Integrated non-cyclical preventive maintenance scheduling and production planning for multi-parallel component production systems with interdependencies-induced degradation

Abstract

Integrating preventive maintenance (PM) scheduling and production planning efficiently remains challenging for researchers and practitioners alike, owing to complex component interdependencies. Existing studies often lack practicality due to oversimplified assumptions. In this paper, we propose a two-stage solution for the integrated PM scheduling and production planning problem within multi-parallel component manufacturing systems, accounting for diverse interdependencies. These interdependencies encompass stochastic, structural, economic, and resource-related interdependencies, all while accounting for the resultant degradation. The primary objective is jointly optimizing costs associated with holding, backorders, production, setup, and maintenance. Initially, we employ an Ordinary Differential System to compute a realistic production capacity. Additionally, we analyze the effects of interdependence-induced degradation on system availability and failures. Subsequently, we formulate the problem as an integer programming model to determine an optimal joint plan. The effectiveness of our approach is validated through numerical examples and sensitivity analysis, providing insightful guidance for efficient PM scheduling and production planning, particularly in large-scale production systems.

Appendix

1.1 A.1 Basic model properties

To guarantee that our model (1) is well posed, we prove its solutions’ nonnegativity and boundedness.

Theorem 7.1

Under the condition (2), for all $t\ge 0$, the solutions of our model (1) exhibit the following non-negative properties: $S(t)\ge 0$, $E(t)\ge 0$, $I(t)\ge 0$, $Q(t)\ge 0$, $H(t)\ge 0$, $R(t)\ge 0$ and $S_t(t)\ge 0$.

Proof

If the condition (2) is realized, the following inequalities can be established from the ten equations in our model (1).

$$\begin{aligned} \frac{dS}{dt}\ge & {} -(\beta _{1}+d)S,\nonumber \\ \frac{dE}{dt}\ge & {} -(\beta _2+\theta +d)E,\nonumber \\ \frac{dI}{dt}\ge & {} -(\delta +\eta +\gamma +d)I,\nonumber \\ \frac{dQ}{dt}\ge & {} -(\Upsilon +\varepsilon +d)Q,\nonumber \\ \frac{dH}{dt}\ge & {} -(r+\upsilon +d)H,\nonumber \\ \frac{dR}{dt}\ge & {} -(d+b)R,\nonumber \\ \frac{dS_t}{dt}\ge & {} -(\zeta +d)S_t. \end{aligned}$$

(25)

When we apply the theory of differential inequalities [93, 94], we find:

$$\begin{aligned} S(t)\ge & {} \exp (-(\beta _{1}+d)S),\\ E(t)\ge & {} \exp (-(\beta _2+\theta +d)E),\\ I(t)\ge & {} \exp (-(\delta +\eta +\gamma +d)I),\\ Q(t)\ge & {} \exp (-(\Upsilon +\varepsilon +d)Q),\\ H(t)\ge & {} \exp (-(r+\upsilon +d)H),\\ R(t)\ge & {} \exp (-(d+b)R),\\ S_t(t)\ge & {} \exp (-(\zeta +d)S_t). \end{aligned}$$

$\square$

Theorem 7.2

The solution of our model (1), with the initial conditions specified in (2), is positive in the following set:

$$\begin{aligned} \mathcal {B} =\left\{ (S, E,I, Q,H, R,S_t)\in \mathbb {R}^{7}_{+} : N(t) \le \frac{\Lambda }{d}\right\} , \end{aligned}$$

(26)

where $N=S+E+I+Q+H+R+S_t.$

Proof

Letting $N=S+E+I+Q+H+R+S_t,$ then

$$\begin{aligned} \frac{dN}{dt}+dN=m. \end{aligned}$$

(27)

Then, the solution of differential equation (27) is:

$N(t)= N(0)e^{-dt}+\frac{m}{d}\left( 1-e^{-dt}\right) .$ If $t\rightarrow \infty$, we have $N\le \frac{\Lambda }{d}$. $\square$

1.2 A.2 Equilibrium points and stability analysis

This section determines the equilibrium point $P_0$ in the absence of failure and the equilibrium point $P^*$ in the presence of failure with failure. Finally, we discuss the local stability of $P_0$ and $P^*$.

We need to resolve the following system of equations to obtain the coordinates of equilibrium $P_0(S_0,E_0, I_0,Q_0,H_0,$ $R_0,St_0)$:

$$\begin{aligned} \frac{dS}{dt}=\frac{dE}{dt}=\frac{dI}{dt}=\frac{dQ}{dt}=\frac{dH}{dt}=\frac{dR}{dt}=\frac{dS_t}{dt}=0. \end{aligned}$$

(28)

The equilibrium point $P_0$ is achieved if $E_0=I_0=Q_0=H_0=R_0=St_0=0$. In the absence of failure, we have $\beta _1=0$. As a result, the first system (1) is: $\frac{dS}{dt}=m-dS_0=0$. Therefore, the system (1) admit failure-free equilibrium point $P_{0}(\frac{m}{d},0,0,0,0,0,0)$.

