Mean-Risk Analysis of Wholesale Price Contracts with Stochastic Price-Dependent Demand

Abstract

Given that risk is a pertinent issue in designing supply chain contracts with stochastic demand, Chap. 3 is devoted to developing a mean-risk analysis for the commonly adopted wholesale price contract. The research incorporates contract value risk into the wholesale price contract model. Regarding the contract value risk, it actually relates to the uncertainty in the true value of the contract and arises from various uncertainty sources inherent in the supply chain, such as demand uncertainty, price uncertainty, etc. In addition, given that the supply chain agents with different risk preferences will have different risk attitudes towards the contract value risk, which in turn affects their contracting decisions, the research also considers the degree of supply chain agents risk-aversion towards the contract value risk. This chapter makes the first attempt to assess the efficiency of wholesale price contracts, incorporating contract value risk and risk preferences attached to it; thereby some interesting managerial and academic insights are generated for supply chain contracts.

The research of this chapter is based on Zhao et al. (2014a).

Appendix: Proofs of the Main Results

Proof of Lemma 3.4.1.

First, to 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 $p_{H} = a_{H} -\delta q$, the retailer’s problem can be formulated as $\max \limits _{0\leq q\leq Q}q(a_{H} -\delta q)$. Solving this problem, the solution, denoted by $\overline{q}_{wH}$, is obtained as follows:

$$\displaystyle{ \overline{q}_{wH} = \left \{\!\!\begin{array}{llll} \frac{a_{H}} {2\delta },\ \ \mathrm{if}\ \frac{a_{H}} {2\delta } \leq Q;\\ Q,\ \ \ \mathrm{otherwise }. \end{array} \right. }$$

(3.8.1)

Likewise, when the realization of the demand curve at time 1 is $p_{L} = a_{L} -\delta q$, the retailer’s problem can be formulated as $\max \limits _{0\leq q\leq Q}q(a_{L} -\delta q)$. Solving this problem, the solution, denoted by $\overline{q}_{wL}$, is obtained as follows:

$$\displaystyle{ \overline{q}_{wL} = \left \{\!\!\begin{array}{llll} \frac{a_{L}} {2\delta },\ \ \mathrm{if}\ \frac{a_{L}} {2\delta } \leq Q;\\ Q,\ \ \ \mathrm{otherwise }. \end{array} \right. }$$

(3.8.2)

The following analyzes the retailer’s optimal order quantity at time 0, and this problem can be solved based on three cases of Q:

Case (i):

$0 \leq Q \leq \frac{a_{L}} {2\delta }$. For this case, by (3.8.1) and (3.8.2), the retailer’s expected profit can be obtained as

$$\displaystyle{ \begin{array}{lll} E\Pi _{wr1}(Q) =\alpha Q(a_{H} -\delta Q) + (1-\alpha )Q(a_{L} -\delta Q) - wQ,\end{array} }$$

(3.8.3)

and the corresponding variance of the profit can be obtained as

$$\displaystyle{ \begin{array}{lll} \sigma _{wr1}^{2}(Q)& =&\alpha [Q(a_{H} -\delta Q) - wQ - E\Pi _{wr1}(Q)]^{2} \\ & & + (1-\alpha )[Q(a_{L} -\delta Q) - wQ - E\Pi _{wr1}(Q)]^{2} \\ & =&\alpha [Q(a_{H} -\delta Q) - wQ -\alpha Q(a_{H} -\delta Q) - (1-\alpha )Q(a_{L} -\delta Q) + wQ]^{2} \\ & & + (1-\alpha )[Q(a_{L} -\delta Q) - wQ -\alpha Q(a_{H} -\delta Q) - (1-\alpha )Q(a_{L} -\delta Q) + wQ]^{2} \\ & =&\alpha (1-\alpha )Q^{2}(a_{H} - a_{L})^{2}.\end{array} }$$

(3.8.4)

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

$$\displaystyle{ \begin{array}{lll} \sigma _{wr1}(Q) = \sqrt{\alpha (1-\alpha )}Q(a_{H} - a_{L}).\end{array} }$$

(3.8.5)

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

$$\displaystyle{ \begin{array}{lll} \mathrm{P_{3A.1}:}\ \ \ &&\max \limits _{0\leq Q\leq \frac{a_{L}} {2\delta } }ED_{wr1}(Q) = E\Pi _{wr1}(Q) -\eta \sigma _{wr1}(Q),\end{array} }$$

(3.8.6)

where η indicates the retailer’s degree of risk-aversion towards the value risk of the wholesale price contract. Solving problem P_3A.1, we obtain the solution denoted by $\overline{Q}_{w1}$ as follows:

$$\displaystyle{ \overline{Q}_{w1}(\eta ) = \left \{\!\!\begin{array}{llll} \frac{a_{L}} {2\delta },\ \ \ \ \ \ \mathrm{if}\ \eta \leq \frac{\alpha (a_{H}-a_{L})-w} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})}; \\ \frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} {2\delta },\ \ \ \ \mathrm{if}\ \frac{\alpha (a_{H}-a_{L})-w} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} <\eta \leq \frac{\overline{A}-w} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})}; \\ 0,\ \ \ \ \ \ \ \ \mathrm{otherwise}.\end{array} \right. }$$

(3.8.7)

Case (ii):

$\frac{a_{L}} {2\delta } \leq Q \leq \frac{a_{H}} {2\delta }$. For this case, by (3.8.1) and (3.8.2), the retailer’s expected profit can be obtained as

$$\displaystyle{ \begin{array}{lll} E\Pi _{wr2}(Q) =\alpha Q(a_{H} -\delta Q) + (1-\alpha )\frac{a_{L}} {2\delta } (a_{L} -\delta \frac{a_{L}} {2\delta } ) - wQ,\end{array} }$$

(3.8.8)

and the corresponding variance of the profit can be obtained as

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

(3.8.9)

Since $Q(a_{H} -\delta Q) \geq \frac{a_{L}} {2\delta } (a_{H} -\delta \frac{a_{L}} {2\delta } ) > \frac{a_{L}} {2\delta } (a_{L} -\delta \frac{a_{L}} {2\delta } ) = \frac{a_{L}^{2}} {4\delta }$ for all $\frac{a_{L}} {2\delta } \leq Q \leq \frac{a_{H}} {2\delta }$, the SD of the retailer’s profit is given by

$$\displaystyle{ \begin{array}{lll} \sigma _{wr2}(Q) = \sqrt{\alpha (1-\alpha )}[Q(a_{H} -\delta Q) -\frac{a_{L}^{2}} {4\delta } ].\end{array} }$$

(3.8.10)

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

$$\displaystyle{ \begin{array}{lll} \mathrm{P_{3A.2}:}\ \ \ &&\max \limits _{\frac{a_{L}} {2\delta } \leq Q\leq \frac{a_{H}} {2\delta } }ED_{wr2} = E\Pi _{wr2}(Q) -\eta \sigma _{wr2}(Q).\\ \end{array} }$$

