Mean-risk analysis of wholesale price contracts with stochastic price-dependent demand

Abstract

In supply chain management, it is prevalent to design contract for coordination or proper risk-sharing in the supply chain. However, when a supply chain contract is developed based on the concept of expectation (e.g., expected profit), there is uncertainty risk with respect to the contract value which arises from various uncertainties inherent in the supply chain, such as demand uncertainty, price uncertainty, etc. We call such uncertainty risk associated with the contract value risk since it relates to the true value of the contract. Value risk is obviously an important factor in the design and analysis of a supply chain contract. In addition, individual supply chain agents with different risk preferences will have different risk attitudes towards the contract value risk, which affects their decisions under the contract. Motivated by the above, we conduct in this paper a mean-risk analysis for the commonly adopted wholesale price contracts with a supplier–retailer supply chain facing a stochastic price-dependent downward-sloping demand curve, taking into account contract value risk and the retailer’s degree of risk-aversion. Formulating the problem under study as a supplier-led Stackelberg game, we characterize the wholesale price contract model analytically in terms of only the retailer’s risk preference structure, and derive all the results in closed-form. This study makes the first attempt to assess the efficiency of wholesale price contracts incorporating contract value risk, and thereby some interesting managerial and academic insights are obtained. Our research provides a new perspective of looking at the performance of a supply chain contract.

Appendices

Appendix 1: Proofs of the main results

Proof of Lemma 1

We first examine the retailer’s optimal strategy at time 1, given its order quantity $Q$ at time 0. When the realized demand curve at time 1 is $p_H=a_H-\delta q$, we formulate the retailer’s problem as $\max \limits _{0\le q\le Q}q(a_H-\delta q)$. Solving this problem, we obtain the solution denoted by $\overline{q}_{wH}$ as follows:

$$\begin{aligned} \overline{q}_{wH}=\left\{ \!\! \begin{array}{llll} \frac{a_H}{2\delta }, \ \ \mathrm{{if}}\ \frac{a_H}{2\delta }\le Q;\\ Q,\ \ \ \mathrm{{otherwise}}. \end{array} \right. \end{aligned}$$

(33)

Likewise, when the realization of the demand curve at time 1 is $p_L=a_L-\delta q$, we formulate the retailer’s problem as $\max \limits _{0\le q\le Q}q(a_L-\delta q)$. Solving this problem, we obtain the solution denoted by $\overline{q}_{wL}$ as follows:

$$\begin{aligned} \overline{q}_{wL}=\left\{ \!\! \begin{array}{llll} \frac{a_L}{2\delta }, \ \ \mathrm{{if}}\ \frac{a_L}{2\delta }\le Q;\\ Q,\ \ \ \mathrm{{otherwise}}. \end{array} \right. \end{aligned}$$

(34)

In what follows we analyze the retailer’s optimal order quantity at time 0. We solve this problem based on three cases of $Q$:

Case (i) $0\le Q\le \frac{a_L}{2\delta }$.

For this case, by (33) and (34), we obtain the retailer’s expected profit as

$$\begin{aligned} \begin{array}{lll} E\Pi _{wr1}(Q)=\alpha Q(a_H-\delta Q)+(1-\alpha )Q(a_L-\delta Q)-wQ, \end{array} \end{aligned}$$

(35)

and the corresponding variance of the profit as

$$\begin{aligned} \begin{array}{lll} \sigma _{wr1}^2(Q)&{}=&{}\alpha [Q(a_H-\delta Q)-wQ-E\Pi _{wr1}(Q)]^2 +(1-\alpha )[Q(a_L-\delta Q)\\ &{}&{}-wQ-E\Pi _{wr1}(Q)]^2\\ &{}=&{}\alpha [Q(a_H-\delta Q)\!-\!wQ-\alpha Q(a_H-\delta Q)\!-\!(1\!-\!\alpha )Q(a_L-\delta Q)+wQ]^2\\ &{}&{} +(1\!-\!\alpha )[Q(a_L\!-\!\delta Q)\!-\!wQ\!-\!\alpha Q(a_H\!-\!\delta Q)\!-\!(1\!-\!\alpha )Q(a_L\!-\!\delta Q)\!+\!wQ]^2.\\ &{}=&{}\alpha (1-\alpha )Q^2(a_H-a_L)^2. \end{array}\quad \quad \end{aligned}$$

(36)

Therefore, the standard deviation (SD) of the retailer’s profit is given by

$$\begin{aligned} \begin{array}{lll} \sigma _{wr1}(Q)=\sqrt{\alpha (1-\alpha )}Q(a_H-a_L). \end{array} \end{aligned}$$

(37)

Thus, the problem faced by the retailer at time 0 is formulated as

$$\begin{aligned} \begin{array}{lll} {\hbox {P}_\mathrm{A1}:}\ \ \&\,&\max \limits _{0\le Q\le \frac{a_L}{2\delta }}ED_{wr1}(Q)=E\Pi _{wr1}(Q)-\eta \sigma _{wr1}(Q), \end{array} \end{aligned}$$

(38)

where $\eta$ indicates the retailer’s degree of risk-aversion towards the value risk of the wholesale price contract. Solving problem ${\hbox {P}_\mathrm{A1}}$, we obtain the solution denoted by $\overline{Q}_{w1}$ as follows:

$$\begin{aligned} \overline{Q}_{w1}(\eta )=\left\{ \!\! \begin{array}{llll} \frac{a_L}{2\delta }, \ \ \ \ \ \ \qquad \qquad \qquad \qquad \qquad \mathrm{{if}}\ \eta \le \frac{\alpha (a_H-a_L)-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)};\\ \frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)}{2\delta },\ \ \ \,\,\, \mathrm{{if}}\ \frac{\alpha (a_H-a_L)-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)}<\eta \le \frac{\overline{A}-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)};\\ 0,\ \ \ \ \ \ \ \ \qquad \qquad \qquad \qquad \qquad \mathrm{{otherwise}}. \end{array} \right. \end{aligned}$$

(39)

Case (ii) $\frac{a_L}{2\delta }\le Q\le \frac{a_H}{2\delta }$. For this case, by (33) and (34), we obtain the retailer’s expected profit as

$$\begin{aligned} \begin{array}{lll} E\Pi _{wr2}(Q)=\alpha Q(a_H-\delta Q)+(1-\alpha )\frac{a_L}{2\delta }(a_L-\delta \frac{a_L}{2\delta })-wQ, \end{array}\end{aligned}$$

(40)

and the corresponding variance of the profit as

$$\begin{aligned} \begin{array}{lll} \sigma _{wr2}^2(Q)&{}=&{}\alpha [Q(a_H-\delta Q)-wQ-E\Pi _{wr2}(Q)]^2 +(1-\alpha )[\frac{a_L}{2\delta }(a_L-\delta \frac{a_L}{2\delta })\\ &{}&{}-wQ-E\Pi _{wr2}(Q)]^2\\ &{}=&{}\alpha (1-\alpha )[Q(a_H-\delta Q)-\frac{a_L^2}{4\delta }]^2. \end{array} \end{aligned}$$

(41)

Since $Q(a_H-\delta Q)\ge \frac{a_L}{2\delta }(a_H-\delta \frac{a_L}{2\delta })> \frac{a_L}{2\delta }(a_L-\delta \frac{a_L}{2\delta })=\frac{a_L^2}{4\delta }$ for all $\frac{a_L}{2\delta }\le Q\le \frac{a_H}{2\delta }$, the SD of the retailer’s profit is given by

$$\begin{aligned} \begin{array}{lll} \sigma _{wr2}(Q)=\sqrt{\alpha (1-\alpha )}[Q(a_H-\delta Q)-\frac{a_L^2}{4\delta }]. \end{array} \end{aligned}$$

(42)

Thus, we formulate the problem faced by the retailer at time 0 as

$$\begin{aligned} \begin{array}{lll} {\hbox {P}_\mathrm{A2}:}\ \ \ &{}&{}\max \limits _{\frac{a_L}{2\delta }\le Q\le \frac{a_H}{2\delta }}ED_{wr2}=E\Pi _{wr2}(Q)-\eta \sigma _{wr2}(Q).\\ \end{array} \end{aligned}$$

(43)

Solving problem ${\hbox {P}_\mathrm{A2}}$, we obtain the solution denoted by $\overline{Q}_{w2}$ as follows:

$$\begin{aligned} \overline{Q}_{w2}(\eta )=\left\{ \!\! \begin{array}{llll} \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H-w}{2\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]},\ \ \ \ \mathrm{{if}}\ \eta \le \frac{\alpha (a_H-a_L)-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)};\\ \frac{a_L}{2\delta }, \ \ \ \ \ \ \ \ \ \mathrm{{if}}\ \frac{\alpha (a_H-a_L)-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)}<\eta \le \sqrt{\frac{\alpha }{1-\alpha }};\\ \frac{a_H}{2\delta }, \ \ \ \ \ \ \ \ \ \mathrm{{if}}\ \sqrt{\frac{\alpha }{1-\alpha }}<\eta . \end{array} \right. \end{aligned}$$

