Preventive Maintenance
Preventive maintenance is a schedule of planned maintenance actions
aimed at the prevention of breakdowns and failures. The primary goal of
preventive maintenance is to prevent the failure of equipment before it
actually occurs. It is designed to preserve and enhance equipment
reliability by replacing worn components before they actually fail.
Preventive maintenance activities include equipment checks, partial or
complete overhauls at specified periods, oil changes, lubrication and so
on. In addition, workers can record equipment deterioration so they know
to replace or repair worn parts before they cause system failure. Recent
technological advances in tools for inspection and diagnosis have
enabled even more accurate and effective equipment maintenance. The
ideal preventive maintenance program would prevent all equipment failure
before it occurs.
Value of Preventive Maintenance
There are multiple misconceptions about preventive maintenance. One such
misconception is that PM is unduly costly. This logic dictates that it
would cost more for regularly scheduled downtime and maintenance than it
would normally cost to operate equipment until repair is absolutely
necessary. This may be true for some components; however, one should
compare not only the costs but the long-term benefits and savings
associated with preventive maintenance. Without preventive maintenance,
for example, costs for lost production time from unscheduled equipment
breakdown will be incurred. Also, preventive maintenance will result in
savings due to an increase of effective system service life.
Long-term benefits of preventive maintenance include:
-
Improved system reliability.
-
Decreased cost of replacement.
-
Decreased system downtime.
-
Better spares inventory management.
Long-term effects and cost comparisons usually favor preventive
maintenance over performing maintenance actions only when the system
fails.
When Does Preventive Maintenance Make Sense
Preventive maintenance is a logical choice if, and only if, the
following two conditions are met:
-
Condition #1: The component in question has an increasing failure
rate. In other words, the failure rate of the component increases
with time, thus implying wear-out. Preventive maintenance of a
component that is assumed to have an exponential distribution (which
implies a constant failure rate) does not make sense!
-
Condition #2: The overall cost of the preventive maintenance action
must be less than the overall cost of a corrective action. (Note: In
the overall cost for a corrective action, one should include
ancillary tangible and/or intangible costs, such as downtime costs,
loss of production costs, lawsuits over the failure of a
safety-critical item, loss of goodwill, etc.)
If
both of these conditions are met, then preventive maintenance makes
sense. Additionally, based on the costs ratios, an optimum time for such
action can be easily computed for a single component. This is detailed
in later sections.
The Fallacy of "Constant Failure Rate" and "Preventive Replacement"
Even though we alluded to the fact in the last section of this on-line
reference, Availability,
it is important to make it explicitly clear that if a component has a
constant failure rate (i.e.
defined by an exponential distribution), then preventive maintenance of
the component will have no effect on the component's failure
occurrences. To illustrate this, consider a component with an MTTF =
100 hours, or λ =
0.01, and with preventive replacement every 50 hours. The reliability
vs. time graph for this case is illustrated in Figure 7.3. In Figure
7.3, the component is replaced every 50 hours, thus the component's
reliability is reset to one. At first glance, it may seem that the
preventive maintenance action is actually maintaining the component at a
higher reliability.
Figure 7.3: Reliability vs. time for
a single component with an MTTF =
100 hours, or λ =
0.01, and with preventive replacement every 50 hours.
However, consider the following cases for a single component:
Case 1: The
component's reliability from 0 to 60 hours:
-
With preventive maintenance, the component was replaced with a new
one at 50 hours so the overall reliability is the reliability based
on the reliability of the new component for 10 hours, R(t =
10) = 90.48%, times the reliability of the previous component, R(t =
50) = 60.65%. The result is R(t =
60) = 54.88%.
-
Without preventive maintenance, the reliability would be the
reliability of the same component operating to 60 hours, or R(t =
60) = 54.88%.
Case 2: The
component's reliability from 50 to 60 hours:
-
With preventive maintenance, the component was replaced at 50 hours
so this is solely based on the reliability of the new component, for
a mission of 10 hours, or R(t =
10) = 90.48%.
-
Without preventive maintenance, the reliability would be the
conditional reliability of the same component operating to 60 hours,
having already survived to 50 hours, or 
.
