Keywords

1 Introduction

Preventive replacement policies have been studied extensively in literatures [1,2,3,4,5,6,7]. Barlow and Proschan [1] have firstly given an age replacement model for a finite operating time span, where the unit operates from installation to a fixed interval caused by external factors, and it is replaced at the end of the interval even if no failure has occurred. When the finite time span becomes a random interval with working cycles, during which, it is impossible to perform maintenance policies, optimal replacement policies with random works have been discussed [3, 6, 8, 9].

When the working cycles are taken into account for planning replacement polices, Zhao and Nakagawa [10] proposed the policies of replacement first and replacement last, that would become alternatives in points of cost rate, reliability and maintainability. Replacement first means the unit is replaced preventively at events such as operating time, number of repairs, or mission numbers, etc, whichever takes place first, while replacement last means the unit is replaced preventively at the above events, whichever takes place last. It has been shown that replacement last could let the unit operate working cycles as longer as possible while replacement first are more easier to save total maintenance cost [10]. More recent models of replacement first and replacement last can be found in [11,12,13,14,15].

In this chapter, the above working cycle is reconsidered as mission interval, and we suppose that the arrival time of a mission is a random variable rather than it begins from installation and lasts for an interval, during which, the unit should provide reliability and no maintenance can be done. The typical example is maintaining a hot spare for a key unit in a working system, in which, the spare unit should be active at the time when the key unit fails and provide system reliability for an interval when the key unit is unavailable. From this viewpoint, this chapter discusses preventive replacement policies for random arrival of missions. For this, an average failure rate is firstly given based on the conditional failure probability and the mean time to failure, given that the unit is still survival at time t. We next formulate and optimize the models of age replacement policies and the periodic policies with minimal repairs in analytical ways. Numerical examples are given when the mission arrival time follows a gamma distribution and the failure time of the unit has a Weibull distribution.

2 Average Failure Rate

It is assumed that a unit has a general failure distribution \(F(t)\equiv \Pr \{X\le t\}\) with a density function \(f(t)\equiv \mathrm {d}F(t)/ \mathrm {d}t\) and a finite mean \(\mu \equiv \int _0^\infty \overline{F}(t)\mathrm {d}t\). The conditional failure probability is given by [2]:

$$\begin{aligned} \lambda (t;x)\equiv \frac{F(t+x)-F(t)}{\overline{F}(t)} \quad (0<x<\infty ), \end{aligned}$$
(1)

which represents the probability that the unit fails in interval \([t,t+x]\), given that it is still survival at time t. Note that \(0\le \lambda (t;x) \le 1\). When \( x\rightarrow 0 \), \(\lambda (t;x) /x\) becomes an instant failure rate:

$$\begin{aligned} h(t)\equiv \frac{f(t)}{\overline{F}(t)}=-\frac{1}{\overline{F}(t)}\frac{\mathrm {d}\overline{F}(t)}{\mathrm {d}t}. \end{aligned}$$
(2)

We usually suppose, in modeling maintenance policies, that h(t) increases with t from \(h(0)=0\) to \(h(\infty )\equiv \lim _{t \rightarrow \infty }h(t)\) that might be infinity, i.e., \(\lambda (t;x)\) increases with t from F(x) to 1.

We next define:

$$\begin{aligned} F(t;x)=\frac{\int _t^{t+x} \overline{F}(u) \mathrm {d}u}{ \overline{F}(t)}, \end{aligned}$$
(3)

which means the mean time to failure, given that the unit is still survival at time t. Obviously, when \(t\rightarrow 0\), F(t; x) becomes \(\int _0^x\overline{F}(u) \mathrm {d}u\), that represents the mean time to replacement when the unit is replaced preventively at time x or correctively at failure, whichever takes place first. When \(t\rightarrow \infty \),

$$\begin{aligned} \lim _{t\rightarrow \infty }\frac{\int _t^{t+x} \overline{F}(u) \mathrm {d}u}{ \overline{F}(t)}= \lim _{t\rightarrow \infty } \frac{F(t+x)-F(t)}{f(t)}= \lim _{t\rightarrow \infty }\frac{\lambda (t;x)}{h(t)}=\frac{1}{h(\infty )}.\nonumber \end{aligned}$$

Differentiating \(\int _t^{t+x} \overline{F}(u) \mathrm {d}u/ \overline{F}(t)\) with t, and noting that

$$\begin{aligned}&h(t) \int _t^{t+x} \overline{F}(t) \mathrm {d}t - [F(t+x)-F(t)] \nonumber \\&\le h(t) \int _t^{t+x} \left[ \frac{f(u)}{h(t)}\right] \mathrm {d}u - [F(t+x)-F(t)] =0.\nonumber \end{aligned}$$

which shows that F(t; x) decreases with t from \(\int _0^x \overline{F}(u)\mathrm {d}u\) to \(1/h(\infty )\).

