Recall management in pharmaceutical industry through supply chain coordination

Abstract

This paper is motivated by a real pharmaceutical supply chain (PSC) of Valsartan, facing two major concerns. First, implementing effective policies for product recall management due to disruption occurring in the manufacturing process, including impurity in raw material and packaging errors. Second, determining the optimal selling price and quality of raw material affecting the quality of the final compound, Valsartan, due to the effect of these decisions on Valsartan demand. In this paper, we analytically evaluate the impacts of production disruption, defective products recall, and decentralization among the members on the performance of the multi-echelon PSC. Moreover, for the first time, we address the context of channel coordination as a risk-mitigating strategy in case of disruption occurrence by proposing an altered revenue-sharing contract. Therefore, decisions are analyzed comparing decentralized, centralized, and coordinated decision-making structures under a range of scenarios. The results reveal that the proposed coordination scheme not only shares the incurred costs by disruption occurrence among all members but also improves their profitability compared to that of the decentralized structure, which leads to implementing effective recall management.

Appendices

Appendix A

$$ \frac{{\partial q_{M} \left( {q_{A} } \right)}}{{\partial q_{A} }} = \frac{{2\theta q_{A}^{\theta - 1} }}{{\left( {1 + q_{A}^{\theta } } \right)^{2} }} $$

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$$ \frac{{\partial^{2} q_{M} \left( {q_{A} } \right)}}{{\partial \left( {q_{A} } \right)^{2} }} = \frac{{2\theta \left( {\left( {\theta - 1} \right)q_{A}^{\theta - 2} } \right)}}{{\left( {1 + q_{A}^{\theta } } \right)^{2} }} - \frac{{\left( {2\theta q_{A}^{\theta - 1} } \right)^{2} }}{{\left( {1 + q_{A}^{\theta } } \right)^{3} }} $$

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It can be inferred that if $(\theta > 1)$, the function $q_{M} \left( {q_{A} } \right)$ is convex and if $(0 < \theta \le 1)$ the function $q_{M} \left( {q_{A} } \right)$ is concave. This means that the product’s quality increases with a decreasing rate if $(0 < \theta \le 1)$ and with an increasing rate if $(\theta > 1)$.

Appendix B

$$ \frac{{\partial E(\pi_{A} \left( {q_{A} )} \right)^{dec} }}{{\partial q_{A} }} = \left[ { - P_{A} \left( {RP} \right)_{A} + \left( {w_{A} - c_{A} } \right)} \right]\left[ {\frac{{3\gamma q_{A}^{2\theta - 1} }}{{\left( {1 + q_{A}^{\theta } } \right)^{2} }}} \right] - \vartheta_{1} q_{A} + g_{A} $$

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$$ \frac{{\partial^{2} E(\pi_{A} \left( {q_{A} } \right))^{dec} }}{{\partial q_{A}^{2} }} = \left[ { - P_{A} \left( {RP} \right)_{A} + \left( {w_{A} - c_{A} } \right)} \right]\left[ {\frac{{3\gamma \left( {2\theta - 1} \right)q_{A}^{2\theta - 2} \left( {1 + q_{A}^{\theta } } \right) - 2q_{A}^{\theta - 1} }}{{\left( {1 + q_{A}^{\theta } } \right)^{3} }}} \right] - \vartheta_{1} \le 0 $$

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It can be concluded that if $\left( {\theta \le 0.5} \right)$ then $\frac{{\partial^{2} E(\pi_{A} \left( {q_{A} } \right))^{dec} }}{{\partial q_{A}^{2} }}$ is negative and $E(\pi_{A} \left( {q_{A} )} \right)^{dec}$ is concave with respect to $\left( {q_{A} } \right)$. In our investigated case, the above-mentioned condition is satisfied.

