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Periodic replacement policies with shortage and excess costs

  • S.I.: Reliability Modeling with Applications Based on Big Data
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Abstract

It has been proposed that if replacement time is planned too early prior to failure, a waste of operation cost, i.e., excess costs, would incur because the system might run for an additional period of time to complete critical operations, and if replacement time is too late after failure, a great failure cost, i.e., shortage cost, is incurred due to the delay in time of the carelessly scheduled replacement. In order to make the preventive replacement policies perform in a more general way, the above two variable types of costs are taken into considerations for periodic replacement policies in this paper. We firstly take up a standard model in which the unit is replaced preventively at periodic times. Secondly, the modeling approaches of whichever occurs first and last are applied into periodic and random models, and replacement first and last policies are discussed to find optimum periodic replacement times for a random working time. Furthermore, optimum working numbers are obtained for the extended models. We give analytical discussions of the above replacement policies, and finally, numerical examples are illustrated.

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Acknowledgements

This work is supported by National Natural Science Foundation of China (No. 71801126), Natural Science Foundation of Jiangsu Province (No. BK20180412), Aeronautical Science Foundation of China (No. 2018ZG52080), and Ministry of Science and Technology of Taiwan (No. MOST107-2221-E-030-011).

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Authors and Affiliations

  1. Nanjing University of Aeronautics and Astronautics, Nanjing, 211106, China

    Xufeng Zhao

  2. Fu Jen Catholic University, New Taipei City, 24205, Taiwan

    Mingchih Chen

  3. Aichi Institute of Technology, Toyota, 470-0392, Japan

    Toshio Nakagawa

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  1. Xufeng Zhao
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  2. Mingchih Chen
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  3. Toshio Nakagawa
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Correspondence to Mingchih Chen.

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Appendix

Appendix

A.1. Prove that

$$\begin{aligned} Q_1(K,1)= \frac{\int _{KT}^{(K+1)T} {\overline{G}}(t) \mathrm {d}t}{ \int _{KT}^{(K+1)T} {\overline{G}}(t){\overline{F}}(t) \mathrm {d}t} \end{aligned}$$

increases strictly with K to \(\infty \).

Proof

Noting that for \(0<T<\infty \),

$$\begin{aligned} \frac{1}{{\overline{F}}(KT)}< Q_1(K,1)< \frac{1}{{\overline{F}}((K+1)T)}, \end{aligned}$$

we obtain

$$\begin{aligned}&\lim _{K\rightarrow \infty } Q_1(K,1)=\infty ,\\&Q(K-1,1)< \frac{1}{{\overline{F}}(KT)}< Q_1(K,1), \end{aligned}$$

which follows that \(Q_1(K,1)\) increases strictly with K to \(\infty \). \(\square \)

A.2. Prove that

$$\begin{aligned} {\widetilde{Q}}_1(K,1) = \frac{\int _{KT}^{(K+1)T }G(t) \mathrm {d}t}{\int _{KT}^{(K+1)T} G(t) {\overline{F}}(t) \mathrm {d}t} \end{aligned}$$

increases strictly with K to \(\infty \).

Proof

Noting that for \(0<T<\infty \),

$$\begin{aligned} \frac{1}{{\overline{F}}(KT)}< {\widetilde{Q}}_1(K,1)< \frac{1}{{\overline{F}}((K+1)T)}, \end{aligned}$$

so that it can be proved that \( {\widetilde{Q}}_1(K,1)\) increases strictly with K to \(\infty \).

A.3. Prove that

$$\begin{aligned} Q_1(K,N)\equiv \frac{\int _{KT}^{(K+1)T } [1- G^{(N)}(t)] \mathrm {d}t }{\int _{KT}^{(K+1)T } [1- G^{(N)}(t)] {\overline{F}}(t) \mathrm {d}t} \end{aligned}$$

increases strictly with K to \(\infty \), and when \(G(t)= 1- \mathrm {e}^{-\theta t}\), it increases strictly with N to Q(K) in (6).

Proof

Noting that for \(0<T<\infty \),

$$\begin{aligned} \frac{1}{{\overline{F}}(KT)}< Q_1(K,N) < \frac{1}{ {\overline{F}}((K+1)T)}, \end{aligned}$$

we obtain

$$\begin{aligned}&\lim _{K\rightarrow \infty }Q_1(K,N)=\infty ,\\&Q_1 (K-1,N)< \frac{1}{{\overline{F}}(KT)}< Q_1(K,N), \end{aligned}$$

which follows that \(Q_1(K,N)\) increases strictly with K to \(\infty \).

