Optimum test planning for heterogeneous inverse Gaussian processes

Abstract

The heterogeneous inverse Gaussian (IG) process is one of the most popular and most considered degradation models for highly reliable products. One difficulty with heterogeneous IG processes is the lack of analytic expressions for the Fisher information matrix (FIM). Thus, it is a challenge to find an optimum test plan using any information-based criteria with decision variables such as the termination time, the number of measurements and sample size. In this article, the FIM of an IG process with random slopes can be derived explicitly in an algebraic expression to reduce uncertainty caused by the numerical approximation. The D- and V-optimum test plans with/without a cost constraint can be obtained by using a profile optimum plan. Sensitivity analysis is studied to elucidate how optimum planning is influenced by the experimental costs and planning values of the model parameters. The theoretical results are illustrated by numerical simulation and case studies. Simulations, technical derivations and auxiliary formulae are available online as supplementary materials.

Appendix

Appendix 1: proof of theorem 2.1.

To prove Theorem 2.1, the following result is needed. From the log-likelihood function of the heterogeneous IG process in (3), a few negative integer moments of $1+\sigma _\mu ^2Y(t)$ are studied for its FIM.

Theorem 6.1

If $Y(t)|\mu \sim \mathcal {IG}(\mu t, \lambda t^2)$ and $\delta = \mu ^{-1} \sim \mathcal {N}(\xi , \sigma _\mu ^2/\lambda )$, then we have

$$\begin{aligned} (i) \,&\mathrm{E}\left( \frac{1}{1+\sigma _\mu ^2Y(t)}\right) = \frac{\xi }{\xi + \sigma _\mu ^2 t}, \\ (ii) \,&\mathrm{E}\left( \frac{1}{(1+\sigma _\mu ^2Y(t))^2}\right) = \frac{\xi ^2}{(\xi + \sigma _\mu ^2 t)^2} + \frac{\sigma _\mu ^4 t}{\lambda (\xi + \sigma _\mu ^2 t)^3}, \\ (iii) \,&\mathrm{E}\left( \frac{1}{(1+\sigma _\mu ^2Y(t))^3}\right) = \frac{\xi ^3}{(\xi + \sigma _\mu ^2 t)^3} + \frac{3\xi \sigma _\mu ^4 t}{\lambda (\xi + \sigma _\mu ^2 t)^4} - \frac{3\sigma _\mu ^6 t}{\lambda ^2(\xi + \sigma _\mu ^2 t)^5}. \end{aligned}$$

Proof

Some auxiliary results given in Supplementary Section 2 are used to facilitate the proof of Theorem 6.1. (i) Since $Y(t)|\mu \sim \mathcal {IG}(\mu t , \lambda t^2)$, then the conditional moment-generating function of $Y(t)|\mu$ is given by

$$\begin{aligned} M_{Y(t)|\mu }(x) = \exp \left( \frac{\lambda t }{\mu } - \frac{\lambda t}{\mu }\sqrt{1-\frac{2\mu ^2 x}{\lambda }}\right) , \end{aligned}$$

where $\delta = \mu ^{-1} \sim \mathcal {N}(\xi , \sigma _\mu ^2/\lambda )$. We have

$$\begin{aligned} \text{ E }\left( \frac{1}{1+\sigma _\mu ^2Y(t)}\right)&{\mathop {=}\limits ^{(1)}} \text{ E }\left( \text{ E }\left( \frac{1}{1+\sigma _\mu ^2Y(t)}\bigg |\delta \right) \right) , \\&{\mathop {=}\limits ^{(2)}} \text{ E }\left( \int _{0}^{\infty }\exp \left( \frac{\lambda t }{\mu } - \frac{\lambda t}{\mu }\sqrt{1 + \frac{2\mu ^2 \sigma _\mu ^2 x}{\lambda }} - x\right) \text{ d }x \right) \\&{\mathop {=}\limits ^{(3)}} 1 - \sqrt{2\pi }\lambda t \int _{-\infty }^{\infty } \exp (C(\delta )^2/2) \phi \left( \frac{\delta -\xi }{\sigma _\mu /\sqrt{\lambda }}\right) \Phi (-C(\delta )) \text{ d }\delta \\&{\mathop {=}\limits ^{(4)}} \frac{\xi }{\xi + \sigma _\mu ^2 t}, \end{aligned}$$

where $C(\delta ) = \sqrt{\lambda }\delta /\sigma _\mu + \sqrt{\lambda }\sigma _\mu t$. Here (1) follows the law of total expectation, (2) from Lemma 1 in Supplementary Section 2, (3) from Corollary 1(i) in Supplementary Section 2, (4) from Lemma 3(i) in Supplementary Section 2. Similar to the proof (i), the proofs of (ii) and (iii) are thus omitted since the details are straightforward but tedious. $\square$

Now, we return to the proof of Theorem 2.1.

