Optimal design for constant-stress accelerated degradation test based on gamma process

ABSTRACT

This paper addresses the optimal design problems for constant-stress accelerated degradation test (CSADT) based on gamma processes with fixed effect and random effect. For three optimization criteria, we prove that optimal CSADT plans with multiple stress levels degenerate to two-stress-level test plans only using the minimum and maximum stress levels under model assumptions. Under each optimization criterion, the optimal sample size allocation proportions for the minimum and maximum stress levels are determined theoretically. The effect of the stress level on the objective functions is also discussed. A numerical example and a simulation study are provided to illustrate the obtained results.

KEYWORDS:

• Constant-stress accelerated degradation test
• Gamma process
• Optimization criterion
• Random effect
• Relative efficiency.

MATHEMATICS SUBJECT CLASSIFICATION:

• 62K05

Acknowledgments

This work was supported by the National Natural Science Foundation of China under Grant No. 11671080, the Jiangsu Provincial Key Laboratory of Networked Collective Intelligence under Grant No. BM2017002, and the Innovation Project for College Graduates of Jiangsu Province of China under Grand No. KYLX16 0183.

Appendix

A.1. Proof of Theorem 1

Proof.

Based on Criterion 1: For fixed N, M, Δt, maximizing det[I(θ)|N₁, …, N_{J − 1}] in (13(13) $$det\left[I\left(\theta \right)|N_1,\dots ,N_{J-1}\right]=\frac{N{M}^3\alpha {\Delta t}^3\left[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)-1\right]}{{\beta }_0^2}\left[N\sum _{k=1}^J{N}_k{s}_k^2-{\left(\sum _{k=1}^J{N}_k{s}_k\right)}^2\right].$$(13) ) is equivalent to maximizing $N{\sum }_{k=1}^J{N}_k{s}_k^2-{\left({\sum }_{k=1}^J{N}_k{s}_k\right)}^2=N^2·[{\sum }_{k=1}^J{P}_k{s}_k^2-{\left({\sum }_{k=1}^J{P}_k{s}_k\right)}^2],$ where P_k = N_k/N is the allocation proportion of units under stress level s_k, and $P_J=1-{\sum }_{k=1}^{J-1}{P}_k$. Denoting $D(P_1,\dots ,P_{J-1})={\sum }_{k=1}^J{P}_k{s}_k^2-{\left({\sum }_{k=1}^J{P}_k{s}_k\right)}^2$, we further have $$\begin{array}{c}\hfill \partial D(P_1,\dots ,P_{J-1})/\partial P_j=(s_j^2-s_J^2)-2\left(\sum _{k=1}^J{P}_k{s}_k\right)(s_j-s_J),j=1,\dots ,J-1.\end{array}$$For j = 1, set ∂D(P₁, …, P_{J − 1})/∂P₁ = 0, then we have $s_1+s_J=2\left({\sum }_{k=1}^J{P}_k{s}_k\right)$. For j = 2, …, J − 1, we have ∂D(P₁, …, P_{J − 1})/∂P_j = (s_j − s_J)(s_j − s₁) < 0. Hence, D(P₁, …, P_{J − 1}) is a monotone decreasing function of P_j for j = 2, …, J − 1. Since 0 ⩽ P_j ⩽ 1, D(P₁, …, P_{J − 1}) is maximized at P*_j = 0 for j = 2, …, J − 1. As a result, the test units will not be arranged at the stress level s_k, k = 2, …, J − 1 with the optimal test plan. That is the optimal CSADT plan is a two-stress-level CSADT plan based on Criterion 1 as claimed. Based on Criterion 2: For fixed N, M, Δt, minimizing $Avar\left({\widehat{t}}_p\right|N_1,\dots ,N_{J-1})$ in (14(14) $$\begin{array}{ccc}\hfill Avar\left({\widehat{t}}_p\right|N_1,\dots ,N_{J-1})& =& \frac{\alpha {\left(\frac{\partial t_p}{\partial \alpha }\right)}^2}{NM\Delta t[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)-1]}+\frac{2{\beta }_0\frac{\partial t_p}{\partial \alpha }\frac{\partial t_p}{\partial {\beta }_0}}{NM\Delta t[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)-1]}\hfill \\ & & +\frac{{\beta }_0^2{\left(\frac{\partial t_p}{\partial {\beta }_0}\right)}^2[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)N{\sum }_{k=1}^J{N}_k{s}_k^2-{\left({\sum }_{k=1}^J{N}_k{s}_k\right)}^2]}{NM\alpha \Delta t[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)-1][N{\sum }_{k=1}^J{N}_k{s}_k^2-{\left({\sum }_{k=1}^J{N}_k{s}_k\right)}^2]}.