Impact of risk levels on optimal selling to heterogeneous retailers under dual uncertainties

Abstract

Consider the optimal selling problem of a supplier who sells the same product to two competing retailers under two types of uncertainty—the selling costs of retailers and external demand. The confidence level is used to characterize the risk caused by the two uncertainties and the profits of the participants in the supply chain channel. Our results demonstrate that higher risk levels correlate with lower belief-degree costs of the two retailers and higher belief-degree sizes of the market. For the vertically integrated channel, the supplier always supplies a larger quantity to the retailer with the low belief-degree cost than to the other retailer. Whereas, for the decentralized channel, the optimal selling decision of the supplier sometimes violates the volume ranking suggested by the quantities of the vertically integrated channel, i.e., when the belief-degree cost differences between the two retailers are not significant and the competitive intensity is high, the supplier supplies more units of the product to the retailer with the high belief-degree cost than to the other retailer. Furthermore, a decrease in the risk borne by the channel or an increase in the competitive intensity often reduces the quantities supplied to the retailers. However, in some cases, increased risk or intense competition increases the quantity supplied to the retailer with the low belief-degree cost or to the other retailer. Finally, we design a contract menu of two-part tariffs with quantity controls such that the optimal quantities supplied and retailer profits are the same for the relaxed and original models in the decentralized channel

Appendix

Proof of Theorem 1

Because the function $\Pi _{i}(q_{l},q_{h}; \, x_{i},y)$ is strictly decreasing with $x_{i}$ and is strictly increasing with y, $M\big \{\Pi _{i}(q_{l},q_{h};\xi _{i},\eta )\geqslant \Pi _{0i}\big \}\geqslant \alpha$ is equivalent to

$$\begin{aligned}\Pi _{0i}-\Pi _{i}\left( q_{l},q_{h};\Phi ^{-1}_{i}(\alpha ),\Psi ^{-1}(1-\alpha )\right) \leqslant 0. \end{aligned}$$

And $\Pi _{mi}(q_{l},q_{h};\alpha )=\max \{\Pi _{0i}\}$, chain i’s maximum profit $\Pi _{mi}(\cdot ,\cdot )$ under the confidence level $\alpha$ is

$$\begin{aligned} \Pi _{mi}(q_{l},q_{h};\alpha ) =\Pi _{i}\left( q_{l},q_{h};\Phi ^{-1}_{i}(\alpha ),\Psi ^{-1}(1-\alpha )\right) . \end{aligned}$$

(8)

By Eq. (8), Model (3) can be equivalently turned into the following one,

$$\begin{aligned} \max \limits _{q_{l},q_{h}\geqslant 0}\left\{ \Pi _{l}\big (q_{l},q_{h};\Phi _{l}^{-1}(\alpha ),\Psi ^{-1}(1-\alpha )\big )+ \Pi _{h}\big (q_{l},q_{h};\Phi _{h}^{-1}(\alpha ),\Psi ^{-1}(1-\alpha )\big )\right\} . \end{aligned}$$

(9)

The proof of the theorem is completed.

Proof of Theorem 2

Since the objective function in Model (4) is concave about the order quantity $q_{i}$, its first-order conditions are $2q_{l}+2\theta q_{h}=s_{l}$ and $2q_{h}+2\theta q_{l}=s_{h}.$ Solving the equation set, the conclusions in Theorem 2 can be obtained.

Proof of Proposition 1

By Theorem 2, we have

$$\begin{aligned}\displaystyle \frac{\partial q_{l}^{*}}{\partial \theta }=\frac{-(\theta ^{2}s_{h}-2\theta s_{l}+s_{h})}{2(1-\theta ^{2})^{2}}, \displaystyle \frac{\partial q_{h}^{*}}{\partial \theta }=\frac{-(\theta ^{2} s_{l}-2\theta s_{h}+s_{l})}{2(1-\theta ^{2})^{2}}.\end{aligned}$$

We can find that $\displaystyle \frac{\partial q_{l}^{*}}{\partial \theta }<0$ if $\displaystyle 0\leqslant \theta <\left( s_{l}-\sqrt{s_{l}^{2}-s_{h}^{2}}\right) /s_{h}$ and $\displaystyle \frac{\partial q_{l}^{*}}{\partial \theta }>0$ if $\displaystyle \left[ s_{l}-\sqrt{s_{l}^{2}-s_{h}^{2}}\right) /s_{h}<\theta <\left( s_{l}+\sqrt{s_{l}^{2}-s_{h}^{2}}\right) /s_{h}$. Similarly, the conclusion $\displaystyle \frac{\partial q_{h}^{*}}{\partial \theta }\leqslant 0$ holds if $\displaystyle 0 \leqslant \theta < s_{h}/s_{l}$. The proof of the proposition is completed.

Proof of Proposition 2

The proof is obvious and omitted here.

Proof of Corollary 1

Substituting $s_{i}=\Psi ^{-1}(1-\alpha )-\Phi _{i}^{-1}(\alpha )-c=(a_{2}-c)-\alpha (a_{2}-a_{1}+c_{i})$ and $t_{i}=a_{2}-a_{1}+c_{i}$ into Theorem 2, we can derive the results of Corollary 1.

Proof of Corollary 2

By Corollary 1, we obtain

$$\begin{aligned} \displaystyle \frac{\partial q_{l}^{*}}{\partial \alpha }=\frac{\theta t_{h}-t_{l}}{2(1-\theta ^{2})}, \displaystyle \frac{\partial q_{h}^{*}}{\partial \alpha }=\frac{\theta t_{l}-t_{h}}{2(1-\theta ^{2})}. \end{aligned}$$

Therefore, when $\displaystyle 0\leqslant \theta <t_{l}/t_{h}$, we have $\displaystyle \frac{\partial q_{l}^{*}}{\partial \alpha }<0$. When $\displaystyle t_{l}/t_{h}\leqslant \theta <\frac{b_{2}-c-\alpha t_{l}}{b_{2}-c-\alpha t_{h}}$, we have $\displaystyle \frac{\partial q_{l}^{*}}{\partial \theta }\geqslant 0$. Similarly, the conclusion $\displaystyle \frac{\partial q_{h}^{*}}{\partial \alpha }<0$ holds if $\displaystyle 0 \leqslant \theta< \theta <\frac{b_{2}-c-\alpha t_{l}}{b_{2}-c-\alpha t_{h}}$. The proof of the corollary is completed.