Appendix C

Proof of Proposition 12

From Eq. (37) we derive: $\frac{\partial p}{\partial \alpha } = \frac{1}{2\beta } + \frac{{\partial q_{A} }}{\partial \alpha }\frac{{\gamma \theta q_{A}^{cen^{\theta - 1}} }}{{\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$. On the other hand, from Eq. (36) we have: $\frac{{\partial q_{A} }}{\partial \alpha }{\text{A}} = \frac{\partial p}{\partial \alpha }$, where $A = \frac{{\left( {B + \frac{{g_{M} }}{\gamma } - \frac{{2q_{A}^{cen^{\theta} } \vartheta_{2} }}{{\gamma \left( {1 + q_{A}^{cen^{\theta} } } \right)}}} \right)\left( {1 - \theta + \frac{{2\theta q_{A}^{cen^{\theta} } }}{{1 + q_{A}^{cen^{\theta} } }}} \right)}}{{q_{A}^{cen} }} + 2\frac{{\vartheta_{2} \theta }}{\gamma }\frac{{q_{A}^{cen^{\theta - 1}} }}{{\left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }} + \frac{{\vartheta_{1} \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}{{\gamma 2\theta q_{A}^{cen^{\theta - 1}} }}$ and $B = p - c_{A} - P_{A} \left( {RP} \right)_{A} - P_{M} \left( {RP} \right)_{M} - P_{A} d - P_{A} S_{RA} - P_{M} S_{RM} - c_{M} - S_{D}$. Note that $A$ and $B$ are positive when $p$ is sufficiently high. Therefore, we obtain $\frac{\partial p}{\partial \alpha } = \frac{{A\left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}{{2A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - 2\gamma \theta q_{A}^{cen^{\theta - 1}} }}$ and $\frac{{\partial q_{A} }}{\partial \alpha } = \frac{{\left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}{{2A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - 2\gamma \theta q_{A}^{cen^{\theta - 1}} }}$, which are positive if $\frac{A\beta }{\gamma } > \frac{{\theta q_{A}^{cen^{\theta - 1}} }}{{\left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$ holds. Similarly, $\frac{\partial p}{\partial \beta } = - \frac{\alpha }{{2\beta^{2} }} - \frac{{\gamma q_{A}^{cen^{\theta} } }}{{\beta^{2} \left( {1 + q_{A}^{cen^{\theta} } } \right)}} + \frac{{\partial q_{A} }}{\partial \beta }\frac{{\gamma \theta q_{A}^{cen^{\theta - 1}} }}{{\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$ and $\frac{{\partial q_{A} }}{\partial \beta }{\text{A}} = \frac{\partial p}{\partial \beta }$ are derived from Eqs. (37) and (36), respectively. Accordingly, we have $\frac{\partial p}{\partial \beta } = - \left( {\frac{\alpha }{{2\beta^{2} }} + \gamma \frac{{q_{A}^{cen^{\theta} } }}{{\beta^{2} \left( {1 + q_{A}^{cen^{\theta} } } \right)}}} \right)\frac{{A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}{{A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - \gamma \theta q_{A}^{cen^{\theta - 1}} }}$ and $\frac{{\partial q_{A} }}{\partial \beta } = - \left( {\frac{\alpha }{{2\beta^{2} }} + \gamma \frac{{q_{A}^{cen^{\theta} } }}{{\beta^{2} \left( {1 + q_{A}^{cen^{\theta} } } \right)}}} \right)\frac{{\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}{{A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - \gamma \theta q_{A}^{cen^{\theta - 1}} }}$, which are negative considering $\frac{A\beta }{\gamma } > \frac{{\theta q_{A}^{cen^{\theta - 1}} }}{{\left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$. Moreover, from the decentralized system we have (see Proposition 7): $\frac{\partial p}{\partial \beta } = - \frac{{\alpha + \gamma \left( {1 - \frac{{1 - q_{A}^{dec^\theta } }}{{1 + q_{A}^{dec^\theta } }}} \right)}}{{2\beta^{2} }}$, which is lower than that of the centralized model, since $\frac{{A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}{{A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - \gamma \theta q_{A}^{cen^{\theta - 1}} }} < 1$ and thus by increasing $\beta$ the reduction rate of $p$ decreases, after centralization. The proof is complete.□

