1. LSM733-PRODUCTION OPERATIONS MANAGEMENT By: OSMAN BIN SAIF LECTURE 26 1

2.  JIT Schedulling  Kanban  Toyota Production System  Lean Operations Summary of last Session 2

3. Summary of last Session(Contd.) CHAPTER : Maintenance and Reliability Operations  Global Company Profile: Orlando Utilities Commission  The Strategic Importance of Maintenance and Reliability  Reliability  Improving Individual Components  Providing Redundancy 3

4. Summary of last Session(Contd.)  Maintenance  Implementing Preventive Maintenance  Increasing Repair Capabilities 4

5. Agenda for this Session  Total Productive Maintenance  Techniques for Enhancing Maintenance 5

6. CHAPTER : Decision Modeling  Decision Making & Models.  Decision Tables. – Decision making under uncertainty. – Decision making under risk. – Expected value of perfect information (EVPI). Agenda for this Session (Contd.) 6

7. 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 7

8. Maintenance Costs Figure 17.4 (a) Total costs Breakdown maintenance costs Costs Maintenance commitment Traditional View Preventive maintenance costs Optimal point (lowest cost maintenance policy) 8

9. Maintenance Costs Figure 17.4 (b) Costs Maintenance commitment Full Cost View Optimal point (lowest cost maintenance policy) Total costs Full cost of breakdowns Preventive maintenance costs 9

10. Maintenance Cost Example Should the firm contract for maintenance on their printers? Number of Breakdowns Number of Months That Breakdowns Occurred 02 18 26 3 4 Total: 20 Total: 20 Average cost of breakdown = $300 10

11. Maintenance Cost Example 1.Compute the expected number of breakdowns Number of Breakdowns Frequency Frequency 0 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 11

12. Maintenance Cost Example 2.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 12

13. Maintenance Cost Example 3.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 13

14. Increasing Repair Capabilities 1.Well-trained personnel 2.Adequate resources 3.Ability to establish repair plan and priorities 4.Ability and authority to do material planning 5.Ability to identify the cause of breakdowns 6.Ability to design ways to extend MTBF 14

15. How Maintenance is Performed Figure 17.5 Operator Maintenance department Manufacturer’s field service Depot service (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 15

16. 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 workers to operate and maintain their own machines 16

17. Establishing Maintenance Policies  Simulation  Computer analysis of complex situations  Model maintenance programs before they are implemented  Physical models can also be used  Expert systems  Computers help users identify problems and select course of action 17

18. CHAPTER : Decision Modeling 18

19. The Decision-Making Process ProblemDecision Quantitative Analysis Logic Historical Data Marketing Research Scientific Analysis Modeling Qualitative Analysis Emotions Intuition Personal Experience and Motivation Rumors 19

20. Models and Scientific Management Can Help Managers to Can Help Managers to: – Gain deeper insights into the business. – Make better decisions! Better assess alternative plans and actions. – Quantify, reduce and understand the uncertainty surrounding business plans and actions. 20

21. Steps to Good Decisions Define problem and influencing factors. Establish decision criteria. Select decision-making tool (model). Identify and evaluate alternatives using decision-making tool (model). Select best alternative. Implement decision. Evaluate the outcome. 21

22. Benefits of Models Allow better and faster decisions. Less expensive and disruptive than experimenting with the real world system. Allow managers to ask “What if…?” questions. Force a consistent and systematic approach to the analysis of problems. – Require managers to be specific about constraints and goals. 22

23. Limitations of Models May be expensive and time-consuming to develop and test. May be unused, misused or misunderstood (and feared!). – Due to mathematical and logical complexity. May downplay the value of qualitative information. May use assumptions that oversimplify the real world. 23

24. Decision Theory Terms: Alternative: Course of action or choice. Decision-maker chooses among alternatives. State of nature: An occurrence over which the decision maker has no control. 24

25. Decision Table States of Nature State 1State 2 Alternative 1Outcome 1Outcome 2 Alternative 2Outcome 3Outcome 4 A-25

26. A firm has two options for expanding production of a product: (1) construct a large plant; or (2) construct a small plant. Whether or not the firm expands, the future market for the product will be either favorable or unfavorable. If a large plant is constructed and the market is favorable, then the result is a profit of $200,000. If a large plant is constructed and the market is unfavorable, then the result is a loss of $180,000. If a small plant is constructed and the market is favorable, then the result is a profit of $100,000. If a small plant is constructed and the market is unfavorable, then the result is a loss of $20,000. Of course, the firm may also choose to “do nothing”, which produces no profit or loss. Example - Decision Making Under Uncertainty 26

