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Analysis of product return rate and price competition in two supply chains

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Abstract

This article compares a normal and a reverse supply chain in the Betrand Competition. Each supply chain consists of a retailer and an exclusive supplier with stable partnership. The two chains compete with each other in three competition structures: the Centralized Competition Game, the Hybrid Competition Game (including two cases), and the Decentralized Competition Game. In different competition structures, we examine how the degree of competition intensity between the two chains and product return rate of the reverse chain influence the equilibrium decision of market price, profits of two chains and the choice of centralization. The article differs from the study of the traditional supply chain model as follows. Firstly, we analyze the normal and the reverse chain related to the same product in the Betrand competition. Secondly, the data show that the market price decreases with the rising of the product return rate and the falling of the competition intensity. Thirdly, it is found that the total profit of the normal chain decreases with the rising of the product return rate, while the total profit of the reverse chain increases with rising of the product return rate. Profits of two chains increase with the rise of the competition intensity. Finally, this article shows that centralization is an optimal strategy for one chain whereas centralization may be the best for the other chain.

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Acknowledgements

This article is financially supported by the National Natural Science Foundation of China (Grant Number 71172194), (Grant Number 71390330), (Grant Number 71390331), (Grant Number 71521061), (Grant Number 71571065), (Grant Number 71390335), and the Program for New Century Excellent Talents in University (Grant Number NCET-13-0193) and the Ministry of Education in China of Humanities and Social Science Project (Grant Number 14YJA630077).

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Correspondence to Tong Shu.

Appendices

Appendix A

This is the proof of Proposition 1.

$$\left\{ {\begin{array}{*{20}c} {\frac{{\partial \pi_{1} }}{{\partial p_{1} }} = 0} \\ {\frac{{\partial \pi_{2} }}{{\partial p_{2} }} = 0} \\ \end{array} } \right. \Rightarrow \left\{ {\begin{array}{*{20}c} {2p_{1} - \varepsilon p_{2} = A + c_{m} } \\ { - \varepsilon \left( {1 + \tau } \right)p_{1} + 2\left( {1 + \tau } \right)p_{2} = c_{m} + \tau c_{r} + \left( {1 + \tau } \right)A} \\ \end{array} } \right.$$
$$\Rightarrow \left\{ {\begin{array}{*{20}c} {p_{1}^{ *} = \frac{{\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)A + \left[ {2\left( {1 + \tau } \right) + \varepsilon } \right]c_{m} + \tau \varepsilon c_{r} }}{{\left( {1 + \tau } \right)\left( {4 - \varepsilon^{2} } \right)}}} \\ {p_{2}^{ *} = \frac{{\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)A + \left[ {\varepsilon \left( {1 + \tau } \right) + 2} \right]c_{m} + 2\tau c_{r} }}{{\left( {1 + \tau } \right)\left( {4 - \varepsilon^{2} } \right)}}} \\ \end{array} } \right.$$

This is the end. □

The proof of Propositions 2 and 3 is similar to the proof of Proposition 1.

Appendix B

This is the proof of Proposition 2.

$$\left\{ {\begin{array}{*{20}c} {\frac{{\partial \pi_{1}^{h} }}{{\partial p_{1} }} = 0} \\ {\frac{{\partial \pi_{2r}^{h} }}{{\partial p_{2} }} = 0} \\ \end{array} } \right. \Rightarrow \left\{ {\begin{array}{*{20}c} {2p_{1} - \varepsilon p_{2} = A + c_{m} } \\ {\varepsilon \left( {1 + \tau } \right)p_{1} - 2\left( {1 + \tau } \right)p_{2} = \tau b - \left( {1 + \tau } \right)\left( {A + w} \right)} \\ \end{array} } \right.$$
$$\Rightarrow p_{1}^{h} (w^{h} ) = \frac{{\left( {1 + \tau } \right)\left[ {\left( {\varepsilon + 2} \right)A + \varepsilon w + 2c_{m} } \right] - \tau b\varepsilon }}{{\left( {1 + \tau } \right)\left( {4 - \varepsilon^{2} } \right)}},\;\;p_{2}^{h} (w^{h} ) = \frac{{\left( {1 + \tau } \right)\left[ {\left( {\varepsilon + 2} \right)A + 2w + \varepsilon c_{m} } \right] - 2\tau b}}{{\left( {1 + \tau } \right)\left( {4 - \varepsilon^{2} } \right)}}.$$