(44)

Case (iii) $Q\ge \frac{a_H}{2\delta }$. Similarly, by (33) and (34), we obtain the retailer’s expected profit as

$$\begin{aligned} \begin{array}{lll} E\Pi _{wr3}(Q) =\alpha \frac{a_H}{2\delta }\left( a_H-\delta \frac{a_H}{2\delta }\right) +(1-\alpha )\frac{a_L}{2\delta }\left( a_L-\delta \frac{a_L}{2\delta }\right) -wQ, \end{array} \end{aligned}$$

(45)

and the corresponding variance of the profit as

$$\begin{aligned} \begin{array}{lll} \sigma _{wr3}^2(Q)&{}=&{}\alpha \left[ \frac{a_H}{2\delta }\left( a_H-\delta \frac{a_H}{2\delta }\right) -wQ-E\Pi _{wr3}(Q)\right] ^2 +(1-\alpha )\left[ \frac{a_L}{2\delta }\left( a_L-\delta \frac{a_L}{2\delta }\right) \right. \\ &{}&{}\quad -\left. wQ-E\Pi _{wr3}(Q)\right] ^2\\ &{}=&{}\alpha (1-\alpha )\left( \frac{a_H^2-a_L^2}{4\delta }\right) ^2\!\!\!. \end{array}\end{aligned}$$

(46)

Therefore, the SD of the retailer’s profit is given by

$$\begin{aligned} \begin{array}{lll} \sigma _{wr3}(Q)=\sqrt{\alpha (1-\alpha )}\left( \frac{a_H^2-a_L^2}{4\delta }\right) . \end{array} \end{aligned}$$

(47)

Thus, we formulate the problem faced by the retailer at time 0 as

$$\begin{aligned} \begin{array}{lll} {\hbox {P}_\mathrm{A3}:}\ \ \ &{}&{}\max \limits _{Q\ge \frac{a_H}{2\delta }}ED_{wr3}=E\Pi _{wr3}(Q)-\eta \sigma _{wr3}(Q).\\ \end{array} \end{aligned}$$

(48)

Solving problem ${\hbox {P}_\mathrm{A3}}$, we obtain the optimal solution denoted by $\overline{Q}_{w3}$ as follows:

$$\begin{aligned} \overline{Q}_{w3}(\eta )=\frac{a_H}{2\delta } \ \mathrm{{for\ all}}\ \eta . \end{aligned}$$

(49)

Summarizing (39), (44), and (49), we obtain

(1) When $\eta \le \frac{\alpha (a_H-a_L)-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)}$, the optimal quantity for the retailer to order at time 0 is determined by

$$\begin{aligned} \begin{array}{lll} \overline{Q}_w(\eta )\!=\!{\mathop {\hbox {arg max}}\limits _{Q}}\left\{ { ED}_{wr1}\left( \frac{a_L}{2\delta }\right) ,\ { ED}_{wr2}\left( \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H-w}{2\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]}\right) ,\ { ED}_{wr3}\left( \frac{a_H}{2\delta }\right) \right\} . \end{array}\end{aligned}$$

(50)

Since

$$\begin{aligned} \begin{array} {lll}&{}&{} { ED}_{wr2}\left( \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H-w}{2\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]}\right) \ge { ED}_{wr2}\left( \frac{a_L}{2\delta }\right) ={ ED}_{wr1}\left( \frac{a_L}{2\delta }\right) ,\\ &{}&{}{ ED}_{wr2}\left( \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H-w}{2\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]}\right) \ge { ED}_{wr2}\left( \frac{a_H}{2\delta }\right) ={ ED}_{wr3}\left( \frac{a_H}{2\delta }\right) , \end{array} \end{aligned}$$

(51)

where the inequalities (51) become equality, if and only if $\eta =\frac{\alpha (a_H-a_L)-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)}$, the optimal order quantity in this case is $\overline{Q}_w(\eta )=\frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}] a_H-w}{2\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]}$.

(2) Since we do not know the magnitudes of $\frac{\overline{A}-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)}$ and $\sqrt{\frac{\alpha }{1-\alpha }}$, we proceed the proof by considering two cases as follows:

Case (i) $\sqrt{\frac{\alpha }{1-\alpha }}\le \frac{\overline{A}-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)}$, i.e., $w\le a_L$. Then by (39), (44), and (49), when $\frac{\alpha (a_H-a_L)-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)}< \eta \le \sqrt{\frac{\alpha }{1-\alpha }}$, the optimal quantity for the retailer to order at time 0 is determined by

$$\begin{aligned} \begin{array}{lll} \overline{Q}_w(\eta )\!=\!\mathop {\hbox {arg max}}\limits _{Q}\left\{ { ED}_{wr1} \left( \frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)}{2\delta }\right) ,\ { ED}_{wr2}\left( \frac{a_L}{2\delta }\right) ,\ { ED}_{wr3}\left( \frac{a_H}{2\delta }\right) \right\} \!. \end{array}\qquad \end{aligned}$$

(52)

Since

$$\begin{aligned} { ED}_{wr1}\left( \frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)}{2\delta }\right)&> { ED}_{wr1}\left( \frac{a_L}{2\delta }\right) ={ED}_{wr2}\left( \frac{a_L}{2\delta }\right) \nonumber \\&\ge { ED}_{wr2}\left( \frac{a_H}{2\delta }\right) \nonumber \\&= { ED}_{wr3}\left( \frac{a_H}{2\delta }\right) , \end{aligned}$$

(53)

the optimal quantity in this case is $\overline{Q}_w(\eta )\!=\!\frac{\overline{A}-w-\eta \sqrt{\alpha (1\!-\!\alpha )} (a_H-a_L)}{2\delta }$. When $\left( \frac{\alpha (a_H-a_L)\!-\!w}{\sqrt{\alpha (1\!-\!\alpha )}(a_H\!-\!a_L)}\!\!<\!\right) \sqrt{\frac{\alpha }{1\!-\!\alpha }}<\eta \le \frac{\overline{A}-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)}$, the optimal quantity for the retailer to order at time 0 is determined by

$$\begin{aligned} \begin{array}{lll} \overline{Q}_w(\eta )\!=\!\mathop {\hbox {arg max}}\limits _{Q}\left\{ \!{ ED}_{wr1} \left( \frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)}{2\delta }\right) ,\ { ED}_{wr2}\left( \frac{a_H}{2\delta }\right) ,\ { ED}_{wr3}\left( \frac{a_H}{2\delta }\right) \!\right\} \!. \end{array}\qquad \end{aligned}$$

(54)

Since

$$\begin{aligned}&{ ED}_{wr1}\left( \frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)}{2\delta }\right) -{ ED}_{wr2}\left( \frac{a_H}{2\delta }\right) \left( ={ ED}_{wr3}\left( \frac{a_H}{2\delta }\right) \right) \nonumber \\&\quad =\frac{[\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)-w]^2}{4\delta }\nonumber \\&\qquad -\left[ \frac{\alpha -\eta \sqrt{\alpha (1-\alpha )}}{4\delta }\left( a_H^2-a_L^2\right) + \frac{1}{4\delta }\left( a_L^2-2wa_H\right) \right] \nonumber \\&\quad =\frac{1}{4\delta }[((\alpha -\eta \sqrt{\alpha (1-\alpha )})(a_H-a_L)-w)^2 \nonumber \\&\qquad + \,2w(a_H-a_L)-(\alpha -\eta \sqrt{\alpha (1-\alpha )})(a_H-a_L)^2] \nonumber \\&\quad > 0\ (\mathrm{{note\ that}}\ \eta >\sqrt{\frac{\alpha }{1-\alpha }}\Rightarrow \alpha -\eta \sqrt{\alpha (1-\alpha )}<0), \end{aligned}$$

(55)

the optimal quantity in this case is $\overline{Q}_w(\eta )=\frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)}{2\delta }$.

