Operations Management

1. Operations Management Lecture 29 Learning Curves and Decision Making Tools PowerPoint presentation to accompany Heizer/Render Principles of Operations Management, 7e Operations Management, 9e

2. Recap • Financial Analysis of Projects • Time value of money • Payback Period • Net Present Value (NPV) • Benefit Cost Ratio (BCR) • Internal Rate of Return (IRR)

3. Outline • Learning Curves • Applying the Learning Curve • Arithmetic Approach • Logarithmic Approach • Learning-Curve Coefficient Approach • Strategic Implications of Learning Curves • Limitations of Learning Curves

4. Outline • The Decision Process in Operations • Fundamentals of Decision Making • Decision Tables and Decision Trees • Types of Decision-Making Environments • Decision Making Under Uncertainty • Decision Making Under Risk • Decision Making Under Certainty • Expected Value of Perfect Information (EVPI)

5. Learning Curves • Based on the premise that people and organizations become better at their tasks as the tasks are repeated • Time to produce a unit decreases as more units are produced

6. Studies have shown that human performance usually improves when a task is repeated. In general, performance improves by a fixed percent each time production doubles. More specifically, each time the output doubles, the worker hours per unit decrease to fixed percentage of their previous value. That percentage is called the learning rate. If an individual requires 10 minutes to accomplish a certain task the first time it is attempted and only 8 minutes the second time, that person is said to have an 80 percent learning rate. If output is doubled again from two to four, we would expect the fourth item to be produced in 8(.8) = 6.4 minutes

7. Cost/time per repetition 0 Number of repetitions (volume) Learning Curve Effect Figure E.1

8. Learning Curves T x Ln = Time required for the nth unit where T = unit cost or unit time of the first unit L = learning curve rate n = number of times T is doubled First unit takes 10 labor-hours 70% learning curve is present Fourth unit will require doubling twice — 1 to 2 to 4 Hours required for unit 4 = 10 x (.7)2 = 4.9 hours

9. Learning Curve Examples Table E.1

10. Learning Curve Examples Table E.1

11. Uses of Learning Curves Internal: labor forecasting, scheduling, establishing costs and budgets External: supply chain negotiations Strategic: evaluation of company and industry performance, including costs and pricing

12. Arithmetic Approach • Simplest approach • Labor cost declines at a constant rate, the learning rate, as production doubles

13. Logarithmic Approach Determine labor for any unit, TN , by TN = T1(Nb) where TN = time for the Nth unit T1 = hours to produce the first unit b = (log of the learning rate)/(log 2) = slope of the learning curve

14. Logarithmic Approach Determine labor for any unit, TN , by TN = T1(Nb) where TN = time for the Nth unit T1 = hours to produce the first unit b = (log of the learning rate)/(log 2) = slope of the learning curve Table E.2

15. Logarithmic Example Learning rate = 80% First unit took 100 hours TN = T1(Nb) T3 = (100 hours)(3b) = (100)(3log .8/log 2) = (100)(3–.322) = 70.2 labor hours

16. Coefficient Approach TN = T1C where TN = number of labor-hours required to produce the Nth unit T1 = number of labor-hours required to produce the first unit C = learning-curve coefficient found in Table E.3

17. 70% 85% Unit Number (N) Unit Time Total Time Unit Time Total Time 1 1.000 1.000 1.000 1.000 2 .700 1.700 .850 1.850 3 .568 2.268 .773 2.623 4 .490 2.758 .723 3.345 5 .437 3.195 .686 4.031 10 .306 4.932 .583 7.116 15 .248 6.274 .530 9.861 20 .214 7.407 .495 12.402 Learning-Curve Coefficients Table E.3

18. Coefficient Example First boat required 125,000 hours Labor cost = $40/hour Learning factor = 85% TN = T1C T4 = (125,000 hours)(.723) = 90,375 hours for the 4th boat 90,375 hours x $40/hour = $3,615,000 TN = T1C T4 = (125,000 hours)(3.345) = 418,125 hours for all four boats

19. 100,000 .773 = 129,366 hours Coefficient Example Third boat required 100,000 hours Learning factor = 85% New estimate for the first boat

20. Limitations of Learning Curves • Learning curves differ from company to company as well as industry to industry so estimates should be developed for each organization • Learning curves are often based on time estimates which must be accurate and should be reevaluated when appropriate

21. Limitations of Learning Curves • Any changes in personnel, design, or procedure can be expected to alter the learning curve • The culture of the workplace, resource availability, and changes in the process may alter the learning curve

22. The Decision Process in Operations Clearly define the problems and the factors that influence it Develop specific and measurable objectives Develop a model Evaluate each alternative solution Select the best alternative Implement the decision and set a timetable for completion

23. Fundamentals of Decision Making • Terms: • Alternative – a course of action or strategy that may be chosen by the decision maker • State of nature – an occurrence or a situation over which the decision maker has little or no control

24. Fundamentals of Decision Making • Symbols used in a decision tree: • – decision node from which one of several alternatives may be selected •  – a state-of-nature node out of which one state of nature will occur

25. A decision node A state of nature node Favorable market Unfavorable market Construct large plant Favorable market Construct small plant Unfavorable market Do nothing Decision Tree Example Figure A.1

26. State of Nature Alternatives Favorable Market Unfavorable Market Construct large plant $200,000 –$180,000 Construct small plant $100,000 –$ 20,000 Do nothing $ 0 $ 0 Decision Table Example Table A.1

