Optimal production strategy for a manufacturing and remanufacturing system with return policy

Abstract

This paper considers a monopolistic producer who produces new product in the first period and provides manufactured and remanufactured products in the second period. A return policy in the first period and the competition between the manufactured and the remanufactured products in the second period are jointly considered to develop the optimal production quantity strategies analytically and numerically. We find that if the two products are substitutable, the quantities of the new products manufactured in the first period and the remanufactured products produced in the second period are non-decreasing in the return policy, but the quantity of the manufactured products in the second period is decreasing in the return policy. In this case, the return rate determines the profit level, but the collection rate has little impact on the profit. If the two products are complementary, there is a critical value of the collection rate. When the collection rate is smaller than this value, the production quantities increase in the return policy; otherwise, they decrease in the return policy. In this scenario, the return policy and the collection rate mutually make a positive impact on the profit. Additionally, although a generous return policy is always profitable, it may lead to less production of manufactured and remanufactured products, especially when they are complementary.

Appendix

Proof of Proposition 1

The first derivatives of the objective function with respect to the production quantities $q_{1}$, $q_{2N}$, and $q_{2R}$ are:

$$ \frac{\partial \pi }{{\partial q_{1} }} = - \frac{2}{1 - r}q_{1} + \frac{a}{1 - r} - c,\quad \frac{\partial \pi }{{\partial q_{2N} }} = a - 2q_{2N} + 2kq_{2R} - c,\quad {\text{and}}\quad \frac{\partial \pi }{{\partial q_{2R} }} = a - 2q_{2R} + 2kq_{2N} - c + s. $$

Therefore, the second derivatives and the cross partial derivatives of the objective function with respect to the production quantities are:

$$ \begin{aligned} \frac{{\partial^{2} \pi }}{{\partial q_{1}^{2} }} & = - \frac{2}{1 - r},\quad \frac{{\partial^{2} \pi }}{{\partial q_{1} \partial q_{2N} }} = 0,\quad \frac{{\partial^{2} \pi }}{{\partial q_{1} \partial q_{2R} }} = 0; \\ \frac{{\partial^{2} \pi }}{{\partial q_{2N}^{2} }} & = - 2,\quad \frac{{\partial^{2} \pi }}{{\partial q_{2N} \partial q_{1} }} = 0,\quad \frac{{\partial^{2} \pi }}{{\partial q_{2N} \partial q_{2R} }} = 2k; \\ \frac{{\partial^{2} \pi }}{{\partial q_{2R}^{2} }} & = - 2,\quad \frac{{\partial^{2} \pi }}{{\partial q_{2R} \partial q_{1} }} = 0,\quad \frac{{\partial^{2} \pi }}{{\partial q_{2R} \partial q_{2N} }} = 2k. \\ \end{aligned} $$

We can get the corresponding Hessian Matrix as

$$ \left[ {\begin{array}{*{20}c} {\frac{{\partial^{2} \pi }}{{\partial q_{1}^{2} }}} & {\frac{{\partial^{2} \pi }}{{\partial q_{1} \partial q_{2N} }}} & {\frac{{\partial^{2} \pi }}{{\partial q_{1} \partial q_{2R} }}} \\ {\frac{{\partial^{2} \pi }}{{\partial q_{2N} \partial q_{1} }}} & {\frac{{\partial^{2} \pi }}{{\partial q_{2N}^{2} }}} & {\frac{{\partial^{2} \pi }}{{\partial q_{2N} \partial q_{2R} }}} \\ {\frac{{\partial^{2} \pi }}{{\partial q_{2R} \partial q_{1} }}} & {\frac{{\partial^{2} \pi }}{{\partial q_{2R} \partial q_{2N} }}} & {\frac{{\partial^{2} \pi }}{{\partial q_{2R}^{2} }}} \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} { - \frac{2}{1 - r}} & \quad 0 & \quad 0 \\ 0 & \quad { - 2} & \quad {2k} \\ 0 & \quad {2k} & \quad { - 2} \\ \end{array} } \right]. $$

