The stochastic opportunistic replacement problem, part II: a two-stage solution approach

Abstract

In Almgren et al. (The opportunistic replacement problem: analysis and case studies, preprint, Department of Mathematical Sciences, Chalmers University of Technology and University of Gothenburg, Göteborg, Sweden, 2011) we studied the opportunistic replacement problem, which is a multi-component maintenance scheduling problem with deterministic component lives. The assumption of deterministic lives is a substantial simplification, but valid in applications where critical components are assigned a technical life after which replacement is enforced. Here, we study the stochastic opportunistic replacement problem, which is a more general setting in which component lives are allowed to be stochastic. We consider a stochastic programming approach for the minimization of the expected cost over the remaining planning horizon. Further, we present a means to compute lower bounds on the recourse function. The lower bounds are used in the construction of a decomposition method which extends the integer L-shaped decomposition method to incorporate stronger optimality cuts. In order to obtain a computationally tractable model, a two-stage sample average approximation scheme is utilized. Numerical experiments on problem instances from the wind power and aviation industry as well as on two test instances are performed. The results show that the decomposition method is faster than solving the deterministic equivalent on all four instances considered. Furthermore, the numerical experiments show that decisions based on the stochastic programming approach compared with simpler maintenance policies yield maintenance decisions with a significantly lower expected total maintenance cost on two out of the four instances tested, and an equivalent maintenance cost compared to the best policy on the remaining two instances.

Appendix A: Proof of Proposition 1

We will present a proof of (a) based on induction over the set of times. We will first prove the induction step for the case in which only one component age differs. The proof uses the fact that the failure probability and thus the expected cost of the component at the next time step increases with age. We then show that the proposition holds in the general case. Thereafter, we will prove (b) by utilizing the possibility of replacing the component which differs in age at the next time step and utilizing the result in (a).

To simplify notation we define p _i =G _i (a _i ) for all $i \in \mathcal {N}$. Further, we define $F^{a,t}: \mathbb {B}^{n} \rightarrow \mathbb {R}$ by

where $a^{x}_{i} = a_{i}(1-x_{i})$, for each component $i \in \mathcal {N}$. Note that F ^a,t(x) is the minimal objective value of (1) given the decision x. For any subset $\mathcal {A}\subseteq \mathcal {N}$, define $F^{*a,t}_{\mathcal {A}} = \min\{F^{a,t}(x) \mid x \in \mathbb {B}^{n}, x_{j}=1$ for $j \in \mathcal {A}\}$. Note that this is equal to f _t (ξ,a), as defined in (1), where ξ _i =1 for $i \in \mathcal {A}$ and ξ _i =0 for $i \in \mathcal {N}\backslash \mathcal {A}$. Introduce the notation $\hat {p}_{i} = G_{i}(\hat {a}_{i})$ for $i \in \mathcal {N}$.

(a) We will proceed by induction over the set of times, starting with T and going backwards. According to the definition, it holds that

$$C_T(\hat {a}) = 0 = C_T(a).$$

The induction assumption for $t\in \mathcal {T}$ is that for all $a, \hat {a}\in \mathbb {R}^{n}_{+}$ such that $\hat {a}\leq a$ the inequality $C_{t+1}(\hat {a}) \leq C_{t+1}(a)$ holds. We will establish that this implies a similar statement at time t.

Assume first that $\hat {a}_{i} = a_{i}$ for $i\in \mathcal {N}\backslash \{ l \}$, and $\hat {a}_{l}<a_{l}$. We have that $p_{i} = \hat {p}_{i}$ for $i \in \mathcal {N}\backslash\{l\}$. Note that the non-decreasing failure rate function yields that $p_{l} \geq \hat {p}_{l}$. This implies that $F^{\hat {a},t+1}(x) \leq F^{a,t+1}(x)$ for all $x\in \mathbb {B}^{n}$, which in turn implies that $F^{* \hat {a}, t+1}_{\mathcal {A}} \leq F^{*a, t+1}_{\mathcal {A}}$ for all $\mathcal {A}\subseteq \mathcal {N}$. Note also that $F^{* a,t+1}_{\mathcal {A}} \leq F^{* a, t+1}_{\mathcal {B}}$ holds whenever $\mathcal {A}\subseteq \mathcal {B}\subseteq \mathcal {N}$. Finally note that for any subset $\mathcal {A}\subseteq \mathcal {N}$ such that $l \in \mathcal {A}$ the relation $F^{*a+1,t+1}_{\mathcal {A}\cup\{l\}} =F^{*\hat {a}+1,t+1}_{\mathcal {A}\cup\{l\}}$ holds. Using these observations we obtain that

Using the definition of C _t (a) and the above results we obtain that

where the first sum loops over all combinations of component failures at time t+1. We have thus proven that $C_{t}(\hat {a}) \leq C_{t}(a)$ holds when $\hat {a}_{i} = a_{i}$ for $i\in \mathcal {N}\backslash \{ l \}$, and $\hat {a}_{l} <a_{l}$.

For a general pair $(a,\hat {a})$ such that $\hat {a}\leq a$ we can construct a sequence $\{a^{j}\}_{j=1}^{j=m}$ such that $\hat {a}\leq a^{1} \leq\ldots\leq a^{m} \leq a$ and where only one entry differs between each consecutive pair in the sequence. This implies that $C_{t}(\hat {a}) \leq C_{t}(a)$ which implies that (a) holds.

(b) For t=T the inequality is trivial. We consider the case when t∈{0,…,T−1}. The non-decreasing failure risk function yields the inequality $\hat {p}_{l} \geq p_{l}$. Let there be given $b,\hat {b}\in \mathbb {R}_{+}^{n}$ such that $b_{i}=\hat {b}_{i}$ for $i \in \mathcal {N}\backslash \{ l \}$ and $b_{l} <\hat {b}_{l}$. Assume $\emptyset\neq \mathcal {A}\subseteq \mathcal {N}$. Let x ^∗ be defined such that $x^{*}_{i}=1$ for $i \in \mathcal {A}$ and $F^{*b,t+1}_{\mathcal {A}} =F^{b,t+1}(x^{*})$. Let $\hat {x}_{i} = x^{*}_{i}$ for $i \in \mathcal {N}\backslash \{ l \}$ and $\hat {x}_{l} = 1$. Then

where we have used that $\hat {b}^{\hat {x}} = b^{\hat {x}} \leq b^{x^{*}}$ as well as the statement in (a). We have thus established that $F^{*\hat {b},t+1}_{\mathcal {A}}\leq F^{*b,t+1}_{\mathcal {A}} + c_{l}$ holds. Similar arguments yield the inequality $F^{*\hat {b},t+1}_{\emptyset} \leq F^{*b,t+1}_{\emptyset} + d + c_{l}$. We also have that $F^{*b,t+1}_{ \mathcal {A}\cup \{ l \}} \leq F^{*b,t+1}(\hat {x}) \leq F^{*b,t+1}_{\mathcal {A}}+ c_{l}$ holds and, similarly, $F^{*b,t+1}_{ \{l\}} \leq F^{*b,t+1}_{\emptyset} + c_{l} +d$ holds. Finally, we note that $F^{*\hat {b},t+1}_{ \mathcal {A}\cup \{ l \}} = F^{*b,t+1}_{ \mathcal {A}\cup \{ l \}}$.

Using the above relations, we can now conclude that if $\emptyset\neq \mathcal {A}\subseteq \mathcal {N}$ then

(8)

Similarly we can show that

This implies that

Returning to the original notation, we conclude that the proposition holds.