Proof

By using partial fractions and Theorem 6.1, the entries of the FIM can be shown as follows:

$$\begin{aligned} \text{ E }\left( -\frac{\partial ^2\mathcal {L}(\varvec{\theta })}{\partial \xi ^2}\right)&= \text{ E }\left( \frac{\lambda Y(t_m)}{1+\sigma _\mu ^2 Y(t_m)}\right) = \frac{\lambda }{\sigma _\mu ^2} - \frac{\lambda }{\sigma _\mu ^2}\text{ E }\left( \frac{1}{1+\sigma _\mu ^2 Y(t_m)}\right) = \frac{\lambda t_m}{\xi + \sigma _\mu ^2 t_m},\\ \text{ E }\left( -\frac{\partial ^2\mathcal {L}(\varvec{\theta })}{\partial \xi \partial \sigma _\mu ^2}\right)&= \text{ E }\left( \frac{\lambda Y(t_m) (t_m - \xi Y(t_m))}{(1+\sigma _\mu ^2 Y(t_m))^2}\right) \\&= -\frac{\lambda \xi }{\sigma _\mu ^4} + \lambda \left( \frac{t_m}{\sigma _\mu ^2} + \frac{2\xi }{\sigma _\mu ^4}\right) \text{ E }\left( \frac{1}{1+\sigma _\mu ^2 Y(t_m)}\right) \\&\quad - \lambda \left( \frac{t_m}{\sigma _\mu ^2} + \frac{\xi }{\sigma _\mu ^4}\right) \text{ E }\left( \frac{1}{(1+\sigma _\mu ^2 Y(t_m))^2}\right) \\&= \frac{-t_m}{(\xi + \sigma _\mu ^2 t_m)^2}, \\ \text{ E }\left( -\frac{\partial ^2\mathcal {L}(\varvec{\theta })}{\partial \xi \partial \lambda }\right)&= \text{ E }\left( \frac{\xi Y(t_m) - t_m}{1+\sigma _\mu ^2 Y(t_m)}\right) = \frac{\xi }{\sigma _\mu ^2}- \left( \frac{\xi }{\sigma _\mu ^2} + t_m\right) \text{ E }\left( \frac{1}{1+\sigma _\mu ^2 Y(t_m)}\right) = 0, \\ \text{ E }\left( -\frac{\partial ^2\mathcal {L}(\varvec{\theta })}{\partial (\sigma _\mu ^2)^2}\right)&= \frac{-1}{2}\text{ E }\left( \frac{Y(t_m)^2}{(1+\sigma _\mu ^2 Y(t_m))^2}\right) + \lambda \text{ E }\left( \frac{Y(t_m)(\xi Y(t_m) - t_m)^2}{(1+\sigma _\mu ^2 Y(t_m))^3}\right) \\&= \frac{\lambda \xi ^2}{\sigma _\mu ^6} - \frac{1}{2\sigma _\mu ^4} + \left( \frac{1}{\sigma _\mu ^4} - \frac{2\xi \lambda t_m}{\sigma _\mu ^4} - \frac{3\xi ^2\lambda }{\sigma _\mu ^6}\right) \text{ E }\left( \frac{1}{1+\sigma _\mu ^2 Y(t_m)}\right) \\&\quad + \left( \frac{\lambda t_m^2}{\sigma _\mu ^2} + \frac{4\xi \lambda t_m}{\sigma _\mu ^4} + \frac{3\xi ^2\lambda }{\sigma _\mu ^6} - \frac{1}{2\sigma _\mu ^4} \right) \text{ E }\left( \frac{1}{(1+\sigma _\mu ^2 Y(t_m))^2}\right) \\&\quad - \lambda \left( \frac{t_m^2}{\sigma _\mu ^2} + \frac{2\xi t_m}{\sigma _\mu ^4} + \frac{\xi ^2}{\sigma _\mu ^6} \right) \text{ E }\left( \frac{1}{(1+\sigma _\mu ^2 Y(t_m))^3}\right) \\ {}&= \frac{t_m(\lambda \sigma _\mu ^2 t_m^2 + \xi \lambda t_m + 5)}{2\lambda (\xi + \sigma _\mu ^2 t_m)^3}, \\ \text{ E }\left( -\frac{\partial ^2\mathcal {L}(\varvec{\theta })}{\partial \sigma _\mu ^2\partial \lambda }\right)&= \frac{-1}{2}\text{ E }\left( \frac{(t_m - \xi Y(t_m))^2}{(1+\sigma _\mu ^2 Y(t_m))^2}\right) \\&= \frac{-\xi ^2}{2\sigma _\mu ^4} + \left( \frac{\xi t_m}{\sigma _\mu ^2} + \frac{\xi ^2}{\sigma _\mu ^4}\right) \text{ E }\left( \frac{1}{1+\sigma _\mu ^2 Y(t_m)}\right) \\&\quad - \frac{1}{2} \left( t_m + \frac{\xi ^2}{\sigma _\mu ^2}\right) ^2 \text{ E }\left( \frac{1}{(1+\sigma _\mu ^2 Y(t_m))^2}\right) \\&= \frac{-t_m}{2\lambda (\xi + \sigma _\mu ^2 t_m)},\\ \text{ E }\left( -\frac{\partial ^2\mathcal {L}(\varvec{\theta })}{\partial \lambda ^2}\right)&= \frac{m}{2\lambda ^2}. \end{aligned}$$

