Analytical comparisons in a three-echelon closed-loop supply chain with price and marketing effort-dependent demand: game theory approaches

Abstract

Over the last few decades, closed-loop supply chain (CLSC) has received increasing attention due to concerns over the environment and social liability. Moreover, recycling and remanufacturing of the used products have been examined because of various factors such as concerns over the environment, lack of resources, government legislations. In this research, we consider a three-echelon closed-loop supply chain consisting of a manufacturer, a third party and a retailer. The manufacturer manipulates both manufacturing from raw materials and remanufacturing from the second hands products collected by third party simultaneously. We assume that the market demand depends on selling price and marketing efforts. First, we modeled the supply chain under centralized and decentralized policies and compared their performances using numerical examples. Then we compared and analyzed the optimal decisions under different scenarios by studying the impacts of marketing efforts and collection rate on the decision variables and concluded that the supply chain profit in centralized scenario is larger than decentralized one. Using a comprehensive sensitivity analysis, managerial insights are provided. Moreover, based on the derived results, from the perspective of remanufacturing process and consumers’ welfare, one can conclude that the manufacture-Stackelberg case is often the most effective scenario in CLSC.

Appendices

Appendix 1: Proof of Proposition 1

The Hessian matrix of Π _C is;

$$H\left( {p,\theta ,\tau } \right) = \left( {\begin{array}{*{20}l} {\frac{{\partial^{2} \varPi_{C} }}{{\partial p^{2} }}} \hfill & {\frac{{\partial^{2} \varPi_{C} }}{\partial p \partial \theta }} \hfill & {\frac{{\partial^{2} \varPi_{C} }}{\partial p \partial \tau }} \hfill \\ {\frac{{\partial^{2} \varPi_{C} }}{\partial \theta \partial p}} \hfill & {\frac{{\partial^{2} \varPi_{C} }}{{\partial \theta^{2} }}} \hfill & {\frac{{\partial^{2} \varPi_{C} }}{\partial \theta \partial \tau }} \hfill \\ {\frac{{\partial^{2} \varPi_{C} }}{\partial \tau \partial p}} \hfill & {\frac{{\partial^{2} \varPi_{C} }}{\partial \tau \partial \theta }} \hfill & {\frac{{\partial^{2} \varPi_{C} }}{{\partial \tau^{2} }}} \hfill \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} { - 2\alpha } & \beta & { - \alpha (\Delta - A)} \\ \beta & { - 2c_{\theta } } & {\beta (\Delta - A)} \\ { - \alpha (\Delta - A)} & {\beta (\Delta - A)} & { - 2c_{\tau } } \\ \end{array} } \right)$$

To ensure Π _C is jointly concave in p, θ and τ, we check that,

$$\left| {H_{1} } \right| = - 2\alpha < 0$$

$$\left| {H_{2} } \right| = 4\alpha c_{\theta } - \beta^{2} > 0,\quad {\text{if}}\;c_{\theta } > \frac{{\beta^{2} }}{4\alpha }$$

$$\left| {H_{3} } \right| = 2\left[ {\alpha^{2} c_{\theta } \left( {\Delta - A} \right)^{2} - c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right)} \right] < 0,\quad {\text{if}}\;c_{\tau } > \frac{{\alpha^{2} c_{\theta } (\Delta - A)^{2} }}{{4\alpha c_{\theta } - \beta^{2} }}$$

Under the above conditions, the Hessian matrix is negative definite and the optimal decision can be obtained through solving the first-order derivatives of Eq. (4) with respect to p, θ and τ.

$$\frac{{\partial \varPi_{C} }}{\partial p} = \phi + \beta \theta - 2\alpha p + \alpha c_{m} - \alpha \tau (\Delta - A) = 0$$

$$\frac{{\partial \varPi_{C} }}{\partial \theta } = \beta p - 2c_{\theta } \theta - \beta c_{m} + \beta \tau (\Delta - A) = 0$$

$$\frac{{\partial \varPi_{C} }}{\partial \tau } = (\Delta - A)(\phi - \alpha p + \beta \theta ) - 2c_{\tau } \tau = 0$$

By solving above equations, Proposition 1 is obtained.

Appendix 2: Proof of Proposition 2

First, we analyze the retailer’s reaction to maximize its own profit. Using Eq. (2), we have the first-order derivatives of Π_R with respect to m and θ and the Hessian matrix of Π_R as follows:

$$\frac{{\partial \varPi_{R} }}{{\partial m^{M} }} = \phi - 2\alpha m^{M} + \beta \theta^{M} - \alpha w^{M} = 0$$

$$\frac{{\partial \varPi_{R} }}{{\partial \theta^{M} }} = \beta m^{M} - 2c_{\theta } \theta^{M} = 0$$

$$H^{R} = \left( {\begin{array}{*{20}c} { - 2\alpha } & \beta \\ \beta & { - 2c_{\theta } } \\ \end{array} } \right)$$