Using \(\lambda (t;x)\) and F(t; x), we define:

$$\begin{aligned} \Lambda (t;x)\equiv \frac{F(t+x)-F(t)}{\int _t^{t+x} \overline{F}(u)\mathrm {d}u}\quad (0<x<\infty ), \end{aligned}$$
(4)

which means the average failure rate, given that the unit is still survival at time t. It can be easily proved that \(\Lambda (t;x)\) increases with t from \(F(x)/\int _0^x \overline{F}(u)\mathrm {d}u\) to \(h(\infty )\), and \(h(t) \le \Lambda (t;x) \le h(t+x)\).

3 Age Replacement

In this section, we apply the above average failure rate into age replacement policies with random arrival of missions. That is, the unit begins to operate after installation, and its failure time \(X~(0<X<\infty )\) has a general distribution \(F(t)\equiv \Pr \{X\le t\}\) with finite mean \(\mu \equiv \int _0^\infty \overline{F}(t) \mathrm {d} t\). In addition, the unit should be active at time \(T_o~(0<T_o<\infty )\) for an interval \([T_o,T_o+t_x]~(0\le t_x<\infty )\) to provide reliability. In this case, \(t_x\) can be considered as a mission interval during which the unit provides reliability in [2].

3.1 Constant \(T_o\)

We plan the unit is replaced preventively at time \(T_o+t_x\) \((0\le t_x\le \infty )\) when it is still survival at time \(T_o~(0\le T_o<\infty )\), or it is replaced correctively at failure time X during \((0,T_o+t_x]\), whichever takes place first.

The probability that the unit is replaced at \(T_o+t_x\) is

$$\begin{aligned} \Pr \{X> T_o+t_x\}= \overline{F}(T_o+t_x), \end{aligned}$$
(5)

and the probability that it is replaced at failure is

$$\begin{aligned} \Pr \{X\le T_o+t_x\}= F(T_o+t_x). \end{aligned}$$
(6)

The mean time from installation to replacement is

$$\begin{aligned} (T_o+t_x) \overline{F}(T_o+t_x) + \int _0^{T_o+t_x} t \mathrm {d} F(t) = \int _0^{T_o+t_x} \overline{F}(t) \mathrm {d}t \end{aligned}$$
(7)

Thus, the expected replacement cost rate is

$$\begin{aligned} C_s(t_x;T_o)= \frac{c_p+(c_f-c_p) F(T_o+t_x) }{\int _0^ {T_o+t_x} \overline{F}(t) \mathrm {d}t}, \end{aligned}$$
(8)

where \(c_f \) and \(c_p~(c_p<c_f)\) are the costs of replacement policies done at failure and at \(T_o+t_x\), respectively.

We find optimum \(t_x^*\) to minimize \(C_s(t_x;T_o)\) in (8). Differentiating \(C_s(t_x;T_o)\) with respect to \(t_x\) and setting it equal to zero,

$$\begin{aligned} h(T_o+t_x) \int _0^{T_o+t_x} \overline{F}(t) \mathrm {d}t - F(T_o+t_x) =\frac{c_p}{c_f-c_p}, \end{aligned}$$
(9)

whose left-hand side increases with \(t_x\) from

$$\begin{aligned} h(T_o)\int _0^{T_o} \overline{F}(t) \mathrm {d}t - F(T_o)\nonumber \end{aligned}$$

to \(h(\infty )/\mu -1\). Thus, if h(t) increases strictly with t to \(h(\infty )=\infty \), then there exists a finite and unique \(t_x^*~(0\le t_x^*<\infty )\) which satisfies (9), and the resulting cost rate is

$$\begin{aligned} C_s(t_x^*;T_o)= (c_f-c_p)h(T_o+t_x^*). \end{aligned}$$
(10)

Noting that the left-hand side of (9) increases with \(T_o\), \(t_x^*\) decreases with \(T_o\) from \(T^*\) to 0, where \(T^*\) is an optimum age replacement time that satisfies

$$\begin{aligned} h(T) \int _0^T \overline{F}(t) \mathrm {d}t -F(T)= \frac{c_p}{c_f-c_p}. \end{aligned}$$
(11)

