Retailer’s optimal CSR investment in closed-loop supply chains: the impacts of supply chain structure and channel power structure

Abstract

This paper studies the retailer’s optimal corporate social responsibility (CSR) investment decisions in manufacturer-collecting closed-loop supply chains (CLSCs). We establish a centralized and two decentralized models with and without considering the retailer’s CSR investment. The impacts of the supply chain structure and channel power structure on the retailer’s CSR investment decisions are analyzed. We find that (i) the retailer increases CSR investment as the remanufacturing cost savings increase or the collection cost reduces; (ii) the retailer is more willing to make a CSR investment in the centralized model than in the decentralized models; (iii) the CSR effort level in the manufacturer-led model is higher than that in the retailer-led model; and (iv) when the unit cost of CSR investment is moderate, retailer’s CSR investment benefits the retailer while harming the manufacturer; however, CSR investment always leads to a “win–win” situation when the retailer is the channel leader. Finally, we carry out numerical studies to investigate the effects of model parameters on supply chain equilibrium.

Appendices

Appendix A

• The Derivation of Centralized Models.

When the retailer makes a CSR investment (Model CY), the optimization for the entire supply chain system is:

$$ \mathop {max}\limits_{p,e,\tau } \pi_{T} = \left( {p - c_{m} + {\Delta }\tau - f} \right)\left( {\alpha - \beta p + \gamma e} \right) - C_{L} \tau^{2} - me^{2} . $$

The corresponding Hessian Matrix is given as below:

$$ \left[ {\begin{array}{*{20}c} {\frac{{\partial^{2} \pi_{T} }}{{\partial p^{2} }}} & {\frac{{\partial^{2} \pi_{T} }}{\partial p\partial e}} & {\frac{{\partial^{2} \pi_{T} }}{\partial p\partial \tau }} \\ {\frac{{\partial^{2} \pi_{T} }}{\partial e\partial p}} & {\frac{{\partial^{2} \pi_{T} }}{{\partial e^{2} }}} & {\frac{{\partial^{2} \pi_{T} }}{\partial e\partial \tau }} \\ {\frac{{\partial^{2} \pi_{T} }}{\partial \tau \partial p}} & {\frac{{\partial^{2} \pi_{T} }}{\partial \tau \partial e}} & {\frac{{\partial^{2} \pi_{T} }}{{\partial \tau^{2} }}} \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} { - 2\beta } & \gamma & { - \beta \Delta } \\ \gamma & { - 2m} & {\gamma \Delta } \\ { - \beta \Delta } & {\gamma \Delta } & { - 2C_{L} } \\ \end{array} } \right]. $$

It can be readily verified that when the condition $C_{L} \ge \frac{{m\beta \Delta \left( {\phi - \beta \left( {c_{m} + f - \Delta } \right)} \right)}}{{4m\beta - \gamma^{2} }}$ is satisfied, $\pi_{T}$ is jointly concave in $p$, $e$ and $\tau$.

Jointly solving $\frac{{\partial \pi_{T} }}{\partial p} = 0$, $\frac{{\partial \pi_{T} }}{\partial e} = 0$ and $\frac{{\partial \pi_{T} }}{\partial \tau } = 0$ yields the optimal solutions for the CLSC, which are given by:

$$ \begin{aligned} p^{CY*} & = \frac{{\left( {2m\phi + \left( {2m\beta - \gamma^{2} } \right)\left( {f + c_{m} } \right)} \right)C_{L} - m\beta \Delta^{2} \phi }}{{\left( {4m\beta - \gamma^{2} } \right)C_{L} - m\beta^{2} \Delta^{2} }},e^{CY*} = \frac{{\gamma \left( {\phi - \beta c_{m} - f\beta } \right)C_{L} }}{{\left( {4m\beta - \gamma^{2} } \right)C_{L} - m\beta^{2} \Delta^{2} }}, \\ \tau^{CY*} & = \frac{{m\beta \Delta \left( {\phi - \beta c_{m} - f\beta } \right)}}{{\left( {4m\beta - \gamma^{2} } \right)C_{L} - m\beta^{2} \Delta^{2} }}. \\ \end{aligned} $$

Then, the equilibrium demand and supply chain profits could be obtained.

If the retailer does not make a CSR investment, we can obtain equilibrium outcomes for Model CN by setting $e = 0$ and $f = 0$. All equilibrium solutions for Model CN and Model CY are shown in Table 3.

• The Derivation of Manufacturer-Led Models.

