A Closed-Loop Supply Chain Equilibrium Model with Random and Price-Sensitive Demand and Return

Abstract

This paper proposes a decentralized closed-loop supply chain network model consisting of raw material suppliers, manufacturers, retailers, and recovery centers. We assume that the demands for the product and the corresponding returns are random and price-sensitive. Retailers and recovery centers face penalties associated with shortage demand and supply, respectively. We derive the optimality conditions of the various decision-makers, and establish that the governing equilibrium conditions can be formulated as a finite-dimensional variational inequality problem. The qualitative properties of the solution to the variational inequality are discussed. Numerical examples are provided to illustrate the effects of demand and return uncertainties on quantity shipments and prices.

Appendices

Appendix A: Proof of Lemma 1

Knowing that $y^{\prime }_{j}(p_{j})<0$, one sees that $-(p_{j}+\delta _{j}) y_{j}^{\prime }(p_{j})r_{j}(z_{j})\ge \frac {1}{2}$ is equivalent to M _j (s _j , p _j )≤0 where $M_{j}(s_{j},p_{j})=\frac {1}{r_{j}(z_{j})}+2(p_{j}+\delta _{j})y_{j}^{\prime }(p_{j})$. It is easy to verify that:

$$\begin{array}{@{}rcl@{}} \frac{\partial M_{j}}{\partial s_{j}} & = &\frac{-r_{j}^{\prime}(z_{j})}{r_{j}(z_{j})^{2}},\\ \frac{\partial M_{j}}{\partial p_{j}} & = &\frac{r_{j}^{\prime}(z_{j})y_{j}^{\prime}(p_{j})}{r_{j}(z_{j})^{2}}+2(p_{j}+\delta_{j})y_{j}^{\prime\prime}(p_{j})+2y_{j}^{\prime}(p_{j}),\\ \frac{\partial^{2} M_{j}}{\partial {s_{j}^{2}}} & = & \frac{2r_{j}^{\prime}(z_{j})^{2}-r_{j}(z_{j})r_{j}^{\prime\prime}(z_{j})}{r_{j}(z_{j})^{3}},\\ \frac{\partial^{2} M_{j}}{\partial s_{j} p_{j}} & = & -y_{j}^{\prime}(p_{j})\frac{2r_{j}^{\prime}(z_{j})^{2}-r_{j}(z_{j})r_{j}^{\prime\prime}(z_{j})}{r_{j}(z_{j})^{3}},\\ \frac{\partial^{2} M_{j}}{\partial {p_{j}^{2}}} & = & y_{j}^{\prime}(p_{j})^{2}\frac{2r_{j}^{\prime}(z_{j})^{2}-r_{j}(z_{j})r_{j}^{\prime\prime}(z_{j})}{r_{j}(z_{j})^{3}}\\&&+y_{j}^{\prime\prime}(p_{j})\frac{r_{j}^{\prime}(z_{j})}{r_{j}(z_{j})^{2}}+4y_{j}^{\prime\prime}(p_{j})+2(p_{j}+\delta_{j})y_{j}^{\prime\prime\prime}(p_{j}). \end{array} $$

Note that $\frac {\partial ^{2} M_{j}}{\partial {s_{j}^{2}}}\ge 0$ if $2(r_{j}^{\prime }(z_{j}))^{2}-r_{j}(z_{j}) r_{j}^{\prime \prime }(z_{j})\ge 0$. The determinant of the hessian of M _j , given by ${\Delta }_{j}=\frac {\partial ^{2} M_{j}}{\partial {s_{j}^{2}}}\frac {\partial ^{2} M_{j}}{\partial {p_{j}^{2}}}-\left (\frac {\partial ^{2} M_{j}}{\partial s_{j}p_{j}}\right )^{2}$ simplifies to

$${\Delta}_{j}=\frac{2r_{j}^{\prime}(z_{j})^{2}-r_{j}(z_{j})r_{j}^{\prime\prime}(z_{j})}{r_{j}(z_{j})^{3}}\left( \! y_{j}^{\prime\prime}(p_{j})\frac{r_{j}^{\prime}(z_{j})}{r_{j}(z_{j})^{2}}+4y_{j}^{\prime\prime}(p_{j})+2(p_{j}+\delta_{j})y^{\prime\prime\prime}(p_{j})\right).$$

From Assumptions 1 and 2a, $2(r_{j}^{\prime }(z_{j}))^{2}-r_{j}(z_{j}) r_{j}^{\prime \prime }(z_{j})\ge 0$, $r_{j}^{\prime }(z_{j})\ge 0$, $2y_{j}^{\prime \prime }(p_{j})+p_{j}y^{\prime \prime \prime }(p_{j})\ge 0$ and $y_{j}^{\prime \prime }(p_{j})\ge 0$ therefore Δ _j has the same sign as $2y_{j}^{\prime \prime }(p_{j})+(p_{j}+\delta _{j})y^{\prime \prime \prime }(p_{j})$. For y ^″′(p _j ) ≥ 0, one notice that $2y_{j}^{\prime \prime }(p_{j})+(p_{j}+\delta _{j})y^{\prime \prime \prime }(p_{j})\ge 0$ Since y ^″(p _j ) ≥ 0 and $p_{j}+\delta _{j}\ge p_{j}-\lambda _{j}^{+}\ge 0$. Next for y ^″′(p _j )≤0, $2y_{j}^{\prime \prime }(p_{j})+(p_{j}+\delta _{j})y^{\prime \prime \prime }(p_{j})\ge 0$ since $2y_{j}^{\prime \prime }(p_{j})+p_{j}y^{\prime \prime \prime }(p_{j})\ge 0$ by Assumption 1 and δ _j y ^″′(p _j ) ≥ 0 since δ _j ≤ 0 by definition. Hence Δ _j ≥ 0 implying that the function M _j is convex which in turn implies that the set ${{\Gamma }^{1}_{j}}$ is convex.

Appendix B: Proof of Theorem 1

First, without loss of generality, assume that $p_{Ij}^{*}=\min \limits _{j} \{p_{ij}^{*}\}$. Optimization problem (6) can then be formulated as follows:

$$\begin{array}{@{}rcl@{}} \max\limits {\Pi}_{j}(\tilde{Q}_{j},s_{j},p_{j}) & = & p_{j} y_{j}(p_{j}) -(p_{j}+\lambda_{j}^{-})e_{j}^{-}(z_{j})+ \lambda_{j}^{+} e_{j}^{+}(z_{j}) \\&&- c_{j} s_{j} -p_{Ij}^{*} s_{j}- \sum\limits_{i=1}^{I-1} (p_{ij}^{*}-p_{Ij}^{*}) q_{ij} \\ & =& {{\Pi}_{j}^{1}}(s_{j},p_{j})+ {{\Pi}_{j}^{2}}(\tilde{Q}_{j}), \end{array} $$

where $\tilde {Q}_{j}=(q_{ij})_{i=1}^{I-1}$, ${{\Pi }_{j}^{1}}(s_{j},p_{j})=p_{j} y_{j}(p_{j}) -(p_{j}+\lambda _{j}^{-}) e_{j}^{-}(z_{j})+ \lambda _{j}^{+} e_{j}^{+}(z_{j}) - c_{j} s_{j} -p_{Ij}^{*} s_{j}$ and ${{\Pi }_{j}^{2}}(\tilde {Q}_{j})=- \sum \limits _{i=1}^{I-1} (p_{ij}^{*}-p_{Ij}^{*}) q_{ij}$.

