1. By Assoc. Prof. Dr. Ahmet ÖZTAŞ Gaziantep University Department of Civil Engineering CHP IV-PRESENT WORTH ANALYSIS CE 533 - ECONOMIC DECISION ANALYSIS IN CONSTRUCTION

2. 2 / 68 CE533 - PW Analysis TOPICS Formulating Alternatives PW of Equal-Life Alternatives PW of Different-Life alternatives Future Worth Analysis Capitalized Cost Analysis Independent projects Payback Period CHP IV-PRESENT WORTH ANALYSIS

3. 3 / 68 CE533 - PW Analysis 4.1 FORMULATING MUTUALLY EXCLUSIVE ALTERNATIVES Viable firms/organizations have the capability to generate potential beneficial projects for potential investment Two types of investment categories Mutually Exclusive Set Independent Project Set

4. 4 / 68 CE533 - PW Analysis 4.1 FORMULATING MUTUALLY EXCLUSIVE ALTERNATIVES Mutually Exclusive set is where a candidate set of alternatives exist (more than one) Objective: Pick one and only one from the set. Once selected, the remaining alternatives are excluded.

5. 5 / 68 CE533 - PW Analysis 4.1 INDEPENDENT PROJECT SET Given a set of alternatives (more than one) The objective is to: Select the best possible combination of projects from the set that will optimize a given criteria. Subjects to constraints More difficult problem than the mutually exclusive approach

6. 6 / 68 CE533 - PW Analysis 4.1 FORMULATING MUTUALLY EXCLUSIVE ALTERNATIVES Mutually exclusive alternatives compete with each other. Independent alternatives may or may not compete with each other The independent project selection problem deals with constraints and may require a mathematical programming or bundling technique to evaluate.

7. 7 / 68 CE533 - PW Analysis 4.1 Type of Alternatives Alternative’s CF are classified as revenue-based or cost-based Revenue/Cost – the alternatives consist of cash inflow and cash outflows Select the alternative with the maximum economic value Service – the alternatives consist mainly of cost elements Select the alternative with the minimum economic value (min. cost alternative)

8. 8 / 68 CE533 - PW Analysis 4.1 Evaluating Alternatives Part of Engineering Economy is the selection and execution of the best alternative from among a set of feasible alternatives Alternatives must be generated from within the organization One of the roles of engineers!

9. 9 / 68 CE533 - PW Analysis 4.1 Evaluating Alternatives In part, the role of the engineer to properly evaluate alternatives from a technical and economic view Must generate a set of feasible alternatives to solve a specific problem/concern

10. 10 / 68 CE533 - PW Analysis 4.1 Alternatives Problem Do Nothing Alt. 1 Alt. 2 Alt. m Analysis Selection Execution If there are m investment proposals, we can form up to 2m mutually exclusive alternatives. This includes DN option.

11. 11 / 68 CE533 - PW Analysis 4.1 Alternatives: The Selected Alternative Problem Alt. Selected Execution Audit and Track Selection is dependent upon the data, life, discount rate, and assumptions made.

12. 12 / 68 CE533 - PW Analysis 4.2 Present Worth Approach Equal-Lifes Simple – Transform all of the current and future estimated cash flow back to a point in time (time t = 0) Have to have a discount rate before the analysis in started Result is in equivalent dollars now!

13. 13 / 68 CE533 - PW Analysis 4.2 THE PRESENT WORTH METHOD At an interest rate usually equal to or greater than the Organization’s established MARR. A process of obtaining the equivalent worth of future cash flows to some point in time – called the Present Worth

14. 14 / 68 CE533 - PW Analysis 4.2 THE PRESENT WORTH METHOD P(i%) = P(+) – P(-). P(i%) = P( + cash flows) + P( - cash flows) OR,...

15. 15 / 68 CE533 - PW Analysis 4.2 THE PRESENT WORTH METHOD If P(i%) > 0 then the project is deemed acceptable. If P(i%) < 0 the project is usually rejected. If P(i%) = 0 Present worth of costs = Present worth of revenues – Indifferent!

16. 16 / 68 CE533 - PW Analysis 4.2 THE PRESENT WORTH METHOD If the present worth of a project turns out to = “0,” that means the project earned exactly the discount rate that was used to discount the cash flows! The interest rate that causes a cash flow’s NPV to equal “0” is called the Rate of Return of the cash flow!

17. 17 / 68 CE533 - PW Analysis 4.2 THE PRESENT WORTH METHOD A positive present worth is a