In addition, the FIM, $\mathcal {I}(\varvec{\theta })$, is positive definite, which can be checked by Sylvester’s criterion. Consequently, the theorem can be established. $\square$

Appendix 2: proof of theorem 3.1.

Proof

(i) The proof is verified easily. (ii) (Necessity) Taking the first derivative of $h(t_m)$ with respect to the termination time $t_m$, we obtain

$$\begin{aligned} \frac{\text{ d } h(t_m)}{\text{ d }t_m} = \frac{3t_m n^3(\lambda \xi \sigma _\mu ^2(m-1)t_m^2 + (\lambda \xi ^2(m-1) - 2m\sigma _\mu ^2) t_m + 2\xi m)}{4\lambda ^2(\sigma _\mu ^2 t_m + \xi )^5}. \end{aligned}$$

(18)

Setting the derivative to zero and solving the equation, we have $t_m = (2m\sigma _\mu ^2 - \lambda \xi ^2(m-1) \pm \sqrt{K_0})/(2\sigma _\mu ^2\lambda \xi (m-1))$. Clearly, the discriminant of the quadratic equation in $t_m$ in the numerator of (18) is $K_0$, i.e.,

$$\begin{aligned} K_0&= (\xi \sqrt{\lambda (m-1)} - 2\sigma _\mu \sqrt{m}- \sqrt{2m}\sigma _\mu )(\xi \sqrt{\lambda (m-1)} - 2\sigma _\mu \sqrt{m}+ \sqrt{2m}\sigma _\mu ) \nonumber \\&\quad \times (\lambda \xi ^2(m-1) + 4\xi \sigma _\mu \sqrt{\lambda m(m-1)} + 2m\sigma _\mu ^2). \end{aligned}$$

(19)

By Vieta’s formula, the product of the two roots is $2m/(\lambda \sigma _\mu ^2(m-1)) > 0$, implying that the signs of the two roots are the same. Hence, if the condition ${\xi \sqrt{\lambda }}/{\sigma _\mu } < (2\sqrt{m} - \sqrt{2m})/\sqrt{m-1}$ holds, then it implies

$$\begin{aligned}&\lambda \xi ^2(m-1) -4\xi \sigma _\mu \sqrt{\lambda m(m-1)} + 2m\sigma _\mu ^2> 0\ \text{(i.e., } K_0> 0)\quad \ \text{ and }\\&\quad 2m\sigma _\mu ^2 - \lambda \xi ^2(m-1) > 0. \end{aligned}$$

This means that the quadratic equation has two distinct positive real roots, $(2m\sigma _\mu ^2 - \lambda \xi ^2(m-1) - \sqrt{K_0})/(2\sigma _\mu ^2\lambda \xi (m-1))$ and $(2m\sigma _\mu ^2 - \lambda \xi ^2(m-1) + \sqrt{K_0})/(2\sigma _\mu ^2\lambda \xi (m-1))$, which are the local maximum and minimum points, respectively. In addition, it is easy to verify that $\lim _{t_m \rightarrow \infty } h(t_m) = n^3(m-1)/(4\lambda \sigma _\mu ^6)$. The remaining part is straightforward and is omitted.

