Green fresh product cost sharing contracts considering freshness-keeping effort

Abstract

Nowadays, along with increased public demands on the high-quality green fresh product, the downstream retailer has to spend in the packaging and cold chain transportation and the upstream farmer needs to invest more in green fresh products producing. In this paper, we investigate a green fresh product supply chain problem, in which the retailer transports and sells green fresh products to the ultimate consumer while the farmer produces green fresh product through spending a greenness improvement investment and sells green fresh products to the retailer. Since the fresh product is perishable, the retailer needs to make a costly freshness-keeping effort. Obviously, the freshness-keeping effort, the price, and the greenness improvement level will affect the demand. Thus, to demonstrate the game structure between the retailer and the farmer, a decentralized model without cost sharing, a decentralized Stackelberg cost sharing model, and a Nash bargaining model with cost sharing are formulated. Results show that: (1) The equilibrium decisions under Stackelberg model with cost sharing are larger than that of the Nash bargaining model with cost sharing, while equilibrium decisions in the decentralized model without cost sharing are the least. (2) Both greenness improvement levels in Stackelberg cost sharing contract and Nash bargaining are greater than that in decentralized model without cost sharing. (3) The retailer’s profits in Stackelberg cost sharing contract and Nash bargaining are larger than that in decentralized case without cost sharing, while the farmer’s profits in Stackelberg cost sharing contract and Nash bargaining are larger than in decentralized model without cost sharing in certain conditions. Meanwhile, a numerical example is given to illustrate the results we obtained in the theoretical analysis process.

Appendix

1.1 Appendix A: Summary of abbreviations

$$\begin{aligned} \begin{aligned} {\hat{\psi }}&=\frac{2KLb+K\beta ^2-\sqrt{K^2L^2b^2-2K^2Lb\beta ^2+K^2\beta ^4+KL^2\alpha ^2b+2KL\alpha ^2\beta ^2}}{K(Lb+2\beta ^2)}\\ A_1&=14KLb\psi ^2+K\beta ^2\psi ^2-16KLb\psi -2K\beta ^2\psi \\&\quad +6L\alpha ^2\psi +2KLb+K\beta ^2-L\alpha ^2\\ A_2&=56K^2L^2b^2\psi ^4-10K^2Lb\beta ^2\psi ^4-K^2\beta ^4\psi ^4-152K^2L^2b^2\psi ^3\\&\quad +22K^2Lb\beta ^2\psi ^3\\&\quad +4K^2\beta ^4\psi ^3- 6KL^2\alpha ^2b\psi ^3-12KL\alpha ^2\beta ^2\psi ^3\\&\quad +120K^2L^2b^2\psi ^2-6K^2Lb\beta ^2\psi ^2\\&\quad - 6K^2\beta ^4 \psi ^2-2KL^2\alpha ^2b\psi ^2+23KL\alpha ^2\beta ^2\psi ^2-9L^2\alpha ^4\psi ^2\\&\quad - 8K^2L^2b^2\psi -14K^2Lb\beta ^2\psi +4K^2\beta ^4\psi -8KL^2\alpha ^2b\psi \\&\quad -10KL\alpha ^2\beta ^2\psi +3L^2\alpha ^4\psi -16K^2L^2b^2\\&\quad +8K^2Lb\beta ^2-K^2\beta ^4+ 16KL^2\alpha ^2b-KL\alpha ^2\beta ^2-4L^2\alpha ^4.\\ A_3&=\sqrt{36K^2L^2b^2-34KL^2\alpha ^2b-5KL\alpha ^2\beta ^2+9L^2\alpha ^4}.\\ A_4&=(-2Kb^2c{\psi ^{\mathrm{B}^{*}}}^2+2bKa{\psi ^{\mathrm{B}^{*}}}^2+4Kb^2c\psi ^{\mathrm{B}^{*}}\\&\quad -2\alpha ^2bc\psi ^{\mathrm{B}^{*}}-4bKa\psi ^{\mathrm{B}^{*}}\\&\quad -2Kb^2c+2a\alpha ^2\psi ^{\mathrm{B}^{*}}+\alpha ^2bc+2bKa-a\alpha ^2)L\\ A_5&=6KLb{\psi ^{\mathrm{B}^{*}}}^2-K\beta ^2{\psi ^{\mathrm{B}^{*}}}^2-12KLb\psi ^{\mathrm{B}^{*}}+2K\beta ^2\psi ^{\mathrm{B}^{*}}\\&\quad +4L\alpha ^2\psi ^{\mathrm{B}^{*}} +6KLb-K\beta ^2-3L\alpha ^2\\ A_6&=\sqrt{64K^2L^2b^2-48KL^2\alpha ^2b-12KL\alpha ^2\beta ^2+9L^2\alpha ^4}\\ A_7&=16\,{K}^{2}{L}^{2}{b}^{2}{\psi ^{\mathrm{B}^{*}}}^3-{K}^{2}{\beta }^{4}{\psi ^{\mathrm{B}^{*}}}^3-8\,K{ L}^{2}{\alpha }^{2}b{\psi ^{\mathrm{B}^{*}}}^3\\&\quad -32\,{K}^{2}{L}^{2}{b}^{2}{\psi ^{\mathrm{B}^{*}}}^2-4\,{ K}^{2}Lb{\beta }^{2}{\psi ^{\mathrm{B}^{*}}}^2\\&\quad +3\,{K}^{2}{\beta }^{4}{\psi ^{\mathrm{B}^{*}}}^2+24\,K{L} ^{2}{\alpha }^{2}b{\psi ^{\mathrm{B}^{*}}}^2+2\,KL{\alpha }^{2}{\beta }^{2}{\psi ^{\mathrm{B}^{*}}}^2-4\, {L}^{2}{\alpha }^{4}{\psi ^{\mathrm{B}^{*}}}^2\\&\quad +16\,{K}^{2}{L}^{2}{b}^{2}\psi ^{\mathrm{B}^{*}}\\&\quad + 8\,{K}^{2} Lb{\beta }^{2}{\psi ^{\mathrm{B}^{*}}}-3\,{K}^{2}{\beta }^{4}{\psi ^{ \mathrm {B}^{*}}}-14\,K{L}^{2}{\alpha }^{2}b {\psi ^{\mathrm{B}^{*}}}-4\,KL{\alpha }^{2}{\beta }^{2}{\psi ^{\mathrm{B}^{*}}}\\&\quad +3 \,{L}^{2}{\alpha }^{4}{\psi ^{\mathrm{B}^{*}}}\\&\quad -4\,{K }^{2}Lb{\beta }^{2}+{K}^{2}{\beta }^{4}+2\,KL{\alpha }^{2}{\beta }^{2}\\ A_8&=64\,{K}^{2}{L}^{2}{b}^{2}{\psi ^{\mathrm{B}^{*}}}^3+16\,{K}^{2}Lb{\beta }^{2}{\psi ^{\mathrm{B}^{*}}}^3 -8\,{K}^{2}{\beta }^{4}{\psi ^{\mathrm{B}^{*}}}^3\\&\quad -54\,K{L}^{2}{\alpha }^{2}b{\psi ^{\mathrm{B}^{*}}}^3- 64\,{K}^{2}{L}^{2}{b}^{2}{\psi ^{\mathrm{B}^{*}}}^2\\&\quad -80\,{K}^{2}Lb{\beta }^{2}{\psi ^{\mathrm{B}^{*}}}^2 +24\,{K}^{2}{\beta }^{4}{\psi ^{\mathrm{B}^{*}}}^2\\&\quad +102\,K{L}^{2}{\alpha }^{2}b{\psi ^{\mathrm{B}^{*}}}^2 +24\,KL{\alpha }^{2}{\beta }^{2}{\psi ^{\mathrm{B}^{*}}}^2\\&\quad -27\,{L}^{2}{\alpha }^{4}{\psi ^{\mathrm{B}^{*}}}^2\\&\quad -64\,{K}^{2}{L}^{2}{b}^{2}{\psi ^{\mathrm{B}^{*}}}+112\,{K}^{2}Lb{\beta }^{2}{\psi ^{\mathrm{B}^{*}}}-24\,{ K}^{2}{\beta }^{4}{\psi ^{\mathrm{B}^{*}}}\\&\quad +30\,K{L}^{2}{\alpha }^{2}b{\psi ^{\mathrm{B}^{*}}}-48\,KL{\alpha }^{2} {\beta }^{2}{\psi ^{\mathrm{B}^{*}}}\\&\quad +64\,{K}^{2}{L}^{2}{b}^{2}-48\,{K}^{2}Lb{\beta }^{2}+8\, {K}^{2}{\beta }^{4}-62\,K{L}^{2}{\alpha }^{2}b\\&\quad +24\,KL{\alpha }^{2}{\beta } ^{2}+15\,{L}^{2}{\alpha }^{4}\\ A_9&=12KLb\psi ^3-3K\beta ^2\psi ^3-24KLb\psi ^2+6K\beta ^2\psi ^2\\&\quad +12KLb\psi -3K\beta ^2\psi -L\alpha ^2\\&\psi ^{\mathrm{B}^{*}}=\frac{8KLb+K\beta ^2-3L\alpha ^2-A_3}{K(14Lb+\beta ^2)}\\ {\underline{K}}&=\frac{(31Lb-12\beta ^2-\sqrt{L^2b^2-24Lb\beta ^2 +24\beta ^4})\alpha ^2L}{8(8L^2b^2-6Lb\beta ^2+\beta ^4)}\\ {\overline{K}}&=\frac{(31Lb-12\beta ^2+\sqrt{L^2b^2-24Lb\beta ^2 +24\beta ^4})\alpha ^2L}{8(8L^2b^2-6Lb\beta ^2+\beta ^4)} \end{aligned} \end{aligned}$$