Clearly, the function ${{\Pi }_{j}^{2}}(\tilde {Q}_{j})$ is concave. We need to prove that ${{\Pi }_{j}^{1}}(s_{j},p_{j})$ is concave. One easily verifies that the first derivative of ${{\Pi }_{j}^{1}}$ with respect to s _j and p _j are given by

$$\begin{array}{@{}rcl@{}} \frac{\partial {{\Pi}_{j}^{1}}}{\partial s_{j}} & = & \lambda_{j}^{-}+p_{j} -c_{j} -p_{Ij}^{*}-\left( p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+}\right)F_{j}(z_{j}). \\ \frac{\partial {{\Pi}_{j}^{1}}}{\partial p_{j}} & = & y_{j}(p_{j})-e_{j}^{-}(z_{j})+y_{j}^{\prime}(p_{j})\left[(p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+})F_{j}(z_{j})-\lambda_{j}^{-}\right]. \end{array} $$

Straightforward computations show that:

$$\begin{array}{@{}rcl@{}} \frac{\partial^{2}{{\Pi}_{j}^{1}}}{\partial {s_{j}^{2}}} &= & - \left( p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+}\right) f_{j}(z_{j}), \\ \frac{\partial^{2}{{\Pi}_{j}^{1}}}{\partial s_{j}\partial p_{j}} & = & y^{\prime}_{j}(p_{j})\left( p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+}\right) f_{j}(z_{j})+1-F_{j}(z_{j}), \\ \frac{\partial^{2}{{\Pi}_{j}^{1}}}{\partial {p_{j}^{2}}} & = & -y_{j}^{\prime}(p_{j})^{2}\left( p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+}\right) f_{j}(z_{j})+2y_{j}^{\prime}(p_{j}) F_{j}(z_{j})\\&&+y^{\prime\prime}_{j}(p_{j}) \left[\left( p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+}\right)F_{j}(z_{j})-\lambda_{j}^{-}\right]. \end{array} $$

Let H _j denotes the hessian matrix associated with ${{\Pi }_{j}^{1}}(s_{j},p_{j})$. The matrix H _j is computed as:

$$\left( \begin{array}{cc} H_{s_{j}s_{j}} & H_{p_{j} s_{j}} \\ H_{p_{j} s_{j}} & H_{p_{j}p_{j}} \end{array}\right), $$

with $H_{s_{j}s_{j}}=- \left (p_{j}+\lambda _{j}^{-}-\lambda _{j}^{+}\right ) f_{j}(z_{j}), H_{p_{j}s_{j}}=H_{s_{j}p_{j}}=y^{\prime }_{j}(p_{j}) \left (p_{j}+\lambda _{j}^{-}-\lambda _{j}^{+}\right ) f_{j}(z_{j})+1-F_{j}(z_{j})$and $H_{p_{j}p_{j}}=-y_{j}^{\prime }(p_{j})^{2}\left (p_{j}+\lambda _{j}^{-}-\lambda _{j}^{+}\right ) f_{j}(z_{j})+2y_{j}^{\prime }(p_{j}) F_{j}(z_{j})+y_{j}^{\prime \prime }(p_{j}) \left [\left (p_{j}+\lambda _{j}^{-}-\lambda _{j}^{+}\right )\right . \left .F_{j}(z_{j})-\lambda _{j}^{-}\right ]$.

We will show that the matrix − H _j is positive definite. Clearly, $-H_{s_{j}s_{j}} =\left (p_{j}+\lambda _{j}^{-}-\lambda _{j}^{+}\right ) f_{j}(z_{j})>0$. If we show that d e t( − H _j ) ≥ 0 then $-H_{p_{j}p_{j}}\geq \frac {\left (H_{p_{j} s_{j}}\right )^{2}}{-H_{s_{j}s_{j}}}\geq 0$. d e t( − H _j ) is calculated as:

$$\begin{array}{@{}rcl@{}} det(-H_{j})&=&-y_{j}^{\prime\prime}(p_{j})(p_{j}+\lambda^{-}_{j} -\lambda^{+}_{j})f_{j}(z_{j})\left[\left( p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+}\right)F_{j}(z_{j})-\lambda_{j}^{-}\right]\\[-.5pt] & & -2(p_{j}+\lambda^{-}_{j} -\lambda^{+}_{j})f_{j}(z_{j})y_{j}^{\prime}(p_{j})F_{j}(z_{j})\\[-.5pt] &&-2(p_{j}+\lambda^{-}_{j} -\lambda^{+}_{j})f_{j}(z_{j})y_{j}^{\prime}(p_{j})(1-F_{j}(z_{j}))\\[-.5pt] & & -(1-F_{j}(z_{j}))^{2}\\[-.5pt] & = & (p_{j}+\lambda^{-}_{j} -\lambda^{+}_{j})f_{j}(z_{j})\left[-2y_{j}^{\prime}(p_{j})F_{j}(z_{j})\right.\\[-.5pt] &&\qquad\qquad\qquad\qquad\qquad\left.-y_{j}^{\prime\prime}(p_{j})\left[\left( p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+}\right)F_{j}(z_{j})-\lambda_{j}^{-}\right]\right]\\ & & + 2 (1-F_{j}(z_{j}))^{2}\left[\frac{-y_{j}^{\prime}(p_{j})(p_{j}+\lambda^{-}_{j} -\lambda^{+}_{j})f_{j}(z_{j})}{1-F_{j}(z_{j})}-\frac{1}{2}\right]\\ & = & (p_{j}+\lambda^{-}_{j} -\lambda^{+}_{j})f_{j}(z_{j})F_{j}(z_{j}) \left[-2y_{j}^{\prime}(p_{j})-p_{j} y_{j}^{\prime\prime}(p_{j})\right]\\ & & +(p_{j}+\lambda^{-}_{j} -\lambda^{+}_{j})f_{j}(z_{j}) y_{j}^{\prime\prime}(p_{j})\left[ \lambda^{-}_{j} (1-F_{j}(z_{j}))+\lambda^{+}_{j} F_{j}(z_{j})\right]\\ & & +2 (1-F_{j}(z_{j}))^{2}\left[-y_{j}^{\prime}(p_{j})(p_{j}+\lambda^{-}_{j} -\lambda^{+}_{j})r_{j}(z_{j})-\frac{1}{2}\right]. \end{array} $$

The first term in the above is nonnegative because $2 y^{\prime }_{j}(p_{j})+ p_{j} y_{j}^{\prime \prime }(p_{j})\leq 0$. The second term is also nonnegative because $y_{j}^{\prime \prime }(p_{j})\geq 0$. The last term is nonnegative since it can be written as $2 (1-F_{j}(z_{j}))^{2}\left [\eta ^{1}_{j}(s_{j},p_{j})-1/2\right ]+2 (1-F_{j}(z_{j}))^{2}\max (\lambda _{j}^{-}-\lambda _{j}^{+},0)(-y_{j}^{\prime }(p_{j})r_{j}(z_{j}))$ and ${\eta ^{1}_{j}}(s_{j},p_{j})\ge 1/2$. Hence, we have d e t( − H _j ) ≥ 0 implying that the matrix − H _j is positive definite. Therefore, ${{\Pi }_{j}^{1}}$ is concave in the set ${{\Gamma }^{1}_{j}}$.

Appendix C: Proof of Proposition 1

We need to show that the optimal vector $(s_{j}^{*},p_{j}^{*})$ belongs to the set ${{\Gamma }^{1}_{j}}$. In other words, we will show that ${\eta _{j}^{1}}(s_{j},p_{j})\ge 1/2$ when $\frac {{\partial {\Pi }_{j}^{1}}}{\partial p_{j}}=0$.

$$\frac{\partial {{\Pi}_{j}^{1}}}{\partial p_{j}}=0 \Longleftrightarrow y_{j}^{\prime}(p_{j})=\frac{e_{j}^{-}(z_{j})-y_{j}(p_{j})}{(p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+})F_{j}(z_{j})-\lambda_{j}^{-}}. $$

Since $y_{j}^{\prime }(p_{j})\le 0$ and $y_{j}(p_{j})-e_{j}^{-}(z_{j})\ge 0$ one see that when $\frac {{\partial {\Pi }_{j}^{1}}}{\partial p_{j}}=0$ one has $(p_{j}+\lambda _{j}^{-}-\lambda _{j}^{+})F_{j}(z_{j})-\lambda _{j}^{-} > 0$.

