i> c m k t e m + 2 k t r m z r m + 2 k t β r s . (5)

The Hessian matrix is now $H\left(p,{r}_{m}\right)=\left[\begin{array}{cc}-2b& k-bt\\ k-bt& 2kt-z\end{array}\right]$; Once again, we use

the inequalities $z>\left[2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}\right]/2b$ and $k,b,t>0$, to find that the first-order principal minors ( $-2b<0$ and $2kt-z<0$ while the) and second-order principal minor ( $2bz-{\left(k+bt\right)}^{2}>0$.). Thus, the profit of the manufacturer can be expressed as a concave function of p and ${r}_{m}$.

Assuming that Equation (4) and Equation (5) can be set equal to 0 (equilibrium solution), the following values can be obtained by solving the resulting simultaneous equations:

${p}^{D}=\frac{a\left(z-kt+b{t}^{2}\right)+\left(w+{c}_{m}+t{e}_{m}\right)\left(bz-{k}^{2}-bkt\right)+\left(k-bt\right)z\beta {r}_{s}}{2bz-{\left(k+bt\right)}^{2}}$,

${r}_{m}^{D}=\frac{\left(k+bt\right)\left(a-bw-b{c}_{m}-bt{e}_{m}+k\beta {r}_{s}+bt\beta {r}_{s}\right)}{2bz-{\left(k+bt\right)}^{2}}$.

After substituting ${p}^{D}$ and ${r}_{m}^{D}$ into the profit function of the supplier, the values of w and ${r}_{s}$ can be derived in the first stage and the corresponding Hessian matrix elucidated:

$\begin{array}{l}H\left(w,{r}_{s}\right)\\ =\left[\begin{array}{cc}-\frac{2{b}^{2}z\left[2bz-{k}^{2}-bkt\left(2+\beta \right)-{b}^{2}{t}^{2}\left(1+\beta \right)\right]}{{\left[2bz-{\left(k+bt\right)}^{2}\right]}^{2}}& -\frac{bz\left\{\left[2bz-{\left(k+bt\right)}^{2}\right]\left[bt-\beta \left(k+bt\right)\right]+2bt{\beta }^{2}{\left(k+bt\right)}^{2}\right\}}{{\left[2bz-{\left(k+bt\right)}^{2}\right]}^{2}}\\ -\frac{bz\left\{\left[2bz-{\left(k+bt\right)}^{2}\right]\left[bt-\beta \left(k+bt\right)\right]+2bt{\beta }^{2}{\left(k+bt\right)}^{2}\right\}}{{\left[2bz-{\left(k+bt\right)}^{2}\right]}^{2}}& \frac{2zbt\beta \left(k+bt\right)\left[2bz-{\left(k+bt\right)}^{2}\left(1-{\beta }^{2}\right)\right]}{{\left[2bz-{\left(k+bt\right)}^{2}\right]}^{2}}-z\end{array}\right]\end{array}$.

As $z>\left[2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}\right]/2b$, $0\le \beta \le 1$, and $k,b,t>0$, the first-order principal minors

$-\left\{2{b}^{2}z\left[2bz-{k}^{2}-bkt\left(2+\beta \right)-{b}^{2}{t}^{2}\left(1+\beta \right)\right]\right\}/\left\{{\left[2bz-{\left(k+bt\right)}^{2}\right]}^{2}\right\}<0$

and

$\left\{2zbt\beta \left(k+bt\right)\left[2bz-{\left(k+bt\right)}^{2}\left(1-{\beta }^{2}\right)\right]\right\}/\left\{{\left[2bz-{\left(k+bt\right)}^{2}\right]}^{2}\right\}-z<0$,

and the second-order principal minor

$\left\{{b}^{2}{z}^{2}\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]\right\}/\left\{{\left[2bz-{\left(k+bt\right)}^{2}\right]}^{2}\right\}>0$

can be calculated. Therefore, the profit of the supplier can be expressed as a concave function of w and ${r}_{s}$.

Overall, we find that:

${w}^{D*}=\frac{\left(a-b{c}_{m}-bt{e}_{m}\right)\left[2bz-{k}^{2}-bt\left(k+bt\right)\left(2+3\beta \right)\right]+b\left({c}_{s}+t{e}_{s}\right)\left[2bz-\left(1+{\beta }^{2}\right){\left(k+bt\right)}^{2}-bt\beta \left(k+bt\right)\right]}{4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)}$,

${r}_{s}^{D*}=\frac{\left[k\beta +bt\left(1+\beta \right)\right]\left(a-b{c}_{m}-b{c}_{s}-bt{e}_{m}-bt{e}_{s}\right)}{4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)}$.

By inversely substituting the equilibrium solution into the second stage, the following equations can then be obtained

${p}^{D\text{*}}=\frac{\left[3bz-{k}^{2}-bkt\left(3+3\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]a+b\left[{c}_{m}+{c}_{s}+t\left({e}_{m}+{e}_{s}\right)\right]\left[bz-{k}^{2}\left(1+{\beta }^{2}\right)-bkt\left(1+\beta +{\beta }^{2}\right)\right]}{4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)}$,

${r}_{m}^{D\text{*}}=\frac{\left(k+bt\right)\left(a-b{c}_{m}-b{c}_{s}-bt{e}_{m}-bt{e}_{s}\right)}{4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)}$.

Furthermore, the equilibrium profits of the supplier, manufacturer, and supply chain can be expressed as:

${\pi }_{s}^{D\text{*}}=\frac{z{\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}^{2}}{2\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}$,

${\pi }_{m}^{D\text{*}}=\frac{z\left[2bz-{\left(k+bt\right)}^{2}\right]{\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}^{2}}{2{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{2}}$,

${\pi }^{D\text{*}}=\frac{z\left[6bz-{k}^{2}\left(3+{\beta }^{2}\right)-2bkt\left(3+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}{\left(2+\beta \right)}^{2}\right]{\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}^{2}}{2{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{2}}$.

The total carbon-emission level of the whole supply chain can be expressed as:

$\begin{array}{l}{E}^{D\text{*}}=\frac{bz\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}{{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }\text{\hspace{0.17em}}×\left\{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-bkt\left(3+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(1+\beta \right)\right]\left({e}_{m}+{e}_{s}\right)-\left(1+\beta \right)\left[k\left(1+\beta \right)+bt\left(2+\beta \right)\right]\left[a-b\left({c}_{m}+{c}_{s}\right)\right]\right\}.\end{array}$

Proposition 2:

1) When $\beta =1$, we have ${r}_{s}^{D\text{*}}>{r}_{m}^{D\text{*}}$; and when $\beta =0$, we have ${r}_{s}^{D\text{*}}<{r}_{m}^{D\text{*}}$; on the precondition of $0<\beta <1$, ${r}_{s}^{D\text{*}}>{r}_{m}^{D\text{*}}$ if $k and ${r}_{s}^{D\text{*}}\le {r}_{m}^{D\text{*}}$ if $k\ge bt\beta /\left(1-\beta \right)$.

2) ${\pi }_{s}^{D\text{*}}>{\pi }_{m}^{D\text{*}}$.

Proposition 2 indicates that for the decentralized decision-making case:

1) When the up-stream and down-stream enterprises overflow fully ( $\beta =1$ ), the emission-reduction effort level of the supplier is higher than that of the manufacturer. On the other hand, the reverse is true when the up-stream and down-stream enterprises do not overflow ( $\beta =0$ ). When there is some overflow ( $0<\beta <1$ ), the supplier (manufacturer) exhibits a higher effort level if the environmental awareness level of the consumers is at a low (high) level.

When the up-stream and down-stream enterprises completely overflow, the influence of the supplier on market demand is the same as that of the manufacturer. However, the manufacturer benefits from late-mover advantage and decreases the cost of emission reduction by free-riding. Therefore, compared to the supplier, the manufacturer has a lower emission-reduction effort level. In contrast, when the up-stream and down-stream enterprises do not overflow, market demand is only influenced by the emission-reduction effort of the manufacturer. Thus, the manufacturer makes a high level of effort in order to obtain a large market share.

In the general case where there is some overflow between the up-stream and down-stream enterprises, the manufacturer will prefer to decrease the cost of emission reduction by free-riding when the environmental awareness of the consumers is low. However, when consumers have high environmental awareness, the manufacturer will prefer to acquire more market share by improving their emission-reduction effort level.

2) Due to the first-mover advantage of the supplier, the profit of the supplier is larger than that of the manufacturer.

Proposition 3:

1) $\frac{\partial {r}_{s}^{D\text{*}}}{\partial \beta }>0,\frac{\partial {r}_{m}^{D\text{*}}}{\partial \beta }>0,\frac{\partial {\pi }_{s}^{D\text{*}}}{\partial \beta }>0,\frac{\partial {\pi }_{m}^{D\text{*}}}{\partial \beta }>0$;

2) $\frac{\partial {r}_{s}^{D\text{*}}}{\partial k}>0,\frac{\partial {r}_{m}^{D\text{*}}}{\partial k}>0,\frac{\partial {\pi }_{s}^{D\text{*}}}{\partial k}>0,\frac{\partial {\pi }_{m}^{D\text{*}}}{\partial k}>0$.