Proof of Lemma 1

The proof is obvious and omitted here.

Proof of Theorem 3

The proof of the theorem is similar to that of Theorem 1.

Proof of Theorem 4

For the sake of brevity, we denote retailer i’s profit $\pi _{i}(T_{i},q_{l},q_{h};\Phi ^{-1}(\alpha ),\Psi ^{-1}(1-\alpha ))$ as $\pi _{i}$, Model (7) can be equivalently transformed into the following one,

$$\begin{aligned} \left\{ \begin{array}{ll} \max \limits _{q_{l},q_{h},T_{l},T_{h}} (s_{l}-q_{l}-\theta q_{h})q_{l}+(s_{h}-q_{h}-\theta q_{l})q_{h}-\pi _{l}-\pi _{h}\\ \text{ subject } \text{ to: }\\ \quad \quad \pi _{l}\geqslant \pi _{h}+(s_{l}-s_{h})q_{h}+\theta q_{h}(q_{l}-q_{h}) &{}(IC_{l})\\ \quad \quad \pi _{h}\geqslant \pi _{l}+(s_{h}-s_{l})q_{l}+\theta q_{l}(q_{h}-q_{l}) &{}(IC_{h})\\ \quad \quad \pi _{l}\geqslant 0 &{}(IR_{l})\\ \quad \quad \pi _{h}\geqslant 0. &{}(IR_{h})\\ \end{array} \right. \end{aligned}$$

(10)

By Eqs. $(IC_{l})$ and $(IC_{h})$, we have $(q_{l}-q_{h})\left[ (s_{l}-s_{h})+\theta (q_{l}-q_{h})\right] \leqslant 0$, and thus $q_{l}\geqslant q_{h}$ or $q_{l}\leqslant q_{h}-(s_{l}-s_{h})/\theta$ because $s_{l}>s_{h}$. We first establish the following two lemmas and then use them to solve Model (10).

Lemma 2

When $q_{l}\geqslant q_{h}$, we obatin $\pi _{h}=0$ and $\pi _{l}=(s_{l}-s_{h})q_{h}+\theta q_{h}(q_{l}-q_{h})$. The retailers’ optimal order quantities are as follows.

• 1. If $s_{l}/s_{h}\geqslant 2$, then $q_{l}=s_{l}/2$ and $q_{h}=0$.

• 2. If $\sqrt{10}-2\leqslant s_{l}/s_{h}<2$, then

  • (a) If $\displaystyle 0\leqslant \theta <\frac{4s_{h}-2s_{l}}{3s_{l}}$, then $\displaystyle q_{l}=\frac{(2+\theta )s_{l}-6\theta s_{h}}{4-4\theta -9\theta ^{2}}$ and $\displaystyle q_{h}=\frac{4s_{h}-(2+3\theta )s_{l}}{4-4\theta -9\theta ^{2}}$;

  • (b) If $\displaystyle \frac{(4s_{h}-2s_{l})}{3s_{l}}\leqslant \theta <1$, then $q_{l}=s_{l}/2$ and $q_{h}=0$.

• 3. If $1<s_{l}/s_{h}<\sqrt{10}-2$, then

  • (a) If $\displaystyle 0\leqslant \theta <\frac{2(s_{l}-s_{h})}{3s_{h}-2s_{l}}$, then $\displaystyle q_{l}=\frac{(2+\theta )s_{l}-6\theta s_{h}}{4-4\theta -9\theta ^{2}}$ and $\displaystyle q_{h}=\frac{4s_{h}-(2+3\theta )s_{l}}{4-4\theta -9\theta ^{2}}$;

  • (b) If $\displaystyle \frac{2(s_{l}-s_{h})}{3s_{h}-2s_{l}}\leqslant \theta \leqslant \frac{2s_{h}^{2}-s_{l}^{2}}{s_{l}^{2}}$, then $\displaystyle q_{l}=\frac{s_{h}}{2(1+\theta )}$ and $\displaystyle q_{h}=\frac{s_{h}}{2(1+\theta )}$;

  • (c) If $\displaystyle \frac{2s_{h}^{2}-s_{l}^{2}}{s_{l}^{2}}<\theta <1$, then $q_{l}=s_{l}/2$ and $q_{h}=0$.

Lemma 3

When $q_{l}\leqslant q_{h}-(s_{l}-s_{h})/\theta$, we obtain $\pi _{l}=\pi _{h}=0$. The retailers’ optimal order quantities are as follows.

• 1. If $s_{l}/s_{h}\geqslant 5/3$, then $q_{l}=0$ and $q_{h}=(s_{l}-s_{h})/\theta$.

• 2. If $1< s_{l}/s_{h}<5/3$, then

  • (a) If $\displaystyle 0<\theta \leqslant \frac{2(s_{l}-s_{h})}{3s_{h}-s_{l}}$, then $q_{l}=0$ and $q_{h}=(s_{l}-s_{h})/\theta$;

  • (b) If $\displaystyle \frac{2(s_{l}-s_{h})}{3s_{h}-s_{l}}<\theta <1$, then $\displaystyle q_{l}=\frac{(2+3\theta )s_{h}-(2+\theta )s_{l}}{4\theta +4\theta ^{2}}$ and $\displaystyle q_{h}=\frac{(2+3\theta )s_{l}-(2+\theta )s_{h}}{4\theta +4\theta ^{2}}$.

