A feasible incentive contract between a manufacturer and his fairness-sensitive retailer engaged in strategic marketing efforts

Abstract

This paper considers a feasible incentive contract between a manufacturer and a fairness-sensitive retailer. The manufacturer (he) dominates the supply chain and determines his wholesale price, while the retailer (she) focuses on marketing and retailing. Thus, the retailer’s decision concerns her marketing efforts and retail price. The market demand is linked directly to the retailer’s marketing efforts and retail price. Therefore, we find that the retailer’s fairness preference leads her to engage in high-level marketing efforts, but causes the manufacturer to set a low wholesale price. A strategic decision-making approach that the manufacturer can employ is to consider the retailer’s fairness sensitivity. We also find that the retailer’s concern for fairness influences the manufacturer’s decisions strongly and affects his expected profit negatively. The manufacturer dominates the supply chain, thereby motivating him to design a feasible incentive contract to maximize his expected profit. Thus, a fairness-embedded profit-sharing contract applied with a Nash bargaining process is proposed to maximize the manufacturer’s expected profit. Both mathematical derivations and numerical studies show that the incentive contract leads to Pareto improvement in the utilities of the manufacturer and the retailer.

Appendix

Proof of Theorem 1

By taking the first derivative of $u_r^f$ with respect to p and $\theta$, we have

$$\begin{aligned} \frac{\partial u_r^f }{\partial p}= & {} \left( {1+\tau } \right) \left( {-2bp+a+\lambda \theta +bw} \right) +\tau \gamma b\left( {w-c} \right) \nonumber \\\end{aligned}$$

(15)

$$\begin{aligned} \frac{\partial u_r^f }{\partial \theta }= & {} \left( {1+\tau } \right) \left[ {\lambda \left( {p-w} \right) -\xi \theta } \right] -\tau \gamma \lambda \left( {w-c} \right) \end{aligned}$$

(16)

Therefore, the corresponding Hessian matrix is

$$\begin{aligned} H\left( {p,\theta } \right)= & {} \left[ \begin{array}{c@{\quad }c} {\frac{\partial ^{2}u_r^f }{\partial p^{2}}}&{} {\frac{\partial ^{2}u_r^f }{\partial p\partial \theta }} \\ {\frac{\partial ^{2}u_r^f }{\partial \theta \partial p}}&{} {\frac{\partial ^{2}u_r^f }{\partial \theta ^{2}}} \\ \end{array} \right] \\= & {} \left[ \begin{array}{c@{\quad }c} {-2b(1+\tau )}&{} {\lambda (1+\tau )} \\ {\lambda (1+\tau )}&{} {-\xi (1+\tau )} \\ \end{array} \right] \end{aligned}$$

The Hessian matrix of $u_r^f$ is a negative definite for all p and $\theta$ because $H\left( {p,\theta } \right)$ satisfies the conditions that $-2b(1+\tau )<0$ and $\left( {2b\xi -\lambda ^{2}} \right) (1+\tau )^{2}>0$. Let Eqs. (15) and (16) equal zero. We obtain

$$\begin{aligned} p^{f{*}}= & {} \frac{\left( {a\xi +bw\xi -\lambda ^{2}w} \right) \left( {1+\tau } \right) +\tau \gamma \left( {w-c} \right) \left( {b\xi -\lambda ^{2}} \right) }{\left( {2b\xi -\lambda ^{2}} \right) \left( {1+\tau } \right) }\nonumber \\\end{aligned}$$

(17)

$$\begin{aligned} \theta ^{f{*}}= & {} \frac{\lambda \left( {a-bw} \right) \left( {1+\tau } \right) -\lambda \tau \gamma b\left( {w-c} \right) }{\left( {2b\xi -\lambda ^{2}} \right) \left( {1+\tau } \right) } \end{aligned}$$

(18)

Introducing Eqs. (17) and (18) into $u_m^f$ and taking the first derivative of $u_m^f$ with respect to w, we obtain

$$\begin{aligned}&\frac{\partial u_m^f }{\partial w}\nonumber \\&\quad =\frac{-\,2b^{2}\left[ {\xi \left( {1+\tau } \right) +\xi \tau \gamma } \right] w+b\xi \left( {a+bc} \right) \left( {1+\tau } \right) +2\xi b^{2}\tau \gamma c}{\left( {2b\xi -\lambda ^{2}} \right) \left( {1+\tau } \right) }\nonumber \\ \end{aligned}$$

(19)

Let Eq. (19) equal zero. We then have

$$\begin{aligned} w^{f{*}}=\frac{\left( {a+bc} \right) \left( {1+\tau } \right) +2\tau \gamma bc}{2b\left[ {\left( {1+\tau } \right) +\tau \gamma } \right] } \end{aligned}$$

(20)

