Stochastic inventory system with lead time flexibility: offered by a manufacturer/transporter

Abstract

This paper studies lead time flexibility in a two-stage continuous review supply chain in which the retailer uses the (R, Q) inventory system: when his inventory position reaches R, the retailer places orders with size Q to the manufacturer, who uses a transportation provider to deliver them with different lead time options. According to the contract, the manufacturer is able to expedite or postpone the delivery if the retailer makes such a request. Hence, the retailer has the flexibility to modify the lead time by using the most up-to-date demand information. The optimal lead time policy is found to be a threshold-type policy. The sensitivity analysis also shows that R is much more sensitive to the change of lead time than Q, and thus, the paper is primarily focused on finding optimal R. We also provide a cost approximation which yields unimodal cost in R. Furthermore, we analyze the order crossing problem and derive an upper bound for the probability of order crossing. Finally, we conduct an extensive sensitivity analysis to illustrate the effects of lead time flexibility on supply chain performance and discuss the managerial insights.

Appendix: Proofs of lemmas

Proof of Lemma 1

For (i), see Zheng (1992) and Zipkin (1986a). (ii) From (1) and ${\mathbbm {E}}[(D(T)-y)^+]=\int _0^{\infty } P (D(T)-y \ge x)\mathrm{d}x=\int _y^{\infty } P (D(T) \ge x)\mathrm{d}x$,

$$\begin{aligned} \nabla _1 G(y,T)=-(h+b)P (D(T) \ge y)+h. \end{aligned}$$

Since D(T) is stochastically increasing in T, $\nabla _1 G(y,T)$ nonincreasing in T: $\nabla _{12} G(y,T) \le 0$. Similarly,

$$\begin{aligned} \nabla _{13} B(R,Q,T) &= \frac{1}{Q} \frac{\partial }{\partial T} \frac{\partial }{\partial R} \int _R^{R+Q} {\mathbbm {E}}[(D(T)-y)^+] \mathrm{d}y \\ &= \frac{1}{Q} \frac{\partial }{\partial T} \Big ( {\mathbbm {E}}[(D(T)-R-Q)^+] - {\mathbbm {E}}[(D(T)-R)^+] \Big ) \\ &= - \frac{1}{Q} \frac{\partial }{\partial T} \int _R^{R+Q} P (D(T) \ge x) \mathrm{d}x \, \le \, 0.\\ \nabla _{23} B(R,Q,T) &= \frac{\partial }{\partial T} \frac{\partial }{\partial Q} \frac{\int _R^{R+Q} {\mathbbm {E}}[(D(T)-y)^+] \mathrm{d}y}{Q}\\ &= \frac{1}{Q^2} \frac{\partial }{\partial T} \Big ( Q{\mathbbm {E}}[(D(T)-R-Q)^+] - \int _R^{R+Q} {\mathbbm {E}}[(D(T)-y)^+] \mathrm{d}y \Big ) \\ &= \frac{1}{Q^2} \int _R^{R+Q} \frac{\partial }{\partial T} \Big ( {\mathbbm {E}}[(D(T)-R-Q)^+] - {\mathbbm {E}}[(D(T)-y)^+] \Big ) \mathrm{d}y \, \le \, 0. \end{aligned}$$

We complete the proof for part (ii). $\quad\square$

(iii) We show that B(R, Q, T) is convex in T by first establishing the convexity of $E[D(T)-y]^+$. Note that

$$\begin{aligned} \frac{\partial E[(D(T+\varepsilon )-y)^{+}]}{\partial T}=\lim _{\delta \rightarrow 0} \left( \frac{E[(D(T+\varepsilon +\delta )-y)^+] - E[(D(T+\varepsilon )-y)^+]}{\delta } \right) . \end{aligned}$$

For arbitrarily given y and $T \ge 0$, locally define

$$\begin{aligned} h(x, \delta ):=[d(T)+d(\delta )+x-y]^+ - [d(T)+x-y]^+ \;\;\; \text{ for } x, \delta >0 \end{aligned}$$

where d(T) and $d(\delta )$ are observations from D(T) and $D(\delta )$. Note that $h(x,\delta )=0$ for $x \le -d(T)-d(\delta )+y$, it increases with a slope of one between $-d(T)-d(\delta )+y$ and $-d(T)+y$, and $h(x,\delta )=d(\delta )$ for $x \ge -d(T)+y$. Consequently, $h(x,\delta )$ is a nondecreasing function which implies that $H(x,\delta )$ below is nondecreasing as well.

$$\begin{aligned} H(x, \delta ):=E([D(T)+D(\delta )+x-y]^+ - [D(T)+x-y]^+) \end{aligned}$$

By using the fact that $D(\varepsilon )$ is stochastically greater than $D(\varepsilon )=0$, we obtain $E H(x=D(\varepsilon ),\delta ) \ge E H(x=0,\delta )$ for every $\epsilon , \delta >0$. Dividing this inequality by $\delta$ and taking limits, we arrive at

$$\begin{aligned} \frac{\partial E[(D(T+\varepsilon )-y)^{+}]}{\partial T}=\lim _{\delta \rightarrow 0} \left( \frac{E H(x=D(\varepsilon ),\delta )}{\delta } \right) \ge \lim _{\delta \rightarrow 0} \left( \frac{E H(x=0,\delta )}{\delta } \right) = \frac{\partial E[(D(T)-y)^{+}]}{\partial T} \end{aligned}$$

Finally, the last inequality is the definition of the convexity of $E(D(T)-y)^+$ in T. This implies the convexity of B(R, Q, T) by (3). $\quad\square$

Proof of Theorem 1

Let $I=R-d(\tau )$ and define $\varDelta (I)$ representing the cost difference per unit time when the lead time $T+\beta$ is used instead of T. It follows from (8) that

$$\begin{aligned} \varDelta (I)=C(I,Q,T-\tau +\beta )-C(I,Q,T-\tau ) \end{aligned}$$

which can be rewritten by using (2)

$$\begin{aligned} \varDelta (I) &= \int _{I}^{I+Q}(G(y,T-\tau +\beta )-G(y,T-\tau ))\mathrm{d}y /Q \\ &= -h\mu \beta +(h+b)\int _{I}^{I+Q}\{E[(D(T-\tau +\beta )-y)^{+}]-E[(D(T-\tau )-y)^{+}]\}\mathrm{d}y/Q . \end{aligned}$$

Then, the first derivative is

$$\begin{aligned} \nabla \varDelta (I)=(h+b)\int _{I}^{I+Q}[P(D(T-\tau +\beta ) \le x)- P(D(T-\tau )\le x)]\mathrm{d}x/Q \end{aligned}$$

(20)

Since the demand during the lead time is stochastically increasing with lead time, the integrand is negative so $\nabla \varDelta (I) \le 0$. Note that $\varDelta (I=-Q)= b\mu \beta > 0$. When inventory level $I \rightarrow \infty$, $\varDelta (I)=-h\mu \beta < 0$. By monotonicity and continuity of $\varDelta (I)$, there is an $I'$ such that $\varDelta (I')=0$. Let $\bar{I}=\min \{I',R\}$. By construction for $R-d(\tau ) \ge \bar{I}$, $C(R-d(\tau ),Q,T-\tau ) \ge C(R-d(\tau ),Q,T-\tau +\beta )$ and the retailer exercises the lead time option by using $T+\beta$ as opposed to T. Similarly, we can argue that there exists $\underline{I}$, such that when $R-d(\tau ) \le \underline{I}$ the retailer should use $T-\alpha$ instead of T.

We now show that $\underline{I} \le \bar{I}$. Suppose to the contrary that $\underline{I} > \bar{I}$. For $\bar{I}< I <\underline{I}$,

$$\begin{aligned} C(I,Q,T-\tau ) \ge C(I,Q,T-\tau +\beta ) \ \text{ and } \ C(I,Q,T-\tau ) \ge C(I,Q,T-\tau -\alpha ) \end{aligned}$$

These inequalities contradict the convexity of C(I, Q, T) in T. $\quad\square$

Proof of Lemma 2

Argument for $\bar{I}$ is presented. Let $\Theta (y):=E[(D(T-\tau +\beta )-y)^+]-E[(D(T-\tau )-y)^+]$. Set $\nabla \varDelta (I)=0$ and rearrange to obtain

$$\begin{aligned} \int _{\bar{I}}^{\bar{I}+Q}\Theta (y)\mathrm{d}y=\frac{h\mu \beta Q}{h+b} \end{aligned}$$

(21)

Taking the implicit derivative of (21) with respect to Q

$$\begin{aligned} \left( \frac{\mathrm{d} \bar{I}}{\mathrm{d} Q}+1 \right) \Theta (\bar{I}+Q)-\frac{\mathrm{d} \bar{I}}{\mathrm{d} Q}\Theta (\bar{I})=\frac{h\mu \beta }{h+b} \end{aligned}$$

Since $\Theta (y)$ is decreasing in y,

$$\begin{aligned} \Theta (\bar{I}+Q) \le h\mu \beta /(h+b) \le \Theta (\bar{I}), \end{aligned}$$

which in combination with the implicit derivative yields

$$\begin{aligned} -1 \le \frac{\mathrm{d} \bar{I}}{\mathrm{d} Q}=\frac{h\mu \beta /(h+b)-\Theta (\bar{I}+Q)}{\Theta (\bar{I}+Q)-\Theta (\bar{I})}\le 0. \end{aligned}$$

$\quad\square$

Proof of Lemma 3

(i) and (ii) follow, respectively, from $\nabla _{13} C(R,Q,T) \le 0$ and $\nabla _{23} C(R,Q,T) \le 0$, and the convexity of C(R, Q, T) in R and Q, see Lemma 1.

(iii) We provide a constructive proof. Let R(T) to denote the optimal reorder point corresponding to lead time T. Consider three (R, Q) inventory systems, parameterized for an arbitrary $\tau \ge 0$ as follows.

System 1: Lead time $T+\tau$ and reorder point $R(T+\tau )$. System 2: Lead time T and reorder point $R(T+\tau )$, but release orders $\tau$ time after IP reaches $R(T+\tau )$. System 3: Lead time T, use reorder point R(T).

Clearly, the costs of system 1 and 2 are the same: $C(R(T+\tau ),Q,T+\tau )$. In system 2, we delay an order by $\tau$ time and release at $IP'=R(T+\tau )-D(\tau )$. $C(IP',Q,T) \ge C(R(T),Q,T)$, because R(T) is optimal with respect to T. Then, by using (4), $C(R(T+\tau ),Q,T+\tau )= E_{D(\tau )}C(R(T+\tau )-D(\tau ),Q,T) \ge C(R(T),Q,T)$ which implies $\min _R C(R,Q,T+\tau ) \ge \min _R C(R,Q,T)$.

(iv) First write the optimality equation for R and Q as

$$\begin{aligned} G(R,T)=G(R+Q,T) \;\; \text{ and } \;\; K \mu + \int _R^{R+Q} G(y,T)\mathrm{d}y = Q G(R+Q,T). \end{aligned}$$

By taking the implicit derivative of the optimality equations for R and Q with respect to T, we obtain two simultaneous equations. After some rearrangement of the implicit derivatives obtained from each equation, we, respectively, obtain

$$\begin{aligned} \nabla _1 G(R,T) \frac{\mathrm{d}R}{\mathrm{d}T} &= \nabla _1 G(R+Q,T) \frac{\mathrm{d}(R+Q)}{\mathrm{d}T} +\nabla _2 G(R+Q,T) - \nabla _2 G(R,T) \;\; \text{ and }\\ \frac{\mathrm{d}(R+Q)}{\mathrm{d}T} &= \frac{(1/Q)\int _R^{R+Q} \{ \nabla _2 G(y,T)-\nabla _2 G(R+Q,T) \} \mathrm{d}y}{\nabla _1 G(R+Q,T)} \end{aligned}$$

Using the last equation, we obtain

$$\begin{aligned} \frac{\mathrm{d}R}{\mathrm{d}T} = \frac{(1/Q)\int _R^{R+Q} \{ \nabla _2 G(y,T)-\nabla _2 G(R,T) \} \mathrm{d}y}{\nabla _1 G(R,T)} \end{aligned}$$

By $\nabla _{1,2} G(y,T) \le 0$ of Lemma 1, $\{ \nabla _2 G(y,T)-\nabla _2 G(R,T) \} \le 0$ and $\{ \nabla _2 G(y,T)-\nabla _2 G(R+Q,T) \} \ge 0$ for $R \le y \le R+Q$. On the other hand, G(y, T) is convex in y and $G(R,T)=G(R+Q,T)$. Then, $\nabla _1 G(R,T) \le 0 \le \nabla _1 G(R+Q,T)$. Thus, we establish that the numerator and the denominator have the same signs in $\mathrm{d}R/\mathrm{d}T$ and $\mathrm{d}(R+Q)/\mathrm{d}T$ expressions above. This gives us the desired result that both R and $R+Q$ are nondecreasing in T. $\quad\square$

Proof of Lemma 4

The first part of the claim follows from the definitions of ${\bar{C}}^1_{\tau }$, ${\bar{C}}^2_{\tau }$ and ${\bar{C}}^3_{\tau }$. For the second part, we need to argue that each of ${\bar{C}}^i_{\tau }$ for $1 \le i \le 4$ is unimodal. Consider Case 1 only. For $I \le \underline{I}$, the approximated cost is $\max \{C(I,Q,T-\tau -\alpha ), C(\underline{I},Q,T-\tau -\alpha )\}$. Since $C(I,Q,T-\tau -\alpha )$ is convex in I, the approximated cost is nonincreasing over $I \le \underline{I}$ and is exactly $C(\underline{I},Q,T-\tau -\alpha )$ at $I=\underline{I}$. For $\underline{I} \le I \le \bar{I}$, the approximated cost is convex in I. For $I \ge \bar{I}$, the approximated cost is $\max \{C(I,Q,T-\tau +\beta ), C(\bar{I},Q,T-\tau +\beta )\}$, and it is nondecreasing and takes the value $C(\bar{I},Q,T-\tau +\beta )$ at $I=\bar{I}$. We see that $C^1_{\tau }(I,Q,T)$ is constructed by patching a nonincreasing, a convex and a nondecreasing function together and it is continuous. Then, $C^1_{\tau }(I,Q,T)$ must be unimodal.

We next consider Case 4 where ${\bar{C}}^4_{\tau }(I)=C_{\tau }(I,Q,T)$. For $I \le \underline{I}$, the cost is $C(I,Q,T-\tau -\alpha )$ and is convex. For $\underline{I} \le I \le \bar{I}$, the cost is $C(I,Q,T-\tau )$ and it is nondecreasing because $I(T-\tau ) \le \underline{I}$. Similarly, for $I \ge \bar{I}$, the cost is $C(I,Q,T-\tau +\beta )$ and it is nondecreasing because $I(T-\tau +\beta ) \le \bar{I}$. The cost is first convex and then nondecreasing in I so it must be unimodal. The proof for Case 2 and Case 3 requires a combination of the arguments for Cases 1 and 4. $\quad\square$

Proof of Lemma 5

By the definition of the approximate cost

$$\begin{aligned} {\bar{C}}_F(R,Q,T;\alpha ,\beta ,\tau ) &= E_{D(\tau )} {\bar{C}}_{\tau }(R-D(\tau ),Q,T) \ge E_{D(\tau )} C_{\tau }(R-D(\tau ),Q,T) =C_F(R,Q,T;\alpha ,\beta ,\tau )\\ {\bar{C}}_F(R,Q,T;\alpha ,\beta ,\tau ) &= E_{D(\tau )} {\bar{C}}_{\tau }(R-D(\tau ),Q,T) \le E_{D(\tau )} C(R-D(\tau ),Q,T) =C(R,Q,T) \end{aligned}$$

In addition, using unimodality of ${\bar{C}}_{\tau }(I,Q,T)$ in I, $PF_2$ property for the demands and Theorem 9.1 of Porteus (2002), we establish the unimodality of ${\bar{C}}_F(R,Q,T;\alpha ,\beta ,\tau )$. $\quad\square$

Proof of Theory 2

In case of (i) and (ii), the first inequality is due to switching from deterministic to stochastic demand. The second and the third inequalities are, respectively, by Lemma 5. The argument for (iii) is the same except that $C_F({\bar{R}}^*,{\bar{Q}}^*,T; \alpha , \beta , \tau ) > \min _{R,Q} C(R,Q,T)$ is very unlikely but possible. Therefore, we have only three inequalities as opposed to four in (i) and (ii). $\quad\square$

Proof of Lemma 6

First recall from (8) that

$$\begin{aligned} C_F(R,Q,T;\alpha ,\tau _1)=E_{D(\tau _1)}\min \left\{ \begin{array}{l} C(R-D(\tau _1),Q,T-\tau _1-\alpha ), \\ C(R-D(\tau _1),Q,T-\tau _1), \\ C(R-D(\tau _1),Q,T-\tau _1+\beta ) \end{array} \right\} . \end{aligned}$$

(22)

Then, using (4),

$$\begin{aligned} C(R-D(\tau _1),Q,T-\tau _1) &= E_{D(\tau _2-\tau _1)} C(R-D(\tau _1)-D(\tau _2-\tau _1),Q,T-\tau _1-(\tau _2-\tau _1))\\\ge & {} E_{D(\tau _2-\tau _1)}\min \left\{ \begin{array}{l} C(R-D(\tau _2),Q,T-\tau _2-\alpha ), \\ C(R-D(\tau _2),Q,T-\tau _2), \\ C(R-D(\tau _2),Q,T-\tau _2+\beta ) \end{array} \right\} \end{aligned}$$

Repeating the last argument for every term in the minimization in (22), we obtain

$$\begin{aligned} \min \left\{ \begin{array}{l} C(R-D(\tau _1),Q,T-\tau _1-\alpha ), \\ C(R-D(\tau _1),Q,T-\tau _1), \\ C(R-D(\tau _1),Q,T-\tau _1+\beta ) \end{array} \right\} \ge E_{D(\tau _2-\tau _1)}\min \left\{ \begin{array}{l} C(R-D(\tau _2),Q,T-\tau _2-\alpha ), \\ C(R-D(\tau _2),Q,T-\tau _2), \\ C(R-D(\tau _2),Q,T-\tau _2+\beta ) \end{array} \right\} \end{aligned}$$

Taking the expected value over $D(\tau _1)$ of the both sides establishes the result. $\quad\square$

Proof of Lemma 7

(i) In order for the nth order to cross the $n+1$st, $t_n \le t_{n+1} \le t_n+\alpha +\beta$. We show that the probability in the statement of the lemma increases in $t_{n+1}$.

For $t_{n+1} \ge t_n+\tau$, define $r_1=R-{\bar{I}}$ and $r_2=R-\underline{I}$, we have:

$$\begin{aligned} P (D(t_n,t_n+\tau )\le & {} r_1, D(t_{n+1},t_{n+1}+\tau ) \ge r_2, D(t_n,t_{n+1}) \ge Q) \\ &= P (D(t_n,t_n+\tau ) \le r_1, D(t_n,t_{n+1}) \ge Q) \;\; P (D(t_{n+1},t_{n+1}+\tau ) \ge r_2) \\ &= P (D(0,\tau ) \le r_1, D(0,t_{n+1}-t_n) \ge Q) \;\; P (D(\tau ) \ge r_2). \end{aligned}$$

It is clear from the last equality that the probability increases in $t_{n+1}$.

Consider $t_{n+1} < t_n+\tau$. For a small and fixed $\epsilon >0$, define the following independent random variables for convenience:

$$\begin{aligned} X_1: &= D(t_n,t_{n+1}); \; X_2:=D(t_{n+1},t_{n+1}+\epsilon ); \; X_3:=D(t_{n+1}+\epsilon ,t_n+\tau );\\ X_4: &= D(t_n+\tau ,t_{n+1}+\tau ); \; X_5:=D(t_{n+1}+\tau ,t_{n+1}+\tau +\epsilon ). \end{aligned}$$

It suffices to argue that

$$\begin{aligned} P (X_1+X_2+X_3 \le r_1, X_2+X_3+X_4 \ge r_2, X_1 \ge Q) \le P (X_1+X_2+X_3 \le r_1, X_3+X_4+X_5 \ge r_2, X_1 \ge Q). \end{aligned}$$

Now condition on $X_1=x_1$, $X_3=x_3$ and $X_4=x_4$.

$$\begin{aligned} P (X_2 \le r_1-x_1-x_3, X_2 \ge r_2-x_3-x_4) &= P (X_2 \le r_1-x_1-x_3)-P (X_2 \le r_2-x_3-x_4) \\\le & {} P (X_2 \le r_1-x_1-x_3)-P (X_2 \le r_1-x_1-x_3) \times \\&\quad P (X_5 \le r_2-x_3-x_4) \\ &= P (X_2 \le r_1-x_1-x_3) \; P (X_5 \ge r_2-x_3-x_4) \\ &= P (X_2 \le r_1-x_1-x_3, X_5 \ge r_2-x_3-x_4), \end{aligned}$$

where we use the fact that $X_2$ and $X_5$ are i.i.d. and $P (X_2 \le r_1-x_1-x_3) \le 1$. Finally taking the expected values over $X_1$, $X_3$ and $X_4$ yields the desired result.

(ii) Using part (i) with $r_1=R-{\bar{I}}$ and $r_2=R-\underline{I}$,

$$\begin{aligned} P_c \le P (D(t_n,t_n+\tau ) \le R-{\bar{I}}, D(t_n+\alpha +\beta ,t_n+\alpha +\beta +\tau ) \ge R-\underline{I}, D(t_n,t_n+\alpha +\beta ) \ge Q) \end{aligned}$$

Note that the demand is stationary and then consider two cases.

Case 1: $0 \le \tau \le \alpha +\beta$.

$$\begin{aligned} P (D(0,\tau )\le & {} R-\bar{I}, D(\alpha +\beta ,\alpha +\beta +\tau ) \ge R-\underline{I}, D(0,\alpha +\beta ) \ge Q) \\ &= P (D(0,\tau ) \le R-\bar{I}, D(0,\alpha +\beta ) \ge Q) \; P(D(\tau ) \ge R-\underline{I}) \\ &= P (D(0,\tau ) \le R-\bar{I}, D(0,\tau )+D(\tau ,\alpha +\beta ) \ge Q) \;\; P (D(\tau ) \ge R-\underline{I}) \\\le & {} P (D(0,\tau ) \le R-\bar{I}, D(\tau ,\alpha +\beta ) \ge Q-R+\bar{I}) \;\; P (D(\tau ) \ge R-\underline{I}) \\ &= P (D(\tau ) \le R-\bar{I}) \;\; P (D(\alpha +\beta -\tau ) \ge Q-R+\bar{I}) \;\; P (D(\tau ) \ge R-\underline{I}) \end{aligned}$$

where the inequality is due to the implication between two events below

$$\begin{aligned}{}[D(0,\tau ) \le R-\bar{I}, D(0,\tau )+D(\tau ,\alpha +\beta ) \ge Q ] \;\; \Rightarrow \;\; [ D(0,\tau ) \le R-\bar{I}, D(\tau ,\alpha +\beta ) \ge Q-R+\bar{I} ]. \end{aligned}$$

Case 2: $0 \le \alpha +\beta \le \tau$.

$$\begin{aligned} P (D(0,\tau )\le & {} R-\bar{I}, D(0,\alpha +\beta ) \ge Q, D(\alpha +\beta ,\alpha +\beta +\tau ) \ge R-\underline{I}) \\\le & {} P (D(0,\tau ) \le R-\bar{I}, D(0,\alpha +\beta ) \ge Q, D(\tau ,2\tau ) \ge R-\underline{I}) \\\le & {} P (D(0,\tau ) \le R-\bar{I}, D(0,\tau ) \ge Q, D(\tau ,2\tau ) \ge R-\underline{I}) \\\le & {} P (Q \le D(0,\tau ) \le R-\bar{I}) \;\; P (D(\tau ,2\tau ) \ge R-\underline{I}) \end{aligned}$$

The first inequality follows from increasing $\alpha +\beta$ up to $\tau$ so that the dependence between the events $[D(0,\tau ) \le R-\bar{I}]$ and $[D(\alpha +\beta ,\alpha +\beta +\tau ) \ge R-\underline{I}]$ is eliminated. This inequality can be argued formally as in part (i) of the lemma. The second inequality follows from the case condition $\alpha +\beta \le \tau$. Combining two cases, we prove (ii). $\quad\square$