3.2 Random \(T_o\)

When \(T_o\) is a random variable and has a general distribution \(Y(t)\equiv \Pr \{T_o\le t\}\) with a density function \(y(t)\equiv \mathrm {d} Y(t) / \mathrm {d} t\) and a finite mean \(\gamma =\int _0^\infty \overline{Y}(t) \mathrm {d} t \), we plan that the unit is replaced preventively at time \(T_o+t_x\) \((0\le t_x\le \infty )\) when it is still survival at a random time \(T_o~(0\le T_o<\infty )\), or it is replaced correctively at failure time X during \((0,T_o+t_x]\), whichever takes place first.

The probability that the unit is replaced at \(T_o+t_x\) is

$$\begin{aligned} \Pr \{X> T_o+t_x\}= \int _0^\infty \overline{F}(t+t_x) \mathrm {d}Y(t), \end{aligned}$$
(12)

and the probability that it is replaced at failure is

$$\begin{aligned} \Pr \{X\le T_o+t_x\}= \int _0^\infty F(t+t_x) \mathrm {d}Y(t). \end{aligned}$$
(13)

The mean time from installation to replacement is

$$\begin{aligned}&\int _0^ \infty (t+t_x) \overline{F}(t+t_x) \mathrm {d}Y(t)+ \int _0^\infty \left[ \int _0^{t+t_x} u\mathrm {d}F(u) \right] \mathrm {d}Y(t)\nonumber \\&=\int _0^\infty \left[ \int _0^{t+t_x} \overline{F}(u) \mathrm {d}u \right] \mathrm {d}Y(t). \end{aligned}$$
(14)

Thus, the expected replacement cost rate is

$$\begin{aligned} C_s(t_x;Y)= \frac{c_p+(c_f-c_p)\int _0^\infty F(t+t_x) \mathrm {d}Y(t) }{\int _0^\infty [ \int _0^{t+t_x} \overline{F}(u) \mathrm {d}u ]\mathrm {d}Y(t) }, \end{aligned}$$
(15)

where \(c_f \) and \(c_p~(c_p<c_f)\) are the costs of replacement policies done at failure and at \(T_o+t_x\), respectively.

Clearly,

$$\begin{aligned} \lim _{t_x\rightarrow \infty }C_s(t_x;Y)=\frac{c_f}{\mu },\nonumber \end{aligned}$$
$$\begin{aligned} \lim _{t_x\rightarrow 0}C_s(t_x;Y)=\frac{c_p+(c_f-c_p) \int _0^\infty F(t) \mathrm {d} Y(t)}{\int _0^\infty \overline{F}(t) \overline{Y}(t) \mathrm {d}t},\nonumber \end{aligned}$$

which agrees with random replacement model in [3].

We find optimum \(t_x^*\) to minimize \(C_s(t_x;Y)\) in (15). Differentiating \(C_s(t_x;Y)\) with respect to \(t_x\) and setting it equal to zero,

$$\begin{aligned} h_s(t_x) \int _0^\infty \left[ \int _0^{t+t_x} \overline{F}(u) \mathrm {d}u \right] \mathrm {d}Y(t)- \int _0^\infty F(t+t_x) \mathrm {d}Y(t)=\frac{c_p}{c_f-c_p}, \end{aligned}$$
(16)

where

$$\begin{aligned} h_s(t_x)\equiv \frac{\int _0^\infty f(t+t_x) \mathrm {d}Y(t)}{\int _0^\infty \overline{F}(t+t_x) \mathrm {d}Y(t)}. \nonumber \end{aligned}$$

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

$$\begin{aligned} h_s(t_x)\equiv \lim _{T\rightarrow \infty }h_f(T;t_x) \equiv \lim _{T\rightarrow \infty } \frac{\int _0^T f(t+t_x) \mathrm {d}Y(t)}{ \int _0^T \overline{F}(t+t_x) \mathrm {d}Y(t)}, \nonumber \end{aligned}$$

and it increases with \(t_x\) from \(h_s(0)= \int _0^\infty f(t)\mathrm {e}^{-\theta t} \mathrm {d}t/\int _0^\infty \overline{F}(t)\mathrm {e}^{-\theta t}\mathrm {d}t\) to \(h(\infty )\). Then, the left-hand side of (16) increases with \(t_x\) to \(\infty \) as \(h(\infty )\rightarrow \infty \). In this case, there exists a finite and unique \(t_x^*~(0\le t_x^*<\infty )\) which satisfies (16), and the resulting cost rate is

$$\begin{aligned} C_s(t_x^*;Y)=(c_f-c_p)h_s(t_x^*). \end{aligned}$$
(17)

