The stochastic opportunistic replacement problem, part I: models incorporating individual component lives

Michael Patriksson; Ann-Brith Strömberg; Adam Wojciechowski

doi:10.1007/s10479-012-1131-4

Annals of Operations Research

January 2015, Volume 224, Issue 1, pp 25–50 | Cite as

The stochastic opportunistic replacement problem, part I: models incorporating individual component lives

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Michael Patriksson
Ann-Brith Strömberg
Adam WojciechowskiEmail author

Article

First Online: 19 April 2012

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Abstract

We consider an extension of the opportunistic replacement problem, which has been studied by Dickman et al. (The Journal of the Operational Research Society of India, 28:165–175, 1991), Andréasson (Optimization of opportunistic replacement activities in deterministic and stochastic multi-component systems, Licentiate thesis, Department of Mathematical Sciences, Chalmers University of Technology and University of Gothenburg, Göteborg, Sweden, 2004), and 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), that allows the individuals of the same component to have non-identical lives. Formulating and solving this problem constitute a first step towards solving the opportunistic replacement problem with uncertain component lives. We show that the problem is NP-hard even with time independent costs, and present two 0–1 integer programming models for the problem. We show that in model I the integrality requirement on a majority of the variables can be relaxed; this is in contrast to model II and the model from Andréasson (Optimization of opportunistic replacement activities in deterministic and stochastic multi-component systems, Licentiate thesis, Department of Mathematical Sciences, Chalmers University of Technology and University of Gothenburg, Göteborg, Sweden, 2004). We remove all superfluous variables and constraints in model I and show that the remaining constraints are facet inducing. We also utilize a linear transformation of model I to obtain a stronger version of model II, i.e., model II⁺, which inherits the polyhedral properties of model I. Numerical experiments show that the solution time of model I is significantly lower than those of both model II and Andréasson’s model. It is also slightly lower than the solution time of model II⁺.

Keywords

Maintenance optimization Complexity analysis Integer programming

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Notes

Acknowledgements

The authors wish to thank docent Peter Damaschke, Chalmers University of Technology, for valuable discussions on complexity analysis, and the MSc-students Hjördís Harðardóttir, Pär Werner and Cornelia Jareteg for their implementation of the numerical evaluations. We also thank an anonymous reviewer for his/her valuable comments, which improved the presentation of the paper.

Appendix A: Complexity proof

Proof of Proposition 1

Given a set of vertices V and a set of edges E consider the following instance of the ORPIL. Let q=2, n=|E|+1, T=2|V|+2, and d=1. The lives of the individuals of component n are given by T _n1=1 and T _n2=T _n=2, and the replacement cost is c _n=2|E|(|E|+2). Assign each component i∈{1,…,|E|} to an edge (j _i,k _i)∈E. The lives of the individuals of component i are given by T _i1=2j _i, T _i2=2(k _i−j _i)+1, and T _i=T−2k _i+1, and the replacement cost is given by c _i=|E|+1.

If component n is always replaced at failure, replacements will occur at the time steps 2t+1, where t∈{0,…,|V|}. A minimum cost maintenance schedule, such that component n is always replaced at failure costs at most $(|V|+1)(c_{n} +d) + \sum_{i=1}^{|E|}2(c_{i} + d) =(|V|+1)(2|E|(|E|+2) +1) + 2|E|(|E|+2)$. A maintenance schedule which replaces any individual of component n before failure will cost at least (|V|+2)(c _n+d)=(|V|+1)(2|E|(|E|+2)+1)+2|E|(|E|+2)+1. Therefore component n in an optimal maintenance schedule will always be replaced at failure.

For components $i \in\mathcal{N} \backslash\{ n \}$ at least 2 replacements are necessary, since $T_{i1} + T_{i2} \not> T$. There is a feasible maintenance schedule with 2 replacements since T _i1+T _i2+T _i=T+2>T. The minimum cost maintenance schedule that utilizes 2 replacements of component i costs no more than $(|V|+1)(c_{n} + d) + \sum_{j=1}^{|E|} (2c_{j} + d) = (|V|+1)(c_{n} + d) +2|E|(|E|+1) +|E|$. A replacement of component i more than twice costs at least $(|V|+1)(c_{n}+d) + c_{i} + \sum_{j=1}^{|E|} 2c_{j} = (|V|+1)(c_{n}+d)+ 2|E|(|E|+1) + |E| +1$. Hence each component $i \in\mathcal{N}\backslash\{ n \}$ will be replaced twice in an optimal maintenance schedule.

Consider again the component $i \in\mathcal{N} \backslash\{ n \}$ and the corresponding edge (j _i,k _i)∈E. The possible maintenance schedules for component i that utilize 2 replacements are as follows: either we replace at time 2j _i and 2k _i, at 2j _i−1 and 2k _i, or at 2j _i and 2k _i−1.

Assume now that we have an optimal solution to the above defined instance of the ORPIL. For each $t \in\mathcal{T}$ let z _t=1 if time t is a maintenance occasion and 0 otherwise. The cost of the maintenance schedule is $f^{*} = (|V|+1)(c_{n} +d) + \sum_{i=1}^{|E|} 2c_{i}+ d\sum_{t=1}^{|V|} z_{2t}$. Construct a set of vertices S={i∈V|z _2i=1}. For each edge e∈E let $i \in\mathcal{N}$ denote the corresponding component such that e=(j _i,k _i). We have that $z_{2j_{i}}=1$ or $z_{2k_{i}}=1$ since, according to the above, the component i is replaced on at least one of the times 2j _i and 2k _i. Hence S is a vertex cover. Assume that there exists a vertex cover S′ such that |S′|<|S|. We can then construct a solution to the ORPIL as follows: For each component $i \in\mathcal{N} \backslash\{n \}$ consider the corresponding edge (j _i,k _i)∈E and assume without loss of generality that j _i<k _i. Make the first replacement of component i at time 2j _i if j _i∈S′ and at time 2j _i−1 otherwise and the second replacement at time 2k _i if k∈S′ and at time 2k _i−1 otherwise. Component n is replaced at failure. The schedule results in a cost of $(|V|+1)(c_{n} +d) + \sum_{i=1}^{|E|} 2c_{i}+ |S'| < (|V|+1)(c_{n} +d) + \sum_{i=1}^{|E|} 2c_{i} + |S| =f^{*}$. This is a contradiction. Hence S is the minimal vertex cover. By solving this instance of the ORPIL we thus obtain a minimal vertex cover. □

Appendix B: Facets

Proposition 5

If T _ir≥2 for all $i \in\mathcal{N}$ and $r \in \mathcal{R}$, and T _i≥2 for all $i \in\mathcal{N}$, then (6b) defines a facet of the polyhedron conv(X).