$$\begin{array}{@{}rcl@{}} {\eta_{j}^{1}}(s_{j},p_{j})\ge 1/2 & \Longleftrightarrow & -(p_{j}+\delta_{j}) y_{j}^{\prime}(p_{j})r_{j}(z_{j})-\frac{1}{2}\ge 0\\ & \Longleftrightarrow & -(p_{j}+\delta_{j}) y_{j}^{\prime}(p_{j})-\frac{1}{2r_{j}(z_{j})}\ge 0\\ & \Longleftrightarrow & \frac{p_{j}+\delta_{j}}{(p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+})F_{j}(z_{j})-\lambda_{j}^{-}}\left( y_{j}(p_{j})-e_{j}^{-}(z_{j})\right) -\frac{1}{2r_{j}(z_{j})}\ge 0\\ & \Longleftrightarrow & \frac{p_{j}+\delta_{j}}{(p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+})F_{j}(z_{j})-\lambda_{j}^{-}}\left[y_{j}(p_{j})-e_{j}^{-}(z_{j}) -\frac{F_{j}(z_{j})}{2r_{j}(z_{j})}\right]\\ & & +\frac{1}{(p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+})F_{j}(z_{j})-\lambda_{j}^{-}}\\&&\times\left[\frac{\lambda_{j}^{-}(1-F_{j}(z_{j}))+\min(\lambda_{j}^{-} ,\lambda_{j}^{+})F_{j}(z_{j})}{2r_{j}(z_{j})}\right]\ge 0. \end{array} $$

The last inequality holds if y _j (p _j ) + κ _j (z _j ) ≥ 0 where $\kappa _{j}(z_{j})=-e^{-}_{j}(z_{j})-\frac {F_{j}(z_{j})}{2r_{j}(z_{j})}.$ Note that κ _j (A _j ) = A _j and $\kappa _{j}^{\prime }(z_{j})=\frac {1-F_{j}(z_{j})}{2}+\frac {r^{\prime }(z_{j})F(z_{j})}{2r_{j}(z_{j})^{2}}\geq 0$ because $r_{j}^{\prime }(z_{j})\geq 0$. This yields κ _j (z _j ) ≥ A _j for all z _j ∈[A _j , B _j ]. Then $y_{j}(p_{j})+\kappa _{j}(z_{j})\ge y_{j}(\bar {p}_{j})+A_{j} =0$. Therefore, ${\eta _{j}^{1}}(s_{j},p_{j})\ge 1/2$ when $\frac {{\partial {\Pi }_{j}^{1}}}{\partial p_{j}}=0$.

Appendix D: Proof of Lemma 2

Knowing that z _j = s _j /y _j (p _j ) ≥ 0, one sees that $\frac {-(p_{j}+\delta _{j}) y_{j}^{\prime }(p_{j})g_{j}(z_{j})}{y_{j}(p_{j})}\ge \frac {1}{2}$ is equivalent to N _j (s _j , p _j )≤0 where $N_{j}(s_{j},p_{j})=\frac {y_{j}(p_{j})}{g_{j}(z_{j})}+2(p_{j}+\delta _{j})y_{j}^{\prime }(p_{j})$. It is easy to check that:

$$\begin{array}{@{}rcl@{}} \frac{\partial N_{j}}{\partial s_{j}} & = &\frac{-g_{j}^{\prime}(z_{j})}{g_{j}(z_{j})^{2}},\\ \frac{\partial N_{j}}{\partial p_{j}} & = &\frac{z_{j}g_{j}^{\prime}(z_{j})y_{j}^{\prime}(p_{j})}{g_{j}(z_{j})^{2}}+\frac{y_{j}(p_{j})}{g_{j}(z_{j})}+ 2(p_{j}+\delta_{j}) y_{j}^{\prime\prime}(p_{j})+2y_{j}^{\prime}(p_{j}),\\ \frac{\partial^{2} N_{j}}{\partial {s_{j}^{2}}} & = & \frac{2g_{j}^{\prime}(z_{j})^{2}-g_{j}(z_{j})g_{j}^{\prime\prime}(z_{j})}{g_{j}(z_{j})^{3} y_{j}(p_{j})},\\ \frac{\partial^{2} N_{j}}{\partial s_{j} p_{j}} & = & -z_{j}y_{j}^{\prime}(p_{j})\frac{2g_{j}^{\prime}(z_{j})^{2}-g_{j}(z_{j})g_{j}^{\prime\prime}(z_{j})}{g_{j}(z_{j})^{3} y_{j}(p_{j})}, \end{array} $$

$$\begin{array}{@{}rcl@{}} \frac{\partial^{2} N_{j}}{\partial {p_{j}^{2}}} & = & {z_{j}^{2}}y_{j}^{\prime}(p_{j})^{2}\frac{2g_{j}^{\prime}(z_{j})^{2}-g_{j}(z_{j})g_{j}^{\prime\prime}(z_{j})}{g_{j}(z_{j})^{3} y_{j}(p_{j})}+ \frac{y_{j}^{\prime\prime}(p_{j})}{g_{j}(z_{j})} \\&&+y_{j}^{\prime\prime}(p_{j})\frac{z_{j}g_{j}^{\prime}(z_{j})}{g_{j}(z_{j})^{2}}+4y_{j}^{\prime\prime}(p_{j})+2(p_{j}+\delta_{j})y_{j}^{\prime\prime\prime}(p_{j}). \end{array} $$

Note that $\frac {\partial ^{2} N_{j}}{\partial {s_{j}^{2}}}\ge 0$ if $2(g_{j}^{\prime }(z_{j}))^{2}-g_{j}(z_{j}) g_{j}^{\prime \prime }(z_{j})\ge 0$. The determinant of the hessian of N _j is given by

$${\Theta}_{j}\,=\,\frac{2g_{j}^{\prime}(z_{j})^{2}\,-\,g_{j}(z_{j})g_{j}^{\prime\prime}(z_{j})}{g_{j}(z_{j})^{3} y_{j}(p_{j})}\left( \!y_{j}^{\prime\prime}(p_{j})\frac{z_{j}g_{j}^{\prime}(z_{j})}{g_{j}(z_{j})^{2}}\,+\, \frac{y_{j}^{\prime\prime}(p_{j})}{g_{j}(z_{j})}\,+\,4y_{j}^{\prime\prime}(p_{j})\,+\,2(p_{j}+\delta_{j})y^{\prime\prime\prime}(p_{j})\!\right).$$

From Assumptions 1 and 2b, $2(g_{j}^{\prime }(z_{j}))^{2}-g_{j}(z_{j}) g_{j}^{\prime \prime }(z_{j})\ge 0$, $g_{j}^{\prime }(z_{j})\ge 0$, $2y_{j}^{\prime \prime }(p_{j})+p_{j}y^{\prime \prime \prime }(p_{j})\ge 0$ and $y_{j}^{\prime \prime }(p_{j})\ge 0$ therefore Θ _j has the same sign as $2y_{j}^{\prime \prime }(p_{j})+(p_{j}+\delta _{j})y^{\prime \prime \prime }(p_{j})$. For y ^″′(p _j ) ≥ 0, one notice that $2y_{j}^{\prime \prime }(p_{j})+(p_{j}+\delta _{j})y^{\prime \prime \prime }(p_{j})\ge 0$ since y ^″(p _j ) ≥ 0 and $p_{j}+\delta _{j}\ge p_{j}-\lambda _{j}^{+}\ge 0$. Next for y ^″′(p _j )≤0, $2y_{j}^{\prime \prime }(p_{j})+(p_{j}+\delta _{j})y^{\prime \prime \prime }(p_{j})\ge 0$ since $2y_{j}^{\prime \prime }(p_{j})+p_{j}y^{\prime \prime \prime }(p_{j})\ge 0$ by Assumption 1 and δ _j y ^″′(p _j ) ≥ 0 since δ _j ≤ 0 by definition. Hence Θ _j ≥ 0 implying that the function N _j is convex. We can then conclude that the set ${{\Gamma }^{2}_{j}}$ is convex.