Substitute p h2 (w), p h2 (w) into the equation of π h2m (w)

$$\pi_{2m}^{h} \left( {w^{h} } \right) = \frac{{\left( {1 + \tau } \right)\left[ {\left( {\varepsilon + 2} \right)A + \varepsilon c_{m} + \left( {\varepsilon^{2} - 2} \right)w} \right] + \tau b\left( {2 - \varepsilon^{2} } \right)}}{{\left( {1 + \tau } \right)\left( {4 - \varepsilon^{2} } \right)}}\left[ {\left( {1 + \tau } \right)w - \left( {c_{m} + \tau c_{r} + \tau b} \right)} \right]$$

Then by solving the first order derivate of the profit π h1m (w h), we can get

$$\frac{{\partial \pi_{2m}^{h} \left( {w^{h} } \right)}}{{\partial w^{h} }} = 0 \Rightarrow \frac{{\begin{array}{*{20}c} {\left( {\varepsilon^{2} - 2} \right)\left[ {\left( {1 + \tau } \right)w - \left( {c_{m} + \tau c_{r} + \tau b} \right)} \right]} \\ { + \left[ {\left( {\varepsilon + 2} \right)A + \varepsilon c_{m} + \left( {\varepsilon^{2} - 2} \right)w} \right] + \tau b\left( {2 - \varepsilon^{2} } \right)} \\ \end{array} }}{{\left( {4 - \varepsilon^{2} } \right)}} = 0$$
$$\Rightarrow w^{h} = \frac{{\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)A + \left[ {\left( {2 - \varepsilon^{2} } \right) + \left( {1 + \tau } \right)\varepsilon } \right]c_{m} + \tau \left( {2 - \varepsilon^{2} } \right)\left( {c_{r} + 2b} \right)}}{{2\left( {1 + \tau } \right)\left( {2 - \varepsilon^{2} } \right)}}$$

Substitute the value of \(w^{h} {\text{into }}p_{2}^{h} \left( {w^{h} } \right),p_{2}^{h} \left( {w^{h} } \right)\)

$$p_{1}^{h*} = \frac{{\begin{array}{*{20}c} {\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)A} \\ { + \left[ {\varepsilon \left( {2 - \varepsilon^{2} } \right) + \left( {1 + \tau } \right)\left( {8 - 3\varepsilon^{2} } \right)} \right]c_{m} - \tau \varepsilon \left( {2 - \varepsilon^{2} } \right)c_{r} } \\ \end{array} }}{{2\left( {1 + \tau } \right)\left( {2 - \varepsilon^{2} } \right)\left( {4 - \varepsilon^{2} } \right)}},$$
$$p_{2}^{h*} = \frac{{\begin{array}{*{20}c} {\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)\left( {3 - \varepsilon^{2} } \right)A} \\ { + \left[ {\left( {2 - \varepsilon^{2} } \right) + \varepsilon \left( {1 + \tau } \right)\left( {3 - \varepsilon^{2} } \right)} \right]c_{m} + \tau \left( {2 - \varepsilon^{2} } \right)c_{r} } \\ \end{array} }}{{\left( {1 + \tau } \right)\left( {2 - \varepsilon^{2} } \right)\left( {4 - \varepsilon^{2} } \right)}}$$

This is the end. □

Appendix C

This is the proof of Remark 1.