As
it can be seen, both cases, with and without preventive maintenance,
yield the same results.
Determining Preventive Replacement Time
As
mentioned earlier, if the component has an increasing failure rate, then
a carefully designed preventive maintenance program is beneficial to
system availability. Otherwise, the costs of preventive maintenance
might actually outweigh the benefits. The objective of a good preventive
maintenance program is to either minimize the overall costs (or
downtime, etc.) or meet a reliability objective. In order to achieve
this, an appropriate interval (time) for scheduled maintenance must be
determined. One way to do that is to use the optimum age replacement
model, as presented next. The model adheres to the conditions discussed
previously, or:
-
The component is exhibiting behavior associated with a wear-out
mode. That is, the failure rate of the component is increasing with
time.
-
The cost for planned replacements is significantly less than the
cost for unplanned replacements.
Figure 7.4: Cost curve for
preventive and corrective replacement.
Figure 7.4 shows the Cost Per Unit Time vs. Time plot. In this figure,
it can be seen that the corrective replacement costs increase as the
replacement interval increases. In other words, the less often you
perform a PM action, the higher your corrective costs will be.
Obviously, the longer we let a component operate, its failure rate
increases to a point that it is more likely to fail, thus requiring more
corrective actions. The opposite is true for the preventive replacement
costs. The longer you wait to perform a PM, the less the costs; while if
you do PM too often, the higher the costs. If we combine both costs, we
can see that there is an optimum point that minimizes the costs. In
other words, one must strike a balance between the risk (costs)
associated with a failure while maximizing the time between PM actions.
Optimum Age Replacement Policy
To
determine the optimum time for such a preventive maintenance action
(replacement), we need to mathematically formulate a model that
describes the associated costs and risks. In developing the model, it is
assumed that if the unit fails before time t,
a corrective action will occur and if it does not fail by time t,
a preventive action will occur. In other words, the unit is replaced
upon failure or after a time of operation, t,
whichever occurs first.
Thus, the optimum replacement time can be found by minimizing the cost
per unit time, CPUT(t). CPUT(t) is
given by:
(5)
Where:
-
R(t) =
reliability at time t.
-
CP =
cost of planned replacement.
-
CU =
cost of unplanned replacement.
The
optimum replacement time interval, t,
is the time that minimizes CPUT(t). This
can be found by solving for t such
that:
(6)
Or
by solving for a t that
satisfies Eqn. (7):
(7)
Interested readers can refer to Barlow and Hunter []
for more details on this model.
Introduction to Repairable Systems Example 2
The
failure distribution of a component is described by a 2-parameter
Weibull distribution, with β =
2.5 and η =
1000 hours.
Estimate the optimum replacement age in order to minimize these costs.
Solution to Introduction to Repairable Systems Example 2
Prior to obtaining an optimum replacement interval for this component,
the assumptions of Eqn. (5) must be checked. The component has an
increasing failure rate, since it follows a Weibull distribution with β greater
than one. Note that if β =
1, then the component has a constant failure rate and if β <
1, it has a decreasing failure rate. If either of these cases exist,
then preventive replacement is unwise. Furthermore, the cost for
preventive replacement is less than the corrective replacement cost.
Thus, the conditions for the optimum age replacement policy have been
met.
Using BlockSim, the failure parameters can be entered in the component's
Block Properties window. Select "Optimum Replacement" from the Block
menu, enter the costs and compute the optimum time, 493.0470. Figure 7.5
illustrates this.
Figure 7.5: Using BlockSim's Optimum
Replacement utility to obtain the results in Example 2.
Figure 7.6 shows a plot illustrating the cost per unit time.
Figure 7.6: Graph of cost vs.
replacement time for Example 2.
Discussion on Introduction to Repairable Systems Example 2
The
effect of the corrective/preventive cost ratio on the optimum
replacement interval is plotted in Figure 7.7. It can be seen that as
the cost ratio increases, the optimum replacement interval decreases.
This is an expected result because the corrective replacement costs are
much greater than the preventive replacement costs. Therefore, it
becomes more cost effective to replace the component more frequently
before it fails.
Figure 7.7: Replacement interval as
a function of the corrective/preventive cost ratio.
