Optimal procurement strategies for contractual assembly systems with fluctuating procurement price

Abstract

We consider a multi-component assembly system that produces a single end product in order to satisfy the one-time demand at a (known) future time with an exogenous selling price. Components can be outsourced from outside suppliers with positive leadtimes under either time-inflexible or time-flexible contracts. One of these components, say component 1, faces an uncertainty in its procurement price, depending on the spot market price that is governed by a geometric Brownian motion, while the prices of other components are constant. Given supply contracts, the assembler needs to determine the procurement strategy to maximize the total expected profit that equals the expected revenue minus procurement cost, holding cost, and tardiness penalty cost. Under time-inflexible contracts, the problem is static and a variant of the classic newsvendor problem. We show that leadtime uncertainty will cause the assembler to be more conservative in procurement quantity, but more aggressive in procurement timing than if the leadtime is deterministic. For time-flexible contracts, we show that under certain conditions, the original problem is equivalent to an optimal single stopping problem whose optimal strategies follow either upward or downward base-price procurement policies. Under the general condition, we propose an efficient Monte Carlo simulation method to calculate the optimal solutions. Numerical studies also provide several interesting insights: first, both procurement quantity and profit are non-monotone in the leadtime length; second, the value of time-flexible contracts compared to time-inflexible contracts is close to zero if the price of component 1 has a decreasing trend, but otherwise significant.

Appendices

Appendix

Time-inflexible contract with uncertain leadtime

We now consider the setting with the uncertain leadtime of component 2 which follows distributions c.d.f G(x) and p.d.f g(x). Without confusion, we retain the notations used in the main text. The formulation is a little bit different in that the terms involving $L_2$ should be taken expectation. Then, the expected tardiness of the end product is

$$\begin{aligned} T_{d}={\mathbb {E}}[\max \{(t_{1}+L_{1}-T)^{+}, (t_{2}+L_{2}-T)^{+}\}]. \end{aligned}$$

And, the expected holding time of component 2 from the arrival of the order to the assemble time (due date plus tardiness, if any) is as follows:

$$\begin{aligned} T_{h,2}&=T_d+T-t_2-{\mathbb {E}}[L_2]. \end{aligned}$$

Consequently, the total expected profit function is

$$\begin{aligned} S(q,t_{1},t_{2})=p{\mathbb {E}}[\min \{q,D\}]-C(q,t_{1},t_{2}), \end{aligned}$$

where $C(q,t_{1},t_{2})=s\theta T_{d}+q(h_{1}T_{h,1}+h_{2}T_{h,2})+q({\mathbb {E}}[c_{1,t_{1}}\mid c_{1,0}]+c_{2}).$

The objective is then to solve the optimization problem stated in (6). Due to uncertain leadtime, the assembler cannot perfectly time the procurement of component 2 to avoid its holding cost. That is, Lemma 1 is no longer available in this setting. This indeed complicates the analysis of the problem. Define $t_1^{u,*}$, $t_2^{u,*}$, and $q^{u,*}$ as the optimal procurement timings and quantity under uncertain leadtime, respectively. The following theorem partially characterizes the optimal solutions for the case with $\mu >0$ an $\mu <0$, respectively.

Theorem 4

(a) :

When $\mu >0$, the following results hold:

(i):

the optimal procurement time of component 1 is

$$\begin{aligned} t_{1}^{u*}=\left\{ \begin{array}{ll}0 &{}\quad \text{ if } \quad \frac{h_{1}}{\mu c_{1,0}}\le 1;\\ \frac{1}{\mu }\ln (\frac{h_{1}}{\mu c_{1,0}}), &{}\quad \text{ if } \quad 1<\frac{h_{1}}{\mu c_{1,0}}\le e^{\mu (T-L_{1})};\\ T-L_{1}, &{}\quad \text{ if } \quad e^{\mu (T-L_{1})}<\frac{h_{1}}{\mu c_{1,0}}. \end{array}\right. \end{aligned}$$

(A1)

(ii):

the optimal procurement quantity and procurement time of component 2 satisfy

$$\begin{aligned} q^{u*}&=F^{-1}\left( \frac{p-h_{1}{\hat{T}}_{h,1}-h_{2}{\hat{T}}_{h,2}-c_{1,0}e^{\mu t^{u*}_{1}}-c_{2}}{p}\right) ;\\ t_2^{u*}&=T-G^{-1}\left( \frac{sd+q^{u*}h_{1}}{sd+q^{u*}(h_{1}+h_{2})}\right) , \end{aligned}$$

where ${\hat{T}}_{h,1}={\mathbb {E}}[(t^{u*}_{2}+L_{2}-T)^{+}]+T-t^{u*}_1-L_1$ and ${\hat{T}}_{h,2}={\mathbb {E}}[(t^{u*}_{2}+L_{2}-T)^{+}]+T-t^{u*}_2-{\mathbb {E}}[L_2]$.

(b) :

When $\mu <0$, the following results hold:

(i):

the optimal procurement time of component 1 is

$$\begin{aligned} t_{1}^{u*}=\left\{ \begin{array}{ll}T-L_1 &{}\quad \text{ if } \quad \frac{sd}{-\mu c_{1,0}q^{u*}}\ge e^{\mu (T-L_1)};\\ \frac{1}{\mu }\ln (\frac{sd}{-\mu c_{1,0}q^{u*}}), &{}\quad \text{ if } \quad e^{\mu T}<\frac{sd}{-\mu c_{1,0}q^{u*}}< e^{\mu (T-L_1)};\\ T, &{}\quad \text{ if } \quad \frac{sd}{-\mu c_{1,0}q^{u*}}\le e^{\mu T}. \end{array}\right. \end{aligned}$$

(A2)

(ii):

the optimal procurement time of component 2 is

$$\begin{aligned} t_2^{u*}=t_1^{u*}+L_1-G^{-1}\left( \frac{sd+q^{u*}h_{1}}{sd+q^{u*}(h_{1}+h_{2})}\right) . \end{aligned}$$

(iii):

the optimal order quantity satisfies

$$\begin{aligned} q^{u*}&=F^{-1}\left( \frac{p-h_{1}{\hat{T}}_{h,1}-h_{2}{\hat{T}}_{h,2}-c_{1,0}e^{\mu t^{u*}_{1}}-c_{2}}{p}\right) , \end{aligned}$$

where ${\hat{T}}_{h,i}={\mathbb {E}}[\max \{(t^{u*}_{1}+L_{1}-T)^{+}, (t^{u*}_{2}+L_{2}-T)^{+}\}]+T-t^{u*}_i-{\mathbb {E}}[L_i]$.