This proposition implies that for decentralized decision making, the emission-reduction effort levels and profits of both the supplier and manufacturer grow when the spillover rates and environmental awareness level of the consumers increase.

This indicates that the vertical spillover rate and environmental awareness level of the consumers favor a reduction in pollution and improvement in profit. Therefore, the up-stream and down-stream enterprises in the supply chain are supposed to share information and increase their knowledge and technology spillovers so as to improve the vertical spillover rate. Moreover, it is necessary that they highlight the importance of environmental protection, widely carry out environmentally-friendly practices, and improve the environmental awareness levels of consumers.

Proposition 4:

Comparing carbon emission under centralized and decentralized decision-making conditions we have:

1) ${e}^{C\text{*}}<{e}^{D\text{*}}$;

2) When the conditions are such that $b{t}^{2}{\left(1+\beta \right)}^{2}, we have ${E}^{C\text{*}}<{E}^{D\text{*}}$; while $z>{z}_{3}$ corresponds to ${E}^{C\text{*}}>{E}^{D\text{*}}$. Here, ${z}_{3}$ refers to the root of the function $F\left(z\right)$ and; (which is cubic with respect to z) such that ${z}_{3}\in \left(kt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)/2b+\text{b}{t}^{2}{\left(1+\beta \right)}^{2},+\infty \right)$, where:

$\begin{array}{c}F\left(z\right)={\left[2bz-2bt\left(k+bt\right){\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)\right]}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×\left\{\left(1+\beta \right)\left[k\left(1+\beta \right)+bt\left(2+\beta \right)\right]\left(a-b{c}_{m}-b{c}_{s}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-bkt\left(3+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(1+\beta \right)\right]\left({e}_{m}+{e}_{s}\right)\right\}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×\left\{\left(k+2bt\right){\left(1+\beta \right)}^{2}\left(a-b{c}_{m}-b{c}_{s}\right)-\left[2bz-bkt{\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)\right]\left({e}_{m}+{e}_{s}\right)\right\}\end{array}$

This proposition illustrates that:

1) The carbon-emission level per unit product in the centralized decision making case is lower than that in the decentralized case.

2) When the emission-reduction cost coefficient is at a low level, the total carbon emission in the centralized decision making case is lower than that in the decentralized case. This is because the quantity difference under centralized and decentralized conditions drops as the emission-reduction cost coefficients decrease ( $\partial \left({q}^{C*}-{q}^{D*}\right)/\partial z>0$ ), while the difference in the carbon-emission levels per unit product increases ( $\partial \left({e}^{D*}-{e}^{C*}\right)/\partial z<0$ ). Therefore, when the emission-reduction cost coefficient is at a low level, the quantity difference is small and the total carbon-emission value is mainly influenced by the carbon-emission level per unit product. This explains why the total carbon emission under centralized decision-making conditions is lower than that under decentralized conditions.

The total carbon emission is primarily influenced by the quantity when the emission-reduction cost coefficient is large. Thus, the total carbon emission under centralized decision making is larger than that under decentralized decision making.

Proposition 5:

Again comparing centralized and decentralized decision making cases, we also have:

1) ${r}_{s}^{C\text{*}}>{r}_{s}^{D\text{*}},{r}_{m}^{C\text{*}}>{r}_{m}^{D\text{*}},{\pi }^{C*}>{\pi }^{D*}$;

2) $\frac{\partial \left({r}_{s}^{C\text{*}}-{r}_{s}^{D\text{*}}\right)}{\partial \beta }>0,\frac{\partial \left({r}_{m}^{C\text{*}}-{r}_{m}^{D\text{*}}\right)}{\partial \beta }>0,\frac{\partial \left({\pi }^{C*}-{\pi }^{D*}\right)}{\partial \beta }>0$;

3) $\frac{\partial \left({r}_{s}^{C\text{*}}-{r}_{s}^{D\text{*}}\right)}{\partial k}>0,\frac{\partial \left({r}_{m}^{C\text{*}}-{r}_{m}^{D\text{*}}\right)}{\partial k}>0,\frac{\partial \left({\pi }^{C*}-{\pi }^{D*}\right)}{\partial k}>0$.

Proposition 5 indicates that:

1) The emission-reduction effort levels of the supplier and manufacturer and the profit of the supply chain in the centralized case are larger than those in the decentralized case.

2) With increasing vertical spillover rate, the difference in the emission-reduction effort levels of the supplier and manufacturer (as well as the profit of the whole supply chain) under centralized and decentralized decision-making conditions grows increasingly larger. That is, vertical spillover lowers the coordination efficiency $\mu ={\pi }^{D*}/{\pi }^{C*}$ of the supply chain. Therefore, it is particularly necessary to coordinate the supply chain if vertical spillover is operating.

3) With increasing environmental awareness of consumers, the difference in the emission-reduction effort levels of the supplier and manufacturer (as well as the profit of the supply chain) under centralized and decentralized decision-making conditions grows increasingly larger. Therefore, consistent with the effect of vertical spillover, environmental awareness also reduces the coordination efficiency of the supply chain.

5.3. The Nash Bargaining Model

It can be seen from the foregoing propositions that the profits in the centralized decision making case are higher than those in the decentralized case. However, it cannot be guaranteed that the members in the supply chain will always make decisions similar to those made in a centralized system. Only when the profit distribution satisfies the profit requirements of both enterprises will the cooperative relationship in the supply chain be able to reach a good state of coordination. By using a Nash bargaining game model, the coordination mechanism of the supply chain can be discussed under the assumption that the supplier and manufacturer negotiate a reasonable distribution ratio $\theta$ for the profit of the whole supply chain.

We assume that $\tau$ ( $0<\tau <1$ ) refers to the bargaining ability of the supplier, so that the bargaining ability of the manufacturer is $1-\tau$; If the profit of the supplier is ${\pi }_{s}^{B}=\left(1-\theta \right){\pi }^{C*}$, then the profit of the manufacturer is ${\pi }_{m}^{B}=\theta {\pi }^{C*}$. The coordination mechanism will only be accepted by the two parties when the Pareto improvement is performed on the supply chain, so it will inevitably satisfy ${\pi }_{s}^{B}\ge {\pi }_{s}^{D\text{*}}$ and ${\pi }_{m}^{B}\ge {\pi }_{m}^{D\text{*}}$. Guided by Nash   , the supplier and manufacturer jointly determine $\theta$ so that it maximizes the following objective function

$\underset{\theta }{\mathrm{max}}f\left(\theta \right)={\left({\pi }_{s}^{B}-{\pi }_{s}^{D\text{*}}\right)}^{\tau }{\left({\pi }_{m}^{B}-{\pi }_{m}^{D\text{*}}\right)}^{1-\tau }$

$\text{s}.\text{t}.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left\{\begin{array}{l}0<\theta <1\\ {\pi }_{s}^{B}\ge {\pi }_{s}^{D\text{*}}\\ {\pi }_{m}^{B}\ge {\pi }_{m}^{D\text{*}}\end{array}$

It can be shown that:

${\theta }^{*}=\frac{\left(1-\tau \right)\left({\pi }^{C*}-{\pi }^{D*}\right)+{\pi }_{m}^{D*}}{{\pi }^{C*}}.$

Substituting the equilibrium profits into the above equation gives

${\theta }^{\text{*}}=\frac{4\left(2-\tau \right){b}^{2}{z}^{2}-Mz+N}{{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{2}}$

where,

$\begin{array}{c}M=2b{k}^{2}\left(4+{\beta }^{2}-2\tau \right)+4{b}^{2}kt\left(4+2\beta +{\beta }^{2}-2\tau \right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+2{b}^{3}{t}^{2}\left(5+4\beta -2\tau -{\beta }^{2}+3\tau {\beta }^{2}\right),\end{array}$

and

$\begin{array}{c}N={k}^{4}\left(2+{\beta }^{2}-\tau \right)+4b{k}^{3}t\left(2+\beta +{\beta }^{2}-\tau \right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{b}^{4}{t}^{4}{\left(1+\beta \right)}^{2}\left[3+2\beta \left(\tau -1\right)+{\beta }^{2}\left(\tau -1\right)-\tau \right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+2{b}^{3}k{t}^{3}\left(1+\beta \right)\left[5+\beta +{\beta }^{2}\left(\tau -1\right)+{\beta }^{3}\left(\tau -1\right)-2\tau +2\beta \tau \right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{b}^{2}{k}^{2}{t}^{2}\left[13+12\beta +{\beta }^{4}\left(\tau -1\right)-6\tau +{\beta }^{2}\left(4+3\tau \right)\right].\end{array}$

Proposition 6:

Comparing the bargaining model and decentralized model we have:

1) $0<{\theta }^{*}<1,{\pi }_{s}^{B\text{*}}>{\pi }_{s}^{D\text{*}},{\pi }_{m}^{B\text{*}}>{\pi }_{m}^{D\text{*}}$;

2) $\frac{\partial {\pi }_{s}^{B\text{*}}}{\partial \beta }>0,\frac{\partial {\pi }_{m}^{B\text{*}}}{\partial \beta }>0,\frac{\partial \left({\pi }_{s}^{B\text{*}}-{\pi }_{s}^{D\text{*}}\right)}{\partial \beta }>0,\frac{\partial \left({\pi }_{m}^{B\text{*}}-{\pi }_{m}^{D\text{*}}\right)}{\partial \beta }>0$;

3) $\frac{\partial {\pi }_{s}^{B\text{*}}}{\partial k}>0,\frac{\partial {\pi }_{m}^{B\text{*}}}{\partial k}>0,\frac{\partial \left({\pi }_{s}^{B\text{*}}-{\pi }_{s}^{D\text{*}}\right)}{\partial k}>0,\frac{\partial \left({\pi }_{m}^{B\text{*}}-{\pi }_{m}^{D\text{*}}\right)}{\partial k}>0$.

