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This paper mainly talks about the relationship between an optimal design of accelerated degradation tests and the degradation performance. When there is a linear relationship between the parameter of distribution and the critical values, a special pattern of accelerated degradation tests is presented by taking the critical values as accelerated variable. And the optimal test plan is obtained by minimizing the asymptotic variance of the MLE of the parameter of distribution. Finally, a numerical example is presented to illustrate the procedures of the test plan.

With the development of technology, more and more products have higher reliability and longer service life. In many cases, accelerated degradation testing is superior to accelerated life testing in assessing product life. As time goes on, accelerated degradation test has been widely used in reliability evaluation of various products.

There are many factors involved in accelerated degradation test. One of the important aspects is the optimal arrangement or design of the test. In recent years, in the research of experimental optimal design, Boulanger, Escobar [

In this paper, our main purpose is to find a method based on degradation performance to plan the accelerated degradation tests. We discuss the relationship between the optimal design of a kind of accelerated degradation test and the selection of its degradation of critical value. On condition that there is a linear relationship between the parameter μ of distribution and the critical values C. A special pattern of accelerated degradation tests is presented by taking the critical values as accelerated variable. And the optimal test plan is obtained by minimizing the asymptotic variance of the MLE of the parameter of distribution.

The present paper is structured as follows. In Section 2, we will discuss the relationship between the parameters of failure distribution and the critical value level. In Section 3, we will establish the optimization design model based on degradation performance. Finally, a numerical example is presented to illustrate the procedures of the test plan in Section 4, and Section 5 concludes this paper.

For some products with high reliability and long service life, when evaluating their reliability indexes, even if the environmental stress is increased, the failure cannot be or rarely can be observed in the normal time interval. At this time, we use degradation test. Select a performance attribute of the sample, which degrades slowly with time. When its index value exceeds (higher or lower than) a preset limit value, the sample is judged to be invalid. The preset limit is called the degradation threshold. According to reference [

In this way, a key problem is how to evaluate the failure law of products with the failure data obtained at different critical value levels. This means that the relationship between the parameters of failure distribution and the critical value level should be found. According to literature [

D i k ( t ) = α 1 k + α 2 k t + α 3 k S i + ε i k (t)

where, D i k ( t ) is the degradation value of the kth sample at the ith acceleration stress to time t.

α 1 k , α 2 k , α 3 k are the coefficient of degradation, ε i k ( t ) is the measurement error. Now we take the critical value of degradation as C j , form D i k ( t ) = C j , We can work out the failure time. Because of C j = α 1 k + α 2 k t + α 3 k S i + ε i k ( t ) , we

obtained t i j k = C j − α 1 k − α 3 k S i α 2 k . This represents the failure time of the kth sample at the ith stress level at the jth critical value. Thus, in case of the numbers of invalid samples are m, the average life of the sample at the jth critical value and the ith stress level shall be:

μ i j = 1 m ∑ k = 1 m t i j k = 1 m ∑ k = 1 m C j − α 1 k − α 3 k S i α 2 k = − 1 m ∑ k = 1 m α 1 k α 2 k − 1 m ∑ k = 1 m α 3 k α 1 k S i + 1 m ∑ k = 1 m 1 α 1 k C j

Set α = − 1 m ∑ k = 1 m α 1 k α 2 k , β = − 1 m ∑ k = 1 m α 3 k α 1 k , γ = 1 m ∑ k = 1 m 1 α 1 k .

Thus, we have μ i j = α + β ⋅ S i + γ ⋅ C j .

Therefore, the average life is related to stress and critical value as follows:

μ = α + β ⋅ S + γ ⋅ C (1)

In this way, as mentioned above, more failure data can be obtained by selecting the critical value level, and the test time can be shortened, so as to accelerate the test.

