Optimal replacement policy for a repairable system with deterioration based on a renewal-geometric process

Abstract

The optimal replacement policy is proposed for a new maintenance model of a repairable deteriorating system to minimize the average cost rate throughout the system life cycle. It is assumed that the system undergoes deterioration with an increasing trend of deterioration probability after each repair. More specifically, a novel maintenance model is first presented based on a new defined renewal-geometric process, which splits the operation process into an early renewal process and a late geometric process to characterize such a special deterioration delay. Then, the average cost rate for the new model is formulated according to the renewal-reward theorem. Next, a theorem is presented to derive the theoretical relationships of optimal replacement policies for the geometric-process maintenance model and the new proposed model, respectively. Finally, numerical examples suggest that the optimum values can be determined to minimize the average cost rates.

Appendix

In this section, some requisite proofs procedures for lemmas and theorem are shown below.

Proof of Lemma 1

Although the proof is trivial, the consequences of this result are of major importance. To avoid repetition of proof, we only need to prove that the common expression formula suffices the conclusion. It needs firstly to invoke the common expression formula C(N). Then it follows that

$$\begin{aligned}&C(N+1)-C(N)= \frac{(c_r+c_w)\sum \nolimits _{i=1}^{N}E(Y_i)+c_f}{\sum \nolimits _{i=1}^{N+1}E(X_i)+\sum \nolimits _{i=1}^{N}E(Y_i)}- \frac{(c_r+c_w)\sum \nolimits _{i=1}^{N-1}E(Y_i)+c_f}{\sum \nolimits _{i=1}^{N}E(X_i)+\sum \nolimits _{i=1}^{N-1}E(Y_i)}\\&= \frac{(c_r+c_w)[E(Y_N)\sum \nolimits _{i=1}^{N}E(X_i)-E(X_{N+1})\sum \nolimits _{i=1}^{N-1}E(Y_i)]-c_f[E(Y_N)+E(X_{N+1})]}{\left[ \sum \nolimits _{i=1}^{N+1}E(X_i)+\sum \nolimits _{i=1}^{N}E(Y_i)\right] \cdot \left[ \sum \nolimits _{i=1}^{N}E(X_i)+\sum \nolimits _{i=1}^{N-1}E(Y_i)\right] }. \end{aligned}$$

It can be easily seen that

$C(N+1)-C(N)>0\Longleftrightarrow (c_r+c_w)[E(Y_N)\sum \nolimits _{i=1}^{N}E(X_i)-E(X_{N+1})\sum \nolimits _{i=1}^{N-1}E(Y_i)]-c_f[E(Y_N)$ $+E(X_{N+1})]>0$.

We are now in a position to get the conclusion: $C(N+1)>C(N)\Longleftrightarrow G(N)>1$.

The remainder of the argument is analogous to that used above, the details of which we omit. In summary, this completes the proof of Lemma 1. $\square$

Proof of Lemma 2

The lemma will be proved by showing an equivalence conclusion: $G(N+1)-G(N)>0$. It is ready to invoke the formula G(N), hence

$$\begin{aligned}&G(N+1)-G(N)\\&\quad =\frac{c_r+c_w}{c_f}\cdot \left[ \frac{E(Y_{N+1})\sum \nolimits _{i=1}^{N+1}E(X_{i})- E(X_{N+2})\sum \nolimits _{i=1}^{N}E(Y_{i})}{E(Y_{N+1})+E(X_{N+2})}\right. \\&\qquad \left. - \frac{E(Y_{N})\sum \nolimits _{i=1}^{N}E(X_{i})- E(X_{N+1})\sum \nolimits _{i=1}^{N-1}E(Y_{i})}{E(Y_{N})+E(X_{N+1})}\right] \\&\quad =\frac{c_r+c_w}{c_f}\cdot \displaystyle \frac{{[E(X_{N+1})E(Y_{N+1}) -E(X_{N+2})E(Y_N)]\cdot }\left[ \sum \nolimits _{i=1}^{N+1}E(X_i) +\sum \nolimits _{i=1}^{N}E(Y_i)\right] }{[E(Y_{N+1})+E(X_{N+2})]\cdot [E(Y_{N})+E(X_{N+1})]}. \end{aligned}$$

Next, the proof will be divided in two aspects.