$$\frac{{\partial p_{1} }}{\partial \tau } = \frac{{\begin{array}{*{20}c} {\left[ {\left( {\varepsilon + 2} \right)A + 2c_{m} + \varepsilon c_{r} } \right]\left( {1 + \tau } \right)} \\ { - \left\{ {\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)A + \left[ {2\left( {1 + \tau } \right) + \varepsilon c_{m} + \varepsilon \tau c_{r} } \right]} \right\}} \\ \end{array} }}{{\left( {1 + \tau } \right)^{2} \left( {4 - \varepsilon^{2} } \right)}} = \frac{{ - \varepsilon \left( {c_{m} - c_{r} } \right)}}{{\left( {1 + \tau } \right)^{2} \left( {4 - \varepsilon^{2} } \right)}} < 0,\;\;\frac{{\partial p_{2} }}{\partial \tau } = \frac{{ - 2\left( {c_{m} - c_{r} } \right)}}{{\left( {1 + \tau } \right)^{2} \left( {4 - \varepsilon^{2} } \right)}} < 0,$$
$$\frac{{\partial p_{1} }}{\partial \varepsilon } = \frac{{\begin{array}{*{20}c} {\left[ {\left( {1 + \tau } \right)A + c_{m} + \tau c_{r} } \right]\left( {4 - \varepsilon^{2} } \right)} \\ { + 2\varepsilon \left\{ {\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)A + \left[ {2\left( {1 + \tau } \right) + \varepsilon } \right]c_{m} + \varepsilon \tau c_{r} } \right\}} \\ \end{array} }}{{\left( {1 + \tau } \right)\left( {4 - \varepsilon^{2} } \right)^{2} }} = \frac{{\begin{array}{*{20}c} {\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)^{2} A} \\ { + [4\varepsilon \left( {1 + \tau } \right) + 4 + \varepsilon^{2} ]c_{m} + \left( {4 + \varepsilon^{2} } \right)c_{r} } \\ \end{array} }}{{\left( {1 + \tau } \right)\left( {4 - \varepsilon^{2} } \right)^{2} }} > 0,$$
$$\frac{{\partial p_{2} }}{\partial \varepsilon } = \frac{{\begin{array}{*{20}c} {\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)^{2} A} \\ { + [\left( {1 + \tau } \right)\left( {4 + \varepsilon^{2} } \right) + 4\varepsilon ]c_{m} + 4\tau \varepsilon c_{r} } \\ \end{array} }}{{\left( {1 + \tau } \right)\left( {4 - \varepsilon^{2} } \right)^{2} }} > 0.$$

This is the end of the proof. □

Appendix D

This is the proof of Proposition 4.

$$\left\{ {\begin{array}{*{20}c} {\frac{{\partial \pi_{1r}^{d} }}{{\partial p_{1} }} = 0} \\ {\frac{{\partial \pi_{2r}^{d} }}{{\partial p_{2} }} = 0} \\ \end{array} } \right. \Rightarrow \left\{ {\begin{array}{*{20}c} {p_{1}^{d} = \frac{{\left( {1 + \tau } \right)\left[ {\left( {\varepsilon + 2} \right)A + 2w_{1} + \varepsilon w_{2} } \right] - \tau b\varepsilon }}{{\left( {1 + \tau } \right)\left( {4 - \varepsilon^{2} } \right)}}} \\ {p_{2}^{d} = \frac{{\left( {1 + \tau } \right)\left[ {\left( {\varepsilon + 2} \right)A + \varepsilon w_{1} + 2w_{2} } \right] - 2\tau b}}{{\left( {1 + \tau } \right)\left( {4 - \varepsilon^{2} } \right)}}} \\ \end{array} } \right.$$

Substitute p d1 (w d1 w d2 ), p d2 (w d1 w d2 ) into the equation \(\pi_{1m}^{d} \left( {w_{1} , w_{2} } \right),\pi_{2m}^{d} \left( {w_{1} , w_{2} } \right)\), and solve the first order derivative of \(\pi_{1m}^{d} \left( {w_{1} , w_{2} } \right), \pi_{2m}^{d} \left( {w_{1} , w_{2} } \right)\)

$$\left\{ {\begin{array}{*{20}c} {\frac{{\partial \pi_{1m}^{d} \left( {w_{1} , w_{2} } \right)}}{{\partial w_{1}^{d} }} = 0} \\ {\frac{{\partial \pi_{2m}^{d} \left( {w_{1} , w_{2} } \right)}}{{\partial w_{2}^{d} }} = 0} \\ \end{array} } \right. \Rightarrow \left\{ {\begin{array}{*{20}c} {w_{1}^{d} = \frac{{\left\{ {\begin{array}{*{20}c} {\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)A} \\ { + \left( {2 - \varepsilon^{2} } \right)\left[ {2\left( {1 + \tau } \right)\left( {2 - \varepsilon^{2} } \right) + \varepsilon } \right]c + \left( {2 - \varepsilon^{2} } \right)\tau c_{r} } \\ \end{array}_{m} } \right\}}}{{\left\{ {\left( {1 + \tau } \right)\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)\left( { - 2\varepsilon^{2} - \varepsilon + 4} \right)} \right\}}}} \\ {w_{2}^{d} = \frac{{\left\{ {\begin{array}{*{20}c} {\begin{array}{*{20}c} {\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)A} \\ { + \left( {2 - \varepsilon^{2} } \right)\left[ {2\left( {2 - \varepsilon^{2} } \right) + \left( {1 + \tau } \right)\varepsilon } \right]c} \\ \end{array}_{m} } \\ { + 2\left( {2 - \varepsilon^{2} } \right)^{2} \tau c_{r} + \tau b\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)\left( { - 2\varepsilon^{2} - \varepsilon + 4} \right)} \\ \end{array} } \right\}}}{{\left\{ {\left( {1 + \tau } \right)\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)\left( { - 2\varepsilon^{2} - \varepsilon + 4} \right)} \right\}}}} \\ \end{array} } \right.$$