(3.8.11)

Solving problem P_3A.2, the solution, denoted by $\overline{Q}_{w2}$, is obtained as follows:

$$\displaystyle{ \overline{Q}_{w2}(\eta ) = \left \{\!\!\begin{array}{llll} \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}-w} {2\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]},\ \ \ \ \mathrm{if}\ \eta \leq \frac{\alpha (a_{H}-a_{L})-w} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})}; \\ \frac{a_{L}} {2\delta },\ \ \ \ \ \ \ \ \ \mathrm{if}\ \frac{\alpha (a_{H}-a_{L})-w} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} <\eta \leq \sqrt{ \frac{\alpha }{1-\alpha }}; \\ \frac{a_{H}} {2\delta },\ \ \ \ \ \ \ \ \ \mathrm{if}\ \sqrt{ \frac{\alpha } {1-\alpha }} <\eta.\end{array} \right. }$$

(3.8.12)

Case (iii):

$Q \geq \frac{a_{H}} {2\delta }$. Similarly, by (3.8.1) and (3.8.2), the retailer’s expected profit can be obtained as

$$\displaystyle{ \begin{array}{lll} E\Pi _{wr3}(Q) =\alpha \frac{a_{H}} {2\delta } (a_{H} -\delta \frac{a_{H}} {2\delta } ) + (1-\alpha )\frac{a_{L}} {2\delta } (a_{L} -\delta \frac{a_{L}} {2\delta } ) - wQ,\end{array} }$$

(3.8.13)

and the corresponding variance of the profit can be obtained as

$$\displaystyle{ \begin{array}{lll} \sigma _{wr3}^{2}(Q)& =&\alpha [\frac{a_{H}} {2\delta } (a_{H} -\delta \frac{a_{H}} {2\delta } ) - wQ - E\Pi _{wr3}(Q)]^{2} \\ & & + (1-\alpha )[\frac{a_{L}} {2\delta } (a_{L} -\delta \frac{a_{L}} {2\delta } ) - wQ - E\Pi _{wr3}(Q)]^{2} \\ & =&\alpha (1-\alpha )(\frac{a_{H}^{2}-a_{ L}^{2}} {4\delta } )^{2}.\end{array} }$$

(3.8.14)

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

$$\displaystyle{ \begin{array}{lll} \sigma _{wr3}(Q) = \sqrt{\alpha (1-\alpha )}(\frac{a_{H}^{2}-a_{ L}^{2}} {4\delta } ).\end{array} }$$

(3.8.15)

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

$$\displaystyle{ \begin{array}{lll} \mathrm{P_{3A.3}:}\ \ \ &&\max \limits _{Q\geq \frac{a_{H}} {2\delta } }ED_{wr3} = E\Pi _{wr3}(Q) -\eta \sigma _{wr3}(Q).\\ \end{array} }$$

(3.8.16)

Solving problem P_3A.3, the optimal solution, denoted by $\overline{Q}_{w3}$, is obtained as follows:

$$\displaystyle{ \overline{Q}_{w3}(\eta ) = \frac{a_{H}} {2\delta } \ \mathrm{for\ all}\ \eta. }$$

(3.8.17)

Summarizing (3.8.7), (3.8.12), and (3.8.17), we obtain

• (1) When $\eta \leq \frac{\alpha (a_{H}-a_{L})-w} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})}$, the optimal quantity for the retailer to order at time 0 is determined by

  $$\displaystyle{ \begin{array}{lll} \overline{Q}_{w}(\eta ) =\mathop{ \mathrm{arg\,max}}\limits \limits _{Q}\{ED_{wr1}(\frac{a_{L}} {2\delta } ),\ ED_{wr2}(\frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}-w} {2\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]} ),\ ED_{wr3}(\frac{a_{H}} {2\delta } )\}.\end{array} }$$

  (3.8.18)

  Since

  $$\displaystyle{ \begin{array}{lll} &&ED_{wr2}(\frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}-w} {2\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]} ) \geq ED_{wr2}(\frac{a_{L}} {2\delta } ) = ED_{wr1}(\frac{a_{L}} {2\delta } ), \\ &&ED_{wr2}(\frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}-w} {2\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]} ) \geq ED_{wr2}(\frac{a_{H}} {2\delta } ) = ED_{wr3}(\frac{a_{H}} {2\delta } ),\end{array} }$$

  (3.8.19)

  where the inequalities in (3.8.19) 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 $\sqrt{ \frac{\alpha }{1-\alpha }}$, the following proceeds the proof by considering two cases as follows:

  Case (i):

  $\sqrt{ \frac{\alpha }{1-\alpha }}\leq \frac{\overline{A}-w} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})}$, i.e., w ≤ a _L . Then by (3.8.7), (3.8.12), and (3.8.17), when $\frac{\alpha (a_{H}-a_{L})-w} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} <\eta \leq \sqrt{ \frac{\alpha }{1-\alpha }}$, the optimal quantity for the retailer to order at time 0 is determined by

  $$\displaystyle{ \begin{array}{lll} \overline{Q}_{w}(\eta ) =\mathop{ \mathrm{arg\,max}}\limits \limits _{Q}\{ED_{wr1}(\frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} {2\delta } ),\ ED_{wr2}(\frac{a_{L}} {2\delta } ),\ ED_{wr3}(\frac{a_{H}} {2\delta } )\}.\end{array} }$$

  (3.8.20)

  Since

  $$\displaystyle\begin{array}{rcl} \begin{array}{lll} ED_{wr1}(\frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} {2\delta } ) > ED_{wr1}(\frac{a_{L}} {2\delta } ) = ED_{wr2}(\frac{a_{L}} {2\delta } ) \geq ED_{wr2}(\frac{a_{H}} {2\delta } ) = ED_{wr3}(\frac{a_{H}} {2\delta } ),\end{array}& & {}\end{array}$$

  (3.8.21)

  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 $( \frac{\alpha (a_{H}-a_{L})-w} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} <)\sqrt{ \frac{\alpha }{1-\alpha }} <\eta \leq \frac{\overline{A}-w} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})}$, the optimal quantity for the retailer to order at time 0 is determined by

  $$\displaystyle{ \begin{array}{lll} \overline{Q}_{w}(\eta ) =\mathop{ \mathrm{arg\,max}}\limits \limits _{Q}\{ED_{wr1}(\frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} {2\delta } ),\ ED_{wr2}(\frac{a_{H}} {2\delta } ),\ ED_{wr3}(\frac{a_{H}} {2\delta } )\}.\end{array} }$$

  (3.8.22)

