Template:Reliability test design

Test Design
Quite often, there is a desire to design reliability demonstration tests that have few or no failures. These tests are often required to demonstrate customer reliability and confidence requirements. While it is desirable to be able to test a large population of units to failure in order to obtain information on a product or design's reliability, time and resource constraints sometimes make this impossible. In cases such as these, a test can be run on a specified number of units, or for a specified amount of time, that will demonstrate that the products have met or exceeded a given reliability at a given confidence level. In order to do so without a large amount of cumulative test time or failure data, it is necessary to make assumptions about the failure distribution of the product. In the final analysis, the actual reliability of the units will of course remain unknown, but the reliability engineer will be able to state that certain specifications have been met.

Constant Failure Rate/Chi-Squared Test Design
Another method for designing tests for products that have an assumed constant failure rate, or exponential life distribution, draws on the chi-squared distribution. These represent the true exponential distribution confidence bounds referred to in Chapter 7. This method only returns the necessary accumulated test time for a demonstrated reliability or $$MTTF$$, not a specific time/test unit combination that is obtained using the cumulative binomial method described above. The accumulated test time is equal to the total amount of time experienced by all of the units on test. Assuming that the units undergo the same amount of test time, this works out to be:


 * $${{T}_{a}}=n\cdot {{t}_{TEST}}$$

where $$n$$  is the number of units on test and  $${{t}_{TEST}}$$  is the test time. The chi-squared equation for test time is:


 * $${{T}_{a}}=\frac{MTTF\cdot \chi _{1-CL;2f+2}^{2}}{2}$$


 * where:


 * $$\begin{align}

& \chi _{1-CL;2f+2}^{2}= & \text{the chi squared distribution} \\ & {{T}_{a}}= & \text{the necessary accumulated test time} \\ & CL= & \text{the confidence level, and} \\ & f= & \text{the number of failures} \end{align}$$

Since this methodology only applies to the exponential distribution, the exponential reliability equation can be rewritten as:


 * math>MTTF=\frac{t}{-ln(R)}

and substituted into Eqn. (expchi1) for developing a test that demonstrates reliability at a given time, rather than $$MTTF$$ :


 * $${{T}_{a}}=\frac{\tfrac{-ln(R)}\cdot \chi _{1-CL;2f+2}^{2}}{2}$$

Exponential Distribution Example
In this example, we desire to design a test to demonstrate a reliability of 85% at $${{t}_{DEMO}}=500$$  hours with a 90% confidence, or  $$CL=0.9,$$  and two allowable failures,  $$f=2$$. The chi-squared value can be determined from tables or the Quick Statistical Reference in Weibull++. In this example, the value is calculated as:


 * $$\chi _{1-CL;2r+2}^{2}=\chi _{0.1;6}^{2}=10.6446$$

Substituting this into Eqn. (expchi1), we obtain:


 * $${{T}_{a}}=\frac{\tfrac{500}{-ln(0.85)}\cdot 10.6446}{2}=16,374\text{ hours}$$

This means that 16,374 hours of total test time need to be accumulated with two failures in order to demonstrate the specified reliability.

This example solved in Weibull++ is shown next.

Given the test time, one can now solve for the number of units using Eqn. (expchipv1). Similarly, if the number of units is given, one can determine the test time from Eqn. (expchipv1).