BRIDGING THE GAP BETWEEN THEORY AND PRACTICE IN

MAINTENANCE

D.N.P. (Pra) MURTHYRESEARCH PROFESSOR

THE UNIVERSITY OF QUEENSLAND

PART-3: BUSINESS FOCUS

OUTLINE

• Framework & modelling

• Case 1: Dragline

• Maintenance Outsourcing

• Case 2: Hydraulic pumps

FRAMEWORK & MODELLING

KEY ELEMENTS

Design / UpgradeFunctional requirement

Production rate Equipment degradation

Maintainability requirements

Maintenance (PM / CM)

Output Operating costs

Revenue Profits Investment

Business Goals

Technical

Commercial

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

DRAGLINE CASE STUDY

[CONTINUATION FROM PART 2]

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

LOAD

DEGRADATION

MAINTENANCE

AVAILABILITY

FAILURE

DUTY CYCLE

YIELD

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

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

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

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

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

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

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

AVAILABILITY

( , )( )

vTA T v

ECL v

1 0

( ) [ { ( ) } ]vTK

v vi ri pm pi

ECL T r x dx

Rel

iab

ilit

y

1

0.95

T1 T0

P.M. Interval (T)

Bucket load V1

Bucket load V0

RISK CONSTRAINT

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

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

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

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

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)

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.

MAINTENANCE OUT-SOURCING

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.

KEY ELEMENTS

OWNER'S DECISIONS

OUTCOMES

SERVICE AGENT'S DECISIONS

OWNER'S OBJECTIVES

SERVICE AGENT'S OBJECTIVES

MAINTENANCE SERVICE CONTRACT

UNKNOWN AND UNCONTROLLABLE

FACTORS

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

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?

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

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

EXCAVATORS CASE STUDY

[Outsourcing Hydraulic Pumps]

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

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

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

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.

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.

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

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.

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

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

WPP PLOTS – DATA AND MODEL[SHAPE PARAMETERS SAME]

WPP PLOTS – DATA AND MODEL[SHAPE PARAMETERS DIFFERENT]

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

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

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

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

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

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,

Thank you

Any Questions?