  Since

  $$\displaystyle{ \begin{array}{lll} ED_{wr1}(\frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} {2\delta } ) - ED_{wr2}(\frac{a_{H}} {2\delta } )(= ED_{wr3}(\frac{a_{H}} {2\delta } )) \\ =\frac{[\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})-w]^{2}} {4\delta } - [\frac{\alpha -\eta \sqrt{\alpha (1-\alpha )}} {4\delta } (a_{H}^{2} - a_{ L}^{2}) + \frac{1} {4\delta }(a_{L}^{2} - 2wa_{ H})] \\ =\frac{1} {4\delta }[((\alpha -\eta \sqrt{\alpha (1-\alpha )})(a_{H} - a_{L}) - w)^{2} + 2w(a_{ H} - a_{L}) \\ & & - (\alpha -\eta \sqrt{\alpha (1-\alpha )})(a_{H} - a_{L})^{2}] \\ >0\ (\mathrm{note\ that}\ \eta > \sqrt{ \frac{\alpha }{1-\alpha }}\Rightarrow \alpha -\eta \sqrt{\alpha (1-\alpha )} < 0),\end{array} }$$

  (3.8.23)

  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 (3.8.7), (3.8.12), and (3.8.17), when $\frac{\alpha (a_{H}-a_{L})-w} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} <\eta \leq \frac{\overline{A}-w} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})}(< \sqrt{ \frac{\alpha }{1-\alpha }})$, the optimal quantity for the retailer to order at time 0 is determined by

  $$\displaystyle{ \begin{array}{lll} \overline{Q}_{w}(\eta ) =\mathop{ \mathrm{arg\,max}}\limits \limits _{Q}\{ED_{wr1}(\frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} {2\delta } ),\ ED_{wr2}(\frac{a_{L}} {2\delta } ),\ ED_{wr3}(\frac{a_{H}} {2\delta } )\}.\end{array} }$$

  (3.8.24)

  Since

  $$\displaystyle\begin{array}{rcl} \begin{array}{lll} ED_{wr1}(\frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} {2\delta } ) > ED_{wr1}(\frac{a_{L}} {2\delta } ) = ED_{wr2}(\frac{a_{L}} {2\delta } ) \geq ED_{wr2}(\frac{a_{H}} {2\delta } ) = ED_{wr3}(\frac{a_{H}} {2\delta } ),\end{array}& & {}\end{array}$$

  (3.8.25)

  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 $( \frac{\alpha (a_{H}-a_{L})-w} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} <) \frac{\overline{A}-w} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} <\eta \leq \sqrt{ \frac{\alpha }{1-\alpha }}$, the optimal quantity for the retailer to order at time 0 is determined by

  $$\displaystyle{ \begin{array}{lll} \overline{Q}_{w}(\eta ) =\mathop{ \mathrm{arg\,max}}\limits \limits _{Q}\{ED_{wr1}(0),\ ED_{wr2}(\frac{a_{L}} {2\delta } ),\ ED_{wr3}(\frac{a_{H}} {2\delta } )\}.\end{array} }$$

  (3.8.26)

  Since

  $$\displaystyle{ \begin{array}{lll} ED_{wr1}(0) > ED_{wr1}(\frac{a_{L}} {2\delta } ) = ED_{wr2}(\frac{a_{L}} {2\delta } ) \geq ED_{wr2}(\frac{a_{H}} {2\delta } ) = ED_{wr3}(\frac{a_{H}} {2\delta } ),\end{array} }$$

  (3.8.27)

  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 (3.8.7), (3.8.12), and (3.8.17), the optimal quantity for the retailer to order at time 0 is determined by

  $$\displaystyle{ \begin{array}{lll} \overline{Q}_{w}(\eta ) =\mathop{ \mathrm{arg\,max}}\limits \limits _{Q}\{ED_{wr1}(0),\ ED_{wr2}(\frac{a_{H}} {2\delta } ),\ ED_{wr3}(\frac{a_{H}} {2\delta } )\}.\end{array} }$$

  (3.8.28)

  Since

  $$\displaystyle{ \begin{array}{lll} ED_{wr2}(\frac{a_{H}} {2\delta } ) = ED_{wr3}(\frac{a_{H}} {2\delta } ) < 0 = ED_{wr1}(0),\end{array} }$$

  (3.8.29)

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

Summarizing the above analysis, we obtain that if w ≤ a _L , the optimal quantity for the retailer to order at time 0 is given by

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

(3.8.30)

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

$$\displaystyle{ \overline{Q}_{w}(\eta ) = \left \{\!\!\begin{array}{ll} \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}-w} {2\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]}, &\mathrm{if}\ \eta \leq \frac{\alpha (a_{H}-a_{L})-w} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})}; \\ \frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} {2\delta },&\mathrm{if}\ \frac{\alpha (a_{H}-a_{L})-w} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} <\eta \leq \frac{\overline{A}-w} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})}; \\ 0, &\mathrm{if}\ \frac{\overline{A}-w} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} <\eta \leq \sqrt{ \frac{\alpha }{1-\alpha }}; \\ 0, &\mathrm{otherwise}. \end{array} \right. }$$

(3.8.31)

Summarizing (3.8.30) and (3.8.31), we obtain

$$\displaystyle{ \overline{Q}_{w}(\eta ) = \left \{\!\!\begin{array}{ll} \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}-w} {2\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]}, &\mathrm{if}\ \eta \leq \frac{\alpha (a_{H}-a_{L})-w} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})}; \\ \frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} {2\delta },&\mathrm{if}\ \frac{\alpha (a_{H}-a_{L})-w} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} <\eta \leq \frac{\overline{A}-w} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})}; \\ 0, &\mathrm{otherwise}. \end{array} \right. }$$

(3.8.32)

Equivalently, we transform (3.8.32) into

$$\displaystyle{ \overline{Q}_{\eta }(w) = \left \{\!\!\begin{array}{ll} \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}-w} {2\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]}, &\mathrm{if}\ w \leq [\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_{H} - a_{L}); \\ \frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} {2\delta },&\mathrm{if}\ [\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_{H} - a_{L}) < w \\ &\qquad \qquad \leq \overline{A} -\eta \sqrt{\alpha (1-\alpha )}(a_{H} - a_{L}); \\ 0, &\mathrm{otherwise}. \end{array} \right. }$$

(3.8.33)

Hence, the proof is completed. □ 

Proof of Theorem 3.4.2.

