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Mining EquipmentMaintenance
Fundamental Concepts
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Old Underground Coal Mines
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Modern Longwall Mine
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The effect of Mechanisation on Productivity
0
5000
10000
15000
20000
25000
1916
1920
1925
1930
1935
1940
1945
1950
1952
1955
1960
1969
1972
1975
1980
1985
1988
1995
2003
Tonnes/Man-Year
Produced in QueenslandCoal Mines
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Maintenance Costs
30-50% of the operating costs Annual bill is about $10 billion
Equally significant is the cost of lostproduction when the machine is down
Every 1% improvement in equipment
availability or productivity improves the
company profits by up to 3.5%
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Objective Function forMaximising Company profits
The Annual Production for a given
investment is a suitable objective function
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Total Annual Production
Is this total annual production ?Tonnes/Hour x Total Hours
More realistic prediction is
Tonnes/Hour x (TotalHours - Downtime) Probably the following is more illuminating
Tonnes/Hour x (TotalHours -
-Scheduled Downtime
- Breakdown maintenance)
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Total Hours of PossibleProduction
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Subtract PlannedMaintenance
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Subtract BreakdownsThis is the time available
for production. Obviously,you want to maximise thistime. That is why the ratioof this time to the totaltime is a very importantKPI.
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Availability
TH - PM - BMA= TH
TH Total HoursPM Planned Maintenance
BM Breakdown Maintenance
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Availability
!Operating TimeTH - PM - BM
A= TH Total Time
Is also expressed in terms of MTBF and MTTR as
MTBFA=MTBF+MTTR
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Availability
!Operating TimeTH - PM - BM
A= TH Total Time
Is also expressed in terms of MTBF and MTTR as
MTBFA=MTBF+MTTR
This is the textbookdefinition of AVAILABILITY
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AvailabilityOperating Time = N x MTBF
Downtime = N x MTTR
Total Time = N x (MTBF+MTTR)
Operating Time MTBFA=
Total Time (MTBF
N x
N )x +MTTR
!
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Maximise the Return to the Company
The aim for the company is to maximise the return from its
productive assets.
Let us keep it simple and express this as maximising theproduction through the year.
To achieve this,
(a) The machine availability must be high
(b) The machine must be producing at a high rate when it is
operating
The solution is a compromise between the two.
We will see this in an example.
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Optimum Bucket SizeThe recommended suspended load for your dragline is 150
tonnes. A competitor put a larger bucket on a similar unitand they are running it at a suspended load of 170 tonnes.
Your Manager wants you to install an even bigger bucket,
taking your suspended load up to 210 tonnes. You know
that the safe static load for this dragline is 250 tonnes.
-Does a larger bucket necessarily mean higher production?
-Why not increase it to 250 tonnes if this is the safe load?
-How would you determine the optimum bucket size?
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Question TH - PM - BMA= THPresent Production Po = 9000 t/h.
Total Mine Operating Time in a Year = 8640h
12-hour planned maintenance shift every month.Breakdown maintenance downtime is expected tovary with the production rate as
3336o
PBM wR RhereP
! " !P is the production rate with the bigger bucket.
Find R that maximises the annual production.
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Solution
L
ostpr
oduction
due
to
Breakd
ownm
aintenanc
e
Optimum Point
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Dependence on MTBF
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Maintenance
Maintenance must be considered in thecontext of asset utilisation
Mining is an asset-rich industry
Optimum utilisation of these assets is the
only way a company will stay competitive
This is a task for both production and
maintenance engineers.
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Some Basic Concepts
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FailureLoss of abilityof an item to
perform itsrequiredfunction
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Failure
Broken teethon shearerdowndrive gear
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Failure
Cost associated with
this failure:
Lost production whenthe machine was down
Replacement gear
Maintenance labour
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Failure Rate, MTBF, MTTRMTTR MTBF MTTR MTBF
0 100 200 Hours
R
epair
Repair
Repair
Failure
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Failure Rate, MTBF, MTTRMTTR MTBF MTTR MTBF
0 100 200 Hours
R
epair
Repair
Repair
MTBF = Mean Time Between Failure (100 h)
Failure Rate = Number of failures per unit time (0.01 h-1)
MTTR = Mean Time To Repair (20 h)
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Historical RecordsFailures occur randomly. The repair time is also not constant.How do we find MTBF and MTTR?