Proof

For $j \in\mathcal{N}, k \in\mathcal{R}$ and $s \in \mathcal{T}_{jk}^{(6b)}$ let $F = \{(\tilde{x},x,z)\in\mathrm{conv}(X) |\tilde{x}_{js}^{k} = \tilde{x}_{js+1}^{k}\}$. Define $\lambda_{js}^{k} =1$, $\lambda_{js+1}^{k}=-1$ and $\lambda_{it}^{r} = 0$ for $(i,t,r) \in\mathcal{A}\backslash\{ (j,s,k),(j,s+1,k) \}$. We will show that if

$$ \sum_{(i,t,r)\in \mathcal{A}} \pi_{it}^r\tilde{x}_{it}^r + \sum_{(i,t)\in\mathcal{N} \times \{q,\ldots,T\}}\rho_{it} x_{it} + \sum_{t\in\mathcal{T}}\sigma_t z_t = \tau $$

(9)

holds for all $(\tilde{x},x,z) \in F$ then (π,ρ,σ,τ)=α(λ,0,0,0).

Let $v=(\tilde{x},x,z) \in F$ be defined by assigning a (0,1)-schedule to every component $i \in\mathcal{N}$, and z _t=1 for all $t\in\mathcal{T}$.

For $\hat{i} \in\mathcal{N}$ and $\hat{t} \in\{q,\ldots,T\}$, define u _x as equal to v except that $x_{\hat{i},\hat{t}}=0$. The assumption that $T_{ \hat{i}} \geq2$ implies that v,u∈F. Lemma 2 gives that $\rho_{\hat{i},\hat{t}} = 0$.

For $\hat{t} \in\{q,\ldots,T\}$ let v _z be defined as equal to v except that $x_{i\hat{t}} = 0$ for all $i \in\mathcal{N}$. Let u _z be defined as equal to v _z except that $z_{\hat{t}}=0$. The assumption T _i≥2 for all $i \in \mathcal{N}$ implies that $v_{z},u_{z}\in\hat{F}$. Lemma 2 gives that $\sigma_{\hat{t}}=0$. Assume instead $\hat{t} \in\{0,\ldots,q-1\}$. Let v _z be defined as equal to v with the following exceptions. If $\hat{t}\neq s$ or $\hat{t}+1 \neq r$ assign a $(\hat{t}+1,\hat{t}+1)$-schedule to every component $i \in \mathcal{N}$. If $\hat{t}=s=r-1=0$ assign a (2,1)-schedule to every component $i \in\mathcal{N}$. Finally, if $\hat{t}=s=r-1\geq1$ assign a $(\hat{t}+1,\hat{t})$-schedule to every component $i \in\mathcal{N}$. u _z is defined as equal to v _z except that $z_{\hat{t}}=0$. The assumptions T _ir≥2 for all $i\in\mathcal{N}$ and $r \in \mathcal{R}$ implies that u _z,v _z∈F. Lemma 2 again gives $\sigma_{\hat{t}}=0$. Hence σ=0.

For $(\hat{i},\hat{r},\hat{t}) \in\mathcal{A}\backslash\{ (j,k,s),(j,k,s+1) \}$, define $v_{\tilde{x}}$ as equal to v except that component $\hat{i}$ is assigned a $( \hat{t}+1,\hat{r})$-schedule. Let $u_{\tilde{x}}$ be defined as equal to $v_{\tilde{x}}$ except that $\tilde{x}_{\hat{i}\hat{t}}^{\hat{r}}=1$. The assumption that $T_{\hat{i}\hat{r}} \geq2$ implies that $v_{ \tilde{x}}, u_{\tilde{x}}\in F$. Lemma 2 gives that $\pi_{\hat{i},\hat{t}}^{\hat{r}}= 0$.

We have shown that ρ=0, σ=0 and $\pi_{i,t}^{r}=0$ for $(i,t,r) \in \mathcal{A}\backslash\{ (j,s,k),(j,s+1,k) \}$. Let v _C be defined as equal to v except that every component $i\in\mathcal{N}$ is assigned a $(\sum_{r=1}^{q} T_{ir},q)$-schedule. Since v _C∈F and $\tilde{x}_{js}^{k}=\tilde{x}_{js+1}^{k}=0$ inserting v _C into (9) implies that τ=0. Finally, inserting v into (9) gives that $\pi_{js}^{k} = - \pi_{js+1}^{k}$. We can conclude that (π,ρ,σ,τ)=α(λ,0,0,0) which, together with Theorem 1 and the full dimensionality of conv(X) shown in Proposition 3, implies that F is a facet of conv(X). □

Proposition 6

If T _ir≥2 for all $i \in\mathcal{N}$ and $r \in \mathcal{R}$, and T _i≥2 for all $i \in\mathcal{N}$, then (6c) defines a facet of the polyhedron conv(X).

Proof

For $j \in\mathcal{N}$, $k \in\mathcal{R}$ and $s \in \mathcal{T}_{jk}^{(6c)}$ let $F = \{(\tilde{x},x,z)\in\mathrm{conv}(X)| \tilde {x}_{js+1}^{k+1} = \tilde{x}_{js}^{k}\}$. Define $\lambda_{js}^{k} =-1$, $\lambda_{js+1}^{k+1}=1$ and $\lambda_{it}^{r} = 0$ for $(i,t,r) \in\mathcal{A}\backslash\{(j,s,k),(j,s+1,k+1) \}$. We will show that if

$$ \sum_{(i,t,r) \in\mathcal{A}} \pi_{it}^r\tilde{x}_{it}^r + \sum_{(i,t) \in\mathcal{N} \times\{q,\ldots,T\}}\rho_{it} x_{it} + \sum_{t \in\mathcal{T}}\sigma_t z_t = \tau $$

(10)

for all $(x,\tilde{x},z) \in F$ then (π,ρ,σ,τ)=α(λ,0,0,0).