Substitute the value of w d1 w d2 into p d1 (w d1 w d2 ), p d2 (w d1 w d2 )

$$p_{1}^{d*} = \frac{{\left\{ {\begin{array}{*{20}c} {\begin{array}{*{20}c} {2\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)\left( {3 - \varepsilon^{2} } \right)\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)A} \\ { + \left( {2 - \varepsilon^{2} } \right)\left[ {\left( {1 + \tau } \right)\left( {8 - 3\varepsilon^{2} } \right) + 2\varepsilon \left( {3 - \varepsilon^{2} } \right)} \right]c} \\ \end{array}_{m} } \\ { + 2\left( {2 - \varepsilon^{2} } \right)\left( {3 - \varepsilon^{2} } \right)\varepsilon \tau c_{r} } \\ \end{array} } \right\}}}{{\left\{ {\left( {1 + \tau } \right)\left( {4 - \varepsilon^{2} } \right)\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)\left( { - 2\varepsilon^{2} - \varepsilon + 4} \right)} \right\}}},p_{2}^{d*} = \frac{{\left\{ {\begin{array}{*{20}c} {\begin{array}{*{20}c} {2\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)\left( {3 - \varepsilon^{2} } \right)\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)A} \\ { + \left( {2 - \varepsilon^{2} } \right)\left[ {\left( {8 - 3\varepsilon^{2} } \right) + 2\varepsilon \left( {1 + \tau } \right)\left( {3 - \varepsilon^{2} } \right)} \right]c} \\ \end{array}_{m} } \\ { + \left( {2 - \varepsilon^{2} } \right)\left( {8 - 3\varepsilon^{2} } \right)\tau c_{r} } \\ \end{array} } \right\}}}{{\left( {1 + \tau } \right)\left( {4 - \varepsilon^{2} } \right)\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)\left( { - 2\varepsilon^{2} - \varepsilon + 4} \right)}}$$

This is the end of the proof. □

Appendix E

This is the proof of Proposition 5.

It is easy to prove Proposition 5a. If we want to obtain the in equation π 1 > π H1 , we need to prove \(\frac{{\pi_{1} }}{{\pi_{1}^{H} }} > 1\). Because \(\frac{{\pi_{1} }}{{\pi_{1}^{H} }} = \frac{{2\left( {2 - \varepsilon^{2} } \right)}}{{3 - \varepsilon^{2} }} > 1\) and 0 < ɛ < 1, therefore \(\frac{{2\left( {2 - \varepsilon^{2} } \right)}}{{3 - \varepsilon^{2} }} > 1\). Then Proposition 5a can be obtained.

The following is the proof of Proposition 5b.

$$p_{1}^{d*} = \frac{{\left\{ {\begin{array}{*{20}c} {\begin{array}{*{20}c} {2\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)\left( {3 - \varepsilon^{2} } \right)\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)A} \\ { + \left( {2 - \varepsilon^{2} } \right)\left[ {\left( {1 + \tau } \right)\left( {8 - 3\varepsilon^{2} } \right) + 2\varepsilon \left( {3 - \varepsilon^{2} } \right)} \right]c} \\ \end{array}_{m} } \\ { + 2\left( {2 - \varepsilon^{2} } \right)\left( {3 - \varepsilon^{2} } \right)\varepsilon \tau c_{r} } \\ \end{array} } \right\}}}{{\left\{ {\left( {1 + \tau } \right)\left( {4 - \varepsilon^{2} } \right)\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)\left( { - 2\varepsilon^{2} - \varepsilon + 4} \right)} \right\}}},$$
$$p_{2}^{d*} = \frac{{\left\{ {\begin{array}{*{20}c} {\begin{array}{*{20}c} {2\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)\left( {3 - \varepsilon^{2} } \right)\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)A} \\ { + \left( {2 - \varepsilon^{2} } \right)\left[ {\left( {8 - 3\varepsilon^{2} } \right) + 2\varepsilon \left( {1 + \tau } \right)\left( {3 - \varepsilon^{2} } \right)} \right]c} \\ \end{array}_{m} } \\ { + \left( {2 - \varepsilon^{2} } \right)\left( {8 - 3\varepsilon^{2} } \right)\tau c_{r} } \\ \end{array} } \right\}}}{{\left( {1 + \tau } \right)\left( {4 - \varepsilon^{2} } \right)\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)\left( { - 2\varepsilon^{2} - \varepsilon + 4} \right)}}$$

Therefore, if we want to compare the value of π 2 with π h2 , it is necessary to look at the coefficient of π 2π h2 .