Appendix E: Proof of Theorem 2

As in the additive case, optimization problem (8) can be reformulated as:

$$\begin{array}{@{}rcl@{}} \max\limits {\Pi}_{j}(\tilde{Q}_{j},s_{j},p_{j}) & = & p_{j} y_{j}(p_{j})(1- e_{j}^{-}(z_{j}))+y_{j}(p_{j})\\&&\times\left( \lambda_{j}^{+} e_{j}^{+}(z_{j}) - \lambda_{j}^{-} e_{j}^{-}(z_{j})\right) - c_{j} s_{j} -p_{Ij}^{*} s_{j} \\ & & - \sum\limits_{i=1}^{I-1} (p_{ij}^{*}-p_{Ij}^{*}) q_{ij} \\ & =& {{\Pi}_{j}^{1}}(s_{j},p_{j})+ {{\Pi}_{j}^{2}}(\tilde{Q}_{j}), \end{array} $$

where $\tilde {Q}_{j}=(q_{ij})_{i=1}^{I-1}$, ${{\Pi }_{j}^{1}}(s_{j},p_{j})=p_{j} y_{j}(p_{j})(1- e_{j}^{-}(z_{j}))+y_{j}(p_{j})\left ( \lambda _{j}^{+} e_{j}^{+}(z_{j}) - \lambda _{j}^{-} e_{j}^{-}(z_{j})\right ) - c_{j} s_{j}-p_{Ij}^{*} s_{j}$and ${{\Pi }_{j}^{2}}(\tilde {Q}_{j})=- \sum \limits _{i=1}^{I-1} (p_{ij}^{*}-p_{Ij}^{*}) q_{ij}$.

Clearly, the function ${{\Pi }_{j}^{2}}(\tilde {Q}_{j})$ is concave. We need to prove that ${{\Pi }_{j}^{1}}(s_{j},p_{j})$ is concave. Let us consider the first derivative of ${{\Pi }_{j}^{1}}$ taken with respect to s _j and p _j :

$$\begin{array}{@{}rcl@{}} \frac{\partial {{\Pi}_{j}^{1}}}{\partial s_{j}} & = & -p_{j}\left( F_{j}(z_{j})-1\right)+\lambda_{j}^{+} F_{j}(z_{j})-\lambda_{j}^{-} \left( F_{j}(z_{j})-1\right)-c_{j} -p_{Ij}^{*} \\ & = & \lambda_{j}^{-}+p_{j} -c_{j} -p_{Ij}^{*}-\left( p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+}\right)F_{j}(z_{j}). \\ \frac{\partial {{\Pi}_{j}^{1}}}{\partial p_{j}} & = & y_{j}(p_{j})(1-e_{j}^{-}(z_{j}))+p_{j}y_{j}^{\prime}(p_{j})(1-e_{j}^{-}(z_{j}))+p_{j} z_{j} y_{j}^{\prime}(p_{j})(F_{j}(z_{j})-1) \\ & & +y_{j}^{\prime}(p_{j})\!\left( \lambda_{j}^{+} e_{j}^{+}(z_{j}) \,-\, \lambda_{j}^{-} e_{j}^{-}(z_{j})\right)\,-\,z_{j} y_{j}^{\prime}(p_{j})\left( \lambda_{j}^{+} F_{j}(z_{j})\,-\,\lambda_{j}^{-} (F_{j}(z_{j})\,-\,1)\!\right)\\ & = & y_{j}(p_{j})(1-e_{j}^{-}(z_{j}))+y_{j}^{\prime}(p_{j})\\&&\times\left[ (p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+})[(1-e_{j}^{-}(z_{j}))+z_{j}(F_{j}(z_{j})-1)]-\lambda_{j}^{-}\right]. \end{array} $$

Straightforward computations show that:

$$\begin{array}{@{}rcl@{}} \frac{\partial^{2}{{\Pi}_{j}^{1}}}{\partial {s_{j}^{2}}} &= & - \frac{1}{y_{j}(p_{j})}\left( p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+}\right) f_{j}(z_{j}),\\ \frac{\partial^{2}{{\Pi}_{j}^{1}}}{\partial s_{j}\partial p_{j}} & = & \frac{z_{j}}{y_{j}(p_{j})}y^{\prime}_{j}(p_{j})\left( p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+}\right) f_{j}(z_{j})+1-F_{j}(z_{j}), \\ \frac{\partial^{2}{{\Pi}_{j}^{1}}}{\partial {p_{j}^{2}}} & = & -\frac{{z_{j}^{2}}y_{j}^{\prime}(p_{j})^{2}}{y_{j}(p_{j})}\!\left( p_{j}+\lambda_{j}^{-}\,-\,\lambda_{j}^{+}\right)\! f_{j}(z_{j})\,+\,2y_{j}^{\prime}(p_{j})\! \left[\!(1\,-\,e_{j}^{-}(z_{j}))+z_{j}(F_{j}(z_{j})\,-\,1)\!\right]\\ & & +y^{\prime\prime}_{j}(p_{j}) \left[\left( p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+}\right)\left( (1-e_{j}^{-}(z_{j}))+z_{j}(F_{j}(z_{j})-1)\right)-\lambda_{j}^{-}\right]. \end{array} $$

Let H _j denotes the hessian matrix associated with ${{\Pi }_{j}^{1}}(s_{j},p_{j})$. The matrix H _j is computed as:

$$\left( \begin{array}{cc} H_{s_{j}s_{j}} & H_{p_{j} s_{j}} \\ H_{p_{j}s_{j}} & H_{p_{j}p_{j}} \end{array}\right), $$

with $H_{s_{j}s_{j}}=- \frac {1}{y_{j}(p_{j})}\left (p_{j}+\lambda _{j}^{-}-\lambda _{j}^{+}\right ) f_{j}(z_{j}), H_{p_{j}s_{j}}=H_{s_{j}p_{j}}=\frac {z_{j}}{y_{j}(p_{j})}y^{\prime }_{j}(p_{j})\left (p_{j}+\lambda _{j}^{-}-\lambda _{j}^{+}\right ) f_{j}(z_{j})+1-F_{j}(z_{j})$ and

$$\begin{array}{@{}rcl@{}} H_{p_{j}p_{j}} & = & -\frac{{z_{j}^{2}}y_{j}^{\prime}(p_{j})^{2}}{y_{j}(p_{j})}\!\left( p_{j}\,+\,\lambda_{j}^{-}\,-\,\lambda_{j}^{+}\right) f_{j}(z_{j})+2y_{j}^{\prime}(p_{j})\! \left[(1\,-\,e_{j}^{-}(z_{j}))+z_{j}(F_{j}(z_{j})\,-\,1)\right]\\ & & +y^{\prime\prime}_{j}(p_{j}) \left[\left( p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+}\right)\left( (1-e_{j}^{-}(z_{j}))+z_{j}(F_{j}(z_{j})-1)\right)-\lambda_{j}^{-}\right]\\ & =& -\frac{{z_{j}^{2}}y_{j}^{\prime}(p_{j})^{2}}{y_{j}(p_{j})}\left( p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+}\right) f_{j}(z_{j})+2y_{j}^{\prime}(p_{j}){\int}_{A_{j}}^{z_{j}} xf_{j}(x) dx\\ & & +y^{\prime\prime}_{j}(p_{j}) \left[\left( p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+}\right){\int}_{A_{j}}^{z_{j}} xf_{j}(x) dx-\lambda_{j}^{-}\right]. \end{array} $$

We will show that the matrix − H _j is positive definite. Clearly, $-H_{s_{j}s_{j}} =\frac {1}{y_{j}(p_{j})}\left (p_{j}+\lambda _{j}^{-}-\lambda _{j}^{+}\right ) f_{j}(z_{j})$ is positive. If d e t( − H _j )>0 then $-H_{p_{j}p_{j}}> \frac {\left (H_{p_{j} s_{j}}\right )^{2}}{-H_{s_{j}s_{j}}}> 0$.

