1. A. BobbioReggio Emilia, June 17-18, 20031 Dependability & Maintainability Theory and Methods 3. Reliability Block Diagrams Andrea Bobbio Dipartimento di Informatica Università del Piemonte Orientale, “A. Avogadro” 15100 Alessandria (Italy) bobbio@unipmn.itbobbio@unipmn.it - http://www.mfn.unipmn.it/~bobbio/IFOA IFOA, Reggio Emilia, June 17-18, 2003

2. A. BobbioReggio Emilia, June 17-18, 20032 Model Types in Dependability Combinatorial models assume that components are statistically independent: poor modeling power coupled with high analytical tractability.  Reliability Block Diagrams, FT, …. State-space models rely on the specification of the whole set of possible states of the system and of the possible transitions among them.  CTMC, Petri nets, ….

3. A. BobbioReggio Emilia, June 17-18, 20033 Reliability Block Diagrams  Each component of the system is represented as a block;  System behavior is represented by connecting the blocks;  Failures of individual components are assumed to be independent;  Combinatorial (non-state space) model type.

4. A. BobbioReggio Emilia, June 17-18, 20034 Reliability Block Diagrams (RBDs)  Schematic representation or model;  Shows reliability structure (logic) of a system;  Can be used to determine dependability measures;  A block can be viewed as a “switch” that is “closed” when the block is operating and “open” when the block is failed;  System is operational if a path of “closed switches” is found from the input to the output of the diagram.

5. A. BobbioReggio Emilia, June 17-18, 20035 Reliability Block Diagrams (RBDs) Can be used to calculate: –Non-repairable system reliability given:  Individual block reliabilities (or failure rates);  Assuming mutually independent failures events. –Repairable system availability given:  Individual block availabilities (or MTTFs and MTTRs);  Assuming mutually independent failure and restoration events;  Availability of each block is modeled as 2-state Markov chain.

6. A. BobbioReggio Emilia, June 17-18, 20036 Series system of n components. Components are statistically independent Define event E i = “component i functions properly.” Series system in RBD A1A1A2A2AnAn P(E i ) is the probability “component i functions properly”  the reliability R i (t) (non repairable)  the availability A i (t) (repairable)

7. A. BobbioReggio Emilia, June 17-18, 20037 Reliability of Series system Series system of n components. Components are statistically independent Define event E i = "component i functions properly.” A1A1A2A2AnAn Denoting by R i (t) the reliability of component i Product law of reliabilities:

8. A. BobbioReggio Emilia, June 17-18, 20038 Series system with time-independent failure rate Let i be the time-independent failure rate of component i. Then: The system reliability Rs(t) becomes: R s (t) = e - s t with s =  i i=1 n R i (t) = e - i t 1 1 MTTF = —— = ———— s  i i=1 n

9. A. BobbioReggio Emilia, June 17-18, 20039 Availability for Series System Assuming independent repair for each component, where A i is the (steady state or transient) availability of component i

10. A. BobbioReggio Emilia, June 17-18, 200310 Series system: an example

11. A. BobbioReggio Emilia, June 17-18, 200311 Series system: an example

12. A. BobbioReggio Emilia, June 17-18, 200312 Improving the Reliability of a Series System Sensitivity analysis:  R s R s S i = ———— = ————  R i R i The optimal gain in system reliability is obtained by improving the least reliable component.

13. A. BobbioReggio Emilia, June 17-18, 200313 The part-count method It is usually applied for computing the reliability of electronic equipment composed of boards with a large number of components. Components are connected in series and with time- independent failure rate.

14. A. BobbioReggio Emilia, June 17-18, 200314 The part-count method

15. A. BobbioReggio Emilia, June 17-18, 200315 Redundant systems When the dependability of a system does not reach the desired (or required) level:  Improve the individual components;  Act at the structure level of the system, resorting to redundant configurations.

16. A. BobbioReggio Emilia, June 17-18, 200316 Parallel redundancy A system consisting of n independent components in parallel. It will fail to function only if all n components have failed. E i = “The component i is functioning” E p = “the parallel system of n component is functioning properly.” A1A1 AnAn............

17. A. BobbioReggio Emilia, June 17-18, 200317 Parallel system Therefore :

18. A. BobbioReggio Emilia, June 17-18, 200318 Parallel redundancy F i (t) = P (E i ) Probability component i is not functioning (unreliability) R i (t) = 1 - F i (t) = P (E i ) Probability component i is functioning (reliability) A1A1 AnAn............ — F p (t) =  F i (t) i=1 n R p (t) = 1 - F p (t) = 1 -  (1 - R i (t)) i=1 n

19. A. BobbioReggio Emilia, June 17-18, 200319 2-component parallel system For a 2-component parallel system: F p (t) = F 1 (t) F 2 (t) R p (t) = 1 – (1 – R 1 (t)) (1 – R 2 (t)) = = R 1 (t) + R 2 (t) – R 1 (t) R 2 (t) A1A1 A2

20. A. BobbioReggio Emilia, June 17-18, 200320 2-component parallel system: constant failure rate For a 2-component parallel system with constant failure rate: R p (t) = A1A1 A2 e - 1 t + e - 2 t – e - ( 1 + 2 ) t 1 1 1 MTTF = —— + —— – ———— 1 2 1 + 2

21. A. BobbioReggio Emilia, June 17-18, 200321 Parallel system: an example

22. A. BobbioReggio Emilia, June 17-18, 200322 Partial redundancy: an example

23. A. BobbioReggio Emilia, June 17-18, 200323 Availability for parallel system Assuming independent repair, where A i is the (steady state or transient) availability of component i.

24. A. BobbioReggio Emilia, June 17-18, 200324 Series-parallel systems

25. A. BobbioReggio Emilia, June 17-18, 200325 System vs component redundancy

26. A. BobbioReggio Emilia, June 17-18, 200326 Component redundant system: an example

27. A. BobbioReggio Emilia, June 17-18, 200327 Is redundancy always useful ?

28. A. BobbioReggio Emilia, June 17-18, 200328 Stand-by redundancy A B The system works continuously during 0 — t if: a)Component A did not fail between 0 — t b)Component A failed at x between 0 — t, and component B survived from x to t. x 0 t A B

29. A. BobbioReggio Emilia, June 17-18, 200329 Stand-by redundancy A B x 0 t A B

30. A. BobbioReggio Emilia, June 17-18, 200330 A B Stand-by redundancy (exponential components)

31. A. BobbioReggio Emilia, June 17-18, 200331 Majority voting redundancy A1A1 A2 A3 Voter

32. A. BobbioReggio Emilia, June 17-18, 200332 2:3 majority voting redundancy A1A1 A2 A3 Voter