Let $v=(\tilde{x},x,z)$ be defined by assigning a (0,1)-schedule to each component $i \in\mathcal{N}$ and letting z _t=1 for all $t\in\mathcal{T}$. For $\hat{i}\in\mathcal{N}$ and $\hat{t}\in\{q,\ldots,T\}$, define $u=( \tilde{x},x,z)$ as equal to v except that $x_{\hat{i},\hat{t}}=0$. Since $T_{\hat{i}} \geq 2$ we have u and v∈F. Lemma 2 gives that $\rho_{\hat{i},\hat{t}} = 0$.

For $\hat{t}\in\mathcal{T}$ define $v_{z} = (\tilde{x},x,z)$ as equal to v with the following exceptions. If $\hat{t}\geq q$ then let $x_{i\hat{t}} = 0$ for all $i \in \mathcal{N}$. If $\hat{t}<q$ and $\hat{t}\neq s+1$ or $\hat{t}\neq k$ then assign a $(\hat{t}+1,\hat{t}+1)$-schedule to every component $i \in\mathcal{N}$. If $\hat{t}=s+1=k$ then assign a $(\hat{t}+1,\hat{t})$-schedule to every component $i \in\mathcal{N}$. Let u _z be defined as equal to v _z except that $z_{\hat{t}}=0$. Since T _ir≥2 for every $i\in\mathcal{N}$ and $r \in\mathcal{R}$, and T _i≥2 for every $i\in\mathcal{N}$ we have v _z,u _z∈F. Lemma 2 gives that $\sigma_{\hat{t}}=0$.

For $(\hat{i},\hat{t},\hat{r}) \in\mathcal{A}\backslash\{(j,s,k),(j,s+1,k+1) \}$ defined $v_{\tilde{x}}$ as equal to v with the following exceptions. Let each individual $r\in\{1,\ldots,\hat{r}-1\}$ of component $\hat {i}$ be replaced at time $\min(\sum_{u=1}^{r} T_{\hat{i}u},\hat{t}- \hat{r}+r)$. Let each individual $r \in\{\hat{r},\ldots,q\}$ of component $\hat {i}$ be replaced at time $\hat{t}+1 +r -\hat{r}$. Let x _jt=0 for $t\leq\hat{t}+1 +q -\hat{r}$ and x _jt=1 for $t>\hat{t}+1 +q-\hat{r}$. Define $u_{\tilde{x}}$ as equal to $v_{\tilde{x}}$ except that $\tilde{x}_{\hat{i} \hat{t}}^{\hat{r}}=1$. Since $T_{\hat{i}\hat{r}} \geq2$ we have $v_{\tilde{x}},u_{\tilde {x}} \in F$. Lemma 2 implies that $\pi_{\hat{i}\hat{t}}^{\hat{r}}=0$.

We have shown that ρ=0, σ=0 and $\pi_{i,t}^{r}=0$ for $(i,t,r) \in \mathcal{A}\backslash\{ (j,s,k),(j,s+1,k+1) \}$. Let v _C be defined as equal to v except that each component $i\in\mathcal{N}$ is assigned a $(\sum_{r=1}^{q} T_{ir},q)$-schedule. Since v _C∈F and $\tilde{x}_{js}^{k} = \tilde{x}_{js+1}^{k+1}=0$ inserting v _C into (10) implies that τ=0. Finally, inserting v into (10) gives that $\pi_{js}^{k} = - \pi_{js+1}^{k+1}$. We can conclude that (π,ρ,σ,τ)=α(λ,0,0,0) which, together with Theorem 1 and the full dimensionality of conv(X) shown in Proposition 3, implies that F is a facet of conv(X). □

Proposition 7

If T _ir≥2 for all $i \in\mathcal{N}$ and $r \in \mathcal{R}$, and T _i≥2 for all $i \in\mathcal{N}$, then (6e) defines a facet of the polyhedron conv(X).

Proof

For $j \in\mathcal{N}$ let $F = \{(x,\tilde{x},z) \in\mathrm{conv}(X)| \tilde{x}_{j0}^{1} = z_{0}\}$. Let $\lambda_{j0}^{1} =1$ and $\lambda_{it}^{r} = 0$ for $(i,t,r) \in\mathcal{A}\backslash\{ (j,0,1)\}$. Let ν ₀=−1 and ν _t=0 for $t \in\mathcal{T} \backslash\{0 \}$. We will show that if

$$ \sum_{(i,t,r) \in\mathcal{A}} \pi_{it}^r\tilde{x}_{it}^r + \sum_{(i,t)\in \mathcal{N} \times\{q,\ldots,T\}}\rho_{it} x_{it} + \sum_{t\in\mathcal{T}}\sigma_t z_t = \tau $$

(11)

holds for all $(x,\tilde{x},z) \in F$ then (π,ρ,σ,τ)=α(λ,0,ν,0).

Let $v=(x,\tilde{x},z) \in F$ be defined by assigning a (0,1)-schedule to every component $i \in\nobreak\mathcal{N}$, and z _t=1 for all $t\in\mathcal{T}$.

For $(\hat{i},\hat{t}) \in\mathcal{N} \times\{q,\ldots, T\}$, define u _x as equal to v except that $x_{\hat{i}\hat{t}}=0$. The assumption $T_{\hat{i}} \geq2$ implies that v,u _x∈F. Lemma 2 gives that $\rho_{\hat{i},\hat{t}} = 0$.