The Hessian matrix of Π_R is a negative definite for all values of m and θ if $c_{\theta } > \frac{{\beta^{2} }}{4\alpha }$. With solving above equations,

$$m^{M} = \frac{{2c_{\theta } \left( {\phi - \alpha w^{M} } \right)}}{{4\alpha c_{\theta } - \beta^{2} }}\quad {\text{and}}\quad \theta^{M} = \frac{{\beta \left( {\phi - \alpha w^{M} } \right)}}{{4\alpha c_{\theta } - \beta^{2} }}$$

Then, we investigate the third party’s reaction to maximize its own profit. Using Eq. (3), from the first-order derivative of Π_3P respect to τ, we have;

$$\frac{{\partial \varPi_{3P} }}{{\partial \tau^{M} }} = \left( {b^{M} - A} \right)\left( {\phi - \alpha \left( {w^{M} + m^{M} } \right) + \beta \theta^{M} } \right) - 2c_{\tau } \tau^{M} = 0$$

$$\frac{{\partial^{2} \varPi_{3P} }}{{\left( {\partial \tau^{R} } \right)^{2} }} = - 2c_{\tau } < 0$$

The optimal collection rate is derived by setting the first-order derivative of profit function respect to collection rate equal to zero as follows:

$$\tau^{M} = \frac{{\left( {b^{M} - A} \right)\left( {\phi - \alpha \left( {w^{M} + m^{M} } \right) + \beta \theta^{M} } \right)}}{{2c_{\tau } }}$$

After taking the best reactions of the retailer and the third party, the manufacturer by substituting m, θ and τ to own profit, determines optimal wholesale price, w, and transfer price, b. The Hessian matrix and the first-order derivatives of Eq. (1) with respect to w and b are;

$$\frac{{\partial \varPi_{M} }}{{\partial w^{M} }} = \frac{{2\alpha c_{\theta } \left[ { - 2\alpha^{2} c_{\theta } \left( {\Delta - b^{M} } \right)\left( {b^{M} - A} \right)\left( {\phi - \alpha w^{M} } \right) + c_{\tau } \phi \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right)\left( {2w^{M} - c_{m} } \right)} \right]}}{{c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right)^{2} }}$$

$$\frac{{\partial \varPi_{M} }}{{\partial b^{M} }} = \frac{{2\alpha^{2} c_{\theta }^{2} \left( {\phi - \alpha w^{M} } \right)^{2} \left[ {\left( {\Delta - b^{M} } \right) - \left( {b^{M} - A} \right)} \right]}}{{c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right)^{2} }}$$

and,

$$H^{M} = \left( {\begin{array}{*{20}c} {\frac{{ - 4\alpha^{2} c_{\theta } \left[ {c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } \left( {\Delta - b^{M} } \right)\left( {b^{M} - A} \right)} \right]}}{{c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right)^{2} }}} & {\frac{{ - 4\alpha^{3} c_{\theta }^{2} \left( {\phi - \alpha w^{M} } \right)\left[ {\left( {\Delta - b^{M} } \right) - \left( {b^{M} - A} \right)} \right]}}{{c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right)^{2} }}} \\ {\frac{{ - 4\alpha^{3} c_{\theta }^{2} \left( {\phi - \alpha w^{M} } \right)\left[ {\left( {\Delta - b^{M} } \right) - \left( {b^{M} - A} \right)} \right]}}{{c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right)^{2} }}} & {\frac{{ - 4\alpha^{2} c_{\theta }^{2} \left( {\phi - \alpha w^{M} } \right)^{2} }}{{c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right)^{2} }}} \\ \end{array} } \right)$$

The Hessian H is negative definite if $c_{\tau } > \frac{{\alpha^{2} c_{\theta } (\Delta - A)^{2} }}{{4(4\alpha c_{\theta } - \beta^{2} )}}$.

By setting the first-order derivatives equal to zero, we obtain the optimal w and b and by substituting them into pervious equations, Proposition (2) is obtained.

Appendix 3: Proof of Proposition 3

Similar to the proof of Proposition (2), the best response of the third party for collection rate, τ, is;

$$\tau^{R} = \frac{{\left( {b^{M} - A} \right)\left( {\phi - \alpha \left( {w^{M} + m^{M} } \right) + \beta \theta^{M} } \right)}}{{2c_{\tau } }}$$

Then, by substituting τ into manufacturer’s profit, we obtain Hessian matrix and the first-order derivative of manufacturer’s profit.

