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2006 Prentice Hall, Inc. 17 1

OperationsManagement

Chapter 17Maintenance andReliability

Presented by:Karla Andrea Barbacena &Risaline Cuaresma


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Learning Objectives

When you complete this chapter youshould be able to:

1. Describe how to improve systemreliability

2. Determine system reliability

3. Determine mean time between failure(MTBF)


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Learning Objectives

When you complete this chapter youshould be able to:

3. Distinguish between preventive andbreakdown maintenance

4. Describe how to improve maintenance

5. Compare preventive and breakdownmaintenance costs


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Outline

Global Company Profile: Orlando

Utilities Commission

The Strategic Importance of

Maintenance and Reliability

Important Tactics

Total Productive Maintenance

Techniques for EnhancingMaintenance


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Orlando Utilities

Commission Maintenance of power generating plants Every year each plant is taken off-line

for 1-3 weeks maintenance

Every three years each plant is takenoff-line for 6-8 weeks for completeoverhaul and turbine inspection

Each overhaul has 1,800 tasks andrequires 72,000 labor hours

OUC performs over 12,000 maintenancetasks each year


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Orlando Utilities

Commission Every day a plant is down costs OUC

$110,000

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

Preventive maintenance discovered acracked rotor blade which could have

destroyed a $27 million piece ofequipment


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Strategic Importance of

Maintenance and Reliability Failure has far reaching effects on a

firms

Operation

Reputation

Profitability

Dissatisfied customers

Idle employees Profits becoming losses

Reduced value of investment in plant andequipment



Maintenance and Reliability

The objective of maintenance andreliability is to maintain thecapability of the system whilecontrolling costs

Maintenance is all activities involvedin keeping a systems equipment in

working orderReliability is the probability that a

machine will function properly for aspecified time



Important Tactics

Reliability

1. Improving individual components

2. Providing redundancy

Maintenance

1. Implementing or improvingpreventive maintenance

2. Increasing repair capability or speed



Maintenance Strategy

Employee Involvement

Information sharingSkill trainingReward system

Employee empowerment

Maintenance and ReliabilityProcedures

Clean and lubricateMonitor and adjustMake minor repairKeep computerized records

Results

Reduced inventoryImproved qualityImproved capacityReputation for qualityContinuous improvement

Reduced variability

Figure 17.1



Reliability

Improving individual components

Rs= R1 x R2 x R3x x Rn

where R1 = reliability of component 1

R2 = reliability of component 2and so on



Overall System Reliability

Reliabilityofthesystem

(percent)

Average reliability of each component (percent)

| | | | | | | | |

100 99 98 97 96

100

80

60

40

20

0

Figure 17.2



Rs

R3

.99

R2

.80

Reliability Example

R1

.90

Reliability of the process is

Rs= R1 x R2 x R3= .90 x .80 x .99 = .713 or 71.3%



Rs

R3

.91

R2

.65

Reliability Exercises

R1

.75

Reliability of the process is

Rs = .443625 or 44.3%



Product Failure Rate (FR)

Basic unit of measure for reliability

FR(%) = x100%Number of failures

Number of units tested

FR(N) =Number of failures

Number of unit-hours of operating time

Mean time between failures

MTBF =1

FR(N)



Failure Rate Example

20 air conditioning units designed for use inNASA space shuttles operated for1,000 hoursOne failed after200 hours and one after600 hours

FR(%)= (100%) = 10%220

FR(N)= = .000106 failure/unit hr2

20,000 - 1,200

MTBF= = 9,434 hrs1

.000106



Failure Rate Example

20 air conditioning units designed for use in

NASA space shuttlesoperated for1,000 hoursOne failed after200 hours and one after600 hours

FR(%)= (100%) = 10%2

20

FR(N)= = .000106 failure/unit hr2

20,000 - 1,200

MTBF= = 9,434 hr1

.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 toincrease reliability

+ x

Probabilityof first

componentworking

Probabilityof needing

secondcomponent

Probabilityof second

componentworking

(.8) + (.8) x (1 - .8)