1.2 Appendix B: Model solving process and proofs of Propositions

The centralized channel case. The whole channel supply chain profit function can be rewritten as

$$\begin{aligned} \max _{p,\zeta ,\theta } \pi _\mathrm{SC}=(p-c)(a-bp+\alpha \theta +\beta \zeta )-\frac{1}{2}L\zeta ^2-\frac{1}{2}K\theta ^2. \end{aligned}$$

(9)

The first-order condition

$$\begin{aligned} \frac{ \partial \pi _\mathrm{SC}}{ \partial p}=\beta \zeta +\alpha \theta -bp+a-(p-c)b, \end{aligned}$$

(10)

$$\begin{aligned} \frac{ \partial \pi _\mathrm{SC}}{ \partial \zeta }=p \beta -c \beta -L \zeta \end{aligned}$$

(11)

and

$$\begin{aligned} \frac{ \partial \pi _\mathrm{SC}}{ \partial \theta }=p \alpha -c \alpha -K \theta . \end{aligned}$$

(12)

The second-order condition

$$\begin{aligned}&\frac{ \partial ^2 \pi _\mathrm{SC}}{ \partial p^2}=-2b, \end{aligned}$$

(13)

$$\begin{aligned}&\frac{ \partial ^2 \pi _\mathrm{SC}}{ \partial \zeta ^2}=-L \end{aligned}$$

(14)

and

$$\begin{aligned} \frac{ \partial ^2 \pi _\mathrm{SC}}{ \partial \theta ^2}=-K. \end{aligned}$$

(15)

Moreover, we can also obtain

$$\begin{aligned} \frac{ \partial ^2 \pi _\mathrm{SC}}{ \partial p\partial \zeta }=\beta ,\quad \frac{ \partial ^2 \pi _\mathrm{SC}}{ \partial p\partial \theta }=\alpha ,\quad \frac{ \partial ^2 \pi _\mathrm{SC}}{ \partial \zeta \partial \theta }=0. \end{aligned}$$

Thus, the Hessian matrix is

$$\begin{aligned} \left[ \begin{array}{ccc} -2b &{}\beta &{}\alpha \\ \beta &{} -L&{}0 \\ \alpha &{}0&{}-K\\ \end{array} \right] . \end{aligned}$$

(16)

Through computing the Hessian Matrix, when the conditions $2bL-\beta ^2>0$ and $-2\,bLK+{\beta }^{2}K+{\alpha }^{2}L<0$ are satisfied, the supply chain profit function is joint concave with respect to the decision variables. Thus, equating the first-order condition to 0 and solving the equations, we obtain

$$\begin{aligned}&p^{*}={\frac{KLbc-K{\beta }^{2}c-L{\alpha }^{2}c+KLa}{2\,bLK-{\beta }^{2}K-{ \alpha }^{2}L}}, \end{aligned}$$

(17)

$$\begin{aligned}&\zeta ^{*}={\frac{K\beta \, \left( -bc+a \right) }{2\,bLK-{\beta }^{2}K-{\alpha }^{ 2}L}}, \end{aligned}$$

(18)

and

$$\begin{aligned} \theta ^{*}={\frac{L\alpha \, \left( -bc+a \right) }{2\,bLK-{\beta }^{2}K-{\alpha }^ {2}L}} . \end{aligned}$$

(19)

Subsequently, the optimal profit of the centralized supply chain is

$$\begin{aligned} \pi ^*_\mathrm{sc}=\frac{KL \left( -bc+a \right) ^{2}}{2(2\,bLK-{\beta }^{2}K-{ \alpha }^{2}L)}. \end{aligned}$$

Proof of Proposition1

Proof

We can easily get the results from the computational process. $\square$

Proof of Corollary 1

Proof

According to the analytical expressions given above, we can easily get the results by implementing differentiation. $\square$

The decentralized channel case. Under this environment, we solve the farmer’s optimization problem first:

$$\begin{aligned} \max _{w,\theta } \pi _\mathrm{F}=(w-c)q-\frac{1}{2}K\theta ^2 . \end{aligned}$$