When 0 < ɛ < 1, then the in equation \(\frac{1}{{\left( {1 + \tau } \right)\left( {4 - \varepsilon^{2} } \right)^{2} }} - \frac{{3 - \varepsilon^{2} }}{{2\left( {1 + \tau } \right)\left( {2 - \varepsilon^{2} } \right)\left( {4 - \varepsilon^{2} } \right)^{2} }} = \frac{{1 - \varepsilon^{2} }}{{2\left( {1 + \tau } \right)\left( {2 - \varepsilon^{2} } \right)\left( {4 - \varepsilon^{2} } \right)^{2} }} > 0\). Therefore, it is clear that π 2 > π h2 .

This is the end of the proof. □

Appendix F

It is easy to prove Proposition 9a. The optimal values of retail price in different situations are as follows.

$$\left\{ {\begin{array}{*{20}c} {p_{1}^{ *} = \frac{{\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)A + \left[ {2\left( {1 + \tau } \right) + \varepsilon } \right]c_{m} + \tau \varepsilon c_{r} }}{{\left( {1 + \tau } \right)\left( {4 - \varepsilon^{2} } \right)}}} \\ {p_{2}^{ *} = \frac{{\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)A + \left[ {\varepsilon \left( {1 + \tau } \right) + 2} \right]c_{m} + 2\tau c_{r} }}{{\left( {1 + \tau } \right)\left( {4 - \varepsilon^{2} } \right)}}} \\ \end{array} } \right.,\left\{ {\begin{array}{*{20}c} {p_{1}^{h*} = \frac{{\begin{array}{*{20}c} {\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)A} \\ { + \left[ {\varepsilon \left( {2 - \varepsilon^{2} } \right) + \left( {1 + \tau } \right)\left( {8 - 3\varepsilon^{2} } \right)} \right]c_{m} - \tau \varepsilon \left( {2 - \varepsilon^{2} } \right)c_{r} } \\ \end{array} }}{{2\left( {1 + \tau } \right)\left( {2 - \varepsilon^{2} } \right)\left( {4 - \varepsilon^{2} } \right)}}} \\ {p_{2}^{h*} = \frac{{\begin{array}{*{20}c} {\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)\left( {3 - \varepsilon^{2} } \right)A} \\ { + \left[ {\left( {2 - \varepsilon^{2} } \right) + \varepsilon \left( {1 + \tau } \right)\left( {3 - \varepsilon^{2} } \right)} \right]c_{m} + \tau \left( {2 - \varepsilon^{2} } \right)c_{r} } \\ \end{array} }}{{\left( {1 + \tau } \right)\left( {2 - \varepsilon^{2} } \right)\left( {4 - \varepsilon^{2} } \right)}}} \\ \end{array} } \right.$$
$$\begin{array}{*{20}c} {p_{1}^{H*} = \frac{{\begin{array}{*{20}c} {\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)\left( {3 - \varepsilon^{2} } \right)A + \varepsilon \tau \left( {3 - \varepsilon^{2} } \right)c_{r} } \\ { + \left[ {\left( {1 + \tau } \right)\left( {2 - \varepsilon^{2} } \right) + \varepsilon \left( {3 - \varepsilon^{2} } \right)} \right]c_{m} } \\ \end{array} }}{{\left( {1 + \tau } \right)\left( {2 - \varepsilon^{2} } \right)\left( {4 - \varepsilon^{2} } \right)}}} \\ {p_{2}^{H*} = \frac{{\begin{array}{*{20}c} {\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)A + \tau \left( {8 - 3\varepsilon^{2} } \right)c_{r} } \\ { + \left[ {\left( {1 + \tau } \right)\varepsilon \left( {2 - \varepsilon^{2} } \right) + \left( {8 - 3\varepsilon^{2} } \right)} \right]c_{m} } \\ \end{array} }}{{2\left( {1 + \tau } \right)\left( {2 - \varepsilon^{2} } \right)\left( {4 - \varepsilon^{2} } \right)}}} \\ \end{array} ,\;\left\{ {\begin{array}{*{20}c} {p_{1}^{d*} = } & {\frac{{\left\{ {\begin{array}{*{20}c} {\begin{array}{*{20}c} {2\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)\left( {3 - \varepsilon^{2} } \right)\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)A} \\ { + \left( {2 - \varepsilon^{2} } \right)\left[ {\left( {1 + \tau } \right)\left( {8 - 3\varepsilon^{2} } \right) + 2\varepsilon \left( {3 - \varepsilon^{2} } \right)} \right]c} \\ \end{array}_{m} } \\ { + 2\left( {2 - \varepsilon^{2} } \right)\left( {3 - \varepsilon^{2} } \right)\varepsilon \tau c_{r} } \\ \end{array} } \right\}}}{{\left\{ {\left( {1 + \tau } \right)\left( {4 - \varepsilon^{2} } \right)\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)\left( { - 2\varepsilon^{2} - \varepsilon + 4} \right)} \right\}}}} \\ {p_{2}^{d*} } & {\frac{{\left\{ {\begin{array}{*{20}c} {\begin{array}{*{20}c} {2\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)\left( {3 - \varepsilon^{2} } \right)\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)A} \\ { + \left( {2 - \varepsilon^{2} } \right)\left[ {\left( {8 - 3\varepsilon^{2} } \right) + 2\varepsilon \left( {1 + \tau } \right)\left( {3 - \varepsilon^{2} } \right)} \right]c} \\ \end{array}_{m} } \\ { + \left( {2 - \varepsilon^{2} } \right)\left( {8 - 3\varepsilon^{2} } \right)\tau c_{r} } \\ \end{array} } \right\}}}{{\left\{ {\left( {1 + \tau } \right)\left( {4 - \varepsilon^{2} } \right)\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)\left( { - 2\varepsilon^{2} - \varepsilon + 4} \right)} \right\}}}} \\ \end{array} } \right.$$

