Example of PM optimization
The predominant failure mode of a critical part of a machine is of a degrading nature (erosion, for
instance) and the existing failure data allows you (after a best-of-fit test) to represent its failure
behaviour by a Weibull probability distribution with parameters: location = 0; shape = 2; scale =
3,000 hours. Costs and down time, respectively, are estimated to be 1,000 € and 5 hours per
preventive action (planned) and 3,000 € and 30 hours per corrective action (unplanned). Suppose
you accept the accuracy to be +/- 250 hours. If you elect: a) minimum cost; b) maximum
availability, as decision criterion, what should the periodicity of PM be?
As you can see in the exhibit below, 2.250 hours is the time interval that minimizes cost (98,47
€/100 operating hours) and 1.250 hours is the time interval that maximizes availability. Sensitivity
analysis can be performed in the software and we obtain the lower and upper limits for any %
allowed of increase of the minimum cost or of decrease of the maximum availability.
Curves of cost or availability varying with PM time intervals can be viewed on next page.
Weibull probability distribution
Periodicity
of PM
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
0
250
500
750
1.000
1.250
1.500
1.750
2.000
2.250
2.500
2.750
3.000
3.250
3.500
3.750
4.000
4.250
4.500
4.750
5.000
Failure
density
function
Cumulative
Instantaneous
probability of
failure rate
failure
0
0
5,517E-05 0,0069204
0,0001081 0,0273955
0,0001566 0,0605869
0,0001989 0,1051607
0,0002335 0,1593763
0,0002596 0,2211992
0,0002767 0,2884274
0,000285 0,3588196
0,0002849 0,4302172
0,0002774 0,5006482
0,0002637 0,5684094
0,0002453 0,6321206
0,0002233 0,6907518
0,0001994 0,7436242
0,0001747 0,7903886
0,0001502 0,8309867
0,0001269 0,8656013
0,0001054 0,8946008
8,605E-05 0,9184831
6,909E-05 0,9378235
Mean failure rate =
MTTF =
Costs
Mean life
0
5,556E-05
125
0,0001111
312
0,0001667
483
0,0002222
649
0,0002778
811
0,0003333
969
0,0003889
1.122
0,0004444
1.269
0,0005
1.411
0,0005556
1.547
0,0006111
1.676
0,0006667
1.796
0,0007222
1.909
0,0007778
2.013
0,0008333
2.109
0,0008889
2.195
0,0009444
2.272
0,001
2.340
0,0010556
2.400
0,0011111
2.451
0,000376 failures/hour
hours
2.659
Availability
PM
€/100 h
CM
€/100 h
PM
CM
406,94
213,16
152,78
125,67
111,75
104,33
100,51
98,84
98,47
98,93
99,89
101,12
102,48
103,87
105,22
106,47
107,61
108,61
109,48
110,21
112,84
112,84
112,84
112,84
112,84
112,84
112,84
112,84
112,84
112,84
112,84
112,84
112,84
112,84
112,84
112,84
112,84
112,84
112,84
112,84
97,97%
98,86%
99,12%
99,21%
99,24%
99,24%
99,23%
99,20%
99,17%
99,14%
99,11%
99,08%
99,05%
99,02%
99,00%
98,98%
98,96%
98,95%
98,93%
98,92%
98,87%
98,87%
98,87%
98,87%
98,87%
98,87%
98,87%
98,87%
98,87%
98,87%
98,87%
98,87%
98,87%
98,87%
98,87%
98,87%
98,87%
98,87%
98,87%
98,87%
Optimal periodicity in the light of minimum cost =
Optimal periodicity in the light of maximum availability =
Sensitivity analysis
Lower limit Optimum
Upper limit
Tolerable increment of the minimum cost =
5%
1750
2250
3500
Tolerable decrement of the maximum availability =
1%
500
1250
Outside
Parameters
t0 =
a=
b=
0
2
3000
hours
hours
Scale factor
Increment =
250
Costs
PM =
CM =
1000
3000
€/action
€/action
Durations
PM =
CM =
5
30
Conclusion
PM recomended
hours
2.250
hours
1.250
hours/action
hours/action
The optimum time interval to perform PM is obtained by minimizing the following equation:
C MP =
Where:

CC .F (t p )  C P . 1  F (t p )

t.F (t p )  t p . 1  F (t p )


tp is the time interval between PM actions;
t is the mean life of the components that fail before the tp;
Cc is the cost of a CM action;
Cp is the cost of a PM action;
Ft is the probability of a failure or the probability of a CM occurs;
(1 – Ft) is the probability of a PM action takes place.
1
F(t)
F(tp)
f(t)
0
0
F(tp)
t
tp
PM time interval (hours)
The mean lifet of the components that fail before tp is calculated solving the integral:
tp
tp
0
0
t =  t.f(t).dt =  R(t).dt
As an alternative, although less accurate, you can use a numerical method by dividing the interval
(0; tp) into N classes. The mean lifet can now be obtained from the equation:
i=N
t=
 ti  ti 1

.F (ti )  F (ti 1 )
2

F (ti )
 
1
The next figure shows how Cost varies with t and the optimum PM time interval t*.
Cost
€/hora
Least
cost
0
t*
PM time interval (hours)
If you prefer maximum availability as the decision criterion instead, then you solve the next
equation for a maximum, where symbols have the same meaning as above. Besides those, TP is the
duration of a PM action and TC is the duration of a CM action (frequently the relation TC  TP
holds).
t.Ft  t.1  F (t p )
D=
t.F (t p )  t p .( 1  Ft )  TC .F (t p )  TP . 1  F (t p )




The next figure shows how Availability varies with t and the optimum PM time interval t**.
Maximum
availability(
%)
0
t**
PM time interval (hours)
In the table above, you find, for instance, the cost 98,47 €/100 hours, by doing the following:
t = {[(250 + 0) / 2 x (0,0069204 – 0)] + … + [(2.250 + 2.000) / 2 x (0,4302172 – 0,3588196)]} /
0,4302172 = 1.411 operating hours
[3.000 x 0,4302172 + 1.000 x (1 – 0,4302172)] / [1.411 x 0,4302172 + 2.250 x (1 – 0,4302172)] x
100 = 98,47 €/100 operating hours
And you find 99,24% of availability this way:
t = {[(250 + 0) / 2 x (0,0069204 – 0)] + … + [(1.250 + 1.000) / 2 x (0,1593763 – 0,1051607)]} /
0,1593763 = 811 operating hours
[811 x 0,1593763 + 1.250 x (1 – 0,1593763)] / {[811 x 0,1593763 + 1.250 x (1 – 0,1593763) + 30 x
0,1593763 + 5 x (1 – 0,1593763]} x 100 = 99,24%
Rui Assis
November/05