1. Engineering Economics Contemporary Engineering Economics, 5 th edition, © 2010

2. Federal Express Nature of Project: Equip 40,000 carriers with Power Pads Save 10 seconds per pickup stop Investment cost: $150 million Expected savings: $20 million per year Federal Express

3. Ultimate Questions Is it worth investing $150 million to save $20 million per year, say over 10 years? How long does it take to recover the initial investment? What kind of interest rate should be used in evaluating business investment opportunities?

4. Mr. John Smith’s Investment Problem Built a hydroelectric plant using his personal savings of $800,000 Power generating capacity of 6 million kwhs Estimated annual power sales after taxes - $120,000 Expected service life of 50 years  Was Smith’s $800,000 investment a wise one?  How long does he have to wait to recover his initial investment, and will he ever make a profit?

5. Mr. John Smith’s Hydro Project

6. Question?? How long does it take to recover the investments made by Mr. John Smith from his hydroelectric project?

7. Bank Loan vs. Project Cash Flows Contemporary Engineering Economics, 5 th edition, © 2010

8. Example 5.1 Describing Project Cash Flows – A Computer-Process Control Project Year (n) Cash Inflows (Benefits) Cash Outflows (Costs) Net Cash Flows 00$650,000-$650,000 1215,50053,000162,500 2215,50053,000162,500 ………… 8215,50053,000162,500

9. Present Worth Analysis of Equal-life Alternatives $650,000 A = $215,500 A = $53,000 $650,000 A = $162,500 =

10. Practice Problem How long does it take to recover the initial investment for the computer process control system project in Example 5.1? Contemporary Engineering Economics, 5 th edition, © 2010

11. Example 5.3 Payback Period Contemporary Engineering Economics, 5 th edition, © 2010 NCash FlowCum. Flow 01234560123456 -$105,000+$20,000 $15,000 $25,000 $35,000 $45,000 $35,000 -$85,000 -$70,000 -$45,000 -$10,000 $35,000 $80,000 $115,000 Payback period should occurs somewhere between N = 3 and N = 4. Replacing an old machine (salvage value =$20,000) with a new machine $105,000 value.

12. Contemporary Engineering Economics, 5 th edition, © 2010 $85,000 $15,000 $25,000 $35,000 $45,000 $35,000 0 123456 Years Annual cash flow -100,000 -50,000 0 50,000 100,000 150,000 0123456 Years (n) 3.2 years Payback period Cumulative cash flow ($)

13. Payback Period Contemporary Engineering Economics, 5 th edition, © 2010  Principle: How fast can I recover my initial investment?  Method: Based on the cumulative cash flow (or accounting profit)  Screening Guideline: If the payback period is less than or equal to some specified bench-mark period, the project would be considered for further analysis.  Weakness: Does not consider the time value of money

14. Contemporary Engineering Economics, 5 th edition, © 2010 Discounted Payback Period Calculation Period (n) Cash Flow (A n ) Cost of Funds (15%)* Ending Cash Balance 0-$85,0000 115,000-$85,000(0.15) = -$12,750-82,750 225,000-$82,750(0.15) = -12,413-70,163 335,000-$70,163(0.15) = -10,524-45,687 445,000-$45,687(0.15) =-6,853-7,540 545,000-$7,540(0.15) = -1,13136,329 635,000$36,329(0.15) = 5,44976,778 * Cost of funds = (Unrecovered beginning balance) X (interest rate)

15. Illustration of Discounted Payback Period Contemporary Engineering Economics, 5 th edition, © 2010

16. Present Worth Analysis Notice that the payback was not using the time value money into consideration When we consider the time value money the techniques are called discounted cash flow such as net present worth PW or net present value PV The net present worth is the difference between the present value of the inflows and the present value of the outflows. The NPV value will help us decide if the investment is acceptable or not.

17. Evaluation of a Single Project Step1: determine interest rate this interest rate is often referred to as Minimum Attractive Rate of Return or MARR Step 2: Estimate the service life of the project Step 3: Estimate the cash inflow for each period over the service life Step 4: Estimate the cash outflow for each period over the service life Step 5: Determine the net cash flows for each period (net cash flow=cash inflow – cash outflow) Step 6: Find the present worth of each net cash flow at the MARR. The sum for periods present worth is defined as the present worth of the project Step 7: The positive PW means the project should be accepted, while if negative should be rejected

18. Net Present Worth Measure  Principle: Compute the equivalent net surplus at n = 0 for a given interest rate of i.  Decision Rule: Accept the project if the net surplus is positive. 2 3 4 5 0 1 Inflow Outflow 0 PW(i) inflow PW(i) outflow Net surplus PW(i) > 0

19. Comparing More than One Alternative The following guidelines should be considered when comparing more than one project: If you need to select the best alternative based on PW we select the highest PW as long as they have the same service lives Some alternatives are based on cost then we will accept the one with lowest PW of cost

20. Example - Tiger Machine Tool Company Net Present Worth – Uneven flows $75,000 $24,400 $27,340 $55,760 0 12 3 outflow inflow Considering acquiring a new metal-cutting machine

