An allocation game model with reciprocal behavior and its applications in supply chain pricing decisions

Abstract

Commonly, people are much nicer in response to friendly actions and much nastier and even brutal in response to hostile actions. In social psychology, such a phenomenon is called reciprocity. As an extension of the standard Stackelberg game, we propose a new allocation game framework with consideration of such reciprocal behavior. Specifically, we consider a situation in which the follower evaluates the leaders intention based on the leaders action and then may take either a positive or negative reciprocal action. We also apply the new framework in a supply chain pricing problem to investigate the impact of the retailer’s reciprocal behavior on pricing decisions, obtaining the following interesting results. First, the supplier should take into account the retailers personality and offer a wholesale price based on it. Second, while the retailer can benefit from his reciprocal behavior, the supplier suffers in most cases. Finally, the retailers reciprocal behavior can help alleviate the double marginalization effect, and thus lead to a performance improvement in the whole supply chain.

Appendix

1.1 Proof of Lemma 1

To maximum the retailer’s profit, we get the first and second derivatives with respect to retail price p about Eq. (17):

$$\begin{aligned} \frac{\partial \pi _{R}}{\partial p}=-2p+w+a, \end{aligned}$$

(31)

and

$$\begin{aligned} \frac{\partial ^{2}\pi _{R}}{\partial p^{2}}=-2<0. \end{aligned}$$

(32)

From Eq. (32) we know that $\pi _{R}$ gets its maximum when $\partial \pi _{R}/\partial p =0$. Letting Eq. (31) equal to 0 we have

$$\begin{aligned} \tilde{p}(w)=\frac{a+w}{2}. \end{aligned}$$

Substituting $\tilde{p}(w)$ into Eqs. (16) and (17), we obtain Eqs. (19) and (20) in Lemma 1.

1.2 Proof of Lemma 2

Given the retailer’s best response shown in Lemma 1, we have the supplier’s profit function in Eq. (20).

Then, the first and second derivatives of $\pi _S$ with respect to wholesale price w are

$$\begin{aligned} \frac{\partial \pi _{S}}{\partial w}=-w+\frac{a}{2} \end{aligned}$$

(33)

and

$$\begin{aligned} \frac{\partial ^{2}\pi _{S}}{\partial w^{2}}=-1, \end{aligned}$$

(34)

respectively.

Since $\frac{\partial ^{2}\pi _{S}}{\partial w^{2}}<0$ we know that $\pi _S$ gets its maximum value at $\frac{\partial \pi _{S}}{\partial w}=0$, i.e.,

$$\begin{aligned} w^*=\frac{a}{2}. \end{aligned}$$

Substituting the above equation into Eqs. (18)–(20), we obtain Eqs. (22) and (23) in Lemma 2.

1.3 Proof of Proposition 3

The reciprocal retailer is to maximize his utility given by Eq. (25).

The first and second derivatives of Eq. (25) with respect to p are

$$\begin{aligned} \frac{\partial U_{R}}{\partial p}=-2(1+\alpha )p+(1+\alpha )(a+w)-\gamma w, \end{aligned}$$

(35)

and

$$\begin{aligned} \frac{\partial ^{2} U_{R}}{\partial p^{2}}=-2(1+\alpha )<0. \end{aligned}$$

(36)

Then, letting Eq. (35) equal to 0 we get the optimal response of the retailer, i.e.,

$$\begin{aligned} p^{r}(w)=\frac{a+w}{2}-\frac{\gamma w}{2(1+\alpha )}. \end{aligned}$$

(37)

(I) When $0\le w\le \frac{a}{1+2\beta _{2}}$, Eq. (37)$<a$, so we just need to compare $p^{r}(w)$ with w. Thus, we have:

(1A) while $\gamma _{2}\ge 2\beta _{2}(1+\alpha )$, and $w\le \frac{a(1+\alpha )}{1+\alpha +\gamma _{2}}$, $p^{r}(w)\ge w$, $U_{R}$ takes its maximum at $p^{r}(w)$;

(1B) while $\gamma _{2}<2\beta _{2}(1+\alpha )$, and $w<\frac{a}{1+2\beta _{2}}$, $p^{r}(w)>w$, $U_{R}$ takes its maximum at $p^{r}(w)$;

(1C) while $\gamma _{2}>2\beta _{2}(1+\alpha )$, and $w>\frac{a(1+\alpha )}{1+\alpha +\gamma _{2}}$, $p^{r}(w)<w$, $U_{R}$ takes its maximum at w.

(II) When $\frac{a}{1+2\beta _{2}}<w<\frac{a}{1+2\beta _{1}}$, $w<p^{r}(w)<a$, $U_{R}$ takes its maximum at $p^{r}(w)$.