Proof of Proposition 13

From Eq. (37) we have: $\frac{\partial p}{\partial \gamma } = \frac{{q_{A}^{\text{cen}^{\theta} } }}{{\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)}} + \frac{{\partial q_{A} }}{\partial \gamma }\frac{{\gamma \theta q_{A}^{cen^{\theta - 1}} }}{{\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$. On the other hand, from Eq. (36) we derive: $A\frac{{\partial q_{A} }}{\partial \gamma } - \frac{B}{\gamma } = \frac{\partial p}{\partial \gamma }$. Therefore, we obtain $\frac{\partial p}{\partial \gamma } = \frac{{q_{A}^{cen^{\theta} } A\left( {1 + q_{A}^{cen^{\theta} } } \right) + \theta q_{A}^{cen^{\theta - 1}} B}}{{A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - \gamma \theta q_{A}^{cen^{\theta - 1}} }}$ and $\frac{{\partial q_{A} }}{\partial \gamma } = \frac{{q_{A}^{cen^{\theta} } \left( {1 + q_{A}^{cen^{\theta} } } \right) + \frac{B\theta }{A}q_{A}^{cen^{\theta - 1}} }}{{A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - \gamma \theta q_{A}^{cen^{\theta - 1}} }} + \frac{B}{\gamma A}$, which are positive considering $\frac{A\beta }{\gamma } > \frac{{\theta q_{A}^{cen^{\theta - 1}} }}{{\left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$. The proof is complete.□

Proof of Proposition 14

From Eq. (37) we derive: $\frac{\partial p}{{\partial S_{D} }} = \frac{1}{2} + \frac{{\partial q_{A} }}{{\partial S_{D} }}\frac{{\gamma \theta q_{A}^{cen^{\theta - 1}} }}{{\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$, $\frac{\partial p}{{\partial c_{A} }} = \frac{1}{2} + \frac{{\partial q_{A} }}{{\partial c_{A} }}\frac{{\gamma \theta q_{A}^{cen^{\theta - 1}} }}{{\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$, and $\frac{\partial p}{{\partial c_{M} }} = \frac{1}{2} + \frac{{\partial q_{A} }}{{\partial c_{M} }}\frac{{\gamma \theta q_{A}^{cen^{\theta - 1}} }}{{\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$. Also, $\frac{\partial p}{{\partial S_{D} }} = A\frac{{\partial q_{A} }}{{\partial S_{D} }} + 1$, $\frac{\partial p}{{\partial c_{A} }} = A\frac{{\partial q_{A} }}{{\partial c_{A} }} + 1$, and $\frac{\partial p}{{\partial c_{M} }} = A\frac{{\partial q_{A} }}{{\partial c_{M} }} + 1$ are derived from Eq. (36). Accordingly, we calculate $\frac{\partial p}{{\partial S_{D} }} = \frac{\partial p}{{\partial c_{A} }} = \frac{\partial p}{{\partial c_{M} }} = \frac{{A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - 2\gamma \theta q_{A}^{cen^{\theta - 1}} }}{{2A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - 2\gamma \theta q_{A}^{cen^{\theta - 1}} }}$, which are positive if $\frac{A\beta }{2\gamma } > \frac{{\theta q_{A}^{cen^{\theta - 1}} }}{{\left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$ holds; and $\frac{{\partial q_{A} }}{{\partial S_{D} }} = \frac{{\partial q_{A} }}{{\partial c_{A} }} = \frac{{\partial q_{A} }}{{\partial c_{M} }} = \frac{{ - \beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}{{2A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - 2\gamma \theta q_{A}^{cen^{\theta - 1}} }}$, which are negative if $\frac{A\beta }{2\gamma } > \frac{{\theta q_{A}^{cen^{\theta - 1}} }}{{\left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$ holds. The proof is complete.□