Proof of Theorem 4

First, taking the first partial derivative of $T_{d}(t_1,t_2)$ respect to $t_{1}$ and $t_{2}$, respectively, we can obtain

$$\begin{aligned}&\frac{\partial T_{d}(t_1,t_2)}{\partial t_{1}}=\left\{ \begin{array}{ll}0, &{}\quad \text{ if } \quad t_{1} \le T-L_{1};\\ G(t_{1}+L_{1}-t_{2}), &{}\quad \text{ if } \quad t_{1} > T-L_{1}.\end{array}\right. \end{aligned}$$

(A3)

$$\begin{aligned}&\frac{\partial T_{d}(t_1,t_2)}{\partial t_{2}}=\left\{ \begin{array}{ll}1-G(T-t_{2}), &{}\quad \text{ if } \quad t_{1} \le T-L_{1};\\ 1-G(t_{1}+L_{1}-t_{2}), &{}\quad \text{ if } \quad t_{1} > T-L_{1}.\end{array}\right. \end{aligned}$$

(A4)

Accordingly, the first partial derivatives of $S(q,t_{1},t_{2})$ respect to q, $t_{1}$ and $t_{2}$ are respectively given as:

$$\begin{aligned} \frac{\partial S(t_{1},t_{2},q)}{\partial q}&=p(1-F(q))-h_{1}T_{h1}-h_{2}T_{h2}-c_{1,0}e^{\mu t_{1}}-c_{2}, \nonumber \\ \frac{\partial S(t_{1},t_{2},q)}{\partial t_{1}}&=\left\{ \begin{array}{ll}-\mu c_{1,0}qe^{\mu t_{1}}+h_{1}q, &{}\quad \text{ if } \quad t_{1} \le T-L_{1};\\ {-\mu c_{1,0}qe^{\mu t_{1}}+h_{1}q-(sd+q(h_{1}+h_{2}))G(t_{1}+L_{1}-t_{2})}, &{}\quad \text{ if } \quad t_{1}> T-L_{1}.\end{array}\right. \end{aligned}$$

(A5)

$$\begin{aligned} \frac{\partial S(t_{1},t_{2},q)}{\partial t_{2}}&=\left\{ \begin{array}{ll}(sd+q(h_{1}+h_{2}))G(T-t_{2})-(h_{1}q+sd), &{}\quad \text{ if } \quad t_{1} \le T-L_{1};\\ (sd+q(h_{1}+h_{2}))G(t_{1}+L_{1}-t_{2})-(h_{1}q+sd), &{}\quad \text{ if } \quad t_{1}> T-L_{1}.\end{array}\right. \end{aligned}$$

(A6)

(a) For the case $\mu >0$, we first prove that it is optimal to order component 1 before time $T-L_1$, i.e., $t_1^{u*}\le T-L_1$. Suppose $t_1^{u*}> T-L_1$, then $t_1^{u*}$ and $t_2^{u*}$ should satisfy

$$\begin{aligned} (sd+q(h_{1}+h_{2}))G(t_{1}^{u*}+L_{1}-t_{2}^{u*})-(h_{1}q+sd)=0, \end{aligned}$$

(A7)

which, together with (A5), implies that $-\,\mu c_{1,0}qe^{\mu t_{1}}-sd=0$. This contradicts with the fact that $\mu >0$. Therefore, $t_1^{u*}\le T-L_1$. Define $\hat{t}:=\frac{1}{\mu }\ln (\frac{h_{1}}{\mu c_{1,0}})$. Refer to (A5). If $\hat{t}\le 0$, i.e., $\frac{h_{1}}{\mu c_{1,0}}\le 1$, then $S(q,t_{1},t_2)$ is decreasing in $t_{1}\in [0, T-L_1]$ for any q and $t_2$, and thus, $t_{1}^{u*}=0$. If $0<\hat{t}\le T-L_1$, i.e., $1<\frac{h_{1}}{\mu c_{1,0}}\le e^{\mu (T-L_{1})}$, then $t_{1}^{u*}=\hat{t}$. If ${\hat{t}} > T-L_1$, i.e., $\frac{h_{1}}{\mu c_{1,0}}> e^{\mu (T-L_{1})}$, then $t_1^{u*}=T-L_1$. Moreover by (A6), the optimal ordering time of component 2 satisfies

$$\begin{aligned} t_2^{u*}=T-G^{-1}\left( \frac{sd+qh_{1}}{sd+q(h_{1}+h_{2})}\right) . \end{aligned}$$

By the first derivative of $S(q_1,t_1,t_2)$ respect to $q_1$, we have

$$\begin{aligned} q^{u*}&=F^{-1}\left( \frac{p-h_{1}{\hat{T}}_{h1}-h_{2}{\hat{T}}_{h2}-c_{1,0}e^{\mu t^{u*}_{1}}-c_{2}}{p}\right) , \end{aligned}$$

(A8)

where ${\hat{T}}_{h1}={\mathbb {E}}[(t^{u*}_{2}+L_{2}-T)^{+}]+T-t^{u*}_1-L_1$ and ${\hat{T}}_{h2}={\mathbb {E}}[(t^{u*}_{2}+L_{2}-T)^{+}]+T-t^{u*}_2-{\mathbb {E}}[L_2]$.

(b) For the case $\mu <0$, it is clear that $S(t_1,t_2,q)$ is increasing in $t_1\le T-L_1$ for any given q. Therefore, it is optimal to order component 1 after time $T-L_1$, i.e., $t_1^{u*}>T-L_1$. Moreover, in this case, by the first optimality condition, $t_1^{u*}$ and $t_2^{u*}$ should satisfy (A7). That is,

$$\begin{aligned} t_1^{u*}+L_1-t_2^{u*}=G^{-1}\left( \frac{sd+qh_{1}}{sd+q(h_{1}+h_{2})}\right) . \end{aligned}$$

Putting it into (A5), we can obtain

$$\begin{aligned} \frac{\partial S(t_{1},t_{2},q)}{\partial t_{1}}= -\,\mu c_{1,0}qe^{\mu t_{1}}-sd. \end{aligned}$$

Let ${\hat{t}}_1=\frac{1}{\mu }\ln (\frac{sd}{-\mu c_{1,0}q}).$ If ${\hat{t}}_1\le T-L_1$, i.e., $\frac{sd}{-\mu c_{1,0}q}\ge e^{\mu (T-L_1)}$, then $t_1^{u*}=T-L_1$; if $T-L_1<{\hat{t}}_1<T$, i.e., $e^{\mu T}<\frac{sd}{-\mu c_{1,0}q}< e^{\mu (T-L_1)}$ then $t_1^{u*}={\hat{t}}_1$; if ${\hat{t}}_1 \ge T$, i.e., $\frac{sd}{-\mu c_{1,0}q}\le e^{\mu T},$ then $t_1^{u*}=T$.