$$\begin{array}{@{}rcl@{}} det(-H_{j})&=&-\frac{y_{j}^{\prime\prime}(p_{j})}{y_{j}(p_{j})}(p_{j}+\lambda^{-}_{j} -\lambda^{+}_{j})f_{j}(z_{j})\left[\left( p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+}\right){\int}_{A_{j}}^{z_{j}} xf_{j}(x) dx-\lambda_{j}^{-}\right]\\ & & -2\frac{y_{j}^{\prime}(p_{j})}{y_{j}(p_{j})}(p_{j}+\lambda^{-}_{j} -\lambda^{+}_{j})f_{j}(z_{j}){\int}_{A_{j}}^{z_{j}} xf_{j}(x) dx\\ & & -2\frac{y_{j}^{\prime}(p_{j})}{y_{j}(p_{j})}(p_{j}+\lambda^{-}_{j} -\lambda^{+}_{j})f_{j}(z_{j})z_{j}\left( 1-F_{j}(z_{j})\right)-(1-F_{j}(z_{j}))^{2}\\ & = & \frac{(p_{j}+\lambda^{-}_{j} -\lambda^{+}_{j})f_{j}(z_{j})}{y_{j}(p_{j})}{\int}_{A_{j}}^{z_{j}} xf_{j}(x) dx \left[ -2y_{j}^{\prime}(p_{j})-p_{j}y_{j}^{\prime\prime}(p_{j})\right]\\ & & + \frac{(p_{j}+\lambda^{-}_{j} -\lambda^{+}_{j})f_{j}(z_{j})}{y_{j}(p_{j})}y^{\prime\prime}_{j}(p_{j})\left[ \lambda_{j}^{-}{\int}_{z_{j}}^{B_{j}} xf_{j}(x) dx+\lambda_{j}^{+}{\int}_{A_{j}}^{z_{j}} xf_{j}(x) dx\right]\\ & & + 2(1-F_{j}(z_{j}))^{2}\left[\frac{-y_{j}^{\prime}(p_{j})(p_{j}+\lambda^{-}_{j} -\lambda^{+}_{j})g_{j}(z_{j})}{y_{j}(p_{j})}-\frac{1}{2}\right]. \end{array} $$

The first term in the above is nonnegative because $2 y^{\prime }_{j}(p_{j})+ p_{j} y_{j}^{\prime \prime }(p_{j})\leq 0$. The second term is also nonnegative because $y_{j}^{\prime \prime }(p_{j})\geq 0$. The last term is nonnegative since it can be written as $2 (1-F_{j}(z_{j}))^{2}\left [\eta ^{2}_{j}(s_{j},p_{j})-1/2\right ]+2 (1-F_{j}(z_{j}))^{2}\max (\lambda _{j}^{-}-\lambda _{j}^{+},0) \frac {-y_{j}^{\prime }(p_{j})g_{j}(z_{j})}{y_{j}(p_{j})}$ and ${\eta ^{2}_{j}}(s_{j},p_{j})\ge 1/2$. Hence, we have d e t( − H _j ) ≥ 0 implying that the matrix − H _j is positive definite. Therefore, ${{\Pi }_{j}^{1}}$ is concave in the set ${{\Gamma }^{2}_{j}}$.

Appendix F: Proof of Proposition 2

We need to show that the optimal vector $(s_{j}^{*},p_{j}^{*})$ belongs to the set ${{\Gamma }^{2}_{j}}$. In other words, we need to show that ${\eta _{j}^{2}}(s_{j},p_{j})\ge 1/2$ when $\frac {{\partial {\Pi }_{j}^{1}}}{\partial p_{j}}=0$.

$$\begin{array}{@{}rcl@{}} y_{j}^{\prime}(p_{j}) & = & \frac{-y_{j}(p_{j})(1-e_{j}^{-}(z_{j}))}{\left( p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+}\right)\left[(1-e_{j}^{-}(z_{j}))+z_{j}(F_{j}(z_{j})-1)\right]-\lambda_{j}^{-}} \\ & = & \frac{-y_{j}(p_{j})\left( {\int}_{A_{j}}^{z_{j}}xf_{j}(x)\,dx+z_{j}(1-F_{j}(z_{j})) \right)}{\left( p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+}\right){\int}_{A_{j}}^{z_{j}}xf_{j}(x)\,dx -\lambda_{j}^{-}}. \end{array} $$

Since $y_{j}^{\prime }(p_{j})\le 0$ and ${\int }_{A_{j}}^{z_{j}}xf_{j}(x)\,dx+z_{j}(1-F_{j}(z_{j}))\ge 0$ one see that when $\frac {{\partial {\Pi }_{j}^{1}}}{\partial p_{j}}=0$ one has $\left (p_{j}+\lambda _{j}^{-}-\lambda _{j}^{+}\right ){\int }_{A_{j}}^{z_{j}}xf_{j}(x)\,dx -\lambda _{j}^{-}> 0$.

$$\begin{array}{@{}rcl@{}} {\eta_{j}^{2}}(s_{j},p_{j})\ge 1/2 & \Longleftrightarrow & \frac{-(p_{j}+\delta_{j}) y_{j}^{\prime}(p_{j})g_{j}(z_{j})}{y_{j}(p_{j})}-\frac{1}{2}\ge 0\\ & \Longleftrightarrow & \frac{-(p_{j}+\delta_{j}) y_{j}^{\prime}(p_{j})}{y_{j}(p_{j})}-\frac{1}{2g_{j}(z_{j})}\ge 0\\ &\Longleftrightarrow & \frac{(p_{j}+\delta_{j})\left[{\int}_{A_{j}}^{z_{j}}xf_{j}(x)\,dx+z_{j}(1\,-\,F_{j}(z_{j}))\,-\,\frac{{\int}_{A_{j}}^{z_{j}}xf_{j}(x)\,dx}{2g_{j}(z_{j})}\right]}{\left( p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+}\right){\int}_{A_{j}}^{z_{j}}xf_{j}(x)\,dx -\lambda_{j}^{-}}\\ & & \quad +\frac{\lambda_{j}^{-}{\int}_{z_{j}}^{B_{j}}xf_{j}(x)\,dx+\min(\lambda_{j}^{-},\lambda_{j}^{+}){\int}_{A_{j}}^{z_{j}}xf_{j}(x)\,dx}{2g_{j}(z_{j})\left[\left( p_{j}+\lambda_{j}^{-}-\lambda_{j}^{+}\right){\int}_{A_{j}}^{z_{j}}xf_{j}(x)\,dx -\lambda_{j}^{-} \right]}\ge 0.\\ \end{array} $$

The last inequality holds if ψ _j (z _j ) ≥ 0 where $\psi _{j}(z_{j})={\int }_{A_{j}}^{z_{j}}xf_{j}(x)\,dx+z_{j}(1-F_{j}(z_{j}))-\frac {{\int }_{A_{j}}^{z_{j}}xf_{j}(x)\,dx}{2g_{j}(z_{j})}.$ Note that ψ _j (A _j ) = A _j and $\psi _{j}^{\prime }(z_{j})=\frac {1-F_{j}(z_{j})}{2}+\frac {g_{j}^{\prime }(z_{j}){\int }_{A_{j}}^{z_{j}}xf_{j}(x)\,dx}{2g_{j}(z_{j})^{2}}\geq 0$ because $g_{j}^{\prime }(z_{j})\geq 0$. This yields ψ _j (z _j ) ≥ A _j for all z _j ∈[A _j , B _j ]. Then ψ _j (z _j ) ≥ 0. Therefore, ${\eta _{j}^{2}}(s_{j},p_{j})\ge 1/2$ when $\frac {{\partial {\Pi }_{j}^{1}}}{\partial p_{j}}=0$.