Theorem 3.4.2 will be shown based on two cases as follows:

Case (i)::

$[\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_{H} - a_{L}) \geq c$, i.e., $\eta \leq \frac{\alpha (a_{H}-a_{L})-c} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})}(\triangleq \eta _{1})$. By Lemma 3.4.1, the supplier’s problem in this case can be formulated as

$$\displaystyle{ \begin{array}{lll} \mathrm{P_{3A.4}:}\ \ \ \ \ \overline{w} =\mathop{ \mathrm{arg\,max}}\limits \limits _{w}\{\overline{H}_{1},\overline{H}_{2}\},\end{array} }$$

(3.8.34)

where

$$\displaystyle{ \begin{array}{lll} &&\overline{H}_{1} =\max \limits _{c\leq w\leq [\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_{ H}-a_{L})}\{\frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}-w} {2\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]} \}(w - c), \\ &&\overline{H}_{2} =\max \limits _{[\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_{ H}-a_{L})<w\leq \overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})}[\frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} {2\delta } ](w - c).\end{array} }$$

(3.8.35)

Solving problem $\overline{H}_{1}$, we obtain the solution denoted by $\overline{w}_{1}(\eta )$ as follows:

$$\displaystyle{ \overline{w}_{1}(\eta ) = \left \{\!\!\begin{array}{llll} \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}+c} {2},\ \ \ \ \ \ \ \ \ \ \mathrm{if}\ c \leq [\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_{H} - 2a_{L}); \\ (\alpha -\eta \sqrt{\alpha (1-\alpha )})(a_{H} - a_{L}),\ \ \ \mathrm{if}\ c > [\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_{H} - 2a_{L}). \end{array} \right. }$$

(3.8.36)

Solving problem $\overline{H}_{2}$, the solution, denoted by $\overline{w}_{2}(\eta )$, is obtained as follows:

$$\displaystyle{ \overline{w}_{2}(\eta ) = \left \{\!\!\begin{array}{llll} [\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_{H} - a_{L}),\ \ \mathrm{if}\ c \leq [\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_{H} - a_{L}) - a_{L}; \\ \frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})+c} {2},\ \ \ \ \ \ \ \ \ \ \mathrm{if}\ c > [\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_{H} - a_{L}) - a_{L}. \end{array} \right. }$$

(3.8.37)

For ease of exposition, we denote

$$\displaystyle{ \begin{array}{lll} c_{1} = [\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_{H} - a_{L}) - a_{L}\ \mathrm{and}\ c_{2} = [\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_{H} - 2a_{L}). \end{array} }$$

(3.8.38)

Since c ₂ ≥ c ₁, summarizing (3.8.36) and (3.8.37), it is obtained that

$$\displaystyle{ \overline{w}(\eta ) = \left \{\!\!\begin{array}{ll} \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}+c} {2}, &\mathrm{if}\ c < c_{1}; \\ \frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})+c} {2},&\mathrm{if}\ c > c_{2}; \\ \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}+c} {2}, &\mathrm{if}\ c_{1} \leq c \leq c_{2}\ \mathrm{and}\ \\ &\overline{H}_{1}(\frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}+c} {2} ) \geq \overline{H}_{2}(\frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})+c} {2} ); \\ \frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})+c} {2},&\mathrm{if}\ c_{1} \leq c \leq c_{2}\ \mathrm{and} \\ &\overline{H}_{1}(\frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}+c} {2} ) < \overline{H}_{2}(\frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})+c} {2} ). \end{array} \right. }$$

(3.8.39)

It is easy to obtain

$$\displaystyle{ \begin{array}{lll} \overline{H}_{1}(\frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}+c} {2} )& =&(\frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}-\frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}+c} {2} } {2\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]} )(\frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}+c} {2} - c) \\ & =&\frac{[(\alpha -\eta \sqrt{\alpha (1-\alpha )})a_{H}-c]^{2}} {8\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]}, \end{array} }$$

(3.8.40)

$$\displaystyle{ \begin{array}{lll} & &\overline{H}_{2}(\frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})+c} {2} ) \\ & =&[\frac{\overline{A}-\frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})+c} {2} -\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} {2\delta } ](\frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})+c} {2} - c) \\ & =&\frac{[\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})-c]^{2}} {8\delta }.\end{array} }$$

(3.8.41)

Therefore,

$$\displaystyle{ \begin{array}{lll} & &\overline{H}_{1}(\frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}+c} {2} ) -\overline{H}_{2}(\frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})+c} {2} ) \\ & =&\frac{[(\alpha -\eta \sqrt{\alpha (1-\alpha )})a_{H}-c]^{2}} {8\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]} -\frac{[\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})-c]^{2}} {8\delta } \\ & =&\frac{1-[\alpha -\eta \sqrt{\alpha (1-\alpha )}]} {8\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]} [c^{2} - 2[\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_{H} - a_{L})c + [(\alpha -\eta \sqrt{\alpha (1-\alpha )})a_{H}]^{2} \\ & & - 2[\alpha -\eta \sqrt{\alpha (1-\alpha )}]^{2}a_{H}a_{L} - [\alpha -\eta \sqrt{\alpha (1-\alpha )}][1 - (\alpha -\eta \sqrt{\alpha (1-\alpha )})]a_{L}^{2}]. \end{array} }$$

(3.8.42)

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

$$\displaystyle\begin{array}{rcl} \begin{array}{lll} &&c^{2} - 2[\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_{H} - a_{L})c + [(\alpha -\eta \sqrt{\alpha (1-\alpha )})a_{H}]^{2} \\ & & - 2[\alpha -\eta \sqrt{\alpha (1-\alpha )}]^{2}a_{H}a_{L} - [\alpha -\eta \sqrt{\alpha (1-\alpha )}][1 - (\alpha -\eta \sqrt{\alpha (1-\alpha )})]a_{L}^{2} \leq 0.\end{array} & &{}\end{array}$$

(3.8.43)

Solving inequality (3.8.43), we obtain that

$$\displaystyle{ \begin{array}{lll} \overline{H}_{1}(\frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}+c} {2} ) \leq \overline{H}_{2}(\frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})+c} {2} )\end{array} }$$

(3.8.44)

holds iff c ∈ [c ₃, c ₄], where

$$\displaystyle{ \begin{array}{lll} c_{3} & =&a_{H}[\alpha -\eta \sqrt{\alpha (1-\alpha )}] - a_{L}[\sqrt{\alpha -\eta \sqrt{\alpha (1-\alpha )}} + (\alpha -\eta \sqrt{\alpha (1-\alpha )})], \\ c_{4} & =&a_{H}[\alpha -\eta \sqrt{\alpha (1-\alpha )}] + a_{L}(\sqrt{\alpha -\eta \sqrt{\alpha (1-\alpha )}} - [\alpha -\eta \sqrt{\alpha (1-\alpha )}]). \end{array} }$$

(3.8.45)

Since c ₁ ≤ c ₃ ≤ c ₂ ≤ c ₄, we can simplify (3.8.39) as

$$\displaystyle{ \overline{w}_{1}(\eta ) = \left \{\!\!\begin{array}{llll} \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}+c} {2},\ \ \ \ \ \ \mathrm{if}\ c \leq c_{3}; \\ \frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})+c} {2},\ \ \ \ \ \ \ \ \ \mathrm{if}\ c > c_{3}. \end{array} \right. }$$

(3.8.46)

