Competing refurbishment in a supply chain with different selling modes

Abstract

This study focuses on the optimal pricing and product refurbishing decisions of an original equipment manufacturer (OEM) and a retailer in a supply chain. The OEM sells the new product through a retailer, which serves as a reseller or selling agency, and then chooses whether to provide the refurbished product to consumers. By constructing four models, we obtain the following results. First, we show that the OEM and the retailer always prefer to sell a refurbished product as a monopoly. Second, competing refurbishments leads to a prisoner’s dilemma in which both parties suffer greater losses compared to the no refurbishment scenario. This implies that competing refurbishments, except for product cannibalization, hinder the refurbishing industry. Finally, with different refurbishers, the OEM has different preferences for a new product’s selling format. However, when the retailer serves as the sole refurbisher, the OEM always prefers agency mode.

Appendix

A Proof of Proposition 3

Using the backward method, we will first calculate the new and the refurbished product’s optimal prices($p_N$ and $p_{RO}$). As the OEM make decision on the price of the refurbished product, the retailer decides the new products price, and their profit functions of the OEM and the retailer are displayed in Eqs. (5) and (6) respectively, we have $\frac{\partial ^2\pi _O}{\partial p_{RO}^2}=-\frac{2}{1-\delta }-\frac{2}{\delta }<0$ and $\frac{\partial ^2\pi _T}{\partial p_{N}^2}=-\frac{2}{1-\delta }<0$. Hence, they are concave in $p_{RO}$ and $p_N$ respectively. Thus by jointly solving $\frac{\partial \pi _O}{\partial p_{RO}}$ and $\frac{\partial \pi _T}{\partial p_{N}}=0$, we have the optimal prices are $p_N=\frac{(2+\delta )w-\delta c_N+c_{RO}+2(1-\delta )}{4-\delta }$ and $p_{RO}=\frac{3\delta w-2(\delta c_N-c_{RO})+\delta (1-\delta )}{4-\delta }$. Substituting them into Eq. (5), we have $\frac{\partial ^2\pi _O}{\partial w^2}=-\frac{2(8+\delta )}{(4-\delta )^2}<0$. Hence, w is concave in profit function shown by Eq. (5). Finally, we will get the optimal wholesale price of the new product ${\hat{w}}=\frac{2(4+\delta )c_N+\delta (\delta -c_{RO})+8}{2(8+\delta )}$ by solving $\frac{\partial \pi _O}{\partial w}=0$. Then we have the sales of the new and refurbished products are ${\hat{D}}_N=\frac{(2+\delta )(1+c_{RO}-c_{N}-\delta )}{(1-\delta )(8+\delta )}$ and ${\hat{D}}_{RO}=\frac{2\delta (1+3c_N)-(8-\delta ^2-\delta )c_{RO}-\delta ^2(1+\delta )}{2\delta (1-\delta )(8+\delta )}.$ To make sure the sales are non-negative, the equilibrium results are under the condition $\frac{(8-\delta ^2-\delta )c_{RO}+\delta ^2(1+\delta )-2\delta }{6\delta }\le c_N\le 1+c_{RO}-\delta$ in this case. Then substituting the ${\hat{w}}$ into relevant functions, we can get the Proposition 3.

