Maintenance optimization in failure-prone systems under imperfect preventive maintenance

Abstract

In the majority of the existing preventive optimization models only costs related to maintenance actions are accounted for, while breakdown and operational costs are usually ignored. Liao et al. (J Intell Manuf 21(6):875–884, 2010) proposed a preventive maintenance model to deal with this shortcoming. In the present paper, we revisit and discuss the results provided in Liao et al. (2010) and point out some inconsistencies in the maintenance optimization model proposed therein. Accordingly, we develop a new maintenance optimization model and discuss some of its main cost components. Furthermore, optimality conditions are also formally investigated and a solution method is provided. Numerical experiments are conducted to illustrate the validity of the proposed approach and results are compared with those provided in the original paper by Liao et al. (2010).

Appendices

Appendix A: Proof of Proposition 1

When the values of $R_{0}$ is fixed, the expected total cost rate function becomes a discrete uni variate function of N and is denoted here simply as ${\mathcal {J}}(N)$. To alleviate the notations, we further write the expected total cost rate function as:

$$\begin{aligned} {\mathcal {J}}(N)=\frac{\alpha +N\gamma +\varphi (N)}{\psi (N)}, \end{aligned}$$

where:

$$\begin{aligned}&\alpha =C_r-C_p,\\&\gamma =C_p+C_b -(C_m+C_b)\ln (R_0),\\&\varphi (N)=\sum _{k=1}^N \delta _k,\text { and } \psi (N)=\sum \limits _{k=1}^{N}{T_{k}}. \end{aligned}$$

A number N of PM is optimal if it satisfies the following two conditions:

• 1. ${\mathcal {J}}(N)<{\mathcal {J}}(N-1),$ and

• 2. ${\mathcal {J}}(N)< {\mathcal {J}}(N+1).$

The first condition implies that:

$$\begin{aligned} {\mathcal {J}}(N)<\frac{\alpha +(N-1)\gamma +\varphi (N-1)}{\psi (N-1)}, \end{aligned}$$

Noting that $\delta _N=\varphi (N)-\varphi (N-1)$ and $\psi (N)=\psi (N-1)+T_N$, then the above equation can equivalently be written as:

$$\begin{aligned} {\mathcal {J}}(N)<\frac{a+N\gamma +\varphi (N)}{\psi (N)-T_{N}}-\frac{\gamma +\delta _N}{\psi (N)-T_{N}}, \end{aligned}$$

and then we have:

$$\begin{aligned} {\mathcal {J}}(N)-\left( \frac{a+N\gamma +\varphi (N)}{\psi (N)}\right) \frac{\psi (N)}{\psi (N)-T_{N}}<-\frac{\gamma +\delta _N}{\psi (N)-T_{N}}, \end{aligned}$$

which in turn leads to:

$$\begin{aligned} {\mathcal {J}}(N)\left( \frac{-T_N}{\psi (N)-T_{N}}\right) <-\frac{\gamma +\delta _N}{\psi (N)-T_{N}}, \end{aligned}$$

From the fact that $\psi (N)>T_N$ (i.e., any PM interval is always smaller than the sum of all PM intervals), the above equation finally gives:

$$\begin{aligned} {\mathcal {J}}(N)>\frac{\gamma +\delta _N}{T_{N}}. \end{aligned}$$

Since $\delta _N\ge 0$, then we get:

$$\begin{aligned} {\mathcal {J}}(N)>\frac{\gamma }{T_{N}}. \end{aligned}$$

(21)

A similar procedure can be followed with condition (2) above to get:

$$\begin{aligned} {\mathcal {J}}(N)<\frac{\gamma }{T_{N+1}}. \end{aligned}$$

(22)

Combining inequalities (21) and (22) gives the result of Proposition (1):

$$\begin{aligned} \frac{1}{T_{N}}<\frac{{\mathcal {J}}(N)}{\gamma }< \frac{1}{T_{N+1}}. \end{aligned}$$