We also consider the case where the retailer makes a CSR investment. Backward induction is used to solve the manufacturer Stackelberg game. The retailer’s optimization problem is:

$$ \mathop {max}\limits_{p,e} \pi_{R}^{DMY} = \left( {p - w - f} \right)\left( {\alpha - \beta p + \gamma e} \right) - me^{2} . $$

The Hessian Matrix is as follows:

$$ \left[ {\begin{array}{*{20}c} {\frac{{\partial^{2} \pi_{R}^{DMY} }}{{\partial p^{2} }}} & {\frac{{\partial^{2} \pi_{R}^{DMY} }}{\partial p\partial e}} \\ {\frac{{\partial^{2} \pi_{R}^{DMY} }}{\partial e\partial p}} & {\frac{{\partial^{2} \pi_{R}^{DMY} }}{{\partial e^{2} }}} \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} { - 2\beta } & \gamma \\ \gamma & { - 2m} \\ \end{array} } \right]. $$

When $4\beta m - \gamma^{2} > 0$, the Hessian Matrix is negative-defined. The retailer’s best response functions are:

$$ p^{DMY} \left( {w,\tau } \right) = \frac{{\left( {f + w} \right)\left( {2m\beta - \gamma^{2} } \right) + 2m\phi }}{{4m\beta - \gamma^{2} }},e^{DMY} \left( {w,\tau } \right) = \frac{{\gamma \left( {\left( {f + w} \right)\beta - \phi } \right)}}{{ - 4m\beta + \gamma^{2} }}. $$

Then, the manufacturer’s optimization problem is determined by:

$$ \mathop {max}\limits_{w,\tau } \pi_{M}^{DMY} = \left( {w - c_{m} + {\Delta }\tau } \right)\left( {\alpha - \beta p^{DMY} \left( {w,\tau } \right) + \gamma e^{DMY} \left( {w,\tau } \right)} \right) - C_{L} \tau^{2} . $$

The Hessian Matrix is:

$$ \left[ {\begin{array}{*{20}c} {\frac{{\partial^{2} \pi_{M}^{DMY} }}{{\partial w^{2} }}} & {\frac{{\partial^{2} \pi_{M}^{DMY} }}{\partial w\partial \tau }} \\ {\frac{{\partial^{2} \pi_{M}^{DMY} }}{\partial \tau \partial w}} & {\frac{{\partial^{2} \pi_{M}^{DMY} }}{{\partial \tau^{2} }}} \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {\frac{{4m\beta^{2} }}{{ - 4m\beta + \gamma^{2} }}} & { - \frac{{2m\beta^{2} \Delta }}{{4m\beta - \gamma^{2} }}} \\ { - \frac{{2m\beta^{2} \Delta }}{{4m\beta - \gamma^{2} }}} & { - 2C_{L} } \\ \end{array} } \right] $$

It can be verified that only when $m\beta^{2} \Delta^{2} + 2\left( { - 4m\beta + \gamma^{2} } \right)C_{L} < 0$ (or equivalently, $C_{L} > \frac{{m\beta^{2} \Delta^{2} }}{{2\left( {4m\beta - \gamma^{2} } \right)}}$), $\pi_{M}^{DMY}$ is jointly concave in $w$ and $\tau$. According to Lemma 1, the condition $C_{L} > \frac{{m\beta^{2} \Delta^{2} }}{{2\left( {4m\beta - \gamma^{2} } \right)}}$ can be verified. Then, jointly solving equations $\frac{{\partial \pi_{M}^{DMY} }}{\partial w} = 0$ and $\frac{{\partial \pi_{M}^{DMY} }}{\partial \tau } = 0$ yields the manufacturer’s optimal decisions.

$$ w^{DMY*} = \frac{{C_{L} \left( {4\beta m - \gamma^{2} } \right)\left( {\beta f - \beta c_{m} - \phi } \right) + \beta^{2} \Delta^{2} m\left( {\phi - \beta f} \right)}}{{\beta \left( {2C_{L} \left( {\gamma^{2} - 4\beta m} \right) + \beta^{2} \Delta^{2} m} \right)}}, \tau^{DMY*} = \frac{{\beta \Delta m\left( {\beta c_{m} + \beta f - \phi } \right)}}{{2C_{L} \left( {\gamma^{2} - 4\beta m} \right) + \beta^{2} \Delta^{2} m}} $$

Then, the equilibrium profits for channel players and entire supply chain system can be derived. By setting $e = 0$ and $f = 0$, the equilibrium solutions for Model DMN are obtained. All equilibrium solutions for Model DMN and Model DMY are shown in Table 3.

• The Derivation of Retailer-Led Models.