Proof of Lemma 2

Since $s_{l}>s_{h}$, $q_{l}\geqslant q_{h}\geqslant 0$, we have $\pi _{l}\geqslant \pi _{h}$ and constraint $(IR_{l})$ is unnecessary by the constraints $(IC_{l})$ and $(IR_{h})$. Therefore, constraint $(IR_{h})$ is binding at the optimal solution, i.e., $\pi _{h}=0$. Note that contract $(IC_{l})$ and $(IC_{h})$, the value $\pi _{l}$ for any values of $q_{l}, q_{h}$ is bounded below only constraint $(IC_{l})$, not by constraint $(IC_{h})$. Thus, we first ignore constraint $(IC_{h})$ and show that the optimal solution satisfies $(IC_{h})$ at the end. According to the objective function, we obtain that constraint $(IC_{l})$ is binding at the optimal solution, i.e., $\pi _{l}=(s_{l}-s_{h})q_{h}+\theta q_{h}(q_{l}-q_{h})$. In view of the two conclusions mentioned above, Model (10) is equivalent to the following one,

$$\begin{aligned} \left\{ \begin{array}{ll} \max \limits _{q_{l},q_{h}} s_{l}q_{l}-3\theta q_{l}q_{h}-q_{l}^{2}+(2s_{h}-s_{l})q_{h}-(1-\theta ) q_{h}^{2}\\ \text{ subject } \text{ to: }\\ \quad \quad q_{l}-q_{h}\geqslant 0 \\ \quad \quad q_{h}\geqslant 0. \end{array} \right. \end{aligned}$$

(11)

The multipliers $\lambda _{1}$ and $\lambda _{2}$ are assigned to the constrains in Model (11), respectively. The optimal quantities satisfy the following conditions,

$$\begin{aligned} \left\{ \begin{array}{ll} s_{l}-2q_{l}-3\theta q_{h}+\lambda _{1}= 0 \\ 2s_{h}-s_{l}-3\theta q_{l}-2(1-\theta )q_{h}-\lambda _{1}+\lambda _{2}= 0 \\ \lambda _{1}(q_{l}-q_{h})=0 \\ \lambda _{2}q_{h}=0 \\ \lambda _{1}\geqslant 0,\lambda _{2}\geqslant 0,(\lambda _{1},\lambda _{2})\ne (0,0), \end{array} \right. \end{aligned}$$

(12)

and we consider the following three cases.

Case 1 If $\lambda _{1}>0, \lambda _{2}>0$, then $q_{l}=q_{h}=0$, $\lambda _{1}=-s_{l}<0$ and $\lambda _{2}=-2s_{h}<0$. Contradiction.

Case 2 If $\lambda _{1}>0, \lambda _{2}=0$, then $\displaystyle \lambda _{1}=\frac{(3\theta +2)s_{h}-(2\theta +2)s_{l}}{2(1+\theta )}$, and the retailer’s order quantities are

$$\begin{aligned} \begin{array}{ll} \displaystyle q_{l}=q_{h}=\frac{s_{h}}{2(1+\theta )}. \end{array} \end{aligned}$$

(13)

When $2(s_{l}-s_{h})/(3s_{h}-s_{l})<\theta <1$ and $s_{l}/s_{h}<5/4$, the multiplier $\lambda _{1}>0$. So Eq. (13) is a candidate optimal solution if

$$\begin{aligned} \begin{array}{ll} 2(s_{l}-s_{h})/(3s_{h}-s_{l})<\theta<1, s_{l}/s_{h}<5/4. \end{array} \end{aligned}$$

(14)

Case 3 If $\lambda _{1}=0, \lambda _{2}>0$, then $\lambda _{2}=(2s_{l}-4a_{h}+3\theta s_{l})/2$ and the retailers’ order quantities are

$$\begin{aligned} \begin{array}{ll} q_{l}=s_{l}/2,q_{h}=0. \end{array} \end{aligned}$$

(15)

The multiplier $\lambda _{2}>0$ if

$$\begin{aligned} \begin{array}{ll} s_{l}/s_{h}\geqslant 2 \quad or \quad s_{l}/s_{h}<2 \quad and \quad \displaystyle \frac{4s_{h}-2s_{l}}{3s_{l}}<\theta <1. \end{array} \end{aligned}$$

(16)

Therefore, Eq. (15) is a candidate optimal solution if condition (16) holds. When both (14) and (16) are violated, the optimal solution is

$$\begin{aligned} \begin{array}{ll} \displaystyle q_{l}=\frac{(2+\theta )s_{l}-6\theta s_{h}}{4-4\theta -9\theta ^{2}}, q_{h}=\frac{4s_{h}-(2+3\theta )s_{l}}{4-4\theta -9\theta ^{2}}. \end{array} \end{aligned}$$

(17)

Comparing conditions (14) and (16), we have

• 1. If $s_{l}/s_{h}\geqslant 2$, Eq. (15) is the optimal solution.

• 2. If $\sqrt{10}-2\leqslant s_{l}/s_{h}<2$, then $\displaystyle \frac{4s_{h}-2s_{l}}{3s_{l}}\leqslant \frac{2(s_{l}-s_{h})}{3s_{h}-2s_{l}}$. Therefore, there are three subcases as follows.

  • (a) When $\displaystyle 0\leqslant \theta <\frac{4s_{h}-2s_{l}}{3s_{l}}$, Eq. (17) is the optimal solution.

  • (b) When $\displaystyle \frac{4s_{h}-2s_{l}}{3s_{l}}\leqslant \theta \leqslant \frac{2(s_{l}-s_{h})}{3s_{h}-2s_{l}}$, Eq. (15) is the optimal solution.