(Sufficiency) If the D-optimum termination time is $t^*_m \in (0, \infty )$, then $K_0$ should be positive since $t^*_m$ is not positive when $K_0 \le 0$. For $K_0>0$, we consider the two cases $\xi \sqrt{\lambda (m-1)}/\sigma _\mu - 2 \sqrt{m} < 0$ and $> 0$. For the first case, $K_0>0$ implies that $\xi \sqrt{\lambda (m-1)}/\sigma _\mu - 2 \sqrt{m} < - \sqrt{2 m}$ from (19). For the second case, $K_0>0$ implies that $\xi \sqrt{\lambda (m-1)}/\sigma _\mu - 2 \sqrt{m} > \sqrt{2 m}$, i.e., ${2m}\sigma _\mu ^2 - \lambda \xi ^2(m-1) < 0$. We get the contradiction $t^*_m < 0$. Hence, $K_0 > 0$ implies that $\xi \sqrt{\lambda (m-1)}/\sigma _\mu - 2 \sqrt{m} < 0$ and $\xi \sqrt{\lambda (m-1)}/\sigma _\mu - 2 \sqrt{m} < -\sqrt{2 m}$, which is the condition (a). In addition, we have $\lim _{t_m \rightarrow \infty } h(t_m) = n^3(m-1)/(4\lambda \sigma _\mu ^6)$ and $\text{ d }h(t_m)/\text{d }t_m > 0$ for sufficiently large $t_m$. For $t^*_m$ to be a maximum point, $h(t^*_m) \ge n^3(m-1)/(4 \lambda \sigma _\mu ^6)$ must hold since there must be larger values of h than $h(t^*_m)$ when $h(t^*_m) < n^3(m-1)/(4\lambda \sigma _\mu ^6)$. Therefore, $t^*_m$ can be the unique maximum point for $0< t_m < \infty$ even when $h(t^*_m) = n^3(m-1)/(4\lambda \sigma _\mu ^6)$. The proof is complete. $\square$

Appendix 3: proof of theorem 3.2.

Proof

• (i) Substituting $n = (C_b-C_{op}t_m)/(C_{mea}m + C_{it})$ into $h_0(t_m, n, m)$ gives the profile objective function:

  $$\begin{aligned} h_{1}(t_m, m) = \left( \frac{C_b-C_{op}t_m}{C_{mea}m + C_{it}}\right) ^3 \frac{(3m + \lambda t_m (m-1)(\sigma _\mu ^2 t_m + \xi ))t_m^2}{4\lambda ^2(\sigma _\mu ^2 t_m + \xi )^4}. \end{aligned}$$

  Taking the first derivative with respect to the number of measurements m, we have

  $$\begin{aligned}&\frac{\partial h_{1}(t_m, m)}{\partial m} \\&\quad = \frac{t^2_m (C_b-C_{op}t_m)^3 (3C_{it}+\lambda t_m(\sigma _\mu ^2 t_m + \xi )(C_{it} + 3C_{mea}) - 2C_{mea}(\lambda t_m(\sigma _\mu ^2 t_m + \xi )+3)m)}{4\lambda ^2(\sigma _\mu ^2 t + \xi )^4(C_{mea}m + C_{it})^4}. \end{aligned}$$

  Solving the equation $\partial h_{1}(t_m,m)/\partial m = 0$ for m and checking the roots for feasibility, we obtain $m(t_m)$ and $n(t_m)$ as shown in (8). Again, substituting $m(t_m)$ into $h_{1}(t_m, m)$ gives the profile objective function in (7). By using Proposition 1(i) in Supplementary Section 3, the result of the D-optimum test plan follows.