Consequently, when $k^{2} < 1$, it is easy to verify that the Hessian matrix is strictly negative definite as $H_{1} < 0$, $H_{2} > 0$, and $H_{3} < 0$, where $H_{1}$, $H_{2}$, and $H_{3}$ are the principal minors that $H_{1} = [ { - \frac{2}{1 - r}, - 2, - 2} ]$, $H_{2} =[ {\frac{4}{1 - r},\frac{4}{1 - r},4 - 4k^{2} } ]$, and $H_{3} = - \frac{2}{1 - r}( {4 - 4k^{2} } )$. Thus, the objective function is strictly concave. We complete the proof.$\square$

Proof of Proposition 2

From Proposition 1, we know that the objective function is strictly concave such that the optimal production quantities could be determined by Karush–Kuhn–Tucker (KKT) conditions. Denoting by $\lambda$ the Lagrangian slack variable, the new objective function can be rewritten as:

$$ \begin{aligned} & L\left( {q_{1} ,q_{2N} ,q_{2R} ,\lambda } \right) \\ & \quad = \left( {\frac{1}{1 - r}\left( {a - q_{1} } \right) - c} \right)q_{1} + \left( {a - q_{2N} + kq_{2R} - c} \right) \cdot q_{2N} + \left( {a - q_{2R} + kq_{2N} - c + s} \right) \cdot q_{2R} + \lambda \left( {\theta q_{1} - q_{2R} } \right) \\ \end{aligned} $$

subject to

$$ \left\{ {\begin{array}{*{20}l} {\lambda \left( {\theta q_{1} - q_{2R} } \right) = 0} \hfill \\ {q_{1} ,q_{2N} ,q_{2R} ,\lambda \ge 0} \hfill \\ \end{array} } \right. $$

By setting

$$ \frac{\partial L}{{\partial q_{1} }} = - \frac{2}{1 - r}q_{1} + \frac{a}{1 - r} - c + \lambda \theta = 0; $$

(4)

$$ \frac{\partial L}{{\partial q_{2N} }} = a - 2q_{2N} + 2kq_{2R} - c = 0; $$

(5)

$$ \frac{\partial L}{{\partial q_{2R} }} = a - 2q_{2R} + 2kq_{2N} - c + s - \lambda = 0; $$

(6)

The optimal solutions could be achieved by considering the following two cases.

Case 1 If $\lambda = 0$, then $\theta q_{1} \ge q_{2R}$, and we have:

$$ \left\{ {\begin{array}{*{20}l} { - {{2q_{1} } \mathord{\left/ {\vphantom {{2q_{1} } {\left( {1 - r} \right)}}} \right. \kern-0pt} {\left( {1 - r} \right)}} + {a \mathord{\left/ {\vphantom {a {\left( {1 - r} \right)}}} \right. \kern-0pt} {\left( {1 - r} \right)}} - c = 0;} \hfill \\ {a - 2q_{2N} + 2kq_{2R} - c = 0;} \hfill \\ {a - 2q_{2R} + 2kq_{2N} - c + s = 0.} \hfill \\ \end{array} } \right. $$

Solving the above equations, we can get:

$$ q_{1}^{*} = \frac{{a - \left( {1 - r} \right)c}}{2},\quad q_{2N}^{*} = \frac{{\left( {a - c} \right)\left( {1 + k} \right) + ks}}{{2\left( {1 - k^{2} } \right)}},\quad {\text{and}}\quad q_{2R}^{*} = \frac{{\left( {a - c} \right)\left( {1 + k} \right) + s}}{{2\left( {1 - k^{2} } \right)}}. $$