When \(G(t)= 1- \mathrm {e}^{-\theta t}\), we firstly prove that

$$\begin{aligned} Q(K,j)\equiv&\frac{\int _{KT}^{(K+1)T} t^j \mathrm {e}^{-\theta t} \mathrm {d}t }{\int _{KT}^{(K+1)T} t^j \mathrm {e}^{-\theta t} {\overline{F}}(t) \mathrm {d}t} \\ =&\frac{\int _0^T (t+KT)^j \mathrm {e}^{-\theta t} \mathrm {d}t}{ \int _0^T (t+KT)^j \mathrm {e}^{-\theta t} {\overline{F}}(t+KT) \mathrm {d}t}\ \ \ (j=0,1,2,\ldots ) \end{aligned}$$

increases strictly with j. Forming \(Q(K,j+1)- Q(K,j)\) and letting

$$\begin{aligned} L(j) \equiv&\int _0^T (t+KT)^{j+1} \mathrm {e}^{-\theta t} \mathrm {d}t \int _0^T (u+ KT)^j \mathrm {e}^{-\theta u}{\overline{F}}(u+KT)\mathrm {d}u\\&- \int _0^T (t+ KT)^{j+1} \mathrm {e}^{-\theta t} {\overline{F}}(t +KT)\mathrm {d}t\int _0^T (u+KT)^j \mathrm {e}^{-\theta u}\mathrm {d}u, \end{aligned}$$

i.e.,

$$\begin{aligned} L(j) =&\int _0^T (t+KT)^{j+1} \mathrm {e}^{-\theta t} \left\{ \int _0^T (u+KT)^j \mathrm {e} ^{-\theta u} [F(t+ KT)-F(u+KT)]\mathrm {d}u \right\} \mathrm {d}t\\ =&\int _0^T (t+KT)^{j+1} \mathrm {e}^{-\theta t} \left\{ \int _0^t (u+KT)^j \mathrm {e} ^{-\theta u} [F(t+KT)-F(u+KT)]\mathrm {d}u \right. \\&\left. +\int _t^T (u+ KT)^j \mathrm {e} ^{-\theta u} [F(t+KT)-F(u+KT)] \mathrm {d}u \right\} \mathrm {d}t. \end{aligned}$$

On the other hand,

$$\begin{aligned}&\int _0^T (t+KT)^{j+1} \mathrm {e}^{-\theta t} \left\{ \int _t^T (u+KT)^j \mathrm {e} ^{-\theta u} [F(t+KT)-F(u+KT)] \mathrm {d}u \right\} \mathrm {d}t\\&=\int _0^T (t+KT)^j \mathrm {e}^{-\theta t}\left\{ \int _0^t (u+KT)^{j+1} \mathrm {e}^{-\theta u} [F(u+KT)-F(t+KT)]\mathrm {d}u \right\} \mathrm {d}t. \end{aligned}$$

Thus, we obtain

$$\begin{aligned} L(j)=&\int _0^T (t+KT)^j \mathrm {e}^{-\theta t}\\&\times \left\{ \int _0^t (u+KT)^N \mathrm {e}^{-\theta u} [F(t+KT)-F(u+KT)](t-u) \mathrm {d}u \right\} \mathrm {d}t>0, \end{aligned}$$

which follows that Q(K, j) increases strictly with j. Thus,

$$\begin{aligned} Q_1(K,N)=\frac{\sum _{j=0}^{N-1} \int _{KT}^{(K+1)T} [(\theta t)^j/j!] \mathrm {e}^{-\theta t} \mathrm {d}t}{\sum _{j=0}^{N-1} \int _{KT}^{(K+1)T} [(\theta t)^j/j!] \mathrm {e}^{-\theta t} {\overline{F}}(t) \mathrm {d}t} \end{aligned}$$

increases strictly with N to Q(K). \(\square \)

A.4. Prove that

$$\begin{aligned} H_1 (N,K)\equiv \frac{\int _0^{KT} (\theta t) ^N \mathrm {e}^{-\theta t} \mathrm {d}t }{ \int _{0}^{KT} (\theta t) ^N \mathrm {e}^{-\theta t} {\overline{F}}(t) \mathrm {d}t} \end{aligned}$$

increases with K from 1 to \(1/ \int _0^\infty \theta [(\theta t)^N/N!] \mathrm {e}^{-\theta t} {\overline{F}}(t) \mathrm {d}t \), and increases strictly with N to \(1/ {\overline{F}}(KT)\).