\hfill \end{array}$$(14) ) is equivalent to minimizing $$\begin{array}{ccc}& & Avar(P_1,\dots ,P_{J-1})\hfill \\ & & =\frac{\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)N{\sum }_{k=1}^J{N}_k{s}_k^2-{\left({\sum }_{k=1}^J{N}_k{s}_k\right)}^2}{N{\sum }_{k=1}^J{N}_k{s}_k^2-{\left({\sum }_{k=1}^J{N}_k{s}_k\right)}^2}=\frac{\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right){\sum }_{k=1}^J{P}_k{s}_k^2-{\left({\sum }_{k=1}^J{P}_k{s}_k\right)}^2}{{\sum }_{k=1}^J{P}_k{s}_k^2-{\left({\sum }_{k=1}^J{P}_k{s}_k\right)}^2}.\hfill \end{array}$$Denote $V(P_1,\dots ,P_{J-1})={\left({\sum }_{k=1}^J{P}_k{s}_k\right)}^2/{\sum }_{k=1}^J{P}_k{s}_k^2$. We further have $$Avar(P_1,\dots ,P_{J-1})=\frac{\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)-V(P_1,\dots ,P_{J-1})}{1-V(P_1,\dots ,P_{J-1})}.$$It can be seen that Avar(P₁, …, P_{J − 1}) is an increasing function of V(P₁, …, P_{J − 1}), since αΔtψ₁(αΔt) > 1. Therefore, minimizing Avar(P₁, …, P_{J − 1}) is equivalent to minimizing V(P₁, …, P_{J − 1}). Furthermore, $$\begin{array}{c}\hfill \frac{\partial V(P_1,\dots ,P_{J-1})}{\partial P_j}=\frac{2\left({\sum }_{k=1}^J{P}_k{s}_k\right)(s_j-s_J)}{{\sum }_{k=1}^J{P}_k{s}_k^2}-{\left(\frac{{\sum }_{k=1}^J{P}_k{s}_k}{{\sum }_{k=1}^J{P}_k{s}_k^2}\right)}^2(s_j^2-s_J^2),j=1,\dots ,J-1.\end{array}$$For j = 1, set ∂V(P₁, …, P_{J − 1})/∂P₁ = 0, then we have $\left({\sum }_{k=1}^J{P}_k{s}_k\right)/\left({\sum }_{k=1}^J{P}_k{s}_k^2\right)=2/\left(s_1+s_J\right)$. For j = 2, …, J − 1, we have ∂V(P₁, …, P_{J − 1})/∂P_j = 4(s_j − s_J)(s₁ − s_j)/(s₁ + s_J)² > 0. Hence, V(P₁, …, P_{J − 1}) is a monotone increasing function of P_j for j = 2, …, J − 1. Since 0 ⩽ P_j ⩽ 1, V(P₁, …, P_{J − 1}) is minimized at P*_j = 0 for j = 2, …, J − 1. As a result, the test units will not be arranged at the stress level s_k, k = 2, …, J − 1 with the optimal test plan. That is the optimal CSADT plan is a two-stress-level CSADT plan based on Criterion 2 as claimed. Based on Criterion 3: For fixed N, M, Δt, minimizing tr[I^{− 1}(θ)|N₁, …, N_{J − 1}] in (15(15) $$\begin{array}{ccc}& & tr\left[I^{-1}\left(\theta \right)\right|N_1,\dots ,N_{J-1}]\hfill \\ & & =\frac{\alpha }{NM\Delta t[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)-1]}\hfill \\ & & +\frac{{\beta }_0^2\left[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)N{\sum }_{k=1}^J{N}_k{s}_k^2-{\left({\sum }_{k=1}^J{N}_k{s}_k\right)}^2\right]+N^2[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)-1]}{NM\alpha \Delta t[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)-1]\left[N{\sum }_{k=1}^J{N}_k{s}_k^2-{\left({\sum }_{k=1}^J{N}_k{s}_k\right)}^2\right]}.\hfill \end{array}$$(15) ) is equivalent to minimizing A(P₁, …, P_{J − 1}): $$\begin{array}{c}\hfill A(P_1,\dots ,P_{J-1})=\frac{{\beta }_0^2[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)\left({\sum }_{k=1}^J{P}_k{s}_k^2\right)-{\left({\sum }_{k=1}^J{P}_k{s}_k\right)}^2]+\left[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)-1\right]}{{\sum }_{k=1}^J{P}_k{s}_k^2-{\left({\sum }_{k=1}^J{P}_k{s}_k\right)}^2}.\end{array}$$