Rewriting as

$$\begin{aligned} \max _{w,\theta } \pi _\mathrm{F}=(w-c)[a-b(w+m)+\alpha \theta +\beta \zeta ]-\frac{1}{2}K\theta ^2 . \end{aligned}$$

The Hessian matrix is

$$\begin{aligned} \left[ \begin{array}{cc} -2b &{}\alpha \\ \alpha &{}-K\\ \end{array} \right] . \end{aligned}$$

(20)

Through computing the Hessian matrix, the farmer’s profit function is joint concave with respect to the decision variables when $2bK-\alpha ^2>0$. The first-order condition is

$$\begin{aligned} \frac{ \partial \pi _\mathrm{F}}{ \partial w}=a-b \left( w+m \right) +\alpha \,\theta +\beta \zeta - \left( w-c \right) b \end{aligned}$$

and

$$\begin{aligned} \frac{\partial \pi _\mathrm{F}}{ \partial \theta }=(w-c)\alpha -K \theta . \end{aligned}$$

Thus, equating the first-order condition to 0 and solving the equations, we get

$$\begin{aligned} \theta (m,\zeta )={\frac{\alpha \, \left( \beta \zeta -bc-bm+a \right) }{2\,Kb-{\alpha }^{2}}} \end{aligned}$$

(21)

and

$$\begin{aligned} w(m,\zeta )={\frac{K\zeta \beta +Kbc-Kbm-{\alpha }^{2}c+Ka}{2\,Kb -{\alpha }^{2}}}. \end{aligned}$$

(22)

We substitute Eqs. (21)–(22) into the retailer’s profit function (1) and derive

$$\begin{aligned} \max _{m,\zeta }\pi _\mathrm{R}=\frac{L\zeta ^2\alpha ^2-2bk[mb(c+m)-m(\beta \zeta +a)+L \zeta ^2]}{4Kb-2\alpha ^2}. \end{aligned}$$

By computing the first-order condition

$$\begin{aligned}&\frac{ \partial \pi _\mathrm{R}}{ \partial m}=\frac{bK[a+\beta \zeta -bc-2mb]}{2Kb-\alpha ^2}, \\&\frac{ \partial \pi _\mathrm{R}}{ \partial \zeta }=\frac{L\zeta \alpha ^2+bk(\beta m-2L\zeta )}{2Kb-\alpha ^2}, \end{aligned}$$

and the second-order condition

$$\begin{aligned}&\frac{ \partial ^2 \pi _\mathrm{R}}{ \partial m^2}=\frac{-2Kb^2}{2Kb-\alpha ^2}, \\&\frac{ \partial ^2 \pi _\mathrm{R}}{ \partial \zeta ^2}=\frac{L\alpha ^2-2LbK}{2Kb-\alpha ^2}, \\&\frac{ \partial ^2 \pi _\mathrm{R}}{ \partial m \partial \zeta }=\frac{\beta bK}{2Kb-\alpha ^2}, \end{aligned}$$

we can write the Hessian matrix of $\pi _\mathrm{R}$ as

$$\begin{aligned} \left[ \begin{array}{cc} \frac{-2Kb^2}{2Kb-\alpha ^2} &{}\frac{\beta bK}{2Kb-\alpha ^2} \\ \frac{\beta bK}{2Kb-\alpha ^2} &{} \frac{L\alpha ^2-2LbK}{2Kb-\alpha ^2} \\ \end{array} \right] . \end{aligned}$$

(23)

Through computing the Hessian matrix, the retailer’s profit function is joint concave with respect to the decision variables when $\frac{b^2K(4LbK-\beta ^2K-2L\alpha ^2)}{(2Kb-\alpha ^2)^2}>0$, i.e., $4LbK-\beta ^2K-2L\alpha ^2>0$. Let the first-order condition be 0. We get

$$\begin{aligned} \zeta ^{\mathrm{D}^{*}}=\frac{\beta K(a-bc)}{4LbK-\beta ^2K-2L\alpha ^2} \end{aligned}$$

and

$$\begin{aligned} m^{\mathrm{D}^{*}}=\frac{L(\alpha ^2 bc+2Kab-a\alpha ^2-2Kb^2c)}{b(4LbK-\beta ^2K-2L\alpha ^2)}. \end{aligned}$$

Then, we have

$$\begin{aligned}&\theta ^{\mathrm{D}^{*}}=\frac{L\alpha (a-bc)}{4LbK-\beta ^2K-2L\alpha ^2}, \\&w^{\mathrm{D}^{*}}=\frac{[L(3bc+a)-c\beta ^2]K-2Lc\alpha ^2}{4LbK-\beta ^2K-2L\alpha ^2}, \\&p^{\mathrm{D}^{*}}=w^{\mathrm{D}^{*}}+m^{\mathrm{D}^{*}}=\frac{KLb^2c-Kb\beta ^2c-L\alpha ^2bc+3KLab-La\alpha ^2}{b(4LbK-\beta ^2K-2L\alpha ^2)}, \\&\pi _\mathrm{R}^{\mathrm{D}^{*}}=\frac{KL(a-bc)^2}{2(4LbK-\beta ^2K-2L\alpha ^2)}, \\&\pi _\mathrm{F}^{\mathrm{D}^{*}}=\frac{KL^2(a-bc)^2(2Kb-\alpha ^2)}{2(4LbK-\beta ^2K-2L\alpha ^2)^2}, \\&\pi _\mathrm{SC}^{\mathrm{D}^{*}}=\frac{KL(-bc+a)^2(6KLb-K\beta ^2-3L\alpha ^2)}{2(4KLb-K\beta ^2-2L\alpha ^2)^2}. \end{aligned}$$

Proof of Proposition2

Proof

We can easily get the results from the computational process. $\square$

Proof of Proposition 3

Proof

The above relationships can be proved easily through algebraic comparison. $\square$

The model with cost sharing. Due to the game structure, we solve for the farmer’s profit function first:

$$\begin{aligned} \max _{w,\theta }\pi _\mathrm{F}=(w-c)q-\frac{1}{2}K(1-\psi )\theta ^2. \end{aligned}$$

The first-order condition

$$\begin{aligned}&\frac{ \partial \pi _\mathrm{F}}{ \partial w}=a+\alpha \theta +\beta \zeta +cb-2bw-bm,\\&\frac{\partial \pi _\mathrm{F}}{ \partial \theta }=(w-c)\alpha -K(1-\psi ) \theta , \end{aligned}$$

and the second-order condition

$$\begin{aligned}&\frac{ \partial ^2 \pi _\mathrm{F}}{ \partial w^2}=-2b,\\&\frac{\partial ^2 \pi _\mathrm{F}}{ \partial \theta ^2}=-K(1-\psi ) ,\\&\frac{ \partial ^2 \pi _\mathrm{F}}{ \partial w\partial \theta }=\alpha . \end{aligned}$$