27. Example - Decision Making Under Uncertainty States of Nature Alternatives Favorable Market Unfavorable Market Construct large plant $200,000-$180,000 Construct small plant $100,000-$20,000 $0 Do nothing 27

28. Decision Making Under Uncertainty - Criteria Maximax - Choose alternative that maximizes the maximum outcome for every alternative (Optimistic criterion). Maximin - Choose alternative that maximizes the minimum outcome for every alternative (Pessimistic criterion). Expected Value - Choose alternative with the highest expected value. 28

29. Example - Maximax States of Nature Alternatives Favorable Market Unfavorable Market Construct large plant $200,000-$180,000 Construct small plant $100,000-$20,000 $0 Do nothing Maximax decision is to construct large plant. 29

30. Example - Maximin Minimum in Row -$180,000 -$20,000 $0 Maximin decision is to do nothing. (Maximum of minimums for each alternative) States of Nature Alternatives Favorable Market Unfavorable Market Construct large plant $200,000-$180,000 Construct small plant $100,000-$20,000 $0 Do nothing 30

31. Probabilistic decision situation. States of nature have probabilities of occurrence. Select alternative with largest expected value (EV). – EV = Average return for alternative if decision were repeated many times. Decision Making Under Risk 31

32. Expected Value Equation Probability of payoff EVAVPV VPVVPVVPV ii i i NN (() ()()() )= =  * = * + * ++ * 1 1122 Number of states of nature Value of Payoff Alternative i... N 32

33. Example - Expected Value Suppose: Probability of favorable market = 0.5 Probability of unfavorable market = 0.5 States of Nature Alternatives Favorable Market Unfavorable Market Construct large plant $200,000-$180,000 Construct small plant $100,000-$20,000 $0 Do nothing Expecte d Value $10,000 $40,000 $0 Decision is to “Construct small plant”. 33

34. Example - Expected Value Suppose: Probability of favorable market = 0.7 Probability of unfavorable market = 0.3 States of Nature Alternatives Favorable Market Unfavorable Market Construct large plant $200,000-$180,000 Construct small plant $100,000-$20,000 $0 Do nothing Expecte d Value $86,000 $64,000 $0 Now, decision is to “Construct large plant”. 34

35. Example - Expected Value Over what range of values for probability of favorable market is “Construct large plant” preferred? Solve for x: 380000x-180000 > 120000x-20000 States of Nature Alternatives Favorable Market Unfavorable Market Construct large plant $200,000-$180,000 Construct small plant $100,000-$20,000 $0 Do nothing Expected Value 380,000x - 180,000 120,000x - 20,000 35

36. Solve for x: 380000x - 180000 > 120000x - 20000 x > 0.6154 So, as long as probability of a favorable market exceeds 0.6154, then “Construct large plant”. Example - Expected Value Over what range of values for probability of favorable market is “Construct large plant” preferred? 36

37. Expected Value of Perfect Information (EVPI) EVPI EVPI places an upper bound on what one would pay for additional information. – EVPI is the maximum you should pay to learn the future. EVPI EVPI is the expected value under certainty (EVUC) minus the maximum EV. EVPI = EVUC - maximum EV 37

38. Expected Value Under Certainty (EVUC) )P(S * j    EVUC n j  where: P(S j ) = The probability of state of nature j. n = Number of states of nature. (Best outcome for the state of nature j) 38

39. Example - EVUC Best outcome for Favorable Market = $200,000 Best outcome for Unfavorable Market = $0 States of Nature Alternatives Favorable Market Unfavorable Market Construct large plant $200,000-$180,000 Construct small plant $100,000-$20,000 $0 Do nothing 39

40. Expected Value of Perfect Information EVPIEV EVPI = EVUC - max(EV) = ($200,000*0.50 + 0*0.50) - $40,000 = $60,000 Thus, you should be willing to pay up to $60,000 to learn whether the market will be favorable or not. Suppose: Probability of favorable market = 0.5 Probability of unfavorable market = 0.5 40

41. Expected Value of Perfect Information EVPIEV EVPI = EVUC - max(EV) = ($200,000*0.70 + 0*0.30) - $86,000 = $54,000 Now, you should be willing to pay up to $54,000 to learn whether the market will be favorable or not. Now suppose: Probability of favorable market = 0.7 Probability of unfavorable market = 0.3 41

42. Summary of the Session  Total Productive Maintenance  Techniques for Enhancing Maintenance 42

43. CHAPTER : Decision Modeling  Decision Making & Models.  Decision Tables. – Decision making under uncertainty. – Decision making under risk. – Expected value of perfect information (EVPI). Summary of the Session (Contd.) 43

44. THANK YOU 44