In addition,

$$\displaystyle{ \begin{array}{lll} c \leq c_{3} & \Longleftrightarrow & (a_{H} - a_{L})[\alpha -\eta \sqrt{\alpha (1-\alpha )}] - a_{L}\sqrt{\alpha -\eta \sqrt{\alpha (1-\alpha )}} - c \geq 0 \\ &\Longleftrightarrow&\sqrt{\alpha -\eta \sqrt{\alpha (1-\alpha )}} \leq \frac{a_{L}-\sqrt{a_{L }^{2 }+4c(a_{H } -a_{L } )}} {2(a_{H}-a_{L})} \ \mathrm{or} \\ &&\sqrt{\alpha -\eta \sqrt{\alpha (1-\alpha )}} \geq \frac{a_{L}+\sqrt{a_{L }^{2 }+4c(a_{H } -a_{L } )}} {2(a_{H}-a_{L})}. \end{array} }$$

(3.8.47)

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

$$\displaystyle{ \begin{array}{lll} c \leq c_{3} & \Longleftrightarrow & \sqrt{\alpha -\eta \sqrt{\alpha (1-\alpha )}} \geq \frac{a_{L}+\sqrt{a_{L }^{2 }+4c(a_{H } -a_{L } )}} {2(a_{H}-a_{L})} \\ & \Longleftrightarrow & \eta \leq \eta _{2},\end{array} }$$

(3.8.48)

where

$$\displaystyle{ \begin{array}{lll} \eta _{2} = \frac{4(a_{H}-a_{L})[\alpha (a_{H}-a_{L})-c]-2a_{L}^{2}-2a_{ L}\sqrt{a_{L }^{2 }+4c(a_{H } -a_{L } )}} {4\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})^{2}}. \end{array} }$$

(3.8.49)

It is easy to obtain that

$$\displaystyle{ \begin{array}{lll} \eta _{1} -\eta _{2} & =& \frac{\alpha (a_{H}-a_{L})-c} {\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} -\frac{4(a_{H}-a_{L})[\alpha (a_{H}-a_{L})-c]-2a_{L}^{2}-2a_{ L}\sqrt{a_{L }^{2 }+4c(a_{H } -a_{L } )}} {4\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})^{2}} \\ & =&\frac{2a_{L}^{2}+2a_{ L}\sqrt{a_{L }^{2 }+4c(a_{H } -a_{L } )}} {4\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})^{2}} > 0, \end{array} }$$

(3.8.50)

i.e., η ₁ > η ₂. Hence, the equilibrium wholesale price in the case of η ≤ η ₁ is given by

$$\displaystyle{ \overline{w}_{1}(\eta ) = \left \{\!\!\begin{array}{llll} \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}+c} {2},\ \ \ \ \ \ \mathrm{if}\ \eta \leq \eta _{2}; \\ \frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})+c} {2},\ \ \ \ \ \ \ \ \ \mathrm{if}\ \eta _{2} <\eta \leq \eta _{1}. \end{array} \right. }$$

(3.8.51)

Case (ii)::

$[\alpha -\eta \sqrt{\alpha (1-\alpha )}](a_{H} - a_{L}) < c$, i.e., η > η ₁. By Lemma 3.4.1, the supplier’s problem in the case can be formulated as

$$\displaystyle{ \begin{array}{lll} \mathrm{(P_{3A.5}):}\ \ \ \ \ \max \limits _{c\leq w\leq \overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_{ H}-a_{L})}[\frac{\overline{A}-w-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})} {2\delta } ](w - c),\end{array} }$$

(3.8.52)

Solving problem (P_3A.5), we obtain the solution denoted by $\overline{w}_{2}$ as

$$\displaystyle{ \begin{array}{lll} \overline{w}_{2}(\eta ) = \frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})+c} {2} \ \mathrm{for\ all}\ \eta \in (\eta _{1},\eta _{max}]. \end{array} }$$

(3.8.53)

Summarizing (3.8.51) and (3.8.53), the equilibrium wholesale price is obtained as

$$\displaystyle{ \overline{w}(\eta ) = \left \{\!\!\begin{array}{llll} \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}+c} {2},\ \ \ \ \ \ \mathrm{if}\ \eta \leq \eta _{2}; \\ \frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})+c} {2},\ \ \ \ \ \ \mathrm{if}\ \eta _{2} <\eta \leq \eta _{max}. \end{array} \right. }$$

(3.8.54)

Substituting (3.8.54) into (3.8.32), the equilibrium order quantity of the retailer is obtained as

$$\displaystyle{ \overline{Q}(\eta ) = \left \{\!\!\begin{array}{llll} \frac{[\alpha -\eta \sqrt{\alpha (1-\alpha )}]a_{H}-c} {4\delta [\alpha -\eta \sqrt{\alpha (1-\alpha )}]},\ \ \ \ \ \ \ \mathrm{if}\ \eta \leq \eta _{2}; \\ \frac{\overline{A}-\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})-c} {4\delta },\ \ \mathrm{if}\ \eta _{2} <\eta \leq \eta _{max}. \end{array} \right. }$$

(3.8.55)

Thus, the proof is completed. □ 

Proof of Theorem 3.4.3.

For ease of exposition, 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. Furthermore, if no confusion, denote $\theta (\eta )$, $\overline{B}(\eta )$, $\overline{w}(\eta ),$ and $\overline{Q}(\eta )$ simply by $\theta$, $\overline{B}$, $\overline{w},$ and $\overline{Q}$, respectively. First to show Theorem 3.4.3(1). By (3.8.54) and (3.8.55), it is obtained that if η ≤ η ₂,

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

(3.8.56)

It is easy to obtain

$$\displaystyle{ \begin{array}{llll} \frac{d\overline{R}_{ws}(\eta )} {d\eta } = -\frac{\sqrt{\alpha (1-\alpha )}(\theta a_{H}+c)(\theta a_{H}-c)} {8\delta \theta ^{2}}.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array} }$$

(3.8.57)

Obviously, (3.8.57) < 0 for all η ≤ η ₂. Therefore, $\overline{R}_{ws}(\eta )$ strictly decreases with η for all η ≤ η ₂. Similarly, if η ₂ < η ≤ η _max , by (3.8.54) and (3.8.55), it is obtained that

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

(3.8.58)

$$\displaystyle{ \begin{array}{llll} &&\frac{d\overline{R}_{ws}(\eta )} {d\eta } = -\frac{\sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})(\overline{B}-c)} {4\delta }.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array} }$$

(3.8.59)

Since $\overline{B} > c$ for all η ∈ (η ₂, η _max ), (3.8.59) < 0 for all η ∈ (η ₂, η _max ). Therefore, $\overline{R}_{ws}(\eta )$ strictly decreases with η on (η ₂, η _max ]. To summarize, the result given in Theorem 3.4.3(1) is derived.