  • (c) When $\displaystyle \frac{2(s_{l}-s_{h})}{3s_{h}-2s_{l}}<\theta <1$, Eq. (13) or Eq. (15) is the optimal solution. Comparing the corresponding objective function values, we obtain that Eq. (13) is the optimal solution if $\displaystyle \theta <\frac{2s_{h}^{2}-s_{l}^{2}}{s_{l}^{2}}$. Otherwise, Eq. (15) is the optimal solution. Since $\displaystyle \frac{2s_{h}^{2}-s_{l}^{2}}{s_{l}^{2}}\leqslant \frac{2(s_{l}-s_{h})}{3s_{h}-2s_{l}}$ when $\sqrt{10}-2\leqslant s_{l}/s_{h}$, Eq. (15) is the optimal solution if $\displaystyle \frac{2(s_{l}-s_{h})}{3s_{h}-2s_{l}}<\theta <1$.

• 3. If $s_{l}/s_{h}<\sqrt{10}-2$, then $\displaystyle \frac{2(s_{l}-s_{h})}{3s_{h}-2s_{l}}< \frac{4s_{h}-2s_{l}}{3s_{l}}$. Therefore, there are three subcases as follows.

  • (a) When $\displaystyle 0\leqslant \theta <\frac{2(s_{l}-s_{h})}{3s_{h}-2s_{l}}$, Eq. (17) is the optimal solution;

  • (b) When $\displaystyle \frac{2(s_{l}-s_{h})}{3s_{h}-2s_{l}}\leqslant \theta \leqslant \frac{4s_{h}-2s_{l}}{3s_{l}}$, Eq. (13) is the optimal solution;

  • (c) When $\displaystyle \frac{4s_{h}-2s_{l}}{3s_{l}}<\theta <1$, Eq. (13) or Eq. (15) is the optimal solution. Since $\displaystyle \frac{2(s_{l}-s_{h})}{3s_{h}-2s_{l}}<\frac{2s_{h}^{2}-s_{l}^{2}}{s_{l}^{2}}$ when $1<s_{l}/s_{h}<\sqrt{10}-2$, Eq. (13) is the optimal solution if $\displaystyle \frac{4s_{h}-2s_{l}}{3s_{l}}<\theta \leqslant \frac{2s_{h}^{2}-s_{l}^{2}}{s_{l}^{2}}$, and Eq. (15) is the optimal solution if $\displaystyle \frac{2s_{h}^{2}-s_{l}^{2}}{s_{l}^{2}}<\theta <1$. The proof of Lemma 2 is completed.

Proof of Lemma 3

The proof of this lemma is similar to that of Lemma 2.

According to Lemma 2 and Lemma 3, we obatin

• 1. If $s_{l}/s_{h}\geqslant 2$, then we need to compare the supplier’s profits corresponding to the first conclusions in Lemmas 2 and 3. The difference between them is $\displaystyle \frac{s_{l}^{2}\theta ^{2}+4(s_{h}^{2}-s_{h}s_{l})\theta +4(s_{l}-s_{h})^{2}}{4\theta ^{2}}>0$. Therefore, the first conclusion of Theorem 4 holds.

• 2. The difference between the supplier’s profits corresponding to cases (iia) or (iiia) in Lemma 2 and (iib) in Lemma 3 is

  $$\begin{aligned}\displaystyle \frac{\left[ 4(s_{h}-s_{l})+\theta ^{2}(5s_{l}-3s_{h})\right] ^{2}}{8\theta ^{2}(1+\theta )[4-4\theta -9\theta ^{2}]}>0, \end{aligned}$$

  so Eq. (17) is the optimal solution if $\displaystyle 0\leqslant \theta <min \left( \frac{2(s_{l}-s_{h})}{3s_{h}-2s_{l}},\frac{4s_{h}-2s_{l}}{3s_{l}}\right)$. Similarly, The difference between the supplier’s profits corresponding to case i), iib) or iiic) in Lemma 2 and case iib) in Lemma 3 is

  $$\begin{aligned} \begin{array}{ll} \displaystyle f_1=\frac{2s_{l}^{2}\theta ^{3}+(s_{l}^{2}-s_{h}^{2}-2s_{l}s_{h})\theta ^{2}+4(s_{l}-s_{h})^{2}\theta +4(s_{l}-s_{h})^{2}}{8\theta ^{2}(1+\theta )}, \end{array} \end{aligned}$$

  (18)

  and $\displaystyle \frac{\partial f_1}{\partial s_{l}}=\frac{2s_{l}\theta ^{3}+(s_{l}-s_{h})\theta ^{2}+4(s_{l}-s_{h})\theta +4(s_{l}-s_{h})}{4\theta ^{2}(1+\theta )}>0.$ Thus,

  $$\begin{aligned}f_1\geqslant f_1|_{\frac{s_{l}}{s_{h}}=\sqrt{10}-2}=\frac{s_{h}^{2}\left[ (28-8\sqrt{10})\theta ^{3}+(17-6\sqrt{10})\theta ^{2}+(76-24\sqrt{10})\theta +76-24\sqrt{10}\right] }{8\theta (1+\theta )}>0. \end{aligned}$$

  Thus, the second conclusion of Theorem 4 holds.