When \(T_o\) has a gamma distribution with a density function \(y(t)=\theta ^kt^{k-1} \mathrm {e}^{-\theta t}/(k-1)!\) \((k=1,2,\ldots )\), and the failure time X has a Weibull distribution \(F(t)= 1- \mathrm {e}^{-(\alpha t)^\beta }\) \((\alpha>0, \beta >1)\), Table 1 presents optimum \(t_x^*\) and its cost rate \(C_s(t_x^*;Y)\) for k and \(c_p\) when \(\theta =1.0\), \(\alpha =0.1\), \(\beta =2.0\), and \(c_f=100.0\). Table 1 shows that optimum interval \([T_o, T_o+t_x^*]\) decreases with k and increases with \(c_p\). This means that if k becomes large, then the failure rate increases with \(T_o\) and \(t_x^*\) becomes small. On the other hand, if \(c_p~({<}c_f)\) becomes large, then it is unnecessary to replace the unit at a early time and \(t_o^*\) becomes large.

Table 1 Optimum \(t_x^*\) and its cost rate \(C_s(t_x^*;Y)\) when \(\theta =1.0\), \(\alpha =0.1\), \(\beta =2.0\), and \(c_f=100.0\)
Full size table

3.3 Replace at T and \(T_o+t_x\)

In order to prevent early or late arrivals of time \(T_o\), we plan that the unit is replaced preventively at time \(T~(0< T \le \infty )\) or at time \(T_o+t_x~(0\le t_x\le \infty )\), whichever takes place first. However, no replacement can be done preventively during the interval \([T_o,T_o+t_x]\). In this policy, \(t_x\) is constantly given and \(T_o\) is a random variable with a general distribution Y(t).

The probability that the unit is replaced at T is

$$\begin{aligned} \Pr \{X>T, T_o>T\}=\overline{F}(T)\overline{Y}(T), \end{aligned}$$
(18)

the probability that it is replaced at \(T_o+t_x\) is

$$\begin{aligned} \Pr \{X>T_o+t_x, T_o\le T\}= \int _0^T \overline{F}(t+t_x)\mathrm {d}Y(t), \end{aligned}$$
(19)

and the probability that it is replaced at failure is

$$\begin{aligned} \Pr \{X\le T~\mathrm {and}~T_o\ge T, X\le T_o+t_x~\mathrm {and}~T_o<T \}= F(T)\overline{Y}(T)+ \int _0^T F(t+t_x) \mathrm {d}Y(t), \end{aligned}$$
(20)

where note that (18) + (19) + (20) = 1.

The mean time from installation to replacement is

$$\begin{aligned}&T\overline{F}(T)\overline{Y}(T) + \int _0^T (t+t_x) \overline{F}(t+t_x)\mathrm {d}Y(t)+ \overline{Y}(T)\int _0^T t \mathrm {d}F(t) \nonumber \\ {}&+ \int _0^T \left[ \int _0^{t+t_x} u\mathrm {d}F(u) \right] \mathrm {d}Y(t) \nonumber \\&= \overline{Y}(T)\int _0^T \overline{F}(t)\mathrm {d}t + \int _0^T \left[ \int _0^{t+t_x} \overline{F}(u) \mathrm {d}u \right] \mathrm {d}Y(t). \end{aligned}$$
(21)

Thus, the expected replacement cost rate is

$$\begin{aligned} C_f(T;t_x)=\frac{c_p+(c_f-c_p) [F(T)\overline{Y}(T)+ \int _0^T F(t+t_x) \mathrm {d}Y(t)]}{\overline{Y}(T)\int _0^T \overline{F}(t)\mathrm {d}t + \int _0^T [\int _0^{t+t_x} \overline{F}(u) \mathrm {d}u ]\mathrm {d}Y(t)}, \end{aligned}$$
(22)

Note that when \(t_x\rightarrow \infty \), \(\lim _{t_x\rightarrow \infty }C_f(T;t_x)\) becomes age replacement model in [2], when \(t_x\rightarrow 0\), \(\lim _{t_x\rightarrow 0}C_f(T;t_x)\) becomes random replacement model in [3], when \(T\rightarrow \infty \), \(\lim _{T\rightarrow \infty }C_f(T;t_x)= C_s(t_x;Y)\) in (15), and when \(T\rightarrow 0\), \(\lim _{T\rightarrow 0}C_f(T;t_x)=\infty \).

