HELP System Reliability

[student]

I need to solve following problem:

There is a system consisting of the #a components a and #2 components b (#a = #b = n). System can fail due to the failure of a or b (independent events). Failure distributions are both Weibull. Given: eta_a; eta_b; beta_a = beta_b. How to calculate fraction of the systems failing due to the failure of a (or b)? Is this value independent of the time?

It is urgent! Please help me!

Thank you!

[Pantelis]

You need to be clearer. Are you saying you have a system with only two components, a and b, and that failure of either a and b will cause the system to fail. Then If this is the case this trivial.

Pf(t) (probability of failure at time t) for either A or B is: (where R(t) is the reliability function)
Pf(t)_A=1-[R(t)_A]
Pf(t)_B=1-[R(t)_B]
For the system this is:
Pf(t)_System=1-[R(t)_A*R(t)_B]

[student]

I will try to say it clearer:

Let consider a circuit, which contains n resistors 1ohm (A type) and p resistors 10ohm (B type). All of them connected in series. The circuit is assumed to fail (open loop) if any of the resistors fail. Failure of the resistors obey a Weibull distribution, but with different parameters for 1ohm and 10ohm resistors:
Given:
eta_A
eta_B
In addidtion let's consider two cases:
X. beta_A = beta_B
Y. beta_A beta_B
(betas are also given)

We can assume for simplicity that n = p = z.

Task:

To calculate predicted fraction of circuits failing due to failure of resistor A.

This below is obvious for serial systems:

F(t)_system=1-(R(t)_A*R(t)_B)^z
F(t)_A=1-(R(t)_A)^z
F(t)_B=1-(R(t)_B)^z

But how many percent of the circuits will fail due to the failure of the resistor A?

[Pantelis]

OK, I see…

So now lets look at it from a single A and B perspective.

The probability that the system will fail due to A and A alone then is

P(SF| A Fail/ B Good)=(1-RA)RB

The probability that the system will fail due to B and B alone then is

P(SF| B Fail/ A Good)=(1-RB)RA


Now since all A’s and B’s are the same -- consider a subsystem composed of A’s and a Subsystem composed of B’s.
Then the reliability of each subsystem is (R_A)^n and (R_B)^n respectively. Using the above formulation the result would now be the probability of failure due to either subsystem, A or B. Failure of either subsystem however is due to failure of one of the identical components that are in it ,thus your answer for probability of failure due to A is PF_A=(1-RA^n)RB^n and so forth.

[student]

That what you have written is absolutely true, but then I will give you parameters from the real data fitted by Weibull.

eta_a = 754.17
eta_b = 3192.8
beta_a = beta_b = 1.21
n = p = 21

Could you calculate please predicted fraction of circuits failed due to failure of resistor a(b)?
Can you do it by using the last statement?

Kind regards

[Pantelis]

You also need a time...