$$H^{M} = \left( {\begin{array}{*{20}c} {\frac{{\alpha^{2} (b - A)(\Delta - b)}}{{c_{\tau } }} - 2\alpha } & {\frac{{ - \alpha \left( {\phi - \alpha \left( {w^{M} + m^{M} } \right) + \beta \theta^{M} } \right)\left[ {\left( {\Delta - b^{R} } \right) - \left( {b^{R} - A} \right)} \right]}}{{c_{\tau } }}} \\ {\frac{{ - \alpha \left( {\phi - \alpha \left( {w^{M} + m^{M} } \right) + \beta \theta^{M} } \right)\left[ {\left( {\Delta - b^{R} } \right) - \left( {b^{R} - A} \right)} \right]}}{{c_{\tau } }}} & {\frac{{ - \left( {\phi - \alpha \left( {w^{M} + m^{M} } \right) + \beta \theta^{M} } \right)^{2} }}{{c_{\tau } }}} \\ \end{array} } \right)$$

$$\begin{array}{*{20}l} {\frac{{\partial \varPi_{M} }}{{\partial w^{R} }} = \phi - \alpha \left( {m^{R} + 2w^{R} - c_{m} } \right) + \beta \theta^{R} - \frac{{\alpha (b - A)(\Delta - b)\left( {\phi - \alpha \left( {m^{R} + w^{R} } \right) + \beta \theta^{R} } \right)}}{{c_{\tau } }} = 0} \hfill \\ {\frac{{\partial \varPi_{M} }}{{\partial b^{M} }} = \frac{{\left( {\phi - \alpha \left( {w^{M} + m^{M} } \right) + \beta \theta^{M} } \right)^{2} \left[ {\left( {\Delta - b^{R} } \right) - \left( {b^{R} - A} \right)} \right]}}{{2c_{\tau } }} = 0} \hfill \\ \end{array}$$

The Hessian matrix H is negative definite if $c_{\tau } > \frac{{\alpha (\Delta - A)^{2} }}{8}$. The optimal w and b, according to the above conditions, are

$$b^{R} = \frac{{\Delta + A}}{2}$$

$$w^{R} = \frac{{4\phi c_{\tau } - 4\alpha c_{\tau } \left( {m^{R} - c_{m} } \right) + 4\beta c_{\tau } \theta^{R} - \left[ {\alpha \left( {\Delta - A} \right)^{2} \left( {\phi - \alpha m^{R} + \beta \theta^{R} } \right)} \right]}}{{8\alpha c_{\tau } - \alpha^{2} (\Delta - A)^{2} }}$$

After getting the reactions of the manufacturer and the third party, the retailer specifies the unit retail margin, m, and marketing effort level, θ. The retailer maximizes its own profit by using the first-order derivatives of Eq. (2) with respect to m and θ and setting them equal to zero. Then by substituting them into pervious Equations, Proposition 3 is obtained.

$$\begin{array}{*{20}l} {\frac{{\partial \varPi_{R} }}{{\partial m^{R} }} = \frac{{4c_{\tau } \left( {\phi - \alpha c_{m} - 2\alpha m^{R} + \beta \theta^{R} } \right)}}{{8c_{\tau } - \alpha (\Delta - A)^{2} }} = 0} \hfill \\ {\frac{{\partial \varPi_{R} }}{{\partial \theta^{R} }} = \frac{{4c_{\tau } \left( {\beta m^{R} - 4c_{\theta } \theta^{R} } \right) + 2\alpha c_{\theta } \theta^{R} (\Delta - A)^{2} }}{{8c_{\tau } - \alpha (\Delta - b)^{2} }} = 0} \hfill \\ \end{array}$$

$$H^{R} = \left( {\begin{array}{*{20}c} {\frac{{ - 8\alpha c_{\tau } }}{{8c_{\tau } - \alpha (\Delta - A)^{2} }}} & {\frac{{4\beta c_{\tau } }}{{8c_{\tau } - \alpha (\Delta - A)^{2} }}} \\ {\frac{{4\beta c_{\tau } }}{{8c_{\tau } - \alpha (\Delta - A)^{2} }}} & { - 2c_{\theta } } \\ \end{array} } \right)$$

And the Hessian matrix is negative definite if $c_{\tau } > \frac{{\alpha (\Delta - A)^{2} }}{8}$ and $c_{\tau } > \frac{{\alpha^{2} c_{\theta } (\Delta - A)^{2} }}{{8\alpha c_{\theta } - \beta^{2} }}$

Appendix 4: Proof of Proposition 4

Similar to the proof of Proposition (2), the best response of the retailer for unit retail margin m and marketing effort level θ is;

$$m^{3P} = \frac{{2c_{\theta } \left( {\phi - \alpha w^{3P} } \right)}}{{4\alpha c_{\theta } - \beta^{2} }}$$

$$\theta^{3P} = \frac{{\beta \left( {\phi - \alpha w^{3P} } \right)}}{{4\alpha c_{\theta } - \beta^{2} }}$$

moreover,

$$\frac{{\partial \varPi_{M} }}{{\partial w^{3P} }} = \frac{{\phi + \alpha \left( {c_{m} - 2w^{3P} } \right) - \alpha \tau^{3P} \left( {\Delta - b^{3P} } \right)}}{{4\alpha c_{\theta } - \beta^{2} }}$$

$$\frac{{\partial^{2} \varPi_{M} }}{{\left( {\partial w^{3P} } \right)^{2} }} = \frac{{ - 4\alpha^{2} c_{\theta } }}{{4\alpha c_{\theta } - \beta^{2} }}$$

