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System Reliability Analysis - Multi State Models and General Configurations

Systems Engineering Program. Department of Engineering Management, Information and Systems. EMIS 7305/5305 Systems Reliability, Supportability and Availability Analysis. System Reliability Analysis - Multi State Models and General Configurations. Dr. Jerrell T. Stracener, SAE Fellow.

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System Reliability Analysis - Multi State Models and General Configurations

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  1. Systems Engineering Program Department of Engineering Management, Information and Systems EMIS 7305/5305 Systems Reliability, Supportability and Availability Analysis System Reliability Analysis - Multi State Models and General Configurations Dr. Jerrell T. Stracener, SAE Fellow Leadership in Engineering

  2. Multi State Reliability Models

  3. Multistate Reliability Models • In all cases previously discussed the element analyzed could assume only one of two mutually exclusive states, success or failure, operational or nonoperational. Many components may exhibit three or more mutually exclusive states. • Conceptually multistate reliability models are merely extensions of two state models. In practice, the mathematical complexities make application of multistate reliability models restricted to specialized cases. 3

  4. E 1 E 2 • Multistate Reliability Models • System Low Resistance High Resistance DIODES • Each diode has three states: good, shorted, and opened. If event x represents good, event xS shorted, and event xO opened, the three states are mutually exclusive and 4

  5. The 9 System States • 1 - X1X2 • 2 - X1X20 • 3 - X1X2S • 4 - X2X10 • 5 - X2X1S • 6 - X1SX2S • 7 - X1SX20 • 8 - X10X2S • 9 - X10X20 - S - S - F - S - F - F - F - F - F Multistate Reliability Models • The system tie sets (success combinations) are x1x2, x1 x20, x2 x10 • System Reliability 5

  6. General Reliability Configuration

  7. System Reliability Evaluation Techniques • In general, the reliability configuration of a system will not • consist of only the special configurations we have considered. • More general techniques are required for developing reliability • models for general reliability configurations • Generally, in system reliability analyses the reliability • configuration will not be only one of the special configurations we • have considered therefore, we need a more general technique for • developing the reliability model for a general system 7

  8. System Reliability Evaluation Techniques • Some of the methods for development of a system reliability • model are: • Direct or inspection method • Decomposition method • Event-space method • Path tracing method • Conditional probability (Bayes Theorem) • State space approach (Markov Processes) • Cut-set and tie-set methods • Delta-star techniques 8

  9. Direct or Inspection method Develop the system reliability equation by following the shortest dependent route through the reliability block diagram.

  10. Direct or Inspection method - example Apply the direct or inspection method to develop the system reliability equation for the following reliability block diagram: R6 R7 R4 R5 R8 R1 R2 R3 R11 R9 R10 R12 R13

  11. Direct or Inspection method - example solution For the system to function: 1. R1R2R3 must function R1R2R3 2. R12R13 may or may not function R1R2R3 [R12R13 + (1 - R12R13) 3. If R12R13 fails R11 must function R1R2R3[R12R13 + (1 - R12R13)(R11) 4. R9R10 may or may not function R1R2R3[R12R13 + (1 - R12R13)(R11)[R9R10 + (1 - R9R10)

  12. Direct or Inspection method - example solution 5. If R9R10 fails R4R5 must function R1R2R3[R12R13 + (1 - R12R13)(R11)[R9R10 + (1 - R9R10)(R4R5) 6. R8 may or may not function R1R2R3[R12R13 + (1 - R12R13)(R11)[R9R10 + (1 - R9R10)(R4R5)[(R8+(1-R8) 7. If R8 fails R6R7 must function R(t) = R1R2R3[R12R13 + (1 - R12R13)(R11)[R9R10 + (1 - R9R10)(R4R5)[(R8+(1-R8)(R6R7))]]]

  13. Decomposition into Series and Parallel • Decompose the configuration into series and • parallel configurations • At each decomposition the reliability models for • the series or parallel configuration are utilized • The system reliability model is obtained by • taking the product of the reliability models for the • series and parallel configurations

  14. Decomposition into Series and Parallel - example The reliability block diagram for a system consisting of seven elements A, B1, C, D, E, and E2 is below B1 A B2 D C E1 E2

  15. Decomposition into Series and Parallel - example The associated element reliabilities are: P(A) = 0.95 P(B1) = P(B2) = 0.95 P(C) = 0.98 P(D) = 0.90 P(E1) = P(E2) = 0.90 Find the system reliability.

  16. Decomposition into Series and Parallel - solution step 1 where: P(B) = P(B1B2) = P(B1) + P(B2) - P(B1)P(B2) B A D C E1 E2

  17. Decomposition into Series and Parallel - solution step 2 where: P(E) = P(E1)P(E2) B A D C E

  18. Decomposition into Series and Parallel - solution step 3 where: P(F) = P(D) + P(E) - P(D)P(E) B A C F

  19. Decomposition into Series and Parallel - solution step 4 where: P(S1) = P(A)P(B) P(S2) = P(C)P(D) S1 S2

  20. Decomposition into Series and Parallel - solution step 5 where: P(S) = P(S1) + P(S2) - P(S1)P(S2) S

  21. Decomposition into Series and Parallel - solution Therefore, R(t) = P(S) = P(S1) + P(S2) - P(S1)P(S2) = P(A)P(B) + P(C)P(F) - P(A)P(B)P(C)P(D) = P(A)P(B) + P(C)[P(D) + P(E) - P(D)P(E)] - P(A)P(B)P(C)[P(D) + P(E) - P(D)P(E)] = P(A)P(B) + P(C)P(D) + P(C)P(E) - P(C)P(D)P(E) - P(A)P(B)P(C)P(D) - P(A)P(B)P(C)P(E) + P(A)P(B)P(C)P(D)P(E)