Case (ii) $\sqrt{\frac{\alpha }{1-\alpha }}>\frac{\overline{A}-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)}$, i.e., $w>a_L$. Then by (39), (44), and (49), when $\frac{\alpha (a_H-a_L)-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)}< \eta \le \frac{\overline{A}-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)} \Big (<\sqrt{\frac{\alpha }{1-\alpha }}\Big )$, the optimal quantity for the retailer to order at time 0 is determined by

$$\begin{aligned} \begin{array}{lll} \overline{Q}_w(\eta )\!=\!\mathop {\hbox {arg max}}\limits _{Q}\left\{ \!{ ED}_{wr1} \left( \frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)}{2\delta }\right) ,\ { ED}_{wr2}\left( \frac{a_L}{2\delta }\right) ,\ { ED}_{wr3}\left( \frac{a_H}{2\delta }\right) \right\} \!. \end{array}\qquad \end{aligned}$$

(56)

Since

$$\begin{aligned} { ED}_{wr1}\left( \frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)}{2\delta }\right)&> { ED}_{wr1}\left( \frac{a_L}{2\delta }\right) ={ ED}_{wr2}\left( \frac{a_L}{2\delta }\right) \nonumber \\&\ge { ED}_{wr2}\left( \frac{a_H}{2\delta }\right) \nonumber \\&= { ED}_{wr3}\left( \frac{a_H}{2\delta }\right) , \end{aligned}$$

(57)

the optimal quantity in this case is $\overline{Q}_w(\eta )\!\!=\!\!\frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)}{2\delta }$. When $\Big (\!\frac{\alpha (a_H-a_L)-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)}\!\!<\!\!\Big ) \frac{\overline{A}-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)}\!<\!\eta \! \le \!\sqrt{\frac{\alpha }{1-\alpha }}$, the optimal quantity for the retailer to order at time 0 is determined by

$$\begin{aligned} \begin{array}{lll} \overline{Q}_w(\eta )=\mathop {\hbox {arg max}}\limits _{Q}\left\{ { ED}_{wr1}(0),\ { ED}_{wr2}\left( \frac{a_L}{2\delta }\right) ,\ { ED}_{wr3}\left( \frac{a_H}{2\delta }\right) \right\} . \end{array} \end{aligned}$$

(58)

Since

$$\begin{aligned} \begin{array}{lll} { ED}_{wr1}(0)>{ ED}_{wr1}\left( \frac{a_L}{2\delta }\right) ={ ED}_{wr2}\left( \frac{a_L}{2\delta }\right) \ge { ED}_{wr2}\left( \frac{a_H}{2\delta }\right) ={ ED}_{wr3}\left( \frac{a_H}{2\delta }\right) , \end{array} \end{aligned}$$

(59)

the optimal quantity in this case is $\overline{Q}_w(\eta )=0$.

(3) When $\eta >\max \{\frac{\overline{A}-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)}, \sqrt{\frac{\alpha }{1-\alpha }}\}$, by (39), (44) and (49), the optimal quantity for the retailer to order at time 0 is determined by

$$\begin{aligned} \begin{array}{lll} \overline{Q}_w(\eta )=\mathop {\hbox {arg max}}\limits _{Q}\left\{ { ED}_{wr1}(0),\ { ED}_{wr2}\left( \frac{a_H}{2\delta }\right) ,\ { ED}_{wr3}\left( \frac{a_H}{2\delta }\right) \right\} . \end{array} \end{aligned}$$

(60)

Since

$$\begin{aligned} \begin{array}{lll} { ED}_{wr2}\left( \frac{a_H}{2\delta }\right) ={ ED}_{wr3}\left( \frac{a_H}{2\delta }\right) <0={ ED}_{wr1}(0), \end{array} \end{aligned}$$

(61)

the optimal quantity in this case is $\overline{Q}_w(\eta )=0$.

Summarizing the above analysis, we obtain that if $w\le a_L$, the optimal quantity for the retailer to order at time 0 is given by

$$\begin{aligned} \overline{Q}_{w}(\eta )=\left\{ \!\! \begin{array}{llll} \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H-w}{2\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]}, \ \ \ \ \ \ \ \mathrm{{if}}\ \eta \le \frac{\alpha (a_H-a_L)-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)};\\ \frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)}{2\delta },\; \mathrm{{if}}\ \frac{\alpha (a_H-a_L)-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)}< \eta \le \sqrt{\frac{\alpha }{1-\alpha }};\\ \frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)}{2\delta },\ \ \mathrm{{if}}\ \sqrt{\frac{\alpha }{1-\alpha }}<\eta \le \frac{\overline{A}-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)};\\ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{{otherwise}}; \end{array} \right. \end{aligned}$$

(62)

if $w>a_L$, the optimal quantity for the retailer to order at time 0 is given by

$$\begin{aligned} \overline{Q}_{w}(\eta )=\left\{ \!\! \begin{array}{llll} \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H-w}{2\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]}, \ \ \ \ \ \ \ \mathrm{{if}}\ \eta \le \frac{\alpha (a_H-a_L)-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)};\\ \frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)}{2\delta },\; \mathrm{{if}}\ \frac{\alpha (a_H-a_L)-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)}< \eta \le \frac{\overline{A}-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)};\\ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{{if}}\ \frac{\overline{A}-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)}<\eta \le \sqrt{\frac{\alpha }{1-\alpha }};\\ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{{otherwise}}. \end{array} \right. \end{aligned}$$

(63)

Summarizing (62) and (63), we obtain

$$\begin{aligned} \overline{Q}_{w}(\eta )=\left\{ \!\! \begin{array}{llll} \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H-w}{2\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]}, \ \ \ \ \ \ \ \mathrm{{if}}\ \eta \le \frac{\alpha (a_H-a_L)-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)};\\ \frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)}{2\delta },\ \ \mathrm{{if}}\ \frac{\alpha (a_H-a_L)-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)}< \eta \le \frac{\overline{A}-w}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)};\\ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{{otherwise}}; \end{array} \right. \end{aligned}$$

(64)

Equivalently, we transform (64) into

$$\begin{aligned} \overline{Q}_{\eta }(w)\!\!=\!\!\left\{ \!\! \begin{array}{llll} \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H-w}{2\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]}, \ \ \ \ \ \ \ \mathrm{{if}}\ w\le [\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_H-a_L);\\ \frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)}{2\delta },\ \ \mathrm{{if}}\ [\alpha \!-\!\eta \sqrt{\alpha (1\!-\!\alpha )}](a_H\!-\!a_L)\!<\! w\!\le \!\overline{A}\!-\!\eta \sqrt{\alpha (1\!-\!\alpha )}(a_H\!-\!a_L);\\ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{{otherwise}}. \end{array} \right. \end{aligned}$$

(65)

Hence, the proof is completed. $\square$

Proof of Theorem 1

We show Theorem 1 based on two cases as follows:

Case (i): $[\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_H-a_L)\ge c$, i.e., $\eta \le \frac{\alpha (a_H-a_L)-c}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)}(\triangleq \eta _1)$. By Lemma 1, the supplier’s problem in this case can be formulated as

$$\begin{aligned} \begin{array}{lll} {\hbox {P}_\mathrm{Aw1}:}\ \ \ \ \ \overline{w}=\mathop {\hbox {arg max}}\limits _{w}\{\overline{H}_{1}, \overline{H}_{2}\}, \end{array} \end{aligned}$$

(66)

where

$$\begin{aligned} \begin{array}{lll} &{}&{}\overline{H}_1=\max \limits _{c\le w\le [\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_H-a_L)} \left\{ \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H-w}{2\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]}\right\} (w-c),\\ &{}&{}\overline{H}_2\!=\!\max \limits _{\left[ \alpha -\eta \sqrt{\alpha (1-\alpha )}\right] (a_H-a_L)< w\le \overline{A}-\eta \sqrt{\alpha (1-\alpha )} (a_H-a_L)}\left[ \frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)}{2\delta }\right] (w\!-\!c).\\ \end{array}\qquad \end{aligned}$$

(67)

Solving problem $\overline{H}_1$, we obtain the solution denoted by $\overline{w}_1(\eta )$ as follows:

$$\begin{aligned} \overline{w}_1(\eta )=\left\{ \!\! \begin{array}{llll} \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H+c}{2}, \ \ \ \ \ \ \ \ \ \ \mathrm{{if}}\ c\le [\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_H-2a_L);\\ (\alpha -\eta \sqrt{\alpha (1-\alpha )})(a_H-a_L),\ \ \ \mathrm{{if}}\ c>[\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_H-2a_L). \end{array} \right. \end{aligned}$$

(68)

Solving problem $\overline{H}_2$, we obtain the solution denoted by $\overline{w}_2(\eta )$ as follows:

$$\begin{aligned} \overline{w}_2(\eta )\!=\!\left\{ \!\! \begin{array}{llll} [\alpha \!-\!\eta \sqrt{\alpha (1\!-\!\alpha )}](a_H-a_L), \ \ \mathrm{{if}}\ c\le [\alpha \!-\!\eta \sqrt{\alpha (1-\alpha )}](a_H-a_L)-a_L;\\ \frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)+c}{2},\ \ \ \ \ \ \ \ \ \ \mathrm{{if}}\ c>[\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_H-a_L)-a_L. \end{array} \right. \end{aligned}$$

(69)

For ease of exposition, we denote

$$\begin{aligned} \begin{array}{lll} c_1=[\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_H-a_L)-a_L \ \mathrm{{and}}\ c_2=[\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_H-2a_L). \end{array} \end{aligned}$$

