Supply chain pricing and effort decisions with the participants’ belief under the uncertain demand

Abstract

This paper addresses the pricing and effort decisions of a supply chain consisting of a manufacturer and a retailer. All the parties make optimal decisions to maximize their profits with uncertainty of demand under their confidence levels. Taking this into account, Stackelberg models are formulated to study the impact of the confidence levels on pricing and effort decisions for the decentralized and centralized supply chains. We obtain that the confidence levels of participants have a significant impact on the pricing and effort decisions. Specifically, when the retailer’s confidence level is increasing, the retail price, the wholesales price, the sales effort, the profit of each member and the total profit of supply chain are all increasing. However, the manufacturer’s confidence level is not independent of the power structure, i.e., there are different characteristics under the different power structures. The power structure has an outstanding effect on the profit of each member in the supply chain. The leader’s profit is always more than that of the follower, and the profit of upstream is more than that of the downstream when they have the same power. We use numerical experiments to verify the validity of the model.

Appendix

Proof of Proposition 1

Since ${\mathcal {M}}\{T(p,e)\ge T_0\}$ is strictly decreasing with $T_0$ the maximum profit $T_m(p,e,\gamma )$ of the consolidated company’s profit at an acceptable confidence level $\gamma$ is

$$\begin{aligned} {\mathcal {M}}\{T(p,e)\ge T_m(p,e,\gamma )\}=\gamma . \end{aligned}$$

(31)

According to uncertainty theory, we have

$$\begin{aligned} {\mathcal {M}}\{T(p,e)< T_m(p,e,\gamma )\}=1-\gamma . \end{aligned}$$

(32)

Because the equation ${\mathcal {M}}\{T(p,e) = T_m(p,e,\gamma )\}=0$ always hold, we get

$$\begin{aligned} {\mathcal {M}}\{T(p,e)\le T_m(p,e,\gamma )\}=1-\gamma . \end{aligned}$$

(33)

Therefore, model (9) is equivalently transformed into the planning problem given the confidence level $\gamma$ below

$$\begin{aligned} {\left\{ \begin{array}{ll} \max \ \{(p-\Phi ^{-1}(1-\gamma ))(a-p+\Psi ^{-1}(\gamma )e)\\ -\frac{(1-\Psi ^{-1}(\gamma ))^2}{2} e^2\}\\ {\mathrm{s. t.}} {\left\{ \begin{array}{ll} a-p+\Psi ^{-1}(\gamma )e>0\\ p-\Psi ^{-1}(\gamma )\ge 0. \end{array}\right. } \end{array}\right. } \end{aligned}$$

(34)

Using KT conditions to solve the above model, we get

$$\begin{aligned} e^*= & {} \frac{\Psi ^{-1}(\gamma )(a-\Phi ^{-1}(1-\gamma ))}{2+(\Psi ^{-1}(\gamma ))^2-4\Psi ^{-1}(\gamma )}, \end{aligned}$$

(35)

$$\begin{aligned} p^*= & {} \frac{(a+\Phi ^{-1}(1-\gamma ))(1-2\Psi ^{-1}(\gamma ))+a(\Psi ^{-1}(\gamma ))^2}{2+(\Psi ^{-1}(\gamma ))^2-4\Psi ^{-1}(\gamma )}.\nonumber \\ \end{aligned}$$

(36)

The maximum profit of the integrated enterprise is

$$\begin{aligned} \begin{aligned} T_m^*=\frac{(a-\Phi ^{-1}(1-\gamma ))^2(1-\Psi ^{-1}(\gamma ))^2)}{2(2+(\Psi ^{-1}(\gamma ))^2-4\Psi ^{-1}(\gamma ))}. \end{aligned} \end{aligned}$$

(37)

Proof of Proposition 2

It should be noted that when the manufacturer’s bargaining power is greater than that of the retailer, according to the Stackelberg model theory, the manufacturer is the leader and the retailer is the follower. The leader considers the follower’s maximum sales price and sales effort as given to maximize their own profits.