$\left. {\frac{\partial ^{2}u_m^f }{\partial w^{2}}} \right| _{w=w^{f{*}}} <0.$ Therefore, we conclude that $w^{f{*}}$ is the unique optimal solution for this problem. Introducing Eq. (20) into Eqs. (17) and (18), we obtain

$$\begin{aligned} p^{f{*}}=\frac{3ab\xi +b^{2}c\xi -a\lambda ^{2}-bc\lambda ^{2}}{2b\left( {2b\xi -\lambda ^{2}} \right) }, \quad \theta ^{f{*}}=\frac{\lambda \left( {a-bc} \right) }{2\left( {2b\xi -\lambda ^{2}} \right) } \end{aligned}$$

We introduce $w^{f{*}}$, $p^{f{*}}$, and $\theta ^{f{*}}$ into $u_r^f$ and $u_m^f$, and obtain the retailer’s and the manufacturer’s equilibrium utilities, respectively.

$$\begin{aligned} u_r^{f{*}}= & {} \frac{\xi \left( {a-bc} \right) ^{2}\left( {1+\tau } \right) }{8\left( {2b\xi -\lambda ^{2}} \right) }, \quad \\ u_m^{f{*}}= & {} \frac{\xi \left( {a-bc} \right) ^{2}\left( {1+\tau } \right) }{4\left( {2b\xi -\lambda ^{2}} \right) \left[ {\left( {1+\tau } \right) +\tau \gamma } \right] } \end{aligned}$$

Thus, the utility of the entire supply chain is

$$\begin{aligned} u_{sc}^{f{*}} =\frac{\xi \left( {a-bc} \right) ^{2}\left( {1+\tau } \right) \left( {3+\tau +\tau \gamma } \right) }{8\left( {2b\xi -\lambda ^{2}} \right) \left[ {\left( {1+\tau } \right) +\tau \gamma } \right] } \end{aligned}$$

$\square$

Proof of Corollary 1

On the basis of the functions of $w^{f{*}}$ with $w^{n{*}}$, we obtain $w^{f{*}}-w^{n{*}}=\frac{\left( {bc-a} \right) \tau \gamma }{2b\left[ {\left( {1+\tau } \right) +\tau \gamma } \right] }$. The marketing effort level $\theta$ is a non-negative value. Therefore, we have $a>bc$. Thus,$w^{f{*}}-w^{n{*}}<0$ and $\frac{\partial w^{f{*}}}{\partial \tau }=-\frac{\left( {a-bc} \right) \gamma }{2b\left( {1+\tau +\tau \gamma } \right) ^{2}}<0$. $\square$

Proof of Corollary 2

By observing the optimal marketing effort level and retail price in two target patterns, we have $\theta ^{f{*}}=\theta ^{n{*}}$ and $p^{f{*}}=p^{n{*}}$. Thus, we find that the retail prices and market effort level hold. $\square$

Proof of Corollary 3

On the basis of the solutions of $u_r^{f{*}}$ and $u_r^{n{*}}$, we find that $u_r^{f{*}} -u_{_r }^{n{*}} >0$ and

$$\begin{aligned} \frac{\partial u_r^{f{*}} }{\partial \tau }=\frac{\xi \left( {a-bc} \right) ^{2}}{8\left( {2b\xi -\lambda ^{2}} \right) }>0. \end{aligned}$$

$\square$

Proof of Corollary 4

On the basis of the solutions of $u_m^{f{*}}$ and $u_{_m }^{n{*}}$, we obtain $u_m^{f{*}} -u_{_m }^{n{*}} =-\frac{\xi \left( {a-bc} \right) ^{2}\tau \gamma }{4\left( {2b\xi -\lambda ^{2}} \right) \left[ {\left( {1+\tau } \right) +\tau \gamma } \right] }<0$ and $\frac{\partial u_m^{f{*}} }{\partial \tau }<0$, respectively. $\square$

Proof of Corollary 5

From the solutions of $u_{_{sc} }^{n{*}}$ and $u_{sc}^{f{*}}$, we obtain $u_{sc}^{f*} -u_{_{sc} }^{n*} =\frac{\xi \left( {a-bc} \right) ^{2}}{8\left( {2b\xi -\lambda ^{2}} \right) }\left[ {\frac{\tau ^{2}\left( {1+\gamma } \right) +\tau \left( {1-2\gamma } \right) }{1+\tau +\tau \gamma }} \right]$. For mathematical convenience, we set $f\left( \tau \right) =\tau ^{2}\left( {1+\gamma } \right) +\tau \left( {1-2\gamma } \right)$. We find $f\left( \tau \right)$ is the quadratic function with an unfairness aversion coefficient $\tau$. When $f\left( \tau \right) =0$, we obtain $\tau _1 =0$ and $\tau _2 =\frac{2\gamma -1}{1+\gamma }$.