• 1. When discussing the geometric process, it follows that

  $$\begin{aligned} E(X_{N+1})>E(X_{N+2}), E(Y_{N+1})>E(Y_{N}). \end{aligned}$$

  Then the inequality $E(X_{N+1})E(Y_{N+1})-E(X_{N+2})E(Y_{N})>0$ is obvious. Hence, the conclusion can be drawn: $G_1(N+1)-G_1(N)>0$.

• 2. When discussing the renewal-geometric process, the following three cases should be studied, respectively.

  • (a) If N is less than $m-1$, according to the model assumptions, the following equalities are obtained:

    $$\begin{aligned} E(X_{N+1})=E(X_{N+2}), E(Y_{N+1})=E(Y_{N}). \end{aligned}$$

    Hence, the equality $E(X_{N+1})E(Y_{N+1})-E(X_{N+2})E(Y_{N})=0$ is established. That is, the conclusion is certified: $G_2(N+1)-G_2(N)=0$.

  • (b) If N is equal to $m-1$, the following formulae are derived:

    $$\begin{aligned} E(X_{m})>E(X_{m+1}), E(Y_{m})=E(Y_{m-1}). \end{aligned}$$

    Hence, the inequality $E(X_{m})E(Y_{m})-E(X_{m+1})E(Y_{m-1})>0$ is established. Therefore, $G_2(m)-G_2(m-1)>0$ is obvious.

  • (c) If N is greater than $m-1$, the following inequalities can be gained:

    $$\begin{aligned} E(X_{N+1})>E(X_{N+2}), E(Y_{N+1})>E(Y_{N}). \end{aligned}$$

    The inequality $E(X_{N+1})E(Y_{N+1})-E(X_{N+2})E(Y_{N})>0$ is established. Then, the conclusion can be certified: $G_2(N+1)-G_2(N)>0$.

Hence, the proof of Lemma 2 is completed by combining 1 and 2. $\square$

Proof of Lemma 3

As the sufficient conditions for existence of optimal strategies in the two model, we need only verify the third conclusion of Lemma 2. The proof will be omitted.

Proof of Theorem 1

The procedure of proof should be divided in two aspects by m.

If m is equal to 1, the renewal-geometric process maintenance model will reduce into a corresponding geometric process maintenance model, then the inequality $N^{**} \le N^{*}+m-1$ is obvious.

If m is greater than 1, recalling the renewal-reward theorem, the average cost rate for geometric process maintenance model can be obtained easily:

$$\begin{aligned} C_1(N)= & {} \frac{c_r \sum \nolimits _{i=1}^{N-1}\frac{\mu }{b^{i-1}}+c_f-c_w\sum \nolimits _{i=1}^{N} \displaystyle \frac{\lambda }{a^{i-1}}}{\sum \nolimits _{i=1}^{N}\frac{\lambda }{a^{i-1}}+\sum \nolimits _{i=1}^{N-1}\displaystyle \frac{\mu }{b^{i-1}}}\\= & {} \frac{c_r\displaystyle \frac{b^{N-1}-1}{b^{N-1}-b^{N-2}}\mu +c_f-c_w\displaystyle \frac{a^{N}-1}{a^{N}-a^{N-1}}\lambda }{\displaystyle \frac{a^{N}-1}{a^{N}-a^{N-1}}\lambda +\displaystyle \frac{b^{N-1}-1}{b^{N-1}-b^{N-2}}\mu }. \end{aligned}$$