Since ${\mathcal {M}}\{u(w,p,e)\ge u_0\}$ is strictly decreasing, the maximum profit of the manufacturer at an acceptable confidence level $\alpha$ is

$$\begin{aligned} \begin{aligned} {\mathcal {M}}\{u(w,p,e)\ge u_m(p,e,\alpha )\}=\alpha . \end{aligned} \end{aligned}$$

(38)

According to uncertainty theory and the equation $M\{u(w,p,e)= u_m(p,e,\alpha )\}=0$, we can get

$$\begin{aligned} \begin{aligned} {\mathcal {M}}\{u(w,p,e)\le u_m(p,e,\alpha )\}=1-\alpha . \end{aligned} \end{aligned}$$

(39)

Similarly, the retailer’s maximum profit at an acceptable confidence level $\beta$ is

$$\begin{aligned} {\mathcal {M}}\{v(w,p,e)\le v_m(p,e,\beta )\}=1-\beta . \end{aligned}$$

(40)

Then, the MS model can be equivalently transformed into the following planning problems with given confidence levels

$$\begin{aligned} {\left\{ \begin{array}{ll} \max \ \{(w-\Psi ^{-1}(1-\alpha ))(a-p^*_u+\Psi ^{-1}(\alpha )e^*_u)\}\\ {\mathrm{s. t.}} {\left\{ \begin{array}{ll} \max \ \{(p-w)(a-p+\Psi ^{-1}(\beta )e)-\frac{(1-\Psi ^{-1}(\beta ))^2}{2}e^2\}\\ {\mathrm{s. t.}} {\left\{ \begin{array}{ll} a-p+\Psi ^{-1}(\beta )e>0\\ p-w>0. \end{array}\right. } \end{array}\right. } \end{array}\right. } \end{aligned}$$

(41)

Using KT conditions to solve the above model, we have

$$\begin{aligned} e_u= & {} \frac{(a-w)\Psi ^{-1}(\beta )}{2+(\Psi ^{-1}(\beta ))^2-4\Psi ^{-1}(\beta )}, \end{aligned}$$

(42)

$$\begin{aligned} p_u= & {} \frac{w+a-2a\Psi ^{-1}(\beta )-2w\Psi ^{-1}(\beta )+a(\Psi ^{-1}(\beta ))^2}{2+(\Psi ^{-1}(\beta ))^2-4\Psi ^{-1}(\beta )}.\nonumber \\ \end{aligned}$$

(43)

The unit retail price p and the sales effort e are linear functions of the unit wholesale price w. When the retailer’s reaction function is known, the manufacturer will maximize his profit by selecting the unit wholesale price w by calculating the reaction function.

Let $u(w,p^*_u,e^*_u)$ represent the manufacturer’s profit, then we have the following conclusions:

• (1) $u(w,p^*_u,e^*_u)$ is a concave function of w.

• (2) The optimal wholesale price of the manufacturer is

  $$\begin{aligned} w^*_u=\frac{a+\Phi ^{-1}(1-\alpha )}{2}. \end{aligned}$$

  (44)

  Bring $w^*_u$ into $e_u$ and $p_u$, we get

  $$\begin{aligned} e^*_u= & {} \frac{(a-\Phi ^{-1}(1-\alpha ))\Psi ^{-1}(\beta )}{2(2+(\Psi ^{-1}(\beta ))^2-4\Psi ^{-1}(\beta ))}, \end{aligned}$$

  (45)
  $$\begin{aligned} p^*_u= & {} \frac{(3a+\Phi ^{-1}(1-\alpha ))(1-2\Psi ^{-1}(\beta ))+2a(\Psi ^{-1}(\beta ))^2}{2(2+(\Psi ^{-1}(\beta ))^2-4\Psi ^{-1}(\beta )}.\nonumber \\ \end{aligned}$$

  (46)

• (3) The maximum profits of the manufacturer and the retailer are

  $$\begin{aligned} u_m^*= & {} w^*_ua-w^*_up^*_u+\Psi ^{-1}(\alpha )w^*_ue^*_u-a\Phi ^{-1}(1-\alpha )\nonumber \\&+p^*_u\Phi ^{-1}(1-\alpha )-e^*_u\Phi ^{-1}(1-\alpha )\Psi ^{-1}(\alpha ) \nonumber \\= & {} \frac{(a-\Phi ^{-1}(1-\alpha ))^2(1-2\Psi ^{-1}(\beta )+\Psi ^{-1}(\beta )\Psi ^{-1}(\alpha ))}{4(2+(\Psi ^{-1}(\beta ))^2-4\Psi ^{-1}(\beta ))},\nonumber \\ \end{aligned}$$