(70)

Since $c_2\ge c_1$, summarizing (68) and (69), we have

$$\begin{aligned} \overline{w}(\eta )\!=\!\left\{ \!\! \begin{array}{llll} \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H+c}{2}, \ \ \ \ \ \ \ \ \ \!\!\mathrm{{if}}\ c<c_1;\\ \frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)+c}{2},\ \ \ \ \!\!\!\!\mathrm{{if}}\ c>c_2;\\ \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H+c}{2}, \ \ \ \ \ \ \ \ \ \!\! \mathrm{{if}}\ c_1\le c\le c_2\ \mathrm{{and}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \qquad \overline{H}_1\left( \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H+c}{2}\right) \ge \overline{H}_2\left( \frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)+c}{2}\right) ;\\ \frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)+c}{2},\ \ \ \ \mathrm{{if}}\ c_1\le c\le c_2\ \mathrm{{and}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \qquad \overline{H}_1\left( \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H+c}{2}\right) < \overline{H}_2\left( \frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)+c}{2}\right) . \end{array} \right. \end{aligned}$$

(71)

It is easy to obtain that

$$\begin{aligned}&\begin{array}{lll} \overline{H}_1\left( \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H+c}{2}\right) \!&{}=&{}\!\left( \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H- \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H+c}{2}}{2\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]}\right) \\ &{}&{}\quad \times \left( \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H+c}{2}\!-\!c\right) \\ &{}=&{}\frac{\left[ (\alpha -\eta \sqrt{\alpha (1-\alpha )})a_H-c\right] ^2}{8\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]}, \end{array}\end{aligned}$$

(72)

$$\begin{aligned}&\begin{array}{lll} \overline{H}_2\left( \frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)+c}{2}\right) &{}=&{}\left[ \frac{\overline{A}-\frac{\overline{A}- \eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)+c}{2}- \eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)}{2\delta }\right] \\ &{}&{}\quad \times \left( \frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)+c}{2}-c\right) \\ &{}=&{}\frac{[\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)-c]^2}{8\delta }. \end{array} \end{aligned}$$

(73)

Therefore,

$$\begin{aligned} \begin{array}{lll} &{}&{}\overline{H}_{1}\left( \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H+c}{2}\right) -\overline{H}_{2}\left( \frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)+c}{2}\right) \\ &{}&{}\quad =\frac{[(\alpha -\eta \sqrt{\alpha (1-\alpha )})a_H-c]^2}{8\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]} -\frac{[\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)-c]^2}{8\delta }\\ &{}&{}\quad =\frac{1-[\alpha -\eta \sqrt{\alpha (1-\alpha )}]}{8\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]}[c^2-2 [\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_H-a_L)c +[(\alpha -\eta \sqrt{\alpha (1-\alpha )})a_H]^2\\ &{}&{}\quad -2[\alpha -\eta \sqrt{\alpha (1-\alpha )}]^2a_Ha_L-[\alpha -\eta \sqrt{\alpha (1-\alpha )}] [1-(\alpha -\eta \sqrt{\alpha (1-\alpha )})]a_L^2]. \end{array} \end{aligned}$$

(74)

Since $0<\alpha -\eta \sqrt{\alpha (1-\alpha )}\le \alpha <1$ for all $0\le \eta \le \eta _1$, $\frac{1-[\alpha -\eta \sqrt{\alpha (1-\alpha )}]}{8\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]}>0$ for all $0\le \eta \le \eta _1$. Therefore, the inequality $\overline{H}_{1}(\frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H+c}{2})\le \overline{H}_{2}(\frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)+c}{2})$ is equivalent to

$$\begin{aligned} \begin{array}{lll} &{}&{}c^2-2 [\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_H-a_L)c +[(\alpha -\eta \sqrt{\alpha (1-\alpha )})a_H]^2\\ &{}&{} \!-\!2[\alpha \!-\!\eta \sqrt{\alpha (1-\alpha )}]^2a_Ha_L-[\alpha -\eta \sqrt{\alpha (1-\alpha )}] [1\!-\!(\alpha -\eta \sqrt{\alpha (1-\alpha )})]a_L^2\le 0. \end{array} \end{aligned}$$

(75)

Solving inequality (75), we obtain that $\overline{H}_{1}(\frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H+c}{2})\le \overline{H}_{2}(\frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)+c}{2})$ holds iff $c\in [c_3,c_4]$, where

$$\begin{aligned} \begin{array}{lll} c_3&{}=&{} a_H[\alpha -\eta \sqrt{\alpha (1-\alpha )}] -a_L[\sqrt{\alpha -\eta \sqrt{\alpha (1-\alpha )}} +(\alpha -\eta \sqrt{\alpha (1-\alpha )})],\\ c_4&{}=&{}a_H [\alpha -\eta \sqrt{\alpha (1-\alpha )}] +a_L(\sqrt{\alpha -\eta \sqrt{\alpha (1-\alpha )}} -[\alpha -\eta \sqrt{\alpha (1-\alpha )}]). \end{array} \end{aligned}$$

(76)

Since $c_1\le c_3\le c_2\le c_4$, we can simplify (71) as

$$\begin{aligned} \overline{w}_1(\eta )=\left\{ \!\! \begin{array}{llll} \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H+c}{2}, \ \ \ \ \ \qquad \quad \mathrm{{if}}\ c\le c_3;\\ \frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)+c}{2},\ \ \ \ \ \ \ \ \mathrm{{if}}\ c>c_3. \end{array} \right. \end{aligned}$$

(77)

In addition,

$$\begin{aligned} \begin{array}{lll} c\le c_3&{}\Longleftrightarrow &{}(a_H-a_L) [\alpha -\eta \sqrt{\alpha (1-\alpha )}] -a_L\sqrt{\alpha -\eta \sqrt{\alpha (1-\alpha )}}-c\ge 0\\ &{}\Longleftrightarrow &{}\sqrt{\alpha -\eta \sqrt{\alpha (1-\alpha )}} \le \frac{a_L-\sqrt{a_L^2+4c(a_H-a_L)}}{2(a_H-a_L)}\ \mathrm{{or}}\ \sqrt{\alpha -\eta \sqrt{\alpha (1-\alpha )}}\\ &{}\ge &{} \frac{a_L+\sqrt{a_L^2+4c(a_H-a_L)}}{2(a_H-a_L)}. \end{array} \end{aligned}$$

(78)

Since $\sqrt{\alpha -\eta \sqrt{\alpha (1-\alpha )}}>0$ while $\frac{a_L-\sqrt{a_L^2+4c(a_H-a_L)}}{2(a_H-a_L)}<0$,

$$\begin{aligned} \begin{array}{lll} c\le c_3&{}\Longleftrightarrow &{}\sqrt{\alpha -\eta \sqrt{\alpha (1-\alpha )}} \ge \frac{a_L+\sqrt{a_L^2+4c(a_H-a_L)}}{2(a_H-a_L)}\\ &{}\Longleftrightarrow &{}\eta \le \eta _2, \end{array} \end{aligned}$$

(79)

where

$$\begin{aligned} \begin{array}{lll} \eta _2=\frac{4(a_H-a_L)[\alpha (a_H-a_L)-c]-2a_L^2-2a_L \sqrt{a_L^2+4c(a_H-a_L)}}{4\sqrt{\alpha (1-\alpha )}(a_H-a_L)^2}. \end{array} \end{aligned}$$

(80)

It is easy to obtain that

$$\begin{aligned} \begin{array}{lll} \eta _1-\eta _2&{}=&{}\frac{\alpha (a_H-a_L)-c}{\sqrt{\alpha (1-\alpha )}(a_H-a_L)}- \frac{4(a_H-a_L)[\alpha (a_H-a_L)-c]-2a_L^2-2a_L \sqrt{a_L^2+4c(a_H-a_L)}}{4\sqrt{\alpha (1-\alpha )}(a_H-a_L)^2}\\ &{}=&{}\frac{2a_L^2+2a_L\sqrt{a_L^2+4c(a_H-a_L)}}{4\sqrt{\alpha (1-\alpha )}(a_H-a_L)^2}>0, \end{array} \end{aligned}$$

(81)

i.e., $\eta _1>\eta _2$. Hence, the equilibrium wholesale price in the case of $\eta \le \eta _1$ is given by

$$\begin{aligned} \overline{w}_1(\eta )=\left\{ \!\! \begin{array}{llll} \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H+c}{2}, \ \ \ \ \ \ \qquad \quad \mathrm{{if}}\ \eta \le \eta _2;\\ \frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)+c}{2},\ \ \ \ \ \ \ \ \ \mathrm{{if}}\ \eta _2<\eta \le \eta _1. \end{array} \right. \end{aligned}$$

(82)