so,

$$w^{3P} = \frac{{\phi + \alpha c_{m} - \alpha \tau^{3P} \left( {\Delta - b^{3P} } \right)}}{2\alpha }$$

eventually, the third party gets reactions of the manufacturer and the retailer and determines collection rate, τ, and transfer price, b. So,

$$\begin{array}{*{20}l} {\frac{{\partial \varPi_{3P} }}{{\partial \tau^{3P} }} = \frac{{\alpha c_{\theta } \left( {b^{3P} - A} \right)\left( {\phi - \alpha c_{m} } \right) + 2\alpha^{2} c_{\theta } \tau^{3P} \left( {b^{3P} - A} \right)\left( {\Delta - b^{3P} } \right) - 2c_{\tau } \tau^{3P} \left( {4\alpha c_{\theta } - \beta^{2} } \right)}}{{4\alpha c_{\theta } - \beta^{2} }}} \hfill \\ {\frac{{\partial \varPi_{3P} }}{{\partial b^{3P} }} = \frac{{\alpha c_{\theta } \tau^{3P} \left[ {\left( {\phi - \alpha c_{m} } \right) + \alpha \tau^{3P} \left[ {\left( {\Delta - b^{3P} } \right) - \left( {b^{3P} - A} \right)} \right]} \right]}}{{4\alpha c_{\theta } - \beta^{2} }}} \hfill \\ \end{array}$$

The Hessian matrix is negative definite if $c_{\tau } > \frac{{\alpha^{2} c_{\theta } (\Delta - A)^{2} }}{{4\left( {4\alpha c_{\theta } - \beta^{2} } \right)}}$.

$$H^{3P} = \left( {\begin{array}{*{20}c} {\frac{{2\left[ {\alpha^{2} c_{\theta } \left( {b^{3P} - A} \right)\left( {\Delta - b^{3P} } \right) - c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right)} \right]}}{{4\alpha c_{\theta } - \beta^{2} }}} & {\frac{{\alpha c_{\theta } \left[ {\left( {\phi - \alpha c_{m} } \right) + 2\alpha \tau^{3P} \left[ {\left( {\Delta - b^{3P} } \right) - \left( {b^{3P} - A} \right)} \right]} \right]}}{{4\alpha c_{\theta } - \beta^{2} }}} \\ {\frac{{\alpha c_{\theta } \left[ {\left( {\phi - \alpha c_{m} } \right) + 2\alpha \tau^{3P} \left[ {\left( {\Delta - b^{3P} } \right) - \left( {b^{3P} - A} \right)} \right]} \right]}}{{4\alpha c_{\theta } - \beta^{2} }}} & {\frac{{2\alpha^{2} \tau^{2} c_{\theta } }}{{4\alpha c_{\theta } - \beta^{2} }}} \\ \end{array} } \right)$$

Finally, by substituting τ and b into pervious equations, Proposition 4 is obtained.

Appendix 5: Proof of Proposition 5

The proof of Proposition (5) is similar to Proposition (4) by this difference that the manufacturer determines transfer price b, so we omit this proof.

Appendix 6: Proof of Lemma 2

From Propositions (2), (4) and (5), we conclude that $\tau^{M} = \tau_{1}^{3P} = \tau_{2}^{3P}$. Also from “Appendix 1”, we have $c_{\theta } > \frac{{\beta^{2} }}{4\alpha }$ and $c_{\tau } > \frac{{\alpha^{2} c_{\theta } (\Delta - A)^{2} }}{{4\alpha c_{\theta } - \beta^{2} }}$. Furthermore, we have $\phi - \alpha p > 0$ and $p > c_{m}$, so $\phi - \alpha c_{m} > 0$. By substituting these inequalities in each equation, we can prove subsequent Lemmas easily.

$$\tau^{C} - \tau^{M} = \frac{{3\alpha c_{\theta } c_{\tau } \left( {\phi - \alpha c_{m} } \right)\left( {\Delta - A} \right)\left( {4\alpha c_{\theta } - \beta^{2} } \right)}}{{\left[ {4c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } \left( {\Delta - A} \right)^{2} } \right]\left[ {c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } \left( {\Delta - A} \right)^{2} } \right]}} > 0 \to \tau^{C} > \tau^{M}$$

$$\begin{aligned} \tau^{C} - \tau^{R} = & \frac{{\alpha c_{\theta } \left( {\phi - \alpha c_{m} } \right)\left( {\Delta - A} \right)\left[ {2c_{\tau } \left( {6\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]}}{{\left[ {2c_{\tau } \left( {8\alpha c_{\theta } - \beta^{2} } \right) - 2\alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]\left[ {c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]}} \\ \ge & \frac{{\alpha c_{\theta } \left( {\phi - \alpha c_{m} } \right)(\Delta - A)\left[ {\alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]\left[ {2\left( {\frac{{6\alpha c_{\theta } - \beta^{2} }}{{4\alpha c_{\theta } - \beta^{2} }}} \right) - 1} \right]}}{{\left[ {2c_{\tau } \left( {8\alpha c_{\theta } - \beta^{2} } \right) - 2\alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]\left[ {c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]}} > 0 \to \tau^{C} > \tau^{R} . \\ \end{aligned}$$