We can write the Hessian matrix as

$$\begin{aligned} \left[ \begin{array}{cc} -2b &{}\alpha \\ \alpha &{} -K(1-\psi ) \\ \end{array} \right] . \end{aligned}$$

Since

$$\begin{aligned} \frac{\partial ^2 \pi _\mathrm{F}}{ \partial w^2}=-2b<0, \end{aligned}$$

then the Hessian matrix is negative definite if

$$\begin{aligned} K(1-\psi )\cdot 2b -\alpha \cdot \alpha >0. \end{aligned}$$

Thus, the farmer’s profit function is jointly concave in w and $\theta$. Let the first condition be zero, we obtain

$$\begin{aligned}&\theta (\zeta ,m)=-{\frac{\alpha \, \left( \beta \zeta -bc-bm+a \right) }{2\,Kb\psi -2\,Kb+{\alpha }^{2}}}, \end{aligned}$$

(24)

$$\begin{aligned}&w(\zeta ,m)=\frac{K(1-\psi )(bm-bc-\beta \zeta -a)+\alpha ^2c}{2Kb\psi -2Kb+\alpha ^2}. \end{aligned}$$

(25)

Substituting Eqs. (24–25) into (5), we can derive

$$\begin{aligned} \begin{aligned}&\max _{m,\zeta }\pi _\mathrm{R}\\&\quad = m\left( a-b \left( {\frac{ ( ( c-m ) b+\beta \zeta +a)( \psi -1) K+{\alpha }^{2} c}{2\,Kb\psi -2Kb+{\alpha }^{2}}}+m \right) \right. \\&\qquad \left. -\,{\frac{{\alpha }^{2} ( \beta \zeta -bc-bm+a) }{2\,Kb\psi -2\,K b+{\alpha }^{2}}}+\beta \zeta \right) \\&\qquad -\,\frac{1}{2}L{\zeta }^ {2}-\frac{1}{2}{\frac{K{\alpha }^{2} ( \beta \zeta -bc -bm+a ) ^{2}\psi }{ ( 2\,Kb\psi -2\,Kb+{\alpha }^{2}) ^ {2}}}. \end{aligned} \end{aligned}$$

(26)

According to the first-order condition

$$\begin{aligned} \begin{aligned} \frac{\partial \pi _\mathrm{R}}{ \partial m}&=a-b \left( {\frac{ \left( \left( c-m \right) b+\beta \, \left( \zeta \right) +a \right) \left( \psi -1 \right) K+{\alpha }^{2}c}{2\,Kb\psi - 2\,Kb+{\alpha }^{2}}}+m \right) \\&\quad -{\frac{{\alpha }^{2} \left( \beta \, \left( \zeta \right) -bc-bm+a \right) }{2\,Kb\psi -2\,Kb+{\alpha }^{2}} }+\beta \zeta \\&\quad +m \left( -b \left( -{\frac{b \left( \psi -1 \right) K}{2\,Kb\psi -2\,Kb+{\alpha }^{2}}}+1 \right) \right. \\&\quad +\left. {\frac{{ \alpha }^{2}b}{2\,Kb\psi -2\,Kb+{\alpha }^{2}}} \right) \\&\quad +{\frac{K{\alpha }^{2} \left( \beta \, \left( \zeta \right) -bc-bm+a \right) \psi \,b}{ \left( 2\,Kb\psi -2\,Kb+{\alpha }^{2} \right) ^{2}}},\\ \frac{\partial \pi _\mathrm{R}}{ \partial \zeta }&= m\left( -\frac{b\beta (\psi -1)K}{2Kb\psi -2Kb+\alpha ^2}\right. \\&\left. \quad -\frac{\alpha ^2\beta }{2Kb\psi -2Kb+\alpha ^2}+\beta \right) \\&\quad -L\zeta -\frac{K\alpha ^2(\beta \zeta -bc-bm+a)\psi \beta }{(2Kb\psi -2Kb+\alpha ^2)^2}, \end{aligned} \end{aligned}$$

and the second-order condition

$$\begin{aligned} \frac{\partial ^2 \pi _\mathrm{R}}{ \partial m^2}= & {} -2b\left( -\frac{b(\psi -1)K}{2Kb\psi -2Kb+\alpha ^2}+1\right) \\&+\frac{2 \alpha ^2b}{2Kb\psi -2Kb+\alpha ^2}\\&-\frac{K\alpha ^2b^2\psi }{(2Kb\psi -2Kb+\alpha ^2)^2},\\ \frac{\partial ^2 \pi _\mathrm{R}}{ \partial \zeta ^2}= & {} -L-{\frac{K{\alpha }^{2}\beta ^2\psi }{ \left( 2\,Kb\psi +{\alpha }^{2}-2\,Kb \right) ^{2}}},\\ \frac{\partial ^2 \pi _\mathrm{R}}{\partial m\partial \zeta }= & {} -\frac{b\beta (\psi -1)K}{2Kb\psi -2Kb+\alpha ^2}\!-\! \frac{\alpha ^2\beta }{2Kb\psi -2Kb+\alpha ^2}+\beta \\&+\frac{K\alpha ^2\beta \psi b}{(2Kb\psi -2Kb+\alpha ^2)^2}. \end{aligned}$$

Then, the Hessian matrix is negative definite if

$$\begin{aligned}&-2b\left( -\frac{b(\psi -1)K}{2Kb\psi -2Kb+\alpha ^2}+1\right) +\frac{2\alpha ^2b}{2Kb \psi -2Kb+\alpha ^2}\\&\quad -\frac{K\alpha ^2b^2\psi }{(2Kb\psi -2Kb+\alpha ^2)^2}<0 \end{aligned}$$

and

$$\begin{aligned} \frac{b^2K(4KLb\psi ^2-K\beta ^2\psi ^2-8KLb\psi +2K\beta ^2\psi +3L\alpha ^2 \psi +4KLb-\beta ^2K-2L\alpha ^2)}{(2Kb\psi -2Kb+\alpha ^2)^2}>0. \end{aligned}$$

Equivalent to

$$\begin{aligned} -\frac{b^2K(4Kb\psi ^2-8Kb\psi +3\alpha ^2\psi +4Kb-2\alpha ^2)}{(2Kb \psi -2Kb+\alpha ^2)^2}<0 \end{aligned}$$

and

$$\begin{aligned}&4KLb\psi ^2-K\beta ^2\psi ^2-8KLb\psi \\&\quad +2K\beta ^2\psi +3L\alpha ^2 \psi +4KLb-\beta ^2K-2L\alpha ^2>0, \end{aligned}$$

which implies

$$\begin{aligned} 4Kb\psi ^2-8Kb\psi +3\alpha ^2\psi +4Kb-2\alpha ^2>0 \end{aligned}$$

and

$$\begin{aligned} L(4Kb\psi ^2-8Kb\psi +3\alpha ^2\psi +4Kb-2\alpha ^2)-K\beta ^2(\psi -1)^2>0. \end{aligned}$$

