1. Class # 7 Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 5-1

2. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 5-2 Developed By: Dr. Don Smith, P.E. Department of Industrial Engineering Texas A&M University College Station, Texas Executive Summary Version Chapter 5 Present Worth Analysis

3. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 5-3 LEARNING OBJECTIVES 1.Formulating alternatives 2.PW of equal-life alternatives 3.PW of different-life alternatives 4.Future worth analysis 5.Capitalized Cost 6.Payback period

4. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 5-4 Sct. 5. 1 Formulating Mutually Exclusive Alternatives  One of the important functions of financial management and engineering is the creation of “alternatives”  If there are no alternatives to consider then there really is no problem to solve!  Given a set of “feasible” alternatives, engineering economy attempts to identify the “best” economic approach to a given problem

5. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 5-5 Types of Economic Proposals  Mutually exclusive alternatives  From a set of feasible alternatives, pick one and only one to execute  Mutually exclusive alternatives compete against each other  Independent projects  From a set of feasible alternatives select as many as can be funded in the current period  The “Do Nothing” (DN) alternative should always be considered

6. Revenue: Alternatives include estimates of costs (cash outflows) and revenues (cash inflows) Two types of cash flow estimates Cost: Alternatives include only costs; revenues and savings assumed equal for all alternatives; also called service alternatives Formulating Alternatives © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

7. Convert all cash flows to PW using MARR Precede costs by minus sign; receipts by plus sign For mutually exclusive alternatives, select one with numerically largest PW For independent projects, select all with PW > 0 PW Analysis of Alternatives © 2008 McGraw-Hill All rights reserved For one project, if PW > 0, it is justified EVALUATION Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

8. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 5-8 Present Worth Approach: Mutually Exclusive Projects  Alternatives must be of equal service, that is, evaluate all alternatives over the same number of years.  For a single project, project is financially viable if PW  0 at the MARR  For 2 or more alternatives, select the one with the (numerically) larger PW value  Examples: PW 1 PW 2 Select $-1500 $-500 Alt 2 +2500 -500 Alt 1 -1200 +25 Alt 2

9. 9 5.2 Example: Three Alternatives  Assume i = 10% per year A1 Electric Power First Cost: -2500 Ann. Op. Cost: -900 Sal. Value: +200 Life: 5 years A2 Gas Power First Cost: -3500 Ann. Op. Cost: -700 Sal. Value: +350 Life: 5 years A3 Solor Power First Cost: -6000 Ann. Op. Cost: -50 Sal. Value: +100 Life: 5 years Which Alternative – if any, Should be selected based upon a present worth analysis?

10. 10 5.2 Example: Cash Flow Diagrams 0 1 2 3 4 5 -2500 -3500 -6000 A = -900/Yr. A = -700/Yr. A = -50/Yr. F SV = 200 F SV = 350 F SV = 100 A 1 : Electric A 2 : Gas A 3 :Solar i = 10%/yr and n = 5

11. 11 5.2 Calculate the Present Worth's  Present Worth's are: 1.PW Elec. = -2500 - 900(P/A,10%,5) + 200(P/F,10%,5) = $-5788 2.PW Gas = -3500 - 700(P/A,10%,5) + 350(P/F,10%,5) = $-5936 3.PW Solar = -6000 - 50(P/A,10%,5) + 100(P/F,10%,5) = $-6127 Select “Electric” which has the min. PW Cost!

12. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 5-12 Sct 5.3 Present Worth Analysis of Different-Life Alternatives  For alternatives with un-equal lives the rule is: PW of the alternatives must be compared over the same number of years  Called “ Equal Service ” requirement Approach – 1 of 2 approaches  LCM -- Evaluate the alternatives over the lowest common multiple of lives, e.g., lives of 4 and 6, use n = 12 and assume re-investment at same cash low estimates  Study period -- assume a planning horizon and evaluate the alternatives over this number of years

13. 13 5.3 PW – LCM Example  Two Location Alternatives, A and B where one can lease one of two locations. Location ALocation B First cost, $ -15,000 -18,000 Annual lease cost, $ per year-3,500-3,100 Deposit return,$ 1,000 2,000 Lease term, years 6 9 Note: The lives are unequal. The LCM of 6 and 9 = 18 years!

14. 14 5.3 LCM Example: Required to Find:  Which option is preferred if the interest rate is 15%/year? Location ALocation B First cost, $ -15,000 -18,000 Annual lease cost, $ per year-3,500-3,100 Deposit return,$ 1,000 2,000 Lease term, years 6 9 For now, assume there is no study life constraint – so apply the LCM approach.

15. 15 5.3 LCM Example where “n” = 18 yrs.  The Cash Flow Diagrams are:

16. 16 5.3 Unequal Lives: 2 Alternatives i = 15% per year A LCM(6,9) = 18 year study period will apply for present worth Cycle 1 for ACycle 2 for ACycle 3 for A Cycle 1 for B Cycle 2 for B 18 years 6 years 9 years B

17. 17 5.3 LCM Example Present Worth's  Since the leases have different terms (lives), compare them over the LCM of 18 years.  For life cycles after the first, the first cost is repeated in year 0 of the new cycle, which is the last year of the previous cycle.  These are years 6 and 12 for location A and year 9 for B.  Calculate PW at 15% over 18 years.