Proof of Proposition 15

From Eq. (37) we derive: $\frac{\partial p}{\partial d} = \frac{{P_{A} }}{2} + \frac{{\partial q_{A} }}{\partial d}\frac{{\gamma \theta q_{A}^{cen^{\theta - 1}} }}{{\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$, $\frac{\partial p}{{\partial S_{RA} }} = \frac{{P_{A} }}{2} + \frac{{\partial q_{A} }}{{\partial S_{RA} }}\frac{{\gamma \theta q_{A}^{cen^{\theta - 1}} }}{{\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$, $\frac{\partial p}{{\partial S_{RM} }} = \frac{{P_{M} }}{2} + \frac{{\partial q_{A} }}{{\partial S_{RM} }}\frac{{\gamma \theta q_{A}^{cen^{\theta - 1}} }}{{\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$, $\frac{\partial p}{{\partial \left( {RP} \right)_{A} }} = \frac{{P_{A} }}{2} + \frac{{\partial q_{A} }}{{\partial \left( {RP} \right)_{A} }}\frac{{\gamma \theta q_{A}^{cen^{\theta - 1}} }}{{\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$, and $\frac{\partial p}{{\partial \left( {RP} \right)_{M} }} = \frac{{P_{M} }}{2} + \frac{{\partial q_{A} }}{{\partial \left( {RP} \right)_{M} }}\frac{{\gamma \theta q_{A}^{cen^{\theta - 1}} }}{{\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$. Additionally, according to Eq. (36) we have: $\frac{\partial p}{\partial d} = A\frac{{\partial q_{A} }}{\partial d} + P_{A}$, $\frac{\partial p}{{\partial S_{RA} }} = A\frac{{\partial q_{A} }}{{\partial S_{RA} }} + P_{A}$, $\frac{\partial p}{{\partial \left( {RP} \right)_{A} }} = A\frac{{\partial q_{A} }}{{\partial \left( {RP} \right)_{A} }} + P_{A}$, $\frac{\partial p}{{\partial S_{RM} }} = A\frac{{\partial q_{A} }}{{\partial S_{RM} }} + P_{M}$, and $\frac{\partial p}{{\partial \left( {RP} \right)_{M} }} = A\frac{{\partial q_{A} }}{{\partial \left( {RP} \right)_{M} }} + P_{M}$. Therefore, we calculate $\frac{\partial p}{\partial d} = \frac{\partial p}{{\partial S_{RA} }} = \frac{\partial p}{{\partial \left( {RP} \right)_{A} }} = P_{A} \frac{{A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - 2\gamma \theta q_{A}^{cen^{\theta - 1}} }}{{2A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - 2\gamma \theta q_{A}^{cen^{\theta - 1}} }}$ and $\frac{\partial p}{{\partial S_{RM} }} = \frac{\partial p}{{\partial \left( {RP} \right)_{M} }} = P_{M} \frac{{A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - 2\gamma \theta q_{A}^{cen^{\theta - 1}} }}{{2A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - 2\gamma \theta q_{A}^{cen^{\theta - 1}} }}$, which are positive when $\frac{A\beta }{2\gamma } > \frac{{\theta q_{A}^{cen^{\theta - 1}} }}{{\left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$ holds. Similarly, we derive: $\frac{{\partial q_{A} }}{\partial d} = \frac{{\partial q_{A} }}{{\partial S_{RA} }} = \frac{{\partial q_{A} }}{{\partial \left( {RP} \right)_{A} }} = - \frac{{\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} P_{A} }}{{2A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - 2\gamma \theta q_{A}^{cen^{\theta - 1}} }}$ and $\frac{{\partial q_{A} }}{{\partial S_{RM} }} = \frac{{\partial q_{A} }}{{\partial \left( {RP} \right)_{M} }} = - \frac{{\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} P_{M} }}{{2A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - 2\gamma \theta q_{A}^{cen^{\theta - 1}} }}$, which are negative when $\frac{A\beta }{\gamma } > \frac{{\theta q_{A}^{cen^{\theta - 1}} }}{{\left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$ holds. Moreover, under the decentralized model we have (see Proposition 8): $\frac{\partial p}{{\partial S_{D} }} = \frac{1}{2}$, $\frac{\partial p}{{\partial S_{RA} }} = \frac{{P_{A} }}{2}$, and $\frac{\partial p}{{\partial S_{RM} }} = \frac{{P_{M} }}{2}$, which are higher than those of the centralized system considering $\frac{A\beta }{\gamma } > \frac{{\theta q_{A}^{cen^{\theta - 1}} }}{{\left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$. The proof is complete.□

