Optimal Control of Production, Remanufacturing and Refurbishing Activities in a Finite Planning Horizon Inventory System

Abstract

In this paper, an inventory system with production, remanufacturing and refurbishing activities is studied, over a finite planning horizon. The demand is assumed time varying. Used products are returned by customers, and after inspection, they can be classified either as “remanufacturable” or as “refurbishable” items. The remanufacturing process brings “remanufacturable” items up to quality standards that are as rigorous as those of new items. The refurbished items are sold to a secondary market at a reduced price. In order to control the system, two types of policies are considered. For these two policies, a procedure is proposed that determines the order and remanufacturing quantities and the inventory level of returned (used) items at the start of the inspection and recovery processes, which minimize the total costs.

Appendix

Firstly, we calculate the area under $I_{1}(t)$ [see (27) and (30)] on the interval $t_{i-1}\le t\le t_{i}$ which help for the derivation of holding cost for used products:

$$\begin{aligned}&\varphi \int _{t_{i-1}}^{t_{i}^{p}}\int _{t_{i-1}}^{t}D(y)\hbox {d}y\hbox {d}t+\int _{t_{i}^{p} }^{t_{i}}\left[ -\int _{t}^{t_{i}}\{\varphi D(y)-p\}\hbox {d}y+\varphi q\int _{t_{i-1} }^{t_{i}}D(y)\hbox {d}y\right] \hbox {d}t \\&\quad =\varphi \int _{t_{i-1}}^{t_{i}^{p}}\left( t_{i}^{p}-t\right) D(t)\hbox {d}t+\int _{t_{i}^{p} }^{t_{i}}(t-t_{i}^{p})\{p-\varphi D(t)\}\hbox {d}t +\varphi q\left( t_{i}-t_{i}^{p}\right) \int _{t_{i-1}}^{t_{i}}D(t)\hbox {d}t \\&\quad =\varphi \int _{t_{i-1}}^{t_{i}}(t_{i}-t)D(t)\hbox {d}t-\varphi \left( t_{i}-t_{i}^{p}\right) \int _{t_{i-1}}^{t_{i}}D(t)\hbox {d}t+p\frac{\left( t_{i}-t_{i}^{p}\right) ^{2}}{2}\\&\qquad +\varphi q\left( t_{i}-t_{i}^{p}\right) \int _{t_{i-1}}^{t_{i}}D(t)\hbox {d}t \\&\quad =\varphi \int _{t_{i-1}}^{t_{i}}(t_{i}-t)D(t)\hbox {d}t-\varphi (1-q)\left( t_{i}-t_{i}^{p}\right) \int _{t_{i-1}}^{t_{i}}D(t)\hbox {d}t +p\frac{\left( t_{i}-t_{i}^{p}\right) ^{2}}{2}\,\text {(using (31))}\\&\quad =\varphi \int _{t_{i-1}}^{t_{i}}(t_{i}-t)D(t)\hbox {d}t-\frac{\varphi ^{2}(1-q)^{2} }{ 2p}\left\{ \int _{t_{i-1}}^{t_{i}}D(t)\hbox {d}t\right\} ^{2}. \end{aligned}$$

If $\varDelta _1$ denotes the area under the function $I_1(t)$ from time 0 to H, then it follows that

$$\begin{aligned} \varDelta _1 := \sum _{i=1}^{n_1} \varphi \int _{t_{i-1}}^{t_{i}}(t_{i}-t)D(t)\hbox {d}t-\frac{ \varphi ^{2}(1-q)^{2} }{2p}\left\{ \int _{t_{i-1}}^{t_{i}}D(t)\hbox {d}t\right\} ^{2} \end{aligned}$$

(60)

For the area under $I_{2}(t)$ on the interval $t_{i-1}\le t\le t_{i}$, we obtain using (33) and (35):