The following proceeds to show Theorem 3.4.3(2). Likewise, by (3.8.54) and (3.8.55), together with checking the proofs of Lemma 3.4.1 and Theorem 3.4.2, it is obtained that if η ≤ η ₂,

$$\displaystyle{ \begin{array}{lll} E_{wr}(\eta )& =&\alpha \overline{Q}(a_{H} -\delta \overline{Q}) + (1-\alpha )\frac{a_{L}} {2\delta } (a_{L} -\delta \frac{a_{L}} {2\delta } ) -\overline{w}\overline{Q} \\ & =&\alpha (\frac{\theta a_{H}-c} {4\delta \theta } )(a_{H} -\delta \frac{\theta a_{H}-c} {4\delta \theta } ) + (1-\alpha )\frac{a_{L}} {2\delta } (a_{L} -\delta \frac{a_{L}} {2\delta } ) - (\frac{\theta a_{H}+c} {2} )(\frac{\theta a_{H}-c} {4\delta \theta } ) \\ & =&\frac{(\theta a_{H}-c)^{2}} {16\delta \theta } + \frac{(1-\theta )a_{L}^{2}} {4\delta } +\eta \sqrt{\alpha (1-\alpha )}[\frac{(\theta a_{H}-c)(3\theta a_{H}+c)} {16\delta \theta ^{2}} -\frac{a_{L}^{2}} {4\delta } ],\end{array} }$$

(3.8.60)

$$\displaystyle{ \begin{array}{lll} SD_{wr}(\eta )& =&\sqrt{ \alpha (1-\alpha )}[\overline{Q}(a_{H} -\delta \overline{Q}) -\frac{a_{L}^{2}} {4\delta } ] \\ & =&\sqrt{ \alpha (1-\alpha )}[\frac{\theta a_{H}-c} {4\delta \theta } (a_{H} -\delta \frac{\theta a_{H}-c} {4\delta \theta } ) -\frac{a_{L}^{2}} {4\delta } ] \\ & =&\sqrt{ \alpha (1-\alpha )}[\frac{(\theta a_{H}-c)(3\theta a_{H}+c)} {16\delta \theta ^{2}} -\frac{a_{L}^{2}} {4\delta } ].\end{array} }$$

(3.8.61)

Since for all η ≤ η ₂,

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

(3.8.62)

SD _wr (η) strictly decreases with η for all η ≤ η ₂. Similarly, if η ₂ < η ≤ η _max , it is obtained

$$\displaystyle{ \begin{array}{lll} E_{wr}(\eta )& =&\alpha \overline{Q}(a_{H} -\delta \overline{Q}) + (1-\alpha )\overline{Q}(a_{L} -\delta \overline{Q}) -\overline{w}\overline{Q} \\ & =&\alpha \frac{\overline{B}-c} {4\delta } (a_{H} -\delta \frac{\overline{B}-c} {4\delta } ) + (1-\alpha )\frac{\overline{B}-c} {4\delta } (a_{L} -\delta \frac{\overline{B}-c} {4\delta } ) - (\frac{\overline{B}+c} {2} )(\frac{\overline{B}-c} {4\delta } ) \\ & =&\frac{(\overline{B}-c)^{2}} {16\delta } + \frac{\eta \sqrt{\alpha (1-\alpha )}} {4\delta } (\overline{B} - c)(a_{H} - a_{L}), \end{array} }$$

(3.8.63)

$$\displaystyle{ \begin{array}{lll} SD_{wr}(\eta )& =&\sqrt{ \alpha (1-\alpha )}\overline{Q}(a_{H} - a_{L}) \\ & =&\frac{\sqrt{\alpha (1-\alpha )}} {4\delta } (\overline{B} - c)(a_{H} - a_{L}).\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array} }$$

(3.8.64)

Since

$$\displaystyle{ \begin{array}{lll} \frac{dSD_{wr}(\eta )} {d\eta } = -\frac{\alpha (1-\alpha )} {4\delta } (a_{H} - a_{L})^{2} < 0, \end{array} }$$

(3.8.65)

SD _wr (η) strictly decreases with η for all η ∈ (η ₂, η _max ]. Thus, the proof is completed. □ 

Proof of Lemma 3.5.1.

Using the marginal production cost c to replace the wholesale price w in the proof of Lemma 3.4.1, the system-wide optimal production quantity for the centralized entity can be obtained in a similar way as follows:

$$\displaystyle{ \overline{Q}_{c}(\eta ) = \left \{\!\!\begin{array}{llll} \frac{\theta a_{H}-c} {2\delta \theta },\ \ \ \ \ \ \ \mathrm{if}\ \eta \leq \eta _{1}; \\ \frac{\overline{B}-c} {2\delta },\ \ \ \ \ \ \ \ \ \ \mathrm{if}\ \eta _{1} <\eta \leq \eta _{max}, \end{array} \right. }$$

(3.8.66)

where

$$\displaystyle{ \begin{array}{lll} \theta =\alpha -\eta \sqrt{\alpha (1-\alpha )},\ \ \overline{B} = \overline{A} -\eta \sqrt{\alpha (1-\alpha )}(a_{H} - a_{L}). \end{array} }$$

(3.8.67)

Putting w = c in (3.8.8) and then substituting (3.8.66) into it, it is obtained that if η ≤ η ₁,

$$\displaystyle{ \begin{array}{lll} E_{c}(\eta )& =&\alpha \overline{Q}_{c}(a_{H} -\delta \overline{Q}_{c}) + (1-\alpha )\frac{a_{L}} {2\delta } (a_{L} -\delta \frac{a_{L}} {2\delta } ) - c\overline{Q}_{c} \\ & =&\alpha \frac{\theta a_{H}-c} {2\delta \theta } (a_{H} -\delta \frac{\theta a_{H}-c} {2\delta \theta } ) + (1-\alpha )\frac{a_{L}} {2\delta } (a_{L} -\delta \frac{a_{L}} {2\delta } ) - c\frac{\theta a_{H}-c} {2\delta \theta } \\ & =&\frac{\theta (\theta a_{H}-c)^{2}+\theta ^{2}(1-\theta )a_{ L}^{2}+\eta \sqrt{\alpha (1-\alpha )}[(\theta a_{ H}-c)(\theta a_{H}+c)-\theta ^{2}a_{ L}^{2}]} {4\delta \theta ^{2}} \\ & =&\frac{H(\eta )} {4\delta \theta ^{2}},\end{array} }$$

(3.8.68)

where, for ease of exposition in the following, it is denoted that

$$\displaystyle{ \begin{array}{lll} \theta (\theta a_{H} - c)^{2} +\theta ^{2}(1-\theta )a_{L}^{2} +\eta \sqrt{\alpha (1-\alpha )}[(\theta a_{H} - c)(\theta a_{H} + c) -\theta ^{2}a_{L}^{2}] \triangleq H(\eta ).\end{array} }$$

(3.8.69)

