1. Maintenance and Reliability17
Maintenance and Reliability
PowerPoint presentation to accompany
Heizer and Render
Operations Management, 10e
Principles of Operations Management, 8e
PowerPoint slides by Jeff Heyl
Additional content from Gerry Cook
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

2. Outline The Strategic Importance of Maintenance and Reliability
Improving Individual Components
Providing Redundancy
Maintenance
Implementing Preventive Maintenance
Total Productive Maintenance
Techniques for Enhancing Maintenance
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

3. Learning Objectives Describe how to improve system reliability
Determine system reliability
Determine mean time between failure (MTBF)
Distinguish between preventive and breakdown maintenance
Describe how to improve maintenance
Compare preventive and breakdown maintenance costs
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

4. 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 taken off-line for 6-8 weeks for complete overhaul and turbine inspection
Each overhaul has 1,800 tasks and requires 72,000 labor hours
OUC performs over 12,000 maintenance tasks each year
Every day a plant is down costs OUC $110,000
Unexpected outages cost between $350,000 and $600,000 per day
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

5. Strategic Importance of Maintenance and Reliability
Failure has far reaching effects on a firm’s
Operation
Reputation
Profitability
Dissatisfied customers
Idle employees
Profits becoming losses
Reduced value of investment in plant and equipment
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

6. Maintenance and Reliability
Reliability is the probability that a machine will function properly for a specified time
Tactics to improve Reliability
Improving individual components
Providing redundancy
Maintenance: activities involved in keeping a system’s equipment in working order
Maintenance Tactics
Implementing or improving preventive maintenance
Increasing repair capability or speed
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

7. Reliability Improving individual components Rs = R1 x R2 x R3 x … x Rn
where R1 = reliability of component 1
R2 = reliability of component 2
and so on
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

8. Overall System Reliability
Reliability of the system (percent)
Average reliability of each component (percent)
| | | | | | | | |
100 –
80 –
60 –
40 –
20 –
0 –
n = 10
n = 1
n = 50
n = 100
n = 200
n = 300
n = 400
Figure 17.2
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

9. Reliability Example R1 .90 R2 .80 R3 .99 Rs
Reliability of the process is
Rs = R1 x R2 x R3 = .90 x .80 x .99 = .713 or 71.3%
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

10. Product Failure Rate (FR)
Basic unit of measure for reliability
FR(%) = x 100%
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)
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

11. Failure Rate Example 2 FR(%) = (100%) = 10% 20 2
20 air conditioning units designed for use in
NASA space shuttles operated for 1,000 hours
One failed after 200 hours and one after 600 hours
FR(%) = (100%) = 10% 2 20
FR(N) = = failure/unit hr 2
20, ,200
MTBF = = 9,434 hrs 1
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

12. Failure Rate Example 2 FR(%) = (100%) = 10% 20 2
20 air conditioning units designed for use in
NASA space shuttles operated for 1,000 hours
One failed after 200 hours and one after 600 hours
FR(%) = (100%) = 10% 2 20
FR(N) = = failure/unit hr 2
20, ,200
MTBF = = 9,434 hrs 1
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

13. Providing Redundancy Provide backup components to increase reliability+ x
Probability of first component working
Probability of needing second component
Probability of second component working
(.8) + x
(1 - .8) = +
= .96
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

14. Reliability has increased from .713 to .94
Redundancy Example
A redundant process is installed to support the earlier example where Rs = .713 R1
0.90 R2
0.80 R3
0.99
Reliability has increased from .713 to .94
= [ (1 - .9)] x [ (1 - .8)] x .99
= [.9 + (.9)(.1)] x [.8 + (.8)(.2)] x .99
= .99 x .96 x .99 = .94
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

15. Maintenance Two types of maintenance
Preventive maintenance – routine inspection and servicing to keep facilities in good repair
Breakdown maintenance – emergency or priority repairs on failed equipment
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

16. Maintenance Costs
The traditional view attempted to balance preventive and breakdown maintenance costs
Typically this approach failed to consider the true total cost of breakdowns
Inventory
Employee morale
Schedule unreliability
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

17. cost maintenance policy) cost maintenance policy)
Maintenance Costs
Total costs
Breakdown maintenance costs
Costs
Maintenance commitment
Traditional View
Preventive maintenance costs
Optimal point (lowest
cost maintenance policy)
Costs
Maintenance commitment
Full Cost View
Optimal point (lowest
cost maintenance policy)
Total costs
Full cost of breakdowns
Preventive maintenance costs
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

18. Maintenance Cost Example
Should the firm contract for maintenance on their printers?
Number of Breakdowns
Number of Months That Breakdowns Occurred 2 1 8 6 3 4
Total : 20
Average cost of breakdown = $300
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

