A Fuzzy Reverse Logistics Inventory System Integrating Economic Order/Production Quantity Models

Abstract

This paper develops a reverse inventory model where the recoverable manufacturing process is affected by the learning theory. We propose the inclusion of the fuzzy demand rate of the serviceable products and the fuzzy collection rate of the recoverable products from customers in the total cost function of the model. Two popular defuzzification methods, namely the signed distance technique, a ranking method for fuzzy numbers, and the graded mean integration representation method are employed to find the estimate of the total cost function per unit time in the fuzzy sense. We provide a comprehensive numerical example to illustrate and compare the results obtained by the two mentioned defuzzification methods. This is one of the only few attempts in the related literature comparing the performance of these methods with the effect of the fuzziness of both of the demand and the collection rate in the presence of the learning simultaneously. The results indicate that deciding on which method could be used depends on the target strategy that could focus on the total cost, ordering lot size, or recovery lot size.

Appendices

Appendix 1

By taking the first derivative of the $\tilde{g}\left( n \right)$ with respect to n, we have

$$\frac{{\partial \tilde{g}\left( n \right) }}{\partial n} = b\left( {\beta^{\prime} + \delta^{\prime}} \right)n^{b - 1} - \frac{{2\alpha^{\prime}}}{{n^{3} }} + \left( {b - 1} \right)\zeta^{\prime}n^{b - 2} - \frac{{\gamma^{\prime}}}{{n^{2} }} > 0,$$

(50)

where

$$\alpha^{\prime} = - \frac{{\left( {nC_{\text{o}} + C_{\text{s}} } \right)H_{\text{s}} \Delta }}{{2C_{\text{o}} }},\;\beta^{\prime} = - \frac{{\left( {b + 1} \right)H_{\text{r}} a\rho^{b} }}{b + 2}d\left( {\tilde{r},\tilde{0}_{1} } \right),\;\gamma^{\prime} = \frac{{H_{\text{s}} }}{2}\Delta ,$$

$$\delta^{\prime} = - \frac{{a\rho^{b} H_{\text{s}} }}{{\left( {b + 2} \right)}}d\left( {\tilde{r},\tilde{0}_{1} } \right),\;\zeta^{\prime} = \left( {\frac{{ab\rho^{b - 1} }}{b + 1}} \right)C_{\text{l}} d\left( {\tilde{r},\tilde{0}_{1} } \right),\;\varepsilon^{\prime} = \frac{1}{2}H_{\text{r}} + \frac{{H_{\text{s}} }}{2}d\left( {\frac{{\tilde{r}}}{{\tilde{k}}},\tilde{0}_{1} } \right),$$

It is positive for all values of $n > 0$. Hence, $\tilde{g}\left( n \right)$ is a strictly increasing function for $0 < n < \infty$. Moreover, we have the following limitations

$$\mathop {\lim }\limits_{n \to + \infty } \tilde{g}\left( n \right) = \varepsilon^{\prime} > 0$$

(51)

$$\mathop {\lim }\limits_{{n \to 0^{ + } }} \tilde{g}\left( n \right) = - \infty$$

(52)

Thus, by the Intermediate Value Theorem, there exists a unique $0 < n^{*} < \infty$ such that $\tilde{g}\left( {n^{*} } \right) = 0$.

Appendix 2

From Theorem 1, it is clear that ($y^{*} ,n^{*}$) is the only critical point. Therefore, to prove the Theorem 2, we should firstly calculate the Hessian Matrix of $d\left( {\tilde{V},\tilde{0}} \right)$ as follows

$$\begin{aligned} \frac{{\partial^{2} d\left( {\tilde{V},\tilde{0}} \right)}}{{\partial y^{2} }} = & \frac{{\partial^{2} V\left( {n,y} \right)}}{{\partial y^{2} }} = \frac{{2\left( {nC_{\text{o}} + C_{\text{s}} } \right)}}{{y^{3} }}d\left( {\tilde{r},\tilde{0}_{1} } \right) - \frac{{b\left( {b + 1} \right)H_{\text{r}} ay^{b - 1} }}{b + 2}d\left( {\tilde{r},\tilde{0}_{1} } \right) \\ & - \frac{{bay^{b - 1} H_{\text{s}} }}{{\left( {b + 2} \right)}}d\left( {\tilde{r},\tilde{0}_{1} } \right) + \left( {\frac{{\left( {b - 1} \right)aby^{b - 2} }}{b + 1}} \right)C_{\text{l}} d\left( {\tilde{r},\tilde{0}_{1} } \right) \\ \end{aligned}$$

(53)

For all $y > 0$, $n > 0$, $\partial^{2} d\left( {\tilde{V},\tilde{0}} \right)/\partial y^{2} > 0$.

$$\frac{{\partial^{2} d\left( {\tilde{V},\tilde{0}} \right)}}{{\partial n^{2} }} = \frac{{yH_{\text{s}} }}{{n^{3} }}\Delta$$