By the first-order condition, we have

$$\begin{aligned}&m=\frac{A_4}{b(4KLb\psi ^2-K\beta ^2\psi ^2-8KLb\psi +2K\beta ^2\psi +3L\alpha ^2 \psi +4KLb-K\beta ^2-2L\alpha ^2)}, \\&\zeta =\frac{K\beta (-bc\psi ^2+a\psi ^2+2bc\psi -2a\psi -bc+a)}{4KLb\psi ^2-K\beta ^2\psi ^2-8KLb\psi +2K\beta ^2\psi +3L\alpha ^2\psi +4KLb-K \beta ^2-2L\alpha ^2}. \end{aligned}$$

Substituting m and $\zeta$ into Eq. (26), we can obtain the retailer’s profit function with respect to $\psi$

$$\begin{aligned} \pi _\mathrm{R}(\psi )=\frac{KL(\psi -1)^2(-bc+a)^2}{2(4KLb\psi ^2-K\beta ^2 \psi ^2-8KLb\psi +2K\beta ^2\psi +3L\alpha ^2\psi +4KLb-\beta ^2K-2L\alpha ^2)}. \end{aligned}$$

Then, the first-order condition

$$\begin{aligned} \frac{\mathrm{d}\pi _\mathrm{R}}{\mathrm{d}\psi }=\frac{KL^2(\psi -1)(-bc+a)^2\alpha ^2(3\psi -1)}{2(4KLb \psi ^2-K\beta ^2\psi ^2-8KLb\psi +2K\beta ^2\psi +3L\alpha ^2 \psi +4KLb-\beta ^2K-2L\alpha ^2)^2} \end{aligned}$$

and the second-order condition

$$\begin{aligned} \frac{\mathrm{d}^2\pi _\mathrm{R}}{\mathrm{d}\psi ^2}=\frac{-KL^2(-bc+a)^2\alpha ^2A_9}{(4KLb\psi ^2-K\beta ^2\psi ^2-8KLb\psi +2K\beta ^2\psi +3L\alpha ^2 \psi +4KLb-\beta ^2K-2L\alpha ^2)^3}. \end{aligned}$$

The second-order condition is negative if $A_9>0.$ Thus, $\pi _\mathrm{R}(\psi )$ is concave in $\psi$ for the above condition. Solving the first-order condition for $\psi$ gives $\psi =\frac{1}{3}$ or $\psi =1$.

When $\psi =1$, $A_9>0$ can be reduced to $-L\alpha ^2>0$, which is a contradiction since $L,K, \alpha ,\beta >0$.

Logically, the equilibrium $\psi ^{\mathrm{CS}^{*}}=\frac{1}{3}$. Substituting the value of $\psi ^{\mathrm{CS}^{*}}=\frac{1}{3}$ into the expressions accordingly, we get

$$\begin{aligned} \zeta ^{\mathrm{CS}^{*}}&= \frac{4K\beta (-bc+a)}{16KLb-4\beta ^2K-9L\alpha ^2}, \\ m^{\mathrm{CS}^{*}}= & {} \frac{(-bc+a)(8Kb-3\alpha ^2)L}{b(16KLb-4\beta ^2K-9L\alpha ^2)}, \\ p^{\mathrm{CS}^{*}}&= w^{\mathrm{CS}^{*}}+m^{\mathrm{CS}^{*}}\\& = \frac{4KLb^2c-4Kb\beta ^2c-6L\alpha ^2bc+12KLab-3La\alpha ^2}{b(16KLb-4\beta ^2K-9L\alpha ^2)}, \\ w^{\mathrm{CS}^{*}}= & {} \frac{12KLbc-4K\beta ^2c-9L\alpha ^2c+4KLa}{16KLb-4\beta ^2K-9L\alpha ^2}, \\ \theta ^{\mathrm{CS}^{*}}&= \frac{6(-bc+a)L\alpha }{16KLb-4\beta ^2K-9L\alpha ^2}, \\ \pi _\mathrm{R}^{\mathrm{CS}^{*}}& = \frac{2KL(-bc+a)^2}{16KLb-4\beta ^2K-9L\alpha ^2}, \\ \pi _\mathrm{F}^{\mathrm{CS}^{*}}= & {} \frac{4(-bc+a)^2(4Kb-3\alpha ^2)L^2K}{(16KLb-4\beta ^2K-9L\alpha ^2)^2}, \\ \pi _\mathrm{SC}^{\mathrm{CS}^{*}}= & {} \frac{2(-bc+a)^2(24KLb-4\beta ^2K-15L\alpha ^2)LK}{(16KLb-4\beta ^2K-9L\alpha ^2)^2}. \end{aligned}$$

Proof of Proposition4

Proof

We can easily get the results from the computational process. $\square$

Proof of Corollary 2

Proof

During the computation process, the

$$\begin{aligned} \theta (\psi )=-\frac{(-bc\psi +a\psi +bc-a)L\alpha }{4KLb\psi ^2-K\beta ^2 \psi ^2-8KLb\psi +2K\beta ^2\psi +3L\alpha ^2\psi +4KLb-\beta ^2K-2L\alpha ^2}. \end{aligned}$$

By computing the first-order derivative, we can obtain the result. $\square$

Proof of Proposition 5

Proof

We prove the result about $\zeta$; the result about w can be proved analogously. During the computation process, the

$$\begin{aligned} \zeta (\psi )=\frac{K\beta (-bc\psi ^2+a\psi ^2+2bc\psi -2a\psi -bc+a)}{4KLb\psi ^2-K\beta ^2\psi ^2-8KLb\psi +2K\beta ^2\psi +3L\alpha ^2\psi +4KLb-\beta ^2K-2L\alpha ^2}. \end{aligned}$$

By computing the first-order derivative with respect to $\psi$ and making analysis, we can obtain the result. $\square$

Proof of Proposition 6

Proof

By computing the first-order derivative with respect to $\psi$ and making analysis, we can obtain the result. $\square$

Proof of Proposition 7

Proof

The result can be proved analogously with Proposition 5. $\square$

Bargaining cost sharing contract model. Since the farmer and the retailer decide the $\psi$ via bargaining, we formulate the objective function as

$$\begin{aligned} \max _{\psi }\pi _\mathrm{F}\pi _\mathrm{R} \end{aligned}$$

(27)

Due to the game structure, we solve the farmer’s optimization problem first:

$$\begin{aligned} \max _{w,\theta }\pi _\mathrm{F}=(w-c)q-\frac{1}{2}K(1-\psi )\theta ^2 \end{aligned}$$

