plotG(data,char,size,clr)

b32135c5-4fca-4d39-9a98-dba3f9d89f2d 36 4 5 decimal

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

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 :

The project . . .

The life cycle starts with n = 10 units undr test for 1500 hours. Several units fail.

What is the probability that th ith unit fail at time Thours. Use Weibull probability

destiny distribution function. The family of rate distribution function is given by =>

a scaled parameter, b shape parameter.

Occasionaly, the varaiable 't' is replaced by t-γ where γ is a 3rd parameter used to

define a suitable zero point.

If β > 1, the rate increases with 't' ... if β < 1 the rate decreases with 't'.

If β = 1 the rate is constant, in which case the Weibull distribution equals the

exponential distribution.

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t R β α / t α / (β 1 -^* ≡

n 10 :

time.test 1500 :

3 appVersion 0.98.6179 ≡

An alternate approach to finding the probability of failure of each unit is to use the remarkably good approximation formula for median rank

T.m F i 0.3 - n 0.4 + / ≡

time of observed failure T 812922120813 mat transpose :

k T length : i 1 T length range : 21 line

Y i : transpose 12313 mat

variables in FORMULA x T ln vectorize : v 11 i 0.3 - n 0.4 + / - / ln ln vectorize : 21 sys

x v 12 mat 6.6995 6.8265 7.0967 31 mat 2.6638 - 1.7233 - 1.202 - 31 mat 12 mat

Weibull shape factor β k x i el v i el * (i 1 k sum (* x i el i 1 k sum (v i el i 1 k sum * - k x i el 2^i 1 k sum (* x i el i 1 k sum (2^- / : 3.406

Weibull scale factor α v i el i 1 k sum x i el (2^i 1 k sum * x i el i 1 k sum x i el v i el * (i 1 k sum * - β k x i el (2^i 1 k sum (* x i el i 1 k sum (2^- (* - / exp : 1671.135

Recapitulate the project . . .

Using equation 4.1 from the source Widget, we can forcast the expected

the number of failures as shown in the figure below.

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points T Y augment o 5 black plotG :

i 15002 time.test * (range : t i : ii 11500 range : tt ii : 41 line

t Num n 1 t α / (β^- exp - (* :

α Num 6.3212

as it looks, all units wiil have failed in about twice the testing duration ... this is a conceptual NON-sense ... better make cabages climb hydro pole

t n 1 t α / ( β ^ - exp - ( * vectorize augment tt n 1 tt α / ( β ^ - exp - ( * vectorize augment points time.test 5 + 50 black 1 5 mat 4 1 sys

Notice that at the time t = α we expect 63.21% of the population to have failed, regardless of the value of β. The estimated failure rate for each individual unit as a function of time is:

t R β α / t α / (β 1 -^* :

3000 R % 0.8

α R % 0.2

t Num 1 t α / (β^- exp - (:

α Num % 63.2

Number of units, failed units

time.test Num % 50

at time

time.test 1500

50% out of

n 10

units under test failed.

The Probability of failure of each unit left is

time.test R % 0.2

At time t = α = 1672.135 we expect the probability of failure

α R % 0.2

Conventional Weibull analysis

Copy/paste in blank sheet; solver is intolerant of

strangers above. It might fail from not virgin creation,

you may have to re-code step by step.

]]> plot(data,char,size,clr)

data char size clr plot 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 :

T 812922120831 mat :

length _T n 3 :

log vraisemblance α β x L n α β * ln * β 1 - (T i el ln i 1 n sum * + (α T i el β^i 1 n sum * - :

partial derivatives and
1rst order conditions. α β x L α diff n α / T i el (β^i 1 n sum - ≡ (0 ≡ α β x L β diff n β / T i el ln i 1 n sum + α T i el (β^(T i el ln * i 1 n sum * - ≡ (0 ≡ 21 sys

system n a / T i el (b^i 1 n sum - 0 ≡ n b / T i el ln i 1 n sum + a T i el (b^(T i el ln * i 1 n sum * - 0 ≡ 21 mat :

init a 1 ≡ b 1 ≡ 21 mat :

a b sol system init FindRoot :

a b 21 mat a b sol : #

sanity n a / T i el (b^i 1 n sum - n b / T i el ln i 1 n sum + a T i el (b^(T i el ln * i 1 n sum * - 21 mat : #

Weibull  log likely
probability PDF x β W 1 β 1 el β 2 el * x β 2 el 1 - (^β 1 el x β 2 el^* - exp * (* - eval :

β 1.0752 0.2334 21 mat :

T 812922120831 mat :

i 1 T rows range Y i el T i el β W eval : for XY T Y augment : data XY o 5 black plot : 31 line

Weibull probability of 'UNARY failure over T days'

data x β W 2 1 sys

XY 812192211208132 mat

The life cycle analysis starts with n=10 units under test for time_test=1208 hours. 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: Time to failure in days.

Weibull  log likely
probability PDF x β W β 1 el β 2 el * x β 2 el 1 - (^β 1 el x β 2 el^* - exp * (* eval :

'N' number of units under test over 'T' N 10 :

i 1 T rows range Y i el N T i el β W (* eval : for XY T Y augment : data XY o 5 black plot : 31 line

rate of survivals of all units under test over the observation period T

data N x β W ( * 2 1 sys

day 1500 T 1500 :

@ 1500 days, 0.0025% of units have survived
thus => 100*0.0025 =  0.25 units of the total. N T β W (* % 0.0025