1. Lecture No. 41 Chapter 12 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010

2. Decision Tree Analysis A graphical tool for describing (1) the actions available to the decision-maker, (2) the events that can occur, and (3) the relationship between the actions and events. Contemporary Engineering Economics, 5th edition, © 2010

3. Constructing a Decision Tree A Company is considering marketing a new product. Once the product is introduced, there is a 70% chance of encountering a competitive product. Two options are available each situation.  Option 1 (with competitive product): Raise your price and see how your competitor responds. If the competitor raises price, your profit will be $60. If they lower the price, you will lose $20.  Option 2 (without competitive product): You still two options: raise your price or lower your price. The conditional profits associated with each event along with the likelihood of each event is shown in the decision tree. Contemporary Engineering Economics, 5th edition, © 2010

4. Rollback Procedure To analyze a decision tree, we begin at the end of the tree and work backward. For each chance node, we calculate the expected monetary value (EMV), and place it in the node to indicate that it is the expected value calculated over all branches emanating from that node. For each decision node, we select the one with the highest EMV (or minimum cost). Then those decision alternatives not selected are eliminated from further consideration. Contemporary Engineering Economics, 5th edition, © 2010

5. Making Sequential Investment Decisions Contemporary Engineering Economics, 5th edition, © 2010

6. Decision Rules  Market the new product.  Whether or not you encounter a competitive product, raise your price.  The expected monetary value associated with marketing the new product is $44. Contemporary Engineering Economics, 5th edition, © 2010

7. Practice Problem A company is considering the purchase of a new labor- saving machine. The machine’s cost will turn out to be $55 per day. Each hour of labor that is saved reduces costs by $5. However, there is some uncertainty over the number of hours that actually will be saved. It is judged that the hours of labor saved per day will be 10, 11, or 12, with probabilities of 0.10, 0.60, 0.30, respectively. Let us define “profit” as the excess of labor-cost savings over the machine cost. Contemporary Engineering Economics, 5th edition, © 2010

8. Construct a Decision Tree Contemporary Engineering Economics, 5th edition, © 2010 Invest Do not invest -$5 0 $5 0 0.10 0.60 0.30 10 11 12 $1 $1.0 EMV = $1.0 Decision: Purchase the equipment

9. Expected Value of Perfect Information (EVPI) What is EVPI? This is equivalent to asking yourself how much you can improve your decision if you had perfect information. Mathematical Relationship: EVPI = EPPI – EMV = EOL where EPPI (Expected profit with perfect information) is the expected profit you could obtain if you had perfect information, and EMV (Expected monetary value) is the expected profit you could obtain based on your own judgment. This is equivalent to expected opportunity loss (EOL). Contemporary Engineering Economics, 5th edition, © 2010

10. Expected Value of Perfect Information (EVPI) Contemporary Engineering Economics, 5th edition, © 2010 State of Nature Best Strategy Maximum Payoff Probability the State of Nature Occurs Expected Payoff or each State 10 11 12 Don’t Buy Indifferent Buy 005005 0.10 0.60 0.30 0 1.5  Expected Profit with Perfect Information (EPPI): (0.10)(0) + (0.60)(0) + (0.30)(5) = $1.5  Expected Value of Perfect Information (EVPI) = EPPI – EMV $1.5 - $1 = $0.5

11. Bill’s Decision Problem – $50,000 to Invest  Decision Problem:  Buying a highly speculative stock (d 1 ) with three potential levels of return – High (50%), Medium (9%), and Low (- 30%).  Buying a very safe U.S. Treasury bond (d 2 ) with a guaranteed 7.5% return.  Seek advice from an expert?  Seek professional advice before making the decision  Do not seek professional advice – do on his own. Contemporary Engineering Economics, 5th edition, © 2010

