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A. BobbioBertinoro, March 10-14, 20031 Dependability Theory and Methods 2. 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 Bertinoro, March 10-14, 2003
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A. BobbioBertinoro, March 10-14, 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, ….
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A. BobbioBertinoro, March 10-14, 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.
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A. BobbioBertinoro, March 10-14, 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.
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A. BobbioBertinoro, March 10-14, 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.
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A. BobbioBertinoro, March 10-14, 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)
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A. BobbioBertinoro, March 10-14, 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:
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A. BobbioBertinoro, March 10-14, 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
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A. BobbioBertinoro, March 10-14, 20039 Availability for Series System Assuming independent repair for each component, where A i is the (steady state or transient) availability of component i
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A. BobbioBertinoro, March 10-14, 200310 Series system: an example
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A. BobbioBertinoro, March 10-14, 200311 Series system: an example
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A. BobbioBertinoro, March 10-14, 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.
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A. BobbioBertinoro, March 10-14, 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.
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A. BobbioBertinoro, March 10-14, 200314 The part-count method
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A. BobbioBertinoro, March 10-14, 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.
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A. BobbioBertinoro, March 10-14, 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............
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A. BobbioBertinoro, March 10-14, 200317 Parallel system Therefore :
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A. BobbioBertinoro, March 10-14, 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
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A. BobbioBertinoro, March 10-14, 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
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A. BobbioBertinoro, March 10-14, 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
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A. BobbioBertinoro, March 10-14, 200321 Parallel system: an example
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A. BobbioBertinoro, March 10-14, 200322 Partial redundancy: an example
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A. BobbioBertinoro, March 10-14, 200323 Availability for parallel system Assuming independent repair, where A i is the (steady state or transient) availability of component i.
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A. BobbioBertinoro, March 10-14, 200324 Series-parallel systems
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A. BobbioBertinoro, March 10-14, 200325 System vs component redundancy
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A. BobbioBertinoro, March 10-14, 200326 Component redundant system: an example
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A. BobbioBertinoro, March 10-14, 200327 Is redundancy always useful ?
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A. BobbioBertinoro, March 10-14, 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
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A. BobbioBertinoro, March 10-14, 200329 Stand-by redundancy A B x 0 t A B
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A. BobbioBertinoro, March 10-14, 200330 A B Stand-by redundancy (exponential components)
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A. BobbioBertinoro, March 10-14, 200331 Majority voting redundancy A1A1 A2 A3 Voter
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A. BobbioBertinoro, March 10-14, 200332 2:3 majority voting redundancy A1A1 A2 A3 Voter
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