Let $p = w + s$, where $s$ denotes the retail margin of the retailer. We first analyze the manufacturer’s optimization problem.

$$ \mathop {max}\limits_{w,\tau } \pi_{M}^{DRY} = \left( {w - c_{m} + {\Delta }\tau } \right)\left( {\alpha - \beta \left( {w + s} \right) + \gamma e} \right) - C_{L} \tau^{2} . $$

The Hessian Matrix of the manufacturer is:

$$ \left[ {\begin{array}{*{20}c} {\frac{{\partial^{2} \pi_{M}^{DRY} }}{{\partial w^{2} }}} & {\frac{{\partial^{2} \pi_{M}^{DRY} }}{\partial w\partial \tau }} \\ {\frac{{\partial^{2} \pi_{M}^{DRY} }}{\partial \tau \partial w}} & {\frac{{\partial^{2} \pi_{M}^{DRY} }}{{\partial \tau^{2} }}} \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} { - 2\beta } & { - \beta \Delta } \\ { - \beta {\Delta }} & { - 2C_{L} } \\ \end{array} } \right]. $$

According to Lemma 1, it can be judged that $4\beta C_{L} - \beta^{2} {\Delta }^{2} > 0$, the Hessian Matrix is negative-defined. Then, the first-order conditions of the manufacturers lead to:

$$ w^{DRY} \left( {p,e} \right) = \frac{{\beta \Delta^{2} \left( {p\beta - e\gamma - \phi } \right) + 2\left( { - p\beta + e\gamma + \phi + \beta c_{m} } \right)C_{L} }}{{2\beta C_{L} }},\tau^{DRY} \left( {p,e} \right) = \frac{{\Delta \left( { - p\beta + e\gamma + \phi } \right)}}{{2C_{L} }}. $$

Then, the retailer’s optimization problem is determined by:

$$ \mathop {max}\limits_{p,e} \pi_{R}^{DRY} = \left( {p - w^{DRY} \left( {p,e} \right) - f} \right)\left( {\alpha - \beta p + \gamma e} \right) - me^{2} . $$

The Hessian Matrix of the retailer is:

$$ \left[ {\begin{array}{*{20}c} {\frac{{\partial^{2} \pi_{R}^{DRY} }}{{\partial p^{2} }}} & {\frac{{\partial^{2} \pi_{R}^{DRY} }}{\partial p\partial e}} \\ {\frac{{\partial^{2} \pi_{R}^{DRY} }}{\partial e\partial p}} & {\frac{{\partial^{2} \pi_{R}^{DRY} }}{{\partial e^{2} }}} \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} {\beta \left( { - 4 + \frac{{\beta \Delta^{2} }}{{C_{L} }}} \right)} & {\gamma \left( {3 - \frac{{\beta \Delta^{2} }}{{C_{L} }}} \right)} \\ {\gamma \left( {3 - \frac{{\beta \Delta^{2} }}{{C_{L} }}} \right)} & { - \frac{{2\left( {m\beta + \gamma^{2} } \right)}}{\beta } + \frac{{\gamma^{2} \Delta^{2} }}{{C_{L} }}} \\ \end{array} } \right]. $$

According to Lemma 1, we can verify that the second-order determinant of the Hessian Matrix $8m\beta - \gamma^{2} - \frac{{2m\beta^{2} \Delta^{2} }}{{C_{L} }} > 0$ is satisfied, and $\pi_{R}^{DRY}$ is jointly concave in $p$ and $e$. Jointly solving equations $\frac{{\partial \pi_{R}^{DRY} }}{\partial p} = 0$ and $\frac{{\partial \pi_{R}^{DRY} }}{\partial e} = 0$ yields the retailer’s optimal decisions:

$$ \begin{aligned} p^{{DRY*}}& = \frac{{\left( {f\beta \left( {2m\beta - \gamma ^{2} } \right) + \left( {6m\beta - \gamma ^{2} } \right)\phi + \beta \left( {2m\beta - \gamma ^{2} } \right)c_{m} } \right)C_{L} - m\beta ^{2} \Delta ^{2} \phi }}{{\beta \left( {2\left( {4m\beta - \gamma ^{2} } \right)C_{L} - m\beta ^{2} \Delta ^{2} } \right)}}, \\ e^{{DRY*}} & = \frac{{\gamma \left( {\phi - f\beta - \beta c_{m} } \right)C_{L} }}{{2\left( {4m\beta - \gamma ^{2} } \right)C_{L} - m\beta ^{2} \Delta ^{2} }} \end{aligned} $$

Then, the equilibrium profits for channel members and entire supply chain in Model DRY can be obtained. Similarly, the equilibrium solutions for Model DRN can be obtained by setting $e = 0$ and $f = 0$. All equilibrium outcomes for Model DRY and Model DRN are shown in Table 4.

Appendix B

Proof of Lemma 1

According to the derivation of the centralized models, we know that there exist unique optimal solutions for the Model CY when $C_{L} \ge \frac{{m\beta \Delta \left( {\phi - \beta \left( {c_{m} + f - \Delta } \right)} \right)}}{{4m\beta - \gamma^{2} }}$. Similarly, in Model CN, we can obtain the condition $C_{L} \ge \frac{{\Delta \left( {\beta \Delta + \phi - \beta c_{m} } \right)}}{4}$. Then, combining the boundaries of $C_{L}$ in Model CY and CY, we have $C_{L} \ge max\left\{ {\frac{{\Delta \left( {\beta \Delta + \phi - \beta c_{m} } \right)}}{4},\frac{{m\beta \Delta \left( {\phi - \beta \left( {c_{m} + f - \Delta } \right)} \right)}}{{4m\beta - \gamma^{2} }} } \right\}$.