Now let us show that if the condition $\frac{\delta _{k+1}}{T_{k+1}}>\frac{\delta _{k+1}}{T_{k+1}}$, then a number N which is solution of the above inequality is unique and finite. To do so, let us return back to conditions (1) and (2) above. From condition (1), it follows that:

$$\begin{aligned} {\mathcal {J}}(N)-{\mathcal {J}}(N-1)<0. \end{aligned}$$

Substituting ${\mathcal {J}}(N)$ and ${\mathcal {J}}(N-1)$ by their respective values and computing the difference ${\mathcal {J}}(N) - {\mathcal {J}}(N-1)$, we have:

$$\begin{aligned} {\mathcal {J}}(N)-{\mathcal {J}}(N-1)= & {} \frac{\alpha +N\gamma +\varphi (N)}{\psi (N)}\nonumber \\&-\frac{\alpha +(N-1)\gamma +\varphi (N-1)}{\psi (N-1)}, \end{aligned}$$

From the fact that $\psi (N-1)=\psi (N)-T_N$ and $\varphi (N-1)=\varphi (N)-\delta _N$ the above equation can be simplified as:

$$\begin{aligned}&{\mathcal {J}}(N)-{\mathcal {J}}(N-1)\nonumber \\&\quad =\frac{\gamma \left( \psi (N)-NT_N\right) +\left( \psi (N)\delta _N-T_N\varphi (N)\right) -\alpha T_N}{\psi (N-1)\psi (N)}.\nonumber \\ \end{aligned}$$

(23)

Since ${\mathcal {J}}(N)-{\mathcal {J}}(N-1)<0$ together with the fact that $\psi (N)$ is positive valued, it follows that the numerator of the above equation is negative. This implies that:

$$\begin{aligned} \xi (N)<\frac{\alpha }{\gamma }, \end{aligned}$$

where the discrete function $\xi (N)$ is defined as:

$$\begin{aligned} \xi (N)=\frac{\gamma \left( \psi (N)-NT_N\right) +\left( \psi (N)\delta _N-T_N\varphi (N)\right) }{\gamma T_N}. \end{aligned}$$

By considering condition (2) and following the same reasoning as done from condition (1), condition (2) is written as:

$$\begin{aligned} \xi (N+1)> \frac{\alpha }{\gamma }. \end{aligned}$$

The function $\xi (N)$ as defined above is increasing in N. Indeed, computing the difference $\xi (N+1)-\xi (N)$, we get:

$$\begin{aligned} \xi (N+1)-\xi (N)= & {} \frac{\psi (N)\left( T_N-T_{N+1}\right) }{T_{N}T_{N+1}}\\&+\,\psi (N)\left( \frac{\delta _{N+1}}{T_{N+1}}-\frac{\delta _{N}}{T_{N}}\right) . \end{aligned}$$

All $T_k$ terms are non-negative. Furthermore, the term $T_{N}-T_{N+1}$ is positive because the imperfect PM model used guarantees that $T_{N}>T_{N+1}$ as, by definition of the hybrid failure rate, each PM interval is shorter than its predecessor. This together with the assumption stating that $\frac{\delta _{k+1}}{T_{k+1}}>\frac{\delta _{k}}{T_{k}}$, it follows that $\xi (N+1)-\xi (N)$ is strictly positive for all values of N. Hence the discrete function $\xi (N)$ is strictly increasing.

Now, it will be shown that the solution is finite as well. On one hand, by definition of the hybrid failure rate, the term $\psi (N)-NT_N$ is positive. On the other hand, the term $\psi (N)\delta _N-T_N\varphi (N)$ is equivalently computed as:

$$\begin{aligned} \psi (N)\delta _N-T_N\varphi (N)&=\sum _{k=1}^{N}(T_k\delta _N-T_N\delta _k)\\&=\sum _{k=1}^{N}T_kT_N\left( \frac{\delta _N}{T_N}-\frac{\delta _k}{T_k}\right) \end{aligned}$$