By the first derivative of $S(q_1,t_1,t_2)$ respect to $q_1$, we have

$$\begin{aligned} q^{u*}&=F^{-1}\left( \frac{p-h_{1}{\hat{T}}_{h1}-h_{2}{\hat{T}}_{h2}-c_{1,0}e^{\mu t^{u*}_{1}}-c_{2}}{p}\right) , \end{aligned}$$

where ${\hat{T}}_{hi}={\mathbb {E}}[\max \{(t^{u*}_{1}+L_{1}-T)^{+}, (t^{u*}_{2}+L_{2}-T)^{+}\}]+T-t^{u*}_i-{\mathbb {E}}[L_i]$. $\square$

Auxiliary proofs

Proof of Lemma 1

We first consider the case with $t_1^*\le T-L_1$. Referring to (2) and (3), we have $T_{d}=\max _{i=2,\ldots ,N}\left\{ (t_{i}+L_{i}-T)^{+}\right\}$ and $T_{hi}=T_d+T-(t_i+L_i)$, $i=2,\ldots ,N$. Clearly, both holding and tardiness costs are minimized at $t_i^*=T-L_i$, $i=2,\ldots ,N$, with the minimal values $T_d=0$ and $T_{hi}=0$. Referring to (5), this result implies that given $t_1^*\le T-L_1$, $S(q,{\mathbf {t}})$ is minimized at $t_i^*=T-L_i$, $i=2,\ldots ,N,$ for any q.

Now turn to the case with $t_1^*\ge T-L_1$. Define ${\mathbf {t}}_{-1}=(t_2,\ldots ,t_N)$. We write $T_d$ and $T_{h2}$ as a function of ${\mathbf {t}}_{-1}$. By the definition, we have for any ${\mathbf {t}}_{-1}$,

$$\begin{aligned} T_d({\mathbf {t}}_{-1})&=\max _{i=2,\ldots ,N}\{(t_{1}^*+L_{1}-T), (t_{i}+L_{i}-T)^{+}\}\ge (t_{1}^*+L_{1}-T); \\ T_{h2}({\mathbf {t}}_{-1})&=T_d({\mathbf {t}}_{-1})+T-(t_2+L_2)\ge 0. \end{aligned}$$

Note that $T_d({\mathbf {t}}_{-1})|_{t_i=t_1^*+L_1-L_i}=(t_{1}^*+L_{1}-T)$ and $T_{h2}({\mathbf {t}}_{-1})|_{t_i=t_1^*+L_1-L_i}=0$ which means that both lower bounds can be achieved at $t_i^*=t_1^*+L_1-L_i$. Therefore, the results hold. $\square$

Proof of Theorem 1

(a) Recall that $c_{1,t}$ follows a Brownian motion (1) and thus, ${\mathbb {E}}[c_{1,t_1}|c_{1,0}]=c_{1,0}e^{\mu t_{1}}$. Then, ${\widehat{C}}(q,t_1)$ can be rewritten as

$$\begin{aligned} {\widehat{C}}(q,t_1)=sd T_{d}+qh_{1}T_{h1}+q\left( c_{1,0}e^{\mu t_{1}}+c_2^N\right) . \end{aligned}$$

The first partial derivatives of ${\widehat{S}}(q,t_{1})$ respect to q and $t_{1}$ are given as follows.

$$\begin{aligned} \frac{\partial {\widehat{S}}(q,t_{1})}{\partial q}&=p (1-F(q))-c_{1,0}e^{\mu t_{1}}-h_{1}(T-L_{1}-t_{1})^{+}-c_{2}^N;\nonumber \\ \frac{\partial {\widehat{S}}(q,t_{1})}{\partial t_{1}}&=\left\{ \begin{array}{ll}-\mu c_{1,0}qe^{\mu t_{1}}+h_{1}q, &{}\quad \text{ if } \quad t_{1} \le T-L_{1};\\ -\,\mu c_{1,0}qe^{\mu t_{1}}-sd, &{}\quad \text{ if } \quad t_{1} > T-L_{1}.\end{array}\right. \end{aligned}$$

(B9)

The second partial derivatives of ${\widehat{S}}(q,t_{1})$ respect to q and $t_{1}$ are given as follows.

$$\begin{aligned} \frac{\partial ^{2} {\widehat{S}}(q,t_{1})}{\partial q^{2}}&=-\,p f(q)\le 0; \\ \frac{\partial ^{2} {\widehat{S}}(q,t_{1})}{\partial t_{1}^{2}}&=-\,\mu ^{2} c_{1,0}qe^{\mu t_{1}}\le 0. \end{aligned}$$

Therefore, for any fixed $t_{1}$, ${\widehat{S}}(q,t_{1})$ is concave in q and for any fixed q, ${\widehat{S}}(q,t_{1})$ is concave in $t_{1}$.

(b) Referring to (B9), we have that if $\mu > 0$, $\frac{\partial {\widehat{S}}(q,t_{1})}{\partial t_{1}} < 0$ for any $t_{1}> T-L_{1}$, which implies that it is more profitable to procure component 1 before or at time $T-L_{1}$, i.e., $0 \le t_{1}^{*}\le T-L_{1}$. Define $\hat{t}\doteq \frac{1}{\mu }\ln (\frac{h_{1}}{\mu c_{1,0}})$. If $\hat{t}\le 0$, i.e., $\frac{h_{1}}{\mu c_{1,0}}\le 1$, then ${\widehat{S}}(q,t_{1})$ is decreasing in $t_{1}\in [0, T-L_1]$ for any q, and thus, $t_{1}^{*}=0$. If $0<\hat{t}\le T-L_1$, i.e., $1<\frac{h_{1}}{\mu c_{1,0}}\le e^{\mu (T-L_{1})}$, then $t_{1}^{*}=\hat{t}$. If ${\hat{t}} > T-L_1$, i.e., $\frac{h_{1}}{\mu c_{1,0}}> e^{\mu (T-L_{1})}$, ${\widehat{S}}(q,t_{1})$ is increasing in $t_{1}\in [0, T-L_1]$ for any q, and thus, $t_1^*=T-L_1$.