The first-order condition

$$\begin{aligned}&\frac{ \partial \pi _\mathrm{F}}{ \partial w}=a-b(w+m)+\alpha \theta +\beta \zeta -(w-c)b,\\&\frac{\partial \pi _\mathrm{F}}{ \partial \theta }=(w-c)\alpha -K\theta (1-\psi ), \end{aligned}$$

and the second-order condition

$$\begin{aligned}&\frac{ \partial ^2 \pi _\mathrm{F}}{ \partial w^2}=-2b, \\&\frac{\partial ^2 \pi _\mathrm{F}}{ \partial \theta ^2}=-K(1-\psi ) , \\&\frac{ \partial ^2 \pi _\mathrm{R}}{ \partial p\partial \zeta }=\alpha . \end{aligned}$$

We can write the Hessian matrix as

$$\begin{aligned} \left[ \begin{array}{cc} -2b &{}\alpha \\ \alpha &{} -K(1-\psi ) \\ \end{array} \right] . \end{aligned}$$

(28)

The Hessian matrix is negative definite if

$$\begin{aligned} 2bK(1-\psi )-\alpha ^2 > 0. \end{aligned}$$

Under this assumption, the farmer’s profit function is joint concave in p and $\zeta$. Let the first-order condition be zero and solve equations for p and $\zeta$. We can obtain

$$\begin{aligned} w(m,\zeta )=\frac{K(1-\psi )(bm-bc-\beta \zeta -a)+\alpha ^2c}{2Kb\psi -2Kb+\alpha ^2} \end{aligned}$$

(29)

and

$$\begin{aligned} \theta (\zeta ,m)=-{\frac{\alpha \, \left( \beta \zeta -bc-bm+a \right) }{2\,Kb\psi -2\,Kb+{\alpha }^{2}}}. \end{aligned}$$

(30)

Substituting Eqs. (29–30) into the retailer’s profit function, we get

$$\begin{aligned} \max _{m,\zeta }\pi _\mathrm{R}\,=\,& {} m\left( a-b \left( {\frac{ ( ( c-m ) b+\beta \zeta +a)( \psi -1) K+{\alpha }^{2} c}{2\,Kb\psi -2Kb+{\alpha }^{2}}}+m \right) \right. \\&\left. -{\frac{{\alpha }^{2} ( \beta \zeta -bc-bm+a) }{2\,Kb\psi -2\,K b+{\alpha }^{2}}}+\beta \zeta \right) \\&-\frac{1}{2}L{\zeta }^ {2}-\frac{1}{2}{\frac{K{\alpha }^{2} ( \beta \zeta -bc -bm+a ) ^{2}\psi }{ ( 2\,Kb\psi -2\,Kb+{\alpha }^{2}) ^ {2}}}. \end{aligned}$$

The first-order condition

$$\begin{aligned} \frac{\partial \pi _\mathrm{R}}{ \partial m}\,=\, & {} a-b \left( {\frac{ \left( \left( c-m \right) b+\beta \, \left( \zeta \right) +a \right) \left( \psi -1 \right) K+{\alpha }^{2}c}{2\,Kb\psi - 2\,Kb+{\alpha }^{2}}}+m \right) \\&- {\frac{{\alpha }^{2} \left( \beta \, \left( \zeta \right) -bc-bm+a \right) }{2\,Kb\psi -2\,Kb+{\alpha }^{2}} }+\beta \zeta \\&+m \left( -b \left( -{\frac{b \left( \psi -1 \right) K}{2\,Kb\psi -2\,Kb+{\alpha }^{2}}}+1 \right) \right. \\&+\left. {\frac{{ \alpha }^{2}b}{2\,Kb\psi -2\,Kb+{\alpha }^{2}}} \right) \\&+{\frac{K{\alpha }^{2} \left( \beta \, \left( \zeta \right) -bc-bm+a \right) \psi \,b}{ \left( 2\,Kb\psi -2\,Kb+{\alpha }^{2} \right) ^{2}}},\\ \frac{\partial \pi _\mathrm{R}}{ \partial \zeta }= & {} m\left( -\frac{b\beta (\psi -1)K}{2Kb\psi -2Kb+\alpha ^2}\right. \\&\left. - \frac{\alpha ^2\beta }{2Kb\psi -2Kb+\alpha ^2}+\beta \right) -L\zeta \\&-\frac{K\alpha ^2(\beta \zeta -bc-bm+a)\psi \beta }{(2Kb\psi -2Kb+\alpha ^2)^2}, \end{aligned}$$

and the second-order condition

$$\begin{aligned} \frac{\partial ^2 \pi _\mathrm{R}}{ \partial m^2}\,=\, & {} -2b\left( -\frac{b(\psi -1)K}{2Kb\psi -2Kb+\alpha ^2}+1\right) \\&+\frac{2\alpha ^2b}{2Kb\psi -2Kb+\alpha ^2}\\&-\frac{K\alpha ^2b^2\psi }{(2Kb\psi -2Kb+\alpha ^2)^2},\\ \frac{\partial ^2 \pi _\mathrm{R}}{ \partial \zeta ^2}= & {} -L-{\frac{K{\alpha }^{2}\beta ^2\psi }{ \left( 2\,Kb\psi +{\alpha }^{2}-2\,Kb \right) ^{2}}},\\ \frac{\partial ^2 \pi _\mathrm{R}}{ \partial m\partial \zeta }= & {} -\frac{b\beta (\psi -1)K}{2Kb\psi -2Kb+\alpha ^2}\\&- \frac{\alpha ^2\beta }{2Kb\psi -2Kb+\alpha ^2}+\beta +\frac{K\alpha ^2\beta \psi b}{(2Kb\psi -2Kb+\alpha ^2)^2}. \end{aligned}$$

Then, the Hessian matrix is negative definite if

$$\begin{aligned}&-2b\left( -\frac{b(\psi -1)K}{2Kb\psi -2Kb+\alpha ^2}+1\right) +\frac{2\alpha ^2b}{2Kb \psi -2Kb+\alpha ^2}\\&\quad -\frac{K\alpha ^2b^2\psi }{(2Kb\psi -2Kb+\alpha ^2)^2}<0 \end{aligned}$$

and

$$\begin{aligned} \frac{b^2K(4KLb\psi ^2-K\beta ^2\psi ^2-8KLb\psi +2K\beta ^2\psi +3L\alpha ^2 \psi +4KLb-\beta ^2K-2L\alpha ^2)}{(2Kb\psi -2Kb+\alpha ^2)^2}>0. \end{aligned}$$

Equivalent to

$$\begin{aligned} -\frac{b^2K(4Kb\psi ^2-8Kb\psi +3\alpha ^2\psi +4Kb-2\alpha ^2)}{(2Kb \psi -2Kb+\alpha ^2)^2}<0 \end{aligned}$$

and

$$\begin{aligned}&4KLb\psi ^2-K\beta ^2\psi ^2-8KLb\psi \\&\quad +2K\beta ^2\psi +3L\alpha ^2 \psi +4KLb-\beta ^2K-2L\alpha ^2>0 \end{aligned}$$

which implies

$$\begin{aligned} 4Kb\psi ^2-8Kb\psi +3\alpha ^2\psi +4Kb-2\alpha ^2>0 \end{aligned}$$

and

$$\begin{aligned} L(4Kb\psi ^2-8Kb\psi +3\alpha ^2\psi +4Kb-2\alpha ^2)-K\beta ^2(\psi -1)^2>0. \end{aligned}$$