• (ii) Substituting $m = 1$ and $n = (C_b-C_{op}t_m)/(C_{mea} + C_{it})$ into $h_0(t_m, n, m)$ gives the objective function

  $$\begin{aligned} \tilde{h}_1(t_m) = \frac{3(C_b-C_{op}t)^3 t_m^2}{4\lambda ^2(\sigma _\mu ^2 t_m + \xi )^4(C_{mea} + C_{it})^3}. \end{aligned}$$

  Taking the first derivative with respect to the termination time $t_m$, we have

  $$\begin{aligned} \frac{\text{ d } \tilde{h}_1(t_m)}{\text{ d }t_m} = -\frac{3t_m(C_b-C_{op}t_m)^2(C_{op}\sigma _\mu ^2 t_m^2 + (5C_{op}\xi + 2C_b\sigma _\mu ^2)t_m - 2C_b\xi )}{4\lambda ^2(\sigma _\mu ^2 t_m + \xi )^5(C_{mea} + C_{it})^3}. \end{aligned}$$

  Solving the equation $\text{ d } \tilde{h}_1(t_m)/\text{d }t_m = 0$, we then obtain four roots as follows: $t_m = 0$, $C_b/C_{op}$, and $\left( -5C_{op}\xi - 2C_b\sigma _\mu ^2 \pm \sqrt{4C_b^2\sigma _\mu ^4 + 28 C_bC_{op}\xi \sigma _\mu ^2 + 25C_{op}^2\xi ^2}\right) /(2C_{op}\sigma _\mu ^2)$. Checking the roots for feasibility with respect to the constraints (i.e., $0< t_m < (C_b-C_{mea} - C_{it})/{C_{op}}$), we have $t_m^*$ and $n^*$ as shown in (10). Therefore, it is easy to check $\text{ d } \tilde{h}_1(t_m)/\text{d }t_m < 0$ for $t_m > t_m^*$ and $\text{ d } \tilde{h}_1(t_m)/\text{d }t_m > 0$ for $0< t_m < t_m^*$. By using Proposition 1(ii) in Supplementary Section 3, if the condition in (9) is satisfied, then the condition is equivalent to $t_m^* < (C_b- C_{mea} - C_{it})/C_{op}$ and $n^{*} > 1$.

• (iii) Substituting $n = 1$ and $m = (C_b - C_{it} - C_{op}t)/C_{mea}$ into $h_0(t_m, n, m)$ gives the objective function in (11). By using Proposition 1(iii) in Supplementary Section 3, the sufficient condition $0< t_D < (C_b - C_{mea} - C_{it})/C_{op}$ is equivalent to $1< m^* < (C_b-C_{it})/C_{mea}$.

The last case (iv) is trivial. The proof is complete. $\square$

Appendix 4: proof of theorem 3.3.

Proof

• (i) Taking the first derivative of $g(t_m)$ with respect to the termination time $t_m$ and setting the derivative to zero, we get $\alpha _{13} t_m^3 + \alpha _{12} t_m^2 - \alpha _{10} = 0$, where $\alpha _{1i}$ for $i = 0, 2, 3$ is defined in Theorem 3.3(i). It is easy to see that $\alpha _{10} > 0$ and $\alpha _{13} > 0$ for $\xi > 0$. If $\alpha _{12} > 0$, then using Lemma 1 in Supplementary Section 5.1 of Peng and Cheng (2021), the unique positive root $t_m$ in (12) can be obtained immediately by checking the feasibility. If $\alpha _{12} < 0$, by Descartes’ rule of signs (refer to Polya and Szegö (1997)), the cubic equation has only one positive real root $t_m^*$ because of the positive discriminant (i.e., $27\alpha _{13}^2\alpha _{10} \ge 4\alpha _{12}^3$).

• (ii) By elementary calculations, we have $\text{ d } g(t_m)/\text{d } t_m = 0$, which is equivalent to the quartic equation defined in Theorem 3.3(ii). In addition, it can be verified that

  $$\begin{aligned} \lim _{t_m \rightarrow \infty } g(t_m) = \frac{2\lambda ^2\zeta _3(2\sigma _\mu ^2\zeta _2 + \lambda \zeta _3) + 2\lambda \sigma _\mu ^4\zeta _2^2 m + (m-1)\sigma _\mu ^2\zeta _1^2}{n\lambda (m-1)}. \end{aligned}$$

  Hence, if the conditions in Theorem 3.3(ii) are satisfied, the result holds immediately. $\square$

Appendix 5: proof of theorem 3.4.