Test description

The number of test samples is n. All samples are subjected to constant stress accelerated degradation test at q different but constant stress levels. The acceleration stress level (or its deformation) is S 1 , S 2 , ⋯ , S q ( S 0 < S 1 < S 2 < ⋯ < S q , S 0 is the normal stress level， S q is the highest stress level). The ratio of samples is π i to the stress level S i . So the Sample number is n i = n ⋅ π i at the stress level S i . The test adopts timing truncation. The closing time is τ i . D represents the degradation amount of sample performance characteristics. We take m critical values for the degradation. Respectively C 1 , C 2 , ⋯ , C m , where C m is an extreme threshold, C 0 represents the critical value of degradation under normal stress.

For discussion purposes, we first standardize the stress levels as follows: Set X i = S i − S q S 0 − S q , i = 0 , 1 , 2 , ⋯ , q , thus, we obtain 1 = X 0 > X 1 > ⋯ > X q = 0 . The

advantage is that the range of stress change is limited to the interval [ 0 , 1 ] . The level of the threshold is then standardized as follows:

Y j = C j − C m C 0 − C m , j = 1 , 2 , ⋯ , m , then Y 0 = 1 , Y m = 0 . And all Y j are between 0 and 1.

Test assumption

1) Failure time follows lognormal distribution L N ( μ , σ 2 ) ;

2) Parameter μ is a linear function of test stress and degradation critical value (from Equation (1)), hence:

μ = μ ( X i , Y j ) = α + β X i + γ Y j (2)

3) The parameters σ do not change with the change of test stress and degradation critical value.

In the above case, we hope to arrange the experiment based on some standards, so that the experiment can be optimized to a certain extent. Here, the experiment is arranged by minimizing the estimated asymptotic variance A V a r ( μ ^ ( 1 , 1 ) ) of the parameters μ in the normal environment (the stress level is taken X 0 = 1 , while the degradation critical value is taken Y 0 = 1 ).

Maximum likelihood estimation of model parameters:

According to the experimental description and assumption, at a sample observation value, its log likelihood function is

L i , j , k = I i , j , k ( − 1 2 ln ( 2 π ) − ln σ − 1 2 z i , j , k 2 ) + ( 1 − I i , j , k ) ln ( 1 − Φ ( η i , j ) )

where z i , j , k = t i , j , k − α − β X i − γ Y j , η i , j = τ i − α − β X i − γ Y j .

When t i , j , k ≤ τ i , I i , j , k = 1 ; otherwise I i , j , k = 0 . Φ ( ⋅ ) is the distribution function of standard normal distribution. According to the description of the test, each sample has m degradation critical value to judge whether it is invalid. Therefore, m ⋅ n observations will be obtained for n samples. From their independence, the log likelihood function of all observations is:

L = ∑ i = 1 q ∑ j = 1 m ∑ k = 1 n i L i , j , k

The maximum likelihood estimation α ^ , β ^ , γ ^ , σ ^ of parameters α , β , γ , σ can be obtained by maximizing L. At this time, the analytical expression cannot be obtained. The numerical results can be obtained by using the nonlinear programming method in reference [

Taking i = j = 0 in the Equation (2), and let X 0 = 1 , Y 0 = 1 , we obtain

μ ( 1 , 1 ) = μ ( X 0 , Y 0 ) = α + β X 0 + γ Y 0 = α + β + γ

Thus, the estimate of μ is

μ ^ ( 1 , 1 ) = α ^ + β ^ + γ ^ (3)

In order to obtain the progressive variance of μ ^ ( 1 , 1 ) , we first find the Fisher information matrix of parameter maximum likelihood estimation.

