Determining the price and refund of products in a supply chain with quality and advertising costs in a fuzzy environment

Abstract

In online direct selling, three effective elements, namely price, refund and quality, affect the increment (or decrement) of demand and product return. This paper considers forward and backward (i.e., return) pricing decisions under uncertainty and develops a fuzzy mathematical model based on the Stackelberg game approach utilizing the proper action and reaction between a manufacturer and a retailer. Moreover, media advertising and manufacturer’s desire for accepting massive payments made us take into account the advertising as another factor influencing the demand. By an agreement between the manufacturer and the retailer, the costs of advertising and raising the level of the product quality are shared by two agreed rates. Two numerical examples are considered and the associated results are analyzed under fuzzy and crisp conditions when customers are sensitive or insensitive to the quality of the product. It is found that incorporation of the quality factor under a fuzzy environment has a better performance compared with the case of ignoring the quality and uncertainty in the parameters.

Appendix

Proof of Theorem 1

The first-order derivatives system of equations of retailer’s profit function (11) is as follows:

$$ \frac{{\partial E\left[ {\pi_{r} } \right]}}{\partial m} = {\kern 1pt} E\left[ {\tilde{\alpha }{\kern 1pt} {\kern 1pt} \tilde{\it K}_{r} } \right] + E\,[\tilde{\gamma }{\kern 1pt} \tilde{\it K}\tilde{\it K}_{r} ]{\kern 1pt} {\kern 1pt} r - \psi {\kern 1pt} w + E\,[\tilde{\omega }{\kern 1pt} {\kern 1pt} \tilde{\it K}_{r} ]{\kern 1pt} \,a - 2{\kern 1pt} {\kern 1pt} {\kern 1pt} \psi {\kern 1pt} m{\kern 1pt} = 0 $$

$$ \frac{{\partial E\left[ {\pi_{r} } \right]}}{\partial a} = m{\kern 1pt} {\kern 1pt} E\,[\tilde{\omega }{\kern 1pt} {\kern 1pt} \tilde{\it K}_{r} ] - 2\,(1 - t_{1} ){\kern 1pt} {\kern 1pt} \lambda {\kern 1pt} {\kern 1pt} a = 0 $$

$$ \frac{{\partial E\left[ {\pi_{r} } \right]}}{\partial r} = E[\tilde{\gamma }{\kern 1pt} \tilde{\it K}_{r} ]{\kern 1pt} {\kern 1pt} m{\kern 1pt} - E\left[ {\tilde{\it Z}} \right] - 2{\kern 1pt} {\kern 1pt} r{\kern 1pt} E\left[ {\tilde{\zeta }} \right] + w{\kern 1pt} E\left[ {\tilde{\zeta }} \right] + E\left[ {\tilde{\upsilon }} \right]{\kern 1pt} {\kern 1pt} {\kern 1pt} q = 0 $$

$$ \frac{{\partial E\left[ {\pi_{r} } \right]}}{\partial q} = E\left[ {\tilde{\upsilon }} \right]\left( {r - w} \right) - 2\left( {1 - t_{2} } \right){\kern 1pt} {\kern 1pt} \mu {\kern 1pt} {\kern 1pt} q = 0 $$

□

Solving the above system of equations, we get optimal values in Eqs. (13) to (16). Now we get the second-order derivatives:

$$ \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{{\partial^{2} r}} = - 2{\kern 1pt} E\left[ {\tilde{\zeta }} \right]\quad \;\frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{{\partial r{\kern 1pt} \,\partial m}} = E\,[\tilde{\gamma }{\kern 1pt} \tilde{\it K}_{r} ]\quad \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{{\partial r{\kern 1pt} \,\partial a}} = 0\quad \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{\partial r\partial q} = E\left[ {\tilde{\upsilon }} \right] $$

$$ \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{{\partial^{2} a}} = - 2{\kern 1pt} {\kern 1pt} \left( {1 - t_{1} } \right)\lambda \quad \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{{\partial a{\kern 1pt} \,\partial m}} = E\,[\tilde{\omega }{\kern 1pt} {\kern 1pt} \tilde{\it K}_{r} ]{\kern 1pt} \quad \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{{\partial a{\kern 1pt} \,\partial r}} = 0\quad \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{{\partial a{\kern 1pt} \,\partial q}} = 0 $$

$$ \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{{\partial^{2} q}} = - 2\left( {1 - t_{2} } \right)\mu \quad \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{\partial q\partial m} = 0\quad \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{\partial q\,\partial r} = E\left[ {\tilde{\upsilon }} \right]\quad \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{{\partial q{\kern 1pt} \,\partial a}} = 0 $$

Then, the Hessian Matrix of the profit function is as follows.

$$ H = \left| \begin{aligned} \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{{\partial^{2} m}}\quad \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{{\partial m{\kern 1pt} \,\partial r}}\quad \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{{\partial m{\kern 1pt} \,\partial a}}\quad \quad \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{{\partial m{\kern 1pt} \,\partial q}} \hfill \\ \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{{\partial r{\kern 1pt} \,\partial m}}\quad \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{{\partial^{2} r}}\quad \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{{\partial r{\kern 1pt} \,\partial a}}\quad \quad \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{\partial r\partial q} \hfill \\ \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{{\partial a{\kern 1pt} \,\partial m}}\quad \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{{\partial a{\kern 1pt} \,\partial r}}\quad \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{{\partial^{2} a}}\quad \quad \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{{\partial a{\kern 1pt} \,\partial q}} \hfill \\ \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{\partial q\partial m}\quad \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{\partial q\,\partial r}\quad \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{{\partial q{\kern 1pt} \,\partial a}}\quad \quad \frac{{\partial^{2} E\left[ {\pi_{r} } \right]}}{{\partial^{2} q}} \hfill \\ \end{aligned} \right| $$