• 3. The difference between the supplier’s profits corresponding to case iiib) in Lemma 2 and case iib) in Lemma 3 is $\displaystyle f_2=\frac{(s_{l}-s_{h})\left[ 4(s_{l}-s_{h})+4\theta (s_{l}-s_{h})-\theta ^{2}(s_{l}+3s_{h})\right] }{8\theta ^{2}(1+\theta )}$, which has only one positive root

  $$\begin{aligned} \theta =\displaystyle \frac{2(s_{l}-s_{h})+2\sqrt{2(s_{l}^{2}-s_{h}^{2})}}{s_{l}+3s_{h}}. \end{aligned}$$

  Since $\displaystyle f_2|_{\theta =\frac{2(s_{l}-s_{h})}{3s_{h}-2s_{l}}}>0$ and $f_2|_{\theta =\frac{2s_{l}^{2}-s_{h}^{2}}{s_{l}^{2}}}<0$, the conclusion corresponding to case iiib) in Theorem 4 holds. If $\displaystyle \frac{2(s_{l}-s_{h})+2\sqrt{2(s_{l}^{2}-s_{h}^{2})}}{s_{l}+3s_{h}}\leqslant \theta \leqslant \frac{2s_{h}^{2}-s_{l}^{2}}{s_{l}^{2}}$, the conclusion corresponding to case iiic) in Theorem 4 holds. If $\displaystyle \frac{2s_{h}^{2}-s_{l}^{2}}{s_{l}^{2}}<\theta$, then thereexits $\displaystyle \overline{\theta }(s_{l},s_{h})\in (\frac{2s_{h}^{2}-s_{l}^{2}}{s_{l}^{2}},\frac{s_{h}}{s_{l}})$ such that $f_1=0$ since $\displaystyle \lim _{\theta \rightarrow 0^{+}}f_1>0$, $\displaystyle f_1|_{\theta =\frac{2s_{h}^{2}-s_{l}^{2}}{s_{l}^{2}}}<0$ and $\displaystyle f_1|_{\theta =\frac{s_{h}}{s_{l}}}>0$. Thus, the conclusions corresponding to cases iiib) and iiic) in Theorem 4 hold.

Proof of Proposition 3

The monotonicity of $\widehat{q}_{l}$ about $\theta$ is provedfrom the following three cases and that of $\widehat{q}_{h}$ about $\theta$ is similarly proved.

• 1. When $\widehat{q}_{l}> \widehat{q}_{h}>0$,we have $\displaystyle \widehat{q}_{l}=\frac{(2+\theta )s_{l}-6\theta s_{h}}{4-4\theta -9\theta ^{2}}$, and thus,

  $$\begin{aligned} \displaystyle \frac{\partial \widehat{q}_{l}}{\partial \theta }=\frac{3\left[ (3s_{l}-18s_{h})\theta ^{2}+12s_{l}\theta +4s_{l}-8s_{h}\right] }{(4-4\theta -9\theta ^{2})^{2}}. \end{aligned}$$

  Let $\Delta =s_{l}^{2}+4s_{l}s_{h}-6s_{h}^{2}$ and we obtain that$\Delta>0$ if $\sqrt{10}-2< s_{l}/s_{h}<2$ and$\Delta <0$ if $1<s_{l}/s_{h}<\sqrt{10}-2$. Particularly, $\displaystyle \frac{\partial \widehat{q}_{l}}{\partial \theta }\leqslant 0$ if $\Delta =0$. Therefore, we have

  • (a) If $\sqrt{10}-2< s_{l}/s_{h}<2$, then Proof of Proposition 3. $\displaystyle \frac{\partial \widehat{q}_{l}}{\partial \theta }=0$ has two positive roots. The larger root is omitted since it does not satisfythe condition $\displaystyle \theta \in \left( 0, \frac{4s_{h}-2s_{l}}{3s_{l}}\right)$. Since the smaller root is $\displaystyle \theta '_{0}=\frac{6s_{l}-2\sqrt{6\Delta }}{18s_{h}-3s_{l}}<\frac{4s_{h}-2s_{l}}{3s_{l}}$ in this case, we have $\displaystyle \frac{\partial \widehat{q}_{l}}{\partial \theta }\leqslant 0$ if $0\leqslant \theta \le \theta '_{0}$ and $\displaystyle \frac{\partial \widehat{q}_{l}}{\partial \theta }>0$ if $\displaystyle \theta '_{0}<\theta <\frac{4s_{h}-2s_{l}}{3s_{l}}$.

  • (b) If $1<s_{l}/s_{h}<\sqrt{10}-2$, then $\displaystyle \frac{\partial \widehat{q}_{l}}{\partial \theta }<0$.

• 2. When $\widehat{q}_{l} = \widehat{q}_{h}>0$, we have $\displaystyle \widehat{q}_{i}=-\frac{s_{h}}{2(1+\theta ^{2})}<0$, and thus, $q_{i}$ is decreasing in $\theta$.

• 3. When $\widehat{q}_{h}> \widehat{q}_{l}>0$, the proof is similar to that of case 1). Summarizing the above conclusions,we can obtain Proposition 3.

Proof of Proposition 4

The proof of this proposition is similar to that of Proposition 3.

Proof of Proposition 5

From the proof of Theorem 4, we obtain $T_{i}=(\Psi ^{-1}(1-\alpha )-\Phi _{i}^{-1}(\alpha )-q_{i}-\theta q_{j})q_{i}-\pi _{i}$ and prove the proposition from the following three cases.

• 1. When $\widehat{q}_{h}>\widehat{q}_{l}>0$, we obtain $\widehat{q}_{l}=\displaystyle \frac{(2+3\theta )s_{h}-(2+\theta )s_{l}}{4\theta +4\theta ^{2}}$, $\widehat{q}_{h}=\displaystyle \frac{(2+3\theta )s_{l}-(2+\theta )s_{h}}{4\theta +4\theta ^{2}}$ into $T_{i}$, and thus,

  $$\begin{aligned} \begin{array}{ll} \displaystyle {T_{l}}-{T_{h}}=c(\widehat{q}_{l}-\widehat{q}_{h})-\frac{s_{l}^{2}-s_{h}^{2}}{4(1+\theta )}<0 \end{array} \end{aligned}$$

  and

  $$\begin{aligned} \begin{array}{ll} \displaystyle \frac{T_{l}}{\widehat{q}_{l}}-\frac{T_{h}}{\widehat{q}_{h}}=(s_{l}-s_{h})+(1-\theta )(\widehat{q}_{h}-\widehat{q}_{l})>0. \end{array} \end{aligned}$$

  That is, $T_{l}<T_h$ and $T_{l}/\widehat{q}_{l}>T_{h}/\widehat{q}_{h}$ if $\widehat{q}_{h}>\widehat{q}_{l}>0$.