Appendix G: Proof of Theorem 3

As in the retailer case, without loss of generality, we assume that $p_{mI}^{*}=\max \limits _{i} \{p_{mi}^{*}\}$. Optimization problem (11) can be then reformulated as:

$$\begin{array}{@{}rcl@{}} \max\limits {\Pi}_{m}(\tilde{Q}_{m},q_{m},p_{m}) & = & \sum\limits_{i=1}^{I-1} (p_{mi}^{*}-p_{mI}^{*})q_{mi} + p_{mI}^{*} q_{m} + s^{rec}y_{m}(p_{m}) \,-\,\left( {c_{m}^{u}}+c^{u} \bar{\chi}_{m}\right)q_{m} \\ & & \quad -p_{m} y_{m}(p_{m})+\lambda_{m}^{+} e_{m}^{+}(z_{m}) - \lambda_{m}^{-} e_{m}^{-}(z_{m}),\\ & = & {{\Pi}_{m}^{1}}(q_{m},p_{m})+{{\Pi}_{m}^{2}}(\tilde{Q}_{m}), \end{array} $$

where $\tilde {Q}_{m}=(q_{mi})_{i=1}^{I-1}$, ${{\Pi }_{m}^{1}}(q_{m},p_{m})=p_{mI}^{*} q_{m} + s^{rec}y_{m}(p_{m}) -\left ({c_{m}^{u}}+c^{u} \bar {\chi }_{m}\right )q_{m} -p_{m} y_{m}(p_{m})+\lambda _{m}^{+} e_{m}^{+}(z_{m}) - \lambda _{m}^{-} e_{m}^{-}(z_{m})$and ${{\Pi }_{m}^{2}}(\tilde {Q}_{m})=\sum \limits _{i=1}^{I-1} (p_{mi}^{*}-p_{mI}^{*})q_{mi}$.

Clearly, the function ${{\Pi }_{m}^{2}}(\tilde {Q}_{m})$ is concave. We need to show that ${{\Pi }_{m}^{1}}(q_{m},p_{m})$ is concave. The first derivative of ${{\Pi }_{m}^{1}}$ taken with respect to q _m and p _m and are given as:

$$\begin{array}{@{}rcl@{}} \frac{\partial {{\Pi}_{m}^{1}}}{\partial q_{m}} & = & p_{mI}^{*} -{c_{m}^{u}}-c^{u} \bar{\chi}_{m}+\frac{\lambda_{m}^{+}}{\chi_{m}} \left( F_{m}(z_{m})-1\right) -\frac{\lambda_{m}^{-}}{\chi_{m}} F_{m}(z_{m}) \\ & = & p_{mI}^{*} -{c_{m}^{u}}-c^{u} \bar{\chi}_{m}-\frac{\lambda_{m}^{+}}{\chi_{m}} -\frac{\left( \lambda_{m}^{-}-\lambda_{m}^{+}\right)}{\chi_{m}}F_{m}(z_{m}).\\ \frac{\partial {{\Pi}_{m}^{1}}}{\partial p_{m}} & = & s^{rec}y_{m}^{\prime}(p_{m})- p_{m} y_{m}^{\prime}(p_{m})- y_{m}(p_{m})\\ & & +y_{m}^{\prime}(p_{m})\lambda_{m}^{+} \left( 1-F_{m}(z_{m})\right) +y_{m}^{\prime}(p_{m})\lambda_{m}^{-} F_{m}(z_{m}) \\ & = & -y_{m}(p_{m})+y_{m}^{\prime}(p_{m})\left( s^{rec}-p_{m}+\lambda_{m}^{+}+ F_{m}(z_{m})( \lambda_{m}^{-}-\lambda_{m}^{+})\right). \end{array} $$

Straightforward computations show that:

$$\begin{array}{@{}rcl@{}} \frac{\partial^{2}{{\Pi}_{m}^{1}}}{\partial {q_{m}^{2}}} &= & -\frac{1}{{\chi_{m}^{2}}} \left( \lambda_{m}^{-}-\lambda_{m}^{+}\right) f_{m}(z_{m}), \\ \frac{\partial^{2}{{\Pi}_{m}^{1}}}{\partial q_{m}\partial p_{m}} & = & \frac{y^{\prime}_{m}(p_{m})}{\chi_{m}}\left( \lambda_{m}^{-}-\lambda_{m}^{+}\right) f_{m}(z_{m}), \\ \frac{\partial^{2}{{\Pi}_{m}^{1}}}{\partial {p_{m}^{2}}} & = & -y_{m}^{\prime}(p_{m})^{2}\left( \lambda_{m}^{-}-\lambda_{m}^{+}\right) f_{m}(z_{m})-2y_{m}^{\prime}(p_{m}) \\ & & +y^{\prime\prime}_{m}(p_{m}) \left[\left( \lambda_{m}^{-}-\lambda_{m}^{+}\right)F_{m}(z_{m})+s^{rec}-p_{m}+\lambda_{m}^{+}\right]. \end{array} $$

Let H _m denotes the hessian matrix associated with ${{\Pi }_{m}^{1}}(q_{m},p_{m})$. The matrix H _m is calculated as:

$$\left( \begin{array}{cc} H_{q_{m}q_{m}} & H_{p_{m} q_{m}} \\ H_{p_{m}q_{m}} & H_{p_{m}p_{m}} \end{array}\right), $$

with $H_{q_{m}q_{m}}=-\frac {1}{{\chi _{m}^{2}}} \left (\lambda _{m}^{-}-\lambda _{m}^{+}\right ) f_{m}(z_{m}), H_{p_{m}q_{m}}=H_{q_{m}p_{m}}=\frac {y^{\prime }_{m}(p_{m})}{\chi _{m}}\left (\lambda _{m}^{-}-\lambda _{m}^{+}\right ) f_{m}(z_{m})$and $H_{p_{m}p_{m}}=-y_{m}^{\prime }(p_{m})^{2}\left (\lambda _{m}^{-}-\lambda _{m}^{+}\right ) f_{m}(z_{m})-2y_{m}^{\prime }(p_{m}) +y^{\prime \prime }_{m}(p_{m}) \left [\left (\lambda _{m}^{-}-\lambda _{m}^{+}\right )F_{m}(z_{m})+s^{rec}-p_{m}+\lambda _{m}^{+}\right ]$.

We will show that the matrix − H _m is positive definite. Clearly, $-H_{q_{m}q_{m}} =\frac {1}{{\chi _{m}^{2}}} \left (\lambda _{m}^{-}-\lambda _{m}^{+}\right ) f_{m}(z_{m})>0.$ If d e t( − H _m ) ≥ 0 then $-H_{p_{m}p_{m}}\ge \frac {\left (H_{p_{m} q_{m}}\right )^{2}}{-H_{q_{m}q_{m}}}\ge 0$.

$$\begin{array}{@{}rcl@{}} {\chi_{m}^{2}} det(-H_{m})&=&-(\lambda^{-}_{m} -\lambda^{+}_{m})f_{m}(z_{m})\left( \left( \lambda_{m}^{-}-\lambda_{m}^{+}\right)F_{m}(z_{m})+s^{rec}-p_{m}+\lambda_{m}^{+}\right)y_{m}^{\prime\prime}(p_{m})\\ & & +2(\lambda^{-}_{m} -\lambda^{-}_{m})f_{m}(z_{m})y_{m}^{\prime}(p_{m})\\ & = & (\lambda^{-}_{m} -\lambda^{+}_{m})f_{m}(z_{m})\left[2y_{m}^{\prime}(p_{m}) + p_{m} y_{m}^{\prime\prime}(p_{m})\right]\\ & & -(\lambda^{-}_{m} -\lambda^{+}_{m})f_{m}(z_{m})y_{m}^{\prime\prime}(p_{m})\left[\lambda_{m}^{-} F_{m}(z_{m})+\lambda_{m}^{+}(1-F_{m}(z_{m}))+s^{rec}\right]. \end{array} $$

The first part of the last term is nonnegative because $2 y^{\prime \prime }_{m}(p_{m})+ p_{m} y_{m}^{\prime \prime }(p_{m})\geq 0$. The second part is also nonnegative because $y_{m}^{\prime \prime }(p_{m})\leq 0$. Hence, we have d e t( − H _m ) ≥ 0 implying that the matrix − H _m is positive definite. Therefore, ${{\Pi }_{m}^{1}}$ is concave.

