rpra5

download rpra5

of 26

Transcript of rpra5

  • 7/28/2019 rpra5

    1/26

    RPRA 5. Data Analysis 1

    Engineering Risk Benefit Analysis1.155, 2.943, 3.577, 6.938, 10.816, 13.621, 16.862, 22.82,

    ESD.72, ESD.721

    RPRA 5. Data Analysis

    George E. Apostolakis

    Massachusetts Institute of Technology

    Spring 2007

  • 7/28/2019 rpra5

    2/26

    RPRA 5. Data Analysis 2

    Statistical Inference

    Theoretical Model Evidence

    Failure distribution, Sample, e.g.,

    e.g., {t1,,tn}

    How do we estimate from the evidence?

    How confident are we in this estimate? Two methods:

    Classical (frequentist) statistics

    Bayesian statistics

    te)t(f

  • 7/28/2019 rpra5

    3/26

    RPRA 5. Data Analysis 3

    Random Samples

    The observed values are independent and the

    underlying distribution is constant.

    Sample mean:

    Sample variance:

    n

    1it

    n

    1t

    n1 2i2 )tt()1n( 1s

  • 7/28/2019 rpra5

    4/26

    RPRA 5. Data Analysis 4

    The Method of Moments:

    Exponential Distribution

    Set the theoretical moments equal to the sample

    moments and determine the values of the

    parameters of the theoretical distribution.

    Exponential distribution:

    Sample: {10.2, 54.0, 23.3, 41.2, 73.2, 28.0} hrs

    t1 =

    32.386

    9.229)282.732.413.23542.10(

    6

    1t =

    1

    hr026.032.38

    1 =;hrs32.38MTTF =

  • 7/28/2019 rpra5

    5/26

    RPRA 5. Data Analysis 5

    The Method of Moments:

    Normal Distribution

    Sample: {5.5, 4.7, 6.7, 5.6, 5.7}

    = 68.55

    4.28

    5

    )7.56.57.67.47.5(x

    032.2)68.57.5(...)68.55.5()xx(5

    1

    222i =

    =

    713.0s

    508.0)15(

    032.2s

    2

  • 7/28/2019 rpra5

    6/26

    RPRA 5. Data Analysis 6

    The Method of Moments:

    Poisson Distribution

    Sample: {r events in t}

    Average number of events: r

    {3 eqs in 7 years}

    t

    rrt =

    1yr43.0

    7

    3 =

  • 7/28/2019 rpra5

    7/26

    RPRA 5. Data Analysis 7

    The Method of Moments:

    Binomial Distribution

    Sample: {k 1s in n trials}

    Average number of 1s: k

    qn = k

    {3 failures to start in 17 tests}

    n

    kq =

    176.017

    3q =

  • 7/28/2019 rpra5

    8/26

    RPRA 5. Data Analysis 8

    Censored Samples and the Exponential

    Distribution

    Complete sample: All n components fail.

    Censored sample: Sampling is terminated at time t0(with k failures observed) or when the rth failureoccurs.

    Define thetotal operational time as:

    It can be shown that:

    Valid for the exponential distribution only (nomemory).

    k1

    0i t)kn(tT r1

    ri t)rn(tT

    T

    ror

    T

    k =

  • 7/28/2019 rpra5

    9/26

    RPRA 5. Data Analysis 9

    Example

    Sample: 15 components are tested and the testis terminated when the 6th failure occurs.

    The observed failure times are:

    {10.2, 23.3, 28.0, 41.2, 54.0, 73.2} hrs

    The total operational time is:

    Therefore

    7.8882.73)615(2.73542.41283.232.10T =13 hr10x75.6

    7.888

    6 =

  • 7/28/2019 rpra5

    10/26

    RPRA 5. Data Analysis 10

    Bayesian Methods

    Recall Bayes Theorem (slide 16, RPRA 2):

    Prior information can be utilized via the priordistribution.

    Evidence other than statistical can be accommodated

    via the likelihood function.

    Posterior

    Probability

    Prior

    Probability

    Likelihood of the

    Evidence

    ( ) ( ) ( )( ) ( )

    =N

    1 ii

    iii

    HPHEP

    HPHEPEHP

  • 7/28/2019 rpra5

    11/26

    RPRA 5. Data Analysis 11

    The Model of the World

    Deterministic, e.g., a mechanistic computer code

    Probabilistic (Aleatory), e.g., R(t/ ) = exp(- t)

    The MOW deals with observable quantities.

    Both deterministic and aleatory models of the world haveassumptions and parameters.

    How confident are we about the validity of theseassumptions and the numerical values of theparameters?