• 2. Similarly, when $\widehat{q}_{l}>\widehat{q}_{h}>0$, we obtain $T_{l}>T_{h}$ and $T_{l}/\widehat{q}_{l}<T_{h}/\widehat{q}_{h}$. This completes the proof of the proposition.

Proof of Corollary 3

Substituting $s_{i}=\Psi ^{-1}(1-\alpha )-\Phi _{i}^{-1}(\alpha )-c=(a_{2}-c)-\alpha (a_{2}-a_{1}+c_{i})$ and $t_{i}=a_{2}-a_{1}+c_{i}$ into Theorem 4, the corollary can be obtained.

Proof of Corollary 4

The proof of the corollary is similar to that of Corollary 2.

Proof of Theorem 5

First, we prove the existence of a menu of two-part tariffs that implements the supplier’s optimal selling mechanism. If retailer i selects tariff $w_{i}q+k_{i}$, her optimal order quantityis given by the following optimization problem,

$$\begin{aligned} \max \limits _{q_i\geqslant 0}\pi _{mi}(w_{i},k_{i},q_i,q_{j};\alpha )= & {} \max \limits _{q_{i}\geqslant 0}\pi _{i}(w_{i},k_{i},q_i,q_{j};\Phi _{i}^{-1}(\alpha ),\Psi ^{-1}(1-\alpha ))\nonumber \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad= & {} \max \limits _{q_i\geqslant 0}(\Psi ^{-1}(1-\alpha )-\Phi _{i}^{-1}(\alpha )-q_i-\theta q_{j})q_i-w_{i}q_i-k_{i}. \end{aligned}$$

(19)

Its first-order condition is $\overline{q}_i=(\Psi ^{-1}(1-\alpha )-\Phi _{i}^{-1}(\alpha )-\theta q_{j}-w_{i})/2$. Further, $|\frac{\partial \overline{q}_i}{\partial q_{j}}|=\theta /2<1$, i.e., retailer i’s best response is a contraction mapping guaranteeing the uniqueness of the equilibrium conditional on she choosing tariff $w_{i}q+k_{i}$. For the optimal contract menu to be implemented, we must obtain $\overline{q}_i=\widehat{q}_{i}$ and $\pi _{mi}(w_{i},k_{i},\overline{q}_i,\overline{q}_j;\alpha )=\pi _{i}$, and thus, we have

$$\begin{aligned} w_{i}= & {} \Psi ^{-1}(1-\alpha )-\Phi _{i}^{-1}(\alpha )-2\widehat{q}_{i}-\theta \widehat{q}_{j},\nonumber \\ k_{l}= & {} \widehat{q}_{l}^{2}-\widehat{q}_{h}(s_{l}-s_{h})+\theta (\widehat{q}_{l}-\widehat{q}_{h}), k_{h}=\widehat{q}_{h}^{2}. \end{aligned}$$

(20)

We next show that retailer i choosing tariffs $w_{i}q+k_{i}$ is an equilibrium by appropriate quantity control. Fixing retailer j’s selection of tariff $w_{j}q+k_{j}$ and quantity $\widehat{q}_{j}$, if retailer ichooses tariff $w_{j}q+k_{j}$, her optimal order quantity is given by the following optimization problem,

$$\begin{aligned} \max \limits _{q_i\geqslant 0}\pi _{mi}(w_{j},k_{j},q_i,\widehat{q}_{j};\alpha )= & {} \max \limits _{q_i\geqslant 0}\pi _{i}\left( w_{j},k_{j},q_i,\widehat{q}_{j};\Phi _{i}^{-1}(\alpha ),\Psi ^{-1}(1-\alpha )\right) \nonumber \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad= & {} \max \limits _{q_i\geqslant 0}(\Psi ^{-1}(1-\alpha )-\Phi _{i}^{-1}(\alpha )-q_i-\theta \widehat{q}_{j})q_i-w_{j}q_i-k_{j}. \end{aligned}$$

(21)

Its first-order condition is $\widetilde{q}_i=\left( \Psi ^{-1}(1-\alpha )-\Phi _{i}^{-1}(\alpha )-\theta \widehat{q}_{j}-w_{j}\right) /2.$ Substituting $w_{j}$ in (20) into expression above, we obtain

$$\begin{aligned} \widetilde{q}_i=\left( s_{i}-s_{j}+\theta \widehat{q}_{i}+(2-\theta )\widehat{q}_{j}\right) /2, \end{aligned}$$

(22)

and retailer i’s optimal deviation profit becomes $\pi _{mi}(w_{j},k_{j},\widetilde{q}_i,\widehat{q}_{j};\alpha )$. The quantity control of all kinds of situations are as follows

• 1. When $\widehat{q}_{h}>\widehat{q}_{l}>0$, substituting $\displaystyle \widehat{q}_{l}=\frac{(2+3\theta )s_{h}-(2+\theta )s_{l}}{4\theta +4\theta ^{2}}$, $\displaystyle \widehat{q}_{h}=\frac{(2+3\theta )s_{l}-(2+\theta )s_{h}}{4\theta +4\theta ^{2}}$ into deviation quantity and deviation profit, we have $\pi _{i}- \pi _{mi}(w_{j},k_{j},\widetilde{q}_i,\widehat{q}_{j};\alpha )=0$. Thus, retailer i choosing tariff $w_{i}q+k_{i}$ is an equilibrium and quantity control is not necessary.

