Low-carbon supply policies and supply chain performance with carbon concerned demand

Abstract

A dilemma between economic growth and environmental deterioration has never been recognized so seriously until the environmental problem has been impacting everyone’s daily life cogently (e.g., the serious fog and haze in China). Simultaneously, the rising environment awareness of consumers leads to the close attention to the product’s carbon performance reflected in the carbon concerned demand, which provides opportunity to rebuild our business ecosystem. The paper expands the environment view to supply chain operations. A Stackelberg-like model is developed to game-theoretically analyze the decentralized decisions of the manufacturer and retailer. Low-carbon effort is brought into decision by both sides. The centralized decision-making is also investigated as a benchmark to evaluate the supply chain performance. Additionally,we propose a carbon-related price–discount sharing-like scheme to achieve the channel coordination and discuss the possibility of Pareto improvement. Several interesting managerial insights on low-carbon factors are concluded.

Appendices

Appendix 1: Proof of Proposition 1

Proof

We solve the dynamic game of the retailer and the manufacturer using the standard backward induction. We assume all parameters are positive.

Step 1: Solve Retailer’s profit maximization problem

The profit function faced by retailer has been defined in Eq. (4).

The first derivation of $\pi _r$ regarding p and $x_r$ respectively is:

$$\begin{aligned} \frac{{\partial {\pi _r}}}{{\partial p}}= & {} \left[ {D - \alpha p + \beta - \beta \left[ {\psi {{\mathbf {e}}^{ - {\lambda _m}{x_m}}} + \left( {1 - \psi } \right) {{\mathbf {e}}^{ - {\lambda _r}{x_r}}}} \right] } \right] - \alpha \left( {p - w - {x_r}} \right) = 0\\ \frac{{\partial {\pi _r}}}{{\partial {x_r}}}= & {} \left( {p - w - {x_r}} \right) \beta {\lambda _r} \left( {1 - \psi } \right) {{\mathbf {e}}^{ - {\lambda _r}{x_r}}}\nonumber \\&- \left[ {D - \alpha p + \beta - \beta \left[ {\psi {{\mathbf {e}}^{ - {\lambda _m}{x_m}}} + \left( {1 - \psi } \right) {{\mathbf {e}}^{ - {\lambda _r}{x_r}}}} \right] } \right] = 0 \end{aligned}$$

The stationary point $(p,x_r)$ of the $\pi _r$ can be calculated through simultaneous equations. And $(p,x_r)$ can be denoted in Eqs. (5) and (6).

It has been verified that the determinant of the Hessian matrix at the stationary point is negative definite:

$$\begin{aligned} A= & {} \frac{{{\partial ^2}{\pi _r}}}{{\partial {p^2}}} = - 2\alpha < 0\\ B= & {} \frac{{{\partial ^2}{\pi _r}}}{{\partial p\partial {x_r}}} = \beta \left( {1 - \psi } \right) {\lambda _r}{{\mathbf {e}}^{ - {\lambda _r}{x_r}}} + \alpha \\ C= & {} \frac{{{\partial ^2}{\pi _r}}}{{\partial x_r^2}} = - 2\beta {\lambda _r}\left( {1 - \psi } \right) {{\mathbf {e}}^{ - {\lambda _r}{x_r}}} - \beta \lambda _r^2\left( {p - w - {x_r}} \right) \left( {1 - \psi } \right) {{\mathbf {e}}^{ - {\lambda _r}{x_r}}}\\ AC - {B^2}= & {} 4\alpha \beta {\lambda _r}\left( {1 - \psi } \right) {{\mathbf {e}}^{ - {\lambda _r}{x_r}}} + 2\alpha \left( {p - w - {x_r}} \right) \beta {\lambda _r}^2\left( {1 - \psi } \right) {{\mathbf {e}}^{ - {\lambda _r}{x_r}}}\nonumber \\&- {\left[ {\beta {\lambda _r}\left( {1 - \psi } \right) {{\mathbf {e}}^{ - {\lambda _r}{x_r}}} + \alpha } \right] ^2}\nonumber \\= & {} 2\alpha \left( {p - w - {x_r}} \right) \beta {\lambda _r}^2\left( {1 - \psi } \right) {{\mathbf {e}}^{ - {\lambda _r}{x_r}}} > 0 \end{aligned}$$

So the $\pi _r$ reaches the maximum at the stationary point.