We find optimum \(T_f^*\) and \(t_{xf}^*\) to minimize \(C_f(T;t_x)\) in (22) for given \(t_x\). Differentiating \(C_f(T;t_x)\) with respect to T and setting it equal to zero,

$$\begin{aligned}&q_f(T;t_x)\left\{ \overline{Y}(T)\int _0^T \overline{F}(t)\mathrm {d}t + \int _0^T \left[ \int _0^{t+t_x} \overline{F}(u) \mathrm {d}u \right] \mathrm {d}Y(t)\right\} \nonumber \\&- \left[ F(T)\overline{Y}(T)+ \int _0^T F(t+t_x) \mathrm {d}Y(t) \right] =\frac{c_p}{c_f-c_p}, \end{aligned}$$
(23)

where

$$\begin{aligned} q_f(T;t_x)\equiv \frac{~~r(T)\lambda (T;t_x) + h(T)~~}{r(T)\frac{\lambda (T;t_x)}{\Lambda (T;t_x)} +1}\quad \mathrm {and} \quad r(T)\equiv \frac{y(T)}{\overline{Y}(T)},\nonumber \end{aligned}$$

and the instant failure rate h(T), the conditional failure probability \(\lambda (T;t_x)\) and the average failure rate \(\Lambda (T;t_x)\) are included in \(q_f(T;t_x)\).

When \(Y(t)=1-\mathrm {e}^{-\theta t}\), \(r(T)=\theta \) and

$$\begin{aligned} q_f(T;t_x)= \frac{\theta [F(T+t_x)-F(T)]+f(T)}{ \theta \int _T^{T+t_x} \overline{F}(t) \mathrm {d} t + \overline{F}(T)}.\nonumber \end{aligned}$$

Note that

$$\begin{aligned} h(T)<\frac{F(T+t_x)-F(T)}{\int _T^{T+t_x} \overline{F}(t) \mathrm {d}t}<h(T+t_x),\nonumber \end{aligned}$$

then \(q_f(T;t_x)\) increases strictly with T to \(\infty \) as \(h(\infty )\rightarrow \infty \), and also increases strictly with \(t_x\) to \(q_f(T;\infty )\). Thus, the left-hand side of (23) increases with T from 0 to \(\infty \) as \(h(\infty )\rightarrow \infty \). In this case, there exists a finite and unique \(T_f^*~(0<T_f^*<\infty )\) which satisfies (23), and the resulting cost rate is

$$\begin{aligned} C_f(T_f^*;t_x)= (c_f-c_p)q_f(T_f^*;t_x). \end{aligned}$$
(24)

In addition, the left-hand side of (23) increases with \(t_x\), then \(T_f^*\) decreases with \(t_x\) from \(T^*\) which satisfies the following random replacement model [3],

$$\begin{aligned} h(T)\int _0^T \mathrm {e}^{-\theta t} \overline{F}(t)\mathrm {d}t -\int _0^T \mathrm {e}^{-\theta t} \mathrm {d} F(t)=\frac{c_p}{c_f-c_p}.\nonumber \end{aligned}$$

Nest, we find optimum \(t_{xf}^*\) for given T. Differentiating \(C_f(T;t_x)\) with respect to \(t_x\) for given T and setting it equal to zero,

$$\begin{aligned}&h_f(T;t_x)\left\{ \overline{Y}(T)\int _0^T \overline{F}(t)\mathrm {d}t + \int _0^T \left[ \int _0^{t+t_x} \overline{F}(u) \mathrm {d}u \right] \mathrm {d}Y(t)\right\} \nonumber \\&- \left[ F(T)\overline{Y}(T)+ \int _0^T F(t+t_x) \mathrm {d}Y(t) \right] = \frac{c_p}{c_f-c_p}, \end{aligned}$$
(25)

where

$$\begin{aligned} h_f(T;t_x) \equiv \frac{\int _0^T f(t+t_x) \mathrm {d}Y(t)}{\int _0^T \overline{F}(t+t_x) \mathrm {d}Y(t)}<h(T+t_x).\nonumber \end{aligned}$$

