Joint optimization of capacity, production and maintenance planning of leased machines

Abstract

This paper deals with capacity design, production and preventive maintenance planning of a production system made of parallel leased machines. The production system must satisfy a random demand over a finite number of periods. The number of machines to be leased, the quantity to produce by each one and customer demand are variable. A joint optimization approach of a maintenance strategy and an economical production plan is developed by considering the influence of usage rates on the degradation of machines. Firstly, we determine a nearly optimal number of machines to be leased as well as the quantities to produce during each period. Secondly, given the obtained production plan, a periodic preventive maintenance policy with minimal repairs at failure is determined for each machine. Branch and bound and random exploration methods are used to obtain an optimal solution.

Appendices

Appendix A. Deterministic transformation

We will aim at demonstrating the following expression needed to transform the stochastic model to a deterministic one:

$$\begin{aligned} &\sum\limits_{k = 1}^{H} {E\left[ {\left( {I_{k} - a_{1} - a_{2} \cdot D_{k} } \right)^{2} } \right]} = \sum\limits_{k = 1}^{H} {\left( {\hat{I}_{k} - a_{1} - a_{2} \cdot \hat{D}_{k} } \right)^{2} }\\ & \quad+ {\left( {1 + a_{2} } \right)^{2} \cdot \sigma_{D}^{2} \cdot H} + a_{2}^{2} \cdot \sigma_{d}^{2} \cdot \frac{H}{2} \cdot \left( {H - 2} \right) \end{aligned} $$

$$ P_{k} = u \times \Delta k \times \sum\limits_{i = 1}^{{M_{k} }} {\left( {X_{ik}^{N} + X_{ik}^{S} } \right)} $$

$$ E\left( {I_{k} } \right) = \hat{I}_{k} ;E\left( {D_{k} } \right) = \mu_{k}^{d} ;E\left( {M_{k} } \right) = M_{k} ;E\left( {P_{k} } \right) = P_{k} ; $$

$$ V\left( {D_{k} } \right) = \sigma_{d}^{2} ;V\left( {P_{k} } \right) = 0. $$

We have:

$$ I_{k} = I_{k - 1} + P_{k} - D_{k} $$

So, $E\left( {I_{k} } \right) = \hat{I}_{k} = \hat{I}_{k - 1} + P_{k} - \mu_{k}^{d}$

$$ \left( {\hat{I}_{k} - I_{k} } \right)^{2} = \left( {\left( {\hat{I}_{k - 1} - I_{k - 1} } \right) - \left( {\mu_{k}^{d} - D_{k} } \right)} \right)^{2} $$

$$ \begin{aligned} E\left( {\hat{I}_{k} - I_{k} } \right)^{2} &= E\left( {\left( {\hat{I}_{k - 1} - I_{k - 1} } \right) - \left( {\mu_{k}^{d} - D_{k} } \right)} \right)^{2} \hfill \\ &= E\left( {\left( {\hat{I}_{k - 1} - I_{k - 1} } \right)^{2} } \right) + E\left( {\left( {\mu_{k}^{d} - D_{k} } \right)^{2} } \right)\hfill \\ &\quad - 2 \times E\left( {\hat{I}_{k - 1} - I_{k - 1} } \right) \times E\left( {\mu_{k}^{d} - D_{k} } \right) \hfill \\ & = \,E\left( {\hat{I}_{k - 1} - I_{k - 1} } \right)^{2} + \,E\left( {\mu_{k}^{d} - D_{k} } \right)^{2} \hfill \\ \end{aligned} $$

Since:

$$ \left\{ \begin{aligned} &E\left( {\hat{I}_{k - 1} - I_{k - 1} } \right)= E\left( {\hat{I}_{k} } \right) - E\left( {I_{k} } \right) = 0 \hfill \\ &E\left( {\mu_{k}^{d} - D_{k} } \right) = \mu_{k}^{d} - E\left( {D_{k} } \right) = 0 \hfill \\ \end{aligned} \right. $$