(54)

For all $y > 0$, $n > 0$, $\partial^{2} d\left( {\tilde{V},\tilde{0}} \right)/\partial n^{2} > 0$.

$$\frac{{\partial^{2} V\left( {n,y} \right)}}{\partial y\partial n} = - \frac{{C_{o} d\left( {\tilde{r},\tilde{0}_{1} } \right)}}{{y^{2} }} - \frac{{H_{\text{s}} }}{{2n^{2} }}\Delta$$

(55)

Substituting $y^{*} = \rho n^{*}$ into Eqs. (53)–(55), and after some simplifications, determinant of the Hessian Matrix of $d\left( {\tilde{V},\tilde{0}} \right)$ at $(y^{*} ,n^{*} )$ could be given as below:

$$\begin{aligned} & \left| {\begin{array}{*{20}l} {\left. {\frac{{\partial^{2} V\left( {n,y} \right)}}{{\partial y^{2} }}} \right|_{{(y^{*} ,n^{*} )}} } & {\left. {\frac{{\partial^{2} V\left( {n,y} \right)}}{\partial y\partial n}} \right|_{{(y^{*} ,n^{*} )}} } \\ {\left. {\frac{{\partial^{2} V\left( {n,y} \right)}}{\partial y\partial n}} \right|_{{(y^{*} ,n^{*} )}} } & {\left. {\frac{{\partial^{2} V\left( {n,y} \right)}}{{\partial n^{2} }}} \right|_{{(y^{*} ,n^{*} )}} } \\ \end{array} } \right| \hfill \\& = \frac{{y^{*} H_{\text{s}} }}{{n^{{*}^{3}} }}\Delta \left[ { - \frac{{b\left( {b + 1} \right)H_{\text{r}} ay^{b - 1} }}{b + 2}d\left( {\tilde{r},\tilde{0}_{1} } \right) - \frac{{bay^{b - 1} H_{\text{s}} }}{{\left( {b + 2} \right)}}d\left( {\tilde{r},\tilde{0}_{1} } \right) + \left( {\frac{{\left( {b - 1} \right)aby^{b - 2} }}{b + 1}} \right)C_{\text{l}} d\left( {\tilde{r},\tilde{0}_{1} } \right)} \right] \hfill \\ &\quad+ \frac{{H_{\text{s}}^{2} \Delta^{2} C_{\text{s}} }}{{n^{{*}^{5}} C_{\text{o}} }} > 0 \hfill \\ \end{aligned}$$

(56)

Hessian Matrix of $d\left( {\tilde{V},\tilde{0}} \right)$ is positive. Hence, $d\left( {\tilde{V},\tilde{0}} \right)$ has a global minimum at point ($y^{*} ,n^{*}$).

Appendix 3

The first derivative of the $\tilde{f}\left( n \right)$ is positive for all value of $n > 0$.

$$\frac{{\partial \tilde{f}\left( n \right)}}{\partial n} = \frac{{2\gamma C_{\text{s}} }}{{6n^{3} C_{\text{o}} }} + \frac{{b\left( {\beta + \delta } \right)n^{b - 1} + \zeta (b - 1)n^{b - 2} }}{6} > 0$$

(57)

Hence, $\tilde{f}\left( n \right)$ is a strictly increasing function for $0 < n < \infty$. Moreover, we have the following limitations

$$\mathop {\lim }\limits_{n \to + \infty } \tilde{f}\left( n \right) = \varepsilon^{} > 0$$

(58)

$$\mathop {\lim }\limits_{{n \to 0^{ + } }} \tilde{f}\left( n \right) = - \infty$$

(59)

Thus, by the Intermediate Value Theorem, there exists a unique $0 < n^{*} < \infty$ such that $\tilde{f}\left( {n^{*} } \right) = 0$.

Appendix 4

From Theorem 3, it is clear that ($y^{*} ,n^{*}$) is the only critical point. Therefore, to prove the Theorem 4, we should firstly calculate the Hessian Matrix of $\varPhi (\tilde{V}\left( {y,n} \right))$ as follows

$$\begin{aligned} \frac{{\partial^{2} \varPhi (\tilde{V}\left( {y,n} \right))}}{{\partial y^{2} }} = & \frac{{2\left( {nC_{\text{o}} + C_{\text{s}} } \right)(6r + \theta_{2} - \theta_{1} )}}{{y^{3} }} - \frac{{ba\left( {b + 1} \right)H_{\text{r}} \left( {6r + \theta_{2} - \theta_{1} } \right)y^{b - 1} }}{b + 2} \\ & - \frac{{bay^{b - 1} H_{\text{s}} \left( {6r + \theta_{2} - \theta_{1} } \right)}}{{\left( {b + 2} \right)}} + \frac{{abC_{\text{l}} (b - 1)(6r + \theta_{2} - \theta_{1} )y^{b - 2} }}{b + 1} \\ \end{aligned}$$

(60)