Observe that for $0\le t_{1}\le T-L_{1}$, the first partial derivative of ${\widehat{S}}(q,t_1)$ respect to q can be reexpressed as

$$\begin{aligned} \frac{\partial {\widehat{S}}(q,t_{1})}{\partial q}=p (1-F(q))-c_{1,0}e^{\mu t_{1}}-h_{1}(T-L_{1}-t_{1})-c_{2}^N. \end{aligned}$$

Thus, the optimal order quantity should be

$$\begin{aligned} q^*=F^{-1}\left( \frac{p-c_{1,0}e^{\mu t_1^*}-h_1 (T-L_1-t_1^*)-c_2^N}{p}\right) . \end{aligned}$$

(c) Again from (B9), if $\mu < 0$, $\frac{\partial {\widehat{S}}(q, t_{1})}{\partial t_{1}} > 0$ for any $t_{1} \le T-L_{1}$. That is, it is optimal to procure component 1 after time $T-L_{1}$, i.e., $t_1^*\ge T-L_1$. From the first derivative respect to q, we have

$$\begin{aligned} q^{*}=F^{-1}\left( \frac{p-c_{1,0}e^{\mu t_{1}^{*}}-c_{2}^N}{p}\right) . \end{aligned}$$

Clearly, $q^{*}$ is increasing in $t_1^*$. Recall that $T-L_1 \le t_1^*\le T$. As a result, ${\underline{q}}\le q^*\le {\overline{q}}$.

By (B9), if $q^*\le \frac{sde^{-\mu (T-L_1)}}{-\mu c_{1,0}}$, then $\frac{\partial {\widehat{S}}(q,t_{1})}{\partial t_{1}}<0$ for any $t_1>T-L_1$ and thus, $t_1^*=T-L_1$. Following a similar logic, one can readily prove the results for other cases. $\square$

Proof of Theorem 2

The results for the scenario with an uncertain leadtime of component 2 are presented in Theorem 4. Consider the case with $\mu >0$. It is clear that $t_1^*=t_1^{u*}$. From the proof of Theorem 1, we know that

$$\begin{aligned} q^*=F^{-1}\left( \frac{p-c_{1,0}e^{\mu t_1^*}-h_1 (T-L_1-t_1^*)-c_2}{p}\right) . \end{aligned}$$

From (A8), in order to prove $q^*\ge q^{u*}$, it suffices to prove $h_{1}{\hat{T}}_{h1}+h_{2}{\hat{T}}_{h2}\ge h_1 (T-L_1-t_1^*).$ This result holds because ${\hat{T}}_{h1}\ge (T-L_1-t_1^*)$.

Now, turn to the case with $\mu <0$. We prove the result by contradiction. Suppose that $q^* < q^{u*}$. Note that from the proof of Theorem 1, $t_1^*$ and $q^*$ also satisfy (A2). Therefore, one can readily prove that $t_1^{u*}\ge t_1^*$. Hence, we can obtain

$$\begin{aligned} q^*&=F^{-1}\left( \frac{p-c_{1,0}e^{\mu t_1^*}-c_2}{p}\right) \nonumber \\&\ge F^{-1}\left( \frac{p-h_{1}{\hat{T}}_{h1}-h_{2}{\hat{T}}_{h2}-c_{1,0}e^{\mu t^{u*}_{1}}-c_{2}}{p}\right) \nonumber \\&=q^{u*}. \end{aligned}$$

(B10)

This result contradicts our hypothesis and $q^{u*}\le q^*$. By (A2), $t_1^{u*}\ge t_1^*$. $\square$

Proof of Lemma 2

(i) We prove by contradiction that for any $m_1<m_2$, $t_{m_1}^*\ge t_{m_2}^*$. Suppose $t_{m_1}^*<t_{m_2}^*$. That is, at time $t_{m_1}^*$, it is optimal to procure component $m_1$ and delay the procurement of component $m_2$. Then, there always exists a small enough time interval dt such that component $m_2$ will be procured after time $t_{m_1}^*+dt$, i.e., $t_{m_2}^*\ge t_{m_1}^*+dt$. Next, we show that the assembler will be better off by procuring component $m_1$ at time $t_{m_1}^*+dt$ than at time $t_{m_1}^*$ with other decisions being identical. It suffices to show that procuring component $m_1$ at time $t_{m_1}^*+dt$ leads to smaller holding and tardiness costs.

Denote by $T_d$ and ${\widehat{T}}_d$ the tardiness time of procuring component $m_1$ at time $t_{m_1}^*$ and $t_{m_1}^*+dt$, respectively. By the definition, we can obtain

$$\begin{aligned} T_{d}&=\max \left\{ \max _{i\in \{1,\ldots ,m\}\setminus \{m_1,m_2\}}\left\{ (t_{i}+L_{i}-T)^{+}\right\} , (t_{m_1}^*+L_{m_1}-T)^{+},(t_{m_2}^*+L_{m_2}-T)^{+}\right\} , \nonumber \\&=\max \left\{ \max _{i\in \{1,\ldots ,m\}\setminus \{m_1,m_2\}}\left\{ (t_{i}+L_{i}-T)^{+}\right\} , (t_{m_1}^*+dt+L_{m_1}-T)^{+},(t_{m_2}^*+L_{m_2}-T)^{+}\right\} , \nonumber \\&={\widehat{T}}_d, \end{aligned}$$

(B11)

where the first equality follows from $t_{m_2}^*\ge t_{m_1}^*+dt$ and $L_{m_2}\ge L_{m_1}$.

As procurement decisions of other components are identical, so are their holding costs. Now, we check the holding cost of component $m_1$. By the definition, $T_{h,m_1}=T_d+T-(t_{m_1}^*+L_{m_1})$ and ${\widehat{T}}_{h,m_1}={\widehat{T}}_d+T-(t_{m_1}^*+dt+L_{m_1})$, where $T_{h,m_1}$ and ${\widehat{T}}_{h,m_1}$ represents the holding costs of procuring component $m_1$ at time $t_{m_1}^*$ and $t_{m_1}^*+dt$, respectively. Clearly, $T_{h,m_1}>{\widehat{T}}_{h,m_1}$. As a result, it is better off by procuring component $m_1$ at $t_{m_1}^*+dt$, which contradicts the optimality of $t_{m_1}^*$.