Thus, using the variance expression, we obtain that:

$$ V\left( {I_{k} } \right) = V\left( {I_{0} } \right) + \sigma_{d}^{2} $$

Assuming that $V\left( {I_{0} } \right) = 0$ and using this equality and by iterations, we can easily proof that:

$$ V\left( {I_{k} } \right) = k \cdot \sigma_{d}^{2} $$

Since the variance expression of I_k can be written as follows:

$$ V\left( {I_{k} } \right) = E\left( {I_{k}^{2} } \right) - \left[ {E\left( {I_{k} } \right)} \right]^{2} $$

So, we obtain:

$$ E\left( {I_{k}^{2} } \right) = V\left( {I_{0} } \right) + k \cdot \sigma_{d}^{2} + \hat{I}_{k}^{2} $$

On the other hand, we have$I_{k} = I_{k - 1} + P_{k} - D_{k}$ is equivalent to $D_{k} = I_{k - 1} - I_{k} + P_{k}$

So:

$$ I_{k} - a_{1} - a_{2} \cdot D_{t} = \left( {1 + a_{2} } \right) \cdot I_{k} - a_{1} - a_{2} \cdot I_{k - 1} - a_{2} \cdot P_{k} $$

And

$$ \hat{I}_{k} - a_{1} - a_{2} \cdot \mu_{k}^{d} = \left( {1 + a_{2} } \right) \cdot \hat{I}_{k} - a_{1} - a_{2} \cdot \hat{I}_{k - 1} - a_{2} \cdot P_{k} $$

Using both equalities above, we can conclude the following expected expression:

$$ \begin{aligned} & E\left[ {\left( {\left( {\hat{I}_{k} - a_{1} - a_{2} \cdot \mu_{k}^{d} } \right) - \left( {I_{k} - a_{1} - a_{2} \cdot D_{k} } \right)} \right)^{2} } \right] \\ & \quad = \left( {1 + a_{2} } \right)^{2} \cdot E\left[ {\left( {\hat{I}_{k} - I_{k} } \right)^{2} } \right] + a_{2}^{2} \cdot E\left[ {\left( {\hat{I}_{k - 1} - I_{k - 1} } \right)^{2} } \right] \\ & \quad = \left( {1 + a_{2} } \right)^{2} \cdot \left( {V\left( {I_{0} } \right) + k \cdot \sigma_{d}^{2} } \right) + a_{2}^{2} \cdot \left( {V\left( {I_{0} } \right) + \left( {k - 1} \right) \cdot \sigma_{d}^{2} } \right) \\ \end{aligned} $$

Also, we can write:

$$ \begin{aligned} & E\left[ {\left( {\left( {\hat{I}_{k} - a_{1} - a_{2} \cdot \mu_{k}^{d} } \right) - \left( {I_{k} - a_{1} - a_{2} \cdot D_{k} } \right)} \right)^{2} } \right] \\ & \quad = V\left( {I_{k} - a_{1} - a_{2} \cdot D_{k} } \right) \\ & \quad = E\left[ {\left( {I_{k} - a_{1} - a_{2} \cdot D_{k} } \right)^{2} } \right] - \left( {E\left[ {I_{k} - a_{1} - a_{2} \cdot D_{k} } \right]} \right)^{2} \\ \end{aligned} $$

Regarding the two above equations, we obtain:

$$ \begin{aligned} E\left[ {\left( {I_{k} - a_{1} - a_{2} \cdot D_{k} } \right)^{2} } \right] &= \left( {\hat{I}_{k} - a_{1} - a_{2} \cdot \mu_{k}^{d} } \right)^{2} \\& \quad + \left( {1 + a_{2} } \right)^{2} \cdot \left( {V(I_{0} ) + k \cdot \sigma_{d}^{2} } \right)\\& \quad + a_{2}^{2} \cdot \left( {V\left( {I_{0} } \right) + \left( {k - 1} \right) \cdot \sigma_{d}^{2} } \right) \end{aligned}$$

