Design of reliability tests

[Arvind]

Hi,

My objective is to come up with a test plan to demonstrate 65% reliability @ 20k cycles of operation with 80% conf level.

Can anyone please suggest me the correct methodolgy to go about this.
Ive tried DRT by using a ref beta value of 2.1 and type II analysis (test until 20k cycles) using MLE, but the sample sizes are very high.

Is there any other alternative test design.

Thanks
Arvind.

[David]

Hi Arvind,

I ran your design with the DRT in Weibull++ 7 and I came up with the number of required units equal to about 4. Is this the answer you were getting? I have attached a picture of the DRT settings that I used. As you can see, I assumed zero failures. In this case, if 4 units operate for 20K cycles without failure then you have met your specified goal.

I hope this helps.

[Arvind]

Hi David,

I got the same, but i also ran a simulation in simumatic pl find attached the spread sheet of my analysis.

I did this analysis based on the reliability demonstration by simulation example,which can be found at this link . http://www.weibull.com/hotwire/issue60/hottopics60.htm

Request you to clarify as to what are the scenarios where in simulation can be used/avoided and which is the correct method.

Regards
Arvind.

[Arvind]

Hi David,

Also i would like to know how relavent isit to use the equation n=Ln(1-confidence level)/Ln(reliability)

By using CL of 80% and reliability of 65% it comes up to 4.

Request your comments on this pl.

[David]

Hi Arvind,

The reason for the difference in the results between simulation and the DRT is that the assumptions made for each method are different. The DRT uses the Cumulative Binomial distribution to determine based on a zero failure test (or multiple failures) how many units are required to meet a specified goal. In this case, a distribution for the failure times is not being assumed. There is not any information available to define the distribution and to prove the goal the units are generally not tested to failure. Now in simulation, information about the distribution is being assumed since the parameters have to be given. In this case, you have a general idea as to what the distribution is and you are trying to prove that this distribution actually fits your product. And therefore, has a certain reliability at a specific point in time. In simulation you are also testing units to failure, whereas in a demonstration test you are not.

The equation that you gave is from the Binomial distribution for zero failures and can be used to determine the sample size to prove a given requirement. I would recommend that if you do not know the distribution associated with your product that you use the DRT. If you have a general idea as to the specific distribution then you can use simulation.

I hope this helps.

[Arvind]

Hi David,

Thanks for your explanation. Although i was able to appreciate you point of view, i still have some doubts. Even for DRT we have to input the Beta value which is either known before hand or estimated based on previous similar products. If iam correct what you are trying to say is even though Beta value is used as an input for DRT its the ETA value which actually represents a distribution,because for the same value of beta there can be several values of eta. Is this the logic behind it.
so when only beta value is known roughly, DRT should be the way.

Pl clarify.
Regards
Arvind.

[David]

Hi Arvind,

If your Test Time is equal to the Required Time then Beta will not affect your results. You can change Beta to anything else for your demonstration test and you will still get about 4 units. Beta comes into play when the Test Time does not equal the Required Time. This is described in detail in a HotWire article which can be found at http://www.weibull.com/hotwire/issue24/relbasics24.htm. Beta simply allows you to calculate the unreliability at the test time as described in the article. Beta is only one part of defining the Weibull distribution. In defining the distribution, Eta is still required. Beta and Eta together define the distribution. Remember, the DRT does not prove that a specified distribution fits your product. The DRT allows you to prove whether or not your product meets a specified goal (reliability). And yes, unless you have an idea about Beta and Eta then I would recommend that you use the DRT as opposed to simulation.

I hope this helps to clear up some of the confusion.

[xerox]

Hi David,

Also i would like to know how relavent isit to use the equation n=Ln(1-confidence level)/Ln(reliability)

By using CL of 80% and reliability of 65% it comes up to 4.

Request your comments on this pl.

Dear Arvind,

This equation can only applied to exponential life distribution which implies beta equals 1. But in this case, beta is supposed to be 2.1, quite different!

[David]

Actually, that equation does not imply the exponential or any other assumed distribution. The equation is a reduced form of the Cumulative Binomial distribution assuming the number of failures equals zero. Such that:

1-CL = R^n

Therefore,

n = ln(1-CL)/ln(R)

where R is the reliability at the test time. If the test time equals the time for the specified goal then it is a straight forward plug-in to solve for n. If these two times are different then you can use the reliability equation for the assumed distribution to determine the reliability at the test time since this is unknown. Additional information on this can be found at http://www.weibull.com/hotwire/issue24/relbasics24.htm. Even in this case it is not just applicable to the exponential distribution. The link above presents an example using the Weibull distribution.

I hope this helps.

[xerox]

Hi, David

Thanks for your correction. It's my mistake! Hope it dosen't cause any confusion!

[David]

No problem. I am glad the explanation helped.

[magampavan]

hello David

this is pavan ... can u help me by explaning the basic difference between reliability and reliability based design .... what is the advantages of using probabilistic based design when compared to other design methodologies ...

[Tarik El-Azzouzi]

In probabilistic based design, design parameters are considered to be of a random nature. All parameters that define the failure governing stress acting on a component during its mission are taken to be distributed variables and all parameters defining the failure governing strength exhibited by the component during its mission are also taken as distributed. This method is far superior to conventional design methods, where mean values of the design variables are used to arrive at a single value of the nominal stress, which is then modified by a variety of multiplicative single-valued factors. Components resulting from such conventional design methods are either over-safe, leading to wastage of resources in terms of material, manufacturing and operational costs; or unsafe, leading to unexpected failures due to the extreme usage of the product.
I recommend the following readings:

Engineering Design by Reliability:
http://www.reliabilitynews.com/v7i2/design.htm
Estimating Reliability Based on Multiple Random Stress Types: http://www.weibull.com/hotwire/issue82/hottopics82.htm