The government subsidy design considering the reference price effect in a green supply chain

Abstract

This paper constructs a green supply chain with a manufacturer and a retailer. Taking into account the reference price effect of consumers based on the mental accounting theory, we investigate the following government incentive policies: R&D (research and development) subsidy, consumption subsidy, and dual subsidy. For manufacturer-led (M-led) and retailer-led (R-led) supply chains, we evaluate the optimal wholesale price, sales price, green degree of product, and the optimal subsidy of the government aiming to improve the environmental benefit or social welfare. We find that the government goal, power structure and reference price effect impact the design of subsidy mechanisms significantly. First, for M-led supply chain, the government concerned with the environmental benefit goal should only provide R&D subsidy for the manufacturer when the reference price effect is low; otherwise, the government would offer subsidy both for the manufacturer and consumers. However, the government will only offer R&D subsidy when the social welfare goal is adopted. Second, for R-led supply chain, the government aiming to improve the environmental benefit prefers dual subsidy when the reference price effect is low; otherwise, consumption subsidy is preferable. Surprisingly, under the social welfare goal, no subsidy for R-led supply chain tends to be the best option. Intriguingly, embracing the social welfare goal can result in more economic and environmental benefits for M-led supply chain, although the subsidy strategy is less effective than the environmental benefit goal. Our research can provide inspirations and references for designing government subsidy mechanisms in practice.

Appendix

Proofs of Table 3

Non-subsidy case, for the M-led supply chain, we can obtain the optimal strategies of the manufacturer and retailer based on the Stackelberg game and optimization theories. We know $\pi_{m0}^{M} = w_{m0} d_{m0} - \frac{1}{2}g_{m0}^{2}$, $\pi_{m0}^{R} = e_{m0} d_{m0}$, $\, d_{m0} = 1 - p_{m0} + g_{m0}$.

• (1) We can obtain $\frac{{\partial \pi_{m0}^{R} }}{{\partial e_{m0} }} = 1 - w_{m0} - 2e_{m0} + g_{m0}$ and $\frac{{\partial^{2} \pi_{m0}^{R} }}{{\partial e_{m0}^{2} }} = - 2 < 0$. Thus $\pi_{m0}^{R}$ is a strictly concave function of $e_{m0}$. Let $\frac{{\partial \pi_{m0}^{R} }}{{\partial e_{m0} }} = 0$, we have $e_{m0}^{\# } = \frac{{1 - w_{m0} + g_{m0} }}{2}$. It is easy to show that $\pi_{m0}^{M} = w_{m0} (1 - w_{m0} - \frac{{1 - w_{m0} + g_{m0} }}{2} + g_{m0} + \theta f_{m0} ) - \frac{1}{2}g_{m0}^{2}$.

• (2) We can obtain $\frac{{\partial \pi_{m0}^{M} }}{{\partial w_{m0} }} = \frac{{1 - 2w_{m0} + g_{m0} }}{2}$, $\frac{{\partial \pi_{m0}^{M} }}{{\partial g_{m0} }} = \frac{{w_{m0} - 2g{}_{m0}}}{2}$ and $H(w_{m0} ,g{}_{m0}) = \left[ {\begin{array}{*{20}c} { - 1} & \frac{1}{2} \\ \frac{1}{2} & 1 \\ \end{array} } \right]$. We can know easily $\frac{{\partial^{2} \pi_{m0}^{M} }}{{\partial w_{m0}^{2} }} < 0$, $\frac{{\partial^{2} \pi_{m0}^{M} }}{{\partial g_{m0}^{2} }} < 0$, $\det H(w_{m0} ,g{}_{m0}) = \frac{3}{4} > 0$. Thus we have $w_{m0}^{\# } = \frac{2}{3}$ and $g_{m0}^{\# } = \frac{1}{3}$ by solving $\frac{{\partial \pi_{m0}^{M} }}{{\partial w_{m0} }} = 0$ and $\frac{{\partial \pi_{m0}^{M} }}{{\partial g_{m0} }} = 0$. Therefore $p_{m0}^{*} = 1$, $e_{m0}^{*} = \frac{1}{3}$, $d_{m0}^{*} = \frac{1}{3}$, $\pi_{m0}^{M*} = \frac{1}{6}$, $\pi_{m0}^{R*} = \frac{1}{9}$.