Step 2: Solve manufacturer’s profit maximization problem

As the Stackelberg leader, the manufacturer anticipates the retailer’s best response, shown by Eqs. (5) and (6), and incorporate it to her profit function, we have Eq. (7).

The first derivation of $\pi _m$ regarding w and $x_m$ respectively is:

$$\begin{aligned} \frac{{\partial {\pi _m}}}{{\partial w}}= & {} \frac{1}{2}\left[ {D + \beta - \beta \psi {{\mathbf {e}}^{ - {\lambda _m}{x_m}}} - \alpha w - \frac{\alpha }{{{\lambda _r}}}\left( {1 + \ln \frac{{\beta {\lambda _r}\left( {1 - \psi } \right) }}{\alpha }} \right) } \right] \nonumber \\&- \frac{\alpha }{2}\left( {w - c - {x_m}} \right) = 0\\ \frac{{\partial {\pi _m}}}{{\partial {x_m}}}= & {} - \frac{1}{2}\left[ {D + \beta - \beta \psi {{\mathbf {e}}^{ - {\lambda _m}{x_m}}} - \alpha w - \frac{\alpha }{{{\lambda _r}}}\left( {1 + \ln \frac{{\beta {\lambda _r}\left( {1 - \psi } \right) }}{\alpha }} \right) } \right] \nonumber \\&+\, \frac{{\beta \psi {\lambda _m}{{\mathbf {e}}^{ - {\lambda _m}{x_m}}}}}{2}\left( {w - c - {x_m}} \right) = 0 \end{aligned}$$

The stationary point $(w,x_m)$ of the $\pi _m$ can be calculated through simultaneous equations, denoted in Eq. (8).

It also has been verified that the determinant of the Hessian matrix at the stationary point is negative definite:

$$\begin{aligned} A= & {} \frac{{{\partial ^2}{\pi _m}}}{{\partial {w^2}}} = - \alpha < 0\\ B= & {} \frac{{{\partial ^2}{\pi _m}}}{{\partial w\partial {x_m}}} = \frac{1}{2}\beta \psi {\lambda _m}{{\mathbf {e}}^{ - {\lambda _m}{x_m}}} + \frac{\alpha }{2}\\ C= & {} \frac{{{\partial ^2}{\pi _m}}}{{\partial {x_m}^2}} = - \beta \psi {\lambda _m}{{\mathbf {e}}^{ - {\lambda _m}{x_m}}} - \frac{1}{2}\beta \psi {\lambda _m}^2{{\mathbf {e}}^{ - {\lambda _m}{x_m}}}\left( {w - c - {x_m}} \right) \\ AC - {B^2}= & {} \alpha \beta \psi {\lambda _m}{{\mathbf {e}}^{ - {\lambda _m}{x_m}}} + \frac{1}{2}\alpha \beta \psi {\lambda _m}^2{{\mathbf {e}}^{ - {\lambda _m}{x_m}}}\left( {w - c - {x_m}} \right) - {\left( {\frac{1}{2}\beta \psi {\lambda _m}{{\mathbf {e}}^{ - {\lambda _m}{x_m}}} + \frac{\alpha }{2}} \right) ^2}\nonumber \\= & {} \frac{1}{2}\alpha \beta \psi {\lambda _m}^2{{\mathbf {e}}^{ - {\lambda _m}{x_m}}}\left( {w - c - {x_m}} \right) > 0 \end{aligned}$$

So the $\pi _m$ reaches the maximum at the stationary point.

Step 3: Solve the optimal solution and its applicable condition

In the Stackelberg game, the optimum decision for the manufacturer and retailer have been shown in Eqs. (8) and (9).