(III) When $\frac{a}{1+2\beta _{1}}\le w\le a$, Eq. (37)$>w$, so we just need to compare $p^{r}(w)$ with a. Thus, we have:

(3A) while $\gamma _{1}\le 2\beta _1(1+\alpha )$, and $\frac{a}{1+2\beta _{1}}\le w\le \frac{a(1+\alpha )}{1+\alpha +\gamma _{1}}$, $w<p^{r}(w)<a$, $U_{R}$ takes its maximum at $p^{r}(w)$;

(3B) while $\gamma _{1}<2\beta _1(1+\alpha )$, and $\frac{a(1+\alpha )}{1+\alpha +\gamma _{1}}<w<a$,$w<p^{r}(w)>a$, $U_{R}$ takes its maximum at a;

(3C) while $\gamma _{1}>2\beta _1(1+\alpha )$, and $\frac{a}{1+2\beta _1}<w<a$,$w<p^{r}(w)>a$, $U_{R}$ takes its maximum at a.

Substituting $p^{r}(w)$ to Eqs. (15), (16), (17) and (25) we obtain Eqs. (28)–(30) and Proposition 3 is proven.

1.4 Proof of Proposition 4

The supplier is to maximize his profit given by Eq. (30). Combine Proposition 3, we have:

(I) When $\gamma _{2}\le 2\beta _{2}(1+\alpha )$ and $0\le w\le \frac{a}{1+2\beta _{2}}$, we have $\pi _{S}^{r}(w)=w\left( \frac{a-w}{2}+\frac{\gamma _{2}w}{2(1+\alpha )}\right)$ in this case. The first and second derivatives of Eq. (30) with respect to w are

$$\begin{aligned} \frac{\partial \pi _{S}^{r}}{\partial w}=\frac{a}{2}-\frac{w(1+\alpha -\gamma _{2})}{1+\alpha }, \end{aligned}$$

(38)

and

$$\begin{aligned} \frac{\partial ^{2}\pi _{S}^{r}}{\partial w^{2}}=-\frac{(1+\alpha -\gamma _{2})}{1+\alpha }. \end{aligned}$$

(39)

Since $\gamma _{2}<\alpha$, we have $\partial ^{2}\pi _{S}^{r}/\partial w^{2}<0$, which means that $\pi _{S}^{r}$ is concave in $0\le w\le \frac{a}{1+2\beta _{2}}$.

Letting Eq. (38) = 0, we have

$$\begin{aligned} \bar{w}_{1}=\frac{a(1+\alpha )}{2(1+\alpha -\gamma _{2})}. \end{aligned}$$

(40)

From $\gamma _{2}<\alpha$ we know $\bar{w}_{1}>0$. Then we just need to compare $\bar{w}_{1}$ with the bound $\frac{a}{1+2\beta _{2}}$.

Thus we have: (i) if $\bar{w}_{1}<\frac{a}{1+2\beta _{2}}$, i.e., $\beta _{2}<\frac{1}{2}-\frac{\gamma _{2}}{1+\alpha }$, $\pi _{S}^{r}$ reaches its maximum value $\frac{(\alpha +1)a^{2}}{8(1+\alpha -\gamma _{2})}$ at $w=\bar{w}_{1}$ in $[0,\frac{a}{1+2\beta _{2}}]$; and (ii) if $\bar{w}_{1}\ge \frac{a}{1+2\beta _{2}}$, i.e., $\beta _{2}\ge \frac{1}{2}-\frac{\gamma _{2}}{1+\alpha }$, $\pi _{S}^{r}$ reaches its maximum value $\frac{a^{2}}{2(1+2\beta _{2})^{2}}(2\beta _{2}+\frac{\gamma _{2}}{1+\alpha })$ at $w=\frac{a}{1+2\beta _{2}}$ in $[0,\frac{a}{1+2\beta _{2}}]$.

(II) When $\gamma _{2}>2\beta _{2}(1+\alpha )$ and $0\le w\le \frac{a(1+\alpha )}{1+\alpha +\gamma _{2}}$, we have $\pi _{S}^{r}(w)=w\left( \frac{a-w}{2}+\frac{\gamma _{2}w}{2(1+\alpha )}\right)$. Similar to case I, we need to compare $\bar{w}_{1}$ with the bound $\frac{a(1+\alpha )}{1+\alpha +\gamma _{2}}$.