B Proof of Proposition 4

In this case, The OEM sets the prices of the new and the refurbished products. As OEM’s profit function is Eq. (14), we optimize $p_N$ and $p_{RO}$ and find the Hessian of $\pi _O$ is $\left( \begin{array}{ccc} -\frac{2(1-\beta )}{1-\delta } &{} \frac{2-\beta }{1-\delta } \\ \frac{2-\beta }{1-\delta } &{} -\frac{2}{1-\delta }-\frac{2}{\delta } \\ \end{array} \right)$. In addition, as $\delta <\frac{4(1-\beta )}{(2-\beta )^2}$, we can get $|H|=\frac{4(1-\beta )-\delta (2-\beta )^2}{\delta (1-\delta )^2}>0$. Thus, $\pi _O$ jointly concave in $p_N$ and $p_{RO}$. By solving the $\frac{\partial \pi _O}{\partial p_N}=0$ and $\frac{\partial \pi _O}{\partial p_{RO}}=0$, we can get the optimal prices are $p_N^*=\frac{(2-2\delta +\beta \delta )c_N-\beta c_{RO}+2(1-\delta )(1-\beta )}{4(1-\beta )-(2-\beta )^2\delta }$ and $p_{RO}^*=\frac{\beta \delta c_N+(\beta \delta -2\beta -2\delta +2)c_{RO}+\delta (1-\delta )(1-\beta )(2-\beta )}{4(1-\beta )-(2-\beta )^2\delta }$. Then substituting the them into demand functions, we have the sales of the two products are $D_{N}^*=\frac{(2-\beta )c_{RO}-2 c_N+2(1-\beta )-\delta (2-\beta )}{4(1-\beta )-(2-\beta )^2\delta }$ and $D_{RO}^*=\frac{\delta (2-\beta ) c_N-(1-\beta )(2c_{RO}-\beta \delta )}{4(1-\beta )-(2-\beta )^2\delta }.$ To avoid the negative sales, It is need to notice that the cases are under the condition $\frac{(1-\beta )(2c_{RO}-\beta \delta )}{\delta (2-\beta )}\le c_N\le \frac{(2-\beta )c_{RO}+2(1-\beta )-\delta (2-\beta )}{2}$. Substituting them into other relevant functions, the proposition is proved.

C Proof of Proposition 5

Using the backward method, we will first calculate the optimal prices of the new and refurbished product which is shown by Eq. (23). Optimizing for $p_N$ and $p_{RT}$, the Hessian Matrix of $\pi _T$ is $\left( \begin{array}{ccc} -\frac{2}{1-\theta \delta } &{} \frac{2}{1-\theta \delta } \\ \frac{2}{1-\theta \delta } &{} -\frac{2}{(1-\theta \delta )\theta \delta }\\ \end{array} \right)$. and $|H|=\frac{2}{(1-\theta \delta )\theta \delta }>0$. Hence we can conclude $p_N$ and $p_{RT}$ are jointly concave in $\pi _T$. From the first order conditions, jointly solving $\frac{\partial \pi _T}{\partial p_{N}}=0$ and $\frac{\partial \pi _T}{\partial p_{RO}}=0$, we can get $p_N=\frac{1+w}{2}$ and $p_{RT}=\frac{\theta \delta +c_{RT}}{2}$. Then substituting them into the Eq. (22), we have $\frac{\partial ^2 \pi _O}{\partial w^2}=-\frac{1}{1-\theta \delta }<0$. By solving $\frac{\partial \pi _O}{\partial w}=0$, we get the optimal $w=\frac{1+c_{N}+c_{RT}-\theta \delta }{2}$, then substituting this into the sales functions, we can the sales of the two products are ${\hat{D}}_N=\frac{1+c_{RT}-c_N-\theta \delta }{4(1-\theta \delta )}$ and ${\hat{D}}_{RT}=\frac{\theta \delta (1+c_N-\theta \delta )-(2-\theta \delta )c_{RT}}{4\theta \delta (1-\theta \delta )}.$ Furthermore, to ensure the sales of both product non-negative, the equilibrium results are under the condition $\frac{(2-\theta \delta )c_{RT}}{\theta \delta }+\theta \delta -1\le c_N \le 1+c_{RT}-\theta \delta$ in this case. Then substituting the function into other functions, we can have Proposition 5.