• 2. When $\widehat{q}_{l}>\widehat{q}_{h}>0$, substituting $\displaystyle \widehat{q}_{l}=\frac{(2+\theta )s_{l}-6\theta s_{h}}{4-4\theta -9\theta ^{2}}$, $\displaystyle \widehat{q}_{h}=\frac{4s_{h}-(2+3\theta )s_{l}}{4-4\theta -9\theta ^{2}}$ into deviation quantity and deviation profit, we have

  $$\begin{aligned}\widetilde{q}_{l}=\frac{(3s_{h}-5s_{l})\theta ^{2}-6\theta s_{l}+4s_{h}}{2(4-4\theta -9\theta ^{2})}, \end{aligned}$$
  $$\begin{aligned}\pi _{l}- \pi _{ml}(w_{h},k_{h},\widetilde{q}_{l},\widehat{q}_{h};\alpha ))=-\frac{\left[ (3s_{h}-5s_{l})\theta ^{2}+4(s_{l}-s_{h})\right] ^{2}}{4(4-4\theta -9\theta ^{2})^{2}}<0.\end{aligned}$$

  Thus, the retailer with a low belief-degree cost chooses tariff $w_{h}q+k_{h}$ and quantity $\widetilde{q}_{l}$. Note that

  $$\begin{aligned}\pi _{ml}(w_{h},k_{h},q_l,\widehat{q}_{h};\alpha )\end{aligned}$$

  is a strictly concave function of $q_l$ and $\widetilde{q}_{l}-\widehat{q}_{h}=(s_{l}-s_{h}+\theta (\widehat{q}_{l}-\widehat{q}_{h}))/2>0$. If the supplier does not allow a retailerchoosing tariff $w_{h}q+k_{h}$ to order more than $\widehat{q}_{h}$, then the optimal order quantity of the retailer with a low belief-degree cost is $\widehat{q}_{h}$, i.e., she choose the high-belief-degree-cost retailer’s contract such that $\pi _{l}- \pi _{ml}(w_{h},k_{h},\widehat{q}_{h},\widehat{q}_{h};\alpha )=0$. Thus, the retailer with the low belief-degree cost does not need to choose retailer j’s contract. On the other hand,

  $$\begin{aligned} \begin{array}{ll} \pi _{h}-\pi _{mh}(w_{l},k_{l},\widehat{q}_{l},\widetilde{q}_{h};\alpha )\\ \displaystyle =-\frac{\left[ (5s_{l}-3s_{h})\theta ^{2}-4(s_{l}-s_{h})\right] \left[ (5s_{l}-3s_{h})\theta ^{2}+(16s_{l}-24s_{h})\theta +12(s_{l}-s_{h})\right] }{4(4-4\theta -9\theta ^{2})^{2}}. \end{array} \end{aligned}$$

  (23)

• (a) When $\sqrt{10}-2\leqslant s_{l}/s_{h}<2$, we obtain $\pi _{h}-\pi _{mh}(w_{l},k_{l},\widehat{q}_{l},\widetilde{q}_{h};\alpha )\geqslant 0$. Thus, the high-belief-degree-cost retailer does not have an incentive to choose tariff $w_{l}q+k_{l}$.

• (b) When $1<s_{l}/s_{h}<\sqrt{10}-2$, we have

  • If $\displaystyle 0\leqslant \theta \leqslant \frac{(12s_{h}-8s_{l})-2\sqrt{s_{l}^{2}-24s_{l}s_{l}+27s_{h}^{2}}}{5s_{l}-3s_{h}}$, then $\pi _{h}-\pi _{mh}(w_{l},k_{l},\widehat{q}_{l},\widetilde{q}_{h};\alpha )\geqslant 0$, and the high-degree-cost retailer does not have an incentive to choosing tariff $w_{l}q+k_{l}$.

  • If $\displaystyle \frac{(12s_{h}-8s_{l})-2\sqrt{s_{l}^{2}-24s_{l}s_{l}+27s_{h}^{2}}}{5s_{l}-3s_{h}}<\theta <\frac{2(s_{l}-s_{h})}{3s_{h}-2s_{l}}$, then $\pi _{h}-\pi _{mh}(w_{l},k_{l},\widehat{q}_{l},\widetilde{q}_{h};\alpha )<0$. Thus, the high-degree-cost retailerchooses $w_{l}q+k_{l}$ and quantity $\widetilde{q}_{h}$. Note that $\pi _{mh}(w_{l},k_{l},\widehat{q}_{l},q_h;\alpha )$ is a strictly concave function of $q_h$ and $\widetilde{q}_{h}-\widehat{q}_{l}=(s_{h}-s_{l}+\theta (\widehat{q}_{h}-\widehat{q}_{l}))/2<0.$ If the supplier does not allow the retailer choosing tariff $w_{l}q+k_{l}$to order less than $\widehat{q}_{l}$, then the optimal order quantity of the retailer with the high belief-degree cost is $\widehat{q}_{l}$, i.e., she chooses the low-belief-degree-cost retailer’s contract such that $\pi _{h}- \pi _{mh}(w_{l},k_{l},\widehat{q}_{l},\widehat{q}_{l};\alpha )>0$. Thus, the retailer with the high belief-degree cost does not have an incentive to choose tariff $w_{l}q+k_{l}$.

• 3. When $\widehat{q}_{l}=\widehat{q}_{h}>0$, substituting $\widehat{q}_{l}=\displaystyle \frac{s_{h}}{2(1+\theta )}$, $\widehat{q}_{h}=\displaystyle \frac{s_{h}}{2(1+\theta )}$ into deviation quantity and deviation profit, we have

  $$\begin{aligned}\pi _{l}- \pi _{ml}(w_{h},k_{h},\widetilde{q}_{l},\widehat{q}_{h};\alpha )=-(s_{l}-s_{h})^{2}/4<0.\end{aligned}$$