$$\begin{aligned}&\int _{t_{i-1}}^{t_{i}^{p}}\left\{ -\int _{t_{i-1}}^{t}D(y)\hbox {d}y+L_{i}\right\} \hbox {d}t +\int _{{t_i}^p}^{t_{i}}\left[ \int _{{t_i}^p}^{t}\{p-D(y)\}\hbox {d}y-\int _{t_{i-1}}^{t_{i}^{p}}D(y)\hbox {d}y+L_{i}\right] \hbox {d}t \nonumber \\&\quad =-\int _{t_{i-1}}^{t_{i}^{p}}(t_{i}^{p}-t)D(t)\hbox {d}t\nonumber \\&\qquad + \int _{{t_i}^p}^{t_{i}} (t_{i}-t)\{p-D(t)\}\hbox {d}t -(t_i-t_i^p)\int _{t_{i-1}}^{t_{i}^{p}}D(t)\hbox {d}t+(t_{i}-t_{i-1}) L_{i} \nonumber \\&\quad =\int _{t_{i-1}}^{t_{i}}(t-t_{i})D(t)\hbox {d}t+p\frac{(t_{i}-t_{i}^{p})^{2}}{2} +(t_{i}-t_{i-1})L_{i} \nonumber \\&\quad =\int _{t_{i-1}}^{t_{i}}(t-t_{i})D(t)\hbox {d}t+\frac{\varphi ^{2}(1-q)^{2}}{2p} \left\{ \int _{t_{i-1}}^{t_{i}}D(t)\hbox {d}t\right\} ^{2}+(t_{i}-t_{i-1})L_{i}. \end{aligned}$$

(61)

Next, we compute the quantity $\sum _{i=1}^{n_{1}-1}(t_{i}-t_{i-1})L_{i}$. This is equal to:

$$\begin{aligned}&=\sum _{i=1}^{n_{1}-2}t_{i}(L_{i}-L_{i+1})+t_{n_{1}-1}L_{n_{1} -1} \nonumber \\&=\left\{ 1-\varphi (1-q)\right\} \sum _{i=1}^{n_{1}-2}t_{i}\int _{t_{i-1} }^{t_{i}}D(t)\hbox {d}t\nonumber \\&\quad +t_{n_{1}-1}\left[ L_{1}-\left\{ 1-\varphi (1-q)\right\} \int _{0} ^{t_{n_{1}-2}}D(t)\hbox {d}t\right] \; \text {(by 38)} \nonumber \\&=\left\{ 1-\varphi (1-q)\right\} \sum _{i=1}^{n_{1}-1}t_{i}\int _{t_{i-1} }^{t_{i}}D(t)\hbox {d}t + t_{n_1-1} \int _{t_{n_1-1}}^{t_{n_1}^p}D(t)\hbox {d}t \; \; \text {(by 37)}. \end{aligned}$$

(62)

If $\varDelta _2$ denotes the area under the function $I_2(t)$ from 0 to H, then it follows from (61) to (62) that $\varDelta _2:=$

$$\begin{aligned}&\sum _{i=1}^{n_1-1} \left[ \int _{t_{i-1}}^{t_{i}}(t-t_{i})D(t)\hbox {d}t + \frac{\varphi ^{2}(1-q)^{2}}{2p}\left\{ \int _{t_{i-1}}^{t_{i}}D(t)\hbox {d}t\right\} ^{2} \right] \nonumber \\&\quad +\left\{ 1-\varphi (1-q)\right\} \sum _{i=1}^{n_{1}-1}t_{i}\int _{t_{i-1}}^{t_{i}}D(t)\hbox {d}t \nonumber \\&\quad + t_{n_1-1} \int _{t_{n_1-1}}^{t_{n_1}^p}D(t)\hbox {d}t +\int _{t_{n_1-1}}^{t_{n_1}^p}(t-t_{n_1-1})D(t)\hbox {d}t + \int _{{t_n^p}}^H (H-t)\{ p-D(t)\}\hbox {d}t. \end{aligned}$$

(63)

The last two terms of the equation refer to the area in the last period. Finally, we get $\varDelta _2=$

$$\begin{aligned}&\sum _{i=1}^{n_{1}-1} \left[ \int _{t_{i-1}}^{t_{i}}(t-t_{i})D(t)\hbox {d}t+ \frac{\varphi ^{2}(1-q)^{2}}{2p}\left\{ \int _{t_{i-1}}^{t_{i}}D(t)\hbox {d}t\right\} ^{2} \right] \nonumber \\&\quad +\left\{ 1-\varphi (1-q)\right\} \sum _{i=1}^{n_{1}-1}t_{i}\int _{t_{i-1} }^{t_{i}}D(t)\hbox {d}t \nonumber \\&\quad +\int _{t_{n_{1}-1}}^{t_{n_1}^p}tD(t)\hbox {d}t + \int _{{t_{n_{1}}^p}}^H (H-t)\{ (p-D(t)\}\hbox {d}t. \end{aligned}$$

(64)