Furthermore, with comparing τ ^M and τ ^R, we extract the following relationships:

$$\tau^{M} - \tau^{R} = \frac{{\alpha c_{\theta } \left( {\phi - \alpha c_{m} } \right)(\Delta - A)\left[ {2c_{\tau } \beta^{2} - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]}}{{2\left[ {4c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]\left[ {c_{\tau } \left( {8\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]}}$$

• 1. $\tau^{R} < \tau^{M} ,\quad {\text{if}}\;\frac{{3\beta^{2} }}{4\alpha } > c_{\theta }$

• 2. $\tau^{R} < \tau^{M} ,\quad {\text{if}}\;c_{\theta } > \frac{{3\beta^{2} }}{4\alpha }\quad {\text{and}}\quad \frac{{\alpha^{2} c_{\theta } (\Delta - A)^{2} }}{{2\beta^{2} }} > c_{\tau }$

• 3. $\tau^{R} < \tau^{M} ,\quad {\text{if}}\;c_{\theta } > \frac{{3\beta^{2} }}{4\alpha }\quad {\text{and}}\quad c_{\tau } > \frac{{\alpha^{2} c_{\theta } (\Delta - A)^{2} }}{{2\beta^{2} }}$

Finally, by summarizing all of above relationships, Lemma (2) is obtained.

Appendix 7: Proof of Lemma 3

From Propositions (2), (3) and (5), we conclude $b^{M} = b^{R} = b_{2}^{3P}$. Also

$$b_{1}^{3P} - b^{M} > \frac{{\alpha^{2} c_{\theta } (\Delta - A)^{2} }}{{4\alpha c_{\theta } - \beta^{2} }}\frac{{2\left( {4\alpha c_{\theta } - \beta^{2} } \right)}}{{\alpha^{2} c_{\theta } (\Delta - A)}} - \frac{{\Delta - A}}{2} = \frac{3}{2}(\Delta - A) > 0 \to b_{1}^{3P} > b^{M}$$

From above inequality, Lemma (3) is concluded.

Appendix 8: Proof of Lemma 4

First, we prove easily, $w_{2}^{3P} < w_{1}^{3P}$, $w^{M} < w_{2}^{3P}$ and $w^{R} < w_{2}^{3P}$, because;

$$w_{1}^{3P} - w_{2}^{3P} = \frac{{\phi - \alpha c_{m} }}{4\alpha } > 0$$

$$w_{2}^{3P} - w^{M} = \frac{{\alpha c_{\theta } (\Delta - A)^{2} \left( {\phi - \alpha c_{m} } \right)}}{{4\left[ {4c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]}} > 0$$

$$w_{2}^{3P} - w^{R} = \frac{{\left( {\phi - \alpha c_{m} } \right)\left[ {8\left( {4\alpha c_{\theta } - \beta^{2} } \right)^{2} c_{\tau }^{2} - \left[ {\alpha^{2} c_{\theta } \left( {16\alpha c_{\theta } - 3\beta^{2} } \right)(\Delta - A)^{2} } \right]c_{\tau } + \alpha^{4} c_{\theta }^{2} (\Delta - A)^{4} } \right]}}{{4\alpha \left[ {4c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]\left[ {c_{\tau } \left( {8\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]}} > 0$$

Then, we compare $w^{M}$ with $w^{R}$ and derive two roots $c_{\tau 1} ,c_{\tau 2}$ from the following equations if $c_{\theta } < \frac{{\beta^{2} }}{2\alpha }$.

$$w^{M} - w^{R} = \frac{{\left( {\phi - \alpha c_{m} } \right)\left[ {4\left( {4\alpha c_{\theta } - \beta^{2} } \right)^{2} c_{\tau }^{2} - \left[ {2\alpha^{2} c_{\theta } \left( {6\alpha c_{\theta } - \beta^{2} } \right)(\Delta - A)^{2} } \right]c_{\tau } + \alpha^{4} c_{\theta }^{2} (\Delta - A)^{4} } \right]}}{{2\alpha \left[ {4c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]\left[ {c_{\tau } \left( {8\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]}}$$

$$\begin{array}{*{20}l} {c_{\tau 1} = \frac{{\alpha^{2} c_{\theta } (\Delta - A)^{2} \left[ {6\alpha c_{\theta } - \beta^{2} - \sqrt {\left( {\beta^{2} - 2\alpha c_{\theta } } \right)\left( {14\alpha c_{\theta } - 3\beta^{2} } \right)} } \right]}}{{4\left( {4\alpha c_{\theta } - \beta^{2} } \right)^{2} }}} \hfill \\ {c_{\tau 2} = \frac{{\alpha^{2} c_{\theta } (\Delta - A)^{2} \left[ {6\alpha c_{\theta } - \beta^{2} + \sqrt {\left( {\beta^{2} - 2\alpha c_{\theta } } \right)\left( {14\alpha c_{\theta } - 3\beta^{2} } \right)} } \right]}}{{4\left( {4\alpha c_{\theta } - \beta^{2} } \right)^{2} }}} \hfill \\ \end{array}$$