Proposition 6 illustrates that:

1) There exists an optimal distribution ratio that makes the profits of the supplier and manufacturer larger than those in the decentralized case. Additionally, as this optimal ratio exists, the two parties enjoy improved benefits and the supply chain becomes coordinated.

2) In the Nash bargaining model, the profits of the supplier and manufacturer increase with increasing spillover rate and the difference compared to the decentralized decision-making case grows increasingly. That is, vertical spillover improves the coordination efficiency of the supply chain under the bargaining mechanism.

3) The effect of the consumers’ environmental awareness is consistent with that of the vertical spillover. That is, growth in environmental awareness promotes the profits of the supplier and manufacturer, and the profit difference between the two models also rises.

6. Discussion

The purpose of the study is to explore the effect of vertical spillover and environmental awareness of consumers on emission-reduction decision making and profits of up-stream and down-stream enterprises in a supply chain. To this end, we considered a monopolistic two-stage supply chain consisting of a single supplier and a single manufacturer each of which strive to reduce emissions in order to reduce their tax burdens and meet the demand for environmental protection from consumers. During emission reduction, they will probably exhibit a positive vertical spillover effect. The effect of vertical spillover and environmental awareness of consumers on the emission-reduction effort levels and profits of node enterprises, as well as the profit of the whole supply chain, is discussed in this work assuming both centralized or decentralized decision-making modes are involved. The two decision-making modes are compared using multiple indices, including the profit of the supply chain, emission-reduction effort levels, and total carbon-emission levels. Finally, in light of the problem of efficiency loss in the decentralized case, a bargaining model is introduced to coordinate the supply chain. Our conclusions are as follows:

1) In the centralized decision-making case, the emission-reduction effort level of the manufacturer is never smaller than that of the supplier. However, in the decentralized case, they have different effort levels depending on the situation: when the up-stream and down-stream enterprises fully overflow, the effort level of the supplier is greater than that of the manufacturer; when they do not overflow, the former is smaller than the latter. Moreover, if there is overflow, but it is not complete, the effort level of the sup-plier is greater than that of the manufacturer, if the consumers have low environmental awareness. Conversely, the former is lower than the latter when the environmental awareness of the consumers is great.

2) For both centralized and decentralized decision making, increasing the vertical spillover rate improves the emission-reduction effort levels and profits of the supplier and manufacturer. This implies that vertical spillover can reduce carbon emission and improve the profits of the enterprises. Therefore, it is essential that up-stream and down-stream enterprises in a supply chain share their information and increase knowledge and technology spillover so as to enhance the vertical spillover rate.

3) The carbon-emission level per unit product in the centralized decision-making case is lower than that in the decentralized case. However, the total carbon-emission level of the supply chain is not always less than that in the decentralized case and it needs to be discussed in the following contexts. When the emission-reduction cost coefficient is small, the total carbon-emission level in the centralized decision-making case is lower than that in the decentralized case. However, the former is higher than the latter when the cost coefficient is large.

4) The emission-reduction effort levels of the supplier and manufacturer and the profit of the whole supply chain in the centralized decision-making case are larger than those in the decentralized case. Moreover, with increasing vertical spillover, the effort levels of the supplier and manufacturer, as well as the profits of the supply chain, in the centralized decision-making case are increasingly different to those in the decentralized case. This shows that vertical spillover lowers the coordination efficiency in the supply chain. Therefore, if vertical spillover is present, it is especially necessary to coordinate the supply chain.

5) A bargaining model is able to coordinate the supply chain. We have found that there is an optimal profit distribution ratio that makes the profits of the supplier and manufacturer greater than those in the decentralized decision-making model. Moreover, the benefits of the two parties improve, and the profit difference between the two parties constantly grows, as the spillover rate increases. That is, vertical spillover improves the coordination efficiency of the supply chain in the bargaining model. Numerical analysis further shows that when the supplier has a favorable bargaining ability, the profit distribution ratio in the supply chain acquired by the manufacturer drops with increasing vertical spillover. However, the contrary behavior occurs with increasing vertical spill-over when the supplier has a poor bargaining ability.

7. Conclusions and Recommendations

We can know that there are many conclusions in this paper through the section six; following on from the above conclusions, the following advice is offered to those tasked with making decisions about emission-reduction in the up-stream and down-stream enterprises in a supply chain.

Firstly, a vertical spillover effect acts to improve the profits of the member enterprises and supply chain. Thus, it is necessary for the members in the supply chain to communicate with each other, share information, and constantly strengthen their partnership so they can take full advantage of the benefits of the vertical spillover. They must not just consider their own benefits and damage the efficiency of the supply chain. Otherwise, not only will the emission-reduction efficiency decline, but also the profits of the supply chain will drop and they will not be able to acquire the optimal benefits available.

Secondly, increasing the environmental awareness of consumers favors reduced carbon emission and increased profit of the enterprises. Thus, the enterprises should create advertisements (public service advertising) that attract the attention of consumers. Additionally, enterprises can follow in the footsteps of Dell, i.e. disclose their carbon information so that consumers become more familiar with their product information. Also, enterprises should be encouraged to take part in environmental-awareness events (e.g. large-scale activities such as environmental and cultural festivals). This will motivate people to actively participate in creating a culture of environmental protection.

Finally, suppliers and manufacturers should be continually improving their manufacturing techniques to enhance the emission-reduction efficiency and lower their emission-reduction cost coefficients. By doing so, the pollution from the whole supply chain will decrease, which favors the creation of a satisfying industrial environment.

It must be said that there are certain deficiencies in the present study, some of which may dictate the direction of our future research. First, only a monopolic supply chain composed of one supplier and one manufacturer is investigated. Therefore, the effect of vertical spillover (and environmental awareness of consumers) on the emission-reduction strategies of competitive supply chains needs to be studied. In this context, it is not just the vertical spillover effect of up-stream and down-stream enterprises, but also the effect of the strength of the competition between sibling enterprises on the game results, which will have to be explored.

Secondly, for two competitive supply chains, horizontal spillover within the same industry needs to be studied as well as the vertical spillover among the longitudinal industries. In addition, emission-reduction decision making and profits can also be investigated when the node enterprises in the supply chains are simultaneously influenced by the two spillover effects.

Thirdly, considering that the government has a part to play in the game, the government could formulate its carbon taxation strategy by setting its objective to be the maximization of social welfare. On this basis, the node enterprises in the supply chain can then formulate their emission-reduction strategies according to these carbon taxes. Under such conditions, the changes that occur in the decision-making process in the supply chains need to be studied further.

Appendix

The proof of Proposition 1

1) Due to

$q=a-bp+k\left({r}_{m}+\beta {r}_{s}\right)=\frac{bz\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}{2bz-2bkt{\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)-2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}$.

To make the demand greater than zero, it requires that

$a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)>0$,

and by $z>\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}$, $0\le \beta \le 1$, $k,b,t>0$, we can get ${r}_{m}^{C\text{*}}-{r}_{s}^{C\text{*}}=\frac{k\left(1-\beta \right)\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}{2bz-2bkt{\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)-2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}\ge 0$, that is, ${r}_{m}^{C\text{*}}\ge {r}_{s}^{C\text{*}}$. Only if $\beta =1$, ${r}_{m}^{C\text{*}}={r}_{s}^{C\text{*}}$.

2) Owing to $a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)>0$,

$z>\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}$,

$0\le \beta \le 1$, and $k,b,t>0$, we obtain that

$\frac{\partial {r}_{m}^{C\text{*}}}{\partial \beta }=\frac{\left[2{k}^{3}\beta +2{b}^{2}tz+2{b}^{3}{t}^{3}{\left(1+\beta \right)}^{2}+b{k}^{2}t\left(3+6\beta +{\beta }^{2}\right)+2k{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}{{\left[2bz-2bkt{\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)-2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}\right]}^{2}}>0$.