Similarly, for the R-led supply chain, we can also the results by this method, so we will not repeat it here.

Proofs of Table 4

Under environmental benefit goal, for the M-led supply chain, we have $\pi_{m1}^{M} = w_{m1} d_{m1} - \frac{1}{2}g_{m1}^{2} + s_{m1} g_{m1}$, $\pi_{m1}^{R} = e_{m1} d_{m1}$, $\, d_{m1} = 1 - p_{m1} + g_{m1} + \theta f_{m1}$.

• (1) We can obtain $\frac{{\partial \pi_{m1}^{R} }}{{\partial e_{m1} }} = 1 - w_{m1} - 2e_{m1} + g_{m1} + \theta f_{m1}$ and $\frac{{\partial^{2} \pi_{m1}^{R} }}{{\partial e_{m1}^{2} }} = - 2 < 0$. Thus $\pi_{m1}^{R}$ is a strictly concave function of $e_{m1}$. Let $\frac{{\partial \pi_{m1}^{R} }}{{\partial e_{m1} }} = 0$, we have $e_{m1}^{\# } = \frac{{1 - w_{m1} + g_{m1} + \theta f_{m1} }}{2}$. It is easy to show that $\pi_{m1}^{M} = w_{m1} (1 - w_{m1} - \frac{{1 - w_{m1} + g_{m1} + \theta f_{m1} }}{2} + g_{m1} + \theta f_{m1} ) - \frac{1}{2}g_{m1}^{2} + s_{m1} g_{m1}$.

• (2) We can obtain $\frac{{\partial \pi_{m1}^{M} }}{{\partial w_{m1} }} = \frac{{1 - 2w_{m1} + g_{m1} + \theta f_{m1} }}{2}$, $\frac{{\partial \pi_{m1}^{M} }}{{\partial g_{m1} }} = \frac{{w_{m1} - 2g{}_{m1} + 2s_{m1} }}{2}$ and $H(w_{m1} ,g{}_{m1}) = \left[ {\begin{array}{*{20}c} { - 1} & \frac{1}{2} \\ \frac{1}{2} & 1 \\ \end{array} } \right]$. We can know easily $\frac{{\partial^{2} \pi_{m1}^{M} }}{{\partial w_{m1}^{2} }} < 0$, $\frac{{\partial^{2} \pi_{m1}^{M} }}{{\partial g_{m1}^{2} }} < 0$, $\det H(w_{m1} ,g{}_{m1}) = \frac{3}{4} > 0$. Thus we have $w_{m1}^{\# } = \frac{{2(1 + s_{m1} + \theta f_{m1} )}}{3}$ and $g_{m1}^{\# } = \frac{{1 + 4s_{m1} + \theta f_{m1} }}{3}$ by solving $\frac{{\partial \pi_{m1}^{M} }}{{\partial w_{m1} }} = 0$ and $\frac{{\partial \pi_{m1}^{M} }}{{\partial g_{m1} }} = 0$.

• (3) Hence, we have the following model, Next we use KKT conditions to solve the model.

$\begin{gathered} Max \, U(s_{m1} ,f_{m1} ) = \frac{{(\theta^{2} - 3\theta )f_{m1}^{2} }}{9} + \frac{{[\theta (2s_{m1} + 2) - 3s_{m1} - 3]f_{m1} }}{9} - \frac{{8s_{m1}^{2} }}{9} + \frac{{2s_{m1} }}{9} + \frac{1}{9} \hfill \\ \begin{array}{*{20}c} {} & {} \\ \end{array} s.t.\left\{ \begin{gathered} s_{m1} \ge 0 \, \lambda_{1} \ge 0 \hfill \\ f_{m1} \ge 0 \, \lambda_{2} \ge 0 \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered}$.

