occurence of a component is exponentially distributed (e - t ) with fault rate equal to ( ), then the probability that the fault-free component at istant t in the interval  t become faulty is equal to: 1 – e -  t

8. Probability property Prob{there is a fault between t e t+  t} = = Prob{there is a fault before t+  t / the component was fault-free at t} = = Prob{there is a faul before t+  t and the component was fault-free at t} Prob{the component was fault-free at t} = Prob{there is a fault before t+  t}  Prob{there is a fault before t} = Prob{the component was fault-free at t} = (1 – e - (t+  t) )  (1 – e - t )  1 – e - (t+  t)  1 + e - t = e - t e - t

9. Probability property = e - t – e - (t+  t) = e - t = e - t _ e - (t+  t) = 1  e -  t e - t e - t If we expand the exponential part we have the following series: 1 – e -  t = 1  1 + (  t) + (  t) 2 + … 2!   t  (  t) 2  … 2! For value of  t << 1, we have the following good approximation: 1 – e -  t   t

10. TMR reliability evaluation: reduced states diagram State (3,1)  (1,1,1,1) State (2,1)  (0,1,1,1) + (1,0,1,1) + (1,1,0,1) State (G)  all the other states Transition probability (in the interval between t and t+  t): · from state (3,1) to state (2,1) -> 3 e  t ; · from state (3,1) to state (G) -> v  t ; from state(2,1) to state (G) -> 2 e  t+ v  t.

11. TMR reliability evaluation Transition probability (in the interval between t and t+  t): · from state (3,1) to state (2,1) -> 3 e  t ; · from state (3,1) to state (G) -> v  t ; from state(2,1) to state (G) -> 2 e  t+ v  t.

12. TMR reliability evaluation Given the Markov process properties, i.e. the probability of being in any given state, s, at some time, t+  t depends both: – the probability that the system was in a state from which it could transit to state state s given that the transition occurs during  t – the probability that the system was in state s at instant t and there was no event in the interval time  t we have that P (3,1) (t+  t) = (1  3 e  t  v  t) P (3,1) (t) P (2,1) (t+  t) = 3 e  t P (3,1) (t) + (1  2 e  t  v  t) P (2,1) (t) P (G) (t+  t) = v  t P (3,1) (t) + (2 e  t + v  t) P (2,1) (t) + P (G) (t)

13. TMR reliability evaluation With algebric operations:  t  0 P (3,1) (t+  t)  P (3,1) (t) =  (3 e + v ) P (3,1) (t) = d P (3,1) (t)  t dt  t  0 P (2,1) (t+  t)  P (2,1) (t) = 3 e P (3,1) (t)  (2 e + v ) P (2,1) (t) = d P (2,1) (t)  t dt  t  0 P (G) (t+  t)  P (G) (t) = v P (3,1) (t) + (2 e + v ) P (2,1) (t) = d P (G) (t)  t dt

14. TMR reliability evaluation i.e: P' 3,1 (t) =  (3 e + v )P 3,1 (t) P' 2,1 (t) = 3 e P 3,1 (t)  (2 e + v )P 2,1 (t) P' G (t) = v P 3,1 (t) + (2 e + v )P 2,1 (t) That in matrix notation can be expressed as:  (t) =  (t) Q(t) dt (P' 3,1 P' 2,1 P' G ) = (P 3,1 P 2,1 P G ) * Q 

15. TMR reliability evaluation the reliability is the probability of being in any fault-free state, i.e, in this case of being in state (3,1) or (2,1). R(t) = P 3,1 (t) + P 2,1 (t) = 1  P G (t) with the initial conditionP 3,1 (0) = 1

16. aaTMR reliability evaluation where:  ( 3 e + v ) 3 e v Q = 0  ( 2 e + v ) (2 e + v ) 0 0 0 P = Q + I  Q = P  I 1  ( 3 e + v ) 3 e v P = 0 1  ( 2 e + v ) (2 e + v ) 0 0 1

17. Properties of Laplace’s transformation

18. Markov Processes for maintenable systems Two kind of events: - fault of a component (module or voter) - repair of the system (of a module or the voter or both) Hypothesis: the maintenance process is exponentially distributed with repair rate equal to 