Then, Lemma 1 is proved. □

Proof of Proposition 1

(1) According to the optimal solutions for Model CN and Model CY, we obtain

$$ \begin{array}{*{20}c} {\pi_{T}^{CY*} - \pi_{T}^{CN*} = C_{L} \left( {\frac{{\left( {\phi - \beta c_{m} } \right)^{2} }}{{\beta \left( {\beta \Delta^{2} - 4C_{L} } \right)}} - \frac{{m\left( {f\beta - \phi + \beta c_{m} } \right)^{2} }}{{m\beta^{2} \Delta^{2} + \left( { - 4m\beta + \gamma^{2} } \right)C_{L} }}} \right),} \\ \end{array} $$

(A3)

it can be verified that when $f \le f^{C}$, $\pi_{T}^{{CY{*}}} - \pi_{T}^{{CN{*}}} \ge 0$; otherwise, $\pi_{T}^{{CY{*}}} - \pi_{T}^{{CN{*}}} < 0$, where $f^{C} = \frac{\phi }{\beta } - c_{m} - \frac{{\left( {\phi - \beta c_{m} } \right)\sqrt {\beta m\left( {4C_{L} - \beta \Delta^{2} } \right)\left( {C_{L} \left( {4\beta m - \gamma^{2} } \right) - \beta^{2} \Delta^{2} m} \right)} }}{{\beta^{2} m\left( {4C_{L} - \beta \Delta^{2} } \right)}}$.

(2) Based on $f^{C}$ in Proposition 1(1), we obtain that

$$ \begin{aligned} \frac{{\partial f^{C} }}{\partial \Delta } & = \frac{{\gamma^{2} \Delta \left( {\phi - \beta c_{m} } \right)C_{L} }}{{\left( {4C_{L} - \beta \Delta^{2} } \right)\sqrt {m\beta \left( {4C_{L} - \beta \Delta^{2} } \right)\left( {\left( {4m\beta - \gamma^{2} } \right)C_{L} - m\beta^{2} \Delta^{2} } \right)} }} > 0, \\ \frac{{\partial f^{C} }}{{\partial C_{L} }} & = - \frac{{\gamma^{2} \Delta^{2} \left( {\phi - \beta c_{m} } \right)}}{{2\left( {4C_{L} - \beta \Delta^{2} } \right)\sqrt {m\beta \left( {4C_{L} - \beta \Delta^{2} } \right)\left( {\left( {4m\beta - \gamma^{2} } \right)C_{L} - m\beta^{2} \Delta^{2} } \right)} }} < 0. \\ \end{aligned} $$

Then, Proposition 1 is proved. □

Proof of Proposition 2

According to Lemma 2, we can obtain

$$ \begin{array}{*{20}c} {e^{CY*} - e^{CY - NR*} = \frac{{m\beta^{2} \gamma \Delta^{2} \left( {\phi - f\beta - \beta c_{m} } \right)}}{{\left( {4m\beta - \gamma^{2} } \right)\left( {\left( {4m\beta - \gamma^{2} } \right)C_{L} - m\beta^{2} \Delta^{2} } \right)}} > 0.} \\ \end{array} $$

(A4)

$$ \begin{array}{*{20}c} {\tau^{CY*} - \tau^{CN*} = \frac{{\Delta \left( {4fm\beta^{2} - \gamma^{2} \phi + \beta \gamma^{2} c_{m} } \right)}}{{\left( {m\beta^{2} \Delta^{2} - \left( {4m\beta - \gamma^{2} } \right)C_{L} } \right)\left( {4C_{L} - \beta \Delta^{2} } \right)}}} \\ \end{array} $$

(A5)

It can be verified that when $f \le f^{\tau c}$, $\tau^{CY*} \ge \tau^{CN*}$; otherwise, $\tau^{CY*} < \tau^{CN*}$, where $f^{C} = \frac{{\gamma^{2} \left( {\phi - \beta c_{m} } \right)C_{L} }}{{m\beta^{2} \left( {4C_{L} - \beta \Delta^{2} } \right)}}$.