(ii) Consider component i in $\Omega ^L$. To prove $t_i^*\ge T-L_i$, we prove that for any time $t_i\in [0, T-L_i)$, the assembler will be better off by procuring component i at time $T-L_i$ than at time $t_i$, with other decisions being identical. By the definition of $T_d$, the tardiness costs under these two strategies are the same and thus, so are the holding costs of other components. However, procuring component i at time $T-L_i$ will lead to a lower holding cost of component i. Therefore, it is better off to procure component i after time $T-L_i$. Recall that component 1 must be procured before the due date and thus arrive before time $T+L_1$. As a result, other components should arrive before time $T+L_1$, which implies $t_i^*\le T-(L_i-L_1)$. $\square$

Proof of Lemma 3

We now prove that the Problems (13) and (14) are equivalent. By Observation 2, we only need to prove ${\widehat{U}}^*(q,c_{1,0})\ge U^*(q,c_{1,0})$. To achieve this, we propose for Problem (13) a feasible strategy as follows: $t_i^*=T-L_i$, for $i=2,\ldots ,m$, and $t_1^*={\widehat{t}}_1^*$, under which the total expected profit of Problem (13) is $U^*(q,c_{1,0})$.

We use a result of Problem (14) from Lemma 4(i) that if $\mu \ge 0$, ${\widehat{t}}_1^*\le T-L_1$. Therefore, under the constructed strategies, all components arrive before due date and thus, $T_d={\widehat{T}}_d=0$. Moreover, as component $i\in \Omega ^L$ arrive at its due date, no holding cost is incurred, i.e., $T_{h,i}=0$, $i\in \Omega ^L$. Finally, $T_{h,1}={\widehat{T}}_{h,1}=T-({\widehat{t}}_1^*+L_1)$. In summary, all the costs in two systems, as well as the expected revenue, are equal. Therefore, under the proposed feasible strategy, Problem (13)’s expected profit is $U^*(q,c_{1,0})$. Consequently, the optimal expected profit of problem (13) satisfies ${\widehat{U}}^*(q,c_{1,0})\ge U^*(q,c_{1,0})$ and the proposed strategy is optimal. $\square$

Proof of Lemma 4

Let ${\widehat{U}}(q, t_1,c_{1,t_{1}})$ be the maximal expected total profit at time $t_1$ with the current price $c_{1,t_{1}}$. Then, the function ${\widehat{U}}(q,t_1,c_{1,t_{1}})$ satisfies the Wald-Bellman equations (Shiryaev 1978):

$$\begin{aligned} {\widehat{U}}(q,t_1,c_{1,t_{1}})=\max \left\{ {\widehat{S}}(q, t_{1},c_{1,t_{1}}),{\mathbb {E}}[{\widehat{U}}(q,t_1+dt,c_{1,t_{1}+dt})\mid c_{1,t_{1}}]\right\} \end{aligned}$$

(B12)

with the boundary condition ${\widehat{U}}(q, T, c_{1,T})={\widehat{S}}(q,T,c_{1,T})$.

The first term in (B12) represents the total profit if the assembler makes the purchase immediately, while the second term represents the expected profit if the assembler waits for another dt units of time.

By Wald-Bellman equations (B12), it is easy to verify that for any time $t_1$,

$$\begin{aligned} {\widehat{U}}(q,t_1,c_{1,t_{1}})\ge {\widehat{S}}(q, t_{1},c_{1,t_{1}}). \end{aligned}$$

(B13)

By Ito’s Lemma, we have

$$\begin{aligned} d{\widehat{S}}(q,t_1,c_{1,t_{1}})=\left( \frac{\partial {\widehat{S}}}{\partial t_{1}}+\mu c_{1,t_{1}}\frac{\partial {\widehat{S}}}{\partial c_{1,t_{1}}} + \frac{1}{2} \sigma ^{2}c_{1,t_{1}}^{2} \frac{\partial ^{2} {\widehat{S}}}{\partial c_{1,t_{1}}^{2}}\right) dt+\sigma c_{1,t_{1}}\frac{\partial {\widehat{S}}}{\partial c_{1,t_{1}}}dz, \end{aligned}$$

and hence,

$$\begin{aligned} {\mathbb {E}}[d{\widehat{S}}(q,t,c_{1,t_{1}})]=\left( \frac{\partial {\widehat{S}}}{\partial t_{1}}+\mu c_{1,t_{1}}\frac{\partial {\widehat{S}}}{\partial c_{1,t_{1}}} + \frac{1}{2} \sigma ^{2}c_{1,t_{1}}^{2} \frac{\partial ^{2} {\widehat{S}}}{\partial c_{1,t_{1}}^{2}}\right) dt. \end{aligned}$$

(B14)

Taking partial derivatives of ${\widehat{S}}(q,t_1,c_{1,t_{1}})$ respect to $t_1$ and $c_{1,t_{1}}$, we can obtain

$$\begin{aligned} \frac{\partial {\widehat{S}}}{\partial t_{1}}&=\left\{ \begin{array}{ll}h_{1}q, &{}\quad \text{ if } \quad 0 \le t_{1} \le T-L_{1};\\ -sd, &{}\quad \text{ if } \quad T\ge t_{1} >T-L_{1};\end{array}\right. \nonumber \\ \frac{\partial {\widehat{S}}}{\partial c_{1,t_{1}}}&=-\,q;\ \ \ \frac{\partial ^{2} {\widehat{S}}}{\partial c_{1,t_{1}}^{2}}=0. \end{aligned}$$

(B15)

Substituting (B15) into (B14), we can obtain

$$\begin{aligned} {\mathbb {E}}[d{\widehat{S}}(q,t,c_{1,t_{1}})]=\left\{ \begin{array}{ll}(h_{1}q-\mu c_{1,t}q)dt , &{}\quad \text{ if } \quad 0\le t_{1}\le T-L_{1};\\ (-\,sd-\mu c_{1,t}q)dt, &{}\quad \text{ if } \quad T-L_1< t_{1} \le T.\end{array}\right. \end{aligned}$$

(B16)

(i) If $\mu < 0$, then for any $t_{1}\le T-L_{1}$, ${\mathbb {E}}[d{\widehat{S}}(q,t_1,c_{1,t_{1}})]>0$. This, together with (B13), implies that

$$\begin{aligned} {\mathbb {E}}[{\widehat{U}}(q,t_1+dt,c_{1,t_{1}+dt})\mid c_{1,t_{1}}]-{\widehat{S}}(q, t_{1},c_{1,t_{1}})\ge {\mathbb {E}}[d{\widehat{S}}(q,t_1,c_{1,t_{1}})]>0. \end{aligned}$$

In other words, it is more profitable for the assembler to make the purchase at least after time $T-L_{1}$. That is, $T-L_{1}\le t_{1}^{*}\le T$.