Because $\frac{\partial {r}_{s}^{C\text{*}}}{\partial \beta }=\frac{f\left(\beta \right)\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}{{\left[2bz-2bkt{\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)-2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}\right]}^{2}}$, where

$f\left(\beta \right)=2bk\left[z+2b{t}^{2}\beta \left(1+\beta \right)\right]+2{b}^{2}t\left[z+b{t}^{2}{\left(1+\beta \right)}^{2}\right]+{k}^{3}\left({\beta }^{2}-1\right)+b{k}^{2}t\left(2\beta +3{\beta }^{2}-3\right)$.

To judge the size of $\frac{\partial {r}_{s}^{C\text{*}}}{\partial \beta }$, we only compare $f\left(\beta \right)$ to 0. Since

${f}^{\prime }\left(\beta \right)=2{k}^{3}\beta +4{b}^{3}{t}^{3}\left(1+\beta \right)+b{k}^{2}t\left(2+6\beta \right)+2bk\left[2b{t}^{2}\beta +2b{t}^{2}\left(1+\beta \right)\right]>0$,

then $f{\left(\beta \right)}_{\mathrm{min}}=f\left(0\right)=2bkz+2{b}^{2}tz+2{b}^{3}{t}^{3}-{k}^{3}-3b{k}^{2}t$, owing to

$z>\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}$,

$0\le \beta \le 1$, and $k,b,t>0$, we get $f\left(0\right)>0$, thereby $\frac{\partial {r}_{s}^{C\text{*}}}{\partial \beta }>0$.

Because $a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)>0$,

$z>\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}$,

$0\le \beta \le 1$, and $k,b,t>0$, we get

$\frac{\partial {\pi }^{C*}}{\partial \beta }=\frac{z\left[{k}^{2}\beta +2bkt\left(1+\beta \right)+2{b}^{2}{t}^{2}\left(1+\beta \right)\right]{\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}^{2}}{{\left[2bz-2bkt{\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)-2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}\right]}^{2}}>0$.

3) Because $a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)>0$,

$z>\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}$,

$0\le \beta \le 1$, and $k,b,t>0$, we get

$\frac{\partial {r}_{m}^{C\text{*}}}{\partial k}=\frac{\left[{k}^{2}\left(1+{\beta }^{2}\right)+2bkt\left(1+\beta +{\beta }^{2}+{\beta }^{3}\right)+2bz+2{b}^{2}{t}^{2}\beta {\left(1+\beta \right)}^{2}\right]\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}{{\left[2bz-2bkt{\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)-2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}\right]}^{2}}>0$,

$\frac{\partial {r}_{s}{}^{C*}}{\partial k}=\frac{\left[2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}+{k}^{2}\beta \left(1+{\beta }^{2}\right)+2bz\beta +2bkt\left(1+\beta +{\beta }^{2}+{\beta }^{3}\right)\right]\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}{{\left[2bz-2bkt{\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)-2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}\right]}^{2}}>0$,

and $\frac{\partial {\pi }^{C*}}{\partial k}=\frac{z\left[bt{\left(1+\beta \right)}^{2}+k\left(1+{\beta }^{2}\right)\right]\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}{{\left[2bz-2bkt{\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)-2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}\right]}^{2}}>0$.

The proof of Proposition 2

1) Owing to

$q=a-bp+k\left({r}_{m}+\beta {r}_{s}\right)=\frac{bz\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}{4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)}$,

and $z>\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}$, $0\le \beta \le 1$, $k,b,t>0$. To make the demand greater than zero, it requires that

$a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)>0$.

Because ${r}_{s}^{D\text{*}}-{r}_{m}^{D\text{*}}=\frac{\left[bt\beta -k\left(1-\beta \right)\right]\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}{4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)}$, to determine the size of ${r}_{s}^{D\text{*}}$ and ${r}_{m}^{D\text{*}}$, only need compare $bt\beta -k\left(1-\beta \right)$ with 0. Visibility, when $\beta =1$, ${r}_{s}^{D\text{*}}>{r}_{m}^{D\text{*}}$; when $\beta =0$, ${r}_{s}^{D\text{*}}<{r}_{m}^{D\text{*}}$; when $0<\beta <1$, if $k<\frac{bt\beta }{1-\beta }$, then ${r}_{s}^{D\text{*}}>{r}_{m}^{D\text{*}}$; if $k\ge \frac{bt\beta }{1-\beta }$, then ${r}_{s}^{D\text{*}}\le {r}_{m}^{D\text{*}}$.

2)

${\pi }_{s}^{D\text{*}}-{\pi }_{m}^{D\text{*}}=\frac{z\left[2bz-2bkt{\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(2+4\beta +{\beta }^{2}\right)\right]{\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}^{2}}{2{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{2}}$.

By $z>\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}$, $0\le \beta \le 1$, and $k,b,t>0$, we get $2bz-2bkt{\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(2+4\beta +{\beta }^{2}\right)>0$, thus ${\pi }_{s}^{D\text{*}}>{\pi }_{m}^{D\text{*}}$.

The proof of Proposition 3

1) Since $a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)>0$,

$z>\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}$,

$0\le \beta \le 1$, $k,b,t>0$, then

$\frac{\partial {r}_{s}^{D\text{*}}}{\partial \beta }=\frac{\left(k+bt\right)\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]\left\{\left[4bz-4bkt-2{k}^{2}\right]+{\left[k\beta +bt\left(1+\beta \right)\right]}^{2}\right\}}{{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{2}}>0$,

$\frac{\partial {r}_{m}^{D\text{*}}}{\partial \beta }=\frac{2{\left(k+bt\right)}^{2}\left[k\beta +bt\left(2+\beta \right)\right]\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}{{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{2}}>0$,

$\frac{\partial {\pi }_{s}^{D\text{*}}}{\partial \beta }=\frac{z\left(k+bt\right)\left[k\beta +bt\left(2+\beta \right)\right]{\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}^{2}}{{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{2}}>0$,

$\frac{\partial {\pi }_{m}^{D\text{*}}}{\partial \beta }=\frac{2z\left(k+bt\right)\left[k\beta +bt\left(2+\beta \right)\right]\left[2bz-{\left(k+bt\right)}^{2}\right]{\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}^{2}}{{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{3}}>0$.

2) Because $a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)>0$,

$z>\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}$,

$0\le \beta \le 1$, $k,b,t>0$, then we can get

$\frac{\partial {r}_{s}^{D\text{*}}}{\partial k}=\frac{\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]\left[4bz\beta +{k}^{2}\beta \left(2+{\beta }^{2}\right)+{b}^{2}{t}^{2}\left(4+5\beta +2{\beta }^{2}+{\beta }^{3}\right)+2bkt\left(2+2\beta +{\beta }^{2}+{\beta }^{3}\right)\right]}{{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{2}}>0$,

$\frac{\partial {r}_{m}^{D*}}{\partial k}=\frac{\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]\left[4bz+{k}^{2}\left(2+{\beta }^{2}\right)+2bkt\left(2+{\beta }^{2}\right)+{b}^{2}{t}^{2}\left(1+{\beta }^{2}\right)\right]}{{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{2}}>0$,

$\frac{\partial {\pi }_{s}^{D*}}{\partial k}=\frac{z\left[k\left(2+{\beta }^{2}\right)+bt\left(2+2\beta +{\beta }^{2}\right)\right]{\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}^{2}}{{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{2}}>0$,

$\frac{\partial {\pi }_{m}^{D*}}{\partial k}=\frac{z{\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}^{2}g\left(z\right)}{{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{3}}$,

where

$g\left(z\right)=\left[4{b}^{2}t{\left(1+\beta \right)}^{2}+4bk\left(1+{\beta }^{2}\right)\right]z-\left(1+{\beta }^{2}\right){\left(k+bt\right)}^{3}-k\left(k+bt\right)\left(k+2bt\right)$,

it is obviously that we need compare $g\left(z\right)$ with 0 to determine the size of $\frac{\partial {\pi }_{m}^{D*}}{\partial k}$. Because $g\left(z\right)$ is monotone increasing about z, and

$z>\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}$,

$0\le \beta \le 1$, $k,b,t>0$, then we have

$g\left(z\right)>g\left(\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}\right)$

$\begin{array}{c}={k}^{3}{\beta }^{2}\left(3+2{\beta }^{2}\right)+3b{k}^{2}t\beta \left(4+3\beta +4{\beta }^{2}+2{\beta }^{3}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{b}^{3}{t}^{3}\left(3+16\beta +23{\beta }^{2}+16{\beta }^{3}+4{\beta }^{4}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{b}^{2}k{t}^{2}\left(3+24\beta +29{\beta }^{2}+24{\beta }^{3}+8{\beta }^{4}\right)\\ >0,\end{array}$

and sequentially $\frac{\partial {\pi }_{m}^{D*}}{\partial k}>0$.