D Proof of Proposition 6

Using the backward method, we will optimize for $p_{RT}$ in Eq. (30). $\frac{\partial ^2 \pi _T}{\partial p_{RT}^2}=-\frac{2}{1-\theta \delta }-\frac{2}{\theta \delta }<0$, hence, it is a concave function of $p_{RT}$. Solving $\frac{\partial \pi _O}{\partial p_{RT}}=0$, the optimum is $p_{RT}=\frac{\theta \delta (1+\beta )p_N+c_{RT}}{2}$. Then substituting $p_{RT}$ into Eq. (29) to calculate the optimal $p_N$. With $\frac{\partial ^2 \pi _O}{\partial p_{N}^2}=-\frac{(1-\beta )(2-\theta \delta -\theta \delta \beta )}{1-\theta \delta }<0$, we can get the optimal $p_N^*=\frac{c_N}{2(1-\beta )}+\frac{c_{RT}+2(1-\theta \delta )}{2(2-\theta \delta -\theta \delta \beta )}$ by solving $\frac{\partial \pi _O}{\partial p_{N}}=0$. Substituting optimal $p_N$ into the demand functions, we have the sales of the two products are $D_N^*=\frac{c_{RT}+2(1-\theta \delta )}{4(1-\theta \delta )} -\frac{(2-\theta \delta -\theta \delta \beta )c_N}{4(1-\theta \delta )(1-\beta )}$ and $D_{RT}^*=\frac{c_N}{4(1-\theta \delta )}-\frac{(4-3\theta \delta -\theta \delta \beta ) c_{RT}}{4\theta \delta (1-\theta \delta )(2-\theta \delta -\theta \delta \beta )}+\frac{1-\beta }{2(2-\theta \delta -\theta \delta \beta )}.$ To ensure the sales are non-negative, we have

$\frac{(4-3\theta \delta -\theta \delta \beta )c_{RT}-2\theta \delta (1-\beta )(1-\theta \delta )}{\theta \delta (2-\theta \delta -\theta \delta \beta )}\le c_N\le \frac{(1-\beta )[c_{RT}+2(1-\theta \delta )]}{2-\theta \delta (1+\beta )}$ under the case, and it also implies $c_{RT}\le \theta \delta (1-\beta )$. Substituting the optimal results into the relevant function, Proposition 6 is proved.

E Proof of Proposition 7

With the same method as other proposition, we will first calculate the retail prices of two products which is showed by Eqs. (35) and (36). As Hessian matrix of the $p_N$ and $p_{RT}$ in $\pi _T$ is $\left( \begin{array}{ccc} -\frac{2}{1-\delta } &{} 0 \\ 0 &{} -\frac{2}{(1-\theta )\delta }-\frac{2}{\theta \delta }\\ \end{array} \right)$. and $|H|=\frac{4}{(1-\theta )(1-\delta )\theta \delta }>0$, Thus, we can conclude they jointly concave in $\pi _T$. In addition, the $\frac{\partial ^2 \pi _O}{\partial p_{RO}^2}=-\frac{2}{1-\delta }-\frac{2}{\delta (1-\theta )}<0$, $p_{RO}$ is concave in $\pi _O$. Then solving the $\frac{\partial \pi _T}{\partial p_{N}}=0$, $\frac{\partial \pi _T}{\partial p_{RT}}=0$ and $\frac{\partial \pi _O}{\partial p_{RO}}=0$ jointly, we can get the optimum $p_N$, $p_{RT}$ and $p_{RO}$. Substituting them into $\pi _O$ to optimize for w, with $\frac{\partial ^2 \pi _O}{\partial w^2}=-\frac{2(8+\delta )-(19\delta +8)\theta +(1+8\delta )\theta ^2}{(4-\theta -\delta -2\theta \delta )^2}$ and $\theta \ge \delta$, we can conclude $\frac{\partial ^2 \pi _O}{\partial w^2}<0$ and w is concave in $\pi _O$. At last, we get optimal ${\hat{w}}=\frac{[(10\theta ^2-23\theta +4)\delta +(4-\theta )^2]c_N-2(1-\theta ) (\theta +\delta -2\theta \delta )c_{RO}}{2[\delta (8\theta ^2-19\theta +2)+(4-\theta )^2]} +\frac{(8+\theta +\delta -10\theta \delta )c_{RT}+(14\theta ^2-7\theta +2)\delta ^2-3 \theta \delta (7-\theta )+(4-\theta )^2}{2[\delta (8\theta ^2-19\theta +2)+(4-\theta )^2]}$ by solving $\frac{\partial \pi _O}{\partial w}=0$. Then substituting ${\hat{w}}$ into other function, we have we will have ${\hat{D}}_{N} =\frac{{\begin{matrix}2[\delta (7\theta ^2-17\theta +4)-\theta ^2-\theta +8]c_ {RO}+(1-\delta )[\delta (17\theta ^2-34\theta +8)\\ +(4-\theta )^2-3\theta c_{RT}]\end{matrix}}}{4(1-\delta )[\delta (8\theta ^2-19\theta +2)+(4-\theta )^2]} -\frac{[\delta (11\theta ^2-28\theta +8)+(4-\theta )^2]c_N}{4(1-\delta ) [\delta (8\theta ^2-19\theta +2)+(4-\theta )^2]},$

