Weibull Distribution

03d41593-e078-44d8-88cc-639492224725 14 4 5 false false 0 decimal

&[DATE] &[TIME] - &[FILENAME]

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^-^* :

m.1 t.f λ β lf.W vectorize sum :

m.2 n t.f rows - (tt λ β lS.W * :

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 sum (n t.f rows - (tt λ β lS.W * + :

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 - eval :

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

time.test T max eval : 499

n 130 :

k T length : 14

i 1 k range :

Y i el i :

T Max 499

LM Algorithm

Levenberg–Marquardt. 1 f\032E\ a\032E\ νo\032E\ εx\032E\ εy\032E\ M\032E\ LM x\0324\ a\032E\ : y\0324\ x\0324\ f\032E\ eval : R\0324\ y\0324\ length : C\0324\ x\0324\ length : 14 mat εx\0324\ εx\032E\ x\0324\ 1 el UnitsOf * : εo\0324\ C\0324\ identity εx\0324\ * : εy\0324\ εy\032E\ y\0324\ 1 el UnitsOf * : 13 mat A\032E\ Diag n\0324\ A\032E\ rows : D\0324\ n\0324\ n\0324\ matrix : 12 mat k\0324\ 1 n\0324\ range D\0324\ k\0324\ k\0324\ el A\032E\ k\0324\ k\0324\ el : for D\0324\ eval 31 line : y\0324\ F\0324\ 12 / y\0324\ k\0324\ el (2^k\0324\ 1 R\0324\ sum * 11 line : x\0324\ y\0324\ Jac r\0324\ 1 R\0324\ range c\0324\ 1 C\0324\ range J\0324\ r\0324\ c\0324\ el x\0324\ εo\0324\ c\0324\ col + f\032E\ eval r\0324\ el y\0324\ r\0324\ el - : for for J\0324\ εx\0324\ / eval 21 line : x\0324\ Y\0324\ Grad c\0324\ 1 C\0324\ range G\0324\ c\0324\ el x\0324\ εo\0324\ c\0324\ col + f\032E\ eval F\0324\ Y\0324\ - : for G\0324\ εx\0324\ / eval 21 line : J\0324\ x\0324\ y\0324\ Jac : λo\0324\ J\0324\ transpose J\0324\ * Diag max eval : Y\0324\ y\0324\ F\0324\ : 13 mat k\0324\ 1 M\032E\ range x\0324\ Y\0324\ Grad (2^sqrt εy\0324\ ≤ break y\0324\ x\0324\ f\032E\ eval : Y\0324\ y\0324\ F\0324\ : J\0324\ x\0324\ y\0324\ Jac : JTJ\0324\ J\0324\ transpose J\0324\ * eval : 14 mat xo\0324\ x\0324\ JTJ\0324\ λo\0324\ JTJ\0324\ Diag * + (1 -^J\0324\ transpose y\0324\ * (* eval - : Yo\0324\ yo\0324\ xo\0324\ f\032E\ eval : F\0324\ : (Y\0324\ < x\0324\ xo\0324\ : y\0324\ yo\0324\ : Y\0324\ Yo\0324\ : λo\0324\ λo\0324\ νo\032E\ / eval : 14 mat λo\0324\ λo\0324\ νo\032E\ * eval : if 31 line if for x\0324\ eval 91 line :

1 f\032E\ a\032E\ LM xxx f\032E\ a\032E\ νo εx εy MaxIters LM 11 line :

MaxIters 100 :

εx 105 -^:

εy 107 -^:

νo 2 :

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

λ β n tt t.f ml.W x f x 1 el x 2 el n tt t.f ll.W 11 mat 11 line : x f λ β 21 mat LM 21 line :

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 :

λ β 12 mat n time.test T guess.W : 223.92 112 mat

λ β 21 mat λ β n time.test T ml.W : 4069.7266 1.0455 21 mat

SMath solution

Mathcad solution

λ β n time.test T ll.W 131.2891 -

3775 1.1 n time.test T ll.W 131.4961 -

x β n time.test T ll.W n x λ β F.W * T Y o 5 red augment 2 1 sys λ x n time.test T ll.W