For $\hat{t}\in\mathcal{T} \backslash\{ 0 \}$ define v _z as equal to v with the following exceptions. If $\hat{t}\geq q$ let $x_{i\hat{t}}=0$ for all $i \in\mathcal{N}$. If $\hat{t} < q$ assign an $(\hat{t}+1,\hat{t}+1)$-schedule to every component $i \in\mathcal{N}$. Define u _z as equal to v _z except that $z_{\hat{t}}=0$. Since T _i≥2 and $T_{i\hat{t}+1} \geq2$ for all $i \in\mathcal{N}$, we have v _z,u _z∈F. Lemma 2 gives that $\sigma_{\hat{t}}=0$.

For $(\hat{i},\hat{t},\hat{r}) \in\mathcal{A}\backslash\{ (j,0,1) \}$ define $v_{\tilde{x}}$ as equal to v with the following exceptions. Assign a $(\hat{t}+1, \hat{r})$-schedule to $\hat{i}$ and, if $\hat{i}=j$, assign a (1,1)-schedule to each component $i\in\mathcal{N} \backslash\{ \hat{i}\}$ and let $z_{0} =\tilde{x}_{j0}^{1}$. Define $u_{\tilde{x}}$ as equal to $v_{\tilde{x}}$ except that $x_{\hat{i} \hat{t}}^{\hat{r}}=1$. Since T _ir≥2 for all $i \in\mathcal{N}$ and $r \in \mathcal{R}$ we have $v_{\tilde{x}},u_{\tilde{x}} \in F$. Lemma 2 implies that $\pi_{\hat{i}\hat{t}}^{\hat{r}} = 0$.

Define v _c by assigning a $(\sum_{r \in\mathcal{R}} T_{ir},q)$-schedule to each component $i \in\mathcal{N}$. Let z ₀=0 and z _t=1 for $t \in\mathcal{T} \backslash\{ 0 \}$. Since v _c∈F we have τ=0.

Inserting v into Eq. (11) finally gives $\pi_{j0}^{1} =-\nu_{0}$. We can conclude that (π,ρ,σ,τ)=α(λ,0,ν,0) which, together with Theorem 1 and the full dimensionality of conv(X) shown in Proposition 3, implies that F is a facet of conv(X). □

Proposition 8

If T _ir≥2 for all $i \in\mathcal{N}$ and $r \in \mathcal{R}$, and T _i≥2 for all $i \in\mathcal{N}$, then (6f) defines a facet of the polyhedron conv(X).

Proof

For $j \in\mathcal{N}$, $k \in\mathcal{R} \backslash\{ q \}$ and $s \in \mathcal{T}_{jk}^{(6f)}$ define $F = \{(x,\tilde{x},z)\in\mathrm{conv}(X) | \tilde {x}_{js}^{k} =\tilde{x}_{js+T_{jk+1}}^{k+1}\}$. Define $\lambda_{js}^{k} =1$, $\lambda_{js+T_{jk+1}}^{k+1}=-1$ and $\lambda_{it}^{r} = 0$ for $(i,t,r)\in\mathcal{A}\backslash\{ (j,k,s),(j,s+T_{jk+1},k+1) \}$. We will show that if

$$ \sum_{(i,t,r) \in\mathcal{A}} \pi_{it}^r\tilde{x}_{it}^r + \sum_{(i,t)\in \mathcal{N} \times\{q,\ldots,T\}}\rho_{it} x_{it} + \sum_{t\in\mathcal{T}}\sigma_t z_t = \tau $$

(12)

for all $(x,\tilde{x},z) \in F$ then (π,ρ,σ,τ)=α(λ,0,0,0).

Let $v=(x,\tilde{x},z) \in F$ be defined by assigning a (0,1)-schedule to every component $i \in\nobreak\mathcal{N}$, and z _t=1 for all $t \in\mathcal{T}$.

For $(\hat{i},\hat{t}) \in\mathcal{N} \times\{q,\ldots, T\}$, define u _x as equal to v except that $x_{\hat{i}\hat{t}}=0$. The assumption $T_{\hat{i}} \geq2$ implies that v,u _x∈F. Lemma 2 gives that $\rho_{\hat{i},\hat{t}} = 0$.

For $\hat{t}\in\mathcal{T}$ define v _z as equal to v with the following exceptions. If $\hat{t}\geq q$ let $x_{i\hat{t}}=0$ for all $i \in\mathcal{N}$. If $\hat{t} < q$ assign an $(\hat{t}+1,\hat{t}+1)$-schedule to all components $i \in\mathcal{N} \backslash\{ j \}$ and, if $\hat{t}= s+T_{jk+1}$ and $\hat{t}+1 = k+1$ assign an $(\hat {t}+1,\hat{t})$-schedule to component j, otherwise assign an $(\hat{t}+1,\hat{t}+1)$-schedule to component j. Define u _z as equal to v _z except that $z_{\hat{t}}=0$. Since T _i≥2 and $T_{i\hat{t}+1} \geq2$ for all $i \in \mathcal{N}$, we have v _z,u _z∈F. Lemma 2 gives that $\sigma_{\hat{t}}=0$.

For $(\hat{i},\hat{t},\hat{r}) \in\mathcal{A}\backslash\{ (\hat {t},\hat{i}, \hat {r}),(s+T_{jk+1},j,k+1) \}$ define $v_{\tilde{x}}$ as equal to v with the following exceptions. If $\hat{i} \neq j$ or $k<\hat{r}$ assigning an $(\hat{t}+1,\hat{r})$-schedule to $\hat{i}$. If $\hat{i}=j$ and $k \geq\hat{r}$ define $v_{\tilde{x}}$ by replacing individual $r\in\mathcal{R}$ of component j at time t _r, where t _r is defined as follows. Let $t_{\hat{r}}=\hat{t}+1$, t _m=t _m−1+T _jm for $m=\hat{r}+1, \ldots,q$, and t _m=max(t _m+1−T _jm+1,m−1) for $m =1,\ldots,\hat{r}-1$. For all t∈{q,…,T}, let x _jt=0 if t≤t _q and x _jt=1 if t≥t _q+1. Define $u_{\tilde{x}}$ as equal to $v_{\tilde{x}}$ except that $\tilde{x}_{\hat{i}\hat{t}}^{\hat{r}}=0$. Using Lemma 2 gives $\pi_{\hat{i}\hat{t}}^{\hat{r}}=0$.