From the assumption stating that $\frac{\delta _{k+1}}{T_{k+1}}>\frac{\delta _{k}}{T_{k}}$ for all $k\ge 1$, it follows that the term $\psi (N)\delta _N-T_N\varphi (N)$ is also positive. Therefore, the discrete function $\xi (N)$ is such that:

$$\begin{aligned} \gamma \xi (N)>\frac{\psi (N)-NT_N}{T_N}. \end{aligned}$$

From the fact that $\psi (N)>T_1+(N-1)T_N$, the above inequality implies that:

$$\begin{aligned} \gamma \xi (N)> \frac{T_1+(N-1)T_N-NT_N}{T_N}, \end{aligned}$$

which can also be written as:

$$\begin{aligned} \gamma \xi (N)> \left( \frac{T_1}{T_N}-1\right) . \end{aligned}$$

We have that $\xi (N)$ is increasing in N, this together with the fact that:

$$\begin{aligned} \lim \limits _{N\rightarrow +\infty }\left( \frac{T_1}{T_N}\right) =\infty , \end{aligned}$$

it follows that,

$$\begin{aligned}\lim \limits _{N\rightarrow +\infty }\xi (N)=+\infty .\end{aligned}$$

In summary, we have that the function $\xi (N)$ is strictly increasing and tends to $+\infty$ as N tends to $+\infty$. Thus, one may conclude that there exist a finite and unique $N^*$ for which conditions (1) and (2) are satisfied. Therefore, the first integer satisfying these two conditions is the global minimum of the expected total cost rate ${\mathcal {J}}(N)$.

Appendix B: Proof of Lemma 1

Let us denote by $H_1(t)=\int _0^t \lambda _1(x)\mathrm {d}x$, the cumulative failure rate of the system at the start of a replacement cycle where a system is new. From Eq. (6), we have that:

$$\begin{aligned} H_1(Y_k)=H_1(b_{k-1}Y_{k-1})- \frac{\ln (R_0)}{B_{k-1}}. \end{aligned}$$

It follows that:

$$\begin{aligned} \frac{\partial H_1(Y_{k})}{\partial Y_k}&=-\frac{\partial }{\partial Y_k}\left( \frac{\ln (R_0)}{B_{k-1}}\right) \nonumber \\&=-\left( \frac{1}{B_{k-1}}\right) \left( \frac{\partial \ln (R_0)}{\partial R_0}\right) \left( \frac{\partial R_0}{\partial Y_k}\right) \nonumber \\&=-\left( \frac{1}{B_{k-1}}\right) \left( \frac{1}{R_0}\right) \left( \frac{\partial R_0}{\partial Y_k}\right) . \end{aligned}$$

(24)

Since $\frac{\partial H_1(Y_{k})}{\partial Y_k}=\lambda _1(Y_k)$, it follows that:

$$\begin{aligned} \lambda _1(Y_k)=-\left( \frac{1}{B_{k-1}}\right) \left( \frac{1}{R_0}\right) \left( \frac{\partial R_0}{\partial Y_k}\right) , \end{aligned}$$

then we get:

$$\begin{aligned} \frac{\partial Y_{k}}{\partial R_0}=\frac{-1}{B_{k-1}\lambda _1(Y_{k})R_{0}} \end{aligned}$$

Appendix C: Proof of Proposition 2

For fixed values of the number N of PM, the optimal value of the reliability threshold $R_{0}^{*}$ is obtained by solving the following partial derivative:

$$\begin{aligned} \frac{\partial {\mathcal {J}}(R_0,N)}{\partial R_{0}}=0. \end{aligned}$$

Before starting the computation of the above partial derivative, one may observe that due to the operational cost structure the total expected operational cost rate $\frac{{\mathbb {E}}[OC]}{{\mathbb {E}}[{\mathcal {T}}]}$ is such that:

$$\begin{aligned} \frac{{\mathbb {E}}[OC]}{{\mathbb {E}}[{\mathcal {T}}]}= & {} \frac{\sum _{k=1}^N \left( C_0 T_k+kC_1 T_K+\frac{C_2}{2}T_k^2\right) }{\sum _{k=1}^N T_k}\\= & {} C_0+\frac{\sum _{k=1}^N \left( kC_1 T_K+\frac{C_2}{2}T_k^2\right) }{\sum _{k=1}^N T_k}. \end{aligned}$$