Proof

• (i) Substituting $n = (C_b-C_{op}t_m)/(C_{mea}m + C_{it})$ into $g_0(t_m, n, m)$ gives the profile objective function:

  $$\begin{aligned} g_{1}(t_m, m)&= (C_{mea}m + C_{it}) \{(\sigma _\mu ^2 t_m + \xi )(2(\zeta _1 + \zeta _2\lambda (\sigma _\mu ^2 t_m + \xi ))^2 \\&\quad + \zeta _1^2(3 + \lambda t_m(\sigma _\mu ^2 t_m + \xi )) )m \\&\quad + \lambda t_m (4\zeta _1\zeta _3\lambda (\sigma _\mu ^2 t_m + \xi ) - \zeta _1^2(\sigma _\mu ^2 t_m + \xi )^2 + 2\zeta _3\lambda ^2(2\zeta _2(\sigma _\mu ^2 t_m + \xi )^2 \\&\quad \{+ \zeta _3(3 + \lambda t_m(\sigma _\mu ^2 t_m + \xi ))))\}\\&\quad \{\lambda t_m (3m + \lambda t_m (m-1)(\sigma _\mu ^2 t_m + \xi ))(C_b-C_{op}t_m)\}. \end{aligned}$$

  By straightforward calculations, we have $\partial g_{1}(t_m,m)/\partial m = 0$, which is equivalent to the following quadratic equation

  $$\begin{aligned}&C_{mea} (\sigma _\mu ^2 t_m + \xi ) (3 + \lambda t_m (\sigma _\mu ^2 t_m + \xi )) ( 2 (\zeta _1 + \zeta _2 \lambda (\sigma _\mu ^2 t_m + \xi ))^2\\&\quad + \zeta _1^2 (3 + \lambda t_m (\sigma _\mu ^2 t_m + \xi ))) m^2 \\&\quad - 2 C_{mea} \lambda t_m (\sigma _\mu ^2 t_m + \xi )^2 ( 2 (\zeta _1 + \zeta _2 \lambda (\sigma _\mu ^2 t_m + \xi ))^2 + \zeta _1^2 (3 + \lambda t_m (\sigma _\mu ^2 t_m + \xi ))) m \\&\quad + \lambda t_m \{C_{mea} \lambda t_m (\sigma _\mu ^2 t_m + \xi )\{ (2\zeta _3 \lambda - \zeta _1 (\sigma _\mu ^2 t_m + \xi ))^2 - 2 \zeta _3 \lambda ^2 (2 \zeta _2 (\sigma _\mu ^2 t_m + \xi )^2 \\&\quad + \zeta _3 (5 + t_m \lambda (\sigma _\mu ^2 t_m + \xi )))\} \\&\quad - 2 C_{it} (3 \zeta _3 \lambda + (\sigma _\mu ^2 t_m + \xi ) (\zeta _1 + \lambda (\zeta _2 \sigma _\mu ^2 t_m + \zeta _3 t_m \lambda + \zeta _2 \xi )))^2\} = 0. \end{aligned}$$

  After some algebraic manipulations, it can be verified that the discriminant of the quadratic equation in m is given by

  $$\begin{aligned}&8 C_{mea} \lambda t_m (\sigma _\mu ^2 t_m + \xi ) (\lambda t_m (C_{mea}+ C_{it})(\sigma _\mu ^2 t_m + \xi ) + 3 C_{it}) \\&\quad \times \{2(\zeta _1 + \zeta _2 \lambda (\sigma _\mu ^2 t_m + \xi ))^2 + \zeta _1^2 (3 + \lambda t_m (\sigma _\mu ^2 t_m + \xi ))\} \\&\quad \times \{3 \zeta _3 \lambda + (\sigma _\mu ^2 t_m + \xi ) (\zeta _1 + \lambda (\zeta _3 \lambda t_m + \zeta _2 (\sigma _\mu ^2 t_m + \xi )))\}^2 > 0. \end{aligned}$$

  Thus it is easy to check the root $m(t_m)$ as shown in (15) for feasibility. Again, substituting $m(t_m)$ into $g_{1}(t_m, m)$ gives the profile objective function in (14). The result of the V-optimum test plan follows using Proposition 1(i) in Supplementary Section 3.

• (ii) Substituting $m = 1$ and $n = (C_b-C_{op}t_m)/(C_{mea} + C_{it})$ into $g_0(t_m, n, m)$ gives the objective function in (16). By using Proposition 1(ii) in Supplementary Section 3, the result follows directly.

• (iii) Substituting $n = 1$ and $m = (C_b - C_{it} - C_{op}t)/C_{mea}$ into $g_0(t_m, n, m)$ gives the objective function in (17). We have the desired result by using Proposition 1(iii) in Supplementary Section 3.

The last case (iv) is trivial. This completes the proof. $\square$