Finally we define v _c as equal to v except that component j is assigned an $(\sum_{r=1}^{q} T_{jr}, q)$-schedule. Since v _c∈F it implies that τ=0. By inserting v into Eq. (12) we obtain $\pi_{js}^{k} = - \pi_{js+T_{k}^{k+1}}^{k+1}$. We can conclude that (π,ρ,σ,τ)=α(λ,0,0,0) which, together with Theorem 1 and the full dimensionality of conv(X) shown in Proposition 3, implies that F defines a facet of conv (X). □

Proposition 9

If T _ir≥2 for all $i \in\mathcal{N}$ and $r \in\mathcal{R}$, and T _i≥2 for all $i \in\mathcal{N}$, then (6g) defines a facet of the polyhedron conv(X).

Proof

For $j \in\mathcal{N}$ and $l \in\mathcal{T}_{j}^{(6g)}$ let $F = \{(\tilde{x},x,z)\in\mathrm{conv}(X) | \sum_{t=l+1}^{l+T_{j}} x_{jt} = \tilde{x}_{jl}^{q}\}$. Define $\lambda_{jl}^{q} =-1$ and $\lambda_{it}^{r} = 0$ for $(i,t,r)\in \mathcal{A}\backslash\{ (j,l,q) \}$. Let ρ _jt=1 for t∈{l+1,…,l+T _j} and ρ _jt=0 for t∈{q,…,T}∖{l+1,…,l+T _j}. We will show that if

$$ \sum_{(i,t,r) \in\mathcal{A}} \pi_{it}^r\tilde{x}_{it}^r + \sum_{(i,t)\in\mathcal{N} \times\{q,\ldots,T\}}\rho_{it} x_{it} + \sum_{t\in\mathcal{T}}\sigma_t z_t = \tau $$

(13)

for all $(x,\tilde{x},z) \in F$ then (π,ρ,σ,τ)=α(λ,γ,0,0).

Define v by assigning a (0,1)-schedule to every component $i \in \mathcal{N}$ except that x _jt=0 for t∈{l+2,…,l+T _j}. Let z _t=1 for all $t \in\mathcal{T}$. Note that v∈F.

For $(\hat{i},\hat{t}) \in (\mathcal{N} \times\{q,\ldots,T\}) \backslash (\{\hat{i}\} \times\{l+1,\ldots,l+T_{j}\} ) $ define v _x as equal to v with the following exceptions. Let $x_{\hat{i}\hat{t}}=0$. If $\hat{i}=j$ and $\hat{t}=l+T_{j}+1$ let x _j,l+1=0 and $x_{j,l+T_{j}}=1$. Define u _x as equal to v _x except that $x_{\hat{i}\hat{t}}=1$. Since $T_{\hat{i}} \geq2$ we have v _x,u _x∈F and Lemma 2 gives $\rho_{\hat{i}\hat{t}}=0$.

For $\hat{t}\in\mathcal{T}$ define v _z as equal to v with the following exceptions. If $\hat{t}\geq q$ let $x_{i\hat{t}}=0$ for all $i\in\mathcal{N}$. If $\hat{t}=l+1$ let x _j,l+2=1. If $\hat{t}=l+T_{j}+1$ let x _j,l+1=0 and $x_{j,l+T_{j}}=1$. If $\hat{t}<q$ and l≠q assign an $(\hat{t}+1,\hat{t}+1)$-schedule to every component $i \in\mathcal{N}$ except that x _jt=0 for all t∈{l+2,…,l+T _j}. If $\hat{t}< q$ and l=q assign an $(\hat{t}+1, \hat{t}+1)$-schedule to every component $i \in\mathcal{N}$ except that x _jt=0 for all t∈{l+1,…,l+T _j}. Let u _x be defined as equal to v _z except that $z_{\hat{t}}=0$. Since T _i≥2 for all $i \in\mathcal{N}$, and T _ir≥2 for all $i\in\mathcal{N} $ and $r \in\mathcal{R}$, we have v _x,u _x∈F and Lemma 2 gives $\sigma_{\hat{t}}=0$.

For $(\hat{i},\hat{t},\hat{r}) \in\mathcal{A}\backslash\{ (j,l,q) \}$ define $v_{\tilde{x}}$ as equal to v with the following exceptions. If $\hat{i}=j$ and $\hat{t}+q-r>l$ assign a $(\hat{t}+1,\hat{r})$-schedule to component $\hat{i}$ except that x _jt=0 for t∈{l+1,…,l+T _j}. If $\hat{i}\neq j$ or $\hat{t}+q-r\leq l$ assign a $(\hat{t}+1,\hat{r})$-schedule to component $\hat{i}$ and let x _jt=0 for t∈{l+2,…,l+T _j}. Define $u_{\tilde{x}}$ as equal to $v_{\tilde{x}}$ except that $\tilde{x}_{\hat{i}\hat{t}}^{\hat{r}}=1$. Since $T_{\hat{i}\hat {r}}\geq2$ we have $v_{\tilde{x}},u_{\tilde{x}} \in F$ and Lemma 2 gives $\pi_{\hat{i}\hat{t}}^{\hat{r}}=0$

Define v _c as equal to v except that component j is assigned a $(\sum_{r \in\mathcal{R}} T_{jr},q)$-schedule with the exception that x _jt=0 for t∈{l+1,…,l+T _j}. Since v _c∈F, inserting it into equation (13) yields that τ=0.

For s∈{l+1,…,l+T _k} define v _s by assigning a (0,1)-schedule to each component $i \in\mathcal{N}$ except that x _jt=0 for t∈{l+1,…,l+T _j}∖{s}. Since v _s∈F, inserting it into Eq. (13) yields $\rho_{js} =- \pi_{jl}^{q}$. We conclude that (π,ρ,σ,τ)=α(λ,γ,0,0) which, together with Theorem 1 and the full dimensionality of conv(X) shown in Proposition 3, implies that F defines a facet of conv(X). □

Proposition 10

If T _ir≥2 for all $i \in\mathcal{N}$ and $r \in \mathcal{R}$, and T _i≥2 for all $i \in\mathcal{N}$, then (6h) defines a facet of the polyhedron conv(X).