Case (ii) $[\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_H-a_L)<c$, i.e., $\eta >\eta _1$. By Lemma 1, the supplier’s problem in this case can be formulated as

$$\begin{aligned} \begin{array}{lll} \mathrm{{P_{Aw2}:}}\ \ \ \ \ \max \limits _{c\le w\le \overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)} \left[ \frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)}{2\delta }\right] (w-c), \end{array} \end{aligned}$$

(83)

Solving problem ${\hbox {P}_\mathrm{Aw2}}$, we obtain the solution denoted by $\overline{w}_2$ as

$$\begin{aligned} \begin{array}{lll} \overline{w}_2(\eta )=\frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)+c}{2} \ \quad \mathrm{{for\ all}}\ \eta \in (\eta _1, \eta _{max}]. \end{array} \end{aligned}$$

(84)

Summarizing (82) and (84), we obtain the equilibrium wholesale price as

$$\begin{aligned} \overline{w}(\eta )=\left\{ \!\! \begin{array}{llll} \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H+c}{2}, \ \ \ \ \ \ \qquad \mathrm{{if}}\ \eta \le \eta _2;\\ \frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)+c}{2},\ \ \ \ \ \ \mathrm{{if}}\ \eta _2<\eta \le \eta _{max}. \end{array} \right. \end{aligned}$$

(85)

Substituting (85) into (64), we obtain the equilibrium order quantity of the retailer as

$$\begin{aligned} \overline{Q}(\eta )=\left\{ \!\! \begin{array}{llll} \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_H-c}{4\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]}, \ \ \ \ \ \ \ \mathrm{{if}}\ \eta \le \eta _2;\\ \frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)-c}{4\delta },\; \mathrm{{if}}\ \eta _2< \eta \le \eta _{max}. \end{array} \right. \end{aligned}$$

(86)

Thus, the proof is completed. $\square$

Proof of Theorem 2

For ease of exposition, we denote $\theta (\eta )=\alpha -\eta \sqrt{\alpha (1-\alpha )}$ and $\overline{B}(\eta )=\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)$ in the following development. Furthermore, if no confusion, we will respectively denote $\theta (\eta )$, $\overline{B}(\eta )$, $\overline{w}(\eta ),$ and $\overline{Q}(\eta )$ simply by $\theta$, $\overline{B}$, $\overline{w},$ and $\overline{Q}.$ We first show Theorem 2(1). By (85) and (86), we obtain that, if $\eta \le \eta _2$,

$$\begin{aligned} \begin{array}{lll} \overline{R}_{ws}(\eta )&= (\overline{w}-c)\overline{Q} =\left( \frac{\theta a_H+c}{2}-c\right) \left( \frac{\theta a_H-c}{4\delta \theta }\right) =\frac{(\theta a_H-c)^2}{8\delta \theta }. \end{array} \end{aligned}$$

(87)

It is easy to obtain

(88)

Obviously, for all $\eta \le \eta _2$. Therefore, $\overline{R}_{ws}(\eta )$ strictly decreases with $\eta$ for all $\eta \le \eta _2$. Similarly, if $\eta _2<\eta \le \eta _{max}$, by (85) and (86), we obtain

$$\begin{aligned} \begin{array}{lll} \overline{R}_{ws}(\eta )&= (\overline{w}-c)\overline{Q}= \left( \frac{\overline{B}+c}{2}-c\right) \left( \frac{\overline{B}-c}{4\delta }\right) = \frac{(\overline{B}-c)^2}{8\delta }, \end{array} \end{aligned}$$

(89)

(90)

Since $\overline{B}>c$ for all $\eta \in (\eta _2, \eta _{max})$, for all $\eta \in (\eta _2, \eta _{max})$. Therefore, $\overline{R}_{ws}(\eta )$ strictly decreases with $\eta$ on $(\eta _2, \eta _{max}]$. To summarize, we derive the desired result in Theorem 2(1).

We proceed to show Theorem 2(2). Likewise, by (85) and (86), together with checking the proofs of Lemma 1 and Theorem 1, we obtain that, if $\eta \le \eta _2$,

$$\begin{aligned}&\begin{array}{lll} E_{wr}(\eta )&{}=&{}\alpha \overline{Q}(a_H-\delta \overline{Q})+(1-\alpha )\frac{a_L}{2\delta }\left( a_L-\delta \frac{a_L}{2\delta }\right) -\overline{w}\overline{Q}\\ &{}=&{}\alpha \left( \frac{\theta a_H-c}{4\delta \theta }\right) \left( a_H-\delta \frac{\theta a_H-c}{4\delta \theta }\right) +(1-\alpha )\frac{a_L}{2\delta }\left( a_L-\delta \frac{a_L}{2\delta }\right) \\ &{}&{}\quad -\left( \frac{\theta a_H+c}{2}\right) \left( \frac{\theta a_H-c}{4\delta \theta }\right) \\ &{}=&{}\frac{(\theta a_H-c)^2}{16\delta \theta }+\frac{(1-\theta )a_L^2}{4\delta } +\eta \sqrt{\alpha (1-\alpha )}\left[ \frac{(\theta a_H-c)(3\theta a_H+c)}{16\delta \theta ^2}-\frac{a_L^2}{4\delta }\right] , \end{array} \end{aligned}$$

(91)

$$\begin{aligned}&\begin{array}{lll} { SD}_{wr}(\eta )&{}=&{}\sqrt{\alpha (1-\alpha )}[\overline{Q}(a_H-\delta \overline{Q})-\frac{a_L^2}{4\delta }]\\ &{}=&{}\sqrt{\alpha (1-\alpha )}[\frac{\theta a_H-c}{4\delta \theta }(a_H-\delta \frac{\theta a_H-c}{4\delta \theta })-\frac{a_L^2}{4\delta }]\\ &{}=&{}\sqrt{\alpha (1-\alpha )}[\frac{(\theta a_H-c)(3\theta a_H+c)}{16\delta \theta ^2}-\frac{a_L^2}{4\delta }]. \end{array} \end{aligned}$$

(92)

Since for all $\eta \le \eta _2$,

$$\begin{aligned} \begin{array}{lll} \frac{dSD_{wr}(\eta )}{d\eta }=- \frac{\alpha (1-\alpha )(\theta a_H+c)c}{8\delta \theta ^3}<0, \end{array} \end{aligned}$$

(93)

${ SD}_{wr}(\eta )$ strictly decreases with $\eta$ for all $\eta \le \eta _2$. Similarly, if $\eta _2<\eta \le \eta _{max}$, we obtain

$$\begin{aligned} \begin{array}{lll} E_{wr}(\eta )&{}=&{}\alpha \overline{Q}(a_H-\delta \overline{Q})+(1-\alpha )\overline{Q}(a_L-\delta \overline{Q})-\overline{w}\overline{Q}\\ &{}=&{}\alpha \frac{\overline{B}-c}{4\delta }\left( a_H-\delta \frac{\overline{B}-c}{4\delta }\right) +(1-\alpha )\frac{\overline{B}-c}{4\delta }\left( a_L-\delta \frac{\overline{B}-c}{4\delta }\right) -\left( \frac{\overline{B}+c}{2}\right) \left( \frac{\overline{B}-c}{4\delta }\right) \\ &{}=&{}\frac{(\overline{B}-c)^2}{16\delta } +\frac{\eta \sqrt{\alpha (1-\alpha )}}{4\delta }(\overline{B}-c)(a_H-a_L), \end{array} \end{aligned}$$

(94)

$$\begin{aligned} \begin{array}{lll} { SD}_{wr}(\eta )&{}=&{}\sqrt{\alpha (1-\alpha )}\overline{Q}(a_H-a_L)\\ &{}=&{}\frac{\sqrt{\alpha (1-\alpha )}}{{4\delta }}(\overline{B}-c)(a_H-a_L).\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array} \end{aligned}$$

(95)

Since

$$\begin{aligned} \begin{array}{lll} \frac{dSD_{wr}(\eta )}{d\eta }=-\frac{\alpha (1-\alpha )}{4\delta }(a_H-a_L)^2<0, \end{array} \end{aligned}$$

(96)

${ SD}_{wr}(\eta )$ strictly decreases with $\eta$ for all $\eta \in (\eta _2, \eta _{max}]$. Thus, the proof is completed. $\square$

Proof of Lemma 2

Using the marginal production cost $c$ to replace the wholesale price $w$ in the proof of Lemma 1, we obtain in a similar way the system-wide optimal production quantity for the centralized entity as follows:

$$\begin{aligned} \overline{Q}_{c}(\eta )=\left\{ \!\! \begin{array}{llll} \frac{\theta a_H-c}{2\delta \theta }, \ \ \ \ \ \ \ \mathrm{{if}}\ \eta \le \eta _1;\\ \frac{\overline{B}-c}{2\delta },\ \ \ \ \ \ \ \ \ \ \mathrm{{if}}\ \eta _1<\eta \le \eta _{max}. \end{array} \right. \end{aligned}$$