To ensure that the retailer’s profit function is concave, the Hessian Matrix should be provided by the following conditions:

$$ H_{1} = \left| { - 2\,\psi } \right| < 0\quad H_{2} = \left| {\begin{array}{*{20}l} { - 2\,\psi } \hfill & {E\,[\tilde{\gamma }{\kern 1pt} {\kern 1pt} \tilde{\it K}_{r} ]} \hfill \\ {E\,[\tilde{\gamma }{\kern 1pt} {\kern 1pt} \tilde{\it K}_{r} ]} \hfill & { - 2E\left[ {\tilde{\zeta }} \right]} \hfill \\ \end{array} } \right| > 0 $$

$$ H_{3} = \left| {\begin{array}{*{20}l} { - 2\,\psi } \hfill & {E\,[\tilde{\gamma }{\kern 1pt} {\kern 1pt} \tilde{\it K}_{r} ]} \hfill & {E\,[\tilde{\omega }{\kern 1pt} {\kern 1pt} \tilde{\it K}_{r} ]} \hfill \\ {E\,[\tilde{\gamma }{\kern 1pt} {\kern 1pt} \tilde{\it K}_{r} ]} \hfill & { - 2E\left[ {\tilde{\zeta }} \right]} \hfill & 0 \hfill \\ {E\,[\tilde{\omega }{\kern 1pt} {\kern 1pt} \tilde{\it K}_{r} ]} \hfill & 0 \hfill & { - 2\left( {1 - t_{1} } \right)\lambda } \hfill \\ \end{array} } \right| < 0 $$

$$ H_{4} = \left| {\begin{array}{*{20}l} { - 2\,\psi } \hfill & {E\,[\tilde{\gamma }{\kern 1pt} {\kern 1pt} \tilde{\it K}_{r} ]} \hfill & {E\,[\tilde{\omega }{\kern 1pt} {\kern 1pt} \tilde{\it K}_{r} ]} \hfill & 0 \hfill \\ {E\,[\tilde{\gamma }{\kern 1pt} {\kern 1pt} \tilde{\it K}_{r} ]} \hfill & { - 2E\left[ {\tilde{\zeta }} \right]} \hfill & 0 \hfill & {E\left[ {\tilde{\upsilon }} \right]} \hfill \\ {E\,[\tilde{\omega }{\kern 1pt} {\kern 1pt} \tilde{\it K}_{r} ]} \hfill & 0 \hfill & { - 2\left( {1 - t_{1} } \right)\lambda } \hfill & 0 \hfill \\ 0 \hfill & {E\left[ {\tilde{\upsilon }} \right]} \hfill & 0 \hfill & { - 2\left( {1 - t_{2} } \right)\mu } \hfill \\ \end{array} } \right| > 0 $$

Therefore, we have the following conditions:

$$ - 2\,\psi < 0 $$

$$ 4\,E\left[ {\tilde{\zeta }} \right]\;\psi \, - E{\kern 1pt}^{2} {\kern 1pt} [\tilde{\gamma }{\kern 1pt} \tilde{\it K}_{r} ] > 0 $$

$$ 2\,E{\kern 1pt}^{2} \,[\tilde{\omega }{\kern 1pt} {\kern 1pt} \tilde{\it K}_{r} ]\;E\left[ {\tilde{\zeta }} \right] - 8\;E\left[ {\tilde{\zeta }} \right]\;\psi \left( {1 - t_{1} } \right)\lambda {\kern 1pt} + 2\;E{\kern 1pt}^{2} {\kern 1pt} [\tilde{\gamma }{\kern 1pt} \tilde{\it K}_{r} ]\left( {1 - t_{1} } \right)\lambda < 0 $$

$$ \begin{aligned} & E^{2} \left[ {\tilde{\upsilon }} \right]\left( {E{\kern 1pt}^{2} \,[\tilde{\omega }{\kern 1pt} {\kern 1pt} \tilde{\it K}_{r} ] - 4{\kern 1pt} {\kern 1pt} \lambda {\kern 1pt} {\kern 1pt} \psi \left( {1 - t_{1} } \right)} \right) - 4{\kern 1pt} E{\kern 1pt}^{2} \,[\tilde{\omega }{\kern 1pt} {\kern 1pt} \tilde{\it K}_{r} ]{\kern 1pt} {\kern 1pt} E\left[ {\tilde{\zeta }} \right]\mu \left( {1 - t_{2} } \right) + 4E[\tilde{\gamma }{\kern 1pt} {\kern 1pt} \tilde{\it K}_{r} ]{\kern 1pt} {\kern 1pt} \mu {\kern 1pt} \lambda \left( {t_{1} + t_{2} - t_{1} t_{2} - 1} \right) \\ & + 16{\kern 1pt} E\left[ {\tilde{\zeta }} \right]\mu {\kern 1pt} \lambda {\kern 1pt} {\kern 1pt} \psi \left( { - t_{1} - t_{2} + t_{1} t_{2} + 1} \right) > 0 \\ \end{aligned} $$