21. ABC 1PeriodCash Flow 20($75,000) 31$24,400 42$27,340 53$55,760 6 7PW(15%)$3,553.46 =NPV(15%,B3:B5)+B2 Excel Solution

22. -30 -20 -10 0 10 20 30 40 0 510152025303540 PW (i) ($ thousands) i = MARR (%) $3553 17.45% Break even interest rate (or rate of return) Accept Reject Present Worth Profile It is crucial to estimate the MARR correctly

23. Present Worth Amounts at Varying Interest Rates i (%)PW(i)i(%)PW(i) 0$32,50020-$3,412 227,74322-5,924 423,30924-8,296 619,16926-10,539 815,29628-12,662 1011,67030-14,673 128,27032-16,580 145,07734-18,360 162,07636-20,110 17.45*038-21,745 18-75140-23,302 *Break even interest rate Indicates that the investment project has a positive NPW if the interest rate is below 17.45%

24. What Factors Should the Company Consider in Selecting a MARR in Project Evaluation? Risk-free return Inflation factor Risk premiums For example if you were to invest $1,000 in risk-free US Treasury bills for a year, you would expect a real rate of return of 2%. Notice that you risk premium is zero but you have to add inflation effect which might be at 4% during that investment period. That is a total of 6% if I am dealing with internet stock then the risk premium will be around 20% that is a total of 26%

25. Guideline for Selecting a MARR Real Return2% Inflation4% Risk premium0% Total expected return 6% Real Return2% Inflation4% Risk premium20 % Total expected return 26 % Risk-free real return Inflation Risk premium U.S. Treasury Bills Amazon.com Very safe Very risky

26. Can you explain what PW really means? 1. Investment Pool Concept: It is defined as the firm’s treasury where all fund transactions are administrated and managed. If not used it will earn interest at the MARR 2. Project Balance Concept: Where the balance of the project is been tracked assuming that the money is been borrowed with interest (i)

27. Investment Pool Concept Suppose the company has $75,000. It has two options. (1)Take the money out and invest it in the project or (2) leave the money in the company. Let’s see what the consequences are for each option.

28. $75,000 0 1 2 3 $24,400 $27,340 $55,760 Investment pool How much would you have if the Investment is made? $24,400(F/P,15%,2) = $32,269 $27,340(F/P,15%,1) = $31,441 $55,760(F/P,15%,0) = $55,760 $119,470 How much would you have if the investment was not made? $75,000(F/P,15%,3) = $114,066 What is the net gain from the investment? $119,470 - $114,066 = $5,404 Project Return to investment pool N = 3 Meaning of Net Present Worth PW(15%) = $5,404(P/F,15%,3) = $3,553

29. Future Worth Criterion  Given: Cash flows and Minimum Attractive Rate of Return (MARR) (i)  Find: The net equivalent worth at the end of project life $75,000 $24,400 $27,340 $55,760 0 12 3 Project life

30. Future Worth Criterion

31. ABC 1PeriodCash Flow 20($75,000) 31$24,400 42$27,340 53$55,760 6PW(15%)$3553.46 7FW(15%)$5,404.38 =FV(15%,3,0,-B6) Excel Solution

32. How To Use Cash Flow Analyzer Cash Flow Input Fields Output or Analysis results Graphical Plots http://www.eng.auburn.edu/~park/cfa.html

33. Solving Example with Cash Flow Analyzer Net Present Worth Net Future Worth Payback period Project Cash Flows

34. Obtaining a NPW Plot Specify the Range of Interest Rate to plot NPW plot Between 0% and 100%

35. Project Balance Concept Suppose that the firm has no internal funds to finance the project, so will borrow the entire investment from a bank at an interest rate of 12%. Then, any proceeds from the project will be used to pay off the bank loan. Then, our interest is to see if how much money would be left over at the end of the project period.

36. Project Balance Concept N0123N0123N0123N0123 Beginning Balance Interest Payment Project Balance -$75,000 -$11,250 +$24,400 -$61,850 -$9,278 +$27,340 -$43,788 -$6,568 +$55,760 +$5,404 Net future worth, FW(15%) PW(15%) = $5,404 (P/F, 15%, 3) = $3,553

37. Project Balance Diagram 60,000 40,000 20,000 0 -20,000 -40,000 -60,000 -80,000 -100,000 -120,000 01230123 -$75,000 -$61,850 -$43,788 $5,404 Year(n) Terminal project balance (net future worth, or project surplus) Discounted payback period Project balance ($)

38. Example 5.8 Project’s Service Life is Extremely Long Contemporary Engineering Economics, 5 th edition, © 2010 Going back to Mr. John Smith’s Investment  Q1: Was Smith's $800,000 investment a wise one?  Q2: How long does he have to wait to recover his initial investment, and will he ever make a profit? Built a hydroelectric plant using his personal savings of $800,000 Power generating capacity of 6 million kwhs Estimated annual power sales after taxes - $120,000 Expected service life of 50 years