19. Maintenance Cost Example
Compute the expected number of breakdowns
Number of Breakdowns
Frequency
2/20 = .1 2
6/20 = .3 1
8/20 = .4 3
4/20 = .2 ∑
Number of breakdowns
Expected number of breakdowns
Corresponding frequency = x
= (0)(.1) + (1)(.4) + (2)(.3) + (3)(.2)
= 1.6 breakdowns per month
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

20. Maintenance Cost Example
Compute the expected breakdown cost per month with no preventive maintenance
Expected breakdown cost
Expected number of breakdowns
Cost per breakdown = x
= (1.6)($300)
= $480 per month
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

21. Maintenance Cost Example
Compute the cost of preventive maintenance
Preventive maintenance cost
Cost of expected breakdowns if service contract signed
Cost of
service contract = +
= (1 breakdown/month)($300) + $150/month
= $450 per month
Hire the service firm; it is less expensive
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

22. Total Productive Maintenance (TPM)
Designing machines that are reliable, easy to operate, and easy to maintain
Emphasizing total cost of ownership when purchasing machines, so that service and maintenance are included in the cost
Developing preventive maintenance plans that utilize the best practices of operators, maintenance departments, and depot service
Training for autonomous maintenance so operators maintain their own machines and partner with maintenance personnel
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

23. Problems With Breakdown Maintenance
“Run it till it breaks”
Might be ok for low criticality equipment or redundant systems
Could be disastrous for mission-critical plant machinery or equipment
Not permissible for systems that could imperil life or limb (like aircraft)
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

24. Problems With Preventive Maintenance
“Fix it whether or not it is broken”
Scheduled replacement or adjustment of parts/equipment with a well-established service life
Typical example – plant relamping
Sometimes misapplied
Replacing old but still good bearings
Over-tightening electrical lugs in switchgear
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

25. Another Maintenance Strategy
Predictive maintenance – Using advanced technology to monitor equipment and predict failures
Using technology to detect and predict imminent equipment failure
Visual inspection and/or scheduled measurements of vibration, temperature, oil and water quality
Measurements are compared to a “healthy” baseline
Equipment that is trending towards failure can be scheduled for repair
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

26. Maintenance Strategy Comparison
Advantages
Disadvantages
Resources/ Technology Required
Application Example
Breakdown
No prior work required
Disruption of production, injury or death
May need labor/parts at odd hours
Office copier
Preventive
Work can be scheduled
Labor cost, may replace healthy components
Need to obtain labor/parts for repairs
Plant relamping, Machine lubrication
Predictive
Impending failures can be detected & work scheduled
Labor costs, costs for detection equipment and services
Vibration, IR analysis equipment or purchased services
Vibration and oil analysis of a large gearbox
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

27. In-Class Problems from the Lecture Guide Practice Problems
California Instruments, Inc., produces 3,000 computer chips per day. Three hundred are tested for a period of 500 operating hours each. During the test, six failed: two after 50 hours, two at 100 hours, one at 300 hours, and one at 400 hours.
Find FR(%) and FR(N).
FR(%) = failures per number tested = 6/300 = 0.02 = 2%
FR(N) = failures per operating time:
Total time = 300 * 500 = 150,000 hours
Downtime = 2(450) + 2(400) + 1(200) + 1(100) = 2,000 hours
Operating time = Total time – Downtime = 150,000 – 2,000 = 148,000
Therefore: FR(N) = 6/148,000 = failures/hour
MTBF = 1/FR(N) = 24,691 hours
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

28. In-Class Problems from the Lecture Guide Practice Problems
If 300 of these chips are used in building a mainframe computer, how many failures of the computer can be expected per month?
Converting the units of FR(N) to months:
FR(N) = * 24 hours/day * 30 days/month = failures/month 
FR(N) for the 300 units:
FR(N) = failures/month * 300 units = failures/month
 MTBF for the mainframe:
MTBF = 1/FR(N) = 1/8.75 = 0.11 month = 0.11 * 30 = 3.4 days
Calculation for MTBF assumes that failure of any one chip brings down entire system.
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

29. In-Class Problems from the Lecture Guide Practice Problems
 Find the reliability of this system:
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)

30. In-Class Problems from the Lecture Guide Practice Problems
Given the probabilities below, calculate the expected breakdown cost.
Assume a cost of $10 per breakdown.
Number of Breakdowns
Daily Frequency 3 1 2
Expected number of breakdowns = (0)(0.3) + (1)(0.2) + (2)(0.2) + (3)(0.3) = = 1.5 breakdowns/day
Expected breakdown cost = Expected number of breakdowns * Cost per breakdown = 1.5 * $10 = $15/day
Number of Breakdowns
Daily Frequency
Probability 3
0.3 1 2
0.2
10: Ch17 - Maintenance & Reliability(MGMT3102: Fall13)