It is apparent that the boundary conditions of the optimal solution must be satisfied and demonstrated as:

$$\begin{aligned} \left\{ \begin{array}{ll} &{}{{x_m}^* = \frac{1}{{{\lambda _m}}}\ln \frac{{\beta {\lambda _m}\psi }}{\alpha } > 0}\\ &{}{{x_r}^* = \frac{1}{{{\lambda _r}}}\ln \frac{{\beta {\lambda _r}\left( {1 - \psi } \right) }}{\alpha }}\\ &{}{{p^*} - {w^*} - x_r^* \ge 0}\\ &{}{{w^*} - x_m^* - c \ge 0} \end{array} \right. \end{aligned}$$

Reducing the above inequality, we have the optimal condition:

$$\begin{aligned} \left\{ \begin{array}{ll} &{}{\beta {\lambda _r}\left( {1 - \psi } \right) > \alpha }\\ &{}{\beta {\lambda _m}\psi > \alpha }\\ &{}{D + \beta - \frac{\alpha }{{{\lambda _m}}} - \frac{\alpha }{{{\lambda _r}}} - \frac{\alpha }{{{\lambda _m}}}\ln \frac{{\beta \psi {\lambda _m}}}{\alpha } - \frac{\alpha }{{{\lambda _r}}}\ln \frac{{\beta \left( {1 - \psi } \right) {\lambda _r}}}{\alpha } - \alpha c > 0} \end{array} \right. \end{aligned}$$

$\square$

Appendix 2: Proof of Corollary 3

Proof

From the Proposition 1, the optimal retail price and wholesale price:

$$\begin{aligned} p^*= & {} \frac{1}{4}\left[ {\frac{{3D}}{\alpha } + \frac{{3\beta }}{\alpha } - 3\left( {\frac{1}{{{\lambda _m}}} + \frac{1}{{{\lambda _r}}}} \right) + x_m^* + x_r^* + c} \right] \nonumber \\= & {} \frac{1}{4}\left[ {\frac{{3D}}{\alpha } + \frac{{3\beta }}{\alpha } - 3\left( {\frac{1}{{{\lambda _m}}} + \frac{1}{{{\lambda _r}}}} \right) + \frac{1}{{{\lambda _m}}}\ln \frac{{\beta {\lambda _m}\psi }}{\alpha } + \frac{1}{{{\lambda _r}}}\ln \frac{{\beta {\lambda _r}\left( {1 - \psi } \right) }}{\alpha } + c} \right] . \\ w^*= & {} \frac{1}{2}\left[ {\frac{D}{\alpha } + \frac{\beta }{\alpha } - \left( {\frac{1}{{{\lambda _m}}} + \frac{1}{{{\lambda _r}}}} \right) + x_m^* - x_r^* + c} \right] \nonumber \\= & {} \frac{1}{2}\left[ {\frac{D}{\alpha } + \frac{\beta }{\alpha } - \left( {\frac{1}{{{\lambda _m}}} + \frac{1}{{{\lambda _r}}}} \right) + \frac{1}{{{\lambda _m}}}\ln \frac{{\beta {\lambda _m}\psi }}{\alpha } - \frac{1}{{{\lambda _r}}}\ln \frac{{\beta {\lambda _r}\left( {1 - \psi } \right) }}{\alpha } + c} \right] \end{aligned}$$

$\frac{{\partial {p^*}}}{{\partial \left( {\beta /\alpha } \right) }} > 0$ which means that the retail price is always increasing to $\beta /\alpha$ While $\frac{{\partial {w^*}}}{{\partial \left( {\beta /\alpha } \right) }} = 1 + \left( {\frac{1}{{{\lambda _m}}} - \frac{1}{{{\lambda _r}}}} \right) \frac{\alpha }{\beta }$, Note $\varLambda = \left( \frac{1}{{{\lambda _r}}} - \frac{1}{{{\lambda _m}}} \right)$ which captures the manufacture’s technological advantage relative to the retailer’s. So the demand characteristic of low-carbon preference compared to price sensitivity $(\beta /\alpha )$ has a piecewise impact on the wholesale price $w^*$, i.e., $w^*$ is increasing in $(\beta /\alpha )$ greater than $\varLambda$, while decreasing in $\beta /\alpha$ less than $\varLambda$.

Obviously, the generalized technological level of carbon reduction of the manufacturer $(\psi \lambda _m)$ positively impacts both pricing decisions, however, that of the retailer $\left( (1-\psi ) \lambda _r\right)$ has a positive impact on the retail price $p^*$ while a negative impact on the wholesale price $w^*$; The retail price $p^*$ is positively correlative to both low-carbon investments respectively, while the wholesale price $w^*$ is positively correlative to $(x_m^*-x_r^*)$, the gap that the manufacturer’s low-carbon investment is larger than the retailer’s. It is not surprising that the more the manufacturer invests, the higher she would charge. $\square$

Appendix 3: Proofs of Propositions and corollaries in Sect. 4.2

1.1 Proof of Proposition 2

Proof

Under the centralized decision-making system, the supply chain’s profit function is Eq. (13).

