How to use test data of subsystems

[Andrew]

I got test data for a system as a whole. This system contains two sub-systems (connected in series). I also got testing data for these two sub-systems separately. Now I want to estimate the MTTF of the System. How can I combine the individual sub-system test results with System test results to obtain a MTTF figure for the system?

Thank You.

[tarik]

Normally, people use subsystems’ data to understand the system’s reliability because of lack of data on the system level due to cost, logistic and difficulty issues related to testing the system as a whole.

In your case, you say you have data for the subsystem level and the system level. Since you can construct a reliability model using, either, the subsystem data or the system data, I recommend you use the two approaches to construct 2 reliability models and compare them as a way to perform a validation check.

1st Approach:
1- Use the data of each subsystem to construct a reliability distribution. For information about how to do life data analysis, visit: http://www.weibull.com/lifedatawebcontents.htm
2- Using the law that describes the reliability of series configuration, you can derive a reliability model that describes your system. For more info, visit: http://www.weibull.com/SystemRelWeb/series_system.htm

2nd Approach:
Simply use your system data to construct a system reliability distribution using life data analysis methods.

Once you have a system reliability model, you can then derive any metric of interest to you such as reliability at a certain time or MTTF.

Is your system (or subsystems) a repairable system?

[Andrew]

Thank You Tarik. They are non-repairable.

Can I combine 1st Sub-system test data along with whole system test data, considering all the failures happened during system testing associated to 2nd sub-system as censored, and then obtain the reliability distribution for 1st sub-system?
Similarly for second subsystem and then multiply the both to get the system reliability distribution?
I believe it will be theoretically possible. But am I going to lose any information from the data, if I do it this way?

Thank You.

[tarik]

Yes, theoretically it’s possible. It will even allow you to obtain results that are more accurate results as your data sets will be larger.

However, If you just use the system’s data, you would get an idea about how the subsystems perform when they are put together in the system as there might be some ‘interactions’ behaviors (ie. the simple series configuration may not be a completely valid assumption) and also the ‘environment’ the subsystems operate in might be different when they are part of the system.

How was your data collected for the subsystems and the system?

[Harry]

If you have the test data of the whole system, why not just use those data to estimate the reliability of the whole system.

The method your simply time the two sub-system together may not correct. It depends on the configuration of your system. If the two subsystems are independend, it is correct. Otherwise, not. The test of the whole system is telling the truth.

You can think of using Bayesian method which can incorperate your subsystem information into the calculation of the whole system. Since the systme is serie and looks sub-system 1 is more important, you can get the beta (assuming weibull distribution) of the sub-system1 and put this information into Bayesian calculation. The Weibull++7 has the nice function.

[Oleg_I]

Hi,
There is an article about how to combine the test results: http://www.reliasoft.com/pubs/2010_RAMS_reliability_estimation_for_one_shot_syst ems.pdf
Has somebody any comments about it?
The tests of all components are not equivalent to the same number of the system tests in this method.
I think it isn't very correctly.

[David]

Hi Oleg,

Assume a system has two serial subsystems A and B. If you test 3 systems and allow for zero failures, using the Cumulative Binomial distribution (which is implemented in Weibull++ via the Design of Reliability Tests), the system reliability is 58.48% at a confidence level (CL) of 80%. However, this equation assumes you treat the system as a whole and you do not know the configuration of the system. Therefore, in this scenario the system reliability is not calculated from from the subsystem reliability. Testing 3 systems is not the same as testing 3 As and 3 Bs independently. If this was the case then the system reliability at a given CL is the same as the any of its subsystem’s reliability at the same CL. This is counter-intuitive. For a serial system, system’s reliability is always lower than its subsystem’s reliability.

As mentioned previously, testing 3 systems is not the same as testing 3 of each of its subsystems independently. For example, if you test 3 of subsystem A and name the 3 units as A1, A2 and A3, and then test 3 of subsystem B and name the test units as B1, B2, B3, you will have 9 combinations of A and B to build a system. However, if you test 3 systems, you only test 3 combinations of A and B. So the system reliability estimated from the 3 system test is different from the system reliability estimated from 3 independent subsystem tests.

More detail is included in the paper along with the assumptions, methods and simulation validation.

I hope this helps.

[Oleg_I]

Hi David,

You are right but I think there is some possibility to increase reliability estimation.

1. In practice the Cumulative Binomial distribution is applied to systems with a lot of components. The aggregate of components is considered in this case. It can be used for tests of separate components when we are considering the tests of aggregates (A1, B1), (A2, B2) and (A3, B3).

2. The Operating Characteristic Curves of these tests are identical:
1-CL = r1^n*r2^n*...*rk^n - for n autonomous tests of all k components
1-CL = (r1*r2*...*rk)^n - for tests of the n systems with k components
Therefore the reliability estimations by these variants of test must be identical.

3. If we estimate Eta of components (Weibull distributions with known betas) instead of the reliability of components, the separate tests of components and the tests of a system as a whole are identical, as they give us same Eta of components.

What do you think about it?

[David]

Those are very good points. There are many different methods on this topic. Depending on the assumptions, different results can be obtained. If you have another method, please let me know.

[Oleg_I]

David,

Yes, I have another method. Presentation LC (http://www.slideshare.net/Oleg_I/lcsimulator)Simulator.ppt (http://www.slideshare.net/Oleg_I/lcsimulator) explains the work of this new tool.
_____________
Oleg

[David]

Thank you for the link. I understand that you have previously discussed this method with someone from our Theoretical group. We will certainly consider it and get back to you if we have any questions.

[Oleg_I]

There is a continuation of this discussion at http://reliability-discussion.com/showthread.php?t=2394.

[Oleg_I]

It seems a case of the paradox "whole - parts".
If we estimate a reliability of each component separately and use the typical (r1*r2*r3) we derive a low reliability of a system.
But if we do not estimate a reliability of each component separately, how we do it for tests of the system we will get a higher estimation of reliability of a system.
Mathematically it can be defined as:
1. Testing of the n systems with three components

1-CL = (r1*r2*r3)^n => rs = (r1*r2*r3) = (1-CL)^(1/n)

2. Testing of all components separately (the n tests of each component):

1-CL = r1^n * r2^n * r3^n => rs = (r1*r2*r3) = (1-CL)^(1/n) - the expression remains the same

3. Testing of all components separately (the n1 tests of 1st component, n2 tests of 2nd component, n3 tests of 3rd component, n1>n3, n2>n3):

1-CL = r1^n1 * r2^n2 * r3^n3 => rs = (r1*r2*r3) = ((1-CL)/r1^(n1-n3)/r2^(n2-n3))^(1/n3) - rs depend on r1, r2 here.

Find rs->min (the worst case) therefore
r1=1
r2=1
rs = r3 = (1-CL)^(1/n3)

For example:
n1=50, n2=60, n3=27, CL=0.9
The method <http://www.reliasoft.com/pubs/2010_RAMS_reliability_estimation_for_one_shot_syst ems.pdf> gives us
rs=0.874 (see attachment)
This method gives us
rs=(1-0.9)^(1/27)=0.918

What is correctly?