  • 7/28/2019 rpra5

    12/26

    RPRA 5. Data Analysis 12

    The Epistemic Model

    Uncertainties in assumptions are not handled routinely.If necessary, sensitivity studies are performed.

    The epistemic model deals with non-observablequantities.

    Parameter uncertainties are reflected on appropriateprobability distributions.

    For the failure rate: ( )d = Pr(the failure rate has avalue in d about )

  • 7/28/2019 rpra5

    13/26

    RPRA 5. Data Analysis 13

    Unconditional (predictive) probability

    d)()/t(R)t(R

  • 7/28/2019 rpra5

    14/26

    RPRA 5. Data Analysis 14

    Communication of Epistemic

    Uncertainties: The discrete case

    Suppose that P( = 10-2) = 0.4 and P( = 10-3) = 0.6

    Then, P(e-0.001t) = 0.6 and P(e-0.01t) = 0.4

    R(t) = 0.6 e-0.001t + 0.4 e-0.01t

    t

    1.0

    exp(-0.001t)

    exp(-0.01t)

    0.6

    0.4

  • 7/28/2019 rpra5

    15/26

    RPRA 5. Data Analysis 15

    Communication of Epistemic

    Uncertainties: The continuous case

  • 7/28/2019 rpra5

    16/26

    RPRA 5. Data Analysis 16

    Risk Curves

    Figure by MIT OCW.

    1,000100

    Public Acute Fatalities

    Probability

    ofExceedence

    101

    1.0E-09

    1.0E-08

    1.0E-07

    1.0E-06

    1.0E-05

    1.0E-04

    95th Percentile

    Mean

    Median

    5th Percentile

  • 7/28/2019 rpra5

    17/26

    RPRA 5. Data Analysis 17

    The Quantification of Judgment

    Where does the epistemic distribution ( )come from?

    Both substantive and normative goodness arerequired.

    Direct assessments of parameters like failurerates should be avoided.

    A reasonable measure of central tendency to

    estimate is the median. Upper and lower percentiles can also be

    estimated.

  • 7/28/2019 rpra5

    18/26

    RPRA 5. Data Analysis 18

    The lognormal distribution

    It is very common to use the lognormal distribution as

    the epistemic distribution of failure rates.

    2

    2

    2

    )(lnexp

    2

    1)(

    UB = = exp( + 1.645 )

    LB = = exp( - 1.645 )

    95

    05

  • 7/28/2019 rpra5

    19/26

    RPRA 5. Data Analysis 19

    Pumps

    Component/Primary

    Failure Modes

    Assessed Values

    Lower Bound Upper Bound

    Valves

    Failure to start, Qd:

    Failure to operate, Qd:

    Failure to operate, Qd:

    Failure to operate, Qd:

    Failure to open, Qd:

    Failure to open, Qd:

    Plug, Qd:

    Plug, Qd:

    Plug, Qd

    :

    Failure to run, o:

    < 3" diameter, o:

    > 3" diameter, o:

    (Normal Environments)

    3 x 10-4/d 3 x 10-3/d

    3 x 10-6/hr 3 x 10-4/hr

    3 x 10-4/d 3 x 10-3/d

    3 x 10-5/d 3 x 10-4/d

    3 x 10-4/d 3 x 10-3/d

    Plug, Qd:

    3 x 10-5/d 3 x 10-4/d

    1 x 10-4/d 1 x 10-3/d

    3 x 10-5/d 3 x 10-4/d

    3 x 10-5/d 3 x 10-4/d

    3 x 10-6/d 3 x 10-5/d

    3 x 10-5

    /d 3 x 10-4

    /d

    3 x 10-11/hr 3 x 10-8/hr

    3 x 10-12/hr 3 x 10-9/hr

    1 x 10-4/d 1 x 10-3/d

    Motor Operated

    Solenoid Operated

    Check

    Relief

    Manual

    Plug/rupture

    Mechanical

    Failure to engage/disengage

    Pipe

    Clutches

    Mechanical Hardware

    Air Operated

    Table by MIT OCW.

    Adapted from Rasmussen, et al."The Reactor Safety Study."WASH-1400, US Nuclear RegulatoryCommission, 1975.

  • 7/28/2019 rpra5

    20/26

    RPRA 5. Data Analysis 20

    Table by MIT OCW.

    Adapted from Rasmussen, et al.The Reactor Safety Study."

    WASH-1400, US Nuclear RegulatoryCommission, 1975.