Furthermore, we introduce $w^{f{*}}$, $p^{f{*}}$, $\theta ^{f{*}}$ into $\pi _r$ and $\pi _m$, and obtain the retailer’s and the manufacturer’s optimal profit, respectively.

$$\begin{aligned} \pi _r^{f*}= & {} \frac{\xi \left( {a-bc} \right) ^{2}\left( {1+\tau +3\tau \gamma } \right) }{8\left( {2b\xi -\lambda ^{2}} \right) \left[ {\left( {1+\tau } \right) +\tau \gamma } \right] }\end{aligned}$$

(21)

$$\begin{aligned} \pi _m^{f*}= & {} \frac{\xi \left( {a-bc} \right) ^{2}\left( {1+\tau } \right) }{4\left( {2b\xi -\lambda ^{2}} \right) \left[ {\left( {1+\tau } \right) +\tau \gamma } \right] } \end{aligned}$$

(22)

In this paper, we assume $0<\pi _r <\gamma \pi _m$. Hence, introducing Eqs. (21) and (22) into $0<\pi _r <\gamma \pi _m$, we obtain $\tau \in \left( {0,\frac{2\gamma -1}{1+\gamma }} \right)$. By a feature of the quadratic function $f\left( \tau \right)$, we obtain $u_{sc}^{f{*}} -u_{_{sc} }^{n{*}} <0$. $\square$

Proof of Corollary 6

Taking the first derivative of $u_r^o$ with respect to p and $\theta$, we have

$$\begin{aligned} \theta ^{o{*}}= & {} \frac{\lambda \left[ {\left( {a-bw} \right) \left[ {\left( {1+\tau } \right) -t\left( {1+\tau +\tau \gamma } \right) } \right] -b\tau \gamma \left( {w-c} \right) } \right] }{\left( {2b\xi -\lambda ^{2}} \right) \left[ {\left( {1+\tau } \right) -t\left( {1+\tau +\tau \gamma } \right) } \right] },\\ p^{o{*}}= & {} \frac{\left( {a\xi -\lambda ^{2}w} \right) \left[ {\left( {1+\tau } \right) -t\left( {1+\tau +\tau \gamma } \right) } \right] +bw\xi (1-t)\left( {1+\tau +\tau \gamma } \right) -\left[ {bc\xi +\lambda ^{2}\left( {w-c} \right) } \right] \tau \gamma }{\left( {2b\xi -\lambda ^{2}} \right) \left[ {\left( {1+\tau } \right) -t\left( {1+\tau +\tau \gamma } \right) } \right] } \end{aligned}$$

The Hessian matrix of $u_r^o$ is written as

$$\begin{aligned}&H\left( {p,\theta } \right) =\left[ {{\begin{array}{c@{\quad }c} {\frac{\partial ^{2}u_r^o }{\partial p^{2}}}&{} {\frac{\partial ^{2}u_r^o }{\partial p\partial \theta }} \\ {\frac{\partial ^{2}u_r^o }{\partial \theta \partial p}}&{} {\frac{\partial ^{2}u_r^o }{\partial \theta ^{2}}} \\ \end{array} }} \right] \nonumber \\&\quad =\left[ {{\begin{array}{c@{\quad }c} {2bt\left( {1+\tau +\tau \gamma } \right) -2b\left( {1+\tau } \right) }&{} {\lambda \left( {1+\tau } \right) -\lambda t\left( {1+\tau +\tau \gamma } \right) } \\ {\lambda \left( {1+\tau } \right) -\lambda t\left( {1+\tau +\tau \gamma } \right) }&{} {\xi t\left( {1+\tau +\tau \gamma } \right) -\xi \left( {1+\tau } \right) } \\ \end{array} }} \right] \end{aligned}$$

If the level of retailer emphasis on the manufacturer’s profit is not too high, i.e., $0<\gamma <\frac{\left( {1-t} \right) \left( {1+\tau } \right) }{t\tau }$, then we can conclude that $2bt\left( {1+\tau +\tau \gamma } \right) -2b\left( {1+\tau } \right) <0$ and $\left( {2b\xi -\lambda ^{2}} \right) \left[ {t\left( {1+\tau +\tau \gamma } \right) -\left( {1+\tau } \right) } \right] ^{2}>0$. Thus, the Hessian matrix of $u_r^o$ is a negative definite for all p and $\theta$. $\square$

Proof of Corollary 7

According to the condition $0\le t\le 1$, we need to prove $\left[ {\underline{t},\overline{t} } \right] \subset \left[ {0,1} \right]$ to illustrate the effectiveness of the proposed incentive contract, which means that we need to ensure that the following three inequalities are always satisfied:

$$\begin{aligned} \underline{t}>0, \quad \overline{t} \le 1, \quad \underline{t}<\overline{t} \end{aligned}$$

First, we can easily prove $\frac{2\left( {1+\tau } \right) k^{2}}{1+\tau +\tau \gamma }>0$. Thus, we have $\underline{t}>0$. In addition, because of $8k<8k+1$, we obtain $\sqrt{\left( {2k+k^{2}} \right) \left( {2k+k^{2}+1} \right) }-\left( {2k+k^{2}} \right) \le \frac{1}{2}$. Thus, we have $\overline{t} \le 1$. Similarly, we obtain $3k^{3}+4k^{2}<2+k$ given that $0\le k\le \frac{1}{2}$. Therefore, we further have $k^{2}<\sqrt{\left( {2k+k^{2}} \right) \left( {2k+k^{2}+1} \right) }-\left( {2k+k^{2}} \right)$. On the basis of the above analysis, we obtain $\underline{t}<\overline{t}$. $\square$