Proof

For $j \in\mathcal{N}$ and $l \in\mathcal{T}_{j}^{ (6h)}$ let $F = \{(\tilde{x},x,z) \in\mathrm{conv}(X) | \sum_{t=l+1}^{l+T_{j}} x_{jt} = 1 \}$. Define γ _jt=1 for t∈{l+1,…,l+T _j} and γ _jt=0 for t∈{q,…,T}∖{l+1,…,l+T _j}. We will show that if

$$ \sum_{(i,t,r) \in\mathcal{A}} \pi_{it}^r\tilde{x}_{it}^r + \sum_{(i,t)\in\mathcal{N} \times\{q,\ldots,T\}}\rho_{it} x_{it} + \sum_{t\in\mathcal{T}}\sigma_t z_t = \tau $$

(14)

for all $(x,\tilde{x},z) \in F$ then (π,ρ,σ,τ)=α(0,γ,0,1).

Define v by assigning a (0,1)-schedule to every component $i \in \mathcal{N}$ except that x _jt=0 for t∈{l+2,…,l+T _j}. Let z _t=1 for all $t \in\mathcal{T}$. Note that v∈F.

For $(\hat{i},\hat{t}) \in (\mathcal{N} \times\{q,\ldots,T\}) \backslash (\{\hat{i}\} \times\{l+1,\ldots,l+T_{j}\} ) $ define v _x equal to v except that, if $\hat{t}= l+T_{j}+1$ and $\hat{i}=j$, let x _jl+1=0, $x_{jl+T_{j}}=1$. Define u _x equal to v _x except that $x_{\hat{i}\hat{t}}=0$. Since $T_{\hat{i}} \geq2$ we have v _x,u _x∈F and Lemma 2 gives $\rho_{\hat{i}\hat{t}}=0$.

For $\hat{t}\in\mathcal{T}$ define v _z as equal to v with the following exceptions. If $\hat{t}\geq q$ let $x_{i\hat{t}}=0$ for all $i \in\mathcal{N}$. If $\hat{t}= l+T_{j}+1$ or $\hat{t}=l+1$ let x _jl+1=0 and $x_{jl+T_{j}}=1$. If $\hat{t}< q$ assign a $(\hat{t}+1,\hat{t}+1)$-schedule to each component $i \in\mathcal{N}$ except x _jt=0 for t∈{l+2,…,l+T _j}. Let u _x be defined as equal to v _x except that $z_{\hat{t}}=0$. Since T _i≥2 for all $i \in\mathcal{N}$ and T _ir≥2 for all $i\in\mathcal{N} $ and $r\in\mathcal{R}$ we have v _x,u _x∈F and Lemma 2 gives $\sigma_{ \hat{t}}=0$.

For $(\hat{i},\hat{t},\hat{r}) \in\mathcal{A}$ define $v_{\tilde {x}}$ by assigning a $( \hat{t}+1,\hat{r})$-schedule to $\hat{i}$ and a (0,1)-schedule to every component $i \in\mathcal{N} \backslash\{\hat{i}\}$ except x _jt=0 for t∈{l+2,…,l+T _j}. Let z _t=1 for all $t \in\mathcal{T}$. Define u _x as equal to $v_{\tilde{x}}$ except that $\tilde{x}_{\hat{i} \hat{t}}^{\hat{r}}=0$. Since $T_{\hat{i}\hat{r}} \geq2$ we have $v_{\tilde{x}},u_{\tilde{x}}\in F$ and Lemma 2 gives $\pi_{\hat{i}\hat{t}}^{\hat{r}}=0$

For each s∈{l+1,…,l+T _k} define v _s by assigning a (0,1)-schedule to each component $i \in\mathcal{N}$ except x _jt=0 for t∈{l+1,…,l+T _j}∖{s}. We have v _s∈F and inserting it into Eq. (14) yields ρ _js=τ. We can conclude that (π,ρ,σ,τ)=α(0,γ,0,1) which, together with Theorem 1 and the full dimensionality of conv(X) shown in Proposition 3, implies that F defines a facet of conv(X). □

Proposition 11

If T _ir≥2 for all $i \in\mathcal{N}$ and $r \in \mathcal{R}$, and T _i≥2 for all $i \in\mathcal{N}$, then (6i) is a facet of the polyhedron conv(X).

Proof

For $j \in\mathcal{N}$ and $s \in\mathcal{T}_{j}^{ (6i)}$ let $F = \{(\tilde{x},x,z) \in\mathrm{conv}(X) | x_{js} = \tilde {x}_{js-1}^{q} \}$. Define γ _js=1 and γ _it=0 for $(i,t)\in (\mathcal{N}\times\{q, \ldots,T\} ) \backslash (\{j\}\times\{s\} ) $. Let $\pi_{js-1}^{q} = -1$ and $\pi_{it}^{r}=0$ for $(i,t,r) \in\mathcal{A} \backslash\{ (j,s-1,q) \}$. We will show that if

$$ \sum_{(i,t,r) \in\mathcal{A}} \pi_{it}^r\tilde{x}_{it}^r + \sum_{(i,t)\in\mathcal{N} \times\{q,\ldots,T\}}\rho_{it} x_{it} + \sum_{t\in\mathcal{T}}\sigma_t z_t = \tau $$

(15)

for all $(x,\tilde{x},z) \in F$ then (π,ρ,σ,τ)=α(λ,γ,0,0).

Define v by assigning a (0,1)-schedule to every component $i \in \mathcal{N}$ and z _t=1 for all $t \in\mathcal{T}$. Note that v∈F.

For $(\hat{i},\hat{t}) \in\mathcal{N} \times\{q,\ldots,T\} \backslash\{(j,s) \}$ define u _x as equal to v except that x _js=0. Since $T_{\hat{i}} \geq2$ we have v,u _x∈F and Lemma 2 gives $\rho_{\hat{i}\hat{t}}=0$.