39. Mr. Smith’s Hydroelectric Project Contemporary Engineering Economics, 5 th edition, © 2010

40. How Would You Find P for a Perpetual Cash Flow Series, A? Contemporary Engineering Economics, 5 th edition, © 2010

41. Capitalized Equivalent Worth  Principle: PW for a project with an annual receipt of A over infinite service life  Equation: CE(i) = A(P/A, i, ) = A/i Contemporary Engineering Economics, 5 th edition, © 2010 A 0 P =CE(i)

42. It is a special case of the PW that is used when the life of the proposed project is perpetual or the planning horizon is long (example 40 years or more) such as dams, bridges, etc. The process of computing the PW for this infinite series is referred to as the capitalization of project cost. The cost is known as the capitalized cost Capitalized Equivalent Worth (CE)

43. Capitalized Cost: Represents the amount of money that must be invested today in order to yield a certain return A at the end of each and every period forever, assuming interest rate i Capitalized Equivalent Worth (CE)

44. Practice Problem Contemporary Engineering Economics, 5 th edition, © 2010 10 $1,000 $2,000 P = CE (10%) = ? 0 Given: i = 10%, N = ∞ Find: P or CE (10%) ∞

45. Solution: Contemporary Engineering Economics, 4 th edition, © 2007 10 $1,000 $2,000 P = CE (10%) = ? 0 ∞

46. A Bridge Construction Project Contemporary Engineering Economics, 5 th edition, © 2010  Construction cost = $2,000,000  Annual Maintenance cost = $50,000  Renovation cost = $500,000 every 15 years  Planning horizon = infinite period  Interest rate = 5%

47. Cash Flow Diagram for the Bridge Construction Project Contemporary Engineering Economics, 5 th edition, © 2010 $500,000 $2,000,000 $50,000 0 15 30 4560 Years

48.  Construction Cost P 1 = $2,000,000  Maintenance Costs P 2 = $50,000/0.05 = $1,000,000  Renovation Costs P 3 = $500,000(P/F, 5%, 15) + $500,000(P/F, 5%, 30) + $500,000(P/F, 5%, 45) + $500,000(P/F, 5%, 60). = {$500,000(A/F, 5%, 15)}/0.05 = $463,423  Total Present Worth P = P 1 + P 2 + P 3 = $3,463,423 Solution:

49. Alternate way to calculate P 3  Concept: Find the effective interest rate per payment period  Effective interest rate for a 15-year cycle i = (1 + 0.05) 15 - 1 = 107.893%  Capitalized equivalent worth P 3 = $500,000/1.0789 = $463,423 Contemporary Engineering Economics, 5 th edition, © 2010 15 30 4560 0 $500,000 Effective interest rate for a 15-year period

50. Comparing Mutually Exclusive Projects Contemporary Engineering Economics, 5 th edition, © 2010  Principle: Projects must be compared over an equal time span.  Rule of Thumb: If the required service period is given, the analysis period should be the same as the required service period.

51. Case 1  Analysis Period Equals Project Lives with different levels of investment  What to do: Compute the NPW for each project over its life and select the project with the largest NPW. Contemporary Engineering Economics, 5 th edition, © 2010

52. Comparing projects Assume that the unused funds will be invested at MARR. Contemporary Engineering Economics, 5 th edition, © 2010 This portion of investment will earn a 10% return on investment.

53. Case 2  Project Lives Longer than the Analysis Period  What to do:  Estimate the salvage value at the end of the required service period.  Compute the NPW for each project over the required service period.  PW(15%) A = -$362  PW(15%) B = -$364 Contemporary Engineering Economics, 5 th edition, © 2010

54. Case 3A – Service Projects  Project Lives Shorter than the Analysis Period (Automatic Mailing System)  What to do:  Come up with replacement projects that match or exceed the required service period.  Compute the NPW for each project over the required service period.  PW(15%) A = -$34,359  PW(15%) B = -$31,031 Contemporary Engineering Economics, 5 th edition, © 2010

55. Case 3B – Revenue Projects  Analysis Period Coincides with the Project with the Longest Life in the Mutually Exclusive Group (Family operated company found oil in the land they have the options to Drill or Lease)  What to do:  Compute the NPW of each project over its analysis period, assuming that no cash flows after the service life for the shorter-lived project.  PW(15%) Drill = $2,208,470  PW(15%) Lease = $2,180,210 Contemporary Engineering Economics, 5 th edition, © 2010

56. Case 4  Analysis Period Is Not Specified  What to do:  Come up with replacement projects that serve out the least common multiple period (LCM).  Compute the NPW for each project over the LCM.  PW(15%) A = -$53,657  PW(15%) B = -$48,534 Contemporary Engineering Economics, 5 th edition, © 2010

57. Summary Present worth is an equivalence method of analysis in which a project’s cash flows are discounted to a lump sum amount at present time. The MARR or minimum attractive rate of return is the interest rate at which a firm can always earn or borrow money. MARR is generally dictated by management and is the rate at which NPW analysis should be conducted. Two measures of investment, the net future worth and the capitalized equivalent worth, are variations to the NPW criterion. The term mutually exclusive means that, when one of several alternatives that meet the same need is selected, the others will be rejected.

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