12. Decision Tree for Bill’s Investment Problem Contemporary Engineering Economics, 5th edition, © 2010

13. (c) 2001 Contemporary Engineering Economics13 EMV = $898 Or, prior optimal decision is Option 2 Option 1: 1)Period 0: (-$50,000 - $100) = -$50,100 Period 1: (+$75,000 - $100) - 0.20($24,800) =$69,940 PW(5%)=-$50,100 + $69,940 (P/F, 5%, 1) = $16,510 2)Period 0: (-$50,000 - $100)= -$50,100 Period 1: (+$54,500 - $100)- (0.20)($4,300) = $53,540 PW(5%) = -$50,100 + $53,540 (P/F, 5%, 1) = $890 3)Period 0: (-$50,000 - $100) = -$50,100 Period 1: (+$35,000 - $100) – (0.20)(-$14,800) = $37,940 PW(5%)= - $50,100 + $37,940 (P/F, 5%, 1) = -$13,967 Option 2: Period 0: (- $50,000 - $150) = -$50,150 Period 1: (+$53,750 - $150) = $53,600 PW (5%)= -$50,150 + $53,600 (P/F, 5%, 1) = $898 Evaluating Options in Bill’s Investment Problem

14. Expected Value of Perfect Information Contemporary Engineering Economics, 5th edition, © 2010 Potential Return Level Probability Decision Option Optimal Choice with Perfect Information Opportunity Loss Associated with Investing in Bonds Option1: Invest in Stock (Prior Optimal) Option 2: Invest in Bonds High (A)0.25$16,510$898Stock$15,612 Medium (B)0.40890898Bond0 Low(C)0.35-13,967898Bond0 EMV -$405 $898 $3,903 EPPI = (0.25)($16,510) + (0.40)($898) + (0.35)($898) = $4,801 EVPI = EPPI – EV = $4,801 - $898 = $3,903 EOL = (0.25)($15,612) + (0.40)(0) + (0.35)(0) = $3,903

15. Bill’s Investment Problem with an Option of Getting Professional Advice Updating Conditional Profit (or Loss) after Paying a Fee to the Expert (Fee = $200) Revised Decision Tree Contemporary Engineering Economics, 5th edition, © 2010

16. Conditional Probabilities of the Expert’s Prediction, Given a Potential Return on the Stock Contemporary Engineering Economics, 5th edition, © 2010 Given Level of Stock Performance What the Report Will Say High (A) Medium (B) Low (C) Favorable (F)0.800.650.20 Unfavorable (UF)0.200.350.80

17. Nature’s Tree: Conditional Probabilities and Joint Probabilities Nature’s Tree Joint & Marginal Probabilities Contemporary Engineering Economics, 5th edition, © 2010 P(A,F) = P(F|A)P(A) = (0.80)(0.25) = 0.20 P(A,UF|A)P(A) = (0.20)(0.25) = 0.05 P(B,F) = P(F|B)P(B) = (0.65)(0.40) = 0.26 P(B,UF) = P(UF|B)P(B) = (0.35)(0.40) = 0.14 P(F) = 0.20 + 0.26 + 0.07 = 0.53 P(UF) = 1 – P(F) = 1 – 0.53 = 0.47

18. Joint and Marginal Probabilities Contemporary Engineering Economics, 5th edition, © 2010 What the Report Will Say Joint Probabilities When Potential Level of Return is Given Favorable (F)Unfavorable (UF) Marginal Probabilities of Return Level High (A)0.200.050.25 Medium (B)0.260.140.40 Low (C)0.070.280.35 Marginal Probabilities 0.530.471.00

19. Determining Revised Probabilities Contemporary Engineering Economics, 5th edition, © 2010 P(A|F) = P(A,F)/P(F) = 0.20/0.53 = 0.38 P(B|F) = P(B,F)/P(F) = 0.26/0.53 = 0.49 P(C|F) = P(C,F)/P(F) = 0.07/0.53 = 0.13 P(A|UF) = P(A,UF)/P(UF) = 0.05/0.47 = 0.11 P(B|UF) + P(B,UF)/P(UF) = 0.14/0.47 = 0.30 P(C|UF) = P(C,UF)/P(UF) = 0.28/0.47 = 0.59

20. Decision Making after Having Imperfect Information Contemporary Engineering Economics, 5th edition, © 2010 - $6,319