Bayesian Reliability Analysis: Statistical Challenges from Science-Based Stockpile Stewardship

Transcription

1 : Statistical Challenges from Science-Based Stockpile Stewardship Alyson G. Wilson, Ph.D. Statistical Sciences Group Los Alamos National Laboratory May 22, 28

2 Acknowledgments Christine Anderson-Cook Mike Hamada Aparna Huzurbazar Harry Martz

3 Outline

4 Science-Based Stockpile Stewardship The goal of science-based stockpile stewardship is the assessment of safety and reliability in aging warheads in the absence of nuclear testing.

5 Surveillance Transformation Develop and advance the analytical capabilities required to perform computational predictions of stockpile performance, reliability, end-of-life, safety, survivability, use control; and to provide risk-based responsiveness for future replacement and refurbishment decisions.

6 Surveillance Transformation Develop and advance the analytical capabilities required to perform computational predictions of stockpile performance, reliability, end-of-life, safety, survivability, use control; and to provide risk-based responsiveness for future replacement and refurbishment decisions. Uncertainty quantification

7 Surveillance Transformation Develop and advance the analytical capabilities required to perform computational predictions of stockpile performance, reliability, end-of-life, safety, survivability, use control; and to provide risk-based responsiveness for future replacement and refurbishment decisions. Uncertainty quantification Planning

8 : Block Diagrams

9 : Block Diagrams

10 System Reliability Equation R SYS = R JK2 R J7 R J4 R J8 R JE1 R J5 R J6 ( R K14(1) R K14(2) R K16(1) R K16(2) R K2(1) R K2(2) R K15(1) R K15(2) R K19(1) R K19(2) R K14(1) R K14(2) R K16(1) R K16(2) R K2(1) R K2(2) R K15(2) R K19(2) R K14(1) R K14(2) R K16(1) R K16(2) R K2(1) R K2(2) R K15(1) R K19(1) R K14(1) R K14(2) R K16(1) R K2(1) R K15(1) R K15(2) R K19(1) R K19(2) R K14(1) R K14(2) R K16(2) R K2(2) R K15(1) R K15(2) R K19(1) R K19(2) +R K14(1) R K14(2) R K16(1) R K2(1) R K15(2) R K19(2) +R K14(1) R K14(2) R K16(2) R K2(2) R K15(1) R K19(1) +R K14(1) R K16(1) R K2(1) R K15(1) R K19(1) +R K14(2) R K16(2) R K2(2) R K15(2) R K19(2) )

11 Overview of Methodology Estimate reliability for each component Point estimate Measure of uncertainty (credible interval) Combine component reliability distributions using Monte Carlo to get system reliability distribution

12 Binary Data No failures in 3513 tests

13 Data on components and collections of components Data J7A: 383 variables measurements. Requirement( 4.5, 4.5) Data J7B: 468 variables measurements. Requirement( 3.8, 3.8) There are 383 tests with both measurements, and 85 with only the measurements of J7B. In addition, we have the 2327 pass/fail tests that insure that both J7A and J7B were within the requirements. This count includes the 468 tests described above.

14 Likelihood 383 i=1 1 exp( 1 σ A 2σA 2 (x i µ A ) 2 ) 468 j=1 1 exp( 1 σ B 2σB 2 (y i µ B ) 2 ) ((Φ(4.5, µ A, σ A ) Φ( 4.5, µ A, σ A )) (Φ(3.8, µ B, σ B ) Φ( 3.8, µ B, σ B ))) 1859

15 Posterior

16 Assurance Example Six previous systems with similar components Each previous system had 2 tests done Temperature ranges from 3 to 7 Chemistry parameter ranges from to 1 Approximately 15% of the data is right-censored Simulated data

17 More on the Example Interest in developing a replacement component Life extension program Two ways to think about lifetime Time since built Time in use Six previous systems with similar components Temperature and chemistry parameter predictive of lifetime

18 Reliability Demonstration and Assurance Example: Using minimal assumptions, to demonstrate that reliability at time t hours is.99, with 9% confidence, requires testing at least 23 units for t hours with zero failures. To have a 8% chance of passing the test, requires that the true reliability be approximately.999. For complicated, expensive systems, traditional reliability demonstration is usually not practical Reliability assurance is the alternative: Use whatever relevant knowledge you have in a principled approach to plan the test

19 What Test Plan? Denote the test plan (n, t, c). How many of the new components (n) should I test? For how long (t )? How many can fail (c)?