27. Decision-Making Environments • Decision making under uncertainty • Complete uncertainty as to which state of nature may occur • Decision making under risk • Several states of nature may occur • Each has a probability of occurring • Decision making under certainty • State of nature is known

28. Uncertainty • Maximax • Find the alternative that maximizes the maximum outcome for every alternative • Pick the outcome with the maximum number • Highest possible gain • This is viewed as an optimistic approach

29. Uncertainty • Maximin • Find the alternative that maximizes the minimum outcome for every alternative • Pick the outcome with the minimum number • Least possible loss • This is viewed as a pessimistic approach

30. Uncertainty • Equally likely • Find the alternative with the highest average outcome • Pick the outcome with the maximum number • Assumes each state of nature is equally likely to occur

31. States of Nature Favorable Unfavorable Maximum Minimum Row Alternatives Market Market in Row in Row Average Constructlarge plant$200,000 -$180,000 $200,000 -$180,000 $10,000 Constructsmall plant$100,000 -$20,000 $100,000 -$20,000 $40,000 Do nothing$0 $0 $0 $0 $0 Maximax Maximin Equally likely Uncertainty Example Maximax choice is to construct a large plant Maximin choice is to do nothing Equally likely choice is to construct a small plant

32. Risk • Each possible state of nature has an assumed probability • States of nature are mutually exclusive • Probabilities must sum to 1 • Determine the expected monetary value (EMV) for each alternative

33. Expected Monetary Value

34. States of Nature Favorable Unfavorable Alternatives Market Market Construct large plant (A1) $200,000 -$180,000 Construct small plant (A2) $100,000 -$20,000 Do nothing (A3) $0 $0 Probabilities .50 .50 EMV Example Table A.3 EMV(A1) = (.5)($200,000) + (.5)(-$180,000) = $10,000 EMV(A2) = (.5)($100,000) + (.5)(-$20,000) = $40,000 EMV(A3) = (.5)($0) + (.5)($0) = $0

35. States of Nature Favorable Unfavorable Alternatives Market Market Construct large plant (A1) $200,000 -$180,000 Construct small plant (A2) $100,000 -$20,000 Do nothing (A3) $0 $0 Probabilities .50 .50 EMV Example Table A.3 EMV(A1) = (.5)($200,000) + (.5)(-$180,000) = $10,000 EMV(A2) = (.5)($100,000) + (.5)(-$20,000) = $40,000 EMV(A3) = (.5)($0) + (.5)($0) = $0 Best Option

36. Certainty • Is the cost of perfect information worth it? • Determine the expected value of perfect information (EVPI)

37. Expected value with perfect information Maximum EMV EVPI = – Expected Value of Perfect Information EVPI is the difference between the payoff under certainty and the payoff under risk

38. Expected value with perfect information = ($200,000)(.50) + ($0)(.50) = $100,000 EVPI Example The best outcome for the state of nature “favorable market” is “build a large facility” with a payoff of $200,000. The best outcome for “unfavorable” is “do nothing” with a payoff of $0.

39. Maximum EMV EVPI = EVwPI – EVPI Example The maximum EMV is $40,000, which is the expected outcome without perfect information. Thus: = $100,000 – $40,000 = $60,000 The most the company should pay for perfect information is $60,000

40. Decision Trees • Information in decision tables can be displayed as decision trees • A decision tree is a graphic display of the decision process that indicates decision alternatives, states of nature and their respective probabilities, and payoffs for each combination of decision alternative and state of nature • Appropriate for showing sequential decisions

41. Tri-products is trying to decide whether to make-or-buy an accessory item for one of their products. It is projected that this item will sell for $10 each. If the item is outsourced, there is virtually no cost other than the $6 per unit that they would pay their supplier. Internally, they have two choices. Process A requires an investment of $120,000 for design and equipment, but results in a $4 per unit cost. Process B requires only a $100,000 investment, but its per unit cost is $5. Regardless of whether the item is subcontracted or produced internally, there is a 50% chance that they will sell 50,000 units, and a 50% chance that they will sell 100,000 units. Draw the decision tree appropriate to the alternatives and outcomes stated. Using decision trees and EMV, what is their best choice? Illustration

43. A company manufactures specialty pollution-sensing devices for the offshore oil industry. One particular device has reached maturity, and the company is considering whether to replace it with a newer model. Technologies have not changed dramatically, so the new device would have similar functionality to the existing one, but would be smaller and lighter in weight. The firm's three choices are: keep the old model; design a replacement device with internal resources; and purchase a new design from a firm that is one of its suppliers. The market for these devices will be either "receptive" or "neutral" of the replacement model. The financial estimates are as follows: Keeping the old design will yield a profit of $6 million dollars. Designing the replacement internally will yield $10 million if the market is "receptive," but a $3 million loss if the market is "neutral." Acquiring the new design from the supplier will profit $4 million under "receptive," $1 million under "neutral." The company feels that the market has a 70 percent chance of being "receptive" and a 30 percent chance of being "neutral." Draw the appropriate decision tree. Calculate expected value for all courses of action. What action yields the highest expected value? illustration

44. The three expected monetary values are: Develop replacement internally: $10,000,000 x .7 -$3,000,000 x .3 = $6.1 million Purchase new design: $4,000,000 x .7 + $1,000,000 x .3 = $3.1 million Stay with current design: $6 million The company should choose the highest value, and develop a replacement product design with internal resources.