We have $\nabla U(s_{m1} ,f_{m1} ) = \left( \begin{gathered} \frac{{(2\theta - 3)f_{m1} }}{9} - \frac{{16s_{m1} }}{9} + \frac{2}{9} \hfill \\ \frac{{2\theta (\theta - 3)f_{m1} }}{9} + \frac{{2\theta (s_{m1} + 2)}}{9} - \frac{{s_{m1} }}{3} - \frac{1}{3} \hfill \\ \end{gathered} \right)$. Then the optimization problem is characterized by the KKT conditions as follows:

$\left\{ \begin{gathered} \frac{{(2\theta - 3)f_{m1} }}{9} - \frac{{16s_{m1} }}{9} + \frac{2}{9} + \lambda_{1} = 0 \hfill \\ \frac{{2\theta (\theta - 3)f_{m1} }}{9} + \frac{{2\theta (s_{m1} + 2)}}{9} - \frac{{s_{m1} }}{3} - \frac{1}{3} + \lambda_{2} = 0 \hfill \\ s_{m1} \ge 0 \, \lambda_{1} \ge 0 \, s_{m1} \lambda_{1} = 0 \hfill \\ f_{m1} \ge 0 \, \lambda_{2} \ge 0 \, f_{m1} \lambda_{2} = 0 \hfill \\ \end{gathered} \right.$.

When $\lambda_{1} = 0$, $\lambda_{2} = 0$, it is easy to show that $s_{m1}^{*} = \frac{ - 1}{{4\theta^{2} - 12\theta + 1}}$, $f_{m1}^{*} = \frac{2(3 - 2\theta )}{{4\theta^{2} - 12\theta + 1}}$, and $\frac{3}{2} \le \theta < 2$. Hence, $w_{m1}^{*} = \frac{ - 4\theta }{{4\theta^{2} - 12\theta + 1}}$, $g_{m1}^{*} = \frac{ - 2\theta - 1}{{4\theta^{2} - 12\theta + 1}}$, $e_{m1}^{*} = \frac{ - 2\theta }{{4\theta^{2} - 12\theta + 1}}$, $d_{m1}^{*} = \frac{ - 2\theta }{{4\theta^{2} - 12\theta + 1}}$, $p_{m1}^{*} = \frac{ - 6\theta }{{4\theta^{2} - 12\theta + 1}}$, $\pi_{m1}^{M*} = \frac{{12\theta^{2} + 1}}{{2\left( {4\theta^{2} - 12\theta + 1} \right)^{2} }}$, and $\pi_{m1}^{R*} = \frac{{4\theta^{2} }}{{\left( {4\theta^{2} - 12\theta + 1} \right)^{2} }}$.

When $\lambda_{1} = 0$, $\lambda_{2} > 0$, namely $f_{m1}^{*} = 0$. It is easy to show that $s_{m1}^{*} = \frac{1}{8}$, $1 \le \theta < \frac{3}{2}$. Hence, $w_{m1}^{*} = \frac{3}{4}$, ${\text{g}}_{m1}^{*} = \frac{1}{2}$, $e_{m1}^{*} = \frac{3}{8}$, $d_{m1}^{*} = \frac{3}{8}$, $p_{m1}^{*} = \frac{9}{8}$, $\pi_{m1}^{M*} = \frac{7}{32}$, and $\pi_{m1}^{R*} = \frac{9}{64}$.

When $\lambda_{1} > 0$, $\lambda_{2} = 0$, namely $s_{m1}^{*} = 0$. It is easy to show that $f_{m1}^{*} = \frac{3 - 2\theta }{{2\theta (\theta - 3)}}$, but $\lambda_{1} = \frac{1}{2\theta (\theta - 3)} < 0$, thus the result is not the optimal solution.

When $\lambda_{1} > 0$, $\lambda_{2} > 0$, namely $s_{m1}^{*} = 0$, $f_{m1}^{*} = 0$. It is easy to show that $\lambda_{1} = - \frac{2}{9} < 0$, $\lambda_{2} = \frac{3 - 2\theta }{9}$, thus the result is not the optimal solution.

Similarly, for the R-led supply chain, we can also the results by this method, so we will not repeat it here.

Proofs of Table 5

Under social welfare goal, for the M-led supply chain, we have $\pi_{m2}^{M} = w_{m2} d_{m2} - \frac{1}{2}g_{m2}^{2} + s_{m2} g_{m2}$, $\pi_{m2}^{R} = e_{m2} d_{m2}$, $\, d_{m2} = 1 - p_{m2} + g_{m2} + \theta f_{m2}$.