Thus, we have: (i) if $\bar{w}_{1}<\frac{a(1+\alpha )}{1+\alpha +\gamma _{2}}$, i.e., $\gamma _{2}<\frac{1+\alpha }{3}$, $\pi _{S}^{r}$ reaches its maximum value $\frac{(\alpha +1)a^{2}}{8(1+\alpha -\gamma _{2})}$ at $w=\bar{w}_{1}$ in $[0,\frac{a(1+\alpha )}{1+\alpha +\gamma _{2}}]$; and (ii) if $\bar{w}_{1}\ge \frac{a}{1+2\beta _{2}}$, i.e., $\gamma _{2}\ge \frac{1+\alpha }{3}$, $\pi _{S}^{r}$ reaches its maximum value $\frac{(\alpha +1)\gamma _{2}a^{2}}{(1+\alpha +\gamma _{2})^{2}}$ at $w=\frac{a(1+\alpha )}{1+\alpha +\gamma _{2}}$in $[0,\frac{a(1+\alpha )}{1+\alpha +\gamma _{2}}]$.

(III) When $\gamma _{2}>2\beta _{2}(1+\alpha )$ and $\frac{a(1+\alpha )}{1+\alpha +\gamma _{2}}<w<\frac{a}{1+2\beta _{2}}$, we have $\pi _{S}^{r}(w)=w(a-w)$, $\pi _{S}^{r}$ reaches its maximum value $\frac{(\alpha +1)\gamma _{2}a^{2}}{(1+\alpha +\gamma _{2})^{2}}$ at $w=\frac{a(1+\alpha )}{1+\alpha +\gamma _{2}}$ in$(\frac{a(1+\alpha )}{1+\alpha +\gamma _{2}},\frac{a}{1+2\beta _{2}})$.

(IV) When $\frac{a}{1+2\beta _{2}}<w<\frac{a}{1+2\beta _{1}}$, we have $\pi _{S}^{r}(w)=w\frac{(a-w)}{2}$ in this case. The first and second derivatives of Eq. (30) with respect to w are

$$\begin{aligned} \frac{\partial \pi _{S}^{r}}{\partial w}=\frac{a}{2}-w, \end{aligned}$$

(41)

and

$$\begin{aligned} \frac{\partial ^{2}\pi _{S}^{r}}{\partial w^{2}}=-1. \end{aligned}$$

(42)

Since $\partial ^{2}\pi _{S}^{r}/\partial w^{2}<0$, we let Eq. (41) = 0 and

$$\begin{aligned} \bar{w}_{2}=\frac{a}{2}. \end{aligned}$$

(43)

Thus, we have: (i) if $\bar{w}_{2}<\frac{a}{1+2\beta _{2}}$, i.e., $\beta _{1}<\beta _{2}<\frac{1}{2}$, $\pi _{S}^{r}$ reaches its maximum value $\frac{a^{2}\beta _{2}}{2(1+2\beta _{2})^{2}}$ at $w=\frac{a}{1+2\beta _{2}}$ in $(\frac{a}{1+2\beta _{2}},\frac{a}{1+2\beta _{1}})$; and (ii) if $\frac{a}{1+2\beta _{2}}<\bar{w}_{2}<\frac{a}{1+2\beta _{1}}$, i.e., $\beta _{1}<\frac{1}{2}<\beta _{2}$, $\pi _{S}^{r}$ reaches its maximum value $\frac{a^{2}}{8}$ at $w=\bar{w}_{2}$; (iii) if $\bar{w}_{2}>\frac{a}{1+2\beta _{1}}$ , i.e., $\beta _{2}>\beta _{1}>\frac{1}{2}$, $\pi _{S}^{r}$ reaches its maximum value $\frac{a^{2}\beta _{1}}{2(1+2\beta _{1})^{2}}$ at $w=\frac{a}{1+2\beta _{1}}$ in $(\frac{a}{1+2\beta _{2}},\frac{a}{1+2\beta _{1}})$.

(IV) When $\gamma _{1}\le 2\beta _{1}(1+\alpha )$ and $\frac{a}{1+2\beta _{1}}\le w\le \frac{a(1+\alpha )}{1+\alpha +\gamma _{1}}$, we have$\pi _{S}^{r}(w)=w\left( \frac{a-w}{2}-\frac{\gamma _{1}w}{2(1+\alpha )}\right)$ in this case. The first and second derivatives of Eq. (30) with respect to w are

$$\begin{aligned} \frac{\partial \pi _{S}^{r}}{\partial w}=\frac{a}{2}-\frac{w(1+\alpha +\gamma _{1})}{1+\alpha }, \end{aligned}$$

(44)

and

$$\begin{aligned} \frac{\partial ^{2}\pi _{S}^{r}}{\partial w^{2}}=-\frac{(1+\alpha +\gamma _{1})}{1+\alpha }. \end{aligned}$$

(45)

We obtain $\partial ^{2}\pi _{S}^{r}/\partial w^{2}<0$ and let Eq. (44) = 0 and

$$\begin{aligned} \bar{w}_{3}=\frac{a(1+\alpha )}{2(1+\alpha +\gamma _{1})}. \end{aligned}$$

(46)

We can know $\pi _{S}^{r}$ reaches its maximum value $\frac{(\alpha +1)a^{2}}{8(1+\alpha +\gamma _{1})}$ at $w=\bar{w}_{3}$ in $[\frac{a}{1+2\beta _{1}},\frac{a(1+\alpha )}{1+\alpha +\gamma _{1}}]$.