We further have $$\begin{array}{ccc}\hfill \frac{\partial A(P_1,\dots ,P_{J-1})}{\partial P_j}& =& \frac{{\beta }_0^2\left[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)(s_j^2-s_J^2)-2\left({\sum }_{k=1}^J{P}_k{s}_k\right)(s_j-s_J)\right]}{{\sum }_{k=1}^J{P}_k{s}_k^2-{\left({\sum }_{k=1}^J{P}_k{s}_k\right)}^2}\hfill \\ & & -\frac{\{{\beta }_0^2[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)\left({\sum }_{k=1}^J{P}_k{s}_k^2\right)-{\left({\sum }_{k=1}^J{P}_k{s}_k\right)}^2]+\left[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)-1\right]\}}{{\left[{\sum }_{k=1}^J{P}_k{s}_k^2-{\left({\sum }_{k=1}^J{P}_k{s}_k\right)}^2\right]}^2}\hfill \\ & & ·\left[\left(s_j^2-s_J^2\right)-2\left(\sum _{k=1}^J{P}_k{s}_k\right)(s_j-s_J)\right],j=1,\dots ,J-1.\hfill \end{array}$$

For j = 1, set ∂A(P₁, …, P_{J − 1})/∂P₁ = 0, then we have (A.1) $$\begin{array}{ccc}& & \frac{{\beta }_0^2\left[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)\left({\sum }_{k=1}^J{P}_k{s}_k^2\right)-{\left({\sum }_{k=1}^J{P}_k{s}_k\right)}^2\right]+\left[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)-1\right]}{{\sum }_{k=1}^J{P}_k{s}_k^2-{\left({\sum }_{k=1}^J{P}_k{s}_k\right)}^2}\hfill \\ & & =\frac{{\beta }_0^2\left[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)(s_1+s_J)-2\left({\sum }_{k=1}^J{P}_k{s}_k\right)\right]}{(s_1+s_J)-2\left({\sum }_{k=1}^J{P}_k{s}_k\right)}.\hfill \end{array}$$(A.1) From (A.1(A.1) $$\begin{array}{ccc}& & \frac{{\beta }_0^2\left[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)\left({\sum }_{k=1}^J{P}_k{s}_k^2\right)-{\left({\sum }_{k=1}^J{P}_k{s}_k\right)}^2\right]+\left[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)-1\right]}{{\sum }_{k=1}^J{P}_k{s}_k^2-{\left({\sum }_{k=1}^J{P}_k{s}_k\right)}^2}\hfill \\ & & =\frac{{\beta }_0^2\left[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)(s_1+s_J)-2\left({\sum }_{k=1}^J{P}_k{s}_k\right)\right]}{(s_1+s_J)-2\left({\sum }_{k=1}^J{P}_k{s}_k\right)}.\hfill \end{array}$$(A.1) ), one can have that $(s_1+s_J)-2\left({\sum }_{k=1}^J{P}_k{s}_k\right)>0$, since the numerator and denominator of the left side of (A.1(A.1) $$\begin{array}{ccc}& & \frac{{\beta }_0^2\left[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)\left({\sum }_{k=1}^J{P}_k{s}_k^2\right)-{\left({\sum }_{k=1}^J{P}_k{s}_k\right)}^2\right]+\left[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)-1\right]}{{\sum }_{k=1}^J{P}_k{s}_k^2-{\left({\sum }_{k=1}^J{P}_k{s}_k\right)}^2}\hfill \\ & & =\frac{{\beta }_0^2\left[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)(s_1+s_J)-2\left({\sum }_{k=1}^J{P}_k{s}_k\right)\right]}{(s_1+s_J)-2\left({\sum }_{k=1}^J{P}_k{s}_k\right)}.\hfill \end{array}$$(A.1) ) is positive, and the numerator is greater than the denominator at both sides of (A.1(A.1) $$\begin{array}{ccc}& & \frac{{\beta }_0^2\left[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)\left({\sum }_{k=1}^J{P}_k{s}_k^2\right)-{\left({\sum }_{k=1}^J{P}_k{s}_k\right)}^2\right]+\left[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)-1\right]}{{\sum }_{k=1}^J{P}_k{s}_k^2-{\left({\sum }_{k=1}^J{P}_k{s}_k\right)}^2}\hfill \\ & & =\frac{{\beta }_0^2\left[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)(s_1+s_J)-2\left({\sum }_{k=1}^J{P}_k{s}_k\right)\right]}{(s_1+s_J)-2\left({\sum }_{k=1}^J{P}_k{s}_k\right)}.\hfill \end{array}$$(A.1) ). For j = 2, …, J − 1, we have $$\begin{array}{c}\hfill \frac{\partial A(P_1,\dots ,P_{J-1})}{\partial P_j}=\frac{2{\beta }_0^2[\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)-1]\left({\sum }_{k=1}^J{P}_k{s}_k\right)(s_j-s_J)(s_1-s_j)}{\left[{\sum }_{k=1}^J{P}_k{s}_k^2-{\left({\sum }_{k=1}^J{P}_k{s}_k\right)}^2\right]\left[s_1+s_J-2\left({\sum }_{k=1}^J{P}_k{s}_k\right)\right]}>0.\end{array}$$Hence, A(P₁, …, P_{J − 1}) is a monotone increasing function of P_j for j = 2, …, J − 1. Since 0 ⩽ P_j ⩽ 1, A(P₁, …, P_{J − 1}) is minimized at P*_j = 0 for j = 2, …, J − 1. As a result, the test units will not be arranged at the stress level s_k, k = 2, …, J − 1 with the optimal test plan. That is the optimal CSADT plan is a two-stress-level CSADT plan based on Criterion 3 as claimed.

