Joint production, maintenance and quality control chart optimization for a degrading manufacturing–reworking system with assignable causes

Abstract

This paper presents a new joint production, maintenance and quality control chart for a manufacturing system subject to random failure and produces both conforming and non-conforming products. The degradation of the manufacturing unit increases with the production rate and the time of its use, which affects its reliability and the product quality. The purposed joint policy aims to establish, firstly, the optimal plan of production during finite horizon that satisfies random demand under a given service level. and secondly, to determine simultaneously the optimal number of imperfect preventive maintenance actions at each production cycle, and the optimal control chart parameters for appropriate quality monitoring in the presence of assignable causes. Furthermore, rework activity is proposed for a fraction of the non-conforming products, which are detected in the presence of assignable causes, in order to improve their quality and sold as a second-rate product. The proposed optimization model minimises the average total costs including production, reworking, storage, rejection, quality and maintenance. Numerical examples and sensitivity analyses are presented to show the effectiveness of the proposed joint approach.

Appendices

Appendix 1. Service rate constraint transformation

In our study, we assumed that demand fluctuations follow the normal distribution of mean μ₀ and standard deviation σ₀.

The mean production, demand, and storage variables are defined as follows:

$$E\left[U(k)\right]=U(k);E\left[d(k)\right]={\mu}_0;E\left[s(k)\right]=s(k)$$

Therefore, the variance variables for these three are shown below:

$$V\left[U(k)\right]=0;V\left[d(k)\right]={\sigma}_0;V\left[s(k)\right]=k\bullet {\sigma}_0.$$

To satisfy a random customer demand d(k), we consider the probabilistically constrained service level shown below:

$$\mathrm{Prob}\left[s(k)>0\right]\ge \partial$$

To solve this type of problem, it must be converted to a deterministic model from a stochastic model:

$$\mathrm{Prob}\left[s(k)>0\right]\ge \partial \Longleftrightarrow U(k)\ge {P}_{\partial \left[\partial, s\left(k-1\right)\right]}$$

with $U(k)\ge {P}_{\partial}\left[\partial, s\left(k-1\right)\right]=\sqrt{k\bullet {\sigma}_0}\bullet {F}^{-1}\left(\partial \right)+{\mu}_0+s\left(k-1\right)$

In effect, the variable s(k) depends on the customer’s demand d(k). Let V be the stock level variance s(k) defined by V(s(k)) = k ∙ σ₀. In this context, consider s(k) as a random variable characterized by a Gaussian distribution defined by:

$s(k)=E\left(s(k)\right)+\varOmega \bullet \sqrt{k\bullet {\sigma}_0}$. With Ω ∝ N(1, 0) is a standard Gaussian deviation.

$$\mathrm{Prob}\left[s(k)>0\right]\ge \partial$$

$$\mathrm{Prob}\left[E\left(s(k)\right)+\varOmega \bullet \sqrt{k\bullet {\sigma}_0}>0\right]\ge \partial$$

$$\mathrm{Prob}\left[E\left(s\left(k-1\right)\right)+E\left(U(k)\right)-E\left(d(k)\right)+\varOmega \bullet \sqrt{k\bullet {\sigma}_0}>0\right]\ge \partial$$

$$\mathrm{Prob}\left[s\left(k-1\right)+U(k)-{\mu}_0+\varOmega \bullet \sqrt{k\bullet {\sigma}_0}>0\right]\ge \partial$$

$$\mathrm{Prob}\left[\varOmega \bullet \sqrt{k\bullet {\sigma}_0}>-s\left(k-1\right)-U(k)+{\mu}_0\right]\ge \partial$$

$$\mathrm{Prob}\left[\varOmega >\frac{-s\left(k-1\right)-U(k)+{\mu}_0}{\sqrt{k\bullet {\sigma}_0}}\right]\ge \partial$$

Let F be the distribution function of Ω. F is strictly increasing and differentiable, so it is invertible.

$$s\left(k-1\right)+U(k)-{\mu}_0\ge \sqrt{k}\bullet {\sigma}_0\bullet {F}^{-1}\left(\partial \right)$$

$$U(k)\ge \sqrt{k}\bullet {\sigma}_0\bullet {F}^{-1}\left(\partial \right)+{\mu}_0+s\left(k-1\right)$$

Appendix 2. Failure rate estimation

In our study, it is assumed that the failure rate of the machine is linearly related to the productivity U(k) and the operating time τ. The IPM action is executed at time k ∙ θ with k (1…T). The failure rate is then expressed as a function of the failure rate at the previous interval multiplied by a factor e^ρ. Therefore, this failure rate decreases at each MPI time, and the system will be in an ‘AGAN’ state.