Also if $\frac{{3\beta^{2} }}{8\alpha } > c_{\theta }$, we infer $c_{\tau 1} < c_{\tau }^{*} = \frac{{\alpha^{2} c_{\theta } (\Delta - A)^{2} }}{{4\alpha c_{\theta } - \beta^{2} }} < c_{\tau 2}$ and according to condition shown in inequality (50), we have c _τ1 < c _τ . So we conclude that w ^M < w ^R if c _τ2 > c _τ and w ^R < w ^M if c _τ  > c _τ2. As well as, if $\frac{{\beta^{2} }}{2\alpha } > c_{\theta } > \frac{{3\beta^{2} }}{8\alpha }$, we infer $c_{\tau 1} < c_{\tau 2} < c_{\tau }^{*} = \frac{{\alpha^{2} c_{\theta } (\Delta - A)^{2} }}{{4\alpha c_{\theta } - \beta^{2} }}$ and according to inequality (50), we conclude that w ^R < w ^M. Otherwise if $c_{\theta } > \frac{{\beta^{2} }}{2\alpha }$, the equation shows the difference between w ^M with w ^R does not have real roots, therefore w ^R < w ^M. Finally, by summarizing all of above results, Lemma (4) is concluded.

Appendix 9: Proof of Lemma 5

First, we prove easily, $p_{1}^{3P} < p_{2}^{3P}$, $p_{2}^{3P} < p^{M}$, $p^{R} < p^{C}$ and $p^{M} < p^{C}$ if $\frac{{\beta^{2} }}{2\alpha } > c_{\theta }$ and $p_{2}^{3P} < p_{1}^{3P}$, $p^{M} < p_{2}^{3P}$, $p^{C} < p^{R}$ and $p^{C} < p^{M}$ if $c_{\theta } > \frac{{\beta^{2} }}{2\alpha }$, because

$$p_{1}^{3P} - p_{2}^{3P} = \frac{{\left( {2\alpha c_{\theta } - \beta^{2} } \right)\left( {\phi - \alpha c_{m} } \right)}}{{4\alpha \left( {4\alpha c_{\theta } - \beta^{2} } \right)}}$$

$$p_{2}^{3P} - p^{M} = \frac{{\alpha c_{\theta } (\Delta - A)^{2} \left( {2\alpha c_{\theta } - \beta^{2} } \right)\left( {\phi - \alpha c_{m} } \right)}}{{4\left( {4\alpha c_{\theta } - \beta^{2} } \right)\left[ {4c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]}}$$

$$p^{R} - p^{C} = \frac{{4c_{\theta } c_{\tau }^{2} \left( {\phi - \alpha c_{m} } \right)\left( {2\alpha c_{\theta } - \beta^{2} } \right)}}{{\left[ {4c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]\left[ {c_{\tau } \left( {8\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]}}$$

$$p^{M} - p^{C} = \frac{{c_{\tau } \left( {\phi - \alpha c_{m} } \right)\left( {2\alpha c_{\theta } - \beta^{2} } \right)\left[ {\alpha^{2} c_{\theta } (\Delta - A)^{2} + 2c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right)} \right]}}{{\alpha \left[ {4c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]\left[ {c_{\tau } \left( {8\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]}}$$

Furthermore, with comparing p ^R and $p_{1}^{3P}$, we extract the following relations:

• 1. $p^{R} < p_{1}^{3P} ,\quad {\text{if}}\;\frac{{3\beta^{2} }}{8\alpha } > c_{\theta }$

• 2. $p_{1}^{3P} < p^{R} ,\quad {\text{if}}\;\frac{{\beta^{2} }}{2\alpha } > c_{\theta } > \frac{{3\beta^{2} }}{8\alpha }$

• 3. $p^{R} < p_{1}^{3P} ,\quad {\text{if}}\;c_{\theta } > \frac{{\beta^{2} }}{2\alpha }$

And also with comparing p ^M and p ^R, we have:

• 1. $p^{R} < p^{M} ,\quad {\text{if}}\;\frac{{\beta^{2} }}{2\alpha } > c_{\theta }$

• 2. $p^{M} < p^{R} ,\quad {\text{if}}\;\frac{{3\beta^{2} }}{4\alpha } > c_{\theta } > \frac{{\beta^{2} }}{2\alpha }$