Proof of Proposition 16

$\frac{\partial p}{{\partial P_{M} }} = \frac{1}{2}\left( {\left( {RP} \right)_{M} + S_{RM} } \right) + \frac{{\partial q_{A} }}{{\partial P_{M} }}\frac{{\gamma \theta q_{A}^{cen^{\theta - 1}} }}{{\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$ and $\frac{\partial p}{{\partial P_{A} }} = \frac{1}{2}\left( {\left( {RP} \right)_{A} + S_{RA} + d} \right) + \frac{{\partial q_{A} }}{{\partial P_{A} }}\frac{{\gamma \theta q_{A}^{cen^{\theta - 1}} }}{{\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$ are derived from Eq. (37). On the other hand, from Eq. (36) we have: $\frac{\partial p}{{\partial P_{A} }} = A\frac{{\partial q_{A} }}{{\partial P_{A} }} + \left( {RP} \right)_{A} + S_{RA} + d$ and $\frac{\partial p}{{\partial P_{M} }} = A\frac{{\partial q_{A} }}{{\partial P_{M} }} + \left( {RP} \right)_{M} + S_{RM}$. Hence, we obtain $\frac{\partial p}{{\partial P_{M} }} = \left( {\left( {RP} \right)_{M} + S_{RM} } \right)\frac{{A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - 2\gamma \theta q_{A}^{cen^{\theta - 1}} }}{{2A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - 2\gamma \theta q_{A}^{cen^{\theta - 1}} }}$ and $\frac{\partial p}{{\partial P_{A} }} = \left( {\left( {RP} \right)_{A} + S_{RA} + d} \right)\frac{{A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - 2\gamma \theta q_{A}^{cen^{\theta - 1}} }}{{2A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - 2\gamma \theta q_{A}^{cen^{\theta - 1}} }}$, which are positive if $\frac{A\beta }{2\gamma } > \frac{{\theta q_{A}^{cen^{\theta - 1}} }}{{\left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$ holds. Similarly, we have: $\frac{{\partial q_{A} }}{{\partial P_{M} }} = - \frac{{\left( {\left( {RP} \right)_{M} + S_{RM} } \right)\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}{{2A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - 2\gamma \theta q_{A}^{cen^{\theta - 1}} }}$ and $\frac{{\partial q_{A} }}{{\partial P_{A} }} = - \frac{{\left( {\left( {RP} \right)_{A} + S_{RA} + d} \right)\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}{{2A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - 2\gamma \theta q_{A}^{cen^{\theta - 1}} }}$, which are negative if $\frac{{\theta q_{A}^{cen^{\theta - 1}} }}{{\left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }} < \frac{A\beta }{\gamma }$ holds. The proof is complete.□

Proof of Proposition 17

$\frac{\partial p}{{\partial g_{M} }} = \frac{{\partial q_{A} }}{{\partial g_{M} }}\frac{{\gamma \theta q_{A}^{cen^{\theta - 1}} }}{{\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$ and $\frac{{\partial q_{A} }}{{\partial g_{M} }}A - \frac{1}{\gamma } = \frac{\partial p}{{\partial g_{M} }}$ are derived from Eqs. (37) and (36), respectively. Therefore, we obtain $\frac{\partial p}{{\partial g_{M} }} = \frac{{\theta q_{A}^{cen^{\theta - 1}} }}{{A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - \gamma \theta q_{A}^{cen^{\theta - 1}} }}$ and $\frac{{\partial q_{A} }}{{\partial g_{M} }} = \frac{{\theta q_{A}^{cen^{\theta - 1}} }}{{A\left( {A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - \gamma \theta q_{A}^{cen^{\theta - 1}} } \right)}} + \frac{1}{\gamma A}$, which are positive considering $\frac{A\beta }{\gamma } > \frac{{\theta q_{A}^{cen^{\theta - 1}} }}{{\left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$. Besides, using a similar method we have $\frac{\partial p}{{\partial g_{A} }} = \frac{{\partial q_{A} }}{{\partial g_{A} }}\frac{{\gamma \theta q_{A}^{cen^{\theta - 1}} }}{{\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$ and $A\frac{{\partial q_{A} }}{{\partial g_{A} }} - \frac{{\left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}{{\gamma 2\theta q_{A}^{cen^{\theta - 1}} }} = \frac{\partial p}{{\partial g_{A} }}$, which gives $\frac{\partial p}{{\partial g_{A} }} = \frac{{\left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}{{2A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - 2\gamma \theta q_{A}^{cen^{\theta - 1}} }} > 0$ and $\frac{{\partial q_{A} }}{{\partial g_{A} }} = \frac{{\left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}{{2A\left( {A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - \gamma \theta q_{A}^{cen^{\theta - 1}} } \right)}} + \frac{{\left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}{{A\gamma 2\theta q_{A}^{cen^{\theta - 1}} }} > 0$. The proof is complete.□