  (47)
  $$\begin{aligned} v_m^*= & {} p^*_ua-(p^*_u)^2 +\Psi ^{-1}(\beta )p^*_ue^*_u-w^*_ua \nonumber \\&+w^*_up^*_u-w^*_ue^*_u\Psi ^{-1}(\beta )-\frac{1}{2}(e^*_u)^2 \nonumber \\&-\frac{1}{2}(e^*_u)^2(\Psi ^{-1}(\beta )^2+(e^*_u)^2\Psi ^{-1}(\beta ) \nonumber \\= & {} \frac{(a-\Phi ^{-1}(1-\alpha ))^2(1-\Psi ^{-1}(\beta ))^2}{8(2+(\Psi ^{-1}(\beta ))^2-4\Psi ^{-1}(\beta ))}. \end{aligned}$$

  (48)

Proof of Proposition 3

Assuming that the manufacturer (follower) has observed the decision of the retailer (leader), the RS model is established as follows:

$$\begin{aligned} {\left\{ \begin{array}{ll} \max \limits _{(p,e)} \Big \{v_0|{\mathcal {M}}\{v(w,p,e)\ge v_0\}\ge \beta \Big \} \\ {\mathrm{s. t.}} {\left\{ \begin{array}{ll} \max \limits _{w} \Big \{u_0|{\mathcal {M}}\{u(w,p,e)\ge u_0\}\ge \alpha \Big \} \\ {\mathrm{s. t.}} {\left\{ \begin{array}{ll} {\mathcal {M}}\{a-p+be>0\}\\ p>w. \end{array}\right. } \end{array}\right. } \end{array}\right. } \end{aligned}$$

(49)

Then the RS model can be equivalently transformed into the following planning problems with given confidence levels:

$$\begin{aligned} {\left\{ \begin{array}{ll} \max \limits _{(p,e)} \{(p-w)(a-p+\Psi ^{-1}(\beta )e)-\frac{(1-\Psi ^{-1}(\beta ))^2}{2}e^2\}\\ {\mathrm{s. t.}} {\left\{ \begin{array}{ll} \max \limits _{w} \{(w-\Psi ^{-1}(1-\alpha ))(a-p^*_u+\Psi ^{-1}(\alpha )e^*_u)\}\\ {\mathrm{s. t.}} {\left\{ \begin{array}{ll} a-p+\Psi ^{-1}(\beta )e>0\\ p-w>0. \end{array}\right. } \end{array}\right. } \end{array}\right. } \end{aligned}$$

(50)

We obtain the manufacturer’s optimal wholesale price is

$$\begin{aligned} w^*_v=a-p+\Phi ^{-1}(1-\alpha )+\Psi ^{-1}(\alpha )e. \end{aligned}$$

(51)

The above equation is the manufacturer’s response function and the retailer will use this information to maximize his profit. Therefore, the retailer’s profit is subject to the following rules:

• (1) $v(w^*_v,p,e)$ is a concave function of p and e.

• (2) $p^*_v$ and $e^*_v$, respectively, represent the retailer’s optimal retail price and the optimal sales effort as follows

  $$\begin{aligned}&p^*_v=\frac{3a+(2\Psi ^{-1}(\beta )+\Psi ^{-1}(\alpha ))e^*_v+\Phi ^{-1}(1-\alpha ))}{4} \nonumber \\&\quad =\frac{3a+\Phi ^{-1}(1-\alpha ))}{4}\nonumber \\&\qquad + \frac{(2\Psi ^{-1}(\beta )+\Psi ^{-1}(\alpha ))(a-\Phi ^{-1}(1-\alpha ))(2\Psi ^{-1}(\alpha )-\Psi ^{-1}(\beta ))}{4(2-\Psi ^{-1}(\alpha ))(2-4\Psi ^{-1}(\beta )+\Psi ^{-1}(\alpha ))},\nonumber \\ \end{aligned}$$

  (52)
  $$\begin{aligned}&e^*_v=\frac{(a-\Phi ^{-1}(1-\alpha ))(2\Psi ^{-1}(\alpha )-\Psi ^{-1}(\beta ))}{(2-\Psi ^{-1}(\alpha ))(2-4\Psi ^{-1}(\beta )+\Psi ^{-1}(\alpha ))}. \end{aligned}$$

  (53)

