Life test sample size determination based on probability of decisions

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In a reliability evaluation test, one could end up with one of decisions of the reliability specification being met, not met, or inconclusive. In this presentation, we present a methodology of sample size determination prior to the test based on the probability of reaching these decisions. The specific results are obtained for the cases of Exponential distribution and Weibull distribution with a known shape parameter.

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Life test sample size determination based on probability of decisions

  1. 1. Life Test Sample Size  p Determination Based on  Probability of Decisions  P b bilit f D i i (基于决策概率的寿命试验样 本容量确定) Jiliang Zhang(张继良) ©2011 ASQ & Presentation Zhang Presented live on Jan 11th, 2012http://reliabilitycalendar.org/The_Reliability_Calendar/Webinars_liability Calendar/Webinars ‐_Chinese/Webinars_‐_Chinese.html
  2. 2. ASQ Reliability Division  ASQ Reliability Division Chinese Webinar Series Chinese Webinar Series One of the monthly webinars  One of the monthly webinars on topics of interest to  reliability engineers. To view recorded webinar (available to ASQ Reliability  Division members only) visit asq.org/reliability ) / To sign up for the free and available to anyone live  webinars visit reliabilitycalendar.org and select English  Webinars to find links to register for upcoming eventshttp://reliabilitycalendar.org/The_Reliability_Calendar/Webinars_liability Calendar/Webinars ‐_Chinese/Webinars_‐_Chinese.html
  3. 3. 16th ISSAT International Conference Washington, D.C. USA Reliability and Quality in Design August 5-7, 2010 5 7,Life Test Sample Size Determination Based on Probability of Decisions Jiliang Zhang Jili Zh Hewlett-Packard Co. jiliang.zhang@hp.com Cell: 208-9549507 2010 ISSAT – Session 5 – Jiliang Zhang 1
  4. 4. General Consideration in Sample SizeDetermination Sample Size = # of Units + Test Duration Test type and objective Reliability growth, evaluation, qualification MTBF versus durability Reliability requirement/specification Statistical confidence Unit variation Cost, schedule, and availability Resources – testing, engineering Phase of product development p p Other alternatives or additional data – internal, external Application pp c o 2010 ISSAT – Session 5 – Jiliang Zhang 2
  5. 5. Effectiveness of Sample Size IncreaseThe effectiveness of sample size reduces over sample size increase 2010 ISSAT – Session 5 – Jiliang Zhang 3
  6. 6. Minimum Sample Size For exponential distribution χ 2 1−α ;2 r + 2 T = MTBF 2 Caution: # of failures are unknown prior to the test, which is related to true performance hi h i l d f 2010 ISSAT – Session 5 – Jiliang Zhang 4
  7. 7. Contents Problem Statement General Methodology Exponential Distribution Case Weibull Distribution with a Known Shape Parameter Case Conclusions 2010 ISSAT – Session 5 – Jiliang Zhang 5
  8. 8. Problem Statement Manager: “What is the sample size should be to determine the reliability specification being met or not? not?” COST! SCHEDULE! Planning purpose Reliability Engineer: “Sample size is related to the true performance that is unknown prior to the test” (with any given sample size, one could possibly end up with one of the following decisions with certain probabilities: “met”, “not met”, or “unknown” With today’s economic condition, this becomes more y challenging and needs better methods 2010 ISSAT – Session 5 – Jiliang Zhang 6
  9. 9. Type of Life Test in Study Consider one situation: from the life test result, the statistical conclusion could be one of following: specification is met if the lower limit ≥ specification at specified confidence level (“Good); specification is not met if the upper limit < specification at p pp p specified confidence level (“Bad”) otherwise, we don t know the specification is met or not due don’t to the limited sample size in the test (“Inconclusive”) The probability of “Inclusive” will always be there and is Inclusive desired to be small but might be acceptable under certain circumstances due to limited budget, consequence of “Inconclusive”, product development phase, or additional information availabilityy 2010 ISSAT – Session 5 – Jiliang Zhang 7
  10. 10. General Methodology Link the sample size to the probability of reaching conclusions of “Good”, “Bad”, and “Inconclusive” if the performance is Good Bad Inconclusive indeed good enough and/or bad enough. Probability of Decision Table: Decision Worse-than- Better-than- specification ifi ti specification ifi ti performance performance “Good” p0G Type II Error p1G “Inconclusive” p0U p1U “Bad” p0B p1B Type I Error Note: this is not an acceptance test in which there are only two conclusions can be drawn: either accept or reject A sample size can be determined if a resulting probability of decision table is acceptable 2010 ISSAT – Session 5 – Jiliang Zhang 8