We assume that V(I₀) = 0, therefore in the objective function we can make the following transformation:

$$ \begin{aligned} & \sum\limits_{k = 1}^{H} {E\left[ {\left( {I_{k} - a_{1} - a_{2} \cdot D_{k} } \right)^{2} } \right]} = \sum\limits_{k = 1}^{H} {\left( {\hat{I}_{k} - a_{1} - a_{2} \cdot \mu_{t}^{d} } \right)^{2} }\\ &\quad\quad{ + \left( {1 + a_{2} } \right)^{2} \cdot \sigma_{D}^{2} \cdot \sum\limits_{k = 1}^{H} k } + a_{2}^{2} \cdot \sigma_{d}^{2} \cdot \sum\limits_{k = 1}^{H} {\left( {k - 1} \right)} \\ & \quad = \sum\limits_{k = 1}^{H} {\left( {\hat{I}_{k} - a_{1} - a_{2} \cdot \mu_{k}^{d} } \right)^{2} + \left( {1 + a_{2} } \right)^{2} \cdot \sigma_{d}^{2} \cdot \frac{H}{2} } \\ & \quad \quad\cdot \left( {H + 1} \right) + a_{2}^{2}\cdot \sigma_{d}^{2} \cdot \frac{H}{2} \cdot \left( {H - 1} \right) \\ \end{aligned} $$

Appendix B. Service rate constraint transformation

In order to ensure the customer satisfaction, we take into account in our model a probabilistic constraint given by:

$$ Prob\left( {I_{k} \ge 0} \right) \ge \beta $$

Looking forward to resolving the model, it was necessary to transform it from a stochastic probabilistic form to a deterministic one. To do this, we transform the service rate constraint announced above as follows:

$$ Prob\left( {I_{k} \ge 0} \right) \ge \beta \to P_{k} \ge P_{\beta } \left( {\beta ,\hat{I}_{k - 1} } \right) $$

where

$$ P_{k} \ge P_{\beta } \left( {\beta ,I_{k - 1} } \right) = \sqrt k \times \left( {\sigma_{d} } \right) \times \varphi^{ - 1} \left( \beta \right) - \hat{I}_{k - 1} + \mu_{k}^{D} $$

In fact: From proof (A), the variance of inventory variable is defined by $V\left( {I_{k} } \right) = k \cdot \sigma_{d}^{2}$. This inventory variable depends linearly on the random demand variation. It’s possible to consider the inventory variable as a random variable follows a normal distribution defined by:

$$ I_{k} = \hat{I}_{k} + X_{k} \times \sqrt {V\left( {I_{k} } \right)} \Leftrightarrow I_{k} = \hat{I}_{k} + X_{k} \times \sqrt k \times \sigma_{d} $$

With $X_{k} \propto N\left( {0,1} \right)$ is a standard Gaussian deviate.

$$ Prob\left( {I_{k} \ge 0} \right) \ge \beta $$

$$ Prob\left[ {\hat{I}_{k} + X_{k} \times \sqrt k \times \sigma_{d} \ge 0} \right] \ge \beta $$

$$ Prob\left[ {\hat{I}_{k - 1} + X_{k} \times \sqrt k \times \sigma_{d} + P_{k} - \mu_{k}^{D} \ge 0} \right] \ge \beta $$

$$ Prob\left[ {X_{k} \times \sqrt k \times \sigma_{d} \ge - \hat{I}_{k - 1} - P_{k} + \mu_{k}^{D} } \right] \ge \beta $$

$$ Prob\left[ {X_{k} \ge \frac{{ - \hat{I}_{k - 1} - P_{k} + \mu_{k}^{D} }}{{\sqrt k \times \sigma_{d} }}} \right] \ge \beta $$

$$ 1 - Prob\left[ {X_{k} \le \frac{{ - \hat{I}_{k - 1} - P_{k} + \mu_{k}^{D} }}{{\sqrt k \times \sigma_{d} }}} \right] \ge \beta $$

Assuming that $\varphi$ is the repartition function of this variable X.

$\varphi$ is strictly increasing, indefinitely differentiable and therefore we conclude that $\varphi$ is invertible:

$$ 1 - \varphi \left[ {\frac{{ - \hat{I}_{k - 1} - P_{k} + \mu_{k}^{D} }}{{\sqrt k \times \sigma_{d} }}} \right] \ge \beta $$

$$ \hat{I}_{k - 1} + P_{k} - \mu_{k}^{D} \ge \sqrt k \times \sigma_{d} \times \varphi^{ - 1} \left( \beta \right) $$

$$ P_{k} \ge \sqrt k \times \sigma_{d} \times \varphi^{ - 1} \left( \beta \right) + \mu_{k}^{D} - \hat{I}_{k - 1}. $$