The first derivation of ${\varPi }$ regarding p, ${x_r}$ and ${x_m}$ respectively is:

$$\begin{aligned} \frac{{\partial {\varPi }}}{{\partial p}}= & {} \left[ {D + \beta - \alpha p - \beta \left[ {\psi {{\mathbf {e}}^{ - {\lambda _m}{x_m}}} + \left( {1 - \psi } \right) {{\mathbf {e}}^{ - {\lambda _r}{x_r}}}} \right] } \right] - \alpha \left( {p - c - {x_r} - {x_m}} \right) = 0\\ \frac{{\partial {\varPi }}}{{\partial {x_r}}}= & {} \left( {p - c - {x_r} - {x_m}} \right) \beta {\lambda _r}\left( {1 - \psi } \right) {{\mathbf {e}}^{ - {\lambda _r}{x_r}}} \\&- \left[ {D + \beta - \alpha p - \beta \left[ {\psi {{\mathbf {e}}^{ - {\lambda _m}{x_m}}} + \left( {1 - \psi } \right) {{\mathbf {e}}^{ - {\lambda _r}{x_r}}}} \right] } \right] = 0 \\ \frac{{\partial {\varPi }}}{{\partial {x_m}}}= & {} \left( {p - c - {x_r} - {x_m}} \right) \beta {\lambda _m}\psi {{\mathbf {e}}^{ - {\lambda _m}{x_m}}} \\&- \left[ {D + \beta - \alpha p - \beta \left[ {\psi {{\mathbf {e}}^{ - {\lambda _m}{x_m}}} + \left( {1 - \psi } \right) {{\mathbf {e}}^{ - {\lambda _r}{x_r}}}} \right] } \right] = 0 \end{aligned}$$

The stationary point $\left( {p,{x_r},{x_m}} \right)$ of the ${\varPi }$ can be calculated through simultaneous equations denoted in Eqs. (14) and (15).

It has been verified that the determinant of the Hessian matrix M at the stationary point is negative definite:

$$\begin{aligned} M = \left[ \begin{array}{lll} {\frac{{{\partial ^2}{\varPi }}}{{\partial {p^2}}}}&{}{\frac{{{\partial ^2}{\varPi }}}{{\partial p\partial {x_m}}}}&{}{\frac{{{\partial ^2}{\varPi }}}{{\partial p\partial {x_r}}}}\\ {\frac{{{\partial ^2}{\varPi }}}{{\partial {x_m}\partial p}}}&{}{\frac{{{\partial ^2}{\varPi }}}{{\partial {x_m}^2}}}&{}{\frac{{{\partial ^2}{\varPi }}}{{\partial {x_m}\partial {x_r}}}}\\ {\frac{{{\partial ^2}{\varPi }}}{{\partial {x_r}\partial p}}}&{}{\frac{{{\partial ^2}{\varPi }}}{{\partial {x_r}\partial {x_m}}}}&{}{\frac{{{\partial ^2}{\varPi }}}{{\partial {x_r}^2}}} \end{array} \right] \end{aligned}$$