If we treat failure as a random event, then we can use the
well-established tools of probability and statistics to modelthe uptime, downtime and availability for our equipment.
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Random Failure?
All nature is but art, unknown to thee
All chance, direction thou canst not see
All discord, harmony not understood
Alexander Pope, Essay on Man
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Poisson ProcessPoisson distribution is commonly used in forecasting to
represent the number of occurrences of a specific event ina given continuous interval.
Ships arriving at a dock on a given day
Traffic accidents on the SE freeway in a month
Mad cow disease breakouts in the world in one year
Typos per page in a long report typed by Hal Gurgenci
Cable shovel failures in one day of operation
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Poisson Distribution
# $( ) !
xte tp x
x
% %&
!
This is the probability distribution of the Poisson random
variable X representing the number of outcomes occurring in
a given time interval t. The parameter % is the averagenumber of outcomes per unit time.
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ReliabilityAssume failure events follow a Poisson distribution.What is the probability of having NO FAILURES in a
given time interval t?
This can be found by substituting x=0 in the Poisson
distribution function:
# $0
(0)0!
t
te t
p e
%
%%& &! !
This is referred to as the survival probability or the
reliability
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Reliability( ) tt e %&!
Reliability is the probability that a product will
operate throughout a specified period without failure
when maintained in accordance with the manufacturer's
instructions; and
when not subjected to the environmental or operational
stresses beyond limits stipulated by the manufacturer
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The value of e
Two centuries ago, a Polish Statistician, Ladislaus
Bortkiewicz, investigated the Prussian army fatalities
caused by horse kicks. According to army reports, the rate
was about one fatality every 1.64 years. Ladislaus
collected the reports for one year. These were 200 reportsand 109 recorded no deaths at all.
Can you estimate the value of
eusing the above data?
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ExampleA major piece of equipment fails twice a day on average.
Consider its reliability over a period of month.
What is the probability of failure at any time during that
month?
R li bilit
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Reliability
2t
e
&
!
Z I
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Zoom In
2t
e&
!
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Failure ProbabilityIf the reliability, R(t), is the probability to survive through
time t, then the probability of failing through that period is
1 R(t)
or
( ) 1 ( ) 1
t
F t R t e
%&
! & ! &Let us plot this
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Failure ProbabilityThis chart gives the probability of
the failure from 0 to time t. In
other words, it is the Cumulative
Distribution Function for failures.
How would we find the probability
density function or p.d.f. This is
sometimes useful.
We differentiate the cumulative
distribution function.
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Failure Distribution FunctionCumulative
Distribution Function ( ) 1 ( ) 1 tF t R t e %&! & ! &
The time-derivative of the c.d.f. gives the Probability
Distribution Function or p.d.f.
( ) tdF
f t e
dt
%% &! !
This form of p.d.f. is called the Exponential Distribution.
It represents the case when the hazard rate or failure rate, ,
is constant over time.
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Failure p.d.f.
The probability of failure through aunit time interval is given by
1
( )t
t
p f t dt'
!
(
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Let us zoom into the p.d.f.The probability of failure through a unit time interval is
given by the area under the curve
1 1 1( ) ( ) 1 ( ) ( )
2 2
t
tp f t dt f t f t f t
'! ) ' " ) ' )(
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Hazard RateThe hazard rate is the conditional probability of failure in a
small time interval (t, t+dt). It is conditional on there being
no failure until t:
( )( )
( )
f th t
t
!
For exponential failure distribution, the hazard rate is
constant:
( )( )( )
t
t
f t eh tR t e
%
%
% %&
&! ! !
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Constant Hazard Rate
The exponentialdistribution corresponds to
a constant failure rate a.k.a.
constant hazard rate
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Is % always constant ?
( ) t
f t e %
% &
!
Failure p.d.f.The constant % is the failure rate(%= 1/MTBF)
So far, we treated % as constant
This is called the exponentialdistribution
Is this always true?
Let us first give an example
Reliability function
( ) tR t e %&!
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Uniformly Increasing RateAssume the failure
rate is increasing by
the formula 0.1twhere t is measured
in days
It starts from zero and
at the end of the
month, the hazard rate
is 3 failures per day.
How do we generate the reliability function for this
component?