According to the above form of log likelihood function L i , j , k , We can get the Fisher information matrix of an observation value as follows

F i , j = [ E ( − ∂ 2 L i , j , k ∂ 2 α ) E ( − ∂ 2 L i , j , k ∂ α ∂ β ) E ( − ∂ 2 L i , j , k ∂ α ∂ γ ) E ( − ∂ 2 L i , j , k ∂ α ∂ σ ) E ( − ∂ 2 L i , j , k ∂ α ∂ β ) E ( − ∂ 2 L i , j , k ∂ 2 β ) E ( − ∂ 2 L i , j , k ∂ β ∂ γ ) E ( − ∂ 2 L i , j , k ∂ β ∂ σ ) E ( − ∂ 2 L i , j , k ∂ α ∂ γ ) E ( − ∂ 2 L i , j , k ∂ β ∂ γ ) E ( − ∂ 2 L i , j , k ∂ 2 γ ) E ( − ∂ 2 L i , j , k ∂ γ ∂ σ ) E ( − ∂ 2 L i , j , k ∂ α ∂ σ ) E ( − ∂ 2 L i , j , k ∂ β ∂ σ ) E ( − ∂ 2 L i , j , k ∂ γ ∂ σ ) E ( − ∂ 2 L i , j , k ∂ 2 σ ) ] = 1 σ 2 [ A i , j X i A i , j Y j A i , j B i , j X i A i , j X i 2 A i , j X i Y j A i , j X i B i , j Y j A i , j X i Y j A i , j Y j 2 A i , j Y j B i , j B i , j X i B i , j Y j B i , j C i , j ]

where,

A i , j = Φ ( η i , j ) − ϕ ( η i , j ) ( η i , j − ϕ ( η i , j ) 1 − Φ ( η i , j ) ) ,

B i , j = − ϕ ( η i , j ) { 1 + η i , j ( η i , j − ϕ ( η i , j ) 1 − Φ ( η i , j ) ) }

C i , j = 2 Φ ( η i , j ) − η i , j ϕ ( η i , j ) ( 1 + η i , j 2 − η i , j ϕ ( η i , j ) 1 − Φ ( η i , j ) ) ,

ϕ ( ⋅ ) is the distribution function of standard normal distribution.

Thus, the Fisher information matrix of all observations is F = ∑ i = q ∑ j = 1 m n π i F i , j .

So, the covariance matrix of the MLE α ^ , β ^ , γ ^ , σ ^ of the parameter α , β , γ , σ can be calculated as

∑ = [ V a r ( α ^ ) C o v ( α ^ , β ^ ) C o v ( α ^ , γ ^ ) C o v ( α ^ , σ ^ ) C o v ( α ^ , β ^ ) V a r ( β ^ ) C o v ( β ^ , γ ^ ) C o v ( β ^ , σ ^ ) C o v ( α ^ , γ ^ ) C o v ( β ^ , γ ^ ) V a r ( γ ^ ) C o v ( γ ^ , σ ) C o v ( α ^ , σ ^ ) C o v ( β ^ , σ ^ ) C o v ( γ ^ , σ ^ ) V a r ( σ ^ ) ] = F − 1

According to the theory of maximum likelihood estimation and (3), we have

h T = ( ∂ μ ^ ( 1 , 1 ) ∂ α , ∂ μ ^ ( 1 , 1 ) ∂ β , ∂ μ ^ ( 1 , 1 ) ∂ γ , ∂ μ ^ ( 1 , 1 ) ∂ σ ) = ( 1 , 1 , 1 , 0 )

Thus, we obtain

A V a r ( μ ^ ( 1 , 1 ) ) = h T ⋅ Σ ⋅ h = ( 1 , 1 , 1 , 0 ) ⋅ Σ ⋅ ( 1 , 1 , 1 , 0 ) T (4)

In general, before the test, the maximum stress that the sample can withstand and the extreme critical value of degradation have been determined. Therefore, when the sample size n is determined, we can select the value of X i , Y j , π i to minimize the above-mentioned progressive variance A V a r ( μ ^ ( 1 , 1 ) ) = ( 1 , 1 , 1 , 0 ) ⋅ Σ ⋅ ( 1 , 1 , 1 , 0 ) T . So the optimal model can be described as follows

Min : A V a r ( μ ^ ( 1 , 1 ) ) = ( 1 , 1 , 1 , 0 ) ⋅ Σ ⋅ ( 1 , 1 , 1 , 0 ) T (5)

where, the decision variables are: X i , Y j , π i , i = 1 , 2 , ⋯ , q − 1 ; j = 1 , 2 , ⋯ , m − 1 , and 0 ≤ X i , Y j , π i ≤ 1 .