• (4) We can obtain $\frac{{\partial \pi_{m2}^{R} }}{{\partial e_{m2} }} = 1 - w_{m2} - 2e_{m2} + g_{m2} + \theta f_{m2}$ and $\frac{{\partial^{2} \pi_{m2}^{R} }}{{\partial e_{m2}^{2} }} = - 2 < 0$. Thus $\pi_{m2}^{R}$ is a strictly concave function of $e_{m2}$. Let $\frac{{\partial \pi_{m2}^{R} }}{{\partial e_{m2} }} = 0$, we have $e_{m2}^{\# } = \frac{{1 - w_{m2} + g_{m2} + \theta f_{m2} }}{2}$. It is easy to show that $\pi_{m2}^{M} = w_{m2} (1 - w_{m2} - \frac{{1 - w_{m2} + g_{m2} + \theta f_{m2} }}{2} + g_{m2} + \theta f_{m2} ) - \frac{1}{2}g_{m2}^{2} + s_{m2} g_{m2}$.

• (5) We can obtain $\frac{{\partial \pi_{m2}^{M} }}{{\partial w_{m2} }} = \frac{{1 - 2w_{m2} + g_{m2} + \theta f_{m2} }}{2}$, $\frac{{\partial \pi_{m2}^{M} }}{{\partial g_{m2} }} = \frac{{w_{m2} - 2g_{m2} + 2s_{m2} }}{2}$ and $H(w_{m2} ,g_{m2} ) = \left[ {\begin{array}{*{20}c} { - 1} & \frac{1}{2} \\ \frac{1}{2} & 1 \\ \end{array} } \right]$. We can know easily $\frac{{\partial^{2} \pi_{m2}^{M} }}{{\partial w_{m2}^{2} }} < 0$, $\frac{{\partial^{2} \pi_{m2}^{M} }}{{\partial g_{m2}^{2} }} < 0$, $\det H(w_{m2} ,g{}_{m2}) = \frac{3}{4} > 0$. Thus we have $w_{m2}^{\# } = \frac{{2(1 + s_{m2} + \theta f_{m2} )}}{3}$ and $g_{m2}^{\# } = \frac{{1 + 4s_{m2} + \theta f_{m2} }}{3}$ by solving $\frac{{\partial \pi_{m2}^{M} }}{{\partial w_{m2} }} = 0$ and $\frac{{\partial \pi_{m2}^{M} }}{{\partial g_{m2} }} = 0$.

• (6) Hence, we have the following model, Next we use KKT conditions to solve the model.

$\begin{gathered} Max \, U(s_{m2} ,f_{m2} ) = \frac{{(8\theta^{2} - 6\theta )f_{m2}^{2} }}{18} + \frac{{8(\theta - \frac{3}{8})(s_{m2} + 1)f}}{9} - \frac{{s_{m2}^{2} }}{18} + \frac{{8s_{m2} }}{9} + \frac{4}{9} \hfill \\ \begin{array}{*{20}c} {} & {} \\ \end{array} s.t.\left\{ \begin{gathered} s_{m2} \ge 0 \, \lambda_{1} \ge 0 \hfill \\ f_{m2} \ge 0 \, \lambda_{2} \ge 0 \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered}$.

We have $\nabla U(s_{m2} ,f_{m2} ) = \left( \begin{gathered} \frac{{(8\theta - 3)f_{m2} }}{9} - \frac{{s_{m2} }}{9} + \frac{8}{9} \hfill \\ \frac{{8f_{m2} \theta^{2} }}{9} + \frac{{( - 6f_{m2} + 8s_{m2} + 8)\theta }}{9} - \frac{{s_{m2} }}{3} - \frac{1}{3} \hfill \\ \end{gathered} \right)$. Then the optimization problem is characterized by the KKT conditions as follows:

$\left\{ \begin{gathered} \frac{{(8\theta - 3)f_{m2} }}{9} - \frac{{s_{m2} }}{9} + \frac{8}{9} + \lambda_{1} = 0 \hfill \\ \frac{{8f_{m2} \theta^{2} }}{9} + \frac{{( - 6f_{m2} + 8s_{m2} + 8)\theta }}{9} - \frac{{s_{m2} }}{3} - \frac{1}{3} + \lambda_{2} = 0 \hfill \\ s_{m2} \ge 0 \, \lambda_{1} \ge 0 \, s_{m2} \lambda_{1} = 0 \hfill \\ f_{m2} \ge 0 \, \lambda_{2} \ge 0 \, f_{m2} \lambda_{2} = 0 \hfill \\ \end{gathered} \right.$.