(ii) For the case with $\mu > 0$, we prove an equivalent result by induction that it is optimal to purchase immediately for any time $t_1\in (T-L_1, T]$ and observed price. The result clearly holds for time T. Suppose it also holds for time $t_1+dt$, i.e.,

$$\begin{aligned} {\widehat{U}}(q, t_1+dt,c_{1,t_{1}+dt})={\widehat{S}}(q, t_{1}+dt,c_{1,t_{1}+dt}). \end{aligned}$$

Then, for time $t_1$, we have that

$$\begin{aligned} {\widehat{U}}(q,t_1,c_{1,t_{1}})&=\max \left\{ {\widehat{S}}(q, t_{1},c_{1,t_{1}}), {\mathbb {E}}[{\widehat{U}}(q,t_1+dt,c_{1,t_{1}+dt})\mid c_{1,t_{1}} ]\right\} , \\&=\max \left\{ {\widehat{S}}(q, t_{1},c_{1,t_{1}}), {\mathbb {E}}[{\widehat{S}}(q,t_1+dt,c_{1,t_{1}+dt})\mid c_{1,t_{1}} ]\right\} . \end{aligned}$$

By (B16) and $\mu >0$, one can easily verify that ${\widehat{S}}(q, t_{1},c_{1,t_{1}})>{\mathbb {E}}[{\widehat{S}}(q,t_1+dt,c_{1,t_{1}+dt})\mid c_{1,t_{1}}]$. That is, it is optimal to purchase immediately and the induction completes. $\square$

Proof of Lemma 5

The main idea in the following is similar to that in the proof of Theorem 4.3 in Jacka and Lynn (1992). For any $t \in [0,T]$, define the continuation value function as follows:

$$\begin{aligned} V(t,x):=\sup \limits _{\tau _t\in (t,T]}{\mathbb {E}}[g(\tau _t,c_{\tau _t})|c_t=x], \end{aligned}$$

which means the optimal value the decision maker can expect if she decides not to stop at time t with $c_t=x$. It is clear that the optimal policy is to stop at t if $g(t,x)\ge V(t,x)$ and not to stop if $V(t,x)>g(t,x)$. For any stopping time $\tau _t\in (t,T]$, we have

$$\begin{aligned} {\mathbb {E}}[g(\tau _t,c_{\tau _t})|c_t]= & {} a {\mathbb {E}}[\tau _t|c_t]+b {\mathbb {E}}[ c_{\tau _t}|c_t] \nonumber \\= & {} a {\mathbb {E}}[\tau _t|c_t]+b {\mathbb {E}}[c_t e^{(\mu -\frac{\sigma ^2}{2})(\tau _t-t)+\sigma z_{\tau _t-t}}|c_t] \nonumber \\= & {} a {\mathbb {E}}[\tau _t|c_t]+b c_t {\mathbb {E}}[e^{\mu (\tau _t-t)}]. \end{aligned}$$

(B17)

Because $\tau _t > t$, we can directly get from (B17) that if $a>0$ and $b\mu >0$,

$$\begin{aligned} {\mathbb {E}}[g(\tau _t,c_{\tau _t})|c_t] > a t +b c_t, \end{aligned}$$

(B18)

and if $a<0$ and $b\mu <0$,

$$\begin{aligned} {\mathbb {E}}[g(\tau _t,c_{\tau _t})|c_t] < a t +b c_t. \end{aligned}$$

(B19)

Note that (B18) implies that

$$\begin{aligned} V(t,c_t) \ge g(t,c_t), \end{aligned}$$

for any $\tau _t\in [0,T]$ and $c_t$. Then the decision maker should always wait until the time horizon T and part (i) is proved. In contrast, (B19) results in $V(t,c_t)<g(t,c_t)$ for any $t\in [0,T]$ and $c_t$, which means the decision maker should immediately stop at $t=0$. Then, part (ii) holds.

Next, we prove parts (iii) and (iv). Note that for any stopping time $\tau _t\in (t,T]$ and $c_t$,

$$\begin{aligned} g(\tau _t, c_{\tau _t})-g(t,c_t)=a(\tau _t-t)+b(c_{\tau _t}-c_t). \end{aligned}$$

Now suppose a stopping time $\tau _t^x \in (t,T]$ is optimal for $c_t=x$, we have

$$\begin{aligned} V(t,x)-g(t,x)= & {} {\mathbb {E}}[g(\tau _t^x,c_{\tau _t^x})|c_t=x]-g(t,x) \nonumber \\= & {} a({\mathbb {E}}[\tau _t^x-t])+b x {\mathbb {E}}[e^{\mu (\tau _t^x-t)}-1]. \end{aligned}$$

(B20)

On the other hand, $\tau _t^x$ may not necessarily be optimal for $c_t=y$ and $y \ne x$, which implies

$$\begin{aligned} V(t,y)-g(t,y)\ge & {} {\mathbb {E}}[g(\tau _t^x,c_{\tau _t^x})|c_t=y]-g(t,y) \nonumber \\= & {} a({\mathbb {E}}[\tau _t^x-t])+b y {\mathbb {E}}[e^{\mu (\tau _t^x-t)}-1], \end{aligned}$$

(B21)

where the last equality is because $\tau _t^x$ has nothing to do with $y \ne x$.

(iii) Assume that the decision maker should wait at $c_t=x$, which means $V(t,x)-g(t,x)>0$. If $b \mu <0$, then for any $y<x$, (B20) and (B21) implies $V(t,y)-g(t,y)>V(t,x)-g(t,x)>0$ and then the decision maker should wait at $c_t=y$, too. Hence, part (iii) holds, where ${C}_{Lt}$ is the lowest price at which it is optimal to stop.

(iv) Assume that the decision maker should wait at $c_t=x$, which means $V(t,x)-g(t,x)> 0$. If $b \mu >0$, then for any $y>x$, (B20) and (B21) implies $V(t,y)-g(t,y)>V(t,x)-g(t,x)>0$ which in turn implies that the decision maker should wait at $c_t=y$, too. Then, part (iv) holds, where ${C}_{Ht}$ is the highest price at which it is optimal to stop. $\square$

Proof of Theorem 3

Lemma 3 has already proved that the original system is essentially equivalent to a standard stopping time problem involving only component 1 for the case with $\mu \ge 0$. Moreover, $t_i^*=T-L_i$, for $i=2,\ldots ,m$. Clearly, when $\mu <0$ and $\Omega ^L$ is an empty set, the original system is also equivalent to an OSS problem. Therefore, we only need to study Problem (14).