We can verify that $q_{1}^{*} > 0$, $q_{2R}^{*} > 0$. From $q_{2N}^{*} > 0$, we need $\left( {a - c} \right)\left( {1 + k} \right) + ks > 0$. Additionally, $p_{1}^{*} - c = \frac{1}{1 - r}\left( {a - q_{1}^{*} } \right) - c = \frac{{a - \left( {1 - r} \right)c}}{{2\left( {1 - r} \right)}} > 0$, $p_{2N}^{*} - c = a - q_{2N}^{*} + kq_{2R}^{*} - c = \frac{a - c}{2} > 0$, and $p_{2R}^{*} - c + s = a - q_{2R}^{*} + kq_{2N}^{*} - c + s = \frac{a - c + s}{2} > 0$, which indicate that the selling prices are acceptable. However, from the constraint $\theta q_{1} \ge q_{2R}$, with some algebra, we have the following condition:

$$ \theta \left( {1 - k^{2} } \right)\left( {a - \left( {1 - r} \right)c} \right) - \left( {1 + k} \right)\left( {a - c} \right) - s \ge 0. $$

Thus, defining $c_{1} = \theta \left( {1 - k^{2} } \right)\left( {a - \left( {1 - r} \right)c} \right) - \left( {1 + k} \right)\left( {a - c} \right) - s$ and $c_{2} = \left( {a - c} \right)\left( {1 + k} \right) + ks$, we get the following conclusion:

(a) In case $c_{1} \ge 0$ and $c_{2} > 0$, the optimal production quantities are:

$$ q_{1}^{*} = \frac{{a - \left( {1 - r} \right)c}}{2},\quad q_{2N}^{*} = \frac{{\left( {a - c} \right)\left( {1 + k} \right) + ks}}{{2\left( {1 - k^{2} } \right)}},\quad {\text{and}}\quad q_{2R}^{*} = \frac{{\left( {a - c} \right)\left( {1 + k} \right) + s}}{{2\left( {1 - k^{2} } \right)}}; $$

(b) In case $c_{1} \ge 0$ and $c_{2} \le 0$, the optimal production quantities are

$$ q_{1}^{*} = \frac{{a - \left( {1 - r} \right)c}}{2},\quad q_{2N}^{*} = 0,\quad {\text{and}}\quad q_{2R}^{*} = \frac{{\left( {a - c} \right)\left( {1 + k} \right) + s}}{{2\left( {1 - k^{2} } \right)}}. $$

Case 2 If $\lambda > 0$, then $\theta q_{1} = q_{2R}$, and we have:

Substituting it into Eq. (5), with some algebra, we have:

$$ q_{2N} = \frac{{a + 2\theta kq_{1} - c}}{2}. $$

(7)

Additionally, with (4) + (6) * θ, we get

$$ \theta \left( {a - 2q_{2R} + 2kq_{2N} - c + s} \right) - \frac{2}{1 - r}q_{1} + \frac{a}{1 - r} - c = 0. $$

(8)

Substituting Eq. (7) into (8), with some algebra, we have:

$$ q_{1}^{*} = \frac{{a - \left( {1 - r} \right)c + \theta \left( {1 - r} \right)\left( {\left( {a - c} \right)\left( {1 + k} \right) + s} \right)}}{{2\left( {1 + \theta^{2} \left( {1 - k^{2} } \right)\left( {1 - r} \right)} \right)}},\quad q_{2N}^{*} = \frac{{a + 2\theta kq_{1}^{*} - c}}{2},\quad {\text{and}}\quad q_{2R}^{*} = \theta q_{1}^{*} . $$

Similarly, we can verify that $q_{1}^{*} > 0$, $q_{2R}^{*} > 0$. From $q_{2N}^{*} > 0$, we need $a + 2\theta kq_{1}^{*} - c > 0$, which indicates that $a - c + \theta k\left( {a - \left( {1 - r} \right)c + \theta \left( {1 - r} \right)s} \right) + \theta^{2} \left( {a - c} \right)\left( {1 + k} \right)\left( {1 - r} \right) > 0$. In addition, the constraints in terms of the selling prices are

