Chapter 8. Calculation of PFD using Markov

Transcription

1 Chapter 8. Calculation of PFD using Markov Mary Ann Lundteigen Marvin Rausand RAMS Group Department of Mechanical and Industrial Engineering NTNU (Version 0.1) Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 1 / 33

2 Learning Objectives Introduction The main learning objectives associated with these slides are to: Study how Markov analysis can be used to calculate the PFD Become familiar with how CCFs and the effects of DU and DD failures are included Understand how Markov model can be used to incorporate the effects of demand rate and demand duration The slides include topics from Chapter 8 in Reliability of Safety-Critical Systems: Theory and Applications. DOI: / Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 2 / 33

3 Introduction Outline of Presentation 1 Introduction 2 About Markov Approach 3 Using Steady State to Calculate PFD 4 Using Time Dependent Solutions to calculate PFD 5 Calculating MTTF S 6 Including Demand Duration 7 Adding C MooN in PDS-Method Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 3 / 33

4 Introduction Markov Approach in Brief Some keywords: Suitable for multistate and dynammic systems Must satisfy the Markov properties Can model system states, beyond failure states Can be used to find analytical formulas and calculate steady state and time-dependent probabilities Can be used to determine mean time to first failure (MTTF S ) Figure: Russian mathematician Andrei Markov ( ) Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 4 / 33

5 About Markov Approach The Markov approach - step by step 1. Define system states (table format) 2. Set up the transition diagram ( Markov model ) 3. Include the transition rates 4. Set up the transition matrix 5. Do your calculations, either in terms of time dependent analysis or in terms of steady state Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 5 / 33

6 About Markov Approach The Markov approach - example Consider a 2oo3 voted system of identical components. Step 1: Set up the system states, first assuming no common cause failures. State State description 0 Three channels are functioning 1 Two channels are functioning and one is failed 2 One channel is functioning and two are failed 3 Three channels are failed It is assumed that repair always restores the system to a fully functional state. Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 6 / 33

7 About Markov Approach The Markov Approach - Example Consider a 2oo3 voted system of identical components. Step 2 and 3: Set up the Markov model, and include the transition rates 3 OK 0 3λ 2λ 2 OK 1 OK 1 F 2 F μ λ 3 F 3 μ 2 μ 3 The failed states of this subsystem are state 2 and state 3. Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 7 / 33

8 About Markov Approach The Markov approach - example Consider a 2oo3 voted system of identical components. Step 4: Set up the transition matrix A = 3λ 3λ 0 0 µ 1 (µ 1 + 2λ) 2λ 0 µ 2 0 (µ 2 + λ) λ µ 3 0 µ 3 Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 8 / 33

9 About Markov Approach The Markov Approach - Example What if CCFs are included? λ (c) 3 OK 0 3λ (i) μ 1 2 OK 1 F 1 λ (c) 2λ (i) 1 OK 2 F 2 λ 3 F 3 μ 2 μ 3 A = (3λ (i) + λ (c) ) 3λ (i) 0 λ (c) µ 1 (µ 1 + 2λ (i) + λ (c) ) 2λ (i) λ (c) µ 2 0 (µ 2 + λ) λ µ µ 3 Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 9 / 33

10 About Markov Approach The Markov Approach - What to Calculate? With basis in the Markov model, it is possible to calculate: Time dependent probabilities ( Probability of being in state i at time t) Steady state probabilities ( Average probability of being in state i, % of time in state i ) Visit frequency to a specific state or a set of states (e.g., into the failed state) Mean time to first entry to a specific state (e.g., mean time to failure) Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 10 / 33

11 About Markov Approach Using Markov to Calculate PFD Let D be the set of states where the voted system is down (e.g., in the failed state). Two calculate PFD avg, we have two options: Option 1: Calculations based on time dependent probabilities: PFD(t) = P i (t) i D The PFD avg becomes: PFD avg = 1 τ τ 0 PFD(t)dt Option 2: Calculations based on steady state probabilities. In this case: PFD avg = i D P i Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 11 / 33

12 Using Steady State to Calculate PFD Using Steady-State Probabilities Consider a single system that may fail due to DU or DD failures. The system states are: State State description 0 The channel is functioning (no DU or DD failures) 1 The channel has a DD fault 2 The channel has a DU fault λ DD λ DU DD 1 OK 0 DU 2 μ DD μ DU Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 12 / 33

13 Parameters Using Steady State to Calculate PFD The Markov transition matrix becomes: A = (λ DD + λ DU ) λ DD λ DU µ DD µ DD 0 µ DU 0 µ DU Parameter Description Comments λ DU Dangerous undetected (DU) failure rate λ DD Dangerous undetected (DD) failure rate µ DU Repair rate of DU failures 1/( τ 2 + MRT ) µ DD Repair rate of DD failures 1/MTTR Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 13 / 33