Reliability with varying
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Reliability with varyinghazard rate
dN Nmtdt ! &
In a sample of N, dN will fail in a time interval dt
dN mtdtN
! &2
ln ln2
tN m C! & '
m=0.1
2 2/ 2 / 2mt mt oN Ce N e
& &! ! Where N is the value of N at t=0
Then the reliability function is 2/ 2mt
o
NeN
&!
This corresponds to a Weibull distribution
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Weibull Reliability
t
R e
*
+, -&. /
0 1!R The probability of surviving through time t* Shape factor
+ Scale factor
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Weibull Distribution Curves
( ) 1 ( ) 1
t
F t R t e
*
+
, -&. /
0 1! & ! &c.d.f.
1
( )
tdF
f t t edt
*
+*
*
*
+
, -&. /&
0 1! !p.d.f.
1( )( )( )
f th t tR t
***+
&! !Hazard Rate
W ib ll s ith diff nt
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Weibull curves with different
hazard functions
t
R e
*
+, -&. /0 1!
1( )h t t***+
&!
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Bathtub CurveInfant
Mortality Useful Working LifeFallingApart
Fai
lur e
Rat e, %
Life in operation, hours
Optimum Point to Replace
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Actual Failure Patterns
These curves are the failurepatterns observed on
aircraft components in a
study completed in 1978 byNowlan and Heap.
This shows that only 4% of
the components go througha bathtub curve.
4%
2%
5%
7%
82%(68% of this with infant mortality)
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RELIABILITY OF SYSTEMS
Series Systems
Parallel Systems (Redundancy)
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Series SystemsA series system is a chain of components. When one of these parts fails,
the entire system fails.
Series Systems BA C
B CR R R R!
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Parallel SystemsThe failure for a parallel system means the failure of each
individual component. The system failure probability is then the
product of individual failure probabilities (1 R).
A
B
C
1 (1 )(1 )(1 )A B CR R R R! & & & &
Most mining machinery
systems are series
systems. In other words,the failure of one
component fails the
entire system. The
redundancy in miningcan be provided by
having multiple systems,
eg spare trucks or
shovels.
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Managing Reliability Optimum utilisation of its capital investment in
equipment is essential for company profits
Equipment reliability plays a major role in this
Therefore, managing reliability is a core business
for a mining company This is a task for both production and maintenance
engineers. In the rest of this presentation, we will
talk about the maintenance function.
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Maintenance Function Preventive Maintenance Prevent failures by performing a set of maintenance
tasks at periodic intervals Service
Inspection
Replacement
Corrective Maintenance Repair after a failure to bring the machine back to an
operating state
Which one delivers higher overall systemavailability?
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Corrective Maintenance We assume that corrective maintenance
brings the system to as new state.
Then it has no effect on system reliability
Its impact on system availability is measure
by the Mean Time To Repair (MTTR)
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Mean Time To Repair Fault Identification What caused the failure? What needs to be repaired?
Set-up time Find and bring the right person to the job
Actual repair
Logistic delays Waiting for the spare part
Restart time Time spent to bring the system back to normal
operation after the fault is repaired
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How to minimise MTTR Identify the failed components quickly.
This is achieved by experienced operators,on-line fault detection tools
For frequent failures have the repair crewwith the right skills on standby
Ditto for the frequently failing spare parts
Design the equipment and the operatingprocedure to minimise re-starting time
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PM Trade-Offs The cost of failure
MTTR
The cost of the repair and the replacement
The cash cost of the planned maintenanceaction (salaries, consumables, etc)
The opportunity cost (lost production)
Preventive Maintenance
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Preventive Maintenance
Issues Service
Effect on System Reliability
Inspection
P-F time (between potential and actual systemfailure)
Replacement
Failure distribution curves
Effect on System Reliability
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Would inspections help?
Failure
P-F Interval
The time when we can recognise
Potential Failure
Time
Scheduled inspections help when
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Scheduled inspections help when
Potential failure condition is clearly defined
The P-F interval is consistent
It is practical to inspect at intervals less than
the P-F interval
The P-F interval is long enough to
implement corrective maintenance action
Scheduled replacements help when
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Scheduled replacements help when
The component breakdown has costly
consequences (eg chain of failures, distance
from the workshop, etc)
The dominant failure mode is age-relatedwith the hazard rate consistently increasing
above an acceptable value at around the setreplacement period
Weibull curves with different
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ff
hazard functions
t
R e
*
+, -&. /0 1!
1( )h t t***+&!
