Maintenance Planning

For components and systems with increasing failure rate, planned replacements and inspections are often used to prevent failures in order to improve system availability and reduce overall maintenance cost. For example, the common used preventive maintenance strategy is to replace an aged item at a given time interval in order to minimize the long term cost per unit time. An optimal time interval exists if the failure rate is creasing and the cost of failure is much higher than the cost of a scheduled replacement. Another alternative is to minimize the short term cost per unit time, such as in one maintenance cycle. A replacement cycle is defined as the time when a failure occurs or at the scheduled replacement if there is no failure. Similarly, periodical inspection is applied to detect a potential failure and to prevent its happening. The objective of inspection also can be to minimize either the long time or the short time cost. In this section, we will discuss how to find the optimal maintenance and inspection intervals in order to minimize the long term and the short term maintenance cost.

Optimal Preventive Replacement Interval

Minimize the Long Term Cost Per Unit Time

This is the most popularly used objective for optimal preventive replacement (maintenance). The theory is discussed and one example is given at reliability modeling for repairable system. We will briefly explain it here. Since the item is either replaced at failure or at the scheduled maintenance interval [math]\displaystyle{ x }[/math], the expected cycle length is:

[math]\displaystyle{ E(\text{Cycle Length})=\int_{0}^{x}{tf(t)dt}+R(x)x=\int_{0}^{x}{R(t)dt} }[/math]

The expected cost for one cycle is

[math]\displaystyle{ E(\text{Cost Per Cycle})={{C}_{P}}R(x)+{{C}_{U}}\left( 1-R(x) \right) }[/math]

The expected long term cost per unit time

[math]\displaystyle{ CPU{{T}_{L}}(t)=\frac{E(\text{Cost Per Cycle})}{E\left( \text{Cycle Length} \right)}=\frac{{{C}_{P}}R(x)+{{C}_{U}}\left( 1-R(x) \right)}{\int_{0}^{x}{R(t)dt}} }[/math]

The optimal preventive maintenance interval will be the value t which minimizes [math]\displaystyle{ CPU{{T}_{L}} }[/math]. In the above equations:

• [math]\displaystyle{ f(x) }[/math] is the probability density function of the failure time
• [math]\displaystyle{ R(x) }[/math] is the reliability function of the failure time
• [math]\displaystyle{ {{C}_{P}} }[/math] is the average cost due to a planned maintenance
• [math]\displaystyle{ {{C}_{U}} }[/math] is the average cost due to a unplanned maintenance

Minimize the Short Term Cost Per Unit Time

The above equations are used for minimizing the long term cost and are based on the reward renewal theory. However, the average cost of planned and unplanned maintenance may change with time, and new versions of the component may be used. The failure time distribution and the cost function need to be updated frequently. Therefore, sometimes an optimal interval to minimize the short term cost is needed. The expected cost per unit time in one replacement cycle is defined as

[math]\displaystyle{ CPU{{T}_{S}}(t)=E\left( \frac{\text{Cost Per Cycle}}{\text{Cycle Length}} \right)=\int_{0}^{x}{\frac{{{C}_{U}}}{t}f(t)dt}+\frac{{{C}_{P}}}{x}\left( 1-F(x) \right) }[/math] For a Weibull distribution with its shape parameter [math]\displaystyle{ \beta \gt 1 }[/math], the optimal interval [math]\displaystyle{ {{x}_{0}} }[/math] is [math]\displaystyle{ {{x}_{o}}=\eta {{\left( \frac{{{C}_{P}}}{\beta \left( {{C}_{U}}-{{C}_{P}} \right)} \right)}^{\frac{1}{\beta }}} }[/math] where [math]\displaystyle{ \eta }[/math] is the scale parameter of the Weibull distribution.

Optimal Inspection Interval

Inspection is often used to test if a device is working properly. For example, for failure due to degradation such as corrosion, inspection can find an oncoming failure and replace the device. If a failure cannot be detected during the inspection, it will be noticed immediately when the failure occurs, and the device will be replaced at the failure. Assume a failure occurs at time [math]\displaystyle{ t }[/math]. If an inspection is conducted within the range of [math]\displaystyle{ [p\times t,t] }[/math], the failure will be detected and the component will be replaced. p is a value between 0 and 1. This is illustrated in the following plot.