Then, Proposition 2 is proved. □

Proof of Proposition 3

Similar to the proof of Proposition 1, this proposition can be easily proved. □

Proof of Observation 1

Based on the retailer's optimal profits in Model DMN and Model DMY, we have

$$ \begin{array}{*{20}c} {\pi_{R}^{DMY*} - \pi_{R}^{DMN*} = C_{L}^{2} \left( { - \frac{{4\left( {\phi - \beta c_{m} } \right)^{2} }}{{\beta \left( {\beta \Delta^{2} - 8C_{L} } \right)^{2} }} + \frac{{m\left( {4m\beta - \gamma^{2} } \right)\left( {f\beta - \phi + \beta c_{m} } \right)^{2} }}{{\left( {m\beta^{2} \Delta^{2} + 2\left( { - 4m\beta + \gamma^{2} } \right)C_{L} } \right)^{2} }}} \right)} \\ \end{array} $$

(A6)

because $\frac{{\partial \pi_{R}^{DMY*} }}{\partial f} = \frac{{2m\beta \left( {4m\beta - \gamma^{2} } \right)\left( {f\beta - \phi + \beta c_{m} } \right)C_{L}^{2} }}{{\left( {m\beta^{2} \Delta^{2} + 2\left( { - 4m\beta + \gamma^{2} } \right)C_{L} } \right)^{2} }} < 0$, ${\uppi }_{R}^{DMY*}$ decreases with $f$, the minimum value of ${\uppi }_{R}^{DMY*}$ is obtained at the point $f^{MD}$ when $f \in \left( {0,f^{MD} } \right]$.

Moreover, based on the optimal profits for the manufacturer in Model DMN and Model DMY, we have

$$ \begin{array}{*{20}c} {\pi_{M}^{DMY*} - \pi_{M}^{DMN*} = C_{L} \left( {\frac{{\left( {\phi - \beta c_{m} } \right)^{2} }}{{\beta \left( {\beta \Delta^{2} - 8C_{L} } \right)}} - \frac{{m\left( {f\beta - \phi + \beta c_{m} } \right)^{2} }}{{m\beta^{2} \Delta^{2} + 2\left( { - 4m\beta + \gamma^{2} } \right)C_{L} }}} \right)} \\ \end{array} $$

(A7)

when $f \in \left( {0,f^{MD1} } \right]$, $\pi_{M}^{DMY*} \ge \pi_{M}^{DMN*}$, where $f^{MD1} = \frac{\phi }{\beta } - c_{m} + \frac{{\sqrt {m\beta^{3} \left( {\phi - \beta c_{m} } \right)^{2} \left( {\beta \Delta^{2} - 8C_{L} } \right)\left( {m\beta^{2} \Delta^{2} + 2\left( { - 4m\beta + \gamma^{2} } \right)C_{L} } \right)} }}{{m\beta^{3} \left( {8C_{L} - \beta \Delta^{2} } \right)}}$.

Then, compared with $f^{MD}$ and $f^{MD1}$, we obtain

$$ \begin{aligned} f^{MD} & = \frac{\phi }{\beta } - c_{m} - \frac{{\sqrt {2\left( {\phi - \beta c_{m} } \right)^{2} \left( {2\left( {4m\beta - \gamma^{2} } \right)C_{L} - m\beta^{2} \Delta^{2} } \right)^{2} m\beta \left( {4m\beta - \gamma^{2} } \right)} }}{{m\beta^{2} \left( {4m\beta - \gamma^{2} } \right)\left( {8C_{L} - \beta \Delta^{2} } \right)}} \\ f^{MD1} & = \frac{\phi }{\beta } - c_{m} - \frac{{\sqrt {m\beta \left( {\phi - \beta c_{m} } \right)^{2} \left( {8C_{L} - \beta \Delta^{2} } \right)\left( {2\left( {4m\beta - \gamma^{2} } \right)C_{L} - m\beta^{2} \Delta^{2} } \right)\left( {4m\beta - \gamma^{2} } \right)^{2} } }}{{m\beta^{2} \left( {8C_{L} - \beta \Delta^{2} } \right)\left( {4m\beta - \gamma^{2} } \right)}} \\ \end{aligned} $$

let $f_{1} = 2\left( {\phi - \beta c_{m} } \right)^{2} \left( {2\left( {4m\beta - \gamma^{2} } \right)C_{L} - m\beta^{2} \Delta^{2} } \right)^{2} m\beta \left( {4m\beta - \gamma^{2} } \right)$, $f_{2} =\break m\beta \left( {\phi - \beta c_{m} } \right)^{2} \left( {8C_{L} - \beta \Delta^{2} } \right)\left( {2\left( {4m\beta - \gamma^{2} } \right)C_{L} - m\beta^{2} \Delta^{2} } \right)\left( {4m\beta - \gamma^{2} } \right)^{2}$, because $f_{1} - f_{2} = \beta \left( { - m} \right)\left( {4\beta m - \gamma^{2} } \right)\left( {\phi - \beta c_{m} } \right)^{2} \left( {2C_{L} \left( {\gamma^{2} - 4\beta m} \right) + \beta^{2} \Delta^{2} m} \right)$ $\left( {4C_{L} \left( {\gamma^{2} - 4\beta m} \right) + \beta \Delta^{2} \left( {2\beta m - \gamma^{2} } \right)} \right) < 0$, we obtain $f^{MD} > f^{MD1}$. Then, Observation 1 is proved. □