18. 18 5.3 PW Calculation for A and B -18 yrs  PW A = -15,000 - 15,000(P/F,15%a,6) + 1000(P/F,15%,6)  - 15,000(P/F,15%,12) + 1000(P/F,15%,12) + 1000(P/F,15%,18) - 3500(P/A,15%,18)  = $-45,036  PW B = -18,000 - 18,000(P/F,15%,9) + 2000(P/F,15%,9) + 2000(P/F,15%,18) - 3100(P/A,15 %,18)  = $-41,384 Select “B”: Lowest PW Cost @ 15%

19. 19 5.3 Example Problem with a 5-yr SP  Assume a 5- year Study Period for both options: For a 5-year study period no cycle repeats are necessary. PW A = -15,000 - 3500(P/A,15%,5) + 1000(P/F,15%,5) = $-26,236 PW B = -18,000- 3100(P/A,15%,5) + 2000(P/F,15%,5) = $-27,397 Location A is now the better choice. Note: The assumptions made for the A and B alternatives! Do not expect the same result with a study period approach vs. the LCM approach!

20. Capitalized Cost (CC) Analysis CC refers to the present worth of a project with a very long life, that is, PW as n becomes infinite Basic equation is: CC = P = A i For example, in order to be able to withdraw $50,000 per year forever at i = 10% per year, the amount of capital required is 50,000/0.10 = $500,000 © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

21. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 5-21 P/A Factor as “n” approaches infinity  The P/A factor is:  On the right hand side, divide both numerator and denominator by (1+i) n  If “n” approaches  the above reduces to:  Let CC represent “capitalized cost”  Annual amount forever is A = Pi = (CC)i

22. © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 Capitalized Cost (CC) Analysis

23. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 5-23 Sct 5.6 Payback Period Analysis  Also call payout analysis  Extension of present worth method  Two forms: 1. With no interest -- i = 0% (no-return payback) 2. With an assumed interest rate -- i > 0% (discounted payback analysis)  Technique: estimates the time n p to recover the initial investment in a project, either with or without interest earned

24. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 5-24 Payback Period Analysis-RULE  Never use payback analysis as the primary means of making an accept/reject decision on an alternative!  Best used as a screening technique or preliminary analysis tool  Historically, this method was a primary analysis tool and often resulted in incorrect selections  To apply, the cash flows must have at least one (+) cash flow in the sequence

25. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 5-25 Basic Formula for Payback Analysis  Determine the number of years n p that it takes for all negative net cash flows to exactly equal all positive cash flows  If i = 0% and all NCF estimates are the same, the payback calculation simplifies to n p = P/NCF

26. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 5-26 Payback - Interpretations and Fallacies  Common managerial philosophy is that a shorter payback is preferred to a longer payback period; this is not a good approach from economic vantage  Not a preferred method for final decision making – better as a screening tool  Ignores all cash flows after the payback time period  May not use all of the cash flows in the cash flow sequence.  Refer to Example 5.7 Don’t use no-return (i = 0%) payback for final decisions. It  Neglects any required return on the investment  Neglects all cash flows after time n p including any positive cash flows that contribute to making a positive return on the investment

27. 27 5.6 Example 1 for Payback  Consider a 5-year cash flow as shown. At a 0% interest rate, how long does it take to recover (pay back) this investment?

28. 28 5.6 Payback Example: 0% Interest  Form the Cumulative Cash Flow Amounts. Compute the cumulative Cash Flow amounts as:

29. 29 5.6 Payback Example: 0% Interest  Form the Cumulative Cash Flow Amounts. Note: The cumulative cash flow Amounts go from (-) to (+) Between years 3 and 4.

30. 30 5.6 Example: 3 ‹ n p ‹ 4  Payback is between 3 and 4 years.  Get “fancy” and interpolate as: Cumulative Cash Flow Amts. Payback Period = 3.375 years (really 4 years!)

31. 31 5.6 Same Problem: Set i = 5%  Perform Discounted Payback at i = 5%.  Cannot simply use the cumulative cash flow.  Have to form the discounted cash flow cumulative amount and look for the first sigh change from (-) to (+).  Example calculations follow.

32. 32 5.6 %-Year Example at 5%  Discounted Payback at 10%: tabulations: P/F,10%, t Factor CF t (P/F,i%,tAccumulated Sum Locate the time periods where the first sign change From (-) to (+) occurs!

33. 33 5.6 %-Year Example at 5%  PB is between 4 and 5 years at 5% P/F,10%, t Factor CF t (P/F,i%,tAccumulated Sum Locate the time periods where the first sign change From (-) to (+) occurs!

34. 34 5.6 Payback Period at 10%  Interpolation of PB time at 5%. CF(t)Accum. DC. CF At 10%, the Payback Time Period approx. 4.76 yrs

35. Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 5-35 Chapter 5 End of Slide Set