When $\lambda_{1} = 0$, $\lambda_{2} = 0$, it is easy to show that $s_{m2}^{*} = \frac{ - 1}{{8\theta^{2} - 6\theta + 1}} < 0$, $f_{m2}^{*} = \frac{ - 8\theta + 3}{{8\theta^{2} - 6\theta + 1}} < 0$, thus the result is not the optimal solution.

When $\lambda_{1} = 0$, $\lambda_{2} > 0$, namely $f_{m2}^{*} = 0$. It is easy to show that $s_{m2}^{*} = 8$, $f_{m2}^{*} = 0$. Hence, $w_{m2}^{*} = 6$, ${\text{g}}_{m2}^{*} = 11$, $e_{m2}^{*} = 3$, $d_{m2}^{*} = 3$, $p_{m2}^{*} = 9$, $\pi_{m2}^{M*} = \frac{91}{2}$, $\pi_{m2}^{R*} = 9$.

When $\lambda_{1} > 0$, $\lambda_{2} = 0$, namely $s_{m2}^{*} = 0$. It is easy to show that $f_{m2}^{*} = \frac{ - 8\theta + 3}{{2\theta (4\theta - 3)}} < 0$, thus the result is not the optimal solution.

When $\lambda_{1} > 0$, $\lambda_{2} > 0$, namely $s_{m2}^{*} = 0$, $f_{m2}^{*} = 0$. It is easy to show that $\lambda_{1} = - \frac{8}{9} < 0$, $\lambda_{2} = = \frac{3 - 8\theta }{9} < 0$, thus the result is not the optimal solution.

Similarly, for the R-led supply chain, we can also the results by this method, so we will not repeat it here.

Proofs of propositions 1 and 2

According to Tables 4 and 5, we can easily know that Propositions 1 and 2 hold, thus we will not repeat them here.

Proofs of Proposition 3

For M-led supply chain, when $1 \le \theta \le \frac{3}{2}$, it is easy to know that $g_{m0}^{*} < g_{m1}^{*} < g_{m2}^{*}$; when $\frac{3}{2} < \theta < 2$, $g_{m2}^{*} - g_{m1}^{*} = \frac{{2(22\theta^{2} - 65\theta + 6)}}{{(2\theta - 3)^{2} - 8}} > 0$, $g_{m1}^{*} - g_{m0}^{*} = - \frac{{2(2\theta^{2} - 3\theta + 2)}}{{3[(2\theta - 3)^{2} - 8]}} > 0$. Hence, $g_{m0}^{*} < g_{m1}^{*} < g_{m2}^{*}$ holds. Similarity, when $1 \le \theta \le \frac{3}{2}$, it is easy to know that $d_{m0}^{*} < d_{m1}^{*} < d_{m2}^{*}$, when $\frac{3}{2} < \theta < 2$, $d_{m2}^{*} - d_{m1}^{*} = \frac{{12\theta^{2} - 34\theta + 3}}{{(2\theta - 3)^{2} - 8}} > 0$, $d_{m1}^{*} - d_{m0}^{*} = - \frac{{4\theta^{2} - 6\theta + 1}}{{3[(2\theta - 3)^{2} - 8]}} > 0$. Hence, $d_{m0}^{*} < d_{m1}^{*} < d_{m2}^{*}$ holds.

For R-led supply chain, when $1 \le \theta < 1 + \frac{\sqrt 3 }{2}$, $g_{r1}^{*} - g_{r2}^{*} = - \frac{{(2\theta - 1)^{2} + 2}}{{2(4\theta^{2} - 8\theta + 1)}} > 0$, thus $g_{r0}^{*} = g_{r2}^{*} < g_{r1}^{*}$. When $1 + \frac{\sqrt 3 }{2} \le \theta < 2$, $d_{r1}^{*} - d_{r2}^{*} = \frac{1 - \theta }{{2(\theta - 2)}} > 0$, we have $d_{r0}^{*} = d_{r2}^{*} < d_{r1}^{*}$.

The other proofs are similar to these, we will not repeat them here.

