#1 2016-06-19 10:55:32

rgeng
Member

PFD avg question

A question concerning PFD avg.

Suppose to make a simple calculation, starting from a base event with the law "Test périodique simple" (Same considerations with other laws based on exponential probability), a gate OR, with this only base event. So you have E001 ad P001.
Suppose to have a mission time = 1 and time test =1 (proof test). Suppose lambda=0.12
The program gives the following results, with XFTA calculation:

Date   Pr/Pfd      Pfdavg
0       0.11308    0.11308
0.01   0.11308    0.11308
.....
1       0.11308    0.11308

These values are strange for the following reasons:

1)  According PFD definition (Qs(T) - ref p.7.2 XFTA manual) it is strange that Pr/Pfd = 0,11308.  Then it is also strange that pfdavg has the same value for the different times (spet times), for the reason  2) below.

2) Second, but most important, PFD avg is the 'average'. So, we could have roughly half of these values, if we appoximate the probability function as linear/ triangular .
  But, more exactly, If we have exponential law P(t) = 1 - e^(-labda*t) , the indefinite integral is: t + 1/lambda*e^(-labda*t) + c.  So, integrating P(t) from 0 to t , where t= proof test time, we have:   PFD average at time t is :   t + 1/lambda*e^(-labda*t) - 1/labda .

So, if we have for example lambda = 0,12,  proof test time = 1,
we should have  PFD avg=0,05767   
(for step 0,01 --> 0,00059976, for step 0,02 --> 0,001199  and so on..)

We could verifiy these values with a simple spreadsheet.

Or, alternatively,  using LOGAN program  (demo version full functioning up to 30 events..)  at link http://loganfta.com/download/download.html   (for example assuming Probability Type Unrevealed , lambda=0.12 (called in this case as 'frequency') and prrof test time =1)


Regards

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#2 2016-06-20 10:52:44

Emmanuel
Administrator

Re: PFD avg question

Hello Roberto,

I tryed your example and, in my case,  all seem to be correct.

Please, could you tell me if the picture hereafter is ok for you ?

741334Capture.png

Regards

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#3 2016-06-20 13:30:01

rgeng
Member

Re: PFD avg question

Hello Emmanuel,
it's strange, maybe I'm doing something wrong, infact Your results seem to be correct.
I've tryed also to change language, from italian, to english and then to french, but my results are the same.

Since I do not know how to attach the files to this message, i've created a link on dropbox
https://www.dropbox.com/sh/x2c4nll6aaf7 … EeuDa?dl=0

What am I doing wrong?

Thank You in advance for Your help        Roberto

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#4 2016-06-20 13:48:57

rgeng
Member

Re: PFD avg question

PROBLEM SOLVED!

I've found my error !

Infact, in the E001  I've previously put  the last value in the cell  "Time" (optional)  equal to 1.

Instead, if I omitt that value (blank), the results are correct.

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#5 2016-06-22 08:27:58

Emmanuel
Administrator

Re: PFD avg question

Great!

If you agree, I will move your topic in the "All questions about Arbre-Analyste" forum.

Please, don't hesitate to come back here if you have any other question.

Regards.

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#6 2016-06-22 10:24:42

rgeng
Member

Re: PFD avg question

Thank You, Emmanuel.

I agree, of course: infact that wasn't a program's bug, but a my mistake.

So I've learned to be care before to input a value, in particualr if 'optional'. Infact, if I've understood well, the optional time is a 'feature' (not a bug!) of the program  that permits to input a fixed value of probability, based on a selected time t, and so the probability is fixed to P(time) at that time also in case of proof test.  In that case the probability is not a PFD avg (PFDavg is infact meaningless with this option) but an imposed probability at that time, and so that value is imposed at every time step.
If this consideration could be useful for someone else...

Anyway, Arbre Analyste  is a excellent software !

Regards      Roberto

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