Appendix H: Proof of Theorem 4

Similar to the additive case, optimization problem (13) can be reformulated as:

$$\begin{array}{@{}rcl@{}} \max\limits {\Pi}_{m}(\tilde{Q}_{m},q_{m},p_{m}) & = & \sum\limits_{i=1}^{I-1} (p_{mi}^{*}-p_{mI}^{*})q_{mi} + p_{mI}^{*} q_{m} + s^{rec}y_{m}(p_{m}) \,-\,\left( {c_{m}^{u}}+c^{u} \bar{\chi}_{m}\right)q_{m} \\ & & \quad -p_{m}y_{m}(p_{m})+y_{m}(p_{m})\left( \lambda_{m}^{+} e_{m}^{+}(z_{m}) - \lambda_{m}^{-} e_{m}^{-}(z_{m})\right)\\ & = & {{\Pi}_{m}^{1}}(q_{m},p_{m})+{{\Pi}_{m}^{2}}(\tilde{Q}_{m}), \end{array} $$

where $\tilde {Q}_{m}=(q_{mi})_{i=1}^{I-1}$, ${{\Pi }_{m}^{1}}(q_{m},p_{m})=p_{mI}^{*} q_{m} + s^{rec}y_{m}(p_{m}) -\left ({c_{m}^{u}}+c^{u} \bar {\chi }_{m}\right )q_{m} -p_{m}y_{m}(p_{m})+y_{m}(p_{m})\left (\lambda _{m}^{+} e_{m}^{+}(z_{m}) - \lambda _{m}^{-} e_{m}^{-}(z_{m})\right )$and ${{\Pi }_{m}^{2}}(\tilde {Q}_{m})=\sum \limits _{i=1}^{I-1} (p_{mi}^{*}-p_{mI}^{*})q_{mi}$.

Clearly, the function ${{\Pi }_{m}^{2}}(\tilde {Q}_{m})$ is concave. We need to show that ${{\Pi }_{m}^{1}}(q_{m},p_{m})$ is concave. The first derivative of ${{\Pi }_{m}^{1}}$ taken with respect to q _m and p _m and are computed as:

$$\begin{array}{@{}rcl@{}} \frac{\partial {{\Pi}_{m}^{1}}}{\partial q_{m}} & = & p_{mI}^{*} -{c_{m}^{u}}-c^{u} \bar{\chi}_{m}+\frac{\lambda_{m}^{+}}{\chi_{m}} \left( F_{m}(z_{m})-1\right) -\frac{\lambda_{m}^{-}}{\chi_{m}} F_{m}(z_{m}) \\ & = & p_{mI}^{*} -{c_{m}^{u}}-c^{u} \bar{\chi}_{m}-\frac{\lambda_{m}^{+}}{\chi_{m}} -\frac{\left( \lambda_{m}^{-}-\lambda_{m}^{+}\right)}{\chi_{m}}F_{m}(z_{m}).\\ \frac{\partial {{\Pi}_{m}^{1}}}{\partial p_{m}} & = & -y_{m}(p_{m}) -y_{m}^{\prime}(p_{m}) p_{m} + y_{m}^{\prime}(p_{m})s^{rec}\\ & & +y_{m}^{\prime}(p_{m})\left[\left( \lambda_{m}^{+} e_{m}^{+}(z_{m}) - \lambda_{m}^{-} e_{m}^{-}(z_{m})\right)-\lambda_{m}^{+} z_{m} (F_{m}(z_{m})-1)+ \lambda_{m}^{-} z_{m} F_{m}(z_{m})\right]\\ & = & -y_{m}(p_{m})+y_{m}^{\prime}(p_{m})\left[s^{rec}\,-\,p_{m} + \lambda_{m}^{-}\,-\,(\lambda_{m}^{-}\,-\,\lambda_{m}^{+})\left( (1+e_{m}^{-}(z_{m}))\!- \!z_{m} F_{m}(z_{m})\right)\right]. \end{array} $$

Straightforward computations show that:

$$\begin{array}{@{}rcl@{}} \frac{\partial^{2}{{\Pi}_{m}^{1}}}{\partial {q_{m}^{2}}} &= & -\frac{1}{{\chi_{m}^{2}} y_{m}(p_{m})} \left( \lambda_{m}^{-}-\lambda_{m}^{+}\right) f_{m}(z_{m}),\\ \frac{\partial^{2}{{\Pi}_{m}^{1}}}{\partial q_{m}\partial p_{m}} & = & \frac{y^{\prime}_{m}(p_{m})z_{m}}{\chi_{m} y_{m}(p_{m})}\left( \lambda_{m}^{-}-\lambda_{m}^{+}\right) f_{m}(z_{m}),\\ \frac{\partial^{2}{{\Pi}_{m}^{1}}}{\partial {p_{m}^{2}}} & = & -\frac{{z_{m}^{2}}y_{m}^{\prime}(p_{m})^{2}}{y_{m}(p_{m})}\left( \lambda_{m}^{-}-\lambda_{m}^{+}\right) f_{m}(z_{m})-2y_{m}^{\prime}(p_{m}) \\ & & +y^{\prime\prime}_{m}(p_{m}) \left[s^{rec}\,-\,p_{m}\,+\,\lambda_{m}^{-}\,-\,(\lambda_{m}^{-}-\lambda_{m}^{+})\left( (1+e_{m}^{-}(z_{m}))-z_{m} F_{m}(z_{m})\right)\right]. \end{array} $$

Let H _m denotes the hessian matrix associated with ${{\Pi }_{m}^{1}}(p_{m},p_{m})$. The matrix H _m is calculated as:

$$\left( \begin{array}{cc} H_{q_{m}q_{m}} & H_{p_{m} q_{m}} \\ H_{p_{m} q_{m}} & H_{p_{m}p_{m}} \end{array}\right), $$

with $H_{q_{m}q_{m}}=-\frac {1}{{\chi _{m}^{2}} y_{m}(p_{m})} \left (\lambda _{m}^{-}-\lambda _{m}^{+}\right ) f_{m}(z_{m}), H_{q_{m}p_{m}}=H_{p_{m}q_{m}}\frac {y^{\prime }_{m}(p_{m})z_{m}}{{\chi _{m}^{2}} y_{m}(p_{m})}\left (\lambda _{m}^{-}-\lambda _{m}^{+}\right ) f_{m}(z_{m})$and $H_{p_{m}p_{m}}=-\frac {{z_{m}^{2}}y_{m}^{\prime }(p_{m})^{2}}{y_{m}(p_{m})}\left (\lambda _{m}^{-}-\lambda _{m}^{+}\right ) f_{m}(z_{m})-2y_{m}^{\prime }(p_{m}) +y^{\prime \prime }_{m}(p_{m}) \left [s^{rec}-p_{m}+\lambda _{m}^{-}-(\lambda _{m}^{-}-\lambda _{m}^{+})\left ((1+e_{m}^{-}(z_{m}))-z_{m} F_{m}(z_{m})\right )\right ]$.

We will show that the matrix − H _m is positive definite. Clearly, $-H_{q_{m}q_{m}} =\frac {1}{{\chi _{m}^{2}} y_{m}(p_{m})} \left (\lambda _{m}^{-}-\lambda _{m}^{+}\right ) f_{m}(z_{m})>0$. If d e t( − H _m )>0 then $-H_{p_{m}p_{m}}\ge \frac {\left (H_{p_{m} q_{m}}\right )^{2}}{-H_{q_{m}q_{m}}} \ge 0$.

