Optimal pricing strategy of a hybrid trade old for new and remanufactured products supply chain

Abstract

Remanufacturing has been widely recognized in practice. However, the low acceptance of remanufactured products deeply hinders the performance of remanufacturing. Recently, trade-in programs are growing in popularity for selling remanufactured products to promote consumptions. In this paper, under the constraint of consumer participation, we investigate the conditions under which a trade-in program for remanufactured products should be adopted and how to optimize the pricing and production decisions. Cannibalization between new and remanufactured products has been analyzed for both primary and replacement markets.

Appendix. Proofs

Proof of Proposition 1

Substituting $D_{1n}^{NR}=1-p_{1n}$, $D_{2n}^{NR}=n\cdot (1-p_{2n})$ and $D_{2n-t}^{NR}=1-\frac{p_{2n}-p_{2t}}{\delta }$ into the profit function given in Eq. (1), we can transform the decision problem as follows:

$$\begin{aligned} \begin{array}{l} \mathop {\max \varPi ^{NR}}\limits _{p_{1n}, p_{2n}, p_{2t}} =\left( {p_{1n} -c_{n}} \right) (1-p_{1n}) +n\left( {p_{2n} -c_{n}} \right) (1-p_{2n}) \\ \qquad \qquad \qquad +\left( {p_{2n}+s -p_{2t}-c_n} \right) \left( 1-\frac{p_{2n}-p_{2t}}{\delta }\right) \\ \text {subject~to}\quad 0\le 1-\frac{p_{2n}-p_{2t}}{\delta }\le (1-\alpha )(1-p_{1n}),\\ \phantom {\text {subject~to}}\quad 1-p_{1n}\ge 0, n (1-p_{2n})\ge 0 \end{array} \end{aligned}$$

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Table 3 Lagrangian multipliers in model NR

The first inequality tells that the trade-in demand can not be larger than the demands of buying new products in period 1, and the trade-in demands must be non-negative. Because the last two non-negative constraints will lead to trivial cases, thus we omit them. Let ${\tilde{\alpha }}=1-\alpha$, the corresponding Lagrangian problem can be formulated as follows:

$$\begin{aligned} \begin{array}{l} \ell ^{NR}({p_{1n}, p_{2n}, p_{2t}}) =\left( {p_{1n} -c_{n}} \right) (1-p_{1n}) +n\left( {p_{2n} -c_{n}} \right) (1-p_{2n}) \\ \qquad \qquad \qquad +\left( {p_{2n}+s -p_{2t}-c_n} \right) \left( 1-\frac{p_{2n}-p_{2t}}{\delta }\right) \\ \qquad \qquad \qquad +l_1({\tilde{\alpha }}(1-p_{1n})-\frac{\delta -p_{2n}+p_{2t}}{\delta })+l_2\left( 1-\frac{p_{2n}-p_{2t}}{\delta }\right) \\ \end{array} \end{aligned}$$

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with

$$\begin{aligned} \frac{\partial \ell ^{NR}}{\partial p_{1n}}= & {} 1-2p_{1n}+c_n-l_1{\tilde{\alpha }}=0\\ \frac{\partial \ell ^{NR}}{\partial p_{2n}}= & {} n(1-2p_{2n}+c_n)-\frac{s-p_{2t}-c_n-\delta -p_{2t}-l_1}{\delta }=0\\ \frac{\partial \ell ^{NR}}{\partial p_{2t}}= & {} -\frac{2p_{2t}-p_{2n}-s+c_n+\delta +l_1}{\delta }=0\\ \frac{\partial \ell ^{NR}}{\partial l_{1}}= & {} {\tilde{\alpha }}(1-p_{1n})-\frac{\delta -p_{2n}+p_{2t}}{\delta }=0 \end{aligned}$$

Because the Hessian of this Lagrangian is positive definite, we have 4 cases to consider (see the following table).

For Case NR-1, it requires $q_{1n} =q_{2n-t}=0$, which is not possible to happen, so it should be omitted. For Case NR-2, by solving the systems of equations, it can be obtained that $p_{1n}=\frac{1+2{\tilde{\alpha }}^2\delta +c_n-{\tilde{\alpha }}\delta +{\tilde{\alpha }} c_n-{\tilde{\alpha }} s}{2(1+{\tilde{\alpha }}^2\delta )}$, $p_{2n}=\frac{1+c_n}{2}$, $p_{2t}=\frac{{\tilde{\alpha }}\delta +{\tilde{\alpha }}^2\delta s-{\tilde{\alpha }}\delta c_n-{\tilde{\alpha }}^2\delta ^2-2\delta +1+\delta {\tilde{\alpha }}^2+c_n}{2(1+{\tilde{\alpha }}^2\delta )}$ and $\ell _1=\frac{(\delta -\tilde{a}\delta +\tilde{a}\delta c_n-c_n+s)}{1+\tilde{a}^2\delta }$. $\ell _1>0$ requires that $s>(1-{\tilde{\alpha }}\delta )c_n-\delta (1-{\tilde{\alpha }})$. In this case, we have $D_{2n-t}=D_{1n}=\frac{1-c_n+{\tilde{\alpha }}\delta -{\tilde{\alpha }} c_n+{\tilde{\alpha }} s}{2(1+{\tilde{\alpha }}^2\delta )}$, and the non-negativity of this demand requires that $s>\frac{c_n-1+{\tilde{\alpha }}(c_n-\delta )}{{\tilde{\alpha }}}$. Because $(1-{\tilde{\alpha }}\delta )c_n-\delta (1-{\tilde{\alpha }}) -\frac{c_n-1+{\tilde{\alpha }}(c_n-\delta )}{{\tilde{\alpha }}}= \frac{({\tilde{\alpha }}^2 \delta +1)(1-c_n)}{{\tilde{\alpha }}}>0$, thus, the existence of Case NR-2 requires that $s>(1-{\tilde{\alpha }}\delta )c_n-\delta (1-{\tilde{\alpha }})$. Similarly we can see that the condition when Case NR-3 exists is $s\le (1-{\tilde{\alpha }}\delta )c_n-\delta (1-{\tilde{\alpha }})$. Note that in Case NR-4, the trade-in demand is 0 with $\ell _2=-\delta +c_n-s$, and $\ell _2>0$ requires $s<-\delta +c_n$. $\square$