20 Test Criteria Interest centers on t = 1 Posterior Producer s Risk: Choose a test plan so that if the test is failed, there is a small probability that the reliability at t = 1 is high Posterior Consumer s Risk: Choose a test plan so that if the test is passed, there is a small probability that the reliability at t = 1 is low Reliable Life Criterion: Choose a test plan so that if the test is passed, there is a high probability that the 1 α quantile of the distribution is greater than t = 1

21 Model Fit the data using a Weibull regression Common shape parameter β Scale parameter λ ij i = 1,..., 6, j = 1,..., 2 log(λij ) = γ + γ 1 Temp ij + γ 2 Chem ij + γ 3 (Temp x Chem) ij + ω i Common regression model with hierarchical random effect ω i for each system ω i N(, τ ω ) Diffuse priors for γ, γ 1, γ 2, γ 3, τ ω, β

22 More on the Example Assume the new system follows the common regression model with its own parameter ω N Choose a test plan (n, t, c) so that given the previous data, if the test is passed, there is a.9 probability that the.5 quantile of the distribution when Temp = 5 and Chem =.5 is greater than t = 1 Choose c = so that the test is passed only if there are no failures q.5 = λ 1/β [ log(.95)] 1/β

23 Computation P(q.5 > t R = 1 TIP, t)

24 Computation P(q.5 > t R = 1 TIP, t) = P(λ < log(.95)t β R TIP, t)

25 Computation P(q.5 > t R = 1 TIP, t) = P(λ < log(.95)t β R TIP, t) = log(.95)t β R f (λ, β TIP, t) dλdβ

26 Computation P(q.5 > t R = 1 TIP, t) = P(λ < log(.95)t β R TIP, t) = = log(.95)t β R log(.95)t β R f (λ, β TIP, t) dλdβ P(TIP λ, β, t)p(λ, β t) P(TIP λ, β, t)p(λ, β t)dλdβ dλdβ

27 Computation P(q.5 > t R = 1 TIP, t) = P(λ < log(.95)t β R TIP, t) = = = log(.95)t β R log(.95)t β R log(.95)t β R f (λ, β TIP, t) dλdβ P(TIP λ, β, t)p(λ, β t) P(TIP λ, β, t)p(λ, β t)dλdβ dλdβ exp( nλt β )p(λ, β t)dλdβ exp( nλt β )p(λ, β t)dλdβ.9

28 Monte Carlo Approximation P(q.5 > 1 TIP, t) = log(.95) exp( nλt β )p(λ, β t)dλdβ exp( nλt β )p(λ, β t)dλdβ N j= exp( nλ(j) t β(j) )I [λ (j) log(.95)] N j= exp( nλ(j) t β(j) ) For each n, solve for t.

29 Test Plans t n

30 Observations Straightforward generalization to c > test plans Similar framework applicable to other test situations: for example, pass/fail or failure count data

31 Science-Based Stockpile Stewardship is a rich source of statistical problems, requiring methodological development in specific problem contexts Example of simultaneous inference within a system reliability representation using multilevel data Reliability assurance methodology provides a formal way to use prior information to develop executable test plans Extension to resource allocation for entire system

32 References Hamada, Wilson, Reese, Martz (28). Reliability, Springer. Wilson and Huzurbazar (27). networks for multilevel system reliability. Reliability Engineering and System Safety 92: Wilson, McNamara, Wilson (27). Information integration for complex systems. Reliability Engineering and System Safety 92:

33 References C. Anderson-Cook et al. (27). stockpile reliability methodology for complex systems with application to a munitions stockpile. Military Operations Research 12: Wilson et al. (26). Advances in system reliability assessment. Statistical Science 21: Hamada et al. (24). A fully approach for combining multilevel failure information in fault tree quantification and corresponding optimal resource allocation. Reliability Engineering and System Safety 86: