Inventory and pricing decisions for a dual-channel supply chain with deteriorating products

Abstract

Dual-channel supply chain structure, i.e., a traditional retail channel added by an online direct channel, is widely adopted by a lot of firms, including some companies selling deteriorating products (e.g. fruits, vegetables and meats, etc.). However, few papers in literature consider deterioration property of products in dual-channel business models. In this paper, a single-retailer-single-vendor dual-channel supply chain model is studied, in which the vendor sells deteriorating products through its direct online channel and the indirect retail channel. In addition to quantity deterioration, quality of the products also drops with time and affects the demand rate in the retail channel. The pricing decisions and the inventory decisions for the two firms are simultaneously studied. Models of centralized (i.e., the two firms make decisions jointly) and decentralized (i.e., the two firms make decisions separately, vendor as the Stackelberg leader) problems are established. Proper algorithms are proposed to obtain the optimal decisions of prices, ordering frequencies and ordering quantities. The results suggest that decentralization of the supply chain not only erodes the two firms’ profit, but also incurs higher wastes comparing to that under centralization. However, a revenue sharing and two part tariff contract can coordinate the supply chain. Under utilizing the contract, each firm’s profit is improved and the total waste rate of the supply chain is reduced. It is also shown that the contract is more efficient for both firms under higher product deterioration rate. Besides, the contract is more efficient for the retailer, while less efficient for the vendor under higher quality dropping rate. In the model extension, online channel delivery time is assumed to be endogenous and linked to demands in both channels. The results show that products’ deterioration rate and quality dropping rate have significant impacts to the firms’ delivery time decisions, as well as the pricing and inventory decisions.

Appendices

Appendix 1: Proof of Lemma 1

Proof

The inventory level in the nth phase in time interval $t \in [(n-1)T, nT]$ satisfies the differential equation $\dot{I}_{vn}(t)=-\theta {I}_{v n}(t)-D_{v},t \in [(n-1)T, nT]$ with boundary condition $I_{vn}(t=nT)=0$.

Solving the equation, the inventory level in the nth phase is

$$\begin{aligned} I_{vn }(t)=\frac{D_{v}}{\theta }(e^{\theta (nT-t)}-1), \quad t \in [(n-1)T,nT]. \end{aligned}$$

(44)

Then, solving the differential equation in the $(n-1)$th phase, i.e., $\dot{I}_{v(n-1)}(t)=-\theta {I}_{v (n-1)}(t)-D_{v},t \in [(n-2)T, (n-1)T]$ with boundary condition $I_{v (n-1)}[t=(n-1)T]-I_{v n}[t=(n-1)T]=Q_{rn}$, the inventory level in time interval $t \in [(n-2)T,(n-1)T]$ can be solved as

$$\begin{aligned} I_{v(n-1)}(t)=\frac{D_{v}}{\theta }(e^{\theta (nT-t)}-1)+Q_{rn}e^{\theta ((n-1)T-t)}, \quad t \in [(n-2)T,(n-1)T]. \end{aligned}$$

(45)

Following the same approach, the inventory level in time interval $t \in [(n-3)T,(n-2)T]$ is

$$\begin{aligned} I_{v(n-2)}(t)=\frac{D_{v}}{\theta }(e^{\theta (nT-t)}-1)+Q_{rn}e^{\theta (n-1)T-t}+Q_{r(n-1)}e^{\theta ((n-2)T-t)}, \quad t \in [(n-3)T,(n-2)T]. \end{aligned}$$

(46)

Finally, the inventory level in time interval $t \in [0,T]$ is

$$\begin{aligned} I_{v1}(t)=\frac{D_{v}}{\theta }(e^{\theta (nT-t)}-1)+Q_{rn}e^{\theta ((n-1)T-t)}+Q_{r(n-1)}e^{\theta ((n-2)T-t)}+ \cdots +Q_{r2}e^{\theta (T-t)}, \quad t \in [0,T]. \end{aligned}$$

(47)

Based on the above analysis, the inventory level can be inducted as

$$\begin{aligned} I_{v j}(t)=\frac{D_{v}}{\theta }(e^{\theta (nT-t)}-1)+\frac{d_{r}}{\theta -\mu }(e^{(\theta -\mu ) nT}-e^{(\theta -\mu ) jT}) e^{-\theta t}, t \in [(j-1)T, jT],\quad j=1,2,\ldots ,n. \end{aligned}$$

(48)

This ends the proof of Lemma 1. $\square$

Appendix 2: Proof of Proposition 1

Proof

Before the proof, some definitions are made:

$$\begin{aligned} X_{1}&= \frac{1-e^{-\mu nT}}{\mu nT},\quad X_{2}=\frac{e^{\theta nT}-\theta nT-1}{\theta ^{2}nT},\quad X_{3}=\frac{1}{(\theta -\mu )\theta nT}\left( e^{(\theta -\mu )nT} (1-e^{-\theta nT})+(e^{\theta T}-1)\frac{1-e^{-\mu nT}}{1-e^{\mu T}}\right) ,\\ X_{4}&= \frac{1}{(\theta -\mu )nT}\left( \frac{e^{\theta T}-1}{\theta }-\frac{e^{\mu T}-1}{\mu }\right) \frac{1-e^{-\mu nT}}{e^{\mu T}-1},\\ X_{5}&= \frac{1}{\theta nT}(e^{\theta nT}-1), ~~~~ X_{6}=\frac{1}{(\theta -\mu )nT}(e^{(\theta -\mu )nT}-1),\\ X_{7}&= \frac{1}{(\theta -\mu )nT}(e^{(\theta -\mu )T}-1) \frac{1-e^{-\mu nT}}{1-e^{-\mu T}}. \end{aligned}$$

The profit function can be rewritten as

$$\begin{aligned} TP_{sc}=p_{v}D_{v}+p_{r}d_{r}X_{1} -h_{v}D_{v}X_{2} -h_{v}d_{r}X_{3} -h_{r}d_{r}X_{4} -c_{v}D_{v} X_{5}-c_{v}d_{r}X_{6} -\frac{A_{r}}{T}-\frac{A_{v}}{nT}. \end{aligned}$$

(49)

(1) When n and T are fixed, taking the second order derivative of $TP_{sc}$ with respect to $p_{v}$ and $p_{r}$, the Hessian matrix can be obtained as

$$\begin{aligned} H = \left( \begin{array}{cc} \frac{\partial ^{2} TP_{sc}}{\partial p_{v}^{2}} & \frac{\partial ^{2}TP_{sc}}{\partial p_{v} \partial p_{r}} \\ \frac{\partial ^{2} TP_{sc}}{\partial p_{r}\partial p_{v} } & \frac{\partial ^{2}TP_{sc}}{\partial p_{r}^{2}} \\ \end{array} \right) = \left( \begin{array}{ccc} -2b &(1+X_{1})r\\ (1+X_{1})r&-2b\\ \end{array} \right). \end{aligned}$$

(50)

When the quality losing rate $\mu$ is not very large and $b>r$, $|H|=4b^{2}-(1+X_{1})^2r^{2}>0$ is satisfied. Also $\frac{\partial ^{2}TP_{sc}}{\partial p_{v}^{2}}=-2b<0$, so $TP_{sc}$ is jointly concave in $p_{v}$ and $p_{r}$. When equating both $\frac{\partial TP_{sc}}{\partial p_{v} }$ and $\frac{\partial TP_{sc}}{\partial p_{r} }$ to zero, the optimal prices can be obtained by solving the equation set, which can be expressed as

$$\begin{aligned} P_{v}^{c*}&= \frac{B_{2}r+2bB_{1}X_{1}+rX_{1}B_{2}-arX_{1}-arX_{1}^{2}+\alpha arX_{1}^{2}-2\alpha a bX_{1}+\alpha arX_{1}}{r^{2}X_{1}^{2}-4b^{2}X_{1}+2r^{2}X_{1}+r^{2}}, \end{aligned}$$

(51)

$$\begin{aligned} P_{r}^{c*}&= \frac{2bB_{2}+rB_{1}+rB_{1}X_{1}-2abX_{1}-\alpha ar+2\alpha abX_{1}-\alpha ar X_{1}}{r^{2}X_{1}^{2}-4b^{2}X_{1}+2r^{2}X_{1}+r^{2}}, \end{aligned}$$

(52)

in which $B_{1}=-h_{v}bX_{2}+h_{v}rX_{3}+h_{r}rX_{4}-c_{v}bX_{5}+c_{v}rX_{6}$, $B_{2}=h_{v}rX_{2}-h_{v}bX_{3}-h_{r}bX_{4}+c_{v}rX_{5}-c_{v}bX_{6}$.

(2) For fixed $p_{v}$, $p_{r}$ and n, there is only one decision parameter T.

Taking the second derivative of the profit function with respect to T, we have

$$\begin{aligned} \frac{\partial ^{2}TP_{sc}}{\partial T^{2}}=p_{r}d_{r}X_{1}^{''}-h_{v}D_{v}X_{2}^{''}-h_{v}d_{r}X_{3}^{''}-h_{r}d_{r}X_{4}^{''} -c_{v}D_{v}X_{5}^{''}-c_{v}d_{r}X_{6}^{''}-\frac{2A_{v}}{nT^{3}}-\frac{2A_{r}}{T^{3}}. \end{aligned}$$

(53)

For $X_{1}^{''}=\frac{e^{-\mu nT} }{T}+ \frac{e^{-\mu nT}-1}{\mu nT^{2}}$. When setting $x=-\mu nT$, $X_{1}^{''}=\frac{xe^{-x}+e^{-x}-1 }{\mu nT^{2}}$. Defining a new function $F(x)=xe^{-x}+e^{-x}-1$. When $x\rightarrow 0$, $X_{1}^{''}=0$. The first derivative of F(x) satisfies $F^{'}(x) =-xe^{-x}<0$. Thus in the interval $T\in (0,+\infty )$, $X_{1}^{''}<0$ holds.

For $X_{2}^{''}=\frac{1}{\theta ^{2}nT^{3}}(\theta ^{2}n^{2}T^{2}e^{\theta nT}-2\theta nTe^{\theta nT}+2e^{\theta nT}-2)$ , set $x=\theta nT$ and define a new function $F(x)=x^{2}e^{x}-2xe^{x}+2e^{x}-2$ in which $x\in (0,+\infty )$. When $x\rightarrow 0$, $F(x)=0$ and the first derivative satisfies $F^{'}(x)=x^{2}e^{x}>0$. Thus, in the interval $T\in (0,+\infty )$, $X_{2}^{''}>0$ always holds.

The proofs of $X_{i}^{''}>0$, ($i=3,4,5,6$) are the same as $X_{2}^{''}>0$. Finally, the second derivative of profit function satisfies $\frac{\partial ^{2}TP_{sc}}{\partial T^{2}}<0$, and $TP_{sc}$ is concave in T.

(3) Taking the second derivative of the profit function with respect to T, we have

$$\begin{aligned} \frac{\partial ^{2}TP_{sc}}{\partial n^{2}}=p_{r}d_{r}X_{1}^{''}-h_{v}D_{v}X_{2}^{''}-h_{v}d_{r}X_{3}^{''}-h_{r}d_{r}X_{4}^{''} -c_{v}D_{v}X_{5}^{''}-c_{v}d_{r}X_{6}^{''}-\frac{2A_{v}}{n^{3}T}. \end{aligned}$$

(54)

For $X_{1}^{''}=\frac{e^{-\mu nT} }{n}+ \frac{e^{-\mu nT}-1}{\mu n^{2}T}$. When setting $x=-\mu nT$, $X_{1}^{''}=\frac{xe^{-x}+e^{-x}-1 }{\mu nT^{2}}$. Defining a new function $F(x)=xe^{-x}+e^{-x}-1$. When $x\rightarrow 0$, $X_{1}^{''}=0$. The first derivative of F(x) satisfies $F^{'}(x) =-xe^{-x}<0$. Thus in the interval $T\in (0,+\infty )$, $X_{1}^{''}<0$ holds.

For $X_{2}^{''}=\frac{1}{\theta ^{2}n^{3}T}(\theta ^{2}n^{2}T^{2}e^{\theta nT}-2\theta nTe^{\theta nT}+2e^{\theta nT}-2)$ , set $x=\theta nT$ and define a new function $F(x)=x^{2}e^{x}-2xe^{x}+2e^{x}-2$, for $x\in (0,+\infty )$. When $x\rightarrow 0$, $F(x)=0$ and the first derivative satisfies $F^{'}(x)=x^{2}e^{x}>0$. Thus in the interval $n\in (0,+\infty )$,$X_{2}^{''}>0$ holds.

The proofs of $X_{i}^{''}>0$, $i=3,4,5,6$ are the same as $X_{2}^{''}>0$. Finally, the second derivative of profit function satisfies $\frac{\partial ^{2}TP_{sc}}{\partial n^{2}}<0$, and $TP_{sc}$ is concave in n.

In the above analysis, we proved the existence and uniqueness for the decision variables. To obtain the analytical results, we approximate the exponential terms and solve equations $\frac{\partial TP_{sc}}{\partial n}=0$ and $\frac{\partial TP_{sc}}{\partial n}=0$, respectively. We use the Taylor expansion to approximate exponential terms. For example, the term $e^{\theta T}-1$ is approximated to $e^{\theta T}-1\approx 1+\theta T+\frac{\theta ^{2}T^{2}}{2}-1=\theta T+\frac{\theta ^{2}T^{2}}{2}$. Thus, we obtain the final results in Proposition 1.

This ends the proof of Proposition 1. $\square$

Appendix 3: Proof of Proposition 2

Proof

The vendor’s profit function can be expressed as

$$\begin{aligned} TP_{v}=p_{v}D_{v} +wd_{r}X_{7}-h_{v}D_{v}X_{2} -h_{v}d_{r}X_{3} -c_{v}D_{v} X_{5}-c_{v}d_{r}X_{6} -\frac{A_{v}}{nT}. \end{aligned}$$

(55)

The retailer’s profit function can be expressed as

$$\begin{aligned} TP_{r}=p_{r}d_{r}X_{1} -h_{r}d_{r}X_{4} -wd_{r}X_{7}-\frac{A_{r}}{T}. \end{aligned}$$

(56)

(1) When T and n are fixed, the value of $X_{i},(i=1,2,3,4,5,6,7)$ are determined. Equating the first derivative of retailer’s profit function to zero, the optimal price can be derived as

$$\begin{aligned} p_{r}(p_{v},w)=\frac{1}{2b}\left((1-\alpha )a+rp_{v}+bh_{r}X_{4}+bX_{7}w\right). \end{aligned}$$

(57)

Substitute it into demand function, we have

$$\begin{aligned} D_{v}(p_{v},w)&= \left(\frac{r(1-\alpha )}{2b}+\alpha \right)a + \left(\frac{r^{2}}{2b}-b\right)p_{v}+\frac{1}{2}rh_{r}X_{4}(T) +\frac{1}{2}rwX_{7}(T), \end{aligned}$$

(58)

$$\begin{aligned} d_{r}(p_{v},w)&= \frac{(1-\alpha )a}{2}+\frac{1}{2}rp_{v} -\frac{1}{2}bh_{r}X_{4}-\frac{1}{2}bwX_{7}. \end{aligned}$$

(59)

Substitute the demand functions into vendor’s profit function, and take the first derivative of $TP_{v}$ with respect to $p_{v}$ and w, there is $\frac{\partial ^{2} TP_{v}}{\partial w^{2}}=-bX_{7}^{2}$, $\frac{\partial ^{2} TP_{v}}{\partial p_{v}^{2}}=\frac{r^{2}}{b}-2b$, $\frac{\partial ^{2} TP_{v}}{\partial w \partial p_{v}}=\frac{\partial ^{2} TP_{v}}{\partial p_{v}\partial w}=rX_{7}$. The Hessian matrix is

$$\begin{aligned} H=\left( \begin{array}{cc} \frac{\partial ^{2} TP_{v}}{\partial w^{2}} &\frac{\partial ^{2} TP_{v}}{\partial w \partial p_{v}}\\ \frac{\partial ^{2} TP_{v}}{\partial p_{v}\partial w} &\frac{\partial ^{2} TP_{v}}{\partial p_{v}^{2}}\\ \end{array} \right) =\left( \begin{array}{cc} -bX_{7}^{2} &rX_{7}\\ rX_{7} &\frac{r^{2}}{b}-2b\\ \end{array} \right) \end{aligned}$$

(60)

For $\frac{\partial ^{2}TP_{v}}{\partial w^{2}}=-bX_{7}^{2}<0$, $|H|=(b^{2}-r^{2})X_{7}^{2}>0$, vendor’s profit function is jointly concave in $p_{v}$ and w. And the optimal solution can be derived from the first order conditions.

$\frac{\partial TP_{v}(p_{v},p_{r}(p_{v},w),w)}{\partial p_{v}}=0, \frac{\partial TP_{v}(p_{v},p_{r}(p_{v},w),w)}{\partial w}=0.$ Then the optimal prices can be determined as

$$\begin{aligned} p_{v}^{d*}&= \frac{2bC_{2}+2rC_{1}-2\alpha ab-arX_{7}+\alpha ar+\alpha arX_{7} }{2(r^{2}X_{7}-2b^{2}+r^{2})}, \end{aligned}$$

(61)

$$\begin{aligned} w^{d*}&= \frac{2b^{2}C_{1}-r^{2}C_{1}-ab^{2}X_{7}+\alpha ab^{2}X_{7}+brX_{7}C_{2}-\alpha abrX_{7}}{X_{7}b(r^{2}X_{7}-2b^{2}+r^{2})}, \end{aligned}$$

(62)

$$\begin{aligned} p_{r}^{d*}&= \frac{1}{2b}((1-\alpha )a+rp_{v}^{d*}+bh_{r}X_{4} +bX_{7}w^{d*}), \end{aligned}$$

(63)

in which $C_{1}=\frac{h_{r}bX_{4}X_{7}}{2}+\frac{h_{v}rX_{2}X_{7}}{2}-\frac{h_{v}bX_{3}X_{7}}{2}+\frac{c_{v}rX_{5}X_{7}}{2}-\frac{c_{v}bX_{6}X_{7}}{2}$, $C_{2}=h_{v}X_{2}(\frac{r^{2}}{2b}-b)+\frac{h_{v}rX_{3}}{2}+c_{v}X_{5}(\frac{r^{2}}{2b}-b)+\frac{c_{v}X_{6}r}{2}-\frac{h_{r}rX_{4}}{2}$

(2)Taking the second derivative of retailer’s profit function with respect to T , we have

$$\begin{aligned} \frac{\partial ^{2}TP_{r}}{\partial T^{2}}=p_{r}d_{r}X_{1}^{''} -h_{r}d_{r}X_{4}^{''}-wd_{r}X_{7}^{''}-\frac{2A_{r}}{T^{3}}. \end{aligned}$$

(64)

As proved in Proposition 1, $X_{1}^{''}<0$ , $X_{4}^{''}>0$ and $X_{7}^{''}>0$ are satisfied. Thus $\frac{\partial ^{2}TP_{r}}{\partial T^{2}}<0$, and retailer’s profit function is concave in T.

(3)Taking the second derivative of vendor’s profit function with respect to n ,

$$\begin{aligned} \frac{\partial ^{2}TP_{v}}{\partial n^{2}}=wd_{r}X_{7}^{''}-h_{v}D_{v}X_{2}^{''}-h_{v}d_{r}X_{3}^{''}-c_{v}D_{v}X_{5}^{''}-c_{v}d_{r}X_{6}^{''}(n)-\frac{2A_{r}}{n^{3}T}. \end{aligned}$$

(65)

As proved in Proposition 1, $X_{i}^{''}>0$.$i=2,3,5,6$. For $X_{7}^{''}=\frac{1}{(\theta -\mu )nT}(e^{(\theta -\mu )T}-1)\frac{-\mu ^{2}T^{2}e^{-\mu nT}}{1-e^{-\mu T}}<0$. Finally, the second derivative of retailer’s profit function satisfies $\frac{\partial ^{2}TP_{v}}{\partial n^{2}}<0$ , thus $TP_{v}$ is concave in n. Although is an integer variable, it is obvious that there exists a unique value of n that maximize the profit function when $p_{v}$, $p_{r}$ and T are constant.

In the above analysis, we proved the existence and uniqueness for the decision variables. To obtain the analytical results, we approximate the exponential terms and solve equations $\frac{\partial TP_{r}}{\partial T}=0$ and $\frac{\partial TP_{v}}{\partial n}=0$, respectively. We use the Taylor expansion to approximate exponential terms. We obtain the final results in Proposition 2.

This ends the proof of Proposition 2. $\square$

Appendix 4: Proof of Proposition 3

Proof

For any given w, $\beta$ and F, the retailer’s retail price $p_{r}$ and ordering cycle T should satisfy

$$\begin{aligned}&\frac{\partial TP_{r}^{co}}{\partial p_{r}}|_{p_{r}=p_{r}^{c}}=0, \end{aligned}$$

(66)

$$\begin{aligned}&\frac{\partial TP_{r}^{co}}{\partial T}|_{T=T^{c}}=0. \end{aligned}$$

(67)

Taking the first order partial derivative of (31) with respect to $p_{r}$ and T and setting them to zero yields

$$\begin{aligned} \frac{\partial TP_{r}^{co}}{\partial p_{r}}&= (1-\beta )(1-\mu nT/2)((1-\alpha )a-2bp_{r}+rp_{v})+h_{r}bT/2+wb(1-(\theta -\mu )nT/2)=0, \end{aligned}$$

(68)

$$\begin{aligned} \frac{\partial TP_{r}^{co}}{\partial T}&= (1-\beta )p_{r}d_{r}(-\mu n/2)-h_{r}d_{r}/2-wd_{r}(-(\theta -\mu )n/2)+A_{r}/T^{2}=0. \end{aligned}$$

(69)

Substituting $p_{r}=p_{r}^{c}$ and $T=T^{c}$ into the two equations and solving $(\beta , w)$, the expressions of $\beta ^{co}$ and $w^{co}$ in Proposition 3 can be obtained. In addition, both parties’ profit should be no less than that without coordination. Thus the lump sum fee need to satisfy

$$\begin{aligned} F \in \{TP_{r}^{co}(\beta ^{co},w^{co},F) \geqslant TP_{r}^{d*}, TP_{v}^{co}(\beta ^{co},w^{co},F) \geqslant TP_{v}^{d*} \}. \end{aligned}$$

(70)

This ends the proof of Proposition 3. $\square$

Appendix 5: Proof of Proposition 4

Proof

Calculating the first and second order derivative of B(F) as follows

$$\begin{aligned} \frac{\partial B(F)}{\partial F}&= \{(1-\gamma ) \left[TP_{v}^{co}(\beta ^{co},w^{co},F=0)-TP_{v}^{d*}-F\right] -\gamma \left[TP_{r}^{co}(\beta ^{co},w^{co},F=0)-TP_{r}^{d*}+F\right]\}\times \nonumber \\&\left[TP_{v}^{co}(\beta ^{co},w^{co},F=0)-TP_{v}^{d*}-F\right]^{\gamma -1} \left[TP_{r}^{co}(\beta ^{co},w^{co},F=0)-TP_{r}^{d*}\right]^{-\gamma }. \end{aligned}$$

(71)

$$\begin{aligned} \frac{\partial ^{2} B(F)}{\partial F^{2}}&= -\gamma (1-\gamma ) \left[TP_{v}^{co}(\beta ^{co},w^{co},F=0)-TP_{v}^{d*}-F\right]^{\gamma -2} \left[TP_{r}^{co}(\beta ^{co},w^{co},F=0)-TP_{r}^{d*}+F\right]^{-\gamma -1}\times \nonumber \\&\left[TP_{v}^{co}(\beta ^{co},w^{co},F=0)-TP_{v}^{d*}+TP_{r}^{co} (\beta ^{co},w^{co},F=0)-TP_{r}^{d*}\right]^2. \end{aligned}$$

(72)

The second order derivative $\frac{\partial ^{2} B(F)}{\partial F^{2}}$ is negative, which means there exists an optimal F that maximize the function. Equating $\frac{\partial B(F)}{\partial F}$ to zero, the optimal F can be obtained as

$$\begin{aligned} F^{*}=\left(TP_{r}^{co}(\beta ^{co},w^{co},F=0)-TP_{v}^{d*}\right)*(1-\gamma ) -\left(TP_{r}^{co}(\beta ^{co},w^{co},F=0)-TP_{r}^{d*}\right)*\gamma. \end{aligned}$$

(73)

This ends the proof of Proposition 4. $\square$