Substituting (3.8.66) into (3.8.10), it is obtained that

$$\displaystyle{ \begin{array}{lll} SD_{c}(\eta )& =&\sqrt{ \alpha (1-\alpha )}[\overline{Q}_{c}(a_{H} -\delta \overline{Q}_{c}) -\frac{a_{L}^{2}} {4\delta } ] \\ & =&\sqrt{ \alpha (1-\alpha )}[\frac{\theta a_{H}-c} {2\delta \theta } (a_{H} -\delta \frac{\theta a_{H}-c} {2\delta \theta } ) -\frac{a_{L}^{2}} {4\delta } ] \\ & =&\sqrt{ \alpha (1-\alpha )}[\frac{(\theta a_{H}-c)(\theta a_{H}+c)-\theta ^{2}a_{ L}^{2}} {4\delta \theta ^{2}} ].\end{array} }$$

(3.8.70)

Similarly, if η ₁ < η ≤ η _max , by setting w = c in (3.8.3) and then substituting (3.8.66) into it, we obtain

$$\displaystyle{ \begin{array}{lll} E_{c}(\eta )& =&\alpha \overline{Q}_{c}(a_{H} -\delta \overline{Q}_{c}) + (1-\alpha )\overline{Q}_{c}(a_{L} -\delta \overline{Q}_{c}) - c\overline{Q}_{c} \\ & =&\alpha \frac{\overline{B}-c} {2\delta } (a_{H} -\delta \frac{\overline{B}-c} {2\delta } ) + (1-\alpha )\frac{\overline{B}-c} {2\delta } (a_{L} -\delta \frac{\overline{B}-c} {2\delta } ) - c\frac{\overline{B}-c} {2\delta } \\ & =&\frac{(\overline{B}-c)^{2}+2\eta \sqrt{\alpha (1-\alpha )}(\overline{B}-c)(a_{ H}-a_{L})} {4\delta }.\end{array} }$$

(3.8.71)

Substituting (3.8.66) into (3.8.5), we have

$$\displaystyle{ \begin{array}{lll} SD_{c}(\eta )& =&\sqrt{ \alpha (1-\alpha )}\overline{Q}_{c}(a_{H} - a_{L}) \\ & =&\frac{\sqrt{\alpha (1-\alpha )}} {2\delta } (\overline{B} - c)(a_{H} - a_{L}).\end{array} }$$

(3.8.72)

To summarize, the proof is completed. □ 

Proof of Theorem 3.5.2.

First to show Theorem 3.5.2(i). It can be obtained by Theorem 3.4.3 that the expected channel profit in the decentralized supply chain  is

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

(3.8.73)

where

$$\displaystyle{ \begin{array}{lll} \theta =\alpha -\eta \sqrt{\alpha (1-\alpha )},\ \ \overline{B} = \overline{A} -\eta \sqrt{\alpha (1-\alpha )}(a_{H} - a_{L}), \end{array} }$$

(3.8.74)

$$\displaystyle{ \begin{array}{lll} H(\eta ) =\theta (\theta a_{H} - c)^{2} +\theta ^{2}(1-\theta )a_{L}^{2} +\eta \sqrt{\alpha (1-\alpha )}[(\theta a_{H} - c)(\theta a_{H} + c) -\theta ^{2}a_{L}^{2}].\end{array} }$$

(3.8.75)

Comparing (3.8.73) with (3.8.68) and (3.8.71), it follows that

$$\displaystyle{ EFFe(\eta ) = \left \{\!\!\begin{array}{llll} 75\,\% + \frac{\theta ^{2}(1-\theta )a_{ L}^{2}-\eta \sqrt{\alpha (1-\alpha )}[2c(\theta a_{ H}-c)+\theta ^{2}a_{ L}^{2}]} {4H(\eta )},\ \ \ \ \ \ \ \ \ \ \ \mathrm{if}\ \eta \leq \eta _{2}; \\ 75\,\% + \frac{3\theta ^{2}(\overline{B}-c)^{2}+4\theta ^{2}\eta \sqrt{\alpha (1-\alpha )}(\overline{B}-c)(a_{ H}-a_{L})-3H(\eta )} {4H(\eta )},\ \ \ \mathrm{if}\ \eta _{2} <\eta \leq \eta _{1}; \\ 75\,\% - \frac{\eta \sqrt{\alpha (1-\alpha )}(\overline{B}-c)(a_{H}-a_{L})} {2[(\overline{B}-c)^{2}+2\eta \sqrt{\alpha (1-\alpha )}(\overline{B}-c)(a_{H}-a_{L})]},\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{if}\eta _{1} <\eta \leq \eta _{max}. \end{array} \right. }$$

(3.8.76)

It can be obtained that

$$\displaystyle{ \begin{array}{lll} & &\theta ^{2}(1-\theta )a_{L}^{2} -\eta \sqrt{\alpha (1-\alpha )}[2c(\theta a_{H} - c) +\theta ^{2}a_{L}^{2}] \\ & =&\theta ^{2}(1-\theta )a_{L}^{2} +\theta [2c(\theta a_{H} - c) +\theta ^{2}a_{L}^{2}] -\alpha [2c(\theta a_{H} - c) +\theta ^{2}a_{L}^{2}] \\ & =&[(1-\alpha )a_{L}^{2} + 2ca_{H}]\theta ^{2} - 2c(\alpha a_{H} + c)\theta + 2\alpha c^{2}. \end{array} }$$

(3.8.77)

Substituting (3.8.77) into (3.8.76) derives (3.5.5) in Theorem 3.5.2(i). For ease of exposition, denote

$$\displaystyle{ \begin{array}{lll} H_{1}(\eta ) = [(1-\alpha )a_{L}^{2} + 2ca_{H}]\theta ^{2} - 2c(\alpha a_{H} + c)\theta + 2\alpha c^{2}. \end{array} }$$

(3.8.78)

Since

$$\displaystyle{ \frac{d^{2}H_{1}(\eta )} {d\eta ^{2}} = 2\alpha (1-\alpha )[(1-\alpha )a_{L}^{2} + 2ca_{ H}] > 0, }$$

(3.8.79)

H ₁(η) is strictly convex in η. Let H ₁(η) = 0. Then if there is no solution for this equation with regard to η, then H ₁(η) > 0 for all η ≤ η ₂; otherwise, solving this equation with regard to η, we obtain its two solutions as

$$\displaystyle{ \begin{array}{lll} \eta _{3} & =&\frac{\alpha (1-\alpha )a_{L}^{2}+(\alpha a_{ H}-c)c-c\sqrt{(\alpha a_{H } -c)^{2 } -2\alpha (1-\alpha )a_{L }^{2}}} {\sqrt{\alpha (1-\alpha )}[(1-\alpha )a_{L}^{2}+2ca_{H}]}, \\ \eta _{4} & =&\frac{\alpha (1-\alpha )a_{L}^{2}+(\alpha a_{ H}-c)c+c\sqrt{(\alpha a_{H } -c)^{2 } -2\alpha (1-\alpha )a_{L }^{2}}} {\sqrt{\alpha (1-\alpha )}[(1-\alpha )a_{L}^{2}+2ca_{H}]}. \end{array} }$$

(3.8.80)