Let the first-order condition be zero and solve equations for m and $\zeta$. We can get

$$\begin{aligned}&m=\frac{A_4}{b(4KLb\psi ^2-K\beta ^2\psi ^2-8KLb\psi +2K\beta ^2\psi +3L\alpha ^2 \psi +4KLb-K\beta ^2-2L\alpha ^2)}, \\&\zeta =\frac{K\beta (-bc\psi ^2+a\psi ^2+2bc\psi -2a\psi -bc+a)}{4KLb\psi ^2-K\beta ^2\psi ^2-8KLb\psi +2K\beta ^2\psi +3L\alpha ^2 \psi +4KLb-K\beta ^2-2L\alpha ^2}. \end{aligned}$$

Substituting m and $\zeta$ into Eqs. (29–30) and the Nash bargaining function (27). By computing the first-order condition and the second-order condition of the bargaining model

$$\begin{aligned} \max _{\psi }\pi _\mathrm{F}\pi _\mathrm{R}. \end{aligned}$$

The first-order condition

$$\begin{aligned} \frac{\mathrm{d}\pi _\mathrm{B}}{\mathrm{d}\psi }=\frac{K^2L^2(\psi -1)^4(-cb+a)^4\alpha ^2A_1}{4(4KLb\psi ^2-K\beta ^2\psi ^2-8KLb\psi +2K\beta ^2\psi +3L\alpha ^2 \psi +4KLb-K\beta ^2-2L\alpha ^2)^4}. \end{aligned}$$

The second-order condition

$$\begin{aligned} \frac{\mathrm{d}^{2} \pi _\mathrm{B}}{\mathrm d\psi ^2}=-A_2\cdot \frac{K^2\alpha ^2L^3(-bc+a)^4}{2(4KLb\psi -2K\beta ^2\psi -4KLb+2K\beta ^2+L\alpha ^2)^5}. \end{aligned}$$

The second-order condition is negative if $A_2<0.$ Thus, the objective function is concave in $\psi$ for the above condition. Let the first-order condition be 0. We obtain $\psi =1$,

$$\begin{aligned} \psi =\frac{8KLb+K\beta ^2-3L\alpha ^2+A_3}{K(14Lb+\beta ^2)}, \end{aligned}$$

or

$$\begin{aligned} \psi =\frac{8KLb+K\beta ^2-3L\alpha ^2-A_3}{K(14Lb+\beta ^2)}. \end{aligned}$$

When $\psi =1$, $2Kb(1-\psi )-\alpha ^2>0$ can be deduced to $-\alpha ^2>0$, which is a contradiction. Meanwhile, if

$$\begin{aligned} \psi =\frac{8KLb+K\beta ^2-3L\alpha ^2+A_3}{K(14Lb+\beta ^2)}, \end{aligned}$$

then the $2Kb(1-\psi )-\alpha ^2>0$ is not satisfied. When

$$\begin{aligned} \psi\,=\,& {} \frac{8KLb+K\beta ^2-3L\alpha ^2+A_3}{K(14Lb+\beta ^2)},\\ 2Kb(1-\psi )-\alpha ^2\,=\, & {} \frac{12KLb^2-8L\alpha ^2b-\alpha ^2\beta ^2-2bA_3}{14Lb+\beta ^2}. \end{aligned}$$

If $12KLb^2-8L\alpha ^2b-\alpha ^2\beta ^2<0$, then we can get $2Kb(1-\psi )-\alpha ^2<0$; if $12KLb^2-8L\alpha ^2b-\alpha ^2\beta ^2>0$, we only need to check $(12KLb^2-8L\alpha ^2b-\alpha ^2\beta ^2)^2-(2bA_3)^2$ is greater than or less than 0. By algebra,

$$\begin{aligned}&(12KLb^2-8L\alpha ^2b-\alpha ^2\beta ^2)^2-(2bA_3)^2\\&\quad =-\alpha ^2(14Lb+ \beta ^2)(4KLb^2-2L\alpha ^2b-\alpha ^2\beta ^2). \end{aligned}$$

The problem transfers to verify whether $4KLb^2-2L\alpha ^2b-\alpha ^2\beta ^2$ is greater or less than 0. Since we have $2KLb-\beta ^2K-L\alpha ^2>0$ and $2Kb-\alpha ^2>0$, $2b(2KLb-\beta ^2K-L\alpha ^2)>0$ and $4KLb^2-2L\alpha ^2b-2\beta ^2Kb>0$. Owing to $2Kb-\alpha ^2>0$, we can derive $4KLb^2-2L\alpha ^2b-\alpha ^2\beta ^2>0$. Thus, $2Kb(1-\psi )-\alpha ^2<0$ when

$$\begin{aligned} \psi =\frac{8KLb+K\beta ^2-3L\alpha ^2+A_3}{K(14Lb+\beta ^2)}. \end{aligned}$$

Obviously, the

$$\begin{aligned} \psi =\frac{8KLb+K\beta ^2-3L\alpha ^2+A_3}{K(14Lb+\beta ^2)} \end{aligned}$$

should be omitted. Analogously, we can also prove that $2Kb(1-\psi )-\alpha ^2>0$ can be satisfied when

$$\begin{aligned} \psi =\frac{8KLb+K\beta ^2-3L\alpha ^2-A_3}{K(14Lb+\beta ^2)}. \end{aligned}$$

Thus, the equilibrium

$$\begin{aligned} \psi ^{\mathrm{B}^{*}}=\frac{8KLb+K\beta ^2-3L\alpha ^2-A_3}{K(14Lb+\beta ^2)}. \end{aligned}$$