Proof of Proposition 18

$\frac{\partial p}{{\partial \vartheta_{1} }} = \frac{{\partial q_{A} }}{{\partial \vartheta_{1} }}\frac{{\gamma \theta q_{A}^{cen^{\theta - 1}} }}{{\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$ and $\frac{\partial p}{{\partial \vartheta_{2} }} = \frac{{\partial q_{A} }}{{\partial \vartheta_{2} }}\frac{{\gamma \theta q_{A}^{cen^{\theta - 1}} }}{{\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$ are derived from Eq. (37). Also, from Eq. (36) we have: $A\frac{{\partial q_{A} }}{{\partial \vartheta_{1} }} + \frac{{\left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}{{2\gamma \theta q_{A}^{cen^{\theta} - 2} }} = \frac{\partial p}{{\partial \vartheta_{1} }}$ and $A\frac{{\partial q_{A} }}{{\partial \vartheta_{2} }} + \frac{{2q_{A}^{cen^{\theta} } }}{{\gamma \left( {1 + q_{A}^{cen^{\theta} } } \right)}} = \frac{\partial p}{{\partial \vartheta_{2} }}$. Accordingly, we calculate $\frac{\partial p}{{\partial \vartheta_{1} }} = \frac{{ - \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} q_{A}^{cen} }}{{2A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - 2\gamma \theta q_{A}^{cen^{\theta - 1}} }}$ and $\frac{\partial p}{{\partial \vartheta_{2} }} = \frac{{ - 2\theta q_{A}^{cen^{2\theta - 1}} }}{{A\beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{3} - \gamma \theta q_{A}^{cen^{\theta - 1}} \left( {1 + q_{A}^{cen^{\theta} } } \right)}}$, which are negative if $\frac{A\beta }{\gamma } > \frac{{\theta q_{A}^{cen^{\theta - 1}} }}{{\left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$ holds. Similarly, we obtain $\frac{{\partial q_{A} }}{{\partial \vartheta_{1} }} = \frac{{ - \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} q_{A}^{cen} }}{{2A^{2} \beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} - 2A\gamma \theta q_{A}^{cen^{\theta - 1}} }} - \frac{{\left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}{{2A\gamma \theta q_{A}^{cen^{\theta} - 2} }}$ and $\frac{{\partial q_{A} }}{{\partial \vartheta_{2} }} = \frac{{ - 2\theta q_{A}^{cen^{2\theta - 1} }}}{{A^{2} \beta \left( {1 + q_{A}^{cen^{\theta} } } \right)^{3} - A\gamma \theta q_{A}^{cen^{\theta - 1}} \left( {1 + q_{A}^{cen^{\theta} } } \right)}} - \frac{{2q_{A}^{cen^{\theta} } }}{{A\gamma \left( {1 + q_{A}^{cen^{\theta} } } \right)}}$, which are negative considering $\frac{A\beta }{\gamma } > \frac{{\theta q_{A}^{cen^{\theta - 1}} }}{{\left( {1 + q_{A}^{cen^{\theta} } } \right)^{2} }}$. The proof is complete.□