(97)

where

$$\begin{aligned} \begin{array}{lll} \theta =\alpha -\eta \sqrt{\alpha (1-\alpha )},\ \ \overline{B}=\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L). \end{array}\end{aligned}$$

(98)

Putting $w=c$ in (40) and then substituting (97) into it, we obtain that, if $\eta \le \eta _1$,

$$\begin{aligned} \begin{array}{lll} E_c(\eta )&{}=&{}\alpha \overline{Q}_c(a_H-\delta \overline{Q}_c) +(1-\alpha )\frac{a_L}{2\delta }\left( a_L-\delta \frac{a_L}{2\delta }\right) -c\overline{Q}_c\\ &{}=&{}\alpha \frac{\theta a_H-c}{2\delta \theta }\left( a_H-\delta \frac{\theta a_H-c}{2\delta \theta }\right) +(1-\alpha )\frac{a_L}{2\delta }\left( a_L-\delta \frac{a_L}{2\delta }\right) -c\frac{\theta a_H-c}{2\delta \theta }\\ &{}=&{}\frac{\theta (\theta a_H-c)^2+ \theta ^2(1-\theta )a_L^2+\eta \sqrt{\alpha (1-\alpha )}[(\theta a_H-c)(\theta a_H+c)-\theta ^2a_L^2]}{4\delta \theta ^2}\\ &{}=&{}\frac{H(\eta )}{4\delta \theta ^2}. \end{array} \end{aligned}$$

(99)

where, for ease of exposition in the following, we denote

$$\begin{aligned} \theta (\theta a_H\!-\!c)^2\!+\! \theta ^2(1\!-\!\theta )a_L^2\!+\!\eta \sqrt{\alpha (1\!-\!\alpha )}\left[ (\theta a_H\!-\!c)(\theta a_H+c)\!-\!\theta ^2a_L^2\right] \triangleq H(\eta ). \end{aligned}$$

(100)

Substituting (97) into (42), we have

$$\begin{aligned} \begin{array}{lll} SD_c(\eta )&{}=&{}\sqrt{\alpha (1-\alpha )}\left[ \overline{Q}_c(a_H-\delta \overline{Q}_c) -\frac{a_L^2}{4\delta }\right] \\ &{}=&{}\sqrt{\alpha (1-\alpha )}\left[ \frac{\theta a_H-c}{2\delta \theta } +\left( a_H-\delta \frac{\theta a_H-c}{2\delta \theta }\right) -\frac{a_L^2}{4\delta }\right] \\ &{}=&{}\sqrt{\alpha (1-\alpha )}\left[ \frac{(\theta a_H-c)(\theta a_H+c)-\theta ^2a_L^2}{4\delta \theta ^2}\right] . \end{array} \end{aligned}$$

(101)

Similarly, if $\eta _1< \eta \le \eta _{max}$, by setting $w=c$ in (35) and then substituting (97) into it, we obtain

$$\begin{aligned} \begin{array}{lll} E_c(\eta )&{}=&{}\alpha \overline{Q}_c(a_H-\delta \overline{Q}_c)+(1-\alpha )\overline{Q}_c\left( a_L-\delta \overline{Q}_c\right) -c\overline{Q}_c\\ &{}=&{}\alpha \frac{\overline{B}-c}{2\delta }\left( a_H-\delta \frac{\overline{B}-c}{2\delta }\right) +(1-\alpha )\frac{\overline{B}-c}{2\delta } \left( a_L-\delta \frac{\overline{B}-c}{2\delta }\right) -c\frac{\overline{B}-c}{2\delta }\\ &{}=&{}\frac{(\overline{B}-c)^2+2\eta \sqrt{\alpha (1-\alpha )} (\overline{B}-c)(a_H-a_L)}{4\delta }. \end{array} \end{aligned}$$

(102)

Substituting (97) into (37), we have

$$\begin{aligned} \begin{array}{lll} SD_c(\eta )&{}=&{}\sqrt{\alpha (1-\alpha )}\overline{Q}_c(a_H-a_L)\\ &{}=&{}\frac{\sqrt{\alpha (1-\alpha )}}{2\delta }(\overline{B}-c)(a_H-a_L). \end{array}\end{aligned}$$

(103)

To summarize, the proof is completed. $\square$

Proof of Theorem 3

We first show Theorem 3(i). It can be obtained by Theorem 2 that the expected channel profit in the decentralized supply chain is

$$\begin{aligned} E_{total}(\eta )&= \overline{R}_{ws}(\eta )+E_{wr}(\eta )\nonumber \\&= \left\{ \!\! \begin{array}{llll} \frac{3H(\eta )+\theta ^2(1-\theta )a_L^2 -\eta \sqrt{\alpha (1-\alpha )}[2c(\theta a_H-c)+\theta ^2a_L^2]}{16\delta \theta ^2},\ \ \mathrm{{if}}\ \eta \le \eta _2;\\ \frac{3(\overline{B}-c)^2+4\eta \sqrt{\alpha (1-\alpha )}(\overline{B}-c) (a_H-a_L)}{16\delta }, \ \ \ \ \ \mathrm{{if}}\ \eta _2<\eta \le \eta _{max}, \end{array} \right. \end{aligned}$$

(104)

where

$$\begin{aligned}&\begin{array}{lll} \theta =\alpha -\eta \sqrt{\alpha (1-\alpha )},\ \ \overline{B}=\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L), \end{array}\end{aligned}$$

(105)

$$\begin{aligned}&\begin{array}{lll} H(\eta )\!=\!\theta (\theta a_H\!-\!c)^2\!+\! \theta ^2(1\!-\!\theta )a_L^2\!+\!\eta \sqrt{\alpha (1\!-\!\alpha )}[(\theta a_H\!-\!c)(\theta a_H\!+\!c)\!-\!\theta ^2a_L^2]. \end{array}\qquad \end{aligned}$$

(106)

Comparing (104) with (99) and (102), we have

$$\begin{aligned} EFF{e}(\eta )\!=\!\left\{ \!\! \begin{array}{llll} 75\,\%+\frac{\theta ^2(1-\theta )a_L^2 -\eta \sqrt{\alpha (1-\alpha )}[2c(\theta a_H-c)+\theta ^2a_L^2]}{4H(\eta )},\ \ \ \ \ \ \ \ \ \ \ \,\,\, \mathrm{{if}}\ \eta \le \eta _2;\\ 75\,\%+\frac{3\theta ^2(\overline{B}-c)^2+4\theta ^2\eta \sqrt{\alpha (1-\alpha )}(\overline{B}-c)(a_H-a_L)-3H(\eta )}{4H(\eta )}, \ \ \ \mathrm{{if}}\ \eta _2\!<\!\eta \!\le \!\eta _1\\ 75\,\%-\frac{\eta \sqrt{\alpha (1-\alpha )}(\overline{B}-c)(a_H-a_L)}{2[(\overline{B}-c)^2 +2\eta \sqrt{\alpha (1-\alpha )}(\overline{B}-c)(a_H-a_L)]}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,\,\,\, \mathrm{{if}} \,\eta _1<\eta \le \eta _{max}. \end{array} \right. \end{aligned}$$

(107)

We can obtain

$$\begin{aligned} \begin{array}{lll} &{}&{}\theta ^2(1-\theta )a_L^2 -\eta \sqrt{\alpha (1-\alpha )}[2c(\theta a_H-c)+\theta ^2a_L^2]\\ &{}&{}\quad =\theta ^2(1-\theta )a_L^2 +\theta [2c(\theta a_H-c)+\theta ^2a_L^2]-\alpha [2c(\theta a_H-c)+\theta ^2a_L^2]\\ &{}&{}\quad =[(1-\alpha )a_L^2+2ca_H]\theta ^2-2c(\alpha a_H+c)\theta +2\alpha c^2. \end{array} \end{aligned}$$

(108)

Substituting (108) into (107), we obtain (57) in Theorem 3(i). For ease of exposition, we denote

$$\begin{aligned} \begin{array}{lll} H_1(\eta )=[(1-\alpha )a_L^2+2ca_H]\theta ^2-2c(\alpha a_H+c)\theta +2\alpha c^2. \end{array} \end{aligned}$$

(109)

Since

$$\begin{aligned} \frac{d^2H_1(\eta )}{d\eta ^2}=2\alpha (1-\alpha )[(1-\alpha )a_L^2+2ca_H]>0, \end{aligned}$$

(110)

