Maintenance and Reliability

BRAVEHEART

Gutierrez, KojiHizon, CarloLiterato, DioniSengson, RichardTongol, Earl

MaintenanceMaintenance of power generating of power generating plantsplants

Every year each plant is taken Every year each plant is taken off-off-line for 1-3 weeks maintenanceline for 1-3 weeks maintenance

Every three years each plant is Every three years each plant is taken off-line for 6-8 weeks taken off-line for 6-8 weeks for for complete overhaul and turbine complete overhaul and turbine inspectioninspection

Each overhaul has Each overhaul has 1,800 tasks 1,800 tasks and and requires requires 72,000 labor hours72,000 labor hours

OUC performs over OUC performs over 12,000 12,000 maintenance tasks maintenance tasks each yeareach year

Every day a plant is down Every day a plant is down costs costs OUC $110,000OUC $110,000

Unexpected outages Unexpected outages cost cost between $350,000 and between $350,000 and $600,000 per day$600,000 per day

Preventive maintenance Preventive maintenance discovered a cracked rotor discovered a cracked rotor blade which could have blade which could have destroyed a $27 million destroyed a $27 million piece of equipmentpiece of equipment

STRATEGIC IMPORTANCE OF MAINTENANCEAND RELIABILITY

Failure has far reaching effects Failure has far reaching effects on a firm’son a firm’s OperationOperation ReputationReputation ProfitabilityProfitability Dissatisfied customersDissatisfied customers Idle employeesIdle employees Profits becoming lossesProfits becoming losses Reduced value of Reduced value of

investment in plant and investment in plant and equipmentequipment

ReliabilityReliability1.1. Improving individual componentsImproving individual components

2.2. Providing redundancyProviding redundancy

MaintenanceMaintenance1.1. Implementing or improving preventive Implementing or improving preventive

maintenancemaintenance

2.2. Increasing repair capability or speedIncreasing repair capability or speed

IMPORTANT TACTICS

Employee InvolvementEmployee Involvement

Information sharingSkill trainingReward systemEmployee empowerment

Maintenance and Reliability Maintenance and Reliability ProceduresProcedures

Clean and lubricateMonitor and adjustMake minor repairKeep computerized records

ResultsResults

Reduced inventoryImproved qualityImproved capacityReputation for qualityContinuous improvementReduced variability

MAINTENANCE STRATEGY

RELIABILITY

Improving individual componentsImproving individual components

RRss = R = R11 x R x R22 x R x R33 x … x R x … x Rnn

wherewhere RR11 = reliability of component 1= reliability of component 1

RR22 = reliability of component 2= reliability of component 2

and so onand so on

RELIABILITY EXAMPLE

RRss

RR33

.99

RR22

.80

RR11

.90

Reliability of the process isReliability of the process is

RRss = R = R11 x R x R22 x R x R33 = .90 x .80 x .99 = .713 or 71.3%= .90 x .80 x .99 = .713 or 71.3%

PRODUCT FAILURE RATE

Basic unit of measure for reliabilityBasic unit of measure for reliability

FRFR((%%) ) = x = x 100%100%Number of failuresNumber of failures

Number of units testedNumber of units tested

FRFR((NN)) = =Number of failuresNumber of failures

Number of unit-hours of operating timeNumber of unit-hours of operating time

Mean time between failuresMean time between failures

MTBF =MTBF = 11FRFR((NN))

PRODUCT FAILURE RATE EXAMPLE

2020 air conditioning units designed for use in air conditioning units designed for use in NASA space shuttles operated for NASA space shuttles operated for 1,0001,000 hours hoursOne failed after One failed after 200 200 hours and one after hours and one after 600600 hours hours

FRFR((%%)) = (100%) = 10%= (100%) = 10%22

2020

FRFR((NN)) = = .000106 = = .000106 failure/unit hrfailure/unit hr2220,000 - 1,20020,000 - 1,200

MTBF MTBF = = 9,434 = = 9,434 hrshrs11.000106.000106

PRODUCT FAILURE RATE EXAMPLE

2020 air conditioning units designed for use in air conditioning units designed for use in NASA space shuttles operated for NASA space shuttles operated for 1,0001,000 hours hoursOne failed after One failed after 200 200 hours and one after hours and one after 600600 hours hours

FRFR((%%)) = (100%) = 10%= (100%) = 10%22

2020

FRFR((NN)) = = .000106 = = .000106 failure/unit hrfailure/unit hr2220,000 - 1,20020,000 - 1,200

MTBF MTBF = = 9,434 = = 9,434 hrshrs11.000106.000106

Failure rate per trip

FR = FR(N)(24 hrs)(6 days/trip)FR = (.000106)(24)(6)FR = .153 failures per trip

PROVIDING REDUNDANCY

Provide backup components to increase Provide backup components to increase reliabilityreliability

++ xx

Probability Probability of first of first

component component workingworking

Probability Probability of needing of needing

second second component component

Probability Probability of second of second

component component workingworking

(.8)(.8) ++ (.8)(.8) xx (1 - .8)(1 - .8)

= .8= .8 ++ .16 = .96.16 = .96

A redundant process is installed to support the A redundant process is installed to support the earlier example where Rearlier example where Rss = .713= .713