To determine the coordinates of the equilibrium point $P^*(S^*,E^*,I^*,Q^*,H^*,R^*,S_t^*)$. We resolve the system (28). Then we get:

$$\begin{aligned} E^*= & {} -\frac{f h k m n s z}{b f h k n s z+c e g j p v z+c e g j r s z+c e g k p q z+c e h i p v z+c e h i r s z+c e h k l s z-a t f h k n s},\nonumber \\ S^*= & {} \frac{\beta _{2}+d+\theta }{\beta _{1}}E^*,\nonumber \\ I^*= & {} \frac{\theta }{\delta +\gamma +d+\eta }E^*,\nonumber \\ Q^*= & {} \frac{\delta }{\epsilon +d+\Upsilon }I^*,\nonumber \\ H^*= & {} \frac{\eta I^*+\varepsilon Q^*}{d+\nu +r},\nonumber \\ R^*= & {} \frac{\delta I^*+rH^*+\mu _2S_t^*}{d+\sigma },\nonumber \\ S_t^*= & {} \frac{\Upsilon Q^*+\nu H^*}{\zeta +d}, \end{aligned}$$

(29)

where $a=\beta _1+d$, $b=\beta _2$, $c=\sigma$, $e=\theta$, $f=\eta +\gamma +d+\delta$, $g=\delta$, $h=\varepsilon +d+\Upsilon$, $i=\eta$, $j=\varepsilon$, $k=\nu +r+d$, $l=\gamma$, $n=d+\sigma$, $p=\zeta$, $q=\Upsilon$, $s=\zeta +d$, $t=b+e+d$ and $z=\beta _1$.

1.3 A.3 Local stability

This section determines the conditions for local stability of the two equilibrium points $P_0$ and $P^*$.

Lemma 7.3

The condition (30) guarantees the local asymptotic stability of the equilibrium points $P_0$ and $P^*$.

$$\begin{aligned} \beta _{1}(\beta _2+\theta +d)+\beta _{2}d+d\theta +d\theta +d^2>\beta _{2}^2. \end{aligned}$$

(30)

Proof

The Jacobian matrix of our system (1) is determined as follows:

$$\begin{aligned} J\!\!=\!\!\left\{ \!\!\begin{array}{ccccccc} -\beta _{1}-d&{}\beta _{2}&{}0&{}0&{}0&{}\sigma &{}0\\ \beta _{1}&{}-(\beta _{2}+d+\theta )&{}0&{}0&{}0&{}0&{}0\\ 0&{}\theta &{}-(\eta +\gamma +d+\delta )&{}0&{}0&{}0&{}0\\ 0&{}0&{}\delta &{}-(\varepsilon +d+\Upsilon )&{}0&{}0&{}0\\ 0&{}0&{}\eta &{}\varepsilon &{}-(d+\nu +r)&{}0&{}0\\ 0&{}0&{}\gamma &{}0&{}r&{}-(d+\sigma )&{}\zeta \\ 0&{}0&{}0&{}\Upsilon &{}\nu &{}0&{}-(\zeta +d)\\ \end{array}\!\!\right\} \!, \end{aligned}$$

(31)

Solving this equation $det(J-\lambda I_7)= 0$ leads to the determination of the eigenvalues of the matrix J. Then:

$$\begin{aligned} \lambda _{1}= & {} -\frac{1}{2}\left( \sqrt{\beta _{1}^2-2(\beta _{1}\beta _{2}\!+\!\beta _{1}\theta \!-\!\beta _2\theta )\!+\!5\beta _{2}^2\!+\!\theta ^2}\!+\!\beta _{1}\!+\!\beta _{2}\!+\!\theta +2d \!\right) \!<\!0,\nonumber \\ \lambda _{2}= & {} -(d+\sigma )<0,\nonumber \\ \lambda _{3}= & {} -(d+\eta +\gamma +\delta )<0,\nonumber \\ \lambda _{4}= & {} -(d+\varepsilon +\Upsilon )<0,\nonumber \\ \lambda _{5}= & {} -(d+\zeta )<0,\nonumber \\ \lambda _{6}= & {} -(d+r+\nu )<0,\nonumber \\ \lambda _{7}= & {} \frac{1}{2}\left( \sqrt{\beta _{1}^2\!-\!2(\beta _{1}\beta _{2}\!+\!\beta _{1}\theta \!-\!\beta _2\theta )\!+\!5\beta _{2}^2\!-\!\theta ^2}\!-\!\beta _{1}\!-\!\beta _{2}\!-\!\theta \!-\!2d \right) <0. \end{aligned}$$

(32)

We have $\lambda _{i}<0$ for $i=1,\dots 6$. If the condition $\beta _1(\beta _2+\theta +d)+\beta _{2}d+d\theta +d\theta +d^2>\beta _{2}^2$ is fulfilled. Then $\lambda _7<0$. Therefore, all eigenvalues are negative real parts. The proof is completed. $\square$