${\hat{D}}_{RO}=\frac{(1-\delta )\{[(3\theta ^2-10\theta +1)\delta +2(4-\theta )]c_{RT} +\delta (1-\theta )[\delta (\theta ^2-9\theta +2)+(1-\theta )(4-\theta )]\}}{2\delta (1-\theta )(1-\delta ) [\delta (8\theta ^2-19\theta +2)+(4-\theta )^2]} +\frac{[(\theta ^3-5\theta +1)\delta ^2+(2\theta ^3-11\theta ^2+14\theta +1)\delta -(2-\theta )(4-\theta )] c_{RO}}{\delta (1-\theta )(1-\delta )[\delta (8\theta ^2-19\theta +2)+(4-\theta )^2]} +\frac{[(4-\theta )(3-\theta )-\theta \delta (11-5\theta )]c_{N}}{2(1-\delta )[\delta (8\theta ^2-19\theta +2) +(4-\theta )^2]}$

and ${\hat{D}}_{RT}=\frac{{\begin{matrix}\theta \{\delta (1-\theta )[20-13\theta \delta -2\delta -5\theta -(4-\theta )c_N]\\ +2[\delta (5\theta ^2-14\theta +3)+2(4-\theta )]c_{RO}\}\end{matrix}}}{4\theta \delta (1-\theta )[\delta (8\theta ^2 -19\theta +2)+(4-\theta )^2]} -\frac{[\delta (23\theta ^2-39\theta +4)+4(\theta ^2-6\theta +8)]c_{RT}}{4\theta \delta (1-\theta ) [\delta (8\theta ^2-19\theta +2)+(4-\theta )^2]}$. To make sure the sales are non-negative, the equilibrium results should be under the condition

$\frac{{\begin{matrix}2[(2 - \theta )(4 - \theta ) - (\theta ^3 - 5\theta + 1)\delta ^2 - (2\theta ^3 - 11\theta ^2 + 14\theta + 1)\delta ]c_{RO} - (1 - \delta )\\ \{[(3\theta ^2 - 10\theta + 1)\delta + 2(4 - \theta )]c_{RT} + \delta (1 - \theta )[(\theta ^2 - 9\theta + 2) \delta + (\theta + 1)(4 - \theta )\}\end{matrix}}}{\delta (1 - \theta )[(4 - \theta )(3 - \theta ) - \theta \delta (11 - 5\theta )]}$

$<c_N<\min \Big \{\frac{{\begin{matrix}2(\delta (7\theta ^2 - 17\theta + 4) - \theta ^2 - \theta + 8)c_{RO} + (1 - \delta )[\delta (17\theta ^2 \\ - 34\theta + 8) + (4 - \theta )^2 - 3\theta c_{RT}]\end{matrix}}}{\delta (11\theta ^2 - 28\theta + 8) + (4 - \theta )^2}$,

$~\frac{{\begin{matrix}\theta \delta (1 - \theta )(-13\delta \theta - 2\delta - 5\theta + 20) + 2\theta [(5\theta ^2 - 14\theta + 3)\delta + 8 - 2\theta ]c_{R O} \\ - [(23\theta ^2 - 39\theta + 4)\delta + 4\theta ^2 - 24\theta + 32] c_{RT}\end{matrix}}}{\theta \delta (1 - \theta )(4 - \theta )}\Big \}$. Then the proposition is improved.