For $\hat{t}\in\mathcal{T}$ define v _z as equal to v with the following exceptions. If $\hat{t}\geq q$ let $x_{i\hat{t}}=0$ for every $i \in\mathcal{N}$. If $\hat{t}= s$ assign a $( \hat{t}+1,q)$-schedule to component j. If $\hat{t}< q$ assigning a $(\hat{t}+1,\hat{t}+1)$-schedule to each component $i \in\mathcal{N}$. Define u _z as equal to v _z except that $z_{\hat{t}}=0$. Since T _ir≥2 for all $i\in\mathcal{N}$ and $r \in \mathcal{R}$ and T _i≥2 for all $i\in\mathcal{N}$ we have v _z,u _z∈F. Lemma 2 gives $\sigma_{\hat{t}}=0$.

For $(\hat{i},\hat{t},\hat{r}) \in\mathcal{A}$ let $v_{\tilde {x}}$ be defined as equal to v except that component $\hat{i}$ is assigned an $(\hat{t}+1,\hat{r})$-schedule. Define $u_{\tilde{x}}$ as equal to $v_{\tilde{x}}$ except that $\tilde{x}_{\hat{i}\hat {t}}^{\hat{r}}=1$. Since $T_{\hat{i}\hat{r}} \geq2$ we have $v_{\tilde{x}},u_{\tilde{x}}\in F$ and Lemma 2 gives $\pi_{\hat{i}\hat{t}}^{\hat{r}}=0$.

Define v _c as equal to v except that component j is assigned an (s,q)-schedule. Since v _c∈F inserting v _c into Eq. (15) implies that τ=0. Inserting v into Eq. (15) implies that $\rho_{js} = - \pi_{js-1}^{q}$. We can now conclude that (π,ρ,σ,τ)=α(λ,γ,0,0) which, together with Theorem 1 and the full dimensionality of conv(X) shown in Proposition 3, implies that F defines a facet of conv(X). □

Proposition 12

If T _ir≥2 for all $i \in\mathcal{N}$ and $r \in \mathcal{R}$, and T _i≥2 for all $i \in\mathcal{N}$, then (6j) defines a facet of the polyhedron conv(X).

Proof

For $j \in\mathcal{N}$, let $\tilde{t}= \min(T,\sum_{r \in \mathcal{R}}T_{jr})$ and $F = \{(\tilde{x},x,z) \in\mathrm{conv}(X) |\tilde{x}_{j\tilde {t}}^{q} = 1 \}$. Let $\lambda_{j\tilde{t}}^{q} = 1$ and $\lambda_{it}^{r}=0$ for $(i,t,r) \in \mathcal{A}\backslash\{(j,\tilde{t},q) \}$. We will show that if

$$ \sum_{(i,t,r) \in\mathcal{A}} \pi_{it}^r\tilde{x}_{it}^r + \sum_{(i,t)\in\mathcal{N} \times\{q,\ldots,T\}}\rho_{it} x_{it} + \sum_{t\in\mathcal{T}}\sigma_t z_t = \tau $$

(16)

for all $(x,\tilde{x},z) \in F$ then (π,ρ,σ,τ)=α(λ,0,0,1).

Define v by assigning a (0,1)-schedule to every component $i \in \mathcal{N}$ and z _t=1 for all $t \in\mathcal{T}$. Note that v∈F.

For $(\hat{i},\hat{t}) \in\mathcal{N} \times\{q,\ldots,T\}$ define u _x as equal to v except that $x_{\hat{i}\hat{t}}=0$. Since $T_{\hat{i}} \geq2$ we have v,u _x∈F and Lemma 2 gives $\rho_{\hat{i}\hat{t}}=0$.

For $\hat{t}\in\mathcal{T}$ define v _z as equal to v with the following exceptions. If $\hat{t}\geq q$ let $x_{i\hat{t}}=0$ for all $i \in\mathcal{N}$. If $\hat{t}<q$ then each component $i\in\mathcal{N}$ is assigned an $(\hat{t}+1,\hat{t}+1)$-schedule. Define u _z as equal to v _z except that $z_{\hat{t}}=0$. Since T _ir≥2 for all $i \in\mathcal{N}$ and $r \in\mathcal{R}$, and T _i≥2 for all $i \in\mathcal{N}$ we have v _z,u _z∈F and Lemma 2 gives σ _z=0.

For $(\hat{i},\hat{t},\hat{r}) \in\mathcal{A}\backslash\{ j,\tilde{t},q\}$ let $v_{\tilde{x}}$ equal v except that component $\hat{i}$ is assigned an $(\hat{t}+1,\hat{r})$-schedule. Define $u_{\tilde{x}}$ as equal to $v_{\tilde{x}}$ except that $\tilde{x}_{\hat{i}\hat{t}}^{\hat{r}}=1$. Since $T_{\hat{i}\hat {r}}\geq2$ we have $v_{\tilde{x}},u_{\tilde{x}} \in F$ and Lemma 2 gives $\pi_{\hat{i}\hat{t}}^{\hat{r}}=0$.

Inserting v into Eq. (16) yields $\pi_{j\tilde {t}}^{q} =\tau$. Hence (π,ρ,σ,τ)=α(λ,0,0,1) which, together with Theorem 1 and the full dimensionality of conv(X) shown in Proposition 3, implies that F defines a facet of conv(X). □

Proposition 13

If T _ir≥2 for all $i \in\mathcal{N}$ and $r \in \mathcal{R}$, and T _i≥3 for all $i \in\mathcal{N}$, then (6k) defines a facet of the polyhedron conv(X).