Proof

Noting that for \(0<T<\infty \),

$$\begin{aligned}&\lim _{K\rightarrow 0} H_1(N,K)=1,\ \ \ \lim _{N\rightarrow \infty } H_1(N,K)=\frac{1}{{\overline{F}}(KT)},\\&\lim _{K\rightarrow \infty } H_1(N,K)= \frac{1}{\int _0^\infty \theta [(\theta t)^N/N!] \mathrm {e}^{-\theta t} {\overline{F}}(t) \mathrm {d}t }. \end{aligned}$$

Differentiating \(H_1(N,1)\) with respect to T,

$$\begin{aligned}&(\theta T)^N \mathrm {e}^{-\theta T}\int _0^T (\theta t )^N \mathrm {e}^{-\theta t} {\overline{F}}(t) \mathrm {d}t - (\theta T)^N \mathrm {e}^{-\theta T} {\overline{F}}(T) \int _0^T (\theta t)^N \mathrm {e}^{-\theta t} \mathrm {d}t\\&\quad =(\theta T)^N \mathrm {e}^{-\theta T} \int _0^T (\theta t)^N \mathrm {e}^{-\theta t} [F(T)-F(t)]\mathrm {d}t >0, \end{aligned}$$

which follows that \(H_1(N,1)\) increases strictly with T. Thus, \(H_1(N,K)\) increases strictly with K from 1 to \(1/ \int _0^\infty \theta [(\theta t)^N/N!] \mathrm {e}^{-\theta t} {\overline{F}}(t) \mathrm {d}t \). Next, forming \(H_1(N+1,1)-H_1(N,1)\) and letting

$$\begin{aligned} L(T)\equiv&\int _0^T (\theta t)^{N+1} \mathrm {e}^{-\theta t} \mathrm {d}t \int _0^T (\theta t)^{N} \mathrm {e}^{-\theta t} {\overline{F}}(t) \mathrm {d}t -\int _0^T (\theta t)^{N} \mathrm {e}^{-\theta t} \mathrm {d}t \int _0^T (\theta t)^{N+1} \mathrm {e}^{-\theta t} {\overline{F}}(t) \mathrm {d}t, \end{aligned}$$

we have \(L(0)=0\) and

$$\begin{aligned} L'(T)=(\theta T)^N \mathrm {e}^{-\theta T} {\overline{F}}(T) \int _0^T (\theta t)^N \mathrm {e}^{-\theta t} {\overline{F}}(t)(\theta T-\theta t)[F(T)-F(t)] \mathrm {d}t>0, \end{aligned}$$

which follows that \(L(T)>0\), i.e., \(H_1(N,K)\) increases strictly with N to \(1/{\overline{F}}(KT)\). \(\square \)

A.5. Prove that

$$\begin{aligned} {\widetilde{Q}}_1(K,N)\equiv \frac{\int _{KT}^{(K+1)T} G^{(N)} (t) \mathrm {d}t}{\int _{KT}^{(K+1)T} G^{(N)} (t) {\overline{F}}(t) \mathrm {d}t } \end{aligned}$$

increases strictly with K to \(\infty \), and when \(G(t)=1- \mathrm {e}^{-\theta t}\), it increases strictly with N from Q(K) to \(1/{\overline{F}}((K+1)T)\).

Proof

Noting that for \(0<T<\infty \),

$$\begin{aligned} \frac{1}{{\overline{F}}(KT)}< {\widetilde{Q}}_1(K,N)< \frac{1}{{\overline{F}}((K+1)T)}, \end{aligned}$$

so that it can be proved that \( {\widetilde{Q}}_1(K,N)\) increases strictly with K to \(\infty \).