    "

    Motors

    Failure to start, Q 1 x 10-3d: 1 x 10-4/d /d

    Failure to run

    (Normal Environments), -5o: 3 x 10-6/hr 3 x 10 /hr

    Transformers

    Open/shorts,

    3 x 10-7

    /hr 3 x 10-6

    o: /hr

    Relays

    Failure to energize, Q -5 -4d: 3 x 10 /d 3 x 10 /d

    Circuit Breaker

    Failure to transfer, Qd: 3 x 10-4/d 3 x 10-3/d

    Limit Switches

    Failure to operate, Q 1 x 10-3d: 1 x 10-4/d /d

    Torque SwitchesFailure to operate, Q : 3 x 10-5/d -4

    d3 x 10 /d

    Pressure Switches

    Failure to operate, Qd: 3 x 10-5/d 3 x 10-4/d

    Manual Switches

    Failure to operate, Q : 3 x 10-6 -5d

    /d 3 x 10 /d

    Battery Power Supplies

    Failure to provideproper output, : 1 x 10-6/hr 1 x 10-5/hr

    s

    Solid State Devices

    Failure to function, o: 3 x 10-7/hr 3 x 10-5/hr

    Diesels (complete plant)

    Failure to start, Q -2d: 1 x 10 /d 1 x 10-1/d

    Failure to run, o: 3 x 10-4/hr 3 x 10-2/hr

    Instrumentation

    Failure to operate, o: 1 x 10-7/hr 1 x 10-5/hr

    Electrical Hardware

    Electrical Clutches

    Failure to operate, Qd: 1 x 10-4/d 1 x 10-3/d

    Component/Primary Assessed Values

    Failure Modes Lower Bound Upper Bound

    a. All values are rounded to the nearest

    half order of magnitude on the exponent.

    b. Derived from averaged data on pumps,

    combining standby and operate time.

    c. Approximated from plugging that was

    detected.

    d. Derived from combined standby and

    operate data.

    e. Derived from standby test on batteries,

    which does not include load.

  • 7/28/2019 rpra5

    21/26

    RPRA 5. Data Analysis 21

    Example

    1350 hr10x3)exp(

    =12

    95 hr10x3)645.1exp(=

    40.1,81.5Then =132 hr10x8)

    2exp(][E =

    14

    05 hr10x3)645.1exp(

    =

    Lognormal prior distribution with median and

    95th percentile given as:

  • 7/28/2019 rpra5

    22/26

    RPRA 5. Data Analysis 22

    Updating Epistemic Distributions

    Bayes Theorem allows us to incorporate newevidence into the epistemic distribution.

    d)()/E(L )()/E(L)E/('

    E l f B i d ti f i t i

  • 7/28/2019 rpra5

    23/26

    RPRA 5. Data Analysis 23

    Example of Bayesian updating of epistemic

    distributions

    Five components were tested for 100 hours each and nofailures were observed.

    Since the reliability of each component is exp(-100 ),the likelihood function is:

    L(E/ ) = P(comp. 1 did not fail AND comp. 2 did notfail AND comp. 5 did not fail) = exp(-100 ) x exp(-100 ) xx exp(-100 ) = exp(-500 )

    L(E/ ) = exp(-500 ) Note: The classical statistics point estimate is zero since no

    failures were observed.

  • 7/28/2019 rpra5

    24/26

    RPRA 5. Data Analysis 24

    Prior ( ) and posterior ( )

    distributions

    1e-4 1e-3 1e-2 1e-1

    Probability

    0.000

    0.001

    0.002

    0.003

    0.004

    0.005

    0.006

    0.007

    0.008

  • 7/28/2019 rpra5

    25/26

    RPRA 5. Data Analysis 25

    Impact of the evidence

    Mean

    (hr-1)

    95th

    (hr-1)

    Median

    (hr-1)

    5th

    (hr-1)

    Prior

    distr.

    8.0x10-3 3.0x10-2 3x10-3 3.0x10-3

    Posterior

    distr.

    1.3x10-3 3.7x10-3 9x10-4 1.5x10-4

  • 7/28/2019 rpra5

    26/26

    RPRA 5. Data Analysis 26

    Selected References

    Proceedings of Workshop on Model Uncertainty: Its Characterization and

    Quantification, A. Mosleh, N. Siu, C. Smidts, and C. Lui, Eds., Center forReliability Engineering, University of Maryland, College Park, MD, 1995.

    Reliability Engineering and System Safety, Special Issue on the Treatment

    of Aleatory and Epistemic Uncertainty, J.C. Helton and D.E. Burmaster,Guest Editors., vol. 54, Nos. 2-3, Elsevier Science, 1996.

    Apostolakis, G., The Distinction between Aleatory and Epistemic

    Uncertainties is Important: An Example from the Inclusion of AgingEffects into PSA,Proceedings of PSA 99, International Topical Meeting

    on Probabilistic Safety Assessment, pp. 135-142, Washington, DC, August

    22 - 26, 1999, American Nuclear Society, La Grange Park, Illinois.