BRIDGING THE GAP BETWEEN THEORY AND PRACTICE IN MAINTENANCE D.N.P. (Pra) MURTHY RESEARCH PROFESSOR...
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Transcript of BRIDGING THE GAP BETWEEN THEORY AND PRACTICE IN MAINTENANCE D.N.P. (Pra) MURTHY RESEARCH PROFESSOR...
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BRIDGING THE GAP BETWEEN THEORY AND PRACTICE IN
MAINTENANCE
D.N.P. (Pra) MURTHYRESEARCH PROFESSOR
THE UNIVERSITY OF QUEENSLAND
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PART-3: BUSINESS FOCUS
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OUTLINE
• Framework & modelling
• Case 1: Dragline
• Maintenance Outsourcing
• Case 2: Hydraulic pumps
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FRAMEWORK & MODELLING
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KEY ELEMENTS
Design / UpgradeFunctional requirement
Production rate Equipment degradation
Maintainability requirements
Maintenance (PM / CM)
Output Operating costs
Revenue Profits Investment
Business Goals
Technical
Commercial
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MODELLING
• The elements that are relevant depends on the decision problem
• Need to model the relevant elements separately
• Link the models to build the model for solving the decision problem
• Data plays a critical part
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DRAGLINE CASE STUDY
[CONTINUATION FROM PART 2]
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DECISION PROBLEM
• Commercial considerations dictate an increase in output
• Idea: Increase bucket size (100 tons to 140?)
• Greater load on components
• Implications for reliability and maintenance
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LOAD
DEGRADATION
MAINTENANCE
AVAILABILITY
FAILURE
DUTY CYCLE
YIELD
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MODELLING
• Modelling system in terms of its major components [Decomposition]
• Modelling degradation of each component
• Modelling effect of bucket load on component and system performance
• Involves reliability science, engineering and mathematics
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SYSTEM PERFORMANCE
• Availability: Depends on up and down times
• Down times: To rectify minor failures and preventive maintenance to avoid major failures
• Up time: Productive time
• Cycle: Time between major maintenance
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SYSTEM PERFORMANCE
• Bucket load affects both these variables
• Need to take into account preventive maintenance schedules for different components [Different time scales]
• Multiple objectives: Study different alternatives
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OBJECTIVES
• Maximise total output per year
• Maximise revenue per year
• Minimise total cost per year
• Maximise yield [dirt moved per unit time]
• Need to take into account various constraints
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SYSTEM FAILURE MODELLING
• System comprised of 25 components
• All components need to be working for the system to be working. System fails whenever a component fails.
• System failure distribution is given by a competing risk model involving the failure distribution of the 25 components
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MODELLING THE SYSTEM
• Failure distribution for the system is given by
• Failure distributions of the individual components was discussed in Part 2.
• Minimal repairs for subsequent failure modelling
25
1
( ) 1 ( ) 1 ( )ii
S T F T F T
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AVAILABILITY
• Cycle Time: Depends on load v the ratio of load to the base load
• Up time: Tv
• Expected downtime (for minor and major preventive maintenance) – obtained from field data
• From this we can obtain availability
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AVAILABILITY
( , )( )
vTA T v
ECL v
1 0
( ) [ { ( ) } ]vTK
v vi ri pm pi
ECL T r x dx
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Rel
iab
ilit
y
1
0.95
T1 T0
P.M. Interval (T)
Bucket load V1
Bucket load V0
RISK CONSTRAINT
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AVAILABILITY vs v
1 1.2 1.4 1.6 1.80.3
0.4
0.5
0.6
0.7
0.8
v
Av
aila
bili
ty(v
) (%
)Availability vs Stress Ratio
v
Ava
ilab
ilit
y
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1 1.2 1.4 1.6 1.80.5
1
1.5
2
2.5
3
3.5
4
4.5x 10
4
v
Tv (
ho
urs
)PM Interval vs Stress Ratio
MAJOR PM INTERVAL vs v
v
Maj
or P
M I
nte
rval
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YIELD vs BUCKET LOAD
1 1.2 1.4 1.6 1.840
45
50
55
60
65
70
v
Yie
ld(v
) (t
on
ne
s/m
inu
te)
Yield vs Stress Ratio
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SENSITIVITY STUDY ()
1 1.2 1.4 1.6 1.830
35
40
45
50
55
60
65
70
75
v
Yie
ld(v
) (t
on
ne
s/m
inu
te)
Yield vs Stress Ratio
90% i
100% i
110% i
90% i
110% i
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CONCLUSIONS• Study revealed increase in output yield
with increase in bucket size• Maximum yield corresponds to v 1.3
(dragline load = 182 tonnes or payload of 116 tonnes) as opposed to current payload of 74 tonnes
• Shutdown interval will need to be reduced from 43680 usage hours to 25000 usage hours (or 4.1 calendar years)
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REFERENCE
• For more details, see Townson, P. Murthy, D.N.P. and Gurgenci, H. (2002), Optimisation of Dragline Load, in Case Studies in Reliability and Maintenance, WR Blischke and DNP Murthy [Editors], Wiley, New York.
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MAINTENANCE OUT-SOURCING
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CONCEPT
Outsourcing of maintenance involves some or all of the maintenance actions (preventive and/or corrective) being carried out by an external service agent under a service contract. The contract specifies the terms of maintenance and the cost issues and can involve penalty and incentive terms.