$$ p_{1}^{*} - c = \frac{1}{1 - r}\left( {a - q_{1}^{*} } \right) - c = \frac{{a - \left( {1 - r} \right)c - \theta \left( {1 - r} \right)s + \theta \left( {1 + k} \right)\left( {1 - r} \right)\left( {2\theta \left( {1 - k} \right)\left( {a - \left( {1 - r} \right)c - s} \right) - a + c} \right)}}{{2\left( {1 + \theta^{2} \left( {1 - k^{2} } \right)\left( {1 - r} \right)} \right)}}, $$

$$ p_{2N}^{*} - c = a - q_{2N}^{*} + kq_{2R}^{*} - c = \frac{a - c}{2}, $$

$$ \begin{aligned} p_{2R}^{*} - c + s & = a - q_{2R}^{*} + kq_{2N}^{*} - c + s \\ & = \frac{{\left( {2 + k + \theta \left( {1 - k^{2} } \right)\left( {\theta \left( {1 - r} \right) - 1} \right)} \right)\left( {a - c} \right) + 2s + \theta \left( {1 - k^{2} } \right)\left( {\theta \left( {1 - r} \right)s + rc} \right)}}{{1 + \theta^{2} \left( {1 - k^{2} } \right)\left( {1 - r} \right)}}. \\ \end{aligned} $$

It is clearly that $p_{2N}^{*} - c > 0$. In terms of $p_{2R}^{*} - c + s$, since $0 \le \theta < 1$, $0 < 1 - k^{2} \le 1$, and $- 1 \le \theta \left( {1 - r} \right) - 1 < 0$, we get $0 < 2 + k + \theta \left( {1 - k^{2} } \right)\left( {\theta \left( {1 - r} \right) - 1} \right)$ such that the numerator of $p_{2R}^{*} - c + s$ is larger than zero. Thus, $p_{2R}^{*} - c + s > 0$. However, unfortunately, it is intractable to determine the sign of $p_{1}^{*} - c$ which is dependent on the values of the parameters.

Hence, defining $c_{3} = a - \left( {1 - r} \right)c - \theta \left( {1 - r} \right)s + \theta \left( {1 + k} \right)\left( {1 - r} \right)\left( {2\theta \left( {1 - k} \right)\left( {a - \left( {1 - r} \right)c - s} \right) - a + c} \right)$ and $c_{4} = a - c + \theta k\left( {a - \left( {1 - r} \right)c + \theta \left( {1 - r} \right)s} \right) + \theta^{2} \left( {a - c} \right)\left( {1 + k} \right)\left( {1 - r} \right)$, the conclusions are summarized as follows:

(c) In case $c_{1} < 0$, $c_{3} > 0$, and $c_{4} > 0$, the optimal production quantities are

$$ q_{1}^{*} = \frac{{a - \left( {1 - r} \right)c + \theta \left( {1 - r} \right)\left( {\left( {a - c} \right)\left( {1 + k} \right) + s} \right)}}{{2\left( {1 + \theta^{2} \left( {1 - k^{2} } \right)\left( {1 - r} \right)} \right)}},\quad q_{2N}^{*} = \frac{{a + 2\theta kq_{1}^{*} - c}}{2},\quad q_{2R}^{*} = \theta q_{1}^{*} ; $$

(d) In case $c_{1} < 0$, $c_{3} > 0$, and $c_{4} \le 0$, the optimal production quantities are

$$ q_{1}^{*} = \frac{{a - \left( {1 - r} \right)c + \theta \left( {1 - r} \right)\left( {\left( {a - c} \right)\left( {1 + k} \right) + s} \right)}}{{2\left( {1 + \theta^{2} \left( {1 - k^{2} } \right)\left( {1 - r} \right)} \right)}},\quad q_{2N}^{*} = 0,\quad q_{2R}^{*} = \theta q_{1}^{*} ; $$

(e) In case $c_{1} < 0$ and $c_{3} \le 0$, the optimal production quantities are

$$ q_{1}^{*} = 0,\quad q_{2N}^{*} = \frac{a - c}{2},\quad q_{2R}^{*} = 0. $$

Summarizing the results of (a)–(e) in subcases 1 and 2, we can conclude the optimal production quantities determined by Proposition 2.$\square$