F Proof of Proposition 8

Using the same method with the last proposition, we will first calculate the optimal $p_{RT}$ in the retailer’s profit function which is showed by Eq. (47). As the $\frac{\partial ^2 \pi _T}{\partial p_{RT}^2}=-\frac{2}{\theta \delta (1-\theta )}<0$, $p_{RT}$ is concave in retailer’s profit function. Thus, we can get the optimum $p_{RT}=\frac{\theta p_{RO}+c_{RT}}{2}$ by solving $\frac{\partial \pi _T}{\partial p_{RT}}=0$. Then substituting the $p_{RT}$ in to the OEM’s profit function to find the optimal $p_N$ and $p_{RO}$, we have the Hessian matrix is$\left( \begin{array}{ccc} -\frac{2(1-\beta )}{1-\delta } &{} \frac{2-\beta }{1-\delta } \\ \frac{2-\beta }{1-\delta } &{} -\frac{2-\theta -\theta \delta }{\delta (1-\delta )(1-\theta )}\\ \end{array} \right)$ and $|H|=\frac{2[(2-\theta )(1-\delta )(1-\beta )-\delta \beta ^2(1-\theta )]}{\delta (1-\delta )^2(1-\theta )}$. With the assumption $\delta <\frac{4(1-\beta )}{(2-\beta )^2}$, we have $|H|>0$ and can conclude the $p_N$ and $p_{RO}$ are jointly concave in $\pi _O$. Hence, we can obtain $p_N^*=\frac{{\begin{matrix}c_N+(2-\theta -\theta \delta )[2(1-\beta )(1-\delta )-\beta c_{RO}]\\ +(1-\delta )(2-\beta )c_{RT}\end{matrix}}}{2[(2-\theta )(1-\delta )(1-\beta )-\delta \beta ^2(1-\theta )]}$ and

$p_{RO}^*=\frac{{\begin{matrix}\delta \beta (1-\theta )c_N+[(2-\theta )(1-\delta -\beta )+\beta \delta ] c_{RO}+(1-\delta )(1-\beta )\\ [(1-\theta )(2-\beta )\delta +c_{RT}]\end{matrix}}}{2(2-\theta )(1-\delta ) (1-\beta )-\delta \beta ^2(1-\theta )}$ by solving $\frac{\partial \pi _O}{\partial p_{N}}=0$ and $\frac{\partial \pi _O}{\partial p_{RO}}=0$ jointly. Substituting them into the demands function and ensure the non-negative sales, it needs to notice that the case should under the condition

$\max \{\frac{(1-\beta )\{[(2-\theta )c_{RO}-\beta \delta (1-\theta )](2-\theta -\theta \delta ) -[\beta \delta (1-\theta )+(2-\theta )(1-\delta )]c_{RT}\}}{\delta (1-\theta )(2-\theta )(2-\beta )},$

$\frac{{\begin{matrix}c_{RT}-\theta \delta (1-\beta )(2-\beta )(1-\theta )\\ (1-\delta )-\theta [(2-\theta )(1-\delta )-(2-\theta -\delta )\beta ]c_{RO}\end{matrix}}}{\theta \beta \delta (1-\theta )}\}\le c_N\le 1-\beta -\delta +\frac{\beta (2\delta -c_{RT})}{2(2-\theta )}+\frac{(2-\beta )c_{RO}}{2}$. Substituting the optimal results into other functions, we can have Proposition 8.