$$\begin{array}{@{}rcl@{}} {\chi_{m}^{2}} det(-H_{m})&=&-(\lambda^{-}_{m} -\lambda^{+}_{m})f_{m}(z_{m})\left( \lambda_{m}^{-}+s^{rec}-p_{m}-(\lambda_{m}^{-}-\lambda_{m}^{+})\right.\\ & &\left. \left( (1+e_{m}^{-}(z_{m}))-z_{m} F_{m}(z_{m})\right)\right)\frac{y_{m}^{\prime\prime}(p_{m})}{y_{m}(p_{m})}+2(\lambda^{-}_{m} -\lambda^{+}_{m})f_{m}(z_{m})\frac{y_{m}^{\prime}(p_{m})}{y_{m}(p_{m})}\\ & = & -(\lambda^{-}_{m} -\lambda^{+}_{m})f_{m}(z_{m})\\ &&\times\left( \lambda_{m}^{-}+s^{rec}-p_{m}-(\lambda_{m}^{-}-\lambda_{m}^{+}){\int}_{z_{m}}^{B_{m}}xf_{m}(x) dx\right)\frac{y_{m}^{\prime\prime}(p_{m})}{y_{m}(p_{m})}\\ & & +2(\lambda^{-}_{m} -\lambda^{+}_{m})f_{m}(z_{m})\frac{y_{m}^{\prime}(p_{m})}{y_{m}(p_{m})}\\ & =& \frac{(\lambda^{-}_{m} -\lambda^{+}_{m})f_{m}(z_{m})}{y_{m}(p_{m})}\left[2 y^{\prime}_{m}(p_{m})+ p_{m} y^{\prime\prime}_{m}(p_{m})\right]\\ & & -\frac{(\lambda^{-}_{m} -\lambda^{+}_{m})f_{m}(z_{m})}{y_{m}(p_{m})}y_{m}^{\prime\prime}(p_{m})\\ &&\times\left[ \lambda_{m}^{-}{\int}_{A_{m}}^{z_{m}}xf_{m}(x)\,dx+\lambda_{m}^{+}{\int}_{z_{m}}^{B_{m}}xf_{m}(x)\,dx+s^{rec}\right]. \end{array} $$

The first part of the last term is nonnegative because $2 y^{\prime \prime }_{m}(p_{m})+ p_{m} y_{m}^{\prime \prime }(p_{m})\geq 0$. The second part is also nonnegative because $y_{m}^{\prime \prime }(p_{m})\leq 0$. Hence, we have d e t( − H _m ) ≥ 0 implying that the matrix − H _m is positive definite. Therefore, ${{\Pi }_{m}^{1}}$ is concave.

Appendix I: Proof of Theorem 9

We will only consider the case with additive demand and returns. The proof is similar in the multiplicative case. The expression $\langle \mathcal {F}(X^{\prime })- \mathcal {F}(X^{\prime \prime }), X^{\prime }-X^{\prime \prime }\rangle \ge 0$ is equivalent to (after some algebraic simplification):

$$\begin{array}{@{}rcl@{}} & & \sum\limits_{n=1}^{N} \sum\limits_{i=1}^{I}\left[\frac{\partial {f_{i}^{r}}\left( {\beta_{i}^{r}},\sum\limits_{n=1}^{N} q_{ni}^{\prime}\right)}{\partial q_{ni}}\,-\,\frac{\partial {f_{i}^{r}}\left( {\beta_{i}^{r}},\sum\limits_{n=1}^{N} q_{ni}^{\prime\prime}\right)}{\partial q_{ni}}+\frac{\partial {f_{n}^{r}}\left( \sum\limits_{i=1}^{I} q_{ni}^{\prime}\right)}{\partial q_{ni}}-\frac{\partial {f_{n}^{r}}\left( \sum\limits_{i=1}^{I} q_{ni}^{\prime\prime}\right)}{\partial q_{ni}}\right.\\ & &\left. \qquad\qquad +\frac{\partial c_{ni}(q_{ni}^{\prime})}{\partial q_{ni}}-\frac{\partial c_{ni}(q_{ni}^{\prime\prime})}{\partial q_{ni}}\right]\times [q_{ni}^{\prime} -q_{ni}^{\prime\prime} ] \end{array} $$

$$\begin{array}{@{}rcl@{}} & + & \sum\limits_{m=1}^{M} \sum\limits_{i=1}^{I}\!\left[\!\frac{\partial {f_{i}^{u}}\left( {\beta_{i}^{u}},\sum\limits_{m=1}^{M} q_{mi}^{\prime}\right)}{\partial q_{mi}}\,-\,\frac{\partial {f_{i}^{u}}\left( {\beta_{i}^{u}},\sum\limits_{m=1}^{M} q_{mi}^{\prime\prime}\right)}{\partial q_{mi}} \,+\, \frac{\partial c_{mi}(q_{mi}^{\prime})}{\partial q_{mi}}\,-\,\frac{\partial c_{mi}(q_{mi}^{\prime\prime})}{\partial q_{mi}}\! \right]\\ &&\qquad\qquad\times \left[q_{mi}^{\prime} -q_{mi}^{\prime\prime} \right]\\ & +& \sum\limits_{i=1}^{I} \sum\limits_{j=1}^{J}\left[\frac{\partial c_{ij}(q_{ij}^{\prime})}{\partial q_{ij}}-\frac{\partial c_{ij}(q_{ij}^{\prime\prime})}{\partial q_{ij}}\right] \times [q_{ij}^{\prime} -q_{ij}^{\prime\prime} ]\\ &+ & \sum\limits_{i=1}^{I} \sum\limits_{j=1}^{J}\left[\left( p_{j}^{\prime}+\lambda_{j}^{-}-\lambda_{j}^{+}\right)F_{j}(z_{j}^{\prime})-\left( p_{j}^{\prime\prime}+\lambda_{j}^{-}-\lambda_{j}^{+}\right)F_{j}(z_{j}^{\prime\prime}) \right] \times [q_{ij}^{\prime} -q_{ij}^{\prime\prime} ] \\ & + & \sum\limits_{j=1}^{J} \left[ -y_{j}(p_{j}^{\prime})+e_{j}^{-}(z_{j}^{\prime}) -y_{j}^{\prime}(p_{j}^{\prime})\left[(p_{j}^{\prime}+\lambda_{j}^{-}-\lambda_{j}^{+})F_{j}(z_{j}^{\prime})\right]\right.\\ & &\left. \qquad -\left( y_{j}(p_{j}^{\prime\prime})+e_{j}^{-}(z_{j}^{\prime\prime}) -y_{j}^{\prime}(p_{j}^{\prime\prime})\left[(p_{j}^{\prime\prime}+\lambda_{j}^{-}-\lambda_{j}^{+})F_{j}(z_{j}^{\prime\prime})\right]\right)\right] \times [p_{j}^{\prime} -p_{j}^{\prime\prime} ] \\ & + & \sum\limits_{m=1}^{M} \sum\limits_{i=1}^{I}\left[\frac{\left( \lambda_{m}^{-}-\lambda_{m}^{+}\right)}{\chi_{m}}F_{m}(z_{m}^{\prime})-\frac{\left( \lambda_{m}^{-}-\lambda_{m}^{+}\right)}{\chi_{m}}F_{m}(z_{m}^{\prime\prime})\right] \times [q_{mi}^{\prime} -q_{mi}^{\prime\prime} ] \\ & + &\sum\limits_{m=1}^{M} \left[ y_{m}(p^{\prime}_{m})-y_{m}^{\prime}(p_{m}^{\prime})\left( s^{rec}-p_{m}^{\prime}+\lambda_{m}^{+}\right)-y_{m}^{\prime}(p_{m}^{\prime}) F_{m}(z_{m}^{\prime})\left( \lambda_{m}^{-}-\lambda_{m}^{+}\right)\right. \\ & & \left.\qquad -\left( y_{m}(p^{\prime\prime}_{m})-y_{m}^{\prime}(p_{m}^{\prime\prime})\left( s^{rec}-p_{m}^{\prime\prime}+\lambda_{m}^{+}\right)-y_{m}^{\prime}(p_{m}^{\prime\prime}) F_{m}(z_{m}^{\prime\prime})\left( \lambda_{m}^{-}-\lambda_{m}^{+}\right)\right)\right]\\ &&\qquad\times \left[p_{m}^{\prime} -p_{m}^{\prime\prime} \right] \geq 0 \end{array} $$

which is equivalent to (I)+(I I)+(I I I)+(I V)+(V)+(V I)+(V I I) ≥ 0. Based on the convexity of the cost functions, ${f^{r}_{n}},{f^{r}_{i}},{f^{u}_{i}},c_{ni},c_{mi}$,and c _{i j} , one can have (I) ≥ 0, (I I) ≥ 0, and (I I I) ≥ 0. The proof of (I V)+(V) ≥ 0 can be derived from Theorem 1 since the function π _j is concave for each j, J = 1,...,J. Similarly, we can also show that (V I)+(V I I) ≥ 0 under the assumptions of Theorem 3. Therefore $\mathcal {F}(x)$ is monotone.