Proof of Proposition 8

Because $c_{r1}-c_{r2}=2\sqrt{-2\mu c_n+\mu +c_n^2\mu }>0$ then we have $c_{r1}-c_{r2}>0$. As $c_{r1}c_{r2}=\mu ^2-(-2\mu c_n+\mu +c_n^2\mu )=\mu [\mu -(c_n-1)^2]=\mu (\sqrt{\mu }+1-c_n)(\sqrt{\mu }-1+c_n)$. Because $D_{2n}^{R}=\frac{n(-1+c_n+\mu -c_r)}{2(\mu -1)}>0$, which tells that $\mu -c_r>1-c_n>0$, and because $0<\mu <1$, then we have $\sqrt{\mu }>\sqrt{\mu }-c_r>\sqrt{\mu }-c_r-1+c_n>\mu -c_r-1+c_n>0$. Thus $c_{r1}c_{r2}=\mu (\sqrt{\mu }+1-c_n)(\sqrt{\mu }-1+c_n)>\mu (\sqrt{\mu }+1-c_n)(\sqrt{\mu }-c_r-1+c_n)>0$ and $c_{r1}>c_{r2}>0$. By solving $\partial (\varPi _{2n}^{R}-\varPi _{2r}^{R})/\partial c_r=0$, we have $c_r=\mu$, where $(\varPi _{2n}^{R}-\varPi _{2r}^{R})$ reaches its maximum point. As $0<\mu <1$, thus, $0<c_{r2}<\mu <1$. The non-negativity of $D_{2r}^{R}$ requires that $\mu -c_r>\mu c_n-c_r>0$, which implies that $\mu >c_r$. Thus, the possible $c_r$ that makes $\varPi _{2n}^{R}-\varPi _{2r}^{R}>0$ or $\varPi _{2n}^{R}-\varPi _{2r}^{R}\le 0$ must satisfy $c_r<\mu$. Because of the fact that $\varPi _{2r}^{R}-\varPi _{2n}^{R}=\frac{(c_r-c_{r2})(c_r-c_{r1})}{\mu (1-\mu )}$, and therefore we have two conditions to consider: (1) when $c_{r2}<c_r$, or equivalently, $c_n\le \frac{c_r+\sqrt{\mu }-\mu }{\sqrt{\mu }}$, the value of $(\varPi _{2n}^{R}-\varPi _{2r}^{R})$ is positive because $c_r-c_{r2}>0$ and $c_r-c_{r1}<0$; (2) when $c_{r2}\ge c_r$, or equivalently, $c_n>\frac{c_r+\sqrt{\mu }-\mu }{\sqrt{\mu }}$, the value of $(\varPi _{2n}^{R}-\varPi _{2r}^{R})$ is negative because $c_r-c_{r2}\le 0$ and $c_r-c_{r1}<0$. $\square$

Proof of Proposition 9

It can be obtained that $c_{r4}-c_{r3}=2\sqrt{\frac{(c_n-1+\delta -s)^2(\mu -\delta )}{1-\delta }}>0$ then we have $c_{r4}>c_{r3}$, where $c_{r3}=\mu -\delta +s-\sqrt{\frac{(c_n-1+\delta -s)^2(\mu -\delta )}{1-\delta }}$. Because $1<\mu <1$, then $c_{r3}=\mu -\delta +s-\sqrt{\frac{(c_n-1+\delta -s)^2(\mu -\delta )}{1-\delta }}> \mu -\delta +s-\sqrt{\frac{(c_n-1+\delta -s)^2(1-\delta )}{1-\delta }}=\mu -\delta +s-\sqrt{(c_n-1+\delta -s)^2}$. As $\mu -\delta +s>\mu -c_r-\delta +s>0$, then we have $\mu -\delta +s+\sqrt{(c_n-1+\delta -s)^2}>0$, and with the assumption that $\delta >s$, we can obtain $(\mu -\delta +s+\sqrt{(c_n-1+\delta -s)^2})(\mu -\delta +s-\sqrt{(c_n-1+\delta -s)^2})=(\mu -1+c_n)(\mu -2\delta +2s+1-c_n)>0$. Thus, we have $c_{r3}>0$. By solving $\partial (\varPi _{2n-t}^{R}-\varPi _{2r-t}^{R})/\partial c_r=0$, we obtain that when $c_r={-\delta +s+\mu }$, $(\varPi _{2n-t}^{R}-\varPi _{2r-t}^{R})$ reaches its maximum point. As $s-\delta <0$, then we have $<0-\delta +s+\mu <1$. Because of the fact that the parabola $(\varPi _{2n-t}^{R}-\varPi _{2r-t}^{R})$ is completely axial symmetry on $c_r$, and then we have two conditions to consider: (1) $c_{r4}=\mu -\delta +s+\sqrt{\frac{(c_n-\xi )^2(\mu -\delta )}{1-\delta }}>1$, in this case. (2) $c_{r4}=\mu -\delta +s+\sqrt{\frac{(c_n-\xi )^2(\mu -\delta )}{1-\delta }}\le 1$, The rest of the proof is similar to that of Proposition x, and we omit to show it. $\square$