• (3) The maximum profits of the manufacturer and the retailer are

  $$\begin{aligned}&u^*_m=w^*_va-w^*_vp^*_v+\Psi ^{-1}(\alpha )w^*_ve^*_v +(p^*_v-a-e^*_v\Psi ^{-1}(\alpha ))\Phi ^{-1}(1-\alpha )\nonumber \\&\quad =\Bigg [\frac{a-\Phi ^{-1}(1-\alpha )}{4}-\frac{\Psi ^{-1}(\alpha )(a-\Phi ^{-1}(1-\alpha ))(2\Psi ^{-1}(\alpha )-\Psi ^{-1}(\beta )}{(\Psi ^{-1}(\alpha )-2)(\Psi ^{-1}(\alpha )-4\Psi ^{-1}(\beta )+2)} \nonumber \\&\qquad +\frac{(a-\Phi ^{-1}(1-\alpha ))(\Psi ^{-1}(\alpha )+2\Psi ^{-1}(\beta ))(2\Psi ^{-1}(\alpha )-\Psi ^{-1}(\beta ))}{4(\Psi ^{-1}(\alpha )-2)(\Psi ^{-1}(\alpha )-4\Psi ^{-1}(\beta )+2)}\Bigg ]^2,\nonumber \\ \end{aligned}$$

  (54)
  $$\begin{aligned}&v^*_r=p^*_va-(p^*_v)^2+\Psi ^{-1}(\beta )p^*_ve^*_v \nonumber \\&\qquad -w^*_va+w^*_vp^*_v-w^*_ve^*_v\Psi ^{-1}(\beta )-\frac{1}{2}(e^*_v)^2-\frac{1}{2}(e^*_v)^2(\Psi ^{-1}(\beta ))^2 \nonumber \\&\qquad +(e^*_v)^2\Psi ^{-1}(\beta ) = \Bigg [\frac{a-\Phi ^{-1}(1-\alpha )}{2} \nonumber \\&\qquad +\frac{\Psi ^{-1}(\alpha )(a-\Phi ^{-1}(1-\alpha ))(2\Psi ^{-1}(\alpha )-\Psi ^{-1}(\beta )}{(\Psi ^{-1}(\alpha )-2)(\Psi ^{-1}(\alpha )-4\Psi ^{-1}(\beta )+2)} \nonumber \\&\qquad -\frac{(a-\Phi ^{-1}(1-\alpha ))(\Psi ^{-1}(\alpha )+2\Psi ^{-1}(\beta ))(2\Psi ^{-1}(\alpha )-\Psi ^{-1}(\beta ))}{2(\Psi ^{-1}(\alpha )-2)(\Psi ^{-1}(\alpha )-4\Psi ^{-1}(\beta )+2)}\Bigg ]\nonumber \\&\qquad *\Bigg [\frac{a-\Phi ^{-1}(1-\alpha )}{4}\nonumber \\&\qquad +\frac{\Psi ^{-1}(\beta )(a-\Phi ^{-1}(1-\alpha ))(2\Psi ^{-1}(\alpha )-\Psi ^{-1}(\beta )}{(\Psi ^{-1}(\alpha )-2)(\Psi ^{-1}(\alpha )-4\Psi ^{-1}(\beta )+2)} \nonumber \\&\qquad +\frac{(a-\Phi ^{-1}(1-\alpha ))(\Psi ^{-1}(\alpha )+2\Psi ^{-1}(\beta ))(2\Psi ^{-1}(\alpha )-\Psi ^{-1}(\beta ))}{4(\Psi ^{-1}(\alpha )-2)(\Psi ^{-1}(\alpha )-4\Psi ^{-1}(\beta )+2)}\Bigg ] \nonumber \\&\qquad -\frac{(a-\Phi ^{-1}(1-\alpha ))^2(\Psi ^{-1}(\beta )-1)^2(2\Psi ^{-1}(\alpha )-\Psi ^{-1}(\beta ))^2}{2(\Psi ^{-1}(\alpha )-2)^2(\Psi ^{-1}(\alpha )-4\Psi ^{-1}(\beta )+2)^2}. \end{aligned}$$

  (55)

Proof of Proposition 4

The VN game model is established as follows

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}\max \limits _{w} \{u_0|{\mathcal {M}}\{u(w,p,e)\ge u_0\} \ge \alpha \},\\ &{}\max \limits _{(p,e)} \{v_0|{\mathcal {M}}\{v(w,p,e)\ge v_0\} \ge \beta \}.\\ \end{array}\right. } \end{aligned}$$

(56)

Then the VN model can be equivalently transformed into the following planning problems with given confidence levels:

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}\max \limits _{w} (w-\Psi ^{-1}(1-\alpha ))(a-p^*_u+\Psi ^{-1}(\alpha )e^*_u),\\ &{}\max \limits _{(p,e)} (p-w)(a-p+\Psi ^{-1}(\beta )e)-\frac{(1-\Psi ^{-1}(\beta ))^2}{2}e^2.\\ \end{array}\right. } \end{aligned}$$