Its principal minors of order k is defined as M, where

$$\begin{aligned} {M_1}= & {} \left. {\left| {\frac{{{\partial ^2}{\varPi }}}{{\partial {p^2}}}} \right. } \right| = - 2\alpha < 0\\ {M_2}= & {} \left. {\left| {\begin{array}{ll} {\frac{{{\partial ^2}{\varPi }}}{{\partial {p^2}}}}&{}{\frac{{{\partial ^2}{\varPi }}}{{\partial p\partial {x_m}}}}\\ {\frac{{{\partial ^2}{\varPi }}}{{\partial {x_m}\partial p}}}&{}{\frac{{{\partial ^2}{\varPi }}}{{\partial {x_m}^2}}} \end{array}} \right. } \right| \nonumber \\= & {} \left. {\left| {\begin{array}{ll} { - 2\alpha }&{}{\beta \psi {\lambda _m}{{\mathbf {e}}^{ - {\lambda _m}{x_m}}} + \alpha }\\ {\beta \psi {\lambda _m}{{\mathbf {e}}^{ - {\lambda _m}{x_m}}} + \alpha }&{}{ - 2\beta \psi {\lambda _m}{{\mathbf {e}}^{ - {\lambda _m}{x_m}}} - \left( {p - c - {x_r} - {x_m}} \right) \beta \psi \lambda _m^2{{\mathbf {e}}^{ - {\lambda _m}{x_m}}}} \end{array}} \right. } \right| \nonumber \\= & {} 4\alpha \beta \psi {\lambda _m}{{\mathbf {e}}^{ - {\lambda _m}{x_m}}} + 2\alpha \left( {p - c - {x_r} - {x_m}} \right) \beta \psi \lambda _m^2{{\mathbf {e}}^{ - {\lambda _m}{x_m}}} - {\left( {\beta \psi {\lambda _m}{{\mathbf {e}}^{ - {\lambda _m}{x_m}}} + \alpha } \right) ^2}\\ {M_2}{\bigg |_{p = {p^o},{x_r} = x_r^o,{x_m} = x_m^o}}= & {} 2{\alpha ^2}{\lambda _m}\left( {{p^o} - c - x_r^o - x_m^o} \right) > 0\\ {M_3}= & {} \left. {\left| {\begin{array}{lll} {\frac{{{\partial ^2}{\varPi }}}{{\partial {p^2}}}}&{}{\frac{{{\partial ^2}{\varPi }}}{{\partial p\partial {x_m}}}}&{}{\frac{{{\partial ^2}{\varPi }}}{{\partial p\partial {x_r}}}}\\ {\frac{{{\partial ^2}{\varPi }}}{{\partial {x_m}\partial p}}}&{}{\frac{{{\partial ^2}{\varPi }}}{{\partial {x_m}^2}}}&{}{\frac{{{\partial ^2}{\varPi }}}{{\partial {x_m}\partial {x_r}}}}\\ {\frac{{{\partial ^2}{\varPi }}}{{\partial {x_r}\partial p}}}&{}{\frac{{{\partial ^2}{\varPi }}}{{\partial {x_r}\partial {x_m}}}}&{}{\frac{{{\partial ^2}{\varPi }}}{{\partial {x_r}^2}}} \end{array}} \right. } \right| \\ {M_3}{\bigg |_{p = {p^o},{x_r} = x_r^o,{x_m} = x_m^o}}= & {} - 2\left( {{p^o} - c - x_r^o - x_m^o} \right) {\lambda _m}{\lambda _r}{\alpha ^3} < 0 \end{aligned}$$

So the ${\varPi }$ reaches the maximum at the stationary point. It is apparent that the boundary conditions of the optimal solution must be satisfied and demonstrated as:

$$\begin{aligned} \left\{ \begin{array}{ll} &{}{x_m^o = \frac{1}{{{\lambda _m}}}\ln \frac{{\beta {\lambda _m}\psi }}{\alpha } > 0}\\ &{}{x_r^o = \frac{1}{{{\lambda _r}}}\ln \frac{{\beta {\lambda _r}\left( {1 - \psi } \right) }}{\alpha }}\\ &{}{p^o - x_r^o - x_m^o > 0} \end{array} \right. \end{aligned}$$

Reducing the above inequality, we have the optimal condition:

$$\begin{aligned} \left\{ \begin{array}{ll} &{}{\beta {\lambda _r}\left( {1 - \psi } \right) > \alpha }\\ &{}{\beta {\lambda _m}\psi > \alpha }\\ &{}{D + \beta - \frac{\alpha }{{{\lambda _m}}} - \frac{\alpha }{{{\lambda _r}}} - \frac{\alpha }{{{\lambda _m}}}\ln \frac{{\beta \psi {\lambda _m}}}{\alpha } - \frac{\alpha }{{{\lambda _r}}}\ln \frac{{\beta \left( {1 - \psi } \right) {\lambda _r}}}{\alpha } - \alpha c > 0} \end{array} \right. \end{aligned}$$

Thus, the channel’s profit in the centralized decision-making system has been shown in Eq. (15). $\square$