Proof of Proposition 4

Similar to the proof of Proposition 3, this proposition is easily proved. □

Proof of Observation 2

$$ \begin{array}{*{20}c} {\pi_{R}^{DRY*} - \pi_{R}^{DRN*} = \frac{1}{2}C_{L} \left( {\frac{{\left( {\phi - \beta c_{m} } \right)^{2} }}{{\beta \left( {\beta \Delta^{2} - 4C_{L} } \right)}} - \frac{{2m\left( {f\beta - \phi + \beta c_{m} } \right)^{2} }}{{2m\beta^{2} \Delta^{2} + \left( { - 8m\beta + \gamma^{2} } \right)C_{L} }}} \right)} \\ \end{array} $$

(A8)

we obtain that $\pi_{R}^{DRY*} \ge \pi_{R}^{DRN*}$ when $f \in \left( {0,f^{RD} } \right]$; otherwise, $\pi_{R}^{DRY*} < \pi_{R}^{DRN*}$, where $f^{RD} = \frac{1}{2}\left( {\frac{2\phi }{\beta } - 2c_{m} - \frac{{\sqrt 2 \sqrt {m\beta \left( {\phi - \beta c_{m} } \right)^{2} \left( {4C_{L} - \beta \Delta^{2} } \right)\left( {\left( {8m\beta - \gamma^{2} } \right)C_{L} - 2m\beta^{2} \Delta^{2} } \right)} }}{{m\beta^{2} \left( {4C_{L} - \beta \Delta^{2} } \right)}}} \right)$.

$$ \begin{array}{*{20}c} {\pi_{M}^{DRY*} - \pi_{M}^{DRN*} = C_{L} \left( {\frac{{\left( {\phi - \beta c_{m} } \right)^{2} }}{{4\beta \left( {\beta \Delta^{2} - 4C_{L} } \right)}} - \frac{{m^{2} \beta \left( {f\beta - \phi + \beta c_{m} } \right)^{2} \left( {\beta \Delta^{2} - 4C_{L} } \right)}}{{\left( {2m\beta^{2} \Delta^{2} - \left( {8m\beta - \gamma^{2} } \right)C_{L} } \right)^{2} }}} \right)} \\ \end{array} $$

(A9)

we obtain that $\pi_{M}^{DRY*} \ge \pi_{M}^{DRN*}$ when $f \in \left( {0,f^{RD1} } \right]$; otherwise, $\pi_{M}^{DRY*} < \pi_{M}^{DRN*}$, where $f^{RD1} = \frac{{\gamma^{2} \left( {\phi - \beta c_{m} } \right)C_{L} }}{{2m\beta^{2} \left( {4C_{L} - \beta \Delta^{2} } \right)}}$.

We can easily prove that $f^{RD1} > f^{RD}$. Hence, when $f \in \left( {0,f^{RD} } \right]$, $\pi_{M}^{DRY*} > \pi_{M}^{DRN*}$ and $\pi_{R}^{DRY*} \ge \pi_{R}^{DRN*}$ are always satisfied. Then, Observation 2 is proved. □

Proof of Proposition 5

(1)

$$ \begin{array}{*{20}c} {e^{CY*} - e^{DMY*} = \frac{{\gamma \left( { - 4m\beta + \gamma^{2} } \right)\left( {f\beta - \phi + \beta c_{m} } \right)C_{L}^{2} }}{{\left( {m\beta^{2} \Delta^{2} + \left( { - 4m\beta + \gamma^{2} } \right)C_{L} } \right)\left( {m\beta^{2} \Delta^{2} + 2\left( { - 4m\beta + \gamma^{2} } \right)C_{L} } \right)}} > 0,} \\ \end{array} $$

(A10)

$$ \begin{array}{*{20}c} {e^{CY*} - e^{DRY*} = \frac{{m\beta \gamma \left( {f\beta - \phi + \beta c_{m} } \right)\left( {\beta \Delta^{2} - 4C_{L} } \right)C_{L} }}{{\left( {2m\beta^{2} \Delta^{2} + \left( { - 8m\beta + \gamma^{2} } \right)C_{L} } \right)\left( {m\beta^{2} \Delta^{2} + \left( { - 4m\beta + \gamma^{2} } \right)C_{L} } \right)}} > 0,} \\ \end{array} $$