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KEY ELEMENTS
OWNER'S DECISIONS
OUTCOMES
SERVICE AGENT'S DECISIONS
OWNER'S OBJECTIVES
SERVICE AGENT'S OBJECTIVES
MAINTENANCE SERVICE CONTRACT
UNKNOWN AND UNCONTROLLABLE
FACTORS
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OVERALL FRAMEWORK
ASSET STATE AT THESTART OF CONTRACT
PAST USAGEPAST
MAINTENANCE
OWNER(CUSTOMER)
SERVICEAGENT
CONTRACT
NOMINATEDUSAGE RATE
NOMINATEDMAINTENANCE
ACTUALUSAGE RATE
ACTUALMAINTENANCE
ASSET DEGRADATIONRATE
ASSET STATE AT THEEND OF CONTRACT
PENALTIES /INCENTIVES
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MAINTENANCE ACTIVITIES
• D-1: What (components) need to be outsourced for maintenance?
• D-2: When should the maintenance be carried out?
• D-3: How should the maintenance be carried out?
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ALTERNATE CONTRACT SCENARIOS
DECISIONS
CUSTOMER SERVICE AGENTSCENARIOS
S-1
S-2
S-3
D-1, D-2
D-1
-
D-3
D-2, D-3
D-1, D-2, D-3
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DECISION PROBLEMS
• From a business perspective
• Well defined objective (or goal)
• Models to evaluate alternate options and for deciding on the optimal option
• Most businesses do not do this and outsource decisions are based on qualitative evaluation
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EXCAVATORS CASE STUDY
[Outsourcing Hydraulic Pumps]
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EXCAVATORS
• Excavators are used in mining to load coal or ore on to dump trucks for transporting
• Hydraulic pumps operate the excavators
• Four pumps per machine
• Mine operator had four machines on site
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MAINTENANCE OUTSOURCING
• The company selected on Scenario 1 where the owner decided on D-1 and D-2
• Outsourcing the maintenance of hydraulic pumps
• PM action if a pump did not fail for 12,000 hours [based on manufacturer recommendation]
• CM action on failure
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MAINTENANCE
• Both CM and PM maintenance results in the reconditioned pump being back to as-good-as new
• Some items were junked based on their condition whilst others were subjected either CM or PM action
• Customer used both new and reconditioned pumps
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DATA ASPECTS
• Customer had failure data for items that failed and censored data (resulting from PM actions or discarding)
• No information on number of times a unit was subjected to maintenance action
• Some other information was also collected.
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DATA ASPECTS
• There was no terms in the contract for the Service Agent to provide the owner with the state of items sent for PM action or the failure mode of items sent for CM action.
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DECISION PROBLEM
• The cost of a CM action >> the cost of a PM action
• The owner was interested in seeing if the age for PM actions can be increased to 15,000 hours so as to reduce the maintenance costs paid to the Service Agent
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DATA COLLECTION
• 6 year window yielded 103 data • 46 failure data and 57 censored data.• For each failure data, additional
information relating to (i) the associated excavator (one of four different excavators), (ii) the pump position (one of four different positions) and, (iii) the engine (one of two) was also collected.
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DATA COLLECTION
• For the 45 pumps that failed the following additional information was obtained.
– 15 are known to be new pumps– 2 are suspected to be new pumps– 8 are known to be reconditioned pumps– 2 are suspected to be reconditioned pumps– 19 are unknown
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MODEL FORMULATION
• Based on WPP plot [Discussed in Part 2]
• The model selected was a mixture model
• Two cases: shape parameters (i) same and (ii) not same
1 2( ) ( ) (1 ) ( )F t pF t p F t
( ) 1 exp{ ( / ) },1 2ii iF t t i
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WPP PLOTS – DATA AND MODEL[SHAPE PARAMETERS SAME]
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WPP PLOTS – DATA AND MODEL[SHAPE PARAMETERS DIFFERENT]
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MODEL PARAMETERS
• Model parameters obtained by least squares fit
• Select the one with the same shape parameters
Parameter p 1 2 1 2 ( )
Case 1 0.925 2.22 2.22 14800 465 0.650
Case 2 0.915 2.48 1.83 14400 566 0.515
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MODEL ANALYSIS
• Two sub-populations• MTTF given by • ;• Around 7.5 – 8.5% of items have early failures• Reasons for early failures:
– Particular machine and location? [some data available to test this]
– Operating environment? [no data available]
(1 1/ ) ,1 2i i i i
2 410 1 13110
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OPTIMAL DECISION
• Optimum age for PM – can be derived using the well known PM policy
• Objective function: Asymptotic maintenance cost per unit time
0
{ / } ( ) [1 ( )]( )( )
( ) [1 ( )]
f p
Tp
C C F T F TJ TJ T
Ctf t dt T F T
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0.000158745
0.00015875
0.000158755
0.00015876
0.000158765
0.00015877
0.000158775
0.00015878
0.000158785
0.00015879
0.000158795
15100 15200 15300 15400 15500 15600 15700 15800 15900 16000 16100
T
J(T)
/ 2f pC C
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IMPLICATIONS
• With current reliability the optimum age for PM is 15,000 hours with
• By proper understanding and identification of the root cause one can eliminate early failures
• In this case the reliability increases and the PM interval can be increased
/ 2f pC C
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REFERENCES
• Murthy, D.N.P., Xie, M. and Jiang, R. (2003), Weibull Models, Wiley, New York. – [Deals with many Weibull based models and the use
of WPP plots for model selection.]
• Murthy, D.N.P. and Jack, N. (2008), Outsourcing of Maintenance, in Complex System Maintenance Handbook, K.A.H. Kobbacy and D.N.P. Murthy (eds), Springer Verlag, London,
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Thank you
Any Questions?