Again, since H ₁(η) is strictly convex in η and $H_{1}(0) = (1-\alpha )\alpha ^{2}a_{L}^{2} > 0,$ it follows that η ₄ ≥ η ₃ > 0. Furthermore,

$$\displaystyle{ H_{1}(\eta )\left \{\!\!\begin{array}{llll} \mathrm{is\ positive\ and\ strictly\ decreases\ with}\ \eta \ \mathrm{on}\ [0,\eta _{3}); \\ \mathrm{is\ nonpositive\ on}\ [\eta _{3},\eta _{4}]; \\ \mathrm{is\ positive\ and\ strictly\ increases\ with}\ \eta \ \mathrm{on}\ (\eta _{4},+\infty ). \end{array} \right. }$$

(3.8.81)

In addition, denote

$$\displaystyle{ \begin{array}{lll} H_{2}(\eta ) = \frac{\eta \sqrt{\alpha (1-\alpha )}(\overline{B}-c)(a_{H}-a_{L})} {2[(\overline{B}-c)^{2}+2\eta \sqrt{\alpha (1-\alpha )}(\overline{B}-c)(a_{H}-a_{L})]}. \end{array} }$$

(3.8.82)

We obtain by calculation that for all η ₁ < η < η _max ,

$$\displaystyle{ \begin{array}{lll} \frac{dH_{2}(\eta )} {d\eta } = \frac{\sqrt{\alpha (1-\alpha )}(\overline{B}-c)^{3}(a_{ H}-a_{L})+\eta \alpha (1-\alpha )(\overline{B}-c)^{2}(a_{ H}-a_{L})^{2}} {2[(\overline{B}-c)^{2}+2\eta \sqrt{\alpha (1-\alpha )}(\overline{B}-c)(a_{H}-a_{L})]^{2}} > 0. \end{array} }$$

(3.8.83)

Therefore, H ₂(η) strictly increases with η on (η ₁, η _max ]. Furthermore, obviously, 0 ≤ H ₂(η) ≤ 25 % = H ₂(η _max ) for all η ₁ < η ≤ η _max . To summarize, Theorem 3.5.2(i) follows.

The following proceeds to show Theorem 3.5.2(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 (3.8.61) and (3.8.64), we have

$$\displaystyle{ D_{total}(\eta ) = \left \{\!\!\begin{array}{llll} \sqrt{\alpha (1-\alpha )}[\frac{(\theta a_{H}-c)(3\theta a_{H}+c)-4\theta ^{2}a_{ L}^{2}} {16\delta \theta ^{2}} ],\ \ \mathrm{if}\ \eta \leq \eta _{2}; \\ \frac{\sqrt{\alpha (1-\alpha )}} {4\delta } (\overline{B} - c)(a_{H} - a_{L}),\ \ \ \ \ \mathrm{if}\ \eta _{2} <\eta \leq \eta _{max}. \end{array} \right. }$$

(3.8.84)

Comparing (3.8.84) with (3.8.70) and (3.8.72), it follows that

$$\displaystyle{ EFF_{sd}(\eta ) = \left \{\!\!\begin{array}{llll} 75\,\% - \frac{2c(\theta a_{H}-c)+\theta ^{2}a_{ L}^{2}} {4[(\theta a_{H}-c)(\theta a_{H}+c)-\theta ^{2}a_{L}^{2}]},\ \ \mathrm{if}\ \eta \leq \eta _{2}; \\ \frac{a_{L}} {a_{H}+a_{L}} + \frac{\theta ^{2}(a_{ H}^{2}-a_{ L}^{2})[\theta (a_{ H}-a_{L})-c]+c^{2}a_{ L}} {[\theta ^{2}(a_{H}^{2}-a_{L}^{2})-c^{2}](a_{H}+a_{L})},\ \mathrm{if}\ \eta _{2} <\eta \leq \eta _{1}; \\ 50\,\%,\ \ \ \ \ \mathrm{if}\ \eta _{1} <\eta \leq \eta _{max}. \end{array} \right. }$$

(3.8.85)

Since $\theta (a_{H} - a_{L}) > c$ for all η ≤ η ₂( < η ₁),

$$\displaystyle{ \begin{array}{lll} \frac{2c(\theta a_{H}-c)+\theta ^{2}a_{ L}^{2}} {4[(\theta a_{H}-c)(\theta a_{H}+c)-\theta ^{2}a_{L}^{2}]} > 0\ \mathrm{for\ all\ }\eta \leq \eta _{2},\end{array} }$$

(3.8.86)

$$\displaystyle{ \begin{array}{lll} \frac{\theta ^{2}(a_{ H}^{2}-a_{ L}^{2})[\theta (a_{ H}-a_{L})-c]+c^{2}a_{ L}} {[\theta ^{2}(a_{H}^{2}-a_{L}^{2})-c^{2}](a_{H}+a_{L})} > 0\ \mathrm{for\ all\ }\eta _{2} <\eta \leq \eta _{1}.\end{array} }$$

(3.8.87)

To summarize, the proof of Theorem 3.5.2 is completed. □ 

Proof of Theorem 3.6.1.

By Theorem 3.4.2, together with checking the proof of Lemma 3.4.1, it is obtained that if η ≤ η ₂,

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

(3.8.88)

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

(3.8.89)

where $\theta =\alpha -\eta \sqrt{\alpha (1-\alpha )}.$ It is easy to see that E _p (η) and SD _p (η) both strictly increase with η for all η ≤ η ₂. Similarly, if η ₂ < η ≤ η _max ,

$$\displaystyle{ \begin{array}{lll} E_{p}(\eta )& =&\alpha (a_{H} -\delta \overline{Q}) + (1-\alpha )(a_{L} -\delta \overline{Q}) \\ & =&\alpha (a_{H} -\delta \frac{\overline{B}-c} {4\delta } ) + (1-\alpha )(a_{L} -\delta \frac{\overline{B}-c} {4\delta } ) \\ & =&\frac{3\overline{A}+\eta \sqrt{\alpha (1-\alpha )}(a_{H}-a_{L})+c} {4}.\end{array}}$$

(3.8.90)

Since $\frac{\theta a_{H}-c} {4\delta \theta } \geq \frac{a_{L}} {2\delta }$ for all η ≤ η ₂ whereas $\frac{\overline{B}-c} {4\delta } \leq \frac{a_{L}} {2\delta }$ for all η ₂ < η ≤ η _max , we see from (3.8.88) and (3.8.90) that

$$\displaystyle{ \frac{\alpha c} {4[(\alpha -\eta _{2}\sqrt{\alpha (1-\alpha )})]} + \frac{3\alpha a_{H} + 2(1-\alpha )a_{L}} {4} \leq \frac{3\overline{A} +\eta _{2}\sqrt{\alpha (1-\alpha )}(a_{H} - a_{L}) + c} {4}. }$$

(3.8.91)

Besides, we have

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

(3.8.92)

Obviously, E _p (η) strictly increases with η on (η ₂, η _max ] and SD _p (η) remains unchanged on (η ₂, η _max ]. To summarize, the proof is completed. □