1.2 Proof of Corollary 4

Proof

The supply chain optimal retail price ${p_1}^o$ can be denoted in Eq. (14), while the optimum decision $p^*$ for the retailer in decentralized decision-making system has been shown in Eq. (9).

$$\begin{aligned} {p^*} - p^o= & {} \frac{{3D + 3\beta - \frac{{3\alpha }}{{{\lambda _m}}} - \frac{{3\alpha }}{{{\lambda _r}}} + \frac{\alpha }{{{\lambda _m}}}\ln \frac{{\beta {\lambda _m}\psi }}{\alpha } + \frac{\alpha }{{{\lambda _r}}}\ln \frac{{\beta {\lambda _r}\left( {1 - \psi } \right) }}{\alpha } + \alpha c}}{{4\alpha }} \nonumber \\&-\frac{{2D + 2\beta - \frac{{2\alpha }}{{{\lambda _m}}} - \frac{{2\alpha }}{{{\lambda _r}}} + \frac{{2\alpha }}{{{\lambda _m}}}\ln \frac{{\beta {\lambda _m}\psi }}{\alpha } + \frac{{2\alpha }}{{{\lambda _r}}}\ln \frac{{\beta {\lambda _r}\left( {1 - \psi } \right) }}{\alpha } + 2\alpha c}}{{4\alpha }}\nonumber \\= & {} \frac{{D + \beta - \frac{\alpha }{{{\lambda _m}}} - \frac{\alpha }{{{\lambda _r}}} - \frac{\alpha }{{{\lambda _m}}}\ln \frac{{\beta {\lambda _m}\psi }}{\alpha } - \frac{\alpha }{{{\lambda _r}}}\ln \frac{{\beta {\lambda _r}\left( {1 - \psi } \right) }}{\alpha } - \alpha c}}{{4\alpha }}\nonumber \\= & {} \frac{{D + \beta - \frac{\alpha }{{{\lambda _m}}} - \frac{\alpha }{{{\lambda _r}}} + \frac{\alpha }{{{\lambda _m}}}\ln \frac{{\beta {\lambda _m}\psi }}{\alpha } - \frac{\alpha }{{{\lambda _r}}}\ln \frac{{\beta {\lambda _r}\left( {1 - \psi } \right) }}{\alpha } + \alpha c}}{{4\alpha }} - \frac{{\frac{\alpha }{{{\lambda _m}}}\ln \frac{{\beta {\lambda _m}\psi }}{\alpha } + \alpha c}}{{2\alpha }}\nonumber \\= & {} \frac{{{w^*}}}{2} - \frac{{x_m^* + c}}{2} > 0\\ \varPi ^o - {\varPi ^*}= & {} \varPi ^o - \left( {\pi _m^* + \pi _r^*} \right) \nonumber \\= & {} \frac{1}{\alpha }{\left( {\frac{{D + \beta - \frac{\alpha }{{{\lambda _m}}} - \frac{\alpha }{{{\lambda _r}}} - \frac{\alpha }{{{\lambda _m}}}\ln \frac{{\beta {\lambda _m}\psi }}{\alpha } - \frac{\alpha }{{{\lambda _r}}}\ln \frac{{\beta {\lambda _r}\left( {1 - \psi } \right) }}{\alpha } - \alpha c}}{2}} \right) ^2} \nonumber \\&- \frac{3}{\alpha }{\left( {\frac{{D + \beta - \frac{\alpha }{{{\lambda _m}}} - \frac{\alpha }{{{\lambda _r}}} - \frac{\alpha }{{{\lambda _m}}}\ln \frac{{\beta \psi {\lambda _m}}}{\alpha } - \frac{\alpha }{{{\lambda _r}}}\ln \frac{{\beta \left( {1 - \psi } \right) {\lambda _r}}}{\alpha } - \alpha c}}{4}} \right) ^2}\nonumber \\= & {} \frac{1}{\alpha }{\left( {\frac{{D + \beta - \frac{\alpha }{{{\lambda _m}}} - \frac{\alpha }{{{\lambda _r}}} - \frac{\alpha }{{{\lambda _m}}}\ln \frac{{\beta \psi {\lambda _m}}}{\alpha } - \frac{\alpha }{{{\lambda _r}}}\ln \frac{{\beta \left( {1 - \psi } \right) {\lambda _r}}}{\alpha } - \alpha c}}{4}} \right) ^2} =\frac{1}{3}{\varPi ^*} \end{aligned}$$