When \(G(t)= 1- \mathrm {e}^{-\theta t}\),

$$\begin{aligned} {\widetilde{Q}}_1(K,N)=\frac{\sum _{j=N}^ \infty \int _{KT}^{(K+1)T} [(\theta t)^j/j!] \mathrm {e}^{-\theta t}\mathrm {d}t }{\sum _{j=N}^ \infty \int _{KT}^{(K+1)T}[(\theta t)^j/j!] \mathrm {e}^{-\theta t} {\overline{F}}(t) \mathrm {d}t }, \end{aligned}$$

and \({\widetilde{Q}}_1(K,0 )= Q(K)\). From the proof in A.3, it can be proved that \({\widetilde{Q}}_1(K,N)\) increases strictly with N from Q(K). \(\square \)

A.6. Prove that

$$\begin{aligned} {\widetilde{H}}_1(N,K)\equiv \frac{\int _{KT}^\infty (\theta t)^N\mathrm {e}^{-\theta t} \mathrm {d}t }{ \int _{KT}^\infty (\theta t)^N\mathrm {e}^{-\theta t} {\overline{F}}(t) \mathrm {d}t } \end{aligned}$$

increases strictly with K from \(1/ \int _0^\infty \theta [(\theta t)^N/N!] \mathrm {e}^{-\theta t} {\overline{F}}(t) \mathrm {d}t \) to \(\infty \) and increases strictly with N from \(1/ \int _0^\infty \theta \mathrm {e}^{-\theta t} {\overline{F}}(t+KT)\mathrm {d}t\) to \(\infty \).

Proof

Noting that

$$\begin{aligned}&\lim _{K\rightarrow 0} {\widetilde{H}}_1(N,K) = \frac{1}{ \int _0^\infty \theta [(\theta t)^N/N!] \mathrm {e}^{-\theta t} {\overline{F}}(t) \mathrm {d}t }, \ \ \ \lim _{K\rightarrow \infty } {\widetilde{H}}_1(N,K) = \infty ,\\&\lim _{N\rightarrow 0}{\widetilde{H}}_1(N,K)= \frac{1}{\int _0^\infty \theta \mathrm {e}^{-\theta t} {\overline{F}}(t+KT)\mathrm {d}t},\ \ \ \lim _{N\rightarrow \infty } {\widetilde{H}}_1(N,K) =\infty . \end{aligned}$$

Differentiating \({\widetilde{H}}_1(N,1)\) with respect to T,

$$\begin{aligned}&-(\theta T)^N \mathrm {e}^{-\theta T}\int _T^\infty (\theta t)^N \mathrm {e}^{-\theta t} {\overline{F}}(t ) \mathrm {d}t + (\theta T)^N \mathrm {e}^{-\theta T} {\overline{F}}(T) \int _T^\infty (\theta t)^N\mathrm {e}^{-\theta t} \mathrm {d}t\\&= (\theta T)^N \mathrm {e}^{-\theta T} \int _T^\infty (\theta t)^N \mathrm {e}^{-\theta t} [F(t)-F(T)] \mathrm {d}t>0, \end{aligned}$$

which follows that \({\widetilde{H}}_1(N,K)\) increases strictly with K from \(1/ \int _0^\infty \theta [(\theta t)^N/N!] \mathrm {e}^{-\theta t} {\overline{F}}(t) \mathrm {d}t\) to \(\infty \).

Next, forming \({\widetilde{H}}_1(N+1,K)- {\widetilde{H}}_1(N,K)\) and letting

$$\begin{aligned} L(T)\equiv&\int _T^\infty (\theta t)^{N+1} \mathrm {e}^{\theta t} \mathrm {d}t \int _T^\infty (\theta t)^N \mathrm {e}^{\theta t} {\overline{F}}(t) \mathrm {d}t- \int _T^\infty (\theta t)^N \mathrm {e}^{\theta t} \mathrm {d}t \int _T^\infty (\theta t)^{N+1} \mathrm {e}^{\theta t} {\overline{F}}(t) \mathrm {d}t, \end{aligned}$$

we have \(L(\infty )=0\) and

$$\begin{aligned} L'(T)= (\theta T)^N \mathrm {e}^{-\theta T} \int _T^\infty (\theta t)^N \mathrm {e}^{-\theta t} (\theta T- \theta t) [F(t)- F(T)]\mathrm {d}t<0, \end{aligned}$$

which follows that \(L(T)>0\), i.e., \({\widetilde{H}}_1(N,K)\) increases strictly with N from \(1/ \int _0^\infty \theta \mathrm {e}^{-\theta t} {\overline{F}}(t+KT)\mathrm {d}t\) to \(\infty \). \(\square \)

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Zhao, X., Chen, M. & Nakagawa, T. Periodic replacement policies with shortage and excess costs. Ann Oper Res 311, 469–487 (2022). https://doi.org/10.1007/s10479-020-03566-z

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