As a special case of optimization model, we set q = 2 , m = 2 to calculate A V a r ( μ ^ ( 1 , 1 ) ) .

In this case, the Fisher information matrix is F = ∑ i = 2 ∑ j = 1 2 n π i F i , j = n σ 2 F 0 , F 0 = ( f i , j ) 4 × 4 is a fourth-order matrix, where:

f 1 , 1 = ∑ i = 1 2 ∑ j = 1 2 π i A i , j , f 2 , 2 = ∑ i = 1 2 ∑ j = 1 2 X i 2 π i A i , j , f 3 , 3 = ∑ i = 1 2 ∑ j = 1 2 Y j 2 π i A i , j ,

f 4 , 4 = ∑ i = 1 2 ∑ j = 1 2 π i C i , j , f 1 , 2 = f 2 , 1 = ∑ i = 1 2 ∑ j = 1 2 X i π i A i , j , f 1 , 3 = f 3 , 1 = ∑ i = 1 2 ∑ j = 1 2 Y j π i A i , j ,

f 1 , 4 = f 4 , 1 = ∑ i = 1 2 ∑ j = 1 2 π i B i , j , f 2 , 3 = f 3 , 2 = ∑ i = 1 2 ∑ j = 1 2 X i Y j π i A i , j ,

f 2 , 4 = f 4 , 2 = ∑ i = 1 2 ∑ j = 1 2 X i π i B i , j , f 3 , 4 = f 4 , 3 = ∑ i = 1 2 ∑ j = 1 2 Y j π i B i , j ,

Substituting the above results into (5), we obtain

A V a r ( μ ^ ( 1 , 1 ) ) = ( 1 , 1 , 1 , 0 ) ⋅ Σ ⋅ ( 1 , 1 , 1 , 0 ) T = ( 1 , 1 , 1 , 0 ) ⋅ F 0 − 1 ⋅ ( 1 , 1 , 1 , 0 ) T = n σ 2 V

where, V = E 1 + E 2 + E 3 + 2 ( E 4 + E 5 + E 6 ) | F 0 | , | F 0 | is the determinant of matrix F 0 .

E 1 = f 2 , 2 f 3 , 3 f 4 , 4 + 2 f 2 , 3 f 3 , 4 f 4 , 2 − f 2 , 4 2 f 3 , 3 − f 3 , 4 2 f 2 , 2 − f 2 , 3 2 f 4 , 4

E 2 = f 1 , 4 f 2 , 4 f 3 , 3 + f 1 , 2 f 3 , 4 2 + f 1 , 3 f 2 , 3 2 f 4 , 4 − f 1 , 2 f 3 , 3 f 4 , 4 − f 1 , 4 f 2 , 3 f 3 , 4 − f 1 , 3 f 2 , 4 f 3 , 4

E 3 = f 1 , 2 f 2 , 3 f 4 , 4 + f 1 , 4 f 2 , 2 f 3 , 4 + f 1 , 3 f 2 , 4 2 − f 1 , 4 f 2 , 3 f 2 , 4 − f 1 , 2 f 2 , 4 f 3 , 4 − f 1 , 3 f 2 , 2 f 4 , 4

E 4 = f 1 , 1 f 3 , 3 f 4 , 4 + 2 f 1 , 3 f 1 , 4 f 3 , 4 − f 1 , 4 2 f 3 , 3 − f 1 , 1 f 3 , 4 2 − f 1 , 3 2 f 4 , 4