Proof of Proposition 3

We prove the proposition as the following:

(1) When $c_{1} > 0$, the first derivatives of the optimal production quantities with respect to the return policy and the cross demand sensitivity are

$$ \frac{{\partial q_{1}^{*} }}{\partial r} = \frac{c}{2} > 0,\quad \frac{{\partial q_{2N}^{*} }}{\partial r} = 0,\quad \frac{{\partial q_{2R}^{*} }}{\partial r} = 0 $$

and

$$ \frac{{\partial q_{1}^{*} }}{\partial k} = 0,\quad \frac{{\partial q_{2N}^{*} }}{\partial k} = \frac{{\left( {1 + k} \right)^{2} \left( {a - c - s} \right) + 2\left( {k^{2} + k + 1} \right)s}}{{2\left( {1 - k^{2} } \right)^{2} }} > 0,\quad \frac{{\partial q_{2R}^{*} }}{\partial k} = \frac{{\left( {a - c} \right)\left( {1 + k} \right)^{2} + 2ks}}{{2\left( {1 - k^{2} } \right)^{2} }}. $$

It is easy to verify that there are two threshold values $k_{1} = \frac{{ - \sqrt {s\left( {2a - 2c + s} \right)} - s}}{a - c} - 1$ and $k_{2} = \frac{{\sqrt {s\left( {2a - 2c + s} \right)} - s}}{a - c} - 1$ such that when $k_{1} < k < k_{2}$, $\frac{{\partial q_{2R}^{*} }}{\partial k} \le 0$; otherwise, $\frac{{\partial q_{2R}^{*} }}{\partial k} > 0$. Additionally, we can verify that $k_{1} < - 1$ and $k_{2} < 1$. Therefore, when $- 1 < k < k_{1}$, we know $q_{2R}^{*}$ is decreasing in $k$; when $k_{2} \le k < 1$, we know $q_{2R}^{*}$ is increasing in $k$.

(2) When $c_{1} \le 0$ and $c_{2} > 0$, the corresponding first derivatives with respect to $r$ are

$$ \frac{{\partial q_{1}^{*} }}{\partial r} = \frac{{c + \theta \left( {\theta \left( {1 + k} \right)\left( {c - ak} \right) - s} \right)}}{{2\left( {1 + \theta^{2} \left( {1 - k^{2} } \right)\left( {1 - r} \right)} \right)^{2} }},\quad \frac{{\partial q_{2N}^{*} }}{\partial r} = \theta k\frac{{\partial q_{1}^{*} }}{\partial r},\quad \frac{{\partial q_{2R}^{*} }}{\partial r} = \theta \frac{{\partial q_{1}^{*} }}{\partial r}. $$

Similarly, we can get two threshold values $k_{1} = \frac{{ - \frac{1}{\theta }\sqrt {\theta^{2} \left( {a + c} \right)^{2} + 4a\left( {c - \theta s} \right)} - \left( {a - c} \right)}}{2a}$ and $k_{2} = \frac{{\frac{1}{\theta }\sqrt {\theta^{2} \left( {a + c} \right)^{2} + 4a\left( {c - \theta s} \right)} - \left( {a - c} \right)}}{2a}$ such that when $k_{1} < k < k_{2}$, $\frac{{\partial q_{1}^{*} }}{\partial r} > 0$; otherwise, $\frac{{\partial q_{1}^{*} }}{\partial r} \le 0$. It is readily to verify that $k_{1} \le \frac{{ - \frac{1}{\theta }\sqrt {\theta^{2} \left( {a + c} \right)^{2} } - \left( {a - c} \right)}}{2a} = - 1$; however, unfortunately, $k_{2}$ may be larger or less than 1, which depends on the values of related parameters. On the other hand, $\frac{{\partial q_{2N}^{*} }}{\partial r}$ has opposite sign to $\frac{{\partial q_{1}^{*} }}{\partial r}$ when $k < 0$. Thus, we have the conclusion described in the second part of the proposition.$\square$