(i) For the case with $\mu > 0$, it is clear that Problem (16) is equivalent to the following problem:

$$\begin{aligned} \sup _{t_1\in [0,T-L_1]}{\mathbb {E}}[h_1 t_1-c_{1,t_{1}}|c_{1,0}]. \end{aligned}$$

(B22)

Problem (B22) corresponds to the case of Lemma 5 with $a=qh_1$ and $b=-\,q$. Then, part (iii) of Lemma 5 implies that there exists a threshold price ${\tilde{C}}_{1,t}$ at which the assembler should procure component 1 at t if and only if $c_{1,t} \ge {\tilde{C}}_{1,t}$. That is, the upward base-price procurement policy is optimal for component 1.

(ii) If $\mu <0$ and $\Omega ^L$ is an empty, Problem (17) is equivalent to the following problem:

$$\begin{aligned} \sup _{t_1\in [T-L_1,T]}{\mathbb {E}}[-\,s d t_1-q c_{1,t_{1}}|c_{1,0}]. \end{aligned}$$

(B23)

Clearly, Problem (B23) corresponds to the case of Lemma 5 with $a=-\,sd$ and $b=-\,q$. Then, part (iv) of Lemma 5 implies there exists a threshold price ${\widehat{C}}_{1,t}$ such that the assembler should procure component 1 at t if and only if $c_{1,t} \le {\widehat{C}}_{1,t}$. That is, the downward base-price procurement policy is optimal for component 1. $\square$

Proof of Proposition 1

(i) In the case with $\mu > 0$, the assembler should order component 1 before time $T-L_1$, i.e., $0 \le t_{1}^{*}\le T-L_{1}$. The original problem is thus reduced to the one stated in (16). In this case, we have

$$\begin{aligned} {\mathbb {E}}[d{\widehat{S}}^1(q,t,c_{1,t_{1}})]=(h_{1}q-\mu c_{1,t_1}q)dt. \end{aligned}$$

Define ${\widehat{U}}^1(q, t_1,c_{1,t_{1}})$ as the maximal expected total profit at time $t_1$ with the current unit price $c_{1,t_{1}}$ for Problem (16). To prove the result, it suffices to prove that when $c_{1,t_1}<\frac{h_{1}}{\mu }$, it is optimal to wait, which means that for any $c_{1,t_1}<\frac{h_1}{\mu }$,

$$\begin{aligned} {\widehat{S}}^1(q, t_{1},c_{1,t_{1}})<{\mathbb {E}}[{\widehat{U}}^1(q,t_1+dt,c_{1,t_{1}+dt})\mid c_{1,t_{1}}]. \end{aligned}$$

Recall that for any time $t_1$, ${\widehat{S}}^1(q, t_{1},c_{1,t_{1}})\le {\widehat{U}}^1(q,t_1,c_{1,t_{1}})$. Therefore, we can obtain

$$\begin{aligned}&{\mathbb {E}}[{\widehat{U}}^1(q,t_1+dt,c_{1,t_{1}+dt})\mid c_{1,t_{1}}]-{\widehat{S}}^1(q, t_{1},c_{1,t_{1}})\\&\ge {\mathbb {E}}[{\widehat{S}}^1(q,t_1+dt,c_{1,t_{1}+dt})\mid c_{1,t_{1}}]-{\widehat{S}}^1(q, t_{1},c_{1,t_{1}}) \\&={\mathbb {E}}[d{\widehat{S}}^1(q,t,c_{1,t_{1}})]\\&=(h_{1}q-\mu c_{1,t}q)dt \\&>0, \end{aligned}$$

which by (B12) implies that when $c_{1,t_1}<\frac{h_{1}}{\mu }$, it is optimal to wait which in turn implies that ${\tilde{C}}_{1,t}\ge \frac{h_{1}}{\mu }$.

(ii) In the case with $\mu < 0$, the assembler should order component 1 after time $T-L_1$, i.e., $T-L_1\le t_{1}^{*}\le T$. The original problem is thus reduced to the one stated in (17). In this case, we have

$$\begin{aligned} {\mathbb {E}}[d{\widehat{S}}^2(q,t,c_{1,t_{1}})]=(-\,sd-\mu c_{1,t_1}q)dt. \end{aligned}$$

Define ${\widehat{U}}^2(q, t_1,c_{1,t_{1}})$ as the maximal expected total profit at time $t_1$ with the current unit price $c_{1,t_{1}}$ for Problem (17). To prove the result, it suffices to prove that when $c_{1,t_1}>\frac{-sd}{\mu q}$, it is optimal to wait, which means that for any $c_{1,t_1}>\frac{-sd}{\mu q}$,

$$\begin{aligned} {\widehat{S}}^2(q, t_{1},c_{1,t_{1}})<{\mathbb {E}}[{\widehat{U}}^2(q,t_1+dt,c_{1,t_{1}+dt})\mid c_{1,t_{1}}]. \end{aligned}$$

Recall that for any time $t_1$, ${\widehat{S}}^2(q, t_{1},c_{1,t_{1}})\le {\widehat{U}}^2(q,t_1,c_{1,t_{1}})$. Therefore, we can obtain

$$\begin{aligned}&{\mathbb {E}}[{\widehat{U}}^2(q,t_1+dt,c_{1,t_{1}+dt})\mid c_{1,t_{1}}]-{\widehat{S}}^2(q, t_{1},c_{1,t_{1}})\\&\ge {\mathbb {E}}[{\widehat{S}}^2(q,t_1+dt,c_{1,t_{1}+dt})\mid c_{1,t_{1}}]-{\widehat{S}}^2(q, t_{1},c_{1,t_{1}}) \\&={\mathbb {E}}[d{\widehat{S}}^2(q,t,c_{1,t_{1}})]\\&=(-\,sd-\mu c_{1,t_1}q)dt \\&>0, \end{aligned}$$

which by (B12) implies that when $c_{1,t_1}>\frac{-sd}{\mu q}$, it is optimal to wait at time $t_1$ which in turn implies that ${\widehat{C}}_{1,t}\le \frac{-sd}{\mu q}$. $\square$

Proof of Proposition 2

Recall that given the procurement quantity q, $t_1^{*}$ can be determined by solving the following optimal stopping problem (see (14)):

$$\begin{aligned} {\widehat{U}}^{*}(q, c_{1,0}):= & {} \sup \limits _{0 \le t_1 \le T}{\mathbb {E}}[{\widehat{S}}(q,t_1,c_{1,t_1})|c_{1,0}] \nonumber \\= & {} {\mathbb {E}}[{\widehat{S}}(q,t_1^{*},c_{1,t_1^{*}})|c_{1,0}]. \end{aligned}$$

Moreover, Lemma 4 implies (i) when $\mu >0$, $0 \le t_1^{*} \le T-L_1$ and $T_d=0$; (ii) when $\mu <0$, $T-L_1 \le t_1^{*} \le T$ and $T_d=t_1^*+L_1-T$; (iii) when $\mu =0$, $t_1^{*}=T-L_1$.

