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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, 2012




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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
General Consideration in Sample Size
Determination
             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
Effectiveness of Sample Size Increase




The effectiveness of sample size reduces over sample size increase
                       2010 ISSAT – Session 5 – Jiliang Zhang   3
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
Contents
 Problem Statement
 General Methodology
 Exponential Distribution Case
 Weibull Distribution with a Known Shape
 Parameter Case
 Conclusions




                2010 ISSAT – Session 5 – Jiliang Zhang   5
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
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
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
Exponential Distribution Case – Criterion
of 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
Exponential Distribution Case – Probability
of 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
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
Weibull Distribution with a Known Shape
Parameter 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
Weibull Distribution Case – Criterion of
Decisions
 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
Weibull Distribution Case – Probability of
Decision 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
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
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
2010 ISSAT – Session 5 – Jiliang Zhang   17

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Life test sample size determination based on probability of decisions

  • 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, 2012 http://reliabilitycalendar.org/The_Re liability_Calendar/Webinars_ liability Calendar/Webinars ‐ _Chinese/Webinars_‐_Chinese.html
  • 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 events http://reliabilitycalendar.org/The_Re liability_Calendar/Webinars_ liability Calendar/Webinars ‐ _Chinese/Webinars_‐_Chinese.html
  • 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. General Consideration in Sample Size Determination 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. Effectiveness of Sample Size Increase The effectiveness of sample size reduces over sample size increase 2010 ISSAT – Session 5 – Jiliang Zhang 3
  • 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. 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. 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. 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. 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. Exponential Distribution Case – Criterion of 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. Exponential Distribution Case – Probability of 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. 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. Weibull Distribution with a Known Shape Parameter 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. Weibull Distribution Case – Criterion of Decisions 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. Weibull Distribution Case – Probability of Decision 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. 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. 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. 2010 ISSAT – Session 5 – Jiliang Zhang 17