(57)

In the VN model, we note that manufacturer’s bargaining power is equal to the retailer’s. Therefore, they will make decision independently. The retailer decides the retail price and the sales effort that are similar to the solve of MS model, and the manufacturer decides the wholesale price that is similar to RS model. And we have the following equations

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}e=\frac{(a-w)\Psi ^{-1}(\beta )}{2+(\Psi ^{-1}(\beta ))^2-4\Psi ^{-1}(\beta )},\\ &{}p=\frac{w+a-2a\Psi ^{-1}(\beta )-2w\Psi ^{-1}(\beta )+a(\Psi ^{-1}(\beta ))^2}{2+(\Psi ^{-1}(\beta ))^2-4\Psi ^{-1}(\beta )}.\\ &{}w=a-p+\Phi ^{-1}(1-\alpha )+\Psi ^{-1}(\alpha )e.\\ \end{array}\right. }\nonumber \\ \end{aligned}$$

(58)

Finally, the three reaction functions jointly determine an equilibrium solution that is the solution of the model VN, which is listed as follows

$$\begin{aligned}&p^*_N = \frac{2a+\Phi ^{-1}(1-\alpha )}{3} \nonumber \\&\quad + \frac{\Psi ^{-1}(\beta )(\Psi ^{-1}(\beta )+\Psi ^{-1}(\alpha ))(a-\Phi ^{-1}(1-\alpha ))}{3(3-6\Psi ^{-1}(\beta )+(\Psi ^{-1}(\beta ))^2+\Psi ^{-1}(\alpha )\Psi ^{-1}(\beta ))},\nonumber \\ \end{aligned}$$

(59)

$$\begin{aligned}&e^*_N=\frac{\Psi ^{-1}(\beta )(a-\Phi ^{-1}(1-\alpha ))}{3-6\Psi ^{-1}(\beta )+(\Psi ^{-1}(\beta ))^2+\Psi ^{-1}(\alpha )\Psi ^{-1}(\beta )},\nonumber \\ \end{aligned}$$

(60)

$$\begin{aligned}&w^*_N =\frac{a+2\Phi ^{-1}(1-\alpha )}{3} \nonumber \\&\quad - \frac{\Psi ^{-1}(\beta )(2\Psi ^{-1}(\alpha )-\Psi ^{-1}(\beta ))(a-\Phi ^{-1}(1-\alpha ))}{3(3-6\Psi ^{-1}(\beta )+(\Psi ^{-1}(\beta ))^2+\Psi ^{-1}(\alpha )\Psi ^{-1}(\beta ))}.\nonumber \\ \end{aligned}$$

(61)

And in the VN model, the maximum profits of manufacturer and retailer are

$$\begin{aligned}&u^*_N=\Big [\frac{a-\Phi ^{-1}(1-\alpha )}{3} \nonumber \\&\quad +\frac{\Psi ^{-1}(\beta )(a-\Phi ^{-1}(1-\alpha ))(2\Psi ^{-1}(\alpha )-\Psi ^{-1}(\beta ))}{3(3-6\Psi ^{-1}(\beta )+(\Psi ^{-1}(\beta ))^2+\Psi ^{-1}(\alpha )\Psi ^{-1}(\beta ))}\Big ]^2,\nonumber \\ \end{aligned}$$

(62)

$$\begin{aligned}&v^*_N=\Big [\frac{a-\Phi ^{-1}(1-\alpha )}{3}\nonumber \\&\quad + \frac{\Psi ^{-1}(\beta )(2\Psi ^{-1}(\beta )-\Psi ^{-1}(\alpha ))(a-\Phi ^{-1}(1-\alpha ))}{3(3-6\Psi ^{-1}(\beta )+(\Psi ^{-1}(\beta ))^2+\Psi ^{-1}(\alpha )\Psi ^{-1}(\beta ))}\Big ]^2 \nonumber \\&\quad -\frac{(\Psi ^{-1}(\beta ))^2(a-\Phi ^{-1}(1-\alpha ))^2(\Psi ^{-1}(\beta )-1)^2}{2(3-6\Psi ^{-1}(\beta )+(\Psi ^{-1}(\beta ))^2+\Psi ^{-1}(\alpha )\Psi ^{-1}(\beta ))^2}.\nonumber \\ \end{aligned}$$

(63)