Parallel system

[Rui Assis]

I have a system composed by three components in parallel that fail according to a Normal distribution with parameters:

A: Mean = 550; SD = 50 hours
B: Mean = 600; SD = 75 hours
C: Mean = 800; SD = 150 hours

I need to know the probabilities of each one of the components to be the first, the second and the last to fail. I know how to calculate the probabilities by simulation but I need to solve this problem analytically. How should I tackle this problem?

Thanks

[Pantelis]

Based on your question I am assuming the following:
The fact that they are in parallel is not needed. What you want is the probability that one will fail before the other.
Then:
To compute the probability that X will fail before Y use equation 33 as given in http://www.weibull.com/LifeDataWeb/tests_of_comparison.htm

Using your A,B and C repeat that for different pair combinations, to get what you need.

You can use Weibull++ to do that if you wish.

[Rui Assis]

Thank you Pantelis,

Yes, you are right when you say “The fact that they are in parallel is not needed”. Being the system a parallel arrangement, it means that it will only fail when the last of the three components fails. But I don’t want either to find out what the reliability of the set is – which actually presents no problem – or to estimate the probability of whether the times-to-failure of one component are better or worse than the times-to-failure of any of the other two (equation 33 according to Gerald G. Brown and Herbert C. Rutemiller).

What I really want is to find out how to calculate the probabilities of each one of the components to be the first, the second and the last to fail. Monte-Carlo simulation provides the following approximate results:

Probabilities of A to be the first = 0,67, the second = 0,28, the third = 0,05
Probabilities of B to be the first = 0,28, the second = 0,63, the third = 0,09
Probabilities of C to be the first = 0,05, the second = 0,09, the third = 0,86

But I need an analytical method for the solution.

[tarik]

Rui,
Pantelis’s suggestion is still valid. Let me illustrate how to use it for the A case:

- The probability that A fails first can be obtained by using the mentioned equation or the comparison tool in Weibull+: P1=P(A<B)*P(A<C)
- Similarly, the probability that A fails 3rd is P3=P(A>B)*P(A>C)
- The probability that A fails 2nd is P2= P(A>B)*P(A<c)+>C)

The attached file lists the results of this analysis based on the Weibull++ comparison tool.
<center><table border=1><tr><td>http://63.227.80.170/discus/icons/mime_msexcel.gif
Book1.xls (http://63.227.80.170/discus/messages/95/Book1-526.xls) (15.4 k)</td></tr></table></center>

[tarik]

the 3rd bulleted line didn't get displayed properly in my previous post, here it is again:

- The probability that A fails 2nd is
P2 = P(A>B)P(A<C)+
P(A<b)p(a>C)

[Rui Assis]

Thank you Tarik. I much appreciated your solution and I am finally going to use it thanks to you and Pantelis.

By the way, do you know that I just met Pantelis personally a couple of hours ago in Lisbon in a seminar held by Reliasoft-Brasil? It was such a nice surprise! We had the chance to discuss this case for a few minutes and I can see now clearly what he meant. Thanks once more.

[guest]

Tarik,

The equations you provided are not valid, but can provide

approximate results. Why? P(A<B) and P(A<C), or any other

combination, are not independent. So, we should not multiply them

directly. In order to get correct answer you should relace:

P1=P(A<B)*P(A<C)

with

P1=P(A<B)*P(A<C |A<B)

This calculation introduces the challenges to our analytical

capabilities.

Further, unless you know the distribution of A, B, C, we cannot

calculate these probability. At present we know only mean and SD.

We have taken it granted that the underlying distribution is

Normal.

Even if we assume normal distribution, we can use your previous

equations (for approx. results) only if A, B, and C are

independent.

Hope this helps.