A.2. Proof of Theorem 2

Proof.

Based on Criterion 1: By the proof of Theorem 1 for Criterion 1, we know that to maximize the objective function of Criterion 1 is equivalent to maximizing s²₁P₁ + s²_J(1 − P₁) − [s₁P₁ + s_J(1 − P₁)]². Setting the first derivative of the objective function with respect to P₁ to zero, it is straightforward to find that the maximal determinant of the Fisher information matrix is reached by setting the allocation proportion as P₁ = N₁/N = 1 − N_J/N = 1/2. Based on Criterion 2: By the proof of Theorem 1 for Criterion 2, we know that to minimize the objective function of Criterion 2 is equivalent to minimizing [s₁P₁ + s_J(1 − P₁)]²/[s²₁P₁ + s²_J(1 − P₁)]. Taking the first derivative of the objective function with respect to P₁ and setting the result to zero, it is straightforward to find that the optimal allocation proportion is P₁ = N₁/N = 1 − N_J/N = s_J/(s₁ + s_J). Based on Criterion 3: By the proof of Theorem 1 for Criterion 3, we know that to minimize the objective function of Criterion 3 is equivalent to minimizing $$\frac{\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)[s_1^2{P}_1+s_J^2(1-P_1)]-{[s_1{P}_1+s_J(1-P_1)]}^2+\frac{\alpha \Delta t{\psi }_1\left(\alpha \Delta t\right)-1}{{\beta }_0^2}}{[s_1^2{P}_1+s_J^2(1-P_1)]-{[s_1{P}_1+s_J(1-P_1)]}^2}.$$Setting the first derivative of the objective function with respect to P₁ to zero, it is straightforward to find that the optimal allocation proportion is $$P_1=\frac{N_1}{N}=1-\frac{N_J}{N}=\frac{(s_J^2+\frac{1}{{\beta }_0^2})-\sqrt{(s_J^2+\frac{1}{{\beta }_0^2})(s_1^2+\frac{1}{{\beta }_0^2})}}{s_J^2-s_1^2}.$$

A.3. Proof of Lemma 1

Proof.

Since X|β₀ ∼ Ga(Cβ₀, A) and β₀ ∼ Ga(γ, δ), the joint PDF of (X, β₀) is $$f(x,{\beta }_0)=f\left(x\right|{\beta }_0)·f\left({\beta }_0\right)=\frac{x^{A-1}{\left(C{\beta }_0\right)}^A{e}^{-C{\beta }_{0}x}}{\Gamma \left(A\right)}·\frac{{\beta }_0^{\delta -1}{\gamma }^{\delta }{e}^{-\gamma {\beta }_0}}{\Gamma \left(\delta \right)}.$$

Then $$\begin{array}{cc}\hfill E\left[\frac{1}{\gamma +CX}\right]& ={\int }_0^{\infty }{\int }_0^{\infty }\frac{1}{\gamma +Cx}·f(x,{\beta }_0)d{\beta }_{0}dx=\frac{C^A{\gamma }^{\delta }\Gamma (A+\delta )}{\Gamma \left(A\right)\Gamma \left(\delta \right)}{\int }_0^{\infty }\frac{x^{A-1}}{{(\gamma +Cx)}^{A+\delta +1}}dx.\hfill \end{array}$$

By using integral transformation, it can be derived that $${\int }_0^{\infty }\frac{x^{A-1}}{{(\gamma +Cx)}^{A+\delta +1}}dx=\frac{B(A,\delta +1)}{C^A{\gamma }^{\delta +1}}=\frac{\Gamma \left(A\right)\Gamma (\delta +1)}{\Gamma (A+\delta +1)C^A{\gamma }^{\delta +1}},$$where B(p, q) = ∫¹₀t^{p − 1}(1 − t)^{q − 1}dt is the beta function, and B(p, q) = Γ(p)Γ(q)/Γ(p + q). Finally, we can obtain that $$E\left[\frac{1}{\gamma +CX}\right]=\frac{\delta }{(A+\delta )\gamma }.$$Other results in Lemma 1 can be derived by using the similar method.

Additional information

Funding

National Natural Science Foundation of China [11671080]; Innovation Project for College Graduates of Jiangsu Province of China [KYLX16_0183]; Jiangsu Provincial Key Laboratory of Networked Collective Intelligence [BM2017002].