The proof of Proposition 4

1) Owing to $e=\left({e}_{m}+{e}_{s}\right)-\left(1+\beta \right)\left({r}_{m}+{r}_{s}\right)$, and by Proposition 5 we know ${r}_{s}^{C*}>{r}_{s}^{D*}$, ${r}_{m}^{C*}>{r}_{m}^{D*}$, thus ${e}^{C*}<{e}^{D*}$.

2)

${E}^{C*}-{E}^{D*}=\frac{bz\left(a-b{c}_{m}-b{c}_{s}-bt{e}_{m}-bt{e}_{s}\right)F\left(z\right)}{{\left[2bz-2bt\left(k+bt\right){\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)\right]}^{2}{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{2}}$,

where

$\begin{array}{c}F\left(z\right)={\left[2bz-2bt\left(k+bt\right){\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)\right]}^{2}\left\{\left(1+\beta \right)\left[k\left(1+\beta \right)+bt\left(2+\beta \right)\right]\left(a-b{c}_{m}-b{c}_{s}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-bkt\left(3+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(1+\beta \right)\right]\left({e}_{m}+{e}_{s}\right)\right\}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×\left\{\left(k+2bt\right){\left(1+\beta \right)}^{2}\left(a-b{c}_{m}-b{c}_{s}\right)-\left[2bz-bkt{\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)\right]\left({e}_{m}+{e}_{s}\right)\right\},\end{array}$

the coefficient of the highest order term ${z}^{3}$ is $16{b}^{3}\left({e}_{m}+{e}_{s}\right)$.

To determine the size of ${E}^{C*}$ and ${E}^{D*}$, we need to determine the size of $F\left(z\right)$ and 0. The solution of the equation $F\left(z\right)=0$ is written by ${z}_{1},{z}_{2},{z}_{3}$. Firstly, the coefficient of the highest order term is $16{b}^{3}\left({e}_{m}+{e}_{s}\right)>0$; Secondly, we use some special points to judge null points of $F\left(z\right)$,

$\begin{array}{l}F\left(\frac{{k}^{2}+2bkt+{b}^{2}{t}^{2}\left(1-{\beta }^{2}\right)}{2b}\right)\\ =-bt\beta \left(1+\beta \right){\left[{k}^{2}{\beta }^{2}+2bkt\beta \left(2+\beta \right)+{b}^{2}{t}^{2}\left(1+4\beta +3{\beta }^{2}\right)\right]}^{2}\left(a-b{c}_{m}-b{c}_{s}-bt{e}_{m}-bt{e}_{s}\right)\\ <0,\end{array}$

$\begin{array}{l}F\left(\frac{{k}^{2}\left(2+{\beta }^{2}\right)+2bkt\left(2+2\beta +{\beta }^{2}\right)+{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)}{4b}\right)\\ =\frac{1}{4}\left(1+\beta \right)\left[k\left(1+\beta \right)+bt\left(2+\beta \right)\right]{\left[{k}^{2}{\beta }^{2}+2bkt\beta \left(2+\beta \right)+{b}^{2}{t}^{2}\left(1+4\beta +3{\beta }^{2}\right)\right]}^{2}\left(a-b{c}_{m}-b{c}_{s}-bt{e}_{m}-bt{e}_{s}\right)\\ >0,\end{array}$

$\begin{array}{l}F\left(\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}\right)\\ =-\left(k+2bt\right){\left(1+\beta \right)}^{2}{\left[{k}^{2}{\beta }^{2}+2bkt\beta \left(2+\beta \right)+{b}^{2}{t}^{2}\left(1+4\beta +3{\beta }^{2}\right)\right]}^{2}\left(a-b{c}_{m}-b{c}_{s}-bt{e}_{m}-bt{e}_{s}\right)\\ <0,\end{array}$

$F\left(+\infty \right)=+\infty$, according to the mean value theorem we have

${z}_{1}\in \left(\frac{{k}^{2}+2bkt+{b}^{2}{t}^{2}-{b}^{2}{t}^{2}{\beta }^{2}}{2b},\frac{{k}^{2}\left(2+{\beta }^{2}\right)+2bkt\left(2+2\beta +{\beta }^{2}\right)+{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)}{4b}\right)$,

${z}_{2}\in \left(\frac{{k}^{2}\left(2+{\beta }^{2}\right)+2bkt\left(2+2\beta +{\beta }^{2}\right)+{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)}{4b},\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}\right)$,

${z}_{3}\in \left(\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b},+\infty \right)$;

Lastly, by $z>\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}$, and the character of cubic function, we get that when

$\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b},

$F\left(z\right)<0$, that is ${E}^{C*}<{E}^{D*}$; when $z>{z}_{3}$, $F\left(z\right)>0$, that is ${E}^{C*}>{E}^{D*}$.

The proof of Proposition 5

1) By $a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)>0$,

$z>\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}$,

$0\le \beta \le 1$, $k,b,t>0$, we can obtain

${r}_{s}^{C\text{*}}-{r}_{s}^{D\text{*}}=\frac{\left[k\beta +bt\left(1+\beta \right)\right]\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]\left[2bz-{\left(k+bt\right)}^{2}+{b}^{2}{t}^{2}{\beta }^{2}\right]}{\left[2bz-2bkt{\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)-2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}\right]\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}>0$.

Make $N=\frac{\left[k+bt\left(1+\beta \right)\right]\left(a-b{c}_{m}-b{c}_{s}-bt{e}_{m}-bt{e}_{s}\right)}{4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)}$,

${r}_{m}^{C\text{*}}-N=\frac{\left[k+bt\left(1+\beta \right)\right]\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]\left[2bz-{\left(k+bt\right)}^{2}+{b}^{2}{t}^{2}{\beta }^{2}\right]}{\left[2bz-2bkt{\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)-2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}\right]\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}>0$,

that is ${r}_{m}^{C\text{*}}>N$, and because

$N-{r}_{m}^{D\text{*}}=\frac{bt\beta \left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}{4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)}\ge 0$,

that is $N\ge {r}_{m}^{D\text{*}}$, thus ${r}_{m}^{C\text{*}}>{r}_{m}^{D\text{*}}$.

${\pi }^{C*}-{\pi }^{D*}=\frac{z{\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}^{2}h\left(z\right)}{2\left[2bz-2bkt{\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)-2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}\right]{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{2}}$,

where

$\begin{array}{l}h\left(z\right)=4{b}^{2}{z}^{2}-z\left[4b{k}^{2}+8{b}^{2}kt+2{b}^{3}{t}^{2}\left(2-3{\beta }^{2}\right)\right]-{b}^{4}{t}^{4}{\left(1+\beta \right)}^{2}\left(-1+2\beta +{\beta }^{2}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }-{b}^{2}{k}^{2}{t}^{2}\left(-6+3{\beta }^{2}+{\beta }^{4}\right)-2{b}^{3}k{t}^{3}\left(-2+3{\beta }^{2}+2{\beta }^{3}+{\beta }^{4}\right)+{k}^{4}+4b{k}^{3}t,\end{array}$

to determine the size of ${\pi }^{C*}$ and ${\pi }^{D*}$, we need determine the size of $h\left(z\right)$ and 0. Owing to $z>\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}$, $0\le \beta \le 1$, $k,b,t>0$, we have ${h}^{\prime }\left(z\right)=b\left[8bz-4{\left(k+bt\right)}^{2}+6{b}^{2}{t}^{2}{\beta }^{2}\right]>0$, that means $h\left(z\right)$ is monotone increasing about z, so

$h\left(z\right)>h\left(\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}\right)={\left[{k}^{2}{\beta }^{2}+2bkt\beta \left(2+\beta \right)+{b}^{2}{t}^{2}\left(1+4\beta +3{\beta }^{2}\right)\right]}^{2}>0$,

thus ${\pi }^{C*}>{\pi }^{D*}$.