Proofs of proposition 4

When $\frac{3}{2} < \theta < 2$,$\frac{{\partial g_{m1}^{*} }}{\partial \theta } = \frac{8(\theta + 2)(\theta - 1) + 2}{{(4\theta^{2} - 12\theta + 1)^{2} }} > 0$, $\frac{{\partial d_{m1}^{*} }}{\partial \theta } = \frac{{8\theta^{2} - 2}}{{(4\theta^{2} - 12\theta + 1)^{2} }} > 0$. When $1 \le \theta < 1 + \frac{\sqrt 3 }{2}$, $\frac{{\partial g_{r1}^{*} }}{\partial \theta } = \frac{{2[(2\theta^{2} + 1)^{2} - 6]}}{{(4\theta^{2} - 8\theta + 1)^{2} }} > 0$, $\frac{{\partial d_{r1}^{*} }}{\partial \theta } = \frac{{8\theta^{2} - 2}}{{(4\theta^{2} - 8\theta + 1)^{2} }} > 0$. When $1 + \frac{\sqrt 3 }{2} \le \theta < 2$, $\frac{{\partial g_{r1}^{*} }}{\partial \theta } = \frac{1}{{2(\theta - 2)^{2} }} > 0$, $\frac{{\partial d_{r1}^{*} }}{\partial \theta } = \frac{1}{{2(\theta - 2)^{2} }} > 0$.

Proofs of proposition 5

When $\frac{3}{2} < \theta < 2$, $\frac{{\partial \pi_{m1}^{M*} }}{\partial \theta } = \frac{{ - 4(12\theta^{3} - \theta - 3)}}{{(4\theta^{2} - 12\theta + 1)^{3} }} > 0$, $\frac{{\partial \pi_{m1}^{R*} }}{\partial \theta } = \frac{{ - 8\theta (4\theta^{2} - 1)}}{{(4\theta^{2} - 12\theta + 1)^{3} }} > 0$. When $1 \le \theta < \frac{1 + \sqrt 3 }{2}$, $\frac{{\partial \pi_{r1}^{M*} }}{\partial \theta } = \frac{{ - 4(4\theta^{3} + \theta - 2)}}{{\left( {4\theta^{2} - 8\theta + 1} \right)^{3} }} > 0$, $\frac{{\partial \pi_{r1}^{R*} }}{\partial \theta } = \frac{{ - 8\theta (4\theta^{2} + 1)}}{{(4\theta^{2} - 8\theta + 1)^{3} }} > 0$. When $1 + \frac{\sqrt 3 }{2} \le \theta < 2$, $\frac{{\partial \pi_{r1}^{R*} }}{\partial \theta } = \frac{1}{{2(2 - \theta )^{3} }} > 0$.

Proofs of proposition 6

According to the proofs of Proposition 4, we can know $\frac{{\partial CS_{m1}^{*} }}{\partial \theta } > 0$ and $\frac{{\partial CS_{r1}^{*} }}{\partial \theta } > 0$ hold.

When $\frac{3}{2} < \theta < 2$, we have $\frac{{\partial EB_{m1}^{*} }}{\partial \theta } = \frac{{8\theta (4\theta^{2} + 3\theta - 4) - 2}}{{ - (4\theta^{2} - 12\theta + 1)^{3} }}$. Let $f(\theta ) = 4\theta^{2} + 3\theta - 4$, since $f^{\prime}(\theta ) = 8\theta + 3 > 0$, $f(\theta ) > f(\frac{3}{2}) = \frac{19}{2}$, we can easily know that $8\theta (4\theta^{2} + 3\theta - 4) - 2 > 76\theta - 2 > 0$. We also know that $- (4\theta^{2} - 12\theta + 1)^{3} > 0$, thus $\frac{{\partial EB_{m1}^{*} }}{\partial \theta } > 0$ holds.

Similarly, we have $\frac{{\partial SW_{m1}^{*} }}{\partial \theta } = \frac{{8\theta (8\theta^{2} + 18\theta - 21)}}{{ - (4\theta^{2} - 12\theta + 1)^{3} }} > 0$. When $1 \le \theta < 1 + \frac{\sqrt 3 }{2}$, $\frac{{\partial EB_{r1}^{*} }}{\partial \theta } = \frac{{8\theta (4\theta^{2} + 3\theta - 3) - 2}}{{ - (4\theta^{2} - 8\theta + 1)^{3} }} > 0$, $\frac{{\partial SW_{r1}^{*} }}{\partial \theta } = \frac{{8\theta (4\theta^{2} + 12\theta - 10)}}{{ - (4\theta^{2} - 8\theta + 1)^{3} }} > 0$. When $1 + \frac{\sqrt 3 }{2} \le \theta < 2$, $\frac{{\partial EB_{r1}^{*} }}{\partial \theta } = \frac{1}{{2(2 - \theta )^{3} }} > 0$, $\frac{{\partial SW_{r1}^{*} }}{\partial \theta } = \frac{{\theta^{2} + 3\theta - 2}}{{2\theta^{2} (2 - \theta )^{3} }} > 0$.