(A11)

$$ \begin{array}{*{20}c} {e^{DMY*} - e^{DRY*} = \frac{{\gamma \left( {f\beta - \phi + \beta c_{m} } \right)C_{L} \left( {m\beta^{2} \Delta^{2} - \gamma^{2} C_{L} } \right)}}{{\left( {2m\beta^{2} \Delta^{2} + \left( { - 8m\beta + \gamma^{2} } \right)C_{L} } \right)\left( {m\beta^{2} \Delta^{2} + 2\left( { - 4m\beta + \gamma^{2} } \right)C_{L} } \right)}},} \\ \end{array} $$

(A12)

hence, we can obtain that $e^{CY*} > e^{DRY*} \ge e^{DMY*}$ when $0 < \gamma \le \frac{\beta \Delta \sqrt m }{{\sqrt {C_{L} } }}$; otherwise, $e^{CY*} > e^{DMY*} > e^{DRY*}$.

(2) When the retailer does not make CSR investment,

$$ \begin{array}{*{20}c} {\tau^{CN*} - \tau^{DRN*} = \frac{{ - \Delta \phi + \beta \Delta c_{m} }}{{2\beta \Delta^{2} - 8C_{L} }} > 0,} \\ \end{array} $$

(A13)

$$ \begin{array}{*{20}c} {\tau^{DRN*} - \tau^{DMN*} = \frac{{\beta \Delta^{3} \left( {\phi - \beta c_{m} } \right)}}{{2\left( {\beta \Delta^{2} - 8C_{L} } \right)\left( {\beta \Delta^{2} - 4C_{L} } \right)}} > 0.} \\ \end{array} $$

(A14)

When the retailer makes CSR investment,

$$ \begin{array}{*{20}c} {\tau^{CY*} - \tau^{DMY*} = - \frac{{m\beta \left( {4m\beta - \gamma^{2} } \right)\Delta \left( {f\beta - \phi + \beta c_{m} } \right)C_{L} }}{{\left( {m\beta^{2} \Delta^{2} + \left( { - 4m\beta + \gamma^{2} } \right)C_{L} } \right)\left( {m\beta^{2} \Delta^{2} + 2\left( { - 4m\beta + \gamma^{2} } \right)C_{L} } \right)}} > 0,} \\ \end{array} $$

(A15)

$$ \begin{array}{*{20}c} {\tau^{CY*} - \tau^{DRY*} = \frac{{m^{2} \beta^{2} \Delta \left( {f\beta - \phi + \beta c_{m} } \right)\left( {\beta \Delta^{2} - 4C_{L} } \right)}}{{\left( {2m\beta^{2} \Delta^{2} + \left( { - 8m\beta + \gamma^{2} } \right)C_{L} } \right)\left( {m\beta^{2} \Delta^{2} + \left( { - 4m\beta + \gamma^{2} } \right)C_{L} } \right)}} > 0,} \\ \end{array} $$

(A16)

$$ \begin{array}{*{20}c} {\tau^{DMY*} - \tau^{DRY*} = \frac{{m\beta \Delta \left( {f\beta - \phi + \beta c_{m} } \right)\left( {m\beta^{2} \Delta^{2} - \gamma^{2} C_{L} } \right)}}{{\left( {2m\beta^{2} \Delta^{2} + \left( { - 8m\beta + \gamma^{2} } \right)C_{L} } \right)\left( {m\beta^{2} \Delta^{2} + 2\left( { - 4m\beta + \gamma^{2} } \right)C_{L} } \right)}},} \\ \end{array} $$

(A17)

hence, we can obtain that $\tau^{CY*} > \tau^{DRY*} \ge \tau^{DMY*}$ if $0 < \gamma \le \frac{\beta \Delta \sqrt m }{{\sqrt {C_{L} } }}$; otherwise, $\tau^{CY*} > \tau^{DMY*} > \tau^{DRY*}$.

Then, Proposition 5 is proved. □

Proof of Proposition 6

Similar to the proof of Proposition 5, we can readily prove Proposition 6. □

Proof of Proposition 7

$$ \begin{array}{*{20}c} {f^{c} = \frac{\phi }{\beta } - c_{m} - \frac{{\left( {\phi - \beta c_{m} } \right)\sqrt {m\beta \left( {4C_{L} - \beta \Delta^{2} } \right)\left( {\left( {4m\beta - \gamma^{2} } \right)C_{L} - m\beta^{2} \Delta^{2} } \right)} }}{{m\beta^{2} \left( {4C_{L} - \beta \Delta^{2} } \right)}}} \\ \end{array} $$

$$ \begin{array}{*{20}c} {f^{MD} = \frac{\phi }{\beta } - c_{m} - \frac{{\left( {\phi - \beta c_{m} } \right)\sqrt {4\left( {2\left( {4m\beta - \gamma^{2} } \right)C_{L} - m\beta^{2} \Delta^{2} } \right)m\beta \left( {4m\beta - \gamma^{2} } \right)} }}{{m\beta^{2} \left( {4m\beta - \gamma^{2} } \right)\left( {8C_{L} - \beta \Delta^{2} } \right)}}} \\ \end{array} $$