  11. 11. Exponential Distribution Case – Criterionof Decisions For a time-censored life test, 2T 2T 2T 2T m L1 = mU 1 = χ α ;2r +2 2 χ 12− α ; 2 r Criterion f d i i C i i of decision with the goal of mG ih h l f “Good” or the specification is met if mL1 ≥ mG; “Bad” or the specification is not met if mU1 < mG; “Inconclusive” if mL1 < mG ≤ mU1U1. Define rG be the maximum number of failures that satisfies mL1 ≥ mG, and rB be the minimum number of failures that satisfies mU1 < mG 2010 ISSAT – Session 5 – Jiliang Zhang 9
  12. 12. Exponential Distribution Case – Probabilityof Decision Table Calculation The number of failures within T follows a Poisson distribution. So, The probability of decision of “Good” is T T r −m rG ( ) e piG = P( X ≤ rG | m = mi ) = ∑ m , i = 0, 1 r =0 r! The probability of decision of “Bad” is Bad T T r −m ∞ ( ) e piB = P( X ≥ rB | m = mi ) = ∑ m , i = 0, 1 r = rB r! The probability of “Inconclusive” is T T r −m rB −1 ( ) e piU = P (rG < X < rB | m = mi ) = ∑ m , i = 0, 1 r = rG +1 r! 2010 ISSAT – Session 5 – Jiliang Zhang 10
  13. 13. Exponential Distribution Case – Example Assume 1 − α = 90%, m0 = mG/2 and m1 = 2mG. Set T/ mG = 7. We , can obtain rG = 3 and rB = 11 from the equations in previous slide 6. So, the following probability of decisions table can be obtained from equations in Slide 7: Decision m0 = mG/2 m1= 2mG “Good’ 0.1% 53.7% “Inconclusive” 17.4% 46.2% “Bad” 82.5% 0.1%Note: In this example, one may feel the probability of “Inconclusive” when m1 = 2mG example Inconclusive is too big. Increasing sample size would reduce this uncertainty. In real case, one can decide to further monitor the performance since this is in the middle of p product development , we have additional information, and the consequence of “Inconclusive” may not be that significant. 2010 ISSAT – Session 5 – Jiliang Zhang 11
  14. 14. Weibull Distribution with a Known ShapeParameter Case – Conversion A two-parameter Weibull distribution, ⎛t ⎞ β tβ −⎜ ⎟ ⎜η ⎟ − ηβ R(t ) = e ⎝ ⎠ =e Let s = t β , we h have s − ηβ R( s) = e So, s follows an exponential distribution with m = η β . Therefore, we can expand the previous methodology in exponential case to the Weibull with a known shape parameter case 2010 ISSAT – Session 5 – Jiliang Zhang 12
  15. 15. Weibull Distribution Case – Criterion ofDecisions Let S = k (t c ) β k is the initial number of units tc is predetermined censoring time (k, i (k tc) is considered as a sample size. id d l i For a time-censored life test, 2S β 2S β η L1 = 2 ηU 1 = 2 χ α ;2r +2 χ 1−α ; 2 r Define rG be the maximum number of failures that satisfies ηL1 ≥ ηG, and rB the minimum number of failures that satisfies ηU1 < ηG 2010 ISSAT – Session 5 – Jiliang Zhang 13
  16. 16. Weibull Distribution Case – Probability ofDecision Table Calculation The probability of decision of “Good” is S − S r ηβ ( β ) e rG η p1G = P ( X ≤ rG | η β = ηiβ ) = ∑ , i = 0, 1 r =0 r! The probability of decision of “Bad” is S − S ηβ ( β )r e ∞ η piB = P( X ≥ rB | η = ηi ) = ∑ β β , i = 0, 1 r = rB r! The probability of “Inconclusive” is S − S r ηβ rB −1 ( χ ) e η piU = P(rG < X < rB | η = ηi ) =β β ∑ r = rG +1 +1 r! , i = 0, 1 2010 ISSAT – Session 5 – Jiliang Zhang 14
  17. 17. Weibull Distribution Case – Example Required reliability for 25,000 hours is 97.5%. Assume 1 − α = 90%. The time-between-failures follows a two-parameter p Weibull distribution with a known β = 2.0. The required reliability can be converted to required ηG = 157,118 hours or y q , β requiredη G = 24,686,000,000. We further assume η β = η G /2 β β 0 and η1 = 2η G , which correspond to the reliability of 95.1% β and 98.7%, respectively. For tc = 3×25,000 = 75,000 hours and k = 30, we can obtain rG = 3 and rB = 11 from equations in Slide 10. So, the following probability of decisions table can be obtained from equations in Slide 11: Decision η0 β η1β “Good” 0% 53.4% “Inconclusive” 6.6% 46.6% “Bad” 93.4% 0% 2010 ISSAT – Session 5 – Jiliang Zhang 15
  18. 18. Conclusions Sample size is linked to the probability of decisions or risks of statistical errors or “inconclusive” Probability of decision table can be used to determine the sample size needed Upper and lower performance as well as values in probability of decision table depends on specific application. Some consideration factors could be Required reliability or failure consequence q y q Cost of the test and budget available Product development p p phase Consequence and the possible action items for “Inconclusive” Additional available information 2010 ISSAT – Session 5 – Jiliang Zhang 16
  19. 19. 2010 ISSAT – Session 5 – Jiliang Zhang 17

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