Integrated preventive maintenance and flow shop scheduling under uncertainty

Abstract

This paper is concerned with stochastic scheduling of production and maintenance activities in a permutation flow shop setting. We present a two-stage stochastic mixed-integer program (SMIP) that adapts the conventional permutation flow shop scheduling problem for incorporating multiple preventive maintenance activities with various meter-based intervals. The model handles uncertainties in both processing times and the duration of maintenance activities. The concept of combining maintenance activities in scheduling problems is introduced and formulated, along with other practical considerations. The objective is to minimize the total expected cost associated with lateness penalties and maintenance resources. We use simulation–optimization (SO) for solving large-scale instances of the problem, and for validating the SMIP model. Through extensive computational experiments, we show that the SO method is superior in terms of efficiency and effectiveness and evaluate its sensitivity to the input data. Finally, a case study in earth-moving operations is presented, followed by managerial implications.

Appendix: Generating the coefficients

Consider the set $\varvec{a} = \left\{ {a_{1} ,a_{2} , \ldots ,a_{l} } \right\}, \forall k = 1, \ldots ,l, a_{k} \in {\mathbb{R}}$. $\varvec{S} = \left\{ {\varvec{s}_{1} ,\varvec{s}_{2} , \ldots ,\varvec{s}_{{2^{\varvec{l}} }} } \right\}$ is the set of all the possible subsets of $\varvec{a}$ that are not null, and $\varvec{b} = \left\{ {b_{1} ,b_{2} , \ldots ,b_{{2^{l} }} } \right\} { \ni } b_{r} = \mathop \sum \nolimits_{{a^{\prime} \in \varvec{s}_{\varvec{r}} }} a^{\prime},\forall r = 1, \ldots ,2^{l}$. We want to find the elements of $\varvec{a}$ such that the elements of $\varvec{b}$ are unique. For example, for $\varvec{ a}_{1} = \{ 1,2,3\} ,\;l = 3$, $\varvec{S}_{1} = \{ \{ 1\} ,\{ 2\} ,\{ 3\} ,\{ 1,2\} ,\{ 1,3\} ,\{ 2,3\} ,\{ 1,2,3\} \} \therefore \;\varvec{b}_{1} = \{ 1,2,3,3,4,5,6\}$, where the members of $\varvec{b}_{1}$ are not unique. But for $\varvec{a}_{2} = \{ 1.1,1.2,1.3\} ,l = 3$, $\varvec{S}_{2} = \{ \{ 1.1\} ,\{ 1.2\} ,\{ 1.3\} ,\{ 1.1,1.2\} ,\{ 1.1,1.3\} ,\{ 1.2,1.3\} ,\{ 1.1,1.2,1.3\} \} \therefore \;\varvec{b}_{2} = \{ 1.1,1.2,1.3,2.3,2.4,2.5,3.6\}$, where the members of $\varvec{b}_{2}$ are unique. The following proposition shows one way of generating $\varvec{a}$.

Proposition 1

If $\varvec{a} = \left\{ {1.1,1.01, \ldots ,1.0\underbrace {0 \ldots 0}_{l - 1}1} \right\},\left| \varvec{a} \right| = l,$ the uniqueness of the elements in $\varvec{b}$ is guaranteed.

Proof

For $\varvec{a} = \left\{ {1.1,1.01, \ldots ,1.\underbrace {00 \ldots 0}_{l - 1}1} \right\},\left| \varvec{a} \right| = l$:

$$\varvec{S} = \left\{ {\left\{ {1.1} \right\},\left\{ {1.01} \right\}, \ldots ,\left\{ {1.\underbrace {00 \ldots 0}_{l - 1}1} \right\},\left\{ {1.1,1.2} \right\}, \ldots ,\left\{ {1.1,1.01, \ldots ,1.\underbrace {00 \ldots 0}_{l - 1}1} \right\}} \right\},\left| \varvec{S} \right| = 2^{l} - 1 \therefore \; \varvec{b} = \left\{ {1.1,1.01, \ldots ,1.\underbrace {00 \ldots 0}_{l - 1}1,2.11,2.101, \ldots ,l.\underbrace {11 \ldots 1}_{l}} \right\},\left| \varvec{b} \right| = 2^{l} - 1.$$

Assume $\exists b_{i} ,b_{j} \in \varvec{b}\,and\,\varvec{s}_{\varvec{i}} ,\varvec{s}_{\varvec{j}} \in \varvec{S}{ \ni }b_{i} = b_{j}\,and\,\varvec{s}_{\varvec{i}} \ne \varvec{s}_{\varvec{j}}$. Because $b_{i} = b_{j}$, the integer and decimal parts of the two numbers are equal. Because the integer parts of all of the elements in $\varvec{a}$ are the same (1), the integer part of $b_{i}$ and $b_{j}$ indicate the number of elements in their respective subsets, i.e. $\varvec{s}_{\varvec{i}}$ and $\varvec{s}_{\varvec{j}}$. Because all elements in each subset have 1 as their integer part and each element has a unique number of zeros in the decimal part, the elements have to be equal for $b_{i}$ and $b_{j}$ to be equal. If the elements of $\varvec{s}_{\varvec{i}}$ and $\varvec{s}_{\varvec{j}}$ are identical, $\varvec{s}_{\varvec{i}} = \varvec{s}_{\varvec{j}}$. which contradicts the assumption. Therefore, for $\varvec{a} = \left\{ {1.1,1.01, \ldots ,1.\underbrace {00 \ldots 0}_{l - 1}1} \right\},\left| \varvec{a} \right| = l$ the elements of $\varvec{b}$ are unique.□