Proofs of proposition 7

When $\frac{3}{2} < \theta < 2$, $\frac{{\partial s_{m1}^{*} }}{\partial \theta } = \frac{4(2\theta - 3)}{{(4\theta^{2} - 12\theta + 1)^{2} }} > 0$, $\frac{{\partial f_{m1}^{*} }}{\partial \theta } = \frac{{(4\theta - 6)^{2} + 32}}{{(4\theta^{2} - 12\theta + 1)^{2} }} > 0$. We have $\frac{{\partial S_{m1}^{*} }}{\partial \theta } = \frac{{8\theta \left( {8\theta^{2} - 15\theta + 15} \right) - 14}}{{ - (4\theta^{2} - 12\theta + 1)^{3} }}$. Let $f(\theta ) = 8\theta^{2} - 15\theta + 15$, since $f^{\prime}(\theta ) = 16\theta - 15 > 0$, $f(\theta ) > f\left( \frac{3}{2} \right) = \frac{21}{2}$, we can easily know that $8\theta \left( {8\theta^{2} - 15\theta + 15} \right) - 14 > 84\theta - 14 > 0$. We also know that $- (4\theta^{2} - 12\theta + 1)^{3} > 0$, thus $\frac{{\partial S_{m1}^{*} }}{\partial \theta } > 0$ holds.

When $1 \le \theta < 1 + \frac{\sqrt 3 }{2}$, $\frac{{\partial s_{r1}^{*} }}{\partial \theta } = \frac{8(\theta - 1)}{{(4\theta^{2} - 8\theta + 1)^{2} }} > 0$, $\frac{{\partial f_{r1}^{*} }}{\partial \theta } = \frac{{4(2\theta - 2)^{2} + 12}}{{(4\theta^{2} - 8\theta + 1)^{2} }} > 0$. We have $\frac{{\partial S_{r1}^{*} }}{\partial \theta } = \frac{{8\theta (8\theta^{2} - 9\theta + 6) - 10}}{{ - (4\theta^{2} - 8\theta + 1)^{3} }}$. Let $f(\theta ) = 8\theta^{2} - 9\theta + 6$, since $f^{\prime}(\theta ) = 16\theta - 9 > 0$, $f(\theta ) > f(1) = 5$, we can easily know that $8\theta (8\theta^{2} - 9\theta + 6) - 10 > 40\theta - 10 > 0$. We also know that $- (4\theta^{2} - 12\theta + 1)^{3} > 0$, thus $\frac{{\partial S_{r1}^{*} }}{\partial \theta } > 0$ holds. When $1 + \frac{\sqrt 3 }{2} \le \theta < 2$, $\frac{{\partial f_{r1}^{*} }}{\partial \theta } = \frac{{(\theta - 1)^{2} + 1}}{{\theta^{2} (\theta - 2)^{2} }} > 0$, $\frac{{\partial S_{r1}^{*} }}{\partial \theta } = \frac{2\theta (\theta - 1) + 2 - \theta }{{2\theta^{2} (2 - \theta )^{3} }} > 0$.