$H_1(\eta )$ is strictly convex in $\eta$. Let $H_1(\eta )=0$. Then if there is no solution for this equation with regard to $\eta$, then $H_1(\eta )>0$ for all $\eta \le \eta _2$; otherwise, solving this equation with regard to $\eta$, we obtain its two solution as

$$\begin{aligned} \begin{array}{lll} \eta _3&{}=&{}\frac{\alpha (1-\alpha )a_L^2+(\alpha a_H-c)c-c\sqrt{(\alpha a_H-c)^2-2\alpha (1-\alpha )a_L^2}}{\sqrt{\alpha (1-\alpha )}[(1-\alpha )a_L^2+2ca_H]},\\ \eta _4&{}=&{}\frac{\alpha (1-\alpha )a_L^2+(\alpha a_H-c)c+c\sqrt{(\alpha a_H-c)^2-2\alpha (1-\alpha )a_L^2}}{\sqrt{\alpha (1-\alpha )}[(1-\alpha )a_L^2+2ca_H]}. \end{array} \end{aligned}$$

(111)

Again, since $H_1(\eta )$ is strictly convex in $\eta$ and $H_1(0)=(1-\alpha )\alpha ^2a_L^2>0,$ we know that $\eta _4\ge \eta _3>0$. Furthermore,

$$\begin{aligned} H_1(\eta )\left\{ \!\! \begin{array}{llll} \hbox {is positive and strictly decreases with} \eta \hbox { on } [0,\eta _3);\\ \hbox {is nonpositive on} [\eta _3, \eta _4];\\ \hbox {is positive and strictly increases with} \eta \hbox {on} (\eta _4, +\infty );\\ \end{array} \right. \end{aligned}$$

(112)

In addition, denote

$$\begin{aligned} \begin{array}{lll} H_2(\eta )=\frac{\eta \sqrt{\alpha (1-\alpha )}(\overline{B}-c)(a_H-a_L)}{2[(\overline{B}-c)^2 +2\eta \sqrt{\alpha (1-\alpha )}(\overline{B}-c)(a_H-a_L)]}. \end{array} \end{aligned}$$

(113)

We obtain by calculation that for all $\eta _1<\eta <\eta _{max}$,

$$\begin{aligned} \begin{array}{lll} \frac{dH_2(\eta )}{d\eta }=\frac{\sqrt{\alpha (1-\alpha )} (\overline{B}-c)^3(a_H-a_L)+\eta \alpha (1-\alpha )(\overline{B}-c)^2(a_H-a_L)^2}{2[(\overline{B}-c)^2 +2\eta \sqrt{\alpha (1-\alpha )}(\overline{B}-c)(a_H-a_L)]^2}>0. \end{array} \end{aligned}$$

(114)

Therefore, $H_2(\eta )$ strictly increases with $\eta$ on $(\eta _1, \eta _{max}]$. Furthermore, obviously, $0\le H_2(\eta )\le 25\,\%=H_2(\eta _{max})$ for all $\eta _1<\eta \le \eta _{max}$. To summarize, we obtain Theorem 3(i).

We proceed to show Theorem 3(ii). Since the profit obtained by the supplier is deterministic, the SD of the channel profit achieved in the decentralized supply chain is determined by the SD of the retailer’s profit. Then, by (92) and (95), we have

$$\begin{aligned} D_{total}(\eta )=\left\{ \!\! \begin{array}{llll} \sqrt{\alpha (1-\alpha )}[\frac{(\theta a_H-c)(3\theta a_H+c)-4\theta ^2a_L^2}{16\delta \theta ^2}],\ \ \mathrm{{if}}\ \eta \le \eta _2;\\ \frac{\sqrt{\alpha (1-\alpha )}}{4\delta }(\overline{B}-c)(a_H-a_L), \ \ \ \ \ \,\, \quad \qquad \mathrm{{if}}\ \eta _2<\eta \le \eta _{max}. \end{array} \right. \end{aligned}$$

(115)

Comparing (115) with (101) and (103), we have

$$\begin{aligned} EFF_{sd}(\eta )=\left\{ \!\! \begin{array}{llll} 75\,\%-\frac{2c(\theta a_H-c)+\theta ^2a_L^2}{4[(\theta a_H-c)(\theta a_H+c)-\theta ^2a_L^2]},\ \ \mathrm{{if}}\ \eta \le \eta _2;\\ \frac{a_L}{a_H+a_L}+\frac{\theta ^2(a_H^2-a_L^2) [\theta (a_H-a_L)-c]+c^2a_L}{[\theta ^2(a_H^2-a_L^2)-c^2](a_H+a_L)}, \ \mathrm{{if}}\ \eta _2<\eta \le \eta _1;\\ 50\,\%, \ \ \ \ \ \mathrm{{if}}\ \eta _1<\eta \le \eta _{max}. \end{array} \right. \end{aligned}$$

(116)

Since $\theta (a_H-a_L)> c$ for all $\eta \le \eta _2(<\eta _1)$,

$$\begin{aligned}&\begin{array}{lll} \frac{2c(\theta a_H-c)+\theta ^2a_L^2}{4[(\theta a_H-c)(\theta a_H+c)-\theta ^2a_L^2]}>0\ \quad \mathrm{{for\ all\ }} \eta \le \eta _2, \end{array}\end{aligned}$$

(117)

$$\begin{aligned}&\begin{array}{lll} \frac{\theta ^2(a_H^2-a_L^2) [\theta (a_H-a_L)-c]+c^2a_L}{[\theta ^2(a_H^2-a_L^2)-c^2](a_H+a_L)}>0\ \quad \mathrm{{for\ all\ }} \eta _2<\eta \le \eta _1. \end{array} \end{aligned}$$

(118)

To summarize, we complete the proof of Theorem 3. $\square$

Proof of Theorem 4

By Theorem 1, together with checking the proof of Lemma 1, we obtain that, if $\eta \le \eta _2$,

$$\begin{aligned}&\begin{array}{lll} E_{p}(\eta )&{}=&{}\alpha (a_H-\delta \overline{Q})+(1-\alpha ) (a_L-\delta \frac{a_L}{2\delta })\\ &{}=&{}\alpha (a_H-\delta \frac{\theta a_H-c}{4\delta \theta }) +(1-\alpha )(a_L-\delta \frac{a_L}{2\delta })\\ &{}=&{}\frac{\alpha c}{4\theta }+\frac{3\alpha a_H+2(1-\alpha )a_L}{4}, \end{array}\end{aligned}$$

(119)

$$\begin{aligned}&\begin{array}{lll} SD_{p}(\eta )&{}=&{}\sqrt{\alpha (a_H-\delta \overline{Q}-E_{p} (\eta ))^2+(1-\alpha )(a_L-\delta \frac{a_L}{2\delta }-E_{p}(\eta ))^2}\\ &{}=&{}\sqrt{\alpha (a_H-\delta \frac{\theta a_H-c}{4\delta \theta } -E_{p}(\eta ))^2+(1-\alpha )(a_L-\delta \frac{a_L}{2\delta }-E_{p}(\eta ))^2}\\ &{}=&{}\sqrt{\alpha (1-\alpha )}(\frac{c}{4\theta }+\frac{3a_H-2a_L}{4}), \end{array} \end{aligned}$$

(120)

where $\theta =\alpha -\eta \sqrt{\alpha (1-\alpha )}.$ It is easy to see that $E_{p}(\eta )$ and $SD_{p}(\eta )$ both strictly increase with $\eta$ for all $\eta \le \eta _2$. Similarly, if $\eta _2<\eta \le \eta _{max}$,

$$\begin{aligned} \begin{array}{lll} E_{p}(\eta )&{}=&{}\alpha (a_H-\delta \overline{Q})+(1-\alpha )(a_L-\delta \overline{Q})\\ &{}=&{}\alpha (a_H-\delta \frac{\overline{B}-c}{4\delta }) +(1-\alpha )(a_L-\delta \frac{\overline{B}-c}{4\delta })\\ &{}=&{}\frac{3\overline{A}+\eta \sqrt{\alpha (1-\alpha )}(a_H-a_L)+c}{4}, \end{array} \end{aligned}$$

(121)

Since $\frac{\theta a_H-c}{4\delta \theta }\ge \frac{a_L}{2\delta }$ for all $\eta \le \eta _2$, whereas $\frac{\overline{B}-c}{4\delta }\le \frac{a_L}{2\delta }$ for all $\eta _2<\eta \le \eta _{max}$, we see from (119) and (121) that

$$\begin{aligned} \frac{\alpha c}{4[(\alpha -\eta _2\sqrt{\alpha (1-\alpha )})]}+\frac{3\alpha a_H+2(1-\alpha )a_L}{4}\le \frac{3\overline{A}+\eta _2 \sqrt{\alpha (1-\alpha )}(a_H-a_L)+c}{4}.\nonumber \\ \end{aligned}$$

(122)

Besides, we have

$$\begin{aligned} \begin{array}{lll} SD_{p}(\eta )&{}=&{}\sqrt{\alpha (a_H-\delta \overline{Q} -E_{p}(\eta ))^2+(1-\alpha )(a_L-\delta \overline{Q}-E_{p}(\eta ))^2}\\ &{}=&{}\sqrt{\alpha (a_H-\delta \frac{\overline{B}-c}{4\delta }-E_{p}(\eta ))^2 +(1-\alpha )(a_L-\delta \frac{\overline{B}-c}{4\delta }-E_{p}(\eta ))^2}\\ &{}=&{}\sqrt{\alpha (1-\alpha )}(a_H-a_L). \end{array} \end{aligned}$$

(123)

Obviously, $E_{p}(\eta )$ strictly increases with $\eta$ on $(\eta _2, \eta _{max}]$ and $SD_{p}(\eta )$ remains unchanged on $(\eta _2, \eta _{max}]$. To summarize, we complete the proof. $\square$

Appendix 2: Formulations related to the numerical experiment in Sect. 7

In Sect. 7, an extension of the model is considered with the intercept $a$ in demand curve (1) following a uniform distribution over $[m, n]$. Denote $f(\cdot )=\frac{1}{n-m}$, which is the probability density function of $a$. We first examine the decentralized supply chain and then the centralized one.