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Periodic replacements - 10.5*! Decreasing Hazard Rate
Scheduled replacementincreases failure
probability
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Periodic replacements - 21*!
Constant Hazard Rate
Scheduled replacementhas no effect on failure
probability
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Periodic replacements - 32*!
Increasing Hazard Rate
Scheduled replacementdecreases failure
probability
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Reliability ResearchA significant part of the academic and research community
has been continuing to develop increasingly complex
mathematical models of the engineering systems and the
expected modes of failure under various loading
assumptions.
While the intellectual rigour in these studies and the amount
of effort that go into them cannot be ignored, the
applications to real manufacturing and mining processes
have been limited primarily for the lack of data needed to
support these models
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Industry Data
Analysis and Representation Tools
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Pareto Analysis Pareto Principle : Significant few and
Insignificant many In any application, a large part of the
failures are due to a small number of causes A Pareto plot helps to identify the most
significant causes
The benefit is incurred only by attending thesignificant issues
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Pareto Chart
Pareto Analysis for LongwallFace Equipment Failures
Shearer Drive Shaft
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Scatter Plot Diagrams A scatter plot is a logarithmic plot of
MTTR against the number of failures N.
Since the total downtime associated with
each failure is NxMTTR, constantdowntime curves appear as lines on
logarithmic axes.log log
DowntimeTTR MTTR C N
N! 2 ! &
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Longwall Scatter Plot
Lines of
Constant
Downtime
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Reliability Analysis Pareto Analysis and Scatter Plots are good tools to
identify the reliability sinks in the equipment
The next step is to calculate the failure probability
distribution curves for all critical components.
The MTTR statistics may also be required if
MTTR is not reasonably constant for each item
This step requires high quality data
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Features of High Quality Data Large enough set to have at least 4-5 failure
events for each target failure mode Cover a sufficiently long time period to
eliminate local effects Uniform operating conditions over this
period
Accuracy free of collectors bias
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Failure History Example
Suppose that we have the failure log for acomponent as 180, 216, 930, 990, 1300 and
1850 hours.
Estimate the MTBF assuming an exponential
probability distributions.
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Failure Data (example)Failure times
180h
216h
930h
990h1300h
1850h
36
714
36
714
60
310
550
60
310
550
36 714 60 310 550334
5MTBF
' ' ' '! !
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Curve FittingThe failure log was 180, 216, 930, 990, 1300 and 1850 hours.The TBF array , TBF = {36, 714, 60, 310, 550}.
100%80%60%40%20%Fi
7145503106036TBF
54321i
This is cumulative probability distribution data. We can then compare it
against exponential or Weibull distributions. For example, use the
following form to try an exponential distribution:
1ln 1 ( )
1 ( )
tt F t e
F t
%% &3 4
! 5 & !6 7&8 9
Estimate the Failure
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Distribution
TBF, h
Rough
Estimate
Median
Estimate
36 20% 10%
60 40% 30%310 60% 50%
550 80% 70%714 100% 90%
Exponential Fit334
t
t% &&
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( ) 1 1F t e e! & ! &
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Longwall Equipment
Failure Probability Distribution
Functions for some Critical Items
AFC Blockage/Overload (2002 only)
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1ln
1 ( )F t
3 46 7&8 9
Exponential
AFC Blockage/Overload (2002 only)
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Weibull
AFC Chain Failures
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Exponential
1ln
1 ( )F t
3 46 7&8 9
AFC Chain Failures
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Weibull
BSL Chain Failures
3 4
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Exponential
1ln
1 ( )F t
3 46 7&8 9
BSL Chain Failures
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Weibull
Shearer Cable Failures
3 4
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Exponential
1ln
1 ( )F t
3 46 7&8 9
Shearer Cable Failures
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Weibull
Wh t i t b d ?
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What is to be done? Increase overall availability Minimise the time spent on PM
Decrease number of breakdowns More effective PM
Condition monitoring with long enough P-F time
Engineering changes
Design changes
Changes to the operating procedure
Decrease MTTR
Do all this while maximising the profit Learn lessons for next equipment purchase
The Reliability Function?
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Failed =6Failed=4Failed =2Failed =0
t=3t=1t=0 t=2
t=4 t=5 t=6 t=7
Failed=8 Failed=10 Failed =12 Failed=14
Is this best represented by a Weibull or an exponential distrubution?