Hence, ${p^*} > p^o$, while $\varPi ^o = \frac{4}{3}{\varPi ^*}$ $\square$

Appendix 4: Proofs of Propositions and corollaries in Sect. 4.3

1.1 Proof of Proposition 3

Proof

Substituting $w = \left( {1 - \phi } \right) p + \phi c$ and $\phi = \frac{{{x_r^*}}}{{{x_m^*} + {x_r^*}}}$ into $\pi _r$ and reducing, we have:

$$\begin{aligned} {\pi _r}= & {} \left( {p - w - {x_r}} \right) \left[ {D + \beta - \alpha p - \beta \left[ {\psi {{\mathbf {e}}^{ - {\lambda _m}{x_m}}} + \left( {1 - \psi } \right) {{\mathbf {e}}^{ - {\lambda _r}{x_r}}}} \right] } \right] = \phi {\varPi ^o}\\ {\pi _m}= & {} (1-\phi ){\varPi ^o} \end{aligned}$$

So the individual profits affine to the channelwide one has been achieved. Therefor, we coordinate the supply chain with a fixed allocation. $\square$

1.2 Proof of Corollary 5

Proof

Under the centralized decision-making system with channel coordination, the retailer’s profit can be denoted as Eq. (17), and the manufacturer’s profit can be denoted as Eq. (18), while under the decentralized decision-making system, the manufacturer’s profit can be denoted as:

$$\begin{aligned} \pi _m^* = \frac{2}{\alpha }{\left( {\frac{{D + \beta - \frac{\alpha }{{{\lambda _m}}} - \frac{\alpha }{{{\lambda _r}}} - \frac{\alpha }{{{\lambda _m}}}\ln \frac{{\beta \psi {\lambda _m}}}{\alpha } - \frac{\alpha }{{{\lambda _r}}}\ln \frac{{\beta \left( {1 - \psi } \right) {\lambda _r}}}{\alpha } - \alpha c}}{4}} \right) ^2} \end{aligned}$$

And the retailer’s profit can be denoted as:

$$\begin{aligned} \pi _r^* = \frac{1}{\alpha }{\left( {\frac{{D + \beta - \frac{\alpha }{{{\lambda _m}}} - \frac{\alpha }{{{\lambda _r}}} - \frac{\alpha }{{{\lambda _m}}}\ln \frac{{\beta \psi {\lambda _m}}}{\alpha } - \frac{\alpha }{{{\lambda _r}}}\ln \frac{{\beta \left( {1 - \psi } \right) {\lambda _r}}}{\alpha } - \alpha c}}{4}} \right) ^2} \end{aligned}$$

If $\pi _r^c > \pi _r^*$, $\pi _m^c > \pi _m^*$:

$$\begin{aligned} \pi _r^c - \pi _r^*= & {} \frac{{3\phi - 1}}{\alpha }{\left( {\frac{{D + \beta - \frac{\alpha }{{{\lambda _m}}} - \frac{\alpha }{{{\lambda _r}}} - \frac{\alpha }{{{\lambda _m}}}\ln \frac{{\beta \psi {\lambda _m}}}{\alpha } - \frac{\alpha }{{{\lambda _r}}}\ln \frac{{\beta \left( {1 - \psi } \right) {\lambda _r}}}{\alpha } - \alpha c}}{4}} \right) ^2} > 0\\ \pi _m^c - \pi _m^*= & {} \frac{{2 - 4\phi }}{\alpha }{\left( {\frac{{D + \beta - \frac{\alpha }{{{\lambda _m}}} - \frac{\alpha }{{{\lambda _r}}} - \frac{\alpha }{{{\lambda _m}}}\ln \frac{{\beta \psi {\lambda _m}}}{\alpha } - \frac{\alpha }{{{\lambda _r}}}\ln \frac{{\beta \left( {1 - \psi } \right) {\lambda _r}}}{\alpha } - \alpha c}}{4}} \right) ^2} > 0 \end{aligned}$$

So $\frac{1}{3} < \phi < \frac{1}{2}$, $\pi _m^c > \pi _m^*$ and $\pi _r^c > \pi _r^*$ are satisfied. $\square$