14 Using Steady State to Calculate PFD Solving Steady State Equations Three states (0,1,2) means that we need three equations to solve for P 0, P 1, and P 2. The approach is: Step 1: Set up the steady state equations from PA = 0 The main approach is to (i) choose two equations (out of the three) from the above equations, preferably the ones with most zeros, plus (ii) the equation P 0 + P 1 + P 2 = 1. The equations then becomes: P 0 + P 1 + P 2 = 1 λ DD P 0 µ DD P 1 = 0 λ DU P 0 µ DU P 2 = 0 Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 14 / 33

15 Using Steady State to Calculate PFD Solving Steady State Equations (cont.) Step 2: Solve for P 0, P 1, and P 2 : By hand-calculations or e.g. MAPLE, we find that: P 0 = 1 λ DD µ DD + λ DU µ DU + 1 P 1 = λ DD µ DD P 0 P 2 = λ DU µ DU P 0 Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 15 / 33

16 Using Steady State to Calculate PFD Using Maple - Code Example Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 16 / 33

17 Using Steady State to Calculate PFD Solving Steady State Equations (cont.) Step 3: Determine PFD avg : Since the failed states are state 1 and state 2, we find that: PFD avg = P 1 + P 2 = λ DDMTTR + λ DU ( τ 2 + MRT ) λ DD MTTR + λ DU ( τ 2 + MRT ) + 1 λ DD MTTR + λ DU ( τ 2 + MRT ) Note that µ DD and µ DU have been replaced by 1/MTTR and 1/( τ 2 + MRT ) The approximation is possible because the denominator is close to 1 with λ DD and λ DU being very small Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 17 / 33

18 Using Steady State to Calculate PFD Solving Steady State Equations (cont.) Step 4: Insert the values of the parameters and calculate the result: using input data is table 7.2 in textbook, we get: The PFD avg without the approximation becomes The PFD avg with the approximation becomes For more examples, visit the textbook. Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 18 / 33

19 Using Time Dependent Solutions to calculate PFD Using Time-Dependent Probabilities Consider a single system that may fail due to DU or DD failures. The system states are: State State description 0 The channel is functioning (no DU or DD failures) 1 The channel has a DD fault 2 The channel has a DU fault Note that we do not need a return from the failed state after a DU failure. We assume that the average calculated for the first proof test interval is equal to the long term average. λ DD λ DU DD 1 OK 0 DU 2 μ DD Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 19 / 33

20 Using Time Dependent Solutions to calculate PFD Solving Time-Dependent Probabilities With the absorbing state, the new transition matrix becomes: A = (λ DD + λ DU ) λ DD λ DU µ DD µ DD With one of the rows having just zeros, we see that P 2 (t) and P 2 (t) disappears from the equation. To solve the equation, we reduce the transition matrix (now named A t :) to include only up-states : ( ) (λdd + λ A t = DU ) λ DD µ DD µ DD Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 20 / 33

21 Using Time Dependent Solutions to calculate PFD Solving Time-Dependent Probabilities (cont.) Step 2: Solve for P 0 (t) and P 1 (t) (P 2 (t) can be found from the two first): The Laplace transform becomes: ( ) (P0 (s), P 1 (s))a (λdd + λ t = DU ) λ DD = (sp µ DD µ 0 (s) 1, sp 1 (s)) DD The LinearAlgebra (for handling matrix operations) and the inttrans packages (with invlaplace command) may be used in MAPLE to solve the equations. The time-dependent state solution becomes: ( ) (λdd + λ (P 0 (t), P 1 (t))a t = DU ) λ DD = ( P µ DD µ 0 (t), P 0 (t)) DD The dsolve command in MAPLE may be used to solve for the time-dependent state probabilities. Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 21 / 33

22 Using Time Dependent Solutions to calculate PFD Solving Time-Dependent Probabilities (cont.) Step 3 Find PFD avg : Once the state probabilities have been found, we can calculate PFD avg as: PFD avg = 1 τ τ 0 i D P i (t)dt = 1 1 τ τ 0 i U P i (t)dt where D are the states that are defined as failed state, and U are the states where the system is functioning (even if degraded). MAPLE may be used for this purpose using the int-function. Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 22 / 33

23 Using Time Dependent Solutions to calculate PFD Solving time-dependent state equations (continued) Step 4: Insert the values of the parameters and calculate the result: Reference to input data is table 7.2 in textbook. The PFD avg without the approximation becomes The PFD avg with the approximation becomes For more examples, visit the textbook. Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 23 / 33

24 Using Time Dependent Solutions to calculate PFD Using Maple - code example Code: Result: Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 24 / 33