19. Availability evaluation of TMR system P 3,1 (t) + P 2,1 (t) + P G (t) = 1P 3,1 (0) = 1 P’ 3,1 (t) =  ( 3 e + v ) P 3,1 (t) +  P 2,1 (t) +  P G (t) P’ 2,1 (t) = 3 e P 3,1 (t)  (2 e + v +  ) P 2,1 (t) P’ G (t) = v P 3,1 (t) + (2 e + v ) P 2,1 (t)   P G (t) d  (t) =  (t) Q(t) dt i.e. (P' 3,1 P' 2,1 P' G ) = (P 3,1 P 2,1 P G ) * Q

20. Availability evaluation of TMR system  ( 3 e + v ) 3 e v Q =  ( 2 e + v +  ) (2 e + v )  0    Q = P  I  P = Q + I 1  ( 3 e + v ) 3 e v P =  1  (2 e + v +  ) (2 e + v )  0 1  

21. Istantaneous Availability evaluation of TMR system the Istantaneous Availability is the probability of being in any fault-free state, i.e, in this case of being in state (3,1) or (2,1). A(t) = P 3,1 (t) + P 2,1 (t) = 1  P G (t) with the initial conditionP 3,1 (0) = 1

22. Limiting or steady state Availability evaluation of TMR system P 3,1 (t) + P 2,1 (t) + P G (t) = 1P 3,1 (0) = 1 with t   we have that  P’(t) = 0 P’ 3,1 (t) = 0 =  ( 3 e + v ) P 3,1 (t) +  P 2,1 (t) +  P G (t) P’ 2,1 (t) = 0 = 3 e P 3,1 (t)  (2 e + v +  ) P 2,1 (t) P’ G (t) = 0 = v P 3,1 (t) + (2 e + v ) P 2,1 (t)   P G (t)

23. Limiting or steady state Availability evaluation of TMR system P 3,1 (t) + P 2,1 (t) + P G (t) = 1P 3,1 (0) = 1 with t   we have that  P’(t) = 0 and P(t) = P P’ 3,1 (t) = 0 =  ( 3 e + v ) P 3,1 +  P 2,1 +  P G P’ 2,1 (t) = 0 = 3 e P 3,1  (2 e + v +  ) P 2,1 P’ G (t) = 0 = v P 3,1 + (2 e + v ) P 2,1 (t)   P G

24. Limiting or steady state Availability evaluation of TMR system P 3,1 + P 2,1 + P G = 1 P 3,1 = P 2,1 = P G =

25. Safety evaluation Four kind of events: - fault of a component (module or voter) correcttly diagnoticated - fault of a component not detected - correct repair of the system (of a module or the voter or both) - uncorrect repair of the system  fault rate  repair rate C g  fault detection coverage factor C r  correct repair coverage factor

26. Single component Safety evaluation 0  fault free state GS  safe fault state GI  unsafe fault state Hypothesis: -if a fault is not well diagnosticated then it will never be detected -If a reconfiguration is not wel done then it will be never detected Therefore GI is an absorbing state

27. Single component Safety evaluation Safety = probability to stay in state 0 or GS P O (t) + P GS (t) = 1  P GI (t)P O (0) = 1 P’ O (t) =  ( (1  C g ) + C g )) P O (t) +  C r P GS (t) P’ GS (t) = C g P O (t)  (  (1  C r )+  C r ) P GS (t) P’ GI (t) = (1  C g ) P O (t) + (  (1  C r )P GS (t)

28. Single component Safety evaluation  C g (1-C g ) Q = CrCr  (1-C r ) 00 0 d  (t) =  (t) Q(t) dt i.e. (P' 3,1 P' 2,1 P' G ) = (P 3,1 P 2,1 P G ) * Q

29. Performability Index taking into account even the performance of the system given its state (related to the number of fault-free components) We will discuss it when we will know how evaluate the performance of a system

30. I O C 21 C 23 C 22 C 111 C 12 C 112 C 11 C1C1 C2C2 R 11 =  R 111. R 112 R 1 = 1 - (1 - R 11 ). (1 - R 12 ) R 2 = 1 - (1 - R 21 ). (1 - R 22 ). (1 - R 23 ) R =  R 1. R 2 Reliability/Availability/Safety evaluation of complex system