1.1 The case of decentralized supply chain

We first examine the retailer’s optimal pricing strategy (or equivalently, the product quantity released to the market) at time 1, given an order quantity $Q$ at time 0. Denote $\xi$ as any realization of $a$ in time 1, that is, the realized demand at time 1 is $q=\frac{(\xi -p)}{\delta }$. Thus, the problem faced by the retailer at time 1 can be formulated as

$$\begin{aligned} \max _{0\le q\le Q}p\frac{(\xi -p)}{\delta }. \end{aligned}$$

(124)

Solving the corresponding unconstrained problem of (124), we obtain the optimal solution $\bar{p}=\frac{\xi }{2}$. Further, it can be observed that if $q=\frac{\xi -\bar{p}}{\delta }\le Q$, i.e., $\xi \le 2\delta Q$, the optimal solution of (124), denoted by $p^*$, is $p^*=\bar{p}=\frac{\xi }{2}$, otherwise, the optimal solution of (124) will be achieved only at the boundary of the constraint, which is $p^*=\xi -\delta Q$. Thus, the profit obtained by the retailer at time 1, given an order quantity of $Q$ at time 0, can be formulated as

$$\begin{aligned} \begin{array}{lll} \pi _{wr}(Q,\xi ) &{}=&{}\left\{ \begin{array}{ll} \frac{\xi ^2}{4\delta }-wQ, &{} \hbox {if } \xi \le 2\delta Q;\\ (\xi -\delta Q-w)Q, &{} \hbox {if } \xi > 2\delta Q. \end{array} \right. \end{array} \end{aligned}$$

(125)

Further, the expected profit obtained by the retailer can be formulated as

$$\begin{aligned} \begin{array}{lll} E[\pi _{wr}(Q)]&{}=&{}\int ^{2\delta Q}_m(\frac{\xi ^2}{4\delta }-wQ)f(\xi )d\xi +\int ^{n}_{2\delta Q}(\xi -\delta Q-w)Qf(\xi )d\xi \\ &{}=&{}\frac{8\delta ^3Q^3-12n\delta ^2Q^2+6n^2\delta Q-m^3}{12\delta (n-m)}-wQ. \end{array} \end{aligned}$$

(126)

The standard deviation (SD) associated with the profit can be formulated as

$$\begin{aligned} \begin{array}{lll} SD[\pi _{wr}(Q)]\!&{}=&{}\!\sqrt{\int ^{2\delta Q}_{m}[\frac{\xi ^2}{4\delta }\!-\!wQ\!-\!E[\pi _{wr}(Q)]]^2f(\xi )d\xi \!+\!\int ^{n}_{2\delta Q}[(\xi \!-\!\delta Q\!-\!w)Q\!-\!E[\pi _{wr}(Q)]]^2f(\xi )d\xi }\\ &{}=&{}\sqrt{\frac{-64\delta ^5Q^5+240n\delta ^4 Q^4-240n^2\delta ^3Q^3+80n^3\delta ^2Q^2-3m^5}{240\delta ^2(n-m)}-A^2}, \end{array}\qquad \end{aligned}$$

(127)

where $A=\frac{8\delta ^3Q^3-12n\delta ^2Q^2+6n^2\delta Q-m^3}{12\delta (n-m)}$.

The retailer’s problem at time 0 is to determine an optimal order quantity for given $w$, which can be formulated as the following problem:

$$\begin{aligned} \max _{Q\ge 0}\{E[\pi _{wr}(Q)]-\eta SD[\pi _{wr}(Q)]\}. \end{aligned}$$

(128)

Denote $U_r(Q)=E[\pi _{wr}(Q)]-\eta SD[\pi _{wr}(Q)]$. Since it is quite challenging to find out whether $\frac{d^2 U_r(Q)}{d Q^2}$ is positive or negative, we deploy a numerical experiment approach to find out the optimal solution of (128) for a given interval of $\eta$ in which a unique solution, denoted by $Q^*(w)$, exists to maximize $U_r(Q)$.

We proceed to explore the equilibrium wholesale price for the supplier. Clearly, the supplier’s problem can be formulated as

$$\begin{aligned} \begin{array}{lll} \max _{w\ge 0}\{\pi _m(w)\}=\max _{w\ge 0}\{(w-c)Q^*(w)\}. \end{array} \end{aligned}$$

(129)

With a numerical study, we can find the equilibrium wholesale price $\bar{w}(\eta )$ that satisfies $\frac{d \pi _m(w)}{d w}|_{w=\bar{w}(\eta )}=0$, which in turn helps to identify the equilibrium order quantity $\bar{Q}(\eta )=\bar{Q}(\bar{w}(\eta ))$. Based on the equilibrium order quantity and wholesale price, we can obtain the (expected) profits received by the retailer and the supplier and the SD associated with the channel profit in equilibrium.

1.2 The case of centralized supply chain

We take the supplier and the retailer as a centralized entity. Then for this centralized entity, the profit obtained at time 1, given a production quantity of $Q$ at time 0, can be formulated as

$$\begin{aligned} \begin{array}{lll} \pi _c(Q,\xi ) &{}=&{}\left\{ \begin{array}{ll} \frac{\xi ^2}{4\delta }-cQ, &{} \hbox { if }\xi \le 2\delta Q;\\ (\xi -\delta Q-c)Q, &{} \hbox {if }\xi > 2\delta Q. \end{array} \right. \end{array}\end{aligned}$$

(130)

Hence, the expected profit obtained by this centralized entity can be formulated as

$$\begin{aligned} \begin{array}{lll} E[\pi _c(Q)]&{}=&{}\int ^{2\delta Q}_m(\frac{\xi ^2}{4\delta }-cQ)f(\xi )d\xi +\int ^{n}_{2\delta Q}(\xi -\delta Q-c)Qf(\xi )d\xi \\ &{}=&{}\frac{8\delta ^3Q^3-12n\delta ^2Q^2+6n^2\delta Q-m^3}{12\delta (n-m)}-cQ. \end{array} \end{aligned}$$

(131)

The SD associated with the profit can be formulated as

$$\begin{aligned} \begin{array}{lll} SD[\pi _c(Q)]\!&{}=&{}\!\sqrt{\int ^{2\delta Q}_{m}[\frac{\xi ^2}{4\delta }\!-\!cQ\!-\!E[\pi _c(Q)]]^2f(\xi )d\xi \!+\!\int ^{n}_{2\delta Q}[(\xi \!-\!\delta Q\!-\!c)Q\!-\!E[\pi _c(Q)]]^2f(\xi )d\xi }\\ &{}=&{}\sqrt{\frac{-64\delta ^5Q^5+240n\delta ^4 Q^4-240n^2\delta ^3Q^3+80n^3\delta ^2Q^2-3m^5}{240\delta ^2(n-m)}-A^2}, \end{array}\qquad \end{aligned}$$

(132)

where $A=\frac{8\delta ^3Q^3-12n\delta ^2Q^2+6n^2\delta Q-m^3}{12\delta (n-m)}$.

The problem faced by this centralized entity at time 0 is to determine an optimal production quantity for the channel, which can be formulated as the following problem:

$$\begin{aligned} \max _{Q\ge 0}\{E[\pi _c(Q)]-\eta SD[\pi _c(Q)]\}. \end{aligned}$$

(133)

Denote $U_c(Q)=E[\pi _c(Q)]-\eta SD[\pi _c(Q)]$. Since it is very challenging to find out whether $\frac{d^2 U_c(Q)}{d Q^2}$ is positive or negative, we deploy a numerical experiment approach to find out the optimal solution of (133) for a given interval of $\eta$ in which a unique solution, denoted by $\bar{Q}_c(\eta )$, exists to maximize $U_c(Q)$. After that, we can obtain the expected profit and the SD of the profit for the centralized entity at optimization.