By Theorem 3.1 of Chen and Liu (2014), we have

$$\begin{aligned} \frac{\partial {\widehat{U}}^{*}(q,c_{1,0})}{\partial q}= & {} E[\frac{\partial {\widehat{S}}}{\partial q}(q,t_1,c_{1,t_1})|c_{1,0},t_1=t_1^{*}] \nonumber \\= & {} p(1-F(q))-{\mathbb {E}}[c_{1,t_1}+c_2^N+h_1 (T+T_d-t_1-L_1)|c_{1,0},t_1=t_1^{*}] \nonumber \\= & {} {\left\{ \begin{array}{ll} p(1-F(q)) - {\mathbb {E}}(c_{1,t_1}+c_2^N+h_1(T-t_1-L_1)|c_{1,0},t_1=t_1^{*}), \text {if} \ \mu >0; \\ p(1-F(q))-{\mathbb {E}}[c_{1,t_1}+c_2^N|c_{1,0},t_1=t_1^{*}], \text {if}\ \mu <0; \\ p(1-F(q))-e^{\mu (T-L_1)}c_{1,0}-c_2^N, \text {if}\ \mu =0. \end{array}\right. } \end{aligned}$$

Hence the optimal $q^{*}$ can be characterized by the following equations: (a) when $\mu >0$,

$$\begin{aligned} q^{*}=F^{-1}\left( \frac{p- E[c_{1,t_1^{*}}+c_2^N+h_1(T-t_1^{*}-L_1)]}{p}\right) ; \end{aligned}$$

(b) when $\mu <0$,

$$\begin{aligned} q^{*}=F^{-1}\left( \frac{p-{\mathbb {E}}[c_{1,t_1^{*}}+c_2^N]}{p}\right) ; \end{aligned}$$

(c) when $\mu =0$,

$$\begin{aligned} q^{*}=F^{-1}\left( \frac{p-e^{\mu (T-L_1)}c_{1,0}-c_2^N}{p}\right) . \end{aligned}$$

To obtain the bounds for $q^{*}$, we first observe the following property based on the dynamics of $c_{1,t}$:

$$\begin{aligned} {\mathbb {E}}[e^{-\mu t_1^{*}}c_{1,t_1^{*}}\mid c_{1,0}]=c_{1,0}. \end{aligned}$$

Then, for $\mu >0$, because $0 \le t_1^{*} \le T-L_1$, we know

$$\begin{aligned} {\mathbb {E}}[e^{-\mu (T-L_1)}c_{1,t_1^{*}}\mid c_{1,0}] \le c_{1,0} \le {\mathbb {E}}[c_{1,t_1^{*}}\mid c_{1,0}], \end{aligned}$$

and hence

$$\begin{aligned} c_{1,0} \le {\mathbb {E}}[c_{1,t_1^{*}}\mid c_{1,0}] \le e^{\mu (T-L_1)} c_{1,0}, \end{aligned}$$

which implies the bounds for $q^{*}$. The bounds for $q^{*}$ when $\mu <0$ can be obtained in a similar way.

Proof of Lemma 6

For notation convenience, we take two components for example, i.e., $\Omega ^L=\{2\}$.

(i) We first prove $T-L_1\le t_1^*\le T$. Given q and $t_i^*$ where $i\in \Omega ^L$, it is easy to prove that $T_d$ and $T_{h,i}$, $i\in \Omega ^L$, is independent of $t_1$ for any $t_1\le T-L_1$, and that the holding costs of component 1 is strictly decreasing in $t_1\le T-L_1$. Note that the price of component 1 has a decreasing trend. Therefore, the expected total costs are decreasing in $t_1\le T-L_1$. That is, the assembler will be better off procuring component 1 at time $T-L_1$ than before time $T-L_1$ which in turn implies $T-L_1\le t_1^*\le T$.

(ii) Now we prove by contradiction that $t_2^*+L_2-L_1\le t_1^*\le T$. If $t_1 < t_2+L_2-L_1$, then $T_d=t_2+L_2-T$, $T_{h1}=t_2+L_2-t_1-L_1$, $T_{h2}=0$, and hence

$$\begin{aligned} {\mathbb {E}}[dS]=(qh_1-\mu q c_{1,t_1})dt. \end{aligned}$$

It is obvious that ${\mathbb {E}}[dS]>0$ which implies that the assembler will never make procurement before $t_2+L_2-L_1$, which verifies (ii).

(iii) Applying Theorem 3.1 of Chen and Liu (2014) to the Problem (11) with $\mu <0$ and $\Omega ^L=\{2\}$, we can get:

$$\begin{aligned} \frac{\partial U^{*}(q,c_{1,0})}{\partial q}=p (1-F(q)) -{\mathbb {E}}[h_1T_{h,1}+h_2T_{h,2}+c_{1,t_1^{*}}+c_2|c_{1,0}], \end{aligned}$$

which implies

$$\begin{aligned} q^{*}=F^{-1}\left( \frac{p-{\mathbb {E}}[h_1T_{h,1}+h_2T_{h,2}+c_{1,t_1^{*}}+c_2|c_{1,0}]}{p}\right) . \end{aligned}$$

Because $h_1T_{h,1} \ge 0$, $h_2T_{h,2} \ge 0$, and ${\mathbb {E}}[c_{1,t_1^{*}}|c_{1,0}]={\mathbb {E}}[c_{1,0}e^{\mu t_1^{*}}|c_{1,0}] \ge e^{\mu T}c_{1,0}$, the bounds for $q^{*}$ are,

$$\begin{aligned} 0 \le q^{*} \le F^{-1}\left( \frac{p-e^{\mu T}c_{1,0}-c_2}{p}\right) . \end{aligned}$$

$\square$