• 3. $p^{R} < p^{M} ,\quad {\text{if}}\;c_{\theta } > \frac{{3\beta^{2} }}{4\alpha }\quad {\text{and}}\quad \frac{{\alpha^{2} c_{\theta } (\Delta - A)^{2} }}{{2\beta^{2} }} > c_{\tau }$

• 4. $p^{M} < p^{R} ,\quad {\text{if}}\;c_{\theta } > \frac{{3\beta^{2} }}{4\alpha }\quad {\text{and}}\quad c_{\tau } > \frac{{\alpha^{2} c_{\theta } (\Delta - A)^{2} }}{{2\beta^{2} }}$

Then, we compare p ^R with $p_{2}^{3P}$ and extract two roots $c_{\tau 3} ,c_{\tau 4}$ from the following equations if $\frac{{19\beta^{2} }}{72\alpha } > c_{\theta }$ or $c_{\theta } > \frac{{3\beta^{2} }}{8\alpha }$.

$$p^{R} - p_{2}^{3P} = \frac{{c_{\tau } \left( {\phi - \alpha c_{m} } \right)\left( {2\alpha c_{\theta } - \beta^{2} } \right)\left[ {8\beta^{2} \left( {4\alpha c_{\theta } - \beta^{2} } \right)c_{\tau }^{2} - \alpha^{2} c_{\theta } (\Delta - A)^{2} \left( {24\alpha c_{\theta } - 5\beta^{2} } \right)c_{\tau } + \alpha^{4} c_{\theta }^{2} (\Delta - A)^{4} } \right]}}{{4\alpha \left[ {4c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]\left[ {c_{\tau } \left( {8\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]}}$$

$$\begin{array}{*{20}l} {c_{\tau 3} = \frac{{\alpha^{2} c_{\theta } (\Delta - A)^{2} \left[ {24\alpha c_{\theta } - 5\beta^{2} - \sqrt {\left( {8\alpha c_{\theta } - 3\beta^{2} } \right)\left( {72\alpha c_{\theta } - 19\beta^{2} } \right)} } \right]}}{{16\beta^{2} \left( {4\alpha c_{\theta } - \beta^{2} } \right)}}} \hfill \\ {c_{\tau 4} = \frac{{\alpha^{2} c_{\theta } (\Delta - A)^{2} \left[ {24\alpha c_{\theta } - 5\beta^{2} + \sqrt {\left( {8\alpha c_{\theta } - 3\beta^{2} } \right)\left( {72\alpha c_{\theta } - 19\beta^{2} } \right)} } \right]}}{{16\beta^{2} \left( {4\alpha c_{\theta } - \beta^{2} } \right)}}} \hfill \\ \end{array}$$

Also $c_{\theta } = \frac{{\beta^{2} }}{2\alpha }$ is another root for above equation. So if $\frac{{19\beta^{2} }}{72\alpha } > c_{\theta }$ or $\frac{{\beta^{2} }}{2\alpha } > c_{\theta } > \frac{{3\beta^{2} }}{8\alpha }$, we infer $c_{\tau 3} < c_{\tau 4} < c_{\tau }^{*} = \frac{{\alpha^{2} c_{\theta } (\Delta - A)^{2} }}{{4\alpha c_{\theta } - \beta^{2} }}$ and according to inequality (50), one can derive that $p^{R} < p_{2}^{3P}$. As well as, if $\frac{{3\beta^{2} }}{5\alpha } > c_{\theta } > \frac{{\beta^{2} }}{2\alpha }$, we conclude $c_{\tau 3} < c_{\tau 4} < c_{\tau }^{*} = \frac{{\alpha^{2} c_{\theta } (\Delta - A)^{2} }}{{4\alpha c_{\theta } - \beta^{2} }}$ and earn $p_{2}^{3P} < p^{R}$ consequently. Also, if $c_{\theta } > \frac{{3\beta^{2} }}{5\alpha }$, we infer $c_{\tau 3} < c_{\tau }^{*} = \frac{{\alpha^{2} c_{\theta } (\Delta - A)^{2} }}{{4\alpha c_{\theta } - \beta^{2} }} < c_{\tau 4}$ and according to inequality (50), we conclude that $p^{R} < p_{2}^{3P}$ if $c_{\tau 4} > c_{\tau }$ and $p_{2}^{3P} < p^{R}$ if $c_{\tau } > c_{\tau 4}$. Otherwise if $\frac{{3\beta^{2} }}{8\alpha } > c_{\theta } > \frac{{19\beta^{2} }}{72\alpha }$, then $c_{\tau 3} ,c_{\tau 4}$ are not real roots and therefore $p^{R} < p_{2}^{3P}$. Finally, by summarizing all of above results, Lemma (5) is proved.