Proof

For $j \in\mathcal{N}$ and s∈{q,…,T} let $F =\{(\tilde{x},x,z) \in\mathrm{conv}(X) | x_{js} = 0 \}$. Define γ _js=1 and γ _it=0 for $(i,t) \in\mathcal {N}\times\{q,\ldots,T\} \backslash\{ (j,s) \}$. We will show that if

$$ \sum_{(i,t,r) \in\mathcal{A}} \pi_{it}^r\tilde{x}_{it}^r + \sum_{(i,t) \in\mathcal{N} \times\{q,\ldots,T\}}\rho_{it} x_{it} + \sum_{t \in\mathcal{T}}\sigma_t z_t = \tau $$

(17)

for all $(x,\tilde{x},z) \in F$ then (π,ρ,σ,τ)=α(0,γ,0,0).

Define v by assigning a (0,1)-schedule to every component $i \in \mathcal{N}$ except x _js=0. Let z _t=1 for all $t \in\mathcal{T}$. Note that v∈F.

For $(\hat{i},\hat{t}) \in\mathcal{N} \times\{q,\ldots,T\}$ define u _x as equal to v except that $x_{\hat{i}\hat{t}}=0$. Since $T_{\hat{i}} \geq3$ we have v,u _x∈F and Lemma 2 gives $\rho_{\hat{i}\hat{t}}=0$.

For $\hat{t}\in\mathcal{T}$ define v _z as equal to v with the following exceptions. If $\hat{t}\geq q$ let $x_{i\hat{t}} =0$ for all $i \in\mathcal{N}$. If $\hat{t}<q$ then each component $i\in\mathcal{N}$ is assigned an $(\hat{t}+1,\hat{t}+1)$-schedule with the exception that x _js=0. Define u _z as equal to v _z except that $z_{\hat{t}}=0$. Since T _ir≥2 for all $i \in\mathcal{N}$ and $r \in \mathcal{R}$, and T _i≥3 for all $i \in\mathcal{N}$ we have v _z,u _z∈F and Lemma 2 gives σ _z=0.

For $(\hat{i},\hat{t},\hat{r}) \in\mathcal{A}$ define $v_{\tilde {x}}$ as equal to v with the following exceptions. Assign an $(\hat{t}+1,\hat{r})$-schedule to $\hat{i}$ and a (0,1)-schedule to every component $i \in\mathcal{N} \backslash\{ \hat{i}\}$ except that x _js=0. Define $u_{\tilde{x}}$ as equal to $v_{\tilde{x}}$ except that $\tilde{x}_{\hat{i}\hat{t}}^{\hat{r}}=1$. Since $T_{\hat{i}\hat {r}} \geq2$ we have $v_{\tilde{x}},u_{\tilde{x}} \in F$ and Lemma 2 gives $\pi_{\hat{i}\hat{t}}^{\hat{r}}=0$.

Inserting v into Eq. (17) yields τ=0. We can conclude that (π,ρ,σ,τ)=α(0,γ,0,0) which, together with Theorem 1 and the full dimensionality of conv(X) shown in Proposition 3, implies that F defines a facet of conv(X). □

Proposition 14

If T _ir≥2 $i \in\mathcal{N}$ and $r \in\mathcal{R}$, and T _i≥2 for all $i \in\mathcal{N}$, then (6l) defines a facet of the polyhedron conv(X).

Proof

For $s \in\mathcal{T}$ let $F = \{(\tilde{x},x,z) \in\mathrm{conv}(X)| z_{s} = 1 \}$. Define ν _s=1 and ν _s=0 for $s\in\mathcal{T} \backslash\{ s \}$. We will show that if

$$ \sum_{(i,t,r) \in\mathcal{A}} \pi_{it}^r\tilde{x}_{it}^r + \sum_{(i,t)\in\mathcal{N} \times\{q,\ldots,T\}}\rho_{it} x_{it} + \sum_{t\in\mathcal{T}}\sigma_t z_t = \tau $$

(18)

for all $(x,\tilde{x},z) \in F$ then (π,ρ,σ,τ)=α(0,0,ν,1).

Define v by assigning a (0,1)-schedule to every component $i \in \mathcal{N}$ and let z _t=1 for all $t \in\mathcal{T}$. Note that v∈F.

For $(\hat{i},\hat{t}) \in\mathcal{N} \times\{q,\ldots,T\}$ define u _x as equal to v except that $x_{\hat{i}\hat{t}}=0$. Since $T_{\hat{i}} \geq2$ we have v,u _x∈F and Lemma 2 gives $\rho_{\hat{i}\hat{t}}=0$.

For $\hat{t}\in\mathcal{T} \backslash\{ s \}$ define v _z as equal to v with the following exceptions. If $\hat{t}\geq q$ let $x_{i\hat{t}}=0$ for all $i \in\mathcal{N}$. If $\hat{t} < q$ assigning an $(\hat{t}+1,\hat{t}+1)$-schedule to all components $i\in\mathcal{N}$. Define u _z as equal to v _z except that $z_{\hat{t}}=0$. Since T _i≥2 and $T_{i\hat{t}+1} \geq2$ for $i\in\mathcal{N}$, we have v _z,u _z∈F. Lemma 2 gives that $\sigma_{\hat{t}}=0$.

For $(\hat{i},\hat{t},\hat{r}) \in\mathcal{A}$ define $v_{\tilde {x}}$ as equal to v except that component $\hat{i}$ is assigned an $(\hat{t}+1,\hat{r})$-schedule. Define $u_{\tilde{x}}$ as equal to $v_{\tilde{x}}$ except that $\tilde{x}_{\hat{i}\hat{t}}^{\hat{r}}=1$. Since $T_{\hat{i}\hat {r}} \geq2$ we have $v_{\tilde{x}},u_{\tilde{x}}\in F$ and Lemma 2 gives $\pi_{\hat{i}\hat{t}}^{\hat{r}}=0$.

Inserting v into Eq. (18) yields that σ _s=τ. We conclude that (π,ρ,σ,τ)=α(0,0,ν,1) which, together with Theorem 1 and the full dimensionality of conv(X) shown in Proposition 3, implies that F is a facet of conv(X). □

References

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Michael Patriksson
- 1
Ann-Brith Strömberg
- 1
Adam Wojciechowski
- 1
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1.Department of Mathematical SciencesChalmers University of Technology and University of GothenburgGothenburgSweden

The stochastic opportunistic replacement problem, part I: models incorporating individual component lives

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