I have given a a more detailed answer on scilab discourse, with maths
stuff correctly typeset and plots comparing approximations:



Le 03/02/2023 à 11:52, Heinz Nabielek a écrit :
This is my latest code version: the 'convoluted secondary failure fraction' is 
nicely below primary failure,  but seems too low.

m=2;         // Weibull modulus in mechanism  #1
k=1E-7;      // corrosion rate(s-1) in mechanism  #1
n=500;       // number of time steps in hourly intervals
t=(1:n)';    // timesteps in hours
ts= 3600*t;  // timesteps in seconds
PHI= 1 - exp(-((k*ts)^m)); // failure fraction in mechanism #1
deriv=m*k*((k*ts)^(m-1)) .* exp(-((k*ts)^m)); //derivative of PHI
phi=3600*deriv;  //derivative of PHI integrated over an hour
f= 1 - exp(-ts/3E5);         // "pure" failure fraction in mechanism #2
F=(convol(phi,f)/n)(1:n);   //"convoluted" failure fraction mechanism #2
plot("nl",t, [PHI phi f F'],'-');xgrid;
legend('PHI primary failure','phi primary failure rate','pure secondary failure 
fraction','convoluted secondary failure fraction',4);
xlabel('time (hours)');
ylabel('failure fraction/ failure rate');

On 03.02.2023, at 11:24, Heinz Nabielek <heinznabie...@gmail.com> wrote:

On 03.02.2023, at 11:13, Stéphane Mottelet <stephane.motte...@utc.fr> wrote:
Thanks for the code.

Just a remark on the notations, you should write :

F(T)  = Int_{0}^{T}  PHI(t) . f(T-t) . dt

i.e. not F(t) since t is mute.

However, you should pay attention to the delay notion associated with 
convolution and the relationships between discrete convolution and continuous 
convolution. I am not sure that the output of conv matches with a given 
discretization of the integral above. Maybe rectangle method, but I am not sure 
at all. Anyway, you should have F(0)=0 which does not seem to be the case in 
your graph.
F(t), of course.

But the formula is fundamentally wrong and one should have seen it already from 
a dimensional consideration.

Correct should be F(t)  = Int_{0}^{T}  (d PHI(t)/dt) . f(T-t) . dt

Hope you can help between conv and convol and possible something else.

Dear SciLab Friends:

I have an object consisting of many (~10,000) small components that can fail in 
a statistical way during long-term operation at extreme conditions.

My primary failure model is described by PHI(t) going monotonically from zero 
to one at times from t=0 to T. In the computer, this is realized in n timesteps.

Mechanism 2: There is a secondary failure mechanism that starts only when a 
component has previously failed by mechanism 1 and occurs after some delay. The 
delay function is f(t) and final expression for the evolution of the mechanism 
2 failure is given by:

F(t)  = Int_{0}^{T}  PHI(t) . f(T-t) . dt

a classical convolution Faltungsintegral.

In Scilab, I write

F= conv(PHI, f, "same")/ n;

and the resulting function F looks reasonable for most of the time. Under some 
conditions, however, I can have


at early stages, but this is physically impossible. Because it comes later, F 
must always be below PHI.
What do I do wrong?

PS: massively simplified code below......

k=1E-7;              // corrosion rate(s-1) in mechanism #1
n=100;               // number of timesteps
t=(1:n)';             // time in days
ts= 3600*t;          // time in seconds
PHI= 1 - exp(-((k*ts)^2));   // failure fraction in mechanism #1
plot("nl", t, PHI, 'r--');
f= 1 - exp(-ts/3E5);         // "pure" failure fraction in mechanism #2
F=conv(PHI,f,"same")/length(t); //"convoluted" failure fraction mechanism #2
legend('failure fraction #1','subsequent failure fraction #2',4);
xlabel('time (hours)');
ylabel('failure fraction');
title('Component failure evolution #1 is followed by #2');

Stéphane Mottelet
Ingénieur de recherche
EA 4297 Transformations Intégrées de la Matière Renouvelable
Département Génie des Procédés Industriels
Sorbonne Universités - Université de Technologie de Compiègne
CS 60319, 60203 Compiègne cedex
Tel : +33(0)344234688

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