25 Calculating MTTF S Mean Time to First Failure (MTTF S ) The mean time to first failure, here called MTTF S, can be solved by setting s = 0 in the Laplace transform equations: Consider the single system previously addressed for time-dependent probabilities. With s=0 in the Laplace transform we get: ( ) [P0 (0), P 1 (0)]A t = [P0 (0), P 1 (0)] (λdd + λ DU ) λ DD = [ 1, 0] µ DD µ DD By using hand-calculation or MAPLE, the result becomes: MTTF S = P 0 (0) + P 1 (0) = 1 + λ DD 1 λ DU µ DD λ DU λ DU (You may verify the approximation by inserting parameter values from table 7.2 in textbook. See also Chapter 5.5.5) Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 25 / 33

26 Including Demand Duration Including Demand Duration Consider a safety-critical system (single) that may fail due to DU failure (we omit DD failures). We assume that the system is operating in the low-demand mode, and that a failure to operate on demand may result in a hazardous state. It is further assumed that the SIF is NOT the ultimate safety barrier, so a restoration is possible. The system states are: State State description 0 The channel is functioning (no DU failure) 1 The channel has a DU fault 2 A demand has occurred 3 The system is in a hazardous state μ de λ de OK 0 μ DU λ DU Demand 2 μ T DU 1 λ DU Hazard λ de 3 Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 26 / 33

27 Including Demand Duration Transition Matrix The Markov transition matrix becomes: A = (λ DU + λ de ) λ DU λ de 0 µ DU (µ DU + λ de ) 0 λ de µ de 0 (µ de + λ DU ) λ DU µ T 0 0 µ T Parameter Description λ DU Dangerous undetected (DU) failure rate λ de Demand rate µ de Restoration rate after demand µ T Restoration rate for Hazardous event Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 27 / 33

28 Including Demand Duration Solving for PFD avg It is assumed that the time dependent probabilities have been found using MAPLE (Inverse laplace transforms or integration). Then: PFD avg = 1 τ τ 0 P 1 (t)dt The resulting equation is rather extensive. For high-reliability channels with short demand duration we have λ DU µ DU µ de. In this case, we get approximately PFD 1,avg λ DU µ de (λ de + µ de )(λ de + µ DU ) When λ de µ de, the following approximation is also adequate PFD 2,avg λ DU λ de + µ DU Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 28 / 33

29 Including Demand Duration Including Demand Duration Consider a safety-critical system (single) that may fail due to DU failure (we omit DD failures). We assume that the system is operating in the low-demand mode, and that a failure to operate on demand may result in a hazardous state. In this case, however, the SIF IS the ultimate safety barrier. The system states are: State State description 0 The channel is functioning (no DU failure) 1 The channel has a DU fault 2 A demand has occurred 3 The system is in a hazardous state μ de λ de OK 0 λ DU Demand 2 DU 1 λ DU Hazard λ de 3 Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 29 / 33

30 Including Demand Duration Solving for PFD avg Since the SIF is the ultimiate safety barrier, state 3 is an absorbing state. The PFD avg becomes: PFD avg = 1 τ τ 0 P 1 (t)dt The resulting equation is, as for the SIF that was not the ultimate safety barrier, rather extensive. Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 30 / 33

31 Including Demand Duration Solving for HEF(t) and HEF avg For both situations, i.e. that the SIF is the ultimate safety barrier or is not the ultimate safety barrier, we can find the hazardous event frequency (HEF): The PFD avg becomes: τ PFD avg = 1 P 1 (t)dt τ 0 The resulting equation is, as for the SIF that was not the ultimate safety barrier, rather extensive. HEF(t) = P 1 (t) λ de + P 2 (t) λ DU The average HEF in the proof test interval (0,τ ) is HEF = 1 τ τ 0 HEF(t) dt = 1 τ τ 0 (P 1 (t) λ de + P 2 (t) λ DU ) dt Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 31 / 33

32 Adding C MooN in PDS-Method PDS Method and C MooN If the PDS method is used, it is necessary to address the C MooN factor for the transitions. C MooN gives the correction for N M + 1..N failures in a MooN voted system This means that C 1oo3 includes 3 failures, while C 2oo4 accounts for 3 and 4 failures Consider a CCF transition between two states. Then (C(N-i+1)ooN C(N-i)ooN) is the correction factor for the case that exactly i out of N components fails Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 32 / 33

33 Adding C MooN in PDS-Method PDS Method and C MooN Consider a subsystem voted 2oo4. In this case, the possible transitions becomes: C1oo4β λ (C3oo4-C2oo4)β λ C1oo2β λ 4λ i 3λ i 2λ i λ (C2oo4-C1oo4)β λ (C2oo3-C1oo3)β λ C1oo3β λ Functioning states if 2oo4 system Failed states if 2oo4 system Lundteigen& Rausand Chapter 8.Calculation of PFD using Markov (Version 0.1) 33 / 33