$$ \begin{array}{*{20}c} {f^{RD} = \frac{1}{2}\left( {\frac{2\phi }{\beta } - 2c_{m} - \frac{{\sqrt 2 \sqrt {m\beta^{3} \left( {\phi - \beta c_{m} } \right)^{2} \left( {\beta \Delta^{2} - 4C_{L} } \right)\left( {2m\beta^{2} \Delta^{2} - \left( {8m\beta - \gamma^{2} } \right)C_{L} } \right)} }}{{m\beta^{3} \left( {4C_{L} - \beta \Delta^{2} } \right)}}} \right)} \\ \end{array} $$

We can observe that $f^{C} > f^{RD}$. Then, let $f_{3} \left( {C_{L} } \right) = m\beta ( 4C_{L} - \beta \Delta^{2} )((4m\beta - \gamma^{2})C_{L} - m\beta^{2} \Delta^{2})(8C_{L} - \beta \Delta^{2})^{2} (4m\beta - \gamma^{2})^{2}$, $f_{4} \left( {C_{L} } \right) = 4\left( {2\left( {4m\beta - \gamma^{2} } \right)C_{L} - m\beta^{2} \Delta^{2} } \right)m\beta \left( {4m\beta - \gamma^{2} } \right)\left( {4C_{L} - \beta \Delta^{2} } \right)^{2}$. $f_{3} \left( {C_{L} } \right) - f_{4} \left( {C_{L} } \right) = - m\beta \left( {4m\beta - \gamma^{2} } \right)( 4C_{L} - \beta \Delta^{2})(4(4C_{L} - \beta \Delta^{2})(2(4m\beta - \gamma^{2})C_{L} - m\beta^{2} \Delta^{2}) - ( (4m\beta - \gamma^{2})C_{L} - m\beta^{2} \Delta^{2}) \times(4m\beta - \gamma^{2})(\beta \Delta^{2} - 8C_{L})^{2})$.

Further, let $f_{5} \left( {C_{L} } \right) = 4\left( {4C_{L} - \beta \Delta^{2} } \right)\left( {2\left( {4m\beta - \gamma^{2} } \right)C_{L} - m\beta^{2} \Delta^{2} } \right.$, $\left. {f_{6} \left( {C_{L} } \right) = \left( {4m\beta - \gamma^{2} } \right)\left( {\beta \Delta^{2} - 8C_{L} } \right)^{2} \left( {\left( {4m\beta - \gamma^{2} } \right)C_{L} - m\beta^{2} \Delta^{2} } \right)} \right)$. Because

$$ \begin{aligned} \frac{{\partial f_{5} \left( {C_{L} } \right)}}{{\partial C_{L} }} & = 8\left( {4m\beta - \gamma^{2} } \right)\left( {4C_{L} - \beta \Delta^{2} } \right) + 16\left( {2\left( {4m\beta - \gamma^{2} } \right)C_{L} - m\beta^{2} \Delta^{2} } \right) > 0, \\ \frac{{\partial f_{6} \left( {C_{L} } \right)}}{{\partial C_{L} }} & = \left( {4m\beta - \gamma^{2} } \right)\left( {8C_{L} - \beta \Delta^{2} } \right)\left( {24\left( {4m\beta - \gamma^{2} } \right)C_{L} - \beta \left( {20m\beta - \gamma^{2} } \right)\Delta^{2} } \right) > 0, \\ \end{aligned} $$

$f_{5} \left( {C_{L} } \right)$ and $f_{6} \left( {C_{L} } \right)$ increase with $C_{L}$. We calculate that $f_{5} \left( {\frac{{\beta \Delta^{2} }}{8}} \right) = \frac{1}{2}\beta^{2} \gamma^{2} \Delta^{4} > 0$, $f_{6} \left( {\frac{{\beta \Delta^{2} }}{8}} \right) = 0$; $f_{5} \left( {\frac{{\beta \Delta^{2} }}{4}} \right) = 0$, $f_{6} \left( {\frac{{\beta \Delta^{2} }}{4}} \right) = - \frac{1}{4}\beta^{3} \gamma^{2} \left( {4m\beta - \gamma^{2} } \right)\Delta^{6} < 0$. Hence, $f_{5} \left( {C_{L} } \right) > f_{6} \left( {C_{L} } \right)$; further, we have $f_{3} \left( {C_{L} } \right) < f_{4} \left( {C_{L} } \right)$ and $f^{C} > f^{MD}$.

The proposition is proved. □

Proof of Proposition 8

Similar to the proof of Proposition 6, this proposition is easily proved. □