From the above equation, one may conclude that the term induced by the fixed operational cost rate $C_0$ has no impact on the computation of the partial derivative $\frac{\partial {\mathcal {J}}(R_0,N)}{\partial R_{0}}$. Accordingly, writing the partial derivative $\frac{\partial {\mathcal {J}}(R_0,N)}{\partial R_{0}}$ and setting it equal to 0 leads to the following equality:

$$\begin{aligned}&\frac{\sum \nolimits _{k=1}^{N-1}{(1-a_k)Y_k}+Y_N}{\sum \nolimits _{k=1}^{N-1}{(1-a_k)\frac{\partial {Y_k}}{\partial {R_0}}}+\frac{\partial {Y_N}}{\partial {R_0}}}\\&\quad =\frac{C_r+(N-1)C_p+NC_b-N(C_m+C_b)\ln (R_0)+ \sum _{k=1}^N \left( kC_1+\frac{C_2}{2}(Y_k-a_{k-1}Y_{k-1})\right) (Y_k-a_{k-1}Y_{k-1})}{\frac{N(C_m+C_b)}{R_0}+\frac{\partial (\sum _{k=1}^N \left( kC_1+\frac{C_2}{2}(Y_k-a_{k-1}Y_{k-1})\right) (Y_k-a_{k-1}Y_{k-1}))}{\partial R_0}}. \end{aligned}$$

In the above equation, substituting the expression of the partial derivative $\frac{\partial Y_{k}}{\partial R_{0}}:$ $\frac{\partial Y_{k}}{\partial R_{0}}=\frac{-1}{B_{k-1}\lambda _1(Y_{k})R_{0}},$ obtained from Lemma 1, yields the following equality:

$$\begin{aligned}&\frac{\sum \nolimits _{k=1}^{N-1}{(1-a_k)Y_k}+Y_N}{\sum \nolimits _{k=1}^{N-1}\frac{1-a_k}{B_{k-1}\lambda _1(Y_{k})}+\frac{1}{B_{N-1}\lambda _1(Y_{N})}}\\&\quad =\frac{C_r+(N-1)C_p+NC_b-N(C_m+C_b)\ln (R_0)+ \sum _{k=1}^N \left( kC_1+\frac{C_2}{2}(Y_k-a_{k-1}Y_{k-1})\right) (Y_k-a_{k-1}Y_{k-1})}{N(C_m+C_b)-\sum \nolimits _{k=1}^{N}\frac{(kC_1+C_2(Y_k-a_{k-1}Y_{k-1}))(a_{k-1}b_{k-1}\lambda _1(Y_k)-\lambda _1(Y_{k-1}))}{B_{k-1}\lambda _1(Y_{k-1})\lambda _1(Y_k)}}, \end{aligned}$$

which is equivalently written as:

$$\begin{aligned}&\left( \frac{\sum \nolimits _{k=1}^{N-1}{(1-a_k)Y_k}+Y_N}{\sum \nolimits _{k=1}^{N-1}\frac{1-a_k}{B_{k-1}\lambda _1(Y_{k})}+\frac{1}{B_{N-1}\lambda _1(Y_{N})}}\right) \left( N(C_m+C_b)\right. \\&\qquad \left. -\sum \nolimits _{k=1}^{N}\frac{(kC_1+C_2(Y_k-a_{k-1}Y_{k-1}))(a_{k-1}b_{k-1}\lambda _1(Y_k)-\lambda _1(Y_{k-1}))}{B_{k-1}\lambda _1(Y_{k-1})\lambda _1(Y_k)}\right) \\&\quad =(C_r+(N-1)C_p+NC_b) \\&\qquad +\,\sum _{k=1}^N \left( kC_1+\frac{C_2}{2}(Y_k-a_{k-1}Y_{k-1})\right) \\&\qquad (Y_k-a_{k-1}Y_{k-1})- N(C_m+C_b)\ln (R_0). \end{aligned}$$

Dividing each side of the above equation by the quantity $N(C_m+C_b)$, and performing some basic algebraic operations, we then obtain the result of the proposition.