Appendix 10: Proof of lemma 6

First, we prove easily, $\theta_{1}^{3P} < \theta^{M}$, $\theta^{M} < \theta^{C}$, $\theta^{R} < \theta^{C}$, $\theta_{1}^{3P} < \theta_{2}^{3P}$ and $\theta_{2}^{3P} < \theta^{M}$, because

$$\theta^{M} - \theta_{1}^{3P} = \frac{{\beta c_{\tau } \left( {\phi - \alpha c_{m} } \right)}}{{4c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} }} > 0$$

$$\theta^{C} - \theta^{M} = \frac{{\beta c_{\tau } \left( {\phi - \alpha c_{m} } \right)\left[ {2c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) + \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]}}{{\left[ {4c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]\left[ {c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]}} > 0$$

$$\theta^{C} - \theta^{R} = \frac{{4\alpha \beta c_{\theta } c_{\tau }^{2} \left( {\phi - \alpha c_{m} } \right)}}{{\left[ {c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]\left[ {c_{\tau } \left( {8\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]}} > 0$$

$$\theta_{2}^{3P} - \theta_{1}^{3P} = \frac{{\beta \left( {\phi - \alpha c_{m} } \right)}}{{4\left( {4\alpha c_{\theta } - \beta^{2} } \right)}} > 0$$

$$\theta^{M} - \theta_{2}^{3P} = \frac{{\alpha^{2} \beta c_{\theta } (\Delta - A)^{2} \left( {\phi - \alpha c_{m} } \right)}}{{4\left( {4\alpha c_{\theta } - \beta^{2} } \right)\left[ {c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]}} > 0$$

Also with comparing θ ^M and θ ^R, we infer following relationship;

• 1. $\theta^{R} < \theta^{M} ,\quad {\text{if}}\;\frac{{3\beta^{2} }}{4\alpha } > c_{\theta }$

• 2. $\theta^{M} < \theta^{R} ,\quad {\text{if}}\;c_{\theta } > \frac{{3\beta^{2} }}{4\alpha }\quad {\text{and}}\quad \frac{{\alpha^{2} c_{\theta } (\Delta - A)^{2} }}{{2\beta^{2} }} > c_{\tau }$

• 3. $\theta^{R} < \theta^{M} ,\quad {\text{if}}\;c_{\theta } > \frac{{3\beta^{2} }}{4\alpha }\quad {\text{and}}\quad c_{\tau } > \frac{{\alpha^{2} c_{\theta } (\Delta - A)^{2} }}{{2\beta^{2} }}$

Furthermore, with comparing $\theta_{1}^{3P}$ and θ ^R, then we have;

• 1. $\theta^{R} < \theta_{1}^{3P} ,\quad {\text{if}}\;\frac{{3\beta^{2} }}{8\alpha } > c_{\theta }$

• 2. $\theta_{1}^{3P} < \theta^{R} ,\quad {\text{if}}\;c_{\theta } > \frac{{3\beta^{2} }}{8\alpha }$

Finally, by summarizing all of above results, Lemma (6) is proved.

Appendix 11: Proof of Lemma 7

We prove easily, $\varPi_{T}^{3P,1} < \varPi_{T}^{M}$, $\varPi_{T}^{3P,2} < \varPi_{T}^{M}$, $\varPi_{T}^{M} < \varPi_{T}^{C}$ and $\varPi_{T}^{R} < \varPi_{T}^{C}$, because

$$\varPi_{T}^{M} - \varPi_{T}^{3P,1} = \frac{{c_{\theta } c_{\tau } \left( {\phi - \alpha c_{m} } \right)^{2} \left[ {5c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right)} \right]}}{{\left[ {4c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]^{2} }} > 0$$

$$\varPi_{T}^{M} - \varPi_{T}^{3P,2} = \frac{{\alpha^{2} c_{\theta }^{2} c_{\tau } \left( {\phi - \alpha c_{m} } \right)^{2} (\Delta - A)^{2} \left[ {16c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) + \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]}}{{16\left( {4\alpha c_{\theta } - \beta^{2} } \right)\left[ {4c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]^{2} }} > 0$$

$$\varPi_{T}^{*} - \varPi_{T}^{M} = \frac{{c_{\theta } c_{\tau }^{2} \left( {\phi - \alpha c_{m} } \right)^{2} \left( {4\alpha c_{\theta } - \beta^{2} } \right)\left[ {4c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) + 5\alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]}}{{\left[ {4c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]^{2} \left[ {c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]}} > 0$$

$$\varPi_{T}^{*} - \varPi_{T}^{R} = \frac{{\alpha^{2} c_{\theta }^{2} c_{\tau } \left( {\phi - \alpha c_{m} } \right)^{2} \left[ {64c_{\theta } c_{\tau }^{2} + (\Delta - A)^{2} \left( {4\alpha c_{\theta } - \beta^{2} } \right)c_{\tau } - \alpha^{2} c_{\theta } (\Delta - A)^{4} } \right]}}{{4\left[ {c_{\tau } \left( {4\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]\left[ {c_{\tau } \left( {8\alpha c_{\theta } - \beta^{2} } \right) - \alpha^{2} c_{\theta } (\Delta - A)^{2} } \right]^{2} }} > 0$$

By summarizing all of above results, Lemma (7) is proved too.