Weibull Distribution

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The survival function -- the probability that a unit survives for at least time t:

t λ β S.W e t λ / (β^-^:

t λ / (β^- exp

e t λ / (β^-^

The logarithm of the survival function:

t λ β lS.W t λ / (β^- :

The cumulative failure function -- the probability that a unit fails at a time T<t:

t λ β F.W 1 t λ β S.W - :

The probability density function of failure a time t:

t λ β f.W t λ / (β^β t / * e t λ / (β^-^* :

The logarithm of the probability density function:

t λ β lf.W β t λ / ln * β t / ln + t λ / (β^- :

Given a population of n units observed for a total time of tt with failures observed at times t the log likelihood function is:

λ β n tt t.f ll.W t.f λ β lf.W vectorize n t.f rows + (tt λ β lS.W * Sum :

A maximum likelihood estimate of the parameters, as a function of the data and guess values for the parameters, is then

β 0 >

λ β dummy λ β n tt t.f ll.W :

λ 0 >

λ β n tt t.f ml.W dummy λ β LevenbergMarquardt :

We need guess values for the parameters. Here is a very simplistic estimate, taking b as just 1:

n tt t.f guess.W tt n ln t.f rows ln - / 112 mat :

Julian Date Calculation День в году Tag des Jahres Day of the year

День в году Tag des Jahres Day of the year year month day dayOfTheYear year 4 mod 0 ≡ (year 100 mod ¬ 0 ≡ SJ 1 : year 400 mod 0 ≡ SJ 1 : SJ 0 : if if SJ 0 : if md 3128 SJ + 31303130313130313031112 mat : Tage 0 : day md 1 month el ≤ i 1 : i month < i i 1 + : Tage Tage md 1 i el + : for Tage Tage day + : Tage 31 line Tage if 41 line :

CALENDAR FUNCTIONS-Adapted from various Internet sources

M D Y JD y Y M 2 ≤ (- : m M 12 M 2 ≤ (* + : s y 100 / Floor : b Y 1582 < (Y 1582 ≡ (M 10 < (M 10 ≡ (D 15 < (& (| (& (| 02 s - s 4 / Floor + if : 365.25 y * Floor 30.6001 m 1 + (* Floor + D + 1720994.5 + b + 51 line :

372008 JD 2.4545 106^*

dbd 8252017 JD 7262017 JD - : 30

2017922 dayOfTheYear 265

M D Y DOW M D Y JD 1.5 + Floor 7 mod 1 + :

Day of week (numeric 1-7, sunday = 1)

1132018 DOW 7

M D Y DOWN Sunday Monday Tueaday Wednesday Thursday Friday Saturday 71 mat M D Y DOW el :

Day of week (name)

322018 DOWN Friday

M D Y Week D M 1 Y DOW + 2 - 7 / Floor 1 + :

Week of month (numeric 1-6)

372008 Week 2

DJ Date a DJ 0.5 + Floor : b a 1867216.25 - 36524.25 / Floor : c DJ 0.5 + a - : d 1524 a 2299161 < a a 1 + b + b 4 / Floor - if + : e d 122.1 - 365.25 / Floor : f 365.25 e * Floor : g d f - 30.6001 / Floor : D d f - 30.6001 g * Floor - c + : M g 14 < g 1 - g 13 - if : Y M 2 > e 4716 - e 4715 - if : M D Y 13 mat transpose 111 line :

Date some days after/before given date:

15111954 JD 23000 + Date 32201831 mat

DJD 15111954 JD 23000 + Date : 32201831 mat

DJD 1 el DJD 2 el DJD 3 el DOWN Friday

Enter the date(s) of failure in the three input vectors, The last date is the end of the test:

days 31323171420272814212325141 mat :

months 778910246789101010141 mat :

years 20052005200520052005200620062006200620062006200620062006141 mat :

DOB 6132005 JD :

T months days years JD vectorize DOB - :

The life cycle analysis starts with n=130 units under test for time_test=499 days. Several units fail. What is the probability that the ith unit will fail at time T hours. Use a Weibull probability destiny distribution function. The family of rate distribution functions is given by

T 20307180116246311379384421458495497499141 mat

Time to failure in days

n 130 :

time.test T max :

k T length : 14

time.test 499

i 1 k range :

Y i el i :

T Max 499

T T 1 csort :

λ β 12 mat n time.test T guess.W :

It works fine up to this point

λ β 12 mat 223.92 112 mat

I tried a couple of the nonlinear solvers, but I am not having much success.

λ β n tt t ml.W dummy λ β LevenbergMarquardt :

balance
Guesses x.0 λ β 21 mat :

Accuracy ε 1015 -^:

L dummy x.0 ε LevenbergMarquardt :

L 4.88 1019 -^* 2.857 1021 -^* 21 mat

λ 223.92

Sould be λ=3775, β=1.1

β 1

data char size clr plotG k 1 data rows range r3 k el char : r4 k el size : r5 k el clr : 31 line for data r3 r4 r5 augment 21 line :

But if I force λ & β to match MC15 results it looks close

λ 3775 :

β 1.1 :

points T Y augment o 5 red plotG :

n x λ β F.W * points 2 1 sys

β

x λ β f.W x λ β S.W /

λ 3774.7 :

β 1.1 :

shift N.shifts N.day N.years N.cycle 51 mat 8122010251 mat :

Hours per shift

Shifts per day

week 5 day * :

Days per year

Enter yeats for evaluation

Number of traverses per cycles

hours.year N.day N.shifts * shift * hr * :

hours.year hr 1760

hours.mission shift N.shifts * N.day * N.years * hr * :

hours.mission hr 17600

Mission time is the number of days running

days.mission N.day N.years * :

days.mission 2200

The number of failures during the mission is the cumulative failure rate times the number of units.

N.mission n days.mission λ β F.W * :

N.mission 55.161

The failure rate per day is the number of failures per mission.

fail.day N.mission days.mission / :

fail.day 0.025

Mean Time Between Failure: The failure data was collected by the day the chain failed. The failure rate then considers the shut down time.

MTFB 1 fail.day / : 39.883

Number of days until last chain fails

DOB Date 613200531 mat

days.left n days rows - (MTFB * : 4626.445

day.final DOB days.left + : 2458160.945

day.final Date 211201831 mat

t λ β f.W t 0 ∞ int #

t t λ β f.W * t 0 ∞ int #

1 t λ β F.w n^- #

t λ β S.W #

1 x λ β F.W ( n ^ - x λ β S.W 2 1 sys