When \(Y(t)=1-\mathrm {e}^{-\theta t}\), \(h_f(T;t_x)\) increases with \(t_x\) to \(h(\infty )\). Then, the left-hand side of (25) increases strictly with \(t_x\) from 0 to \(\infty \) as \(h(\infty )\rightarrow \infty \). In this case, there exists a finite and unique \(t_{xf}^*~(0< t_{xf}^*<\infty )\) which satisfies (25), and the resulting cost rate is

$$\begin{aligned} C_f(T;t_{xf}^*)=(c_f-c_p)h_f(T;t_{xf}^*). \end{aligned}$$
(26)

Note that \(t_{xf}^*\) decreases with T to \(t_x^*\) given in (16), as the left-hand side of (22) increases with T to that of (16).

When \(y(t)=\theta ^kt^{k-1} \mathrm {e}^{-\theta t}/(k-1)!\) \((k=1,2,\ldots )\) and \(F(t)= 1- \mathrm {e}^{-(\alpha t)^\beta }\), \((\alpha >0, \beta \ge 1)\), Table 2 presents optimum \(T_f^*\) and its cost rate \(C_f(T_f^*;t_x)\) for \(t_x\) and \(c_p\) when \(\theta =1.0\), \(k=2\), \(\alpha =0.1\), \(\beta =2.0\), and \(c_f=100.0\), and Table 3 presents optimum \(t_{xf}^*\) and its cost rate \(C_f(T;t_{xf}^*)\) for T and \(c_p\) when \(\theta =1.0\), \(k=2\), \(\alpha =0.1\), \(\beta =2.0\), and \(c_f=100.0\). Table 3 shows that \(T_f^*\) increases with \(c_p\) and decreases with \(t_x\) and \(t_{xf}^*\) increases with \(c_p\) and decreases with T, as shown in the above analytical discussions.

Table 2 Optimum \(T_f^*\) and its cost rate \(C_f(T_f^*;t_x)\) when \(\theta =1.0\), \(k=2\), \(\alpha =0.1\), \(\beta =2.0\), and \(c_f=100.0\)
Full size table
Table 3 Optimum \(t_{xf}^*\) and its cost rate \(C_f(T;t_{xf}^*)\) when \(\theta =1.0\), \(k=2\), \(\alpha =0.1\), \(\beta =2.0\), and \(c_f=100.0\)
Full size table

4 Minimal Repair

It is assumed that the unit undergoes minimal repairs at failures and begins to operate again after repairs, where the time for repairs are negligible and the failure rate remains undisturbed by repairs. In this case, we define

$$\begin{aligned} \Lambda (t;x)=\frac{1}{x}\int _t^{t+x} h(u)\mathrm {d}u, \end{aligned}$$
(27)

which means the average failure rate for an interval \([t,t+x]\). It is obviously to show that \(\Lambda (t;x)\) increases with t from H(x)/x to \(h(\infty )\) and increases with x from h(0) to \(h(\infty )\), and \(h(t)\le \Lambda (t;x) \le h(t+x)\).

4.1 Constant \(T_o\)

In order to prevent an increasing repair cost, we plan that the unit is replaced at time \(T_o+t_x~(0<T_o\le \infty , 0\le t_x<\infty )\). Noting that the expected number of failures during \((0, T_o+t_x]\) is \(H(T_o+t_x)\), the expected cost rate is

$$\begin{aligned} C_s(t_x;T_o)= \frac{c_m H(T_o+t_x) +c_p}{T_o+ t_x}, \end{aligned}$$
(28)

where \(c_m\) is minimal repair cost at failure, and \(c_p\) is given in (8).

We find optimum \(t_x^*\) to minimize \(C_s(t_x;T_o)\) for given \(T_o\). Differentiating \(C_s(t_x;T_o)\) with respect to \(t_x\) and setting it equal to zero,

$$\begin{aligned} h(T_o+t_x) (T_o+t_x)- H(T_o+t_x)= \frac{c_p}{c_m}, \end{aligned}$$
(29)

whose left-hand side increases with \(t_x\) from \(h(T_o)T_o- H(T_o)\) to \(\int _0^\infty [h(\infty )- h(t)] \mathrm {d}t\). Thus, if the failure rate h(t) increases strictly with t to \(h(\infty )=\infty \), then there exists a finite and unique \(t_x^*~(0\le t_x^*<\infty )\) which satisfies (29), and the resulting cost rate is

$$\begin{aligned} C_s(t_x^*;T_o)=c_m h(T_o+t_x^*), \end{aligned}$$
(30)