For all $y > 0$, $n > 0$, $\partial^{2} \varPhi (\tilde{V}\left( {y,n} \right))/\partial y^{2} > 0$

$$\begin{aligned} \frac{{\partial^{2} \varPhi (\tilde{V}\left( {y,n} \right))}}{{\partial n^{2} }} = & \frac{{yH_{\text{s}} }}{{3n^{3} }}\left[ {\frac{{(k - r - \theta_{3} - \theta_{2} )^{2} }}{{2\left( {k + \theta_{4} } \right)\left( {r + \theta_{2} } \right)}} + \frac{{2\left( {k - r} \right)^{2} }}{kr} + \frac{{(k - r + \theta_{4} + \theta_{1} )^{2} }}{{2\left( {k - \theta_{3} } \right)\left( {r - \theta_{1} } \right)}}} \right] \\ = & \frac{{yC_{\text{o}} \left[ {\left( {r - \theta_{1} } \right) + 4r + \left( {r + \theta_{2} } \right)} \right]}}{{3n^{3} \pi^{2} }} \\ \end{aligned}$$

(61)

For all $y > 0$, $n > 0$, $\partial^{2} \varPhi (\tilde{V}\left( {y,n} \right))/\partial n^{2} > 0$.

$$\begin{aligned} \frac{{\partial^{2} V\left( {n,y} \right)}}{\partial y\partial n} = & - \frac{{C_{o} \left( {6r + \theta_{2} - \theta_{1} } \right)}}{{6y^{2} }} - \frac{{H_{\text{s}} }}{{n^{2} 6}}\left[ {\frac{{\left( {k - r - \theta_{3} - \theta_{2} } \right)^{2} }}{{2\left( {k + \theta_{4} } \right)\left( {r + \theta_{2} } \right)}} + \frac{{2\left( {k - r} \right)^{2} }}{kr} + \frac{{\left( {k - r + \theta_{4} + \theta_{1} } \right)^{2} }}{{2\left( {k - \theta_{3} } \right)\left( {r - \theta_{1} } \right)}}} \right] \\ = & - \frac{{C_{\text{o}} \left( {6r + \theta_{2} - \theta_{1} } \right)}}{{6y^{2} }} - \frac{{C_{\text{o}} \left[ {\left( {r - \theta_{1} } \right) + 4r + \left( {r + \theta_{2} } \right)} \right]}}{{6n^{2} \pi^{2} }} \\ \end{aligned}$$

(62)

Substituting $y^{*} = \pi n^{*}$ into Eqs. (60)–(62), and after some manipulations, determinant of the Hessian Matrix of $\varPhi (\tilde{V}\left( {y,n} \right))$ at $(y^{*} ,n^{*} )$ could be given as below:

$$\begin{aligned} & \left| {\begin{array}{*{20}l} {\left. {\frac{{\partial^{2} \varPhi (\tilde{V}\left( {y,n} \right))}}{{\partial y^{2} }}} \right|_{{(y^{*} ,n^{*} )}} } & {\left. {\frac{{\partial^{2} \varPhi (\tilde{V}\left( {y,n} \right))}}{\partial y\partial n}} \right|_{{(y^{*} ,n^{*} )}} } \\ {\left. {\frac{{\partial^{2} \varPhi (\tilde{V}\left( {y,n} \right))}}{\partial y\partial n}} \right|_{{(y^{*} ,n^{*} )}} } & {\left. {\frac{{\partial^{2} \varPhi (\tilde{V}\left( {y,n} \right))}}{{\partial n^{2} }}} \right|_{{(y^{*} ,n^{*} )}} } \\ \end{array} } \right| \hfill \\ &= \frac{{y^{*} C_{\text{o}} \left( {6r + \theta_{2} - \theta_{1} } \right)}}{{3n^{{*}^{3}} \pi^{2} }}\left[ { - \frac{{ba\left( {b + 1} \right)H_{\text{r}} \left( {6r + \theta_{2} - \theta_{1} } \right)y^{b - 1} }}{b + 2}} \right. \hfill \\ & \quad \left. { - \frac{{bay^{b - 1} H_{\text{s}} \left( {6r + \theta_{2} - \theta_{1} } \right)}}{{\left( {b + 2} \right)}} + \frac{{abC_{\text{l}} (b - 1)(6r + \theta_{2} - \theta_{1} )y^{b - 2} }}{b + 1}} \right] \hfill \\ &\quad + \frac{{C_{\text{o}} \left( {6r + \theta_{2} - \theta_{1} } \right)^{2} (5nC_{\text{o}} + 6C_{\text{s}} )}}{{9n^{{*}^{5}} \pi^{4} }} > 0 \hfill \\ \end{aligned}$$

(63)

Hessian Matrix of $\varPhi (\tilde{V}\left( {y,n} \right))$ is positive. Hence, $\varPhi (\tilde{V}\left( {y,n} \right))$ has a global minimum at point ($y^{*} ,n^{*}$).