XXX - picture

In the above plot, the inspection is conducted at every X hours. “No” means an oncoming failure will not be detected by inspection, while “Yes” means it can be detected by inspection since the inspection time is within [math]\displaystyle{ [p\times t,t] }[/math]. In this case, whenever a replacement occurs, a new maintenance cycle starts.

Minimize the Long Term Cost Per Unit Time

Define

XXX- picture

M is the largest inspection interval where there is no overlap between the detection zones of two adjunct inspections. For any failures occur after the Mth inspection, it can always be detected. The expected replacement cycle length is

[math]\displaystyle{ \begin{align} & E(\text{Cycle Length})=\int_{0}^{x}{{{C}_{U}}f(t)dt}+\sum\limits_{k=1}^{M}{\left[ \int\limits_{kx}^{\frac{kx}{p}}{\left[ {{C}_{P}}+k{{C}_{I}} \right]}f(t)dt+\int\limits_{\frac{kx}{p}}^{(k+1)x}{\left[ {{C}_{U}}+k{{C}_{I}} \right]}f(t)dt \right]} \\ & +\int\limits_{\left( M+1 \right)x}^{\frac{(M+1)x}{p}}{\left[ {{C}_{P}}+(M+1){{C}_{I}} \right]}f(t)dt+\sum\limits_{k=M+1}^{\infty }{\int\limits_{\frac{kx}{p}}^{\frac{(k+1)x}{p}}{\left[ {{C}_{P}}+(k+1){{C}_{I}} \right]f(t)dt}} \end{align} }[/math]

The expected cost for one cycle is

[math]\displaystyle{ \begin{align} & E(\text{Cost Per Cycle})=\int_{0}^{x}{tf(t)dt}+\sum\limits_{k=1}^{M}{\left[ \int\limits_{kx}^{\frac{kx}{p}}{\left( kx \right)}f(t)dt+\int\limits_{\frac{kx}{p}}^{(k+1)x}{t}f(t)dt \right]} \\ & +\int\limits_{\left( M+1 \right)x}^{\frac{(M+1)x}{p}}{\left[ (M+1)x \right]}f(t)dt+\sum\limits_{k=M+1}^{\infty }{\int\limits_{\frac{kx}{p}}^{\frac{(k+1)x}{p}}{\left[ (k+1)x \right]f(t)dt}} \end{align} }[/math]

The expected long term cost per unit time

[math]\displaystyle{ CPU{{T}_{L}}(t)=\frac{E(\text{Cost Per Cycle})}{E\left( \text{Cycle Length} \right)} }[/math]

where [math]\displaystyle{ {{C}_{I}} }[/math] is the average cost per inspection.

Minimize the Short Term Cost Per Unit Time

The short term (one replacement cycle) cost per unit time is [math]\displaystyle{ \begin{align} & CPU{{T}_{S}}=E\left( \frac{\text{Cost per Cycle}}{\text{Cycle Length}} \right)=\int_{0}^{x}{\frac{{{C}_{U}}}{t}f(t)dt} \\ & +\sum\limits_{k=1}^{M}{\left[ \int\limits_{kx}^{\frac{kx}{p}}{\frac{{{C}_{P}}+k{{C}_{I}}}{kx}}f(t)dt+\int\limits_{\frac{kx}{p}}^{(k+1)x}{\frac{{{C}_{U}}+k{{C}_{I}}}{t}}f(t)dt \right]} \\ & +\int\limits_{\left( M+1 \right)x}^{\frac{(M+1)x}{p}}{\frac{{{C}_{P}}+(M+1){{C}_{I}}}{(M+1)x}}f(t)dt+\sum\limits_{k=M+1}^{\infty }{\int\limits_{\frac{kx}{p}}^{\frac{(k+1)x}{p}}{\frac{{{C}_{P}}+(k+1){{C}_{I}}}{(k+1)x}f(t)dt}} \\ \end{align} }[/math]