Similarly, the average cost rate for renewal-geometric process maintenance model is also expressed:

$$\begin{aligned} C_2(N)= & {} \displaystyle \frac{c_r\sum \nolimits _{i=1}^{m-1}E(Y_i)-c_w\sum \nolimits _{i=1}^{m-1}E(X_i)}{\sum \nolimits _{i=1}^{N-1}E(Y_i)+\sum \limits _{i=1}^NE(X_i)} +\displaystyle \frac{c_r\sum \nolimits _{i=m}^{N-1}E(Y_i)+c_f-c_w\sum \nolimits _{i=m}^{N}E(X_i)}{\sum \nolimits _{i=1}^{N-1}E(Y_i)+\sum \nolimits _{i=1}^NE(X_i)} \nonumber \\= & {} [1-B(N)]\displaystyle \frac{c_r\sum \nolimits _{i=1}^{m-1}E(Y_i)-c_w\sum \nolimits _{i=1}^{m-1}E(X_i)}{\sum \nolimits _{i=1}^{m-1}E(Y_i)+\sum \nolimits _{i=1}^{m-1}E(X_i)}\\&+B(N)\displaystyle \frac{c_r\sum \nolimits _{i=m}^{N-1}E(Y_i)+c_f-c_w\sum \nolimits _{i=m}^{N}E(X_i)}{\sum \nolimits _{i=1}^{N-1}E(Y_i)+\sum \nolimits _{i=1}^NE(X_i)} \nonumber \\= & {} \displaystyle \frac{c_r\mu -c_w\lambda }{\lambda +\mu } +B(N)\left[ C_1(N-m+1)-\displaystyle \frac{c_r\mu -c_w\lambda }{\lambda +\mu }\right] , \end{aligned}$$

where

$$\begin{aligned} B(N)=\displaystyle \frac{\sum \nolimits _{i=m}^{N-1}E(Y_i)+\sum \nolimits _{i=m}^{N}E(X_i)}{\sum \nolimits _{i=1}^{N-1}E(Y_i)+\sum \nolimits _{i=1}^NE(X_i)}. \end{aligned}$$

Based on the conditions of $\displaystyle \frac{a-1}{a\lambda }<\displaystyle \frac{c_r+c_w}{c_f}<\displaystyle \frac{\lambda +a\mu }{[(a-1)m+1]\lambda \mu }$, the average cost rate functions $C_1(N)$ and $C_2(N)$ both convex with respect to N.

(1) When $C_1(N-m+1)<C_1(N-m+2)$, the following conclusion will be got through a simple transformation of the inequality:

$$\begin{aligned} 0<\displaystyle \frac{C_1(N-m+1)-\displaystyle \frac{c_r\mu -c_w\lambda }{\lambda +\mu }}{C_1(N-m+2)-\displaystyle \frac{c_r\mu -c_w\lambda }{\lambda +\mu }} <1<\displaystyle \frac{B(N+1)}{B(N)}. \end{aligned}$$

Furthermore, it follows that

$$\begin{aligned} B(N+1)\left[ C_1(N-m+2)-\displaystyle \frac{c_r\mu -c_w\lambda }{\lambda +\mu }\right] -B(N) \left[ C_1(N-m+1)-\displaystyle \frac{c_r\mu -c_w\lambda }{\lambda +\mu }\right] >0. \end{aligned}$$

In addition, a formula is expressed as follow:

$$\begin{aligned} C_2(N+1)-C_2(N)= & {} B(N+1)\left[ C_1(N-m+2)-\displaystyle \frac{c_r\mu -c_w\lambda }{\lambda +\mu }\right] \nonumber \\&-B(N)\left[ C_1(N-m+1)-\displaystyle \frac{c_r\mu -c_w\lambda }{\lambda +\mu }\right] \end{aligned}$$

Hence, the result of $C_2(N+1)>C_2(N)$ is easy to get. It means that $C_2$ has been increasing when the corresponding $C_1$ starts to increase.

(2) If $C_2(N+1)-C_2(N)<0$, we can easily get the relationship of $C_1(N-m+1)>C_1(N-m+2)$. It recalls us that $C_1$ has been decreasing when the corresponding $C_2$ starts to decrease.

Combining (1) and (2), the optimal values of the two models suffice the relationship: $N^{**} \le N^{*}+m-1$.

The proof of inequality $C_2(N)<C_1(N)$ will not be included here. $\square$