= .8 + .16 = .96


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Redundancy Example

A redundant process is installed to supportthe earlier example where Rs= .713

R1

0.90

0.90

R2

0.80

0.80

R3

0.99

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

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

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

Reliability hasincreased from

.713 to.94


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Maintenance

Two types of maintenance

Preventive maintenance

routine inspection and servicingto keep facilities in good repair

Breakdown maintenanceemergency or priority repairs on

failed equipment


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Implementing Preventive

Maintenance Need to know when a system requires

service or is likely to fail

High initial failure rates are known asinfant mortality

Once a product settles in, MTBF

generally follows a normal distribution Good reporting and record keeping can

aid the decision on when preventivemaintenance should be performed


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Computerized Maintenance

System

Figure 17.3

Output Reports

Inventory andpurchasing reports

Equipment

parts list

Equipmenthistory reports

Cost analysis(Actual vs. standard)

Work orders Preventive

maintenance Scheduled

downtime Emergency

maintenance

Data entry Work requests

Purchaserequests Time reporting Contract work

Data Files

Personnel datawith skills,wages, etc.

Equipment filewith parts list

Maintenance

and work orderschedule

Inventory ofspare parts

Repairhistory file


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Maintenance Costs

The traditional view attempted tobalance preventive and breakdownmaintenance costs

Typically this approach failed toconsider the true total cost ofbreakdowns

Inventory

Employee morale

Schedule unreliability


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Maintenance Costs

Figure 17.4 (a)

Totalcosts

Breakdownmaintenancecosts

Costs

Maintenance commitment

Traditional View

Preventivemaintenancecosts

Optimal point (lowestcost maintenance policy)


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Maintenance Costs

Figure 17.4 (b)

Costs

Maintenance commitment

Full Cost View

Optimal point (lowestcost maintenance policy)

Totalcosts

Full cost of

breakdowns

Preventivemaintenance

costs


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Maintenance Cost Example

Should the firm contract for maintenanceon their printers?

Number of

Breakdowns

Number of Months That

Breakdowns Occurred

0 2

1 8

2 6

3 4

Total: 20

Average cost of breakdown= $300


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1. Compute the expected number ofbreakdowns

Number of

Breakdowns

Frequency Number of

Breakdowns

Frequency

0 2/20 = .1 2 6/20 = .3

1 8/20 = .4 3 4/20 = .2

Number ofbreakdownsExpected numberof breakdowns Correspondingfrequency= x

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

= 1.6 breakdowns per month


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2. Compute the expected breakdown cost permonth with no preventive maintenance

Expectedbreakdown cost Expected numberof breakdowns Cost perbreakdown= x

= (1.6)($300)

= $480 per month


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3. Compute the cost of preventivemaintenance

Preventivemaintenance cost

Cost of expectedbreakdowns if servicecontract signed

Cost ofservice contract

=

+

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

= $450 per month

Hire the service firm; it is less expensive


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Increasing Repair

Capabilities1. Well-trained personnel

2. Adequate resources

3. Ability to establish repair plan andpriorities

4. Ability and authority to do material

planning5. Ability to identify the cause ofbreakdowns

6. Ability to design ways to extend MTBF


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How Maintenance isPerformed

Figure 17.5

OperatorMaintenancedepartment

Manufacturersfield service

Depot service(return equipment)

Preventivemaintenance costs less andis faster the more we move to the left

Competence is higher as wemove to the right


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Total Productive

Maintenance (TPM) Designing machines that are reliable, easy to

operate, and easy to maintain

Emphasizing total cost of ownership whenpurchasing machines, so that service andmaintenance are included in the cost

Developing preventive maintenance plans that

utilize the best practices of operators,maintenance departments, and depot service

Training workers to operate and maintain theirown machines


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Establishing MaintenancePolicies

Simulation

Computer analysis of complex

situationsModel maintenance programs before

they are implemented

Physical models can also be used

Expert systems

Computers help users identifyproblems and select course of action

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