E 5 = f 1 , 4 2 f 2 , 3 + f 1 , 1 f 2 , 4 f 3 , 4 + f 1 , 2 f 1 , 3 f 4 , 4 − f 1 , 1 f 2 , 3 f 4 , 4 − f 1 , 2 f 1 , 4 f 3 , 4 − f 1 , 4 f 1 , 3 f 2 , 4

E 6 = f 1 , 1 f 2 , 2 f 4 , 4 + 2 f 1 , 2 f 1 , 4 f 2 , 4 − f 1 , 4 2 f 2 , 2 − f 1 , 1 f 2 , 4 2 − f 1 , 2 2 f 4 , 4

Because the total number n of samples and σ 2 are constant, the optimization model is:

Min : V = E 1 + E 2 + E 3 + 2 ( E 4 + E 5 + E 6 ) | F 0 | (6)

where, the decision variables are: X 1 , Y 1 , π 1 , and 0 ≤ X 1 , Y 1 , π 1 ≤ 1 .

This nonlinear programming problem can be solved by the method of reference [

In order to illustrate the above optimization process, we take 50 LEDs for accelerated degradation test. The degraded performance attribute is taken as its luminous intensity. The acceleration variable of the test is taken as voltage. Under normal working conditions, the working current of the sample is 40 mA. According to previous experience, the maximum allowable working current is 330 mA (without changing its failure mechanism). Here we consider accelerated degradation tests at two stress levels. The voltage level is S 1 , S 2 . 40 < S 1 < 330 , S 2 = 330 . When the sample is working under normal conditions, if its luminous intensity is lower than 50% of its initial intensity, it will be judged as invalid. If the initial luminous intensity is set to 1, the critical value under normal conditions is C 0 = 0.5 . Now, we accelerate the degradation threshold at the same time. Set the extreme critical value as C 2 = 0.1 . We optimize the experimental arrangement according to the above model.

The estimation of parameters in the model can be based on previous experience or similar data. Here, we set its estimate as α ^ = 5.5 , β ^ = 6 , γ ^ = 3 , σ ^ = 1 . We take the cut-off time of the test at two stress levels as τ 1 = τ 2 = 8 . Put the above data into the optimization model (6), using the nonlinear programming method, we obtain X 1 = 0.412 , Y 1 = 0.646 , π 1 = 0.557

Thus, we have n 1 = n ⋅ π 1 = 50 × 0.557 ≈ 28 , n 2 = n − n 1 = 22 .

And X 1 = S 1 − S 2 S 0 − S 2 , S 1 = 210.5 ; and Y 1 = C 1 − C 2 C 0 − C 2 , C 1 = 0.24 .

Therefore, the optimal design of the test is: Take 28 samples and put them into 210.5 mA for accelerated degradation test. When the luminous intensity of the test sample—led is lower than 24% of the initial value, we determine that the sample is invalid once. When the luminous intensity of the test sample is lower than 50% of the initial value, we decide that it will fail again. The closing time of the test is t = 8 . In addition, 22 samples are placed at 330 mA for accelerated degradation test. When the luminous intensity of the test sample is lower than 24% of the initial value, we determine that the sample is invalid once. When the luminous intensity of the test sample is lower than 50% of the initial value, we decide that it will fail again. The closing time of the test is t = 8 .

In this work we established an optimization design model based on degradation performance for accelerated degradation test. The model shows that the test can be accelerated by selecting the critical value level. For ease of application, we simplified the model, and presented a numerical example to illustrate the procedures of the test plan.

This research was partially supported by the PhD research startup foundation of Guizhou Normal University (Grant No. GZNUD[

The author declares no conflicts of interest regarding the publication of this paper.

Wu, Y.S. (2019) An Optimal Design of Accelerated Degradation Tests Based on Degradation Performance. Open Journal of Statistics, 9, 686-694. https://doi.org/10.4236/ojs.2019.96044