  Thus, the retailer with a low belief-degree cost chooses $w_{h}q+k_{h}$ and quantity $\widetilde{q}_{l}$. Since $\widetilde{q}_{l}-\widehat{q}_{h}=(s_{l}-s_{h})/2>0$, if the supplier does not allow the retailerchoosing tariff $w_{h}q+k_{h}$ to order more than $\widehat{q}_{h}$, then the optimal ordering quantity of the retailer with the low belief-degree cost is $\widehat{q}_{h}$, i.e., she chooses the high-belief-degree-cost retailer’s contract such that $\pi _{l}- \pi _{ml}(w_{h},k_{h},\widehat{q}_{h},\widehat{q}_{h};\alpha )=0$. Thus, the retailer with the low belief-degree cost does not have an incentive to choose tariff $w_{h}q+k_{l}$. On the other hand,

  $$\begin{aligned}\pi _{h}- \pi _{mh}(w_{l},k_{l},\widehat{q}_{l},\widetilde{q}_{h};\alpha )&=-(s_{l}-s_{h})^{2}/4<0,\\ \widetilde{q}_{h}-\widehat{q}_{l}&=(s_{h}-s_{l})/2<0,\\ \pi _{h}- \pi _{mh}(w_{l},k_{l},\widehat{q}_{l},\widehat{q}_{l};\alpha )&=0. \end{aligned}$$

If the supplier does not allow the retailer choosing tariff $w_{l}q+k_{l}$to order less than $\widehat{q}_{l}$, the retailer with the high belief-degree cost does not need to choose retailer j’s contract. The proof of the theorem is completed.

Proof of Corollary 5

We will show that retailer i selecting tariff $w_{j}q+k_{j}$ is not an equilibrium. If retailer i chooses tariff $w_{j}q+k_{j}$, then her optimal order quantity is given by the following optimization problem,

$$\begin{aligned} \max \limits _{q_i\geqslant 0}\pi _{mi}(w_{j},k_{j},q_i,q_{j};\alpha )= & {} \max \limits _{q_i\geqslant 0}\pi _{i}(w_{j},k_{j},q_i,q_{j};\Phi _{i}^{-1}(\alpha ),\Psi ^{-1}(1-\alpha ))\nonumber \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad= & {} \max \limits _{q_i\geqslant 0}(\Psi ^{-1}(1-\alpha )-\Phi _{i}^{-1}(\alpha )-q_i-\theta q_{j})q_i-w_{j}q_i-k_{j}. \end{aligned}$$

(24)

Its first-order condition is $q^{o}_i=(\Psi ^{-1}(1-\alpha )-\Phi _{i}^{-1}(\alpha )-\theta q_{j}-w_{j})/2$. Further, $|\frac{\partial q^{o}_i}{\partial q_{j}}|=\theta /2<1$, i.e., retailer i’s best response is a contraction mapping guaranteeing the uniqueness of the equilibrium conditional on retailer i choosing tariff $w_{j}q+k_{j}$. Substituting $w_{j}$ in (20) into the expression $q^{o}_i$, we obtain

$$\begin{aligned} \begin{array}{ll} q_{l}^{o}=\widehat{q}_{h}+(s_{l}-s_{h})/(2-\theta ), q_{h}^{o}=\widehat{q}_{l}-(s_{l}-s_{h})/(2-\theta ). \end{array} \end{aligned}$$

(25)

• 1. When $\widehat{q}_{l}=\widehat{q}_{h}=\widehat{q}$, recalling that quantity restrictions apply and using the analysis similar to that mentioned above, we obtain that the unique responses will be $q_{l}^{o}=q_{h}^{o}=\widehat{q}$. Using $w_{i}$, $k_{i}$ in (20) and $\widehat{q}_{l}=\widehat{q}_{h}=s_{h}/[2(1+\theta )]=\widehat{q}$, we have $w_{l}\widehat{q}+k_{l}=s_{h}\widehat{q}/2=w_{j}\widehat{q}+k_{j}$.

• 2. When $\widehat{q}_{l}>\widehat{q}_{h}>0$, the unique responses will be $q_{l}^{o}\leqslant \widehat{q}_{h}$ and $q_{h}^{o}\geqslant \widehat{q}_{l}$. As in the previous case, it is a straightforward argument that the unique best responses will be $q_{l}^{o}=\widehat{q}_{h}$ and $q_{h}^{o}=\widehat{q}_{l}$. For this to be an equilibrium, the retailer with the high belief-degree cost should not have an incentive to deviate to tariff $w_{h}q+k_{h}$ and choose quantity $\widehat{q}_{h}$. Her optimal deviation profit should be $\pi _{mh}(w_{h},k_{h},\widehat{q}_{h},\widehat{q}_{h};\alpha )=(\Psi ^{-1}(1-\alpha )-\Phi ^{-1}_{h}-\widehat{q}_{h}-\theta \widehat{q}_{h})\widehat{q}_{h}-w_{h}\widehat{q}_{h}-t_{h}$. Using (20), we have$\pi _{h}-\pi _{mh}(w_{h},k_{h},\widehat{q}_{h},\widehat{q}_{h};\alpha )=-(s_{l}-s_{h})(\widehat{q}_{l}-\widehat{q}_{h})<0$. Thus, the retailer with the high belief-degree cost has an incentive to deviate and choose tariff $w_{h}q+k_{h}$, i.e., retailer i choosing tariff $w_{j}q+k_{j}$ is not an equilibrium in this case.

• 3. When $\widehat{q}_{h}>\widehat{q}_{l}>0$, the retailers’ order quantities are given by (25). By using (20), the high-belief-degree-cost retailer’s profit in this case $\displaystyle \pi _{mh}(w_{h},k_{h}, q_{l}^{o}, q_{h}^{o};\alpha )=\left( \widehat{q}_{l}-\frac{s_{l}-s_{h}}{2-\theta }\right) ^{2}-\widehat{q}_{l}^{2}<0$. Thus, the retailer with the high belief-degree cost is better off not transacting with the supplier. Therefore, retailer i choosing tariff $w_{j}q+k_{j}$ cannot be an equilibrium in this case neither. The proof of the Corollary is completed.