2)

$\frac{\partial \left({r}_{s}^{C\text{*}}-{r}_{s}^{D\text{*}}\right)}{\partial \beta }=\frac{-\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]y\left(z\right)}{{\left[2bz-2bt\left(k+bt\right){\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)\right]}^{2}{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{2}}$,

where

$\begin{array}{c}y\left(z\right)=-2\left(k+bt\right)\left(k\beta +bt\left(1+\beta \right)\right)\left(k\beta +bt\left(2+\beta \right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×\left(2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2b\left(-z+b{t}^{2}{\left(1+\beta \right)}^{2}\right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×\left({k}^{2}+2bkt+b\left(-2z-b{t}^{2}\left(-1+{\beta }^{2}\right)\right)\right)-2{b}^{2}{t}^{2}\beta \left(k\beta +bt\left(1+\beta \right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×\left(2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2b\left(-z+b{t}^{2}{\left(1+\beta \right)}^{2}\right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×\left({k}^{2}\left(2+{\beta }^{2}\right)+2bkt\left(2+2\beta +{\beta }^{2}\right)+b\left(-4z+b{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right)\right)\end{array}$

$\begin{array}{c}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\left(k\beta +bt\left(1+\beta \right)\right)\left(2{k}^{2}\beta +4bkt\left(1+\beta \right)+4{b}^{2}{t}^{2}\left(1+\beta \right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×\left({k}^{2}+2bkt+b\left(-2z-b{t}^{2}\left(-1+{\beta }^{2}\right)\right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×\left({k}^{2}\left(2+{\beta }^{2}\right)+2bkt\left(2+2\beta +{\beta }^{2}\right)+b\left(-4z+b{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right)\right)\end{array}$

$\begin{array}{c}\text{\hspace{0.17em}}\text{ }\text{ }+\left(k+bt\right)\left(2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2b\left(-z+b{t}^{2}{\left(1+\beta \right)}^{2}\right)\right)\\ \text{\hspace{0.17em}}\text{ }\text{ }×\left({k}^{2}+2bkt+b\left(-2z-b{t}^{2}\left(-1+{\beta }^{2}\right)\right)\right)\\ \text{\hspace{0.17em}}\text{ }\text{ }×\left({k}^{2}\left(2+{\beta }^{2}\right)+2bkt\left(2+2\beta +{\beta }^{2}\right)+b\left(-4z+b{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right)\right),\end{array}$

to determine the size of $\frac{\partial \left({r}_{s}^{C\text{*}}-{r}_{s}^{D\text{*}}\right)}{\partial \beta }$ and 0, we need determine the size of $y\left(z\right)$ and 0. Due to ${y}^{‴}\left(z\right)=-96{b}^{3}\left(k+bt\right)<0$, that means ${y}^{″}\left(z\right)$ is monotone decreasing about z. Because $z>\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}$, $0\le \beta \le 1$, $k,b,t>0$, we have

$\begin{array}{c}{y}^{″}\left(z\right)<{y}^{″}\left(\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}\right)\\ =-8{b}^{2}\left[9{k}^{3}{\beta }^{2}+3b{k}^{2}t\beta \left(10+9\beta \right)+{b}^{3}{t}^{3}\left(15+38\beta +23{\beta }^{2}\right)+{b}^{2}k{t}^{2}\left(15+60\beta +41{\beta }^{2}\right)\right]\\ <0,\end{array}$

that means ${y}^{\prime }\left(z\right)$ is monotone decreasing about z, moreover we can get

$\begin{array}{c}{y}^{\prime }\left(z\right)<{y}^{\prime }\left(\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}\right)\\ =-2b\left[9{k}^{5}{\beta }^{4}+b{k}^{4}t{\beta }^{3}\left(64+45\beta \right)+{b}^{5}{t}^{5}{\left(1+\beta \right)}^{2}\left(17+70\beta +57{\beta }^{2}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+2{b}^{2}{k}^{3}{t}^{2}{\beta }^{2}\left(69+128\beta +59{\beta }^{2}\right)+2{b}^{3}{k}^{2}{t}^{3}\beta \left(48+207\beta +244{\beta }^{2}+87{\beta }^{3}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{b}^{4}k{t}^{4}\left(17+192\beta +490{\beta }^{2}+464{\beta }^{3}+149{\beta }^{4}\right)\right]\\ <0,\end{array}$

that means $y\left(z\right)$ is monotone decreasing about z, so

$\begin{array}{c}y\left(z\right)

thus $\frac{\partial \left({r}_{s}^{C\text{*}}-{r}_{s}^{D\text{*}}\right)}{\partial \beta }>0$.

$\frac{\partial \left({r}_{m}^{C\text{*}}-{r}_{m}^{D\text{*}}\right)}{\partial \beta }=\frac{-\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]x\left(z\right)}{{\left[2bz-2bt\left(k+bt\right){\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)\right]}^{2}{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{2}}$,

where

$\begin{array}{c}x\left(z\right)=2{\left(k+bt\right)}^{2}\left[k\beta +bt\left(2+\beta \right)\right]{\left[2bz-2bt\left(k+bt\right){\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)\right]}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-\left\{2{k}^{3}\beta +2{b}^{3}{t}^{3}{\left(1+\beta \right)}^{2}+b{k}^{2}t\left(3+6\beta +{\beta }^{2}\right)+2{b}^{2}t\left[z+kt\left(3+4\beta +{\beta }^{2}\right)\right]\right\}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{2},\end{array}$

to determine the size of $\frac{\partial \left({r}_{m}^{C\text{*}}-{r}_{m}^{D\text{*}}\right)}{\partial \beta }$ and 0, we need to determine the size of $x\left(z\right)$ and 0. Since ${x}^{‴}\left(z\right)=-192{b}^{4}t<0$, so ${x}^{″}\left(z\right)$ is monotone decreasing about z. By $0\le \beta \le 1$, $k,b,t>0$, $z>\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}$, we have

$\begin{array}{c}{x}^{″}\left(z\right)<{x}^{″}\left(\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}\right)\\ =-16{b}^{2}\left[3{k}^{3}\beta +3b{k}^{2}t\left(2+3\beta +2{\beta }^{2}\right)+{b}^{2}k{t}^{2}\left(12+29\beta +12{\beta }^{2}\right)+{b}^{3}{t}^{3}\left(8+23\beta +14{\beta }^{2}\right)\right]\\ <0,\end{array}$

that means ${x}^{\prime }\left(z\right)$ is monotone decreasing about z, so

$\begin{array}{c}{x}^{\prime }\left(z\right)<{x}^{\prime }\left(\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}\right)\\ =-2b\left[{k}^{2}{\beta }^{2}+2bkt\beta \left(2+\beta \right)+{b}^{2}{t}^{2}\left(1+4\beta +3{\beta }^{2}\right)\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×\left[8{k}^{3}\beta +b{k}^{2}t{\left(4+3\beta \right)}^{2}+2{b}^{2}k{t}^{2}\left(16+26\beta +9{\beta }^{2}\right)+{b}^{3}{t}^{3}\left(17+36\beta +19{\beta }^{2}\right)\right]\\ <0,\end{array}$

that means $x\left(z\right)$ is monotone decreasing about z, so

$\begin{array}{c}x\left(z\right)

thus we can get $\frac{\partial \left({r}_{m}^{C\text{*}}-{r}_{m}^{D\text{*}}\right)}{\partial \beta }>0$.

$\frac{\partial \left({\pi }^{C*}-{\pi }^{D*}\right)}{\partial \beta }=\frac{-z{\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}^{2}r\left(z\right)}{2{\left[2bz-2bt\left(k+bt\right){\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)\right]}^{2}{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{3}}$,

where

$\begin{array}{c}r\left(z\right)=\left[2{k}^{2}\beta +4bkt\left(1+\beta \right)+4{b}^{2}{t}^{2}\left(1+\beta \right)\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{3}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-2\left(k+bt\right)\left[k\beta +bt\left(2+\beta \right)\right]\left[{k}^{2}\left(4+{\beta }^{2}\right)+2bkt\left(4+2\beta +{\beta }^{2}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{b}^{2}{t}^{2}\left(5+4\beta +{\beta }^{2}\right)-8bz\right]{\left[2bz-2bt\left(k+bt\right){\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)\right]}^{2},\end{array}$

to determine the size of $\frac{\partial \left({\pi }^{C*}-{\pi }^{D*}\right)}{\partial \beta }$ and 0, we need to determine the size of $r\left(z\right)$ and 0. Owing to $0\le \beta \le 1$, $k,b,t>0$, we have

${r}^{‴}\left(z\right)=-384{b}^{3}\left[{k}^{2}\beta +2bkt\left(1+\beta \right)+{b}^{2}{t}^{2}\left(2+3\beta \right)\right]<0$,

that means ${r}^{″}\left(z\right)$ is monotone decreasing about z. Since

$z>\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}$,

then

$\begin{array}{c}{r}^{″}\left(z\right)<{r}^{″}\left(\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}\right)\\ =-16{b}^{2}\left[9{k}^{4}{\beta }^{3}+18b{k}^{3}t{\beta }^{2}\left(3+2\beta \right)+{b}^{2}{k}^{2}{t}^{2}\beta \left(81+162\beta +86{\beta }^{2}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+2{b}^{3}k{t}^{3}\left(9+81\beta +125{\beta }^{2}+50{\beta }^{3}\right)+{b}^{4}{t}^{4}\left(18+93\beta +142{\beta }^{2}+65{\beta }^{3}\right)\right]\\ <0,\end{array}$

that means ${r}^{\prime }\left(z\right)$ is monotone decreasing about z, so

$\begin{array}{c}{r}^{\prime }\left(z\right)<{r}^{\prime }\left(\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}\right)\\ =-24b\left[{k}^{2}\beta +2bkt\left(1+\beta \right)+2{b}^{2}{t}^{2}\left(1+\beta \right)\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×{\left[{k}^{2}{\beta }^{2}+2bkt\beta \left(2+\beta \right)+{b}^{2}{t}^{2}\left(1+4\beta +3{\beta }^{2}\right)\right]}^{2}\\ <0,\end{array}$

that means $r\left(z\right)$ is monotone decreasing about z, so

$\begin{array}{c}r\left(z\right)

thus $\frac{\partial \left({\pi }^{C*}-{\pi }^{D*}\right)}{\partial \beta }>0$.