(VII) When $\gamma _{1}\le 2\beta _{1}(1+\alpha )$ and $\frac{a(1+\alpha )}{1+\alpha +\gamma _{1}}<w<a$, we have $\pi _{S}^{r}(w)=0$.

(VIII) When $\gamma _{1}>2\beta _{1}(1+\alpha )$ and $\frac{a}{1+2\beta _{1}}<w<a$, we have $\pi _{S}^{r}(w)=0$.

Therefore, the supplier compares his profits in (I), (II), (III), (IV), (VI), (VII), (VIII) to determine the globally optimal wholesale price.

(1) When $\beta _{1}<\beta _{2}<\frac{\gamma _{2}}{2(1+\alpha )}$and $\gamma _{2}\le \frac{1+\alpha }{3}$, we obtain that $\pi _{S}^{r}$ takes its maximum value at $w=\bar{w}_{1}$;

(2) When $\beta _{1}<\beta _{2}<\frac{\gamma _{2}}{2(1+\alpha )}$and $\gamma _{2}>\frac{1+\alpha }{3}$, $\pi _{S}^{r}$ takes its maximum value at $w=\frac{a(1+\alpha )}{1+\alpha +\gamma _{2}}$;

(3) When $\frac{\gamma _{2}}{2(1+\alpha )}<\beta _{2}<\frac{1}{2}-\frac{\gamma _{2}}{1+\alpha }$ and $\gamma _{2}\le \frac{1+\alpha }{3}$, $\pi _{S}^{r}$ takes its maximum value at $w=\bar{w}_{1}$;

(4) When $\frac{1}{2}-\frac{\gamma _{2}}{1+\alpha }\le \beta _{2}<\frac{1}{2}$, $\pi _{S}^{r}$ takes its maximum value at $w=\frac{a}{1+2\beta _{2}}$;

(5) When $\beta _{1}<\frac{1}{2}<\beta _{2}$, if $\beta _{1}<\frac{1}{2}<\beta _{2}\le A$, $\pi _{S}^{r}$ takes its maximum value at $w=\frac{a}{1+2\beta _{2}}$; if $\beta _{1}<\frac{1}{2}<A<\beta _{2}$, $\pi _{S}^{r}$ takes its maximum value at $w=\bar{w}_{1}$, where $A=\frac{1+\alpha +2\sqrt{\gamma _{2}(1+\alpha )}}{2(1+\alpha )}$;

(6) When $\frac{1}{2}<\beta _{1}\le \frac{1}{2}+\frac{\gamma _{1}}{1+\alpha }$, if $\beta _{1}<\beta _{2}<B$, $\pi _{S}^{r}$ takes its maximum value at $w=\frac{a}{1+2\beta _{2}}$; if $\beta _{2}>B$, $\pi _{S}^{r}$ takes its maximum value at $w=\frac{a}{1+2\beta _{1}}$, where $B=\frac{1+4\beta _{1}^{2}+(1+2\beta _{1})\sqrt{(1-2\beta _{1})^{2}+8\beta _{1}\frac{\gamma _{2}}{1+\alpha }}}{8\beta _{1}}$;

(7) When $\frac{1}{2}+\frac{\gamma _{1}}{1+\alpha }<\beta _{1}<\beta _{2}$, if $\frac{1}{2}+\frac{\gamma _{1}+\sqrt{\gamma _{1}}}{1+\alpha }<\beta _{1}<\beta _{2}<C$, $\pi _{S}^{r}$ takes its maximum value at $w=\frac{a}{1+2\beta _{2}}$; if $\beta _{2}>C$, $\pi _{S}^{r}$ takes its maximum value at $w=\bar{w}_{3}$; if $\beta _{1}<\beta _{2}<B$, $\pi _{S}^{r}$ takes its maximum value at $w=\frac{a}{1+2\beta _{2}}$; if $\beta _{2}>B$, $\pi _{S}^{r}$ takes its maximum value at $w=\frac{a}{1+2\beta _{1}}$, where $C=\frac{1}{2}+\frac{\gamma _{2}+\sqrt{(\gamma _{1}+1+\alpha )(\gamma _{1}+\gamma _{2})}}{(1+\alpha )}$.