The equilibrium decisions and profit functions are

$$\begin{aligned}&\zeta ^{\mathrm{B}^{*}}\,=\,\frac{K\beta (-bc+a)(\psi ^{\mathrm{B}^{*}}-1)^2}{4KLb{\psi ^{\mathrm{B}^{*}}}^2-K\beta ^2{\psi ^{\mathrm{B}^{*}}}^2-8KLb{\psi ^{\mathrm{B}^{*}}}+2K\beta ^2{\psi ^{\mathrm{B}^{*}}}+ 3L\alpha ^2{\psi ^{\mathrm{B}^{*}}}+4KLb-K\beta ^2-2L\alpha ^2}, \\&m^{\mathrm{B}^{*}}\,=\,\frac{A_4}{4KLb{\psi ^{\mathrm{B}^{*}}}^2-K\beta ^2{\psi ^{\mathrm{B}^{*}}}^2-8KLb{\psi ^{\mathrm{B}^{*}}}+2K\beta ^2{\psi ^{\mathrm{B}^{*}}}+ 3L\alpha ^2{\psi ^{\mathrm{B}^{*}}}+4KLb-K\beta ^2-2L\alpha ^2}, \\&\theta ^{\mathrm{B}^{*}}\,=\,\frac{(-bc+a)(1-\psi ^{\mathrm{B}^{*}})}{4KLb{\psi ^{\mathrm{B}^{*}}}^2-K\beta ^2{\psi ^{\mathrm{B}^{*}}}^2-8KLb{\psi ^{\mathrm{B}^{*}}}+2K\beta ^2{\psi ^{\mathrm{B}^{*}}}+ 3L\alpha ^2{\psi ^{\mathrm{B}^{*}}}+4KLb-K\beta ^2-2L\alpha ^2}, \\&w^{\mathrm{B}^{*}}\,=\,\frac{(3KLbc-K\beta ^2c+KLa)(\psi ^{\mathrm{B}^{*}}-1)^2+3L\alpha ^2c(\psi _B^{*}-1)+L\alpha ^2c}{4KLb{\psi ^{\mathrm{B}^{*}}}^2-K\beta ^2{\psi ^{\mathrm{B}^{*}}}^2-8KLb{\psi ^{\mathrm{B}^{*}}}+2K\beta ^2{\psi ^{\mathrm{B}^{*}}}+ 3L\alpha ^2{\psi ^{\mathrm{B}^{*}}}+4KLb-K\beta ^2-2L\alpha ^2}, \\&\pi _\mathrm{R}^{\mathrm{B}^{*}}\,=\,\frac{KL(-bc+a)^2(\psi ^{\mathrm{B}^{*}}-1)^2}{2(4KLb{\psi ^{\mathrm{B}^{*}}}^2-K\beta ^2{\psi ^{\mathrm{B}^{*}}}^2-8KLb{\psi ^{\mathrm{B}^{*}}}+2K\beta ^2{\psi ^{\mathrm{B}^{*}}}+ 3L\alpha ^2{\psi ^{\mathrm{B}^{*}}}+4KLb-K\beta ^2-2L\alpha ^2)}, \\&\pi _\mathrm{F}^{\mathrm{B}^{*}}\,=\,\frac{(\psi ^{\mathrm{B}^{*}}-1)^3(-bc+a)^2(2Kb\psi ^{\mathrm{B}^{*}}-2Kb+\alpha ^2)KL^2}{2(4KLb{\psi ^{\mathrm{B}^{*}}}^2-K\beta ^2{\psi ^{\mathrm{B}^{*}}}^2-8KLb{\psi ^{\mathrm{B}^{*}}}+2K\beta ^2{\psi ^{\mathrm{B}^{*}}}+ 3L\alpha ^2{\psi ^{\mathrm{B}^{*}}}+4KLb-K\beta ^2-2L\alpha ^2)^2}, \\&\pi _\mathrm{SC}^{\mathrm{B}^{*}}\,=\,\frac{(\psi ^{\mathrm{B}^{*}}-1)^2(-bc+a)^2 A_5 KL}{2(4KLb{\psi ^{\mathrm{B}^{*}}}^2-K\beta ^2{\psi ^{\mathrm{B}^{*}}}^2-8KLb{\psi ^{\mathrm{B}^{*}}}+2K\beta ^2{\psi ^{\mathrm{B}^{*}}}+ 3L\alpha ^2{\psi ^{\mathrm{B}^{*}}}+4KLb-K\beta ^2-2L\alpha ^2)^2}. \end{aligned}$$

Proof of Proposition 8

Proof

We can get the results from the computational process. $\square$

Model comparison.

Proof of Proposition 9

Proof

$$\begin{aligned} \psi ^{\mathrm{B}^{*}}-\frac{1}{3}\,=\, & {} \frac{8KLb+K\beta ^2-3L\alpha ^2-A_3}{K(14Lb+\beta ^2)}-\frac{1}{3}\\= & {} \frac{10KLb+2K\beta ^2-9L\alpha ^2-3A_3}{3K(14Lb+\beta ^2)}. \end{aligned}$$

We need to prove $10KLb+2K\beta ^2-9L\alpha ^2-3A_3<0$ which is equivalent to $(10KLb+2K\beta ^2-9L\alpha ^2)^2-(3A_3)^2<0$. By algebra, we need to prove

$$\begin{aligned} -K(14Lb+\beta ^2)(16KLb-4K\beta ^2-9L\alpha ^2)<0, \end{aligned}$$

which is true since $K>0$ and $16KLb-4K\beta ^2-9L\alpha ^2>0$. $\square$

Proof of Proposition 10

Proof

We prove $w^{\mathrm{CS}^{*}}>w^{\mathrm{D}^{*}}$ for example. The proofs of $\theta ^{\mathrm{CS}^{*}}>\theta ^{\mathrm{D}^{*}},\,p^{\mathrm{CS}^{*}}>p^{\mathrm{D}^{*}},\zeta ^{\mathrm{CS}^{*}}>\zeta ^{\mathrm{D}^{*}}$ can be done analogously.

The $w^{\mathrm{CS}^{*}}=\frac{12KLbc-4K\beta ^2c-9L\alpha ^2c+4KLa}{16KLb-4\beta ^2K-9L\alpha ^2}$ and the $w^{\mathrm{D}^{*}}=\frac{[L(3bc+a)-c\beta ^2]K-2Lc\alpha ^2}{4LbK-\beta ^2K-2L\alpha ^2}$. It is equivalent to prove $\frac{w^{\mathrm{CS}^{*}}}{w^{\mathrm{D}^{*}}}>1$.

$$\begin{aligned} \begin{aligned} \frac{w^{\mathrm{CS}^{*}}}{w^{\mathrm{D}^{*}}}>1&\Longleftrightarrow \frac{(12KLbc-4K\beta ^2c-9L\alpha ^2+4KLa)(4KLb-\beta ^2-2L\alpha ^2)}{(16KLb-4K\beta ^2-9L\alpha ^2)(3KLbc-K\beta ^2c-2L\alpha ^2c+KLa)}>1\\&\Longleftrightarrow (12KLbc-4K\beta ^2c-9L\alpha ^2+4KLa)(4KLb-\beta ^2-2L\alpha ^2)\\&\qquad -(16KLb-4K\beta ^2-9L\alpha ^2)(3KLbc-K\beta ^2c-2L\alpha ^2c+KLa)>0\\&\Longleftrightarrow KL^2\alpha ^2(-bc+a)>0. \end{aligned} \end{aligned}$$

Obviously, $a-bc>0$ holds true. $\square$

Proof of Proposition 11

Proof

The proof can be obtained with the first-order derivative. $\square$

Proof of Proposition 12

Proof

The proof of (1) can be derived from Propositions 7 and 9; the proof of (2) is analogous to Proposition 10. $\square$