3)

$\frac{\partial \left({r}_{s}^{C*}-{r}_{s}^{D*}\right)}{\partial k}=\frac{-\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]Y\left(z\right)}{{\left[2bz-2bt\left(k+bt\right){\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)\right]}^{2}{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{2}}$,

where

$\begin{array}{c}Y\left(z\right)=\left(-\left(k\beta +bt\left(1+\beta \right)\right)\left(2k\left(2+{\beta }^{2}\right)+2bt\left(2+2\beta +{\beta }^{2}\right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×\left(2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2b\left(-z+b{t}^{2}{\left(1+\beta \right)}^{2}\right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×\left({k}^{2}+2bkt+b\left(-2z-b{t}^{2}\left(-1+{\beta }^{2}\right)\right)\right)+2\left(k+bt\right)\left(k\beta +bt\left(1+\beta \right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×\left(2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2b\left(-z+b{t}^{2}{\left(1+\beta \right)}^{2}\right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×\left({k}^{2}\left(2+{\beta }^{2}\right)+2bkt\left(2+2\beta +{\beta }^{2}\right)+b\left(-4z+b{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right)\right)\end{array}$

$\begin{array}{c}\text{\hspace{0.17em}}\text{ }\text{ }\text{\hspace{0.17em}}-\left(k\beta +bt\left(1+\beta \right)\right)\left(2bt{\left(1+\beta \right)}^{2}+2k\left(1+{\beta }^{2}\right)\right)\\ \text{\hspace{0.17em}}\text{ }\text{ }\text{\hspace{0.17em}}×\left({k}^{2}+2bkt+b\left(-2z-b{t}^{2}\left(-1+{\beta }^{2}\right)\right)\right)\\ \text{\hspace{0.17em}}\text{ }\text{ }\text{\hspace{0.17em}}×\left({k}^{2}\left(2+{\beta }^{2}\right)+2bkt\left(2+2\beta +{\beta }^{2}\right)+b\left(-4z+b{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right)\right)\end{array}$

$\begin{array}{c}\text{\hspace{0.17em}}\text{ }\text{ }+\beta \left(2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2b\left(-z+b{t}^{2}{\left(1+\beta \right)}^{2}\right)\right)\\ \text{\hspace{0.17em}}\text{ }\text{ }×\left({k}^{2}+2bkt+b\left(-2z-b{t}^{2}\left(-1+{\beta }^{2}\right)\right)\right)\\ \text{\hspace{0.17em}}\text{ }\text{ }×\left({k}^{2}\left(2+{\beta }^{2}\right)+2bkt\left(2+2\beta +{\beta }^{2}\right)+b\left(-4z+b{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right)\right)\right),\end{array}$

to determine the size of $\frac{\partial \left({r}_{s}^{C*}-{r}_{s}^{D*}\right)}{\partial k}$ and 0, we need to determine the size of

$Y\left(z\right)$ and 0. Owing to $0\le \beta \le 1$, $k,b,t>0$, then ${Y}^{‴}\left(z\right)=-96{b}^{3}\beta <0$, that means ${Y}^{″}\left(z\right)$ is monotone decreasing about z. By

$z>\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}$,

we have $\begin{array}{c}{Y}^{″}\left(z\right)<{Y}^{″}\left(\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}\right)\\ =-8{b}^{2}\left[{k}^{2}\beta \left(4+9{\beta }^{2}\right)+2bkt\left(2+4\beta +15{\beta }^{2}+9{\beta }^{3}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{b}^{2}{t}^{2}\left(4+19\beta +30{\beta }^{2}+15{\beta }^{3}\right)\right]\\ <0\end{array}$, that means ${Y}^{\prime }\left(z\right)$ is monotone decreasing about z, so

$\begin{array}{c}{Y}^{\prime }\left(z\right)<{Y}^{\prime }\left(\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}\right)\\ =-2b\left[{k}^{4}{\beta }^{3}\left(8+9{\beta }^{2}\right)+4b{k}^{3}t{\beta }^{2}\left(10+8\beta +16{\beta }^{2}+9{\beta }^{3}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{b}^{4}{t}^{4}{\left(1+\beta \right)}^{2}\left(8+41\beta +62{\beta }^{2}+33{\beta }^{3}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+2{b}^{2}{k}^{2}{t}^{2}\beta \left(20+60\beta +101{\beta }^{2}+96{\beta }^{3}+37{\beta }^{4}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+4{b}^{3}k{t}^{3}\left(2+20\beta +58{\beta }^{2}+85{\beta }^{3}+64{\beta }^{4}+19{\beta }^{5}\right)\right]\\ <0,\end{array}$

that means $Y\left(z\right)$ is monotone decreasing about z, so

$\begin{array}{c}Y\left(z\right)

thus $\frac{\partial \left({r}_{s}^{C*}-{r}_{s}^{D*}\right)}{\partial k}>0$.

$\frac{\partial \left({r}_{m}^{C*}-{r}_{m}^{D*}\right)}{\partial k}=\frac{-\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]X\left(z\right)}{{\left[2bz-2bt\left(k+bt\right){\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)\right]}^{2}{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{2}}$,

where

$\begin{array}{c}X\left(z\right)=\left\{{k}^{2}\left(2+{\beta }^{2}\right)+2bkt\left(2+{\beta }^{2}\right)+b\left[4z+b{t}^{2}\left(1+{\beta }^{2}\right)\right]\right\}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×{\left[2bz-2bt\left(k+bt\right){\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)\right]}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}-\left\{{k}^{2}\left(1+{\beta }^{2}\right)+2bkt\left(1+\beta +{\beta }^{2}+{\beta }^{3}\right)+2b\left[z+b{t}^{2}\beta {\left(1+\beta \right)}^{2}\right]\right\}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{2},\end{array}$

to determine the size of $\frac{\partial \left({r}_{m}^{C*}-{r}_{m}^{D*}\right)}{\partial k}$ and 0, we need to determine the size of

$X\left(z\right)$ and 0. Due to $0\le \beta \le 1$, $k,b,t>0$, we have ${X}^{‴}\left(z\right)=-96{b}^{3}<0$, that means ${X}^{″}\left(z\right)$ is monotone decreasing about z. Since

$z>\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}$,

then

$\begin{array}{c}{X}^{″}\left(z\right)<{X}^{″}\left(\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}\right)\\ =-8{b}^{2}\left[{k}^{2}\left(4+9{\beta }^{2}\right)+2bkt\left(4+16\beta +9{\beta }^{2}+4{\beta }^{3}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{b}^{2}{t}^{2}\left(7+32\beta +31{\beta }^{2}+8{\beta }^{3}\right)\right]\\ <0\end{array}$,

that means ${X}^{\prime }\left(z\right)$ is monotone decreasing about z, so

$\begin{array}{c}{X}^{\prime }\left(z\right)<{X}^{\prime }\left(\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}\right)\\ =-2b\left[{k}^{4}{\beta }^{2}\left(8+9{\beta }^{2}\right)+{b}^{4}{t}^{4}{\left(1+\beta \right)}^{2}\left(9+46\beta +65{\beta }^{2}+24{\beta }^{3}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+4b{k}^{3}t\beta \left(8+8\beta +16{\beta }^{2}+9{\beta }^{3}+2{\beta }^{4}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+4{b}^{3}k{t}^{3}\left(4+32\beta +77{\beta }^{2}+82{\beta }^{3}+43{\beta }^{4}+10{\beta }^{5}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+2{b}^{2}{k}^{2}{t}^{2}\left(4+48\beta +93{\beta }^{2}+96{\beta }^{3}+61{\beta }^{4}+12{\beta }^{5}\right)\right]\\ <0,\end{array}$

that means $X\left(z\right)$ is monotone decreasing about z, so

$\begin{array}{l}X\left(z\right)

thus $\frac{\partial \left({r}_{m}^{C*}-{r}_{m}^{D*}\right)}{\partial k}>0$.

