chap17 pump reliability

7/29/2019 Chap17 Pump Reliability

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Operat ionsManagement

Chapter 17Maintenance and Reliability

2006 Prentice Hall, Inc.

PowerPoint presentation to accompanyHeizer/RenderPrinciples of Operations Management, 6eOperations Management, 8e


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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 and

equipment


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Maintenance and Reliability

The objective of maintenance andreliability is to maintain the capability

of the system while controlling costsMaintenance is all activities involved in

keeping a systems equipment inworking order

Reliability is the probability that amachine will function properly for aspecified time


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Important Tactics

Reliability

1. Improving individual components

2. Providing redundancy

Maintenance

1. Implementing or improving preventivemaintenance

2. Increasing repair capability or speed


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Reliability for a Series

Improving individual components

Rs= R1 x R2 x R3x x Rn

where R1 =reliability of component 1

R2 =reliability of component 2and so on


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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%


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Product Failure Rate (FR)

Basic unit of measure for reliability

FR(% ) = x100%Number of fai lures

Number of un i ts tested

FR(N) =Number of fai lures

Number of uni t -hours of operat ing t ime

Mean t ime between fai lures

MTBF =1

FR(N)


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Failure Rate Example

20 air conditioning units designed for use inNASA space shuttles operated for 1,000 hoursOne failed after 200 hours and one after 600 hours

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

FR(N)= = .000106 failure/un it h r2

20,000 - 1,200

MTBF= = 9,434 hrs1

.000106


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Providing Redundancy

Provide backup components to increasereliability parallel components

+ xProbabi l i tyof f i rstcomponentwork ing

Probabi l i tyof n eedingsecond

component

Probabi l i tyof secondcomponentwork ing

(.8) + (.8) x (1 - .8)= .8 + .16 = .96

R1

0.80

0.80


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Redundancy Example

A redundant process is installed to support theearlier 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

Reliabi l i ty has

inc reased from

.713 to.94


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Maintenance

Two types of maintenance

Preventive maintenance routine

inspection and servicing to keepfacilities 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 as infantmortality

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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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 Cost Example

CPA firm specializing in payroll prep

Past 20 months had following printer

breakdown rate - shown on next slide Each time breakdown, loses an

average of $300

Can hire PM firm for $150/month Still expect to experience one

breakdown per month

Should they hire PM firm?


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Should the firm contract for maintenance ontheir 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 num berof breakdown s Correspondingfrequency= x

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

= 1.6 breakdow ns per mon th


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

Expectedbreakdown co st

Expected num berof breakdown s

Cost perbreakdown= x

= (1.6)($300)

= $480 per month


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3. Compute the cost of preventivemaintenance

Prevent ivemaintenance cos t

Cost of expectedbreakdown s i f serv icecon tract sign ed

Cost ofservice con tract

=

+

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

= $450 per month

Hire the service f irm ; i t is less expensive

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