  22. Decomposition into Series and Parallel - solution Therefore, R(t) = P(A)P(B) + P(C)P(D) + P(C)P(E1)P(E2) - P(C)P(D)P(E1)P(E2) - P(A)P(B)P(C)P(D) - P(A)P(B)P(C)P(E1)P(E2) + P(A)P(B)P(C)P(D)P(E1)P(E2) = P(A)[P(B1) + P(B2) - P(B1)P(B2)] + P(C)P(D) + P(C)P(E1)P(E2) + P(C)P(D)P(E1)P(E2) - P(A)P(C)P(D)[P(B1) + P(B2) - P(B1)P(B2)] - P(A)P(C)P(E1)P(E2)[P(B1) + P(B2) - P(B1)P(B2)] + P(A)P(C)P(D)P(E1)P(E2)[P(B1) + P(B2) - P(B1)P(B2)] R(t) = 0.9981

  23. Event Space Method • List all possible logical occurrences in the system • Separated the list into favorable and unfavorable • events. The list should be prepared so that all • events are mutually exclusive. • The probability of success, i.e., the reliability, is • merely the sum of the occurrence probability of • each successful event. • As an alternative approach the reliability could be • computed by first finding the probability of failure, • which is given by the sum of the occurrence • probabilities of each of the unsuccessful events

  24. Event Space Method - Example Find the reliability of a system with a complex structure whose reliability block diagram is given by A D B E C The element reliability is P(A) = P(B) = P(C) = P(D) = P(E) = p

  25. Event Space Method - Example Solution List all 25 = 32 system states (the event space) using the number of combinations of n-things-taken-r-at-a- time formula: r = 1 E2 = ABCDE E3 = ABCDE E4 = ABCDE E5 = ABCDE E6 = ABCDE r = 0 E1 = ABCDE

  26. Event Space Method - Example Solution r = 2 E7 = ABCDE E8 = ABCDE E9 = ABCDE E10 = ABCDE E11 = ABCDE E12 = ABCDE E13 = ABCDE E14 = ABCDE E15 = ABCDE E16 = ABCDE r = 3 E17 = ABCDE E18 = ABCDE E19 = ABCDE E20 = ABCDE E21 = ABCDE E22 = ABCDE E23 = ABCDE E24 = ABCDE E25 = ABCDE E26 = ABCDE r = 4 E27 = ABCDE E28 = ABCDE E29 = ABCDE E30 = ABCDE E31 = ABCDE r = 5 E32 = ABCDE

  27. Event Space Method - Example Solution Determine the impact on system reliability of each event by inspecting the reliability block diagram and construct lists of favorable and unfavorable events Favorable Events E1 E9 E22 E2 E10 E23 E3 E11 E25 E4 E12 E5 E13 E6 E14 E7 E15 E8 E19 Unfavorable Events E16 E24 E29 E17 E26 E30 E18 E27 E31 E20 E28 E32 E21

  28. Event Space Method - Example Solution Probability of System Success R(t) = P(E1 E2 E3 E4 E5 E6 E7 E8 E9 E10 E11 E12 E13 E14 E15 E19 E22 E23E25) = p5 + 5p4(1 - p) + 9p3(1 - p)2 + 4p2(1 - p)3 = p5 - p4 - 3p3 + 4p2 OR R(t) = 1 - P(E16 E17 E18 E21 E24 E26 E27 E28 E29 E30 E31 E32) = 1 -[p3(1-p)2 + 6p2(1-p)3 + 5p(1-p)4 + (1-p)3] = p5 - p4 - 3p3 + 4p2

  29. Path Tracing Method • Identify all successful paths • Each successful path forms a favorable event • The system reliability is then the probability of at • least one successful path • The Path Tracing method is simpler than the event • tracing method in that it eliminates the lengthy • tabulation of the Event Space method

  30. Path Tracing Method • Disadvantages of the Path Tracing method include: • - All possible success paths must be identified • - The events are not generally mutually • exclusive • - Complex systems result in complicated • mathematical expressions which may be • difficult to simplify

  31. Path Tracing Method - Example Given the following reliability block diagram E1 E2 E3 E4 E5 a. Find the system reliability using the path tracing method, if pi = Ri(t), for i = 1, …, 5 b. Find RS(t) = hS(t), and MTBF, if pi = Ri(t) = e-t, for i = 1, …, 5

  32. Path Tracing Method - Example Solution a. Success PathEvent 1 2 A1 4 5 A2 1 3 5 A3 4 3 2 A4 RS(t) P(A1) + P(A2) + P(A3) + P(A4) - P(A1A2) - P(A1A3) - P(A1A4) - P(A2A3) - P(A2A4) - P(A3A4) + P(A1A2A3) + P(A1A2A4) + P(A2A3A4) + P(A1A3A4) - P(A1A2A3A4)

  33. Path Tracing Method - Example Solution = P1P2 + P4P5 + P1P3P5 + P2P3P4 + P1P2P4P5 - P1P2P3P5 - P1P2P3P4 - P1P3P4P5 - P2P3P4P5 - P1P2P3P4P5 + P1P2P3P4P5 + P1P2P3P4P5 + P1P2P3P4P5 + P1P2P3P4P5 - P1P2P3P4P5 = P1P2 + P4P5 + P1P3P5 + P2P3P4 - P1P2P4P5 - P1P2P3P5 - P1P2P3P4 - P1P3P4P5 - P2P3P4P5 - P1P2P3P4P5

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