Therefore, the failure rate in the case of an MPI action can be expressed as follows:

$${\lambda}_k(t)={\lambda}_{k-1}\left(\tau \right)\left(1-\left\lfloor \frac{k-1}{\left(\left\lfloor \frac{k-2}{\theta}\right\rfloor +1\right)\theta}\right\rfloor \right)+\frac{U(k)}{U_{\mathrm{max}}}\ {\lambda}_n(t)+\left\lfloor \frac{k-1}{\left(\left\lfloor \frac{k-2}{\theta}\right\rfloor +1\right)\theta}\right\rfloor \bullet \lambda \left(t=0\right){e}^{\left\lfloor \frac{k}{\theta}\right\rfloor \rho }$$

It is assumed that at time t = 0, λ(t) = λ₀

For k = 1

$${\lambda}_1(t)={\lambda}_0\left(\tau \right)+\frac{U(1)}{U_{\mathrm{max}}}\ {\lambda}_n(t)$$

For k = 2

$${\lambda}_2(t)={\lambda}_1\left(\tau \right)\left(1-\left\lfloor \frac{1}{\theta}\right\rfloor \right)+\frac{U(2)}{U_{\mathrm{max}}}\ {\lambda}_n(t)+\left\lfloor \frac{1}{\theta}\right\rfloor \bullet {\lambda}_0\bullet {e}^{\left\lfloor \frac{2}{\theta}\right\rfloor \rho }=\left({\lambda}_0\left(\tau \right)+\frac{U(1)}{U_{\mathrm{max}}}\ {\lambda}_n(t)\right)\bullet \left(1-\left\lfloor \frac{1}{\theta}\right\rfloor \right)+\frac{U(2)}{U_{\mathrm{max}}}\ {\lambda}_n(t)+\frac{1}{\theta}\bullet {\lambda}_0\bullet {e}^{\left\lfloor \frac{2}{\theta}\right\rfloor \rho }$$

For k = 3

$${\lambda}_3(t)={\lambda}_2\left(\tau \right)\left(1-\left\lfloor \frac{2}{\left(\left\lfloor \frac{1}{\theta}\right\rfloor +1\right)\theta}\right\rfloor \right)+\frac{U(3)}{U_{\mathrm{max}}}\ {\lambda}_n(t)+\left\lfloor \frac{2}{\left(\left\lfloor \frac{1}{\theta}\right\rfloor +1\right)\theta}\right\rfloor \bullet {\lambda}_0\bullet {e}^{\left\lfloor \frac{k}{\theta}\right\rfloor \rho }=\left(\left({\lambda}_0\left(\tau \right)+\frac{U(1)}{U_{\mathrm{max}}}\ {\lambda}_n(t)\right)\bullet \left(1-\left\lfloor \frac{1}{\theta}\right\rfloor \right)+\frac{U(2)}{U_{\mathrm{max}}}\ {\lambda}_n(t)+\left\lfloor \frac{1}{\theta}\right\rfloor \bullet {\lambda}_0\bullet {e}^{\left\lfloor \frac{2}{\theta}\right\rfloor \rho}\right)\bullet \left(1-\left\lfloor \frac{2}{\left(\left\lfloor \frac{1}{\theta}\right\rfloor +1\right)\theta}\right\rfloor \right)+\frac{U(3)}{U_{\mathrm{max}}}\ {\lambda}_n(t)+\left\lfloor \frac{2}{\left(\left\lfloor \frac{1}{\theta}\right\rfloor +1\right)\theta}\right\rfloor \bullet {\lambda}_0\bullet {e}^{\left\lfloor \frac{3}{\theta}\right\rfloor \rho }$$

$$\vdots$$

In order to be general, λ_k(t) can be expressed as follows:

$${\lambda}_k(t)=\prod_{i=1}^k\left(1-\left\lfloor \frac{k-1}{\left(\left\lfloor \frac{k-2}{\theta}\right\rfloor +1\right)\theta}\right\rfloor \right)\times {\lambda}_0+\sum_{i=1}^{k-1}\left(\frac{U(i)}{U_{\mathrm{max}}}\times {\lambda}_n\left(\tau \right)\right)+\frac{U(k)}{U_{\mathrm{max}}}\times {\lambda}_n(t)+\sum_{i=1}^k\left(\left\lfloor \frac{k-1}{\left(\left\lfloor \frac{k-2}{\theta}\right\rfloor +1\right)\theta}\right\rfloor \times {e}^{\left\lfloor \frac{i}{\theta}\right\rfloor \times \rho}\right)\times {\lambda}_0\forall t\in \left\{0,\tau \right\}$$