$\frac{\partial \left({\pi }^{C*}-{\pi }^{D*}\right)}{\partial k}=\frac{-z{\left[a-b\left({c}_{m}+{c}_{s}\right)-bt\left({e}_{m}+{e}_{s}\right)\right]}^{2}R\left(z\right)}{2{\left[2bz-2bt\left(k+bt\right){\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)\right]}^{2}{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{3}}$,

where

$\begin{array}{c}R\left(z\right)=2\left(-2\left(k\left(2+{\beta }^{2}\right)+bt\left(2+2\beta +{\beta }^{2}\right)\right)\left({k}^{2}\left(3+{\beta }^{2}\right)+2bkt\left(3+2\beta +{\beta }^{2}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+b\left(-6z+b{t}^{2}{\left(2+\beta \right)}^{2}\right)\right)+\left(k\left(3+{\beta }^{2}\right)+bt\left(3+2\beta +{\beta }^{2}\right)\right)\end{array}$

$\begin{array}{c}\text{\hspace{0.17em}}\text{\hspace{0.17em}}×\left({k}^{2}\left(2+{\beta }^{2}\right)+2bkt\left(2+2\beta +{\beta }^{2}\right)+b\left(-4z+b{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right)\right)\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×{\left[2bz-2bt\left(k+bt\right){\left(1+\beta \right)}^{2}-{k}^{2}\left(1+{\beta }^{2}\right)\right]}^{2}-2\left[bt{\left(1+\beta \right)}^{2}+k\left(1+{\beta }^{2}\right)\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}×{\left[4bz-{k}^{2}\left(2+{\beta }^{2}\right)-2bkt\left(2+2\beta +{\beta }^{2}\right)-{b}^{2}{t}^{2}\left(3+4\beta +{\beta }^{2}\right)\right]}^{3},\end{array}$

to determine the size of $\frac{\partial \left({\pi }^{C*}-{\pi }^{D*}\right)}{\partial k}$ and 0, we need to determine the size of $R\left(z\right)$ and 0. Since $0\le \beta \le 1$, $k,b,t>0$, then

${R}^{‴}\left(z\right)=-192{b}^{3}\left[k+2k{\beta }^{2}+bt\left(1+4\beta +2{\beta }^{2}\right)\right]<0$,

that means ${R}^{″}\left(z\right)$ is monotone decreasing about z. By

$z>\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}$,

we have

$\begin{array}{c}{R}^{″}\left(z\right)<{R}^{″}\left(\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}\right)\\ =-16{b}^{2}\left[{k}^{3}{\beta }^{2}\left(7+9{\beta }^{2}\right)+b{k}^{2}t\beta \left(28+21\beta +54{\beta }^{2}+27{\beta }^{3}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{b}^{3}{t}^{3}\left(7+46\beta +106{\beta }^{2}+94{\beta }^{3}+29{\beta }^{4}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{b}^{2}k{t}^{2}\left(7+56\beta +120{\beta }^{2}+108{\beta }^{3}+47{\beta }^{4}\right)\right]\\ <0,\end{array}$

that means ${R}^{\prime }\left(z\right)$ is monotone decreasing about z, so

$\begin{array}{c}{R}^{\prime }\left(z\right)<{R}^{\prime }\left(\frac{2bkt{\left(1+\beta \right)}^{2}+{k}^{2}\left(1+{\beta }^{2}\right)+2{b}^{2}{t}^{2}{\left(1+\beta \right)}^{2}}{2b}\right)\\ =-24b\left[bt{\left(1+\beta \right)}^{2}+k\left(1+{\beta }^{2}\right)\right]{\left[{k}^{2}{\beta }^{2}+2bkt\beta \left(2+\beta \right)+{b}^{2}{t}^{2}\left(1+4\beta +3{\beta }^{2}\right)\right]}^{2}\\ <0,\end{array}$

that means $R\left(z\right)$ is monotone decreasing about z, so

$\begin{array}{c}R\left(z\right)

thus $\frac{\partial \left({\pi }^{C*}-{\pi }^{D*}\right)}{\partial k}>0$.

The proof of Proposition 6

1) Due to $0<\tau <1$, ${\pi }^{C*}>{\pi }^{D*}>{\pi }_{s}^{D\text{*}}>{\pi }_{m}^{D\text{*}}$, it is clear that ${\theta }^{*}>0$. Because ${\pi }^{C*}-{\pi }_{m}^{D\text{*}}>{\pi }^{C*}-{\pi }^{D*}>\left(1-\tau \right)\left({\pi }^{C*}-{\pi }^{D*}\right)$, so ${\pi }^{C*}>\left(1-\tau \right)\left({\pi }^{C*}-{\pi }^{D*}\right)+{\pi }_{m}^{D\text{*}}$, thereby ${\theta }^{*}<1$.

We know ${\pi }_{s}^{B\text{*}}=\left(1-{\theta }^{*}\right){\pi }^{C*}=\tau \left({\pi }^{C*}-{\pi }^{D*}\right)+{\pi }_{s}^{D\text{*}}$, because of $0<\tau <1$, and ${\pi }^{C*}>{\pi }^{D*}$, then we get ${\pi }_{s}^{B\text{*}}>{\pi }_{s}^{D\text{*}}$. Similarly,

${\pi }_{m}^{B\text{*}}={\theta }^{*}{\pi }^{C*}=\left(1-\tau \right)\left({\pi }^{C*}-{\pi }^{D*}\right)+{\pi }_{m}^{D\text{*}}>{\pi }_{m}^{D\text{*}}$.

2) Owing to ${\pi }_{s}^{B\text{*}}=\left(1-{\theta }^{*}\right){\pi }^{C*}=\tau \left({\pi }^{C*}-{\pi }^{D*}\right)+{\pi }_{s}^{D\text{*}}$, then

$\frac{\partial {\pi }_{s}^{B\text{*}}}{\partial \beta }=\tau \frac{\partial \left({\pi }^{C*}-{\pi }^{D*}\right)}{\partial \beta }+\frac{\partial {\pi }_{s}^{D\text{*}}}{\partial \beta }$.

By Proposition 3 and Proposition 5, we get $\frac{\partial {\pi }_{s}^{B\text{*}}}{\partial \beta }>0$. Similarly,

${\pi }_{m}^{B\text{*}}={\theta }^{*}{\pi }^{C*}=\left(1-\tau \right)\left({\pi }^{C*}-{\pi }^{D*}\right)+{\pi }_{m}^{D\text{*}}$,

then

$\frac{\partial {\pi }_{m}^{B\text{*}}}{\partial \beta }=\left(1-\tau \right)\frac{\partial \left({\pi }^{C*}-{\pi }^{D*}\right)}{\partial \beta }+\frac{\partial {\pi }_{m}^{D\text{*}}}{\partial \beta }>0$.

Since ${\pi }_{s}^{B\text{*}}-{\pi }_{s}^{D\text{*}}=\tau \left({\pi }^{C*}-{\pi }^{D*}\right)$, so $\frac{\partial \left({\pi }_{s}^{B\text{*}}-{\pi }_{s}^{D\text{*}}\right)}{\partial \beta }=\tau \frac{\partial \left({\pi }^{C*}-{\pi }^{D*}\right)}{\partial \beta }$. By Proposition 5, we get $\frac{\partial \left({\pi }_{s}^{B\text{*}}-{\pi }_{s}^{D\text{*}}\right)}{\partial \beta }>0$. Similarly, because of

${\pi }_{m}^{B\text{*}}-{\pi }_{m}^{D\text{*}}=\left(1-\tau \right)\left({\pi }^{C*}-{\pi }^{D*}\right)$,

we have $\frac{\partial \left({\pi }_{m}^{B\text{*}}-{\pi }_{m}^{D\text{*}}\right)}{\partial \beta }=\left(1-\tau \right)\frac{\partial \left({\pi }^{C*}-{\pi }^{D*}\right)}{\partial \beta }>0$.

3) Since $\frac{\partial {\pi }_{s}^{B\text{*}}}{\partial k}=\tau \frac{\partial \left({\pi }^{C*}-{\pi }^{D*}\right)}{\partial k}+\frac{\partial {\pi }_{s}^{D\text{*}}}{\partial k}$, and by Proposition 3 and Proposition 5, we get $\frac{\partial {\pi }_{s}^{B\text{*}}}{\partial k}>0$. Similarly, $\frac{\partial {\pi }_{m}^{B\text{*}}}{\partial k}=\left(1-\tau \right)\frac{\partial \left({\pi }^{C*}-{\pi }^{D*}\right)}{\partial k}+\frac{\partial {\pi }_{m}^{D\text{*}}}{\partial k}>0$.

Due to $\frac{\partial \left({\pi }_{s}^{B\text{*}}-{\pi }_{s}^{D\text{*}}\right)}{\partial k}=\tau \frac{\partial \left({\pi }^{C*}-{\pi }^{D*}\right)}{\partial k}$, and by Proposition 3 and Proposition 5, we get $\frac{\partial \left({\pi }_{s}^{B\text{*}}-{\pi }_{s}^{D\text{*}}\right)}{\partial k}>0$. Similarly,

$\frac{\partial \left({\pi }_{m}^{B\text{*}}-{\pi }_{m}^{D\text{*}}\right)}{\partial k}=\left(1-\tau \right)\frac{\partial \left({\pi }^{C*}-{\pi }^{D*}\right)}{\partial k}>0$.

Cite this paper
Han, G. (2019) Making Carbon-Emission Reduction Decisions in Supply Chains Based on Vertical Spillover and Environmental Awareness of Consumers. Open Journal of Business and Management, 7, 1657-1689. doi: 10.4236/ojbm.2019.74116.
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