RR11

0.90

0.90

RR22

0.80

0.80

RR33

0.99

= [.9 + .9(1 - .9)] x [.8 + .8(1 - .8)] x .99= [.9 + .9(1 - .9)] x [.8 + .8(1 - .8)] x .99

= [.9 + (.9)(.1)] x [.8 + (.8)(.2)] x .99= [.9 + (.9)(.1)] x [.8 + (.8)(.2)] x .99

= .99 x .96 x .99 = .94= .99 x .96 x .99 = .94

Reliability has Reliability has increased increased

from from .713.713 to to .94.94

PROVIDING REDUNDANCY EXAMPLE

Two types of maintenanceTwo types of maintenance Preventive maintenance – routine Preventive maintenance – routine

inspection and servicing to keep inspection and servicing to keep facilities in good repairfacilities in good repair

Breakdown maintenance – emergency Breakdown maintenance – emergency or priority repairs on failed equipmentor priority repairs on failed equipment

MAINTENANCE

Output ReportsOutput Reports

Inventory and purchasing reports

Equipment parts list

Equipment history reports

Cost analysis (Actual vs. standard)

Work orders– Preventive

maintenance– Scheduled

downtime– Emergency

maintenance

Data entry– Work requests– Purchase

requests– Time reporting– Contract work

Data FilesData Files

Personnel data with skills, wages, etc.

Equipment file with parts list

Maintenanceand work order

schedule

Inventory of spare parts

Repair history file

COMPUTERIZED MAINTENANCE SYSTEM

MAINTENANCE COSTS

TRADITIONAL VIEW

Total Total costscosts

Breakdown Breakdown maintenance maintenance costscosts

Co

sts

Co

sts

Maintenance commitmentMaintenance commitment

Preventive Preventive maintenance maintenance costscosts

Optimal point (lowestOptimal point (lowestcost maintenance policy)cost maintenance policy)

Co

sts

Co

sts

Maintenance commitmentMaintenance commitmentOptimal point (lowestOptimal point (lowest

cost maintenance policy)cost maintenance policy)

Total Total costscosts

Full cost of Full cost of breakdownsbreakdowns

Preventive Preventive maintenance maintenance costscosts

MAINTENANCE COSTS

FULL COST VIEW

MAINTENANCE COST EXAMPLE

1.1. Compute the expected number of Compute the expected number of breakdownsbreakdowns

Number of Number of BreakdownsBreakdowns

FrequencyFrequency Number of Number of BreakdownsBreakdowns

FrequencyFrequency

00 2/20 = .12/20 = .1 22 6/20 = .36/20 = .3

11 8/20 = .48/20 = .4 33 4/20 = .24/20 = .2

∑∑ Number of Number of breakdownsbreakdowns

Expected number Expected number of breakdownsof breakdowns

Corresponding Corresponding frequencyfrequency== xx

= (0)(.1) + (1)(.4) + (2)(.3) + (3)(.2)= (0)(.1) + (1)(.4) + (2)(.3) + (3)(.2)

= 1.6= 1.6 breakdowns per month breakdowns per month

2.2. Compute the expected breakdown cost per month Compute the expected breakdown cost per month with no preventive maintenancewith no preventive maintenance

Expected Expected breakdown costbreakdown cost

Expected number Expected number of breakdownsof breakdowns

Cost per Cost per breakdownbreakdown== xx

= (1.6)($300)= (1.6)($300)

= $480= $480 per month per month

MAINTENANCE COST EXAMPLE

3.3. Compute the cost of preventive maintenanceCompute the cost of preventive maintenance

Preventive Preventive maintenance costmaintenance cost

Cost of expected Cost of expected breakdowns if service breakdowns if service contract signedcontract signed

Cost of Cost of service contractservice contract

==

++

= (1= (1 breakdown/month breakdown/month)($300) + $150)($300) + $150/month/month= $450= $450 per month per month

Hire the service firm; it is less expensive

MAINTENANCE COST EXAMPLE

1.1. Well-trained personnelWell-trained personnel

2.2. Adequate resourcesAdequate resources

3.3. Ability to establish repair plan Ability to establish repair plan and prioritiesand priorities

4.4. Ability and authority to do Ability and authority to do material planningmaterial planning

5.5. Ability to identify the cause of Ability to identify the cause of breakdownsbreakdowns

6.6. Ability to design ways to Ability to design ways to extend MTBFextend MTBF

INCREASING REPAIR CAPABILITIES

OperatorOperator Maintenance Maintenance departmentdepartment

Manufacturer’s Manufacturer’s field servicefield service

Depot serviceDepot service(return equipment)(return equipment)

Preventive maintenance costs less and is faster the more we move to the left

Competence is higher as we move to the right

HOW MAINTENANCE IS PERFORMED

SimulationSimulation Computer analysis of complex situationsComputer analysis of complex situations

Model maintenance programs before they Model maintenance programs before they are implementedare implemented

Physical models can also be usedPhysical models can also be used

Expert systemsExpert systems Computers help users identify problems Computers help users identify problems

and select course of actionand select course of action

ESTABLISHING MAINTENANCE POLICIES

THANK YOU!

koji.carlo.dioni.richard.earlBRAVEHEART