Optimal Design for Accelerated Degradation Test Based on D-Optimality

Abstract

Today, with improvements in industrial methods, many products are designed to function for a long time before they fail. Since failure data play an important role in assessing the reliability of products, some alternative ways need to be found. Collecting degradation data can provide useful information in such cases. Using stochastic processes to model degradation data is common. Due to its favorable properties in this respect, inverse Gaussian process has attracted a lot of attention recently. In this paper, at first a constant-stress accelerated degradation test is planned when the degradation path follows the inverse Gaussian process and then the corresponding optimal design is obtained. To reach this aim, we employ the D-optimality criterion and prove that the optimal CSADT plan with multiple stress levels degenerates to two-stress-level test plans only using the minimum and maximum stress levels under model assumptions. Then, the optimal sample allocation for each stress level is obtained and the effect of stress levels on the objective function is determined. Finally, a real example and simulation studies are presented for more illustration.

Appendix A

For $s_{1} < s_{2} < \ldots < s_{m}$, we simply can write following relations

$$e^{{\alpha /s_{1} }} > e^{{\alpha /s_{2} }} > \cdots > e^{{\alpha /s_{m} }} ,$$

(20)

$$\frac{1}{{s_{1} }}e^{{\alpha /s_{1} }} > \frac{1}{{s_{2} }}e^{{\alpha /s_{2} }} > \cdots > \frac{1}{{s_{m} }}e^{{\alpha /s_{m} }} ,$$

(21)

$$\frac{1}{{s_{1}^{2} }}e^{{\alpha /s_{1} }} > \frac{1}{{s_{2}^{2} }}e^{{\alpha /s_{2} }} > \cdots > \frac{1}{{s_{m}^{2} }}e^{{\alpha /s_{m} }} .$$

(22)

Then, we have:

$$\begin{aligned} D\left( {p_{1} ,p_{2} , \ldots p_{m} } \right) & = \left( {\sum\limits_{i = 1}^{m} {\frac{1}{{s_{i}^{2} }}e^{{\frac{\alpha }{{s_{i} }}}} } } \right)\left( {\sum\limits_{i = 1}^{m} {p_{i} e^{{\frac{\alpha }{{s_{i} }}}} } } \right) - \left( {\sum\limits_{i = 1}^{m} {p_{i} \frac{1}{{s_{i} }}e^{{\frac{\alpha }{{s_{i} }}}} } } \right)^{2} \\ & \quad \le \left( {\left( {1 - p_{m} } \right)\frac{1}{{s_{1}^{2} }}e^{{\frac{\alpha }{{s_{1} }}}} + p_{m} \frac{1}{{s_{m}^{2} }}e^{{\frac{\alpha }{{s_{m} }}}} } \right)\left( {\left( {1 - p_{m} } \right)e^{{\frac{\alpha }{{s_{1} }}}} + p_{m} e^{{\frac{\alpha }{{s_{m} }}}} } \right) - \left( {p_{1} \frac{1}{{s_{1} }}e^{{\frac{\alpha }{{s_{1} }}}} + \left( {1 - p_{1} } \right)\frac{1}{{s_{m} }}e^{{\frac{\alpha }{{s_{m} }}}} } \right)^{2} \\ \end{aligned}$$

(23)

If we show that $p_{1} = p_{m} = 0.5$ maximizes the right side of (23), then (16) is proved. As $p_{1} + p_{m} \le 1$, then to minimize the function

$$g\left( {p_{1} ,p_{m} } \right) = \left( {\left( {1 - p_{m} } \right)\frac{1}{{s_{1}^{2} }}e^{{\frac{\alpha }{{s_{1} }}}} + p_{m} \frac{1}{{s_{m}^{2} }}e^{{\frac{\alpha }{{s_{m} }}}} } \right)\left( {\left( {1 - p_{m} } \right)e^{{\frac{\alpha }{{s_{1} }}}} + p_{m} e^{{\frac{\alpha }{{s_{m} }}}} } \right) - \left( {p_{1} \frac{1}{{s_{1} }}e^{{\frac{\alpha }{{s_{1} }}}} + \left( {1 - p_{1} } \right)\frac{1}{{s_{m} }}e^{{\frac{\alpha }{{s_{m} }}}} } \right)^{2} ,$$

(24)

we define:

$$h\left( {p_{1} ,p_{m} } \right) = g\left( {p_{1} ,p_{m} } \right) + \lambda \left( {1 - p_{1} - p_{m} } \right).$$

(25)

Taking the derivative with respect to $p_{1}$ and $p_{m}$, we have:

$$\frac{{\partial h\left( {p_{1} ,p_{m} } \right)}}{{\partial p_{1} }} = - 2p_{1} \frac{1}{{s_{1}^{2} }}e^{{2\alpha /s_{1} }} + 2\left( {1 - p_{1} } \right)\frac{1}{{s_{m}^{2} }}e^{{2\alpha /s_{m} }} - 2\left( {1 - 2p_{1} } \right)\frac{1}{{s_{1} s_{m} }}e^{{\alpha /s_{1} }} e^{{\alpha /s_{m} }} - \lambda ,$$

(26)

$$\frac{{\partial h\left( {p_{1} ,p_{m} } \right)}}{{\partial p_{m} }} = - 2\left( {1 - p_{m} } \right)\frac{1}{{s_{1}^{2} }}e^{{2\alpha /s_{1} }} + 2p_{m} \frac{1}{{s_{m}^{2} }}e^{{2\alpha /s_{m} }} - \left( {2p_{m} - 1} \right)\left( {\frac{1}{{s_{1}^{2} }} + \frac{1}{{s_{m}^{2} }}} \right)e^{{\alpha /s_{1} }} e^{{\alpha /s_{m} }} - \lambda .$$

(27)

With solving the above two equations for $\lambda$ and setting the two expression equal to each other, we can find the optimal values as $p_{1}^{*} = p_{m}^{*} = 0.5$.