Proofs of proposition 8

When $\frac{3}{2} < \theta \le 2$, we have $\frac{{\partial \Lambda_{m1}^{*} }}{\partial \theta } = \frac{{8\{ (\theta - 1)[\theta (32\theta^{3} - 124\theta^{2} + 124\theta - 15) + 12] + 8\} }}{{ - 9(8\theta^{2} - 10\theta + 1)^{2} }}$. Let $f(\theta ) = 32\theta^{3} - 124\theta^{2} + 124\theta - 15$, then $f^{\prime}(\theta ) = 96\theta^{2} - 248\theta + 124 < 0$. When $\frac{3}{2} < \theta \le \frac{{31 + \sqrt {217} }}{24}$, $f^{\prime}(\theta ) < 0$; when $\frac{{31 + \sqrt {217} }}{24} < \theta \le 2$, $f^{\prime}(\theta ) > 0$, thus $f(\theta ) \ge f\left( {\frac{{31 + \sqrt {217} }}{24}} \right) > - 8$, we can obtain $(\theta - 1)[\theta (32\theta^{3} - 124\theta^{2} + 124\theta - 15) + 12] + 8 > - 8\theta^{2} + 20\theta - 4 > 0$. Thus $\frac{{\partial \Lambda^{*}_{m1} }}{\partial \theta } < 0$ holds. Similarly, when $1 \le \theta < 1 + \frac{\sqrt 3 }{2}$, $\frac{{\partial \Lambda^{*}_{r1} }}{\partial \theta } = \frac{{8(16\theta^{5} - 50\theta^{4} + 52\theta^{3} - 21\theta^{2} + 6\theta ) - 9}}{{ - 2(8\theta^{2} - 6\theta + 1)^{2} }} < 0$ holds. When $1 + \frac{\sqrt 3 }{2} \le \theta \le 2$, $\frac{{\partial \Lambda^{*}_{r1} }}{\partial \theta } = \frac{3}{2} - \theta < 0$.

When $\frac{3}{2} < \theta \le 2$, we have $\frac{{\partial \Omega_{m1}^{*} }}{\partial \theta } = \frac{834\theta + 265}{{72(8\theta^{2} - 10\theta + 1)^{2} }} + \frac{185}{{72(8\theta^{2} - 10\theta + 1)}} + \frac{95}{{36}} - \frac{10\theta }{9}$, we can easily know that $\frac{10\theta }{9} \le \frac{20}{9} < \frac{95}{{36}}$, $\frac{95}{{36}} - \frac{10\theta }{9} > \frac{95}{{36}} - \frac{20}{9} > 0$, $8\theta^{2} - 10\theta + 1 > 0$, thus $\frac{{\partial \Omega_{m1}^{*} }}{\partial \theta } > 0$ holds. When $1 \le \theta < 1 + \frac{\sqrt 3 }{2}$, $\frac{{\partial \Omega^{*}_{r1} }}{\partial \theta } = \frac{{ - 384\theta^{5} + 1200\theta^{4} - 1248\theta^{3} + 600\theta^{2} - 176\theta + 27}}{{4(8\theta^{2} - 6\theta + 1)^{2} }}$, let $\mathop \theta \limits^{ \wedge }$ meets $- 384\theta^{5} + 1200\theta^{4} - 1248\theta^{3} + 600\theta^{2} - 176\theta + 27 = 0$, when $1 \le \theta < \mathop \theta \limits^{ \wedge }$, $\frac{{\partial \Omega^{*}_{r1} }}{\partial \theta } \ge \frac{19}{{36}} > 0$, when $\mathop \theta \limits^{ \wedge } \le \theta < 1 + \frac{\sqrt 3 }{2}$, $\frac{{\partial \Omega^{*}_{r1} }}{\partial \theta } < 0$. When $1 + \frac{\sqrt 3 }{2} \le \theta \le 2$, $\frac{{\partial \Omega^{*}_{r1} }}{\partial \theta } = \frac{9 - 6\theta }{4} < 0$ (Tables 9, 10, 11, 12, 13, 14, and 15).

Table 9 The optimal strategies and profits without subsidy considering consumer green preference

Table 10 The optimal strategies and profits under the environmental benefit goal considering consumer green preference

Where $\hat{\theta } = \frac{{8 - 3k^{2} - k + 2\sqrt {(4 - k^{2} )(4 - 2k^{2} - k)} }}{{(k + 1)^{2} }}$.

Table 11 The optimal strategies and profits under the social welfare goal considering consumer green preference

Where $\Psi = (k^{2} - 8)^{2} - 48$ and $\Upsilon = 2 - k^{2}$.

Table 12 The performance of M-led supply chain under different subsidy scenarios considering consumer green preference

Table 13 The performance of R-led supply chain under different subsidy scenarios considering consumer green preference

Table 14 Comparison results under different scenarios considering consumer green preference

Table 15 The price transfer and demand difference with and without R&D subsidy