Therefore, the following inequations can be calculated easily.

$$p_{1}^{ *} - p_{1}^{h *} = \frac{{\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)A + \left[ {2\left( {1 + \tau } \right) + \varepsilon } \right]c_{m} + \tau \varepsilon c_{r} }}{{\left( {1 + \tau } \right)\left( {4 - \varepsilon^{2} } \right)}} - \frac{{\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)A + \left[ {\varepsilon \left( {2 - \varepsilon^{2} } \right) + \left( {1 + \tau } \right)\left( {8 - 3\varepsilon^{2} } \right)} \right]c_{m} + \tau \varepsilon \left( {2 - \varepsilon^{2} } \right)c_{r} }}{{2\left( {1 + \tau } \right)\left( {2 - \varepsilon^{2} } \right)\left( {4 - \varepsilon^{2} } \right)}};$$
$$= \frac{{ - \varepsilon \left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)A - \left[ {\left( {1 + \tau } \right)\varepsilon - \left( {2 - \varepsilon^{2} } \right)} \right]\varepsilon c_{m} + \varepsilon \tau \left( {2 - \varepsilon^{2} } \right)c_{r} }}{{2\left( {1 + \tau } \right)\left( {2 - \varepsilon^{2} } \right)\left( {4 - \varepsilon^{2} } \right)}} < 0$$
$$p_{1}^{h*} - p_{1}^{d*} = \frac{{\left( {3\varepsilon^{2} - 8} \right)\left\{ {\begin{array}{*{20}c} {\left( {1 + \tau } \right)\left( {\varepsilon + 2} \right)\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)A} \\ { + \varepsilon \left[ {\left( {1 + \tau } \right)\varepsilon + \left( {2 - \varepsilon^{2} } \right)} \right]c_{m} + \left( {2 - \varepsilon^{2} } \right)\varepsilon \tau c_{r} } \\ \end{array} } \right\} }}{{\left\{ {2\left( {1 + \tau } \right)\left( {2 - \varepsilon^{2} } \right)\left( {4 - \varepsilon^{2} } \right)\left( { - 2\varepsilon^{2} + \varepsilon + 4} \right)\left( { - 2\varepsilon^{2} - \varepsilon + 4} \right)} \right\}}} < 0;$$

In the same way, we can get \(p_{1}^{H*} - p_{1}^{d*} < 0\).

It is similar to proof of \(p_{2}^{ *} - p_{2}^{h *} < 0,p_{2}^{h} - p_{1}^{H*} < 0\).

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Zheng, Y., Shu, T., Wang, S. et al. Analysis of product return rate and price competition in two supply chains. Oper Res Int J 18, 469–496 (2018). https://doi.org/10.1007/s12351-016-0273-6

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