Noting that the left-hand side of (29) increases with \(T_o\), \(t_x^*\) decreases with \(T_o\) from \(T^*\) to 0, where \(T^*\) is an optimum periodic replacement time that satisfies

$$\begin{aligned} h(T)T- H(T)=\frac{c_p}{c_m}. \end{aligned}$$
(31)

4.2 Random \(T_o\)

We plan that the unit is replaced at time \(T_o+t_x~(0\le t_x<\infty )\), where \(T_o\) is a random variable with distribution Y(t). Then, the expected cost rate is

$$\begin{aligned} C_s(t_x;Y)=\frac{c_m\int _0^\infty H(t+ t_x) \mathrm {d} Y(t)+c_p}{\int _0^\infty (t+ t_x) \mathrm {d}Y(t)}, \end{aligned}$$
(32)

where \(c_m\) is minimal repair cost at failure, and \(c_p\) is given in (15).

Clearly, \(\lim _{t_x\rightarrow \infty } C_s(t_x;Y)\rightarrow \infty \) and

$$\begin{aligned} \lim _{t_x\rightarrow 0}C_s(t_x;Y)=\frac{c_m\int _0^\infty H(t) \mathrm {d} Y(t)+c_p}{\int _0^\infty t \mathrm {d}Y(t)},\nonumber \end{aligned}$$

which agrees with random replacement model [3].

If there exists an optimum \(t_x^*\) to minimize \(C_s(t_x;Y)\) in (32), it satisfies

$$\begin{aligned} \int _0^\infty (t+t_x) \mathrm {d}Y(t)\int _0^\infty h(t+t_x)\mathrm {d} Y(t) - \int _0^\infty H(t+ t_x) \mathrm {d} Y(t) =\frac{c_p}{c_m}, \end{aligned}$$
(33)

whose left-hand side increases with \(t_x\) to \(\infty \) as \(h(\infty )\rightarrow \infty \). In this case, the resulting cost rate is

$$\begin{aligned} C_s(t_x^*;Y)= c_m\int _0^\infty h(t+t_x^*) \mathrm {d}Y(t). \end{aligned}$$
(34)

When \(y(t)=\theta ^kt^{k-1} \mathrm {e}^{-\theta t}/ \Gamma (k)\) and \(F(t)= 1- \mathrm {e}^{-(\alpha t)^\beta }\), Table 4 presents optimum \(t_{x}^*\) and its cost rate \(C_s(t_{x}^*;Y)\) for k and \(c_m\) when \(\theta =1.0\), \(\alpha =1.0\), \(\beta =2.0\), and \(c_p=100.0\). Table 4 shows that optimum interval \([T_o,T_o+t_x^*]\) decreases when \(c_m\) increases and \(T_o\) arrives at a late time due to the total increasing repair cost. Note that when \(k=5\), \(t_x^*\rightarrow 0\) for all of \(c_m\).

Table 4 Optimum \(t_x^*\) and its cost rate \(C_s(t_x^*;Y)\) when \(\theta =1.0\), \(\alpha =1.0\), \(\beta =2.0\), and \(c_p=100.0\)
Full size table

4.3 Replace at T and \(T_o+t_x\)

We plan that the unit is replaced at time \(T~(0< T \le \infty )\) or at time \(T_o+t_x~(0\le t_x\le \infty )\), whichever takes place first; however, only minimal repairs can be done during the interval \([T_o,T_o+t_x]\). Then, the expected number of repairs between replacement policies is

$$\begin{aligned} H(T)\overline{Y}(T)+ \int _0^T H(t+t_x) \mathrm {d}Y(t), \end{aligned}$$
(35)

and the mean time from installation to replacement is

$$\begin{aligned} T\overline{Y}(T)+ \int _0^T (t+t_x) \mathrm {d}Y(t) = t_xY(T)+ \int _0^T \overline{Y}(t) \mathrm {d}t. \end{aligned}$$
(36)

Thus, the expected replacement cost rate is

$$\begin{aligned} C_f(T; t_x) =\frac{c_m[H(T)\overline{Y}(T)+ \int _0^T H(t+t_x) \mathrm {d}Y(t)]+ c_p}{ t_xY(T)+ \int _0^T \overline{Y}(t) \mathrm {d}t}. \end{aligned}$$
(37)

Differentiating \(C_f(T; t_x)\) with respect to T and setting it equal to zero,

$$\begin{aligned} q_f(T;t_x)\left[ t_xY(T)+ \int _0^T \overline{Y}(t) \mathrm {d}t \right] -\left[ H(T)\overline{Y}(T)+ \int _0^T H(t+t_x) \mathrm {d}Y(t) \right] =\frac{c_p}{c_m}, \end{aligned}$$
(38)

where

$$\begin{aligned} q_f(T;t_x)\equiv \frac{r(T) \Lambda (T;t_x)+ h(T)/t_x }{r(T)+1/t_x}. \end{aligned}$$
(39)

When \(Y(t)=1-\mathrm {e}^{-\theta t}\), \(q_f(T;t_x)\) increases with T to \(h(\infty )/(\theta t_x+1)\). Then, the left-hand side of (38) increases with T from 0 to \(\infty \) as \(h(\infty )\rightarrow \infty \). Therefore, there exits a finite and unique \(T_f^*~(0<T_f^*<\infty )\) which satisfies (38), and the resulting cost rate is

$$\begin{aligned} C_f(T_f^*; t_x) = c_m \frac{ \theta \Lambda (T_f^*;t_x)+ h(T_f^*)/t_x}{\theta +1/t_x}. \end{aligned}$$
(40)

Next, differentiating \(C_f(T; t_x) \) with respect to \(t_x\) and setting it equal to zero,

$$\begin{aligned}&\frac{\int _0^T h(t+t_x) \mathrm {d} Y(t)}{Y(T)} \left[ t_xY(T)+ \int _0^T \overline{Y}(t) \mathrm {d}t \right] \nonumber \\&-\left[ H(T)\overline{Y}(T)+ \int _0^T H(t+t_x) \mathrm {d}Y(t)\right] = \frac{c_p}{c_m}, \end{aligned}$$
(41)

whose left-hand side increases with \(t_x\) to \(\infty \) as \(h(\infty )\rightarrow \infty \). Therefore, there exists a finite and unique \(t_{xf}^*~(0\le t_{xf}^*<\infty )\) which satisfies (40), and the resulting cost rate is

$$\begin{aligned} C_f(T; t_{xf}^*)=c_m \frac{\int _0^T h(t+t_{xf}^*) \mathrm {d} Y(t)}{Y(T)}. \end{aligned}$$
(42)
Table 5 Optimum \(T_f^*\) and its cost rate \(C_f(T_f^*;t_x)\) when \(\theta =1.0\), \(\alpha =1.0\), \(\beta =2.0\), \(t_x=1.0\), and \(c_p=100.0\)
Full size table
Table 6 Optimum \(t_{xf}^*\) and its cost rate \(C_f(T;t_{xf}^*)\) when \(\theta =1.0\), \(\alpha =1.0\), \(\beta =2.0\), \(T=1.0\) and \(c_p=100.0\)
Full size table

When \(y(t)=\theta ^kt^{k-1} \mathrm {e}^{-\theta t}/ \Gamma (k)\) and \(F(t)= 1- \mathrm {e}^{-(\alpha t)^\beta }\), Table 5 presents optimum \(T_f^*\) and its cost rate \(C_f(T_f^*;t_x)\) for \(t_x\) and \(c_m\) when \(\theta =1.0\), \(\alpha =1.0\), \(\beta =2.0\), \(t_x=1.0\), and \(c_p=100.0\), and Table 6 presents optimum \(t_{xf}^*\) and its cost rate \(C_f(T;t_{xf}^*)\) for k and \(c_m\) when \(\theta =1.0\), \(\alpha =1.0\), \(\beta =2.0\), \(T=1.0\), and \(c_p=100.0\).

5 Conclusions

We have firstly obtained a definition of average failure rate, i.e., \(\Lambda (t;x)\), that is based on the conditional failure probability and the mean time to failure given that the unit is still survival at time t. The mathematical monotonicity of \(\Lambda (t;x)\) has been proved analytically. Next, the average failure rate has been applied into preventive replacement policies when the arrival time of a mission is a random variable and lasts for an interval, during which, the unit provides reliability and no maintenance can be done. Optimum replacement time and mission interval have been discussed respectively for the models of age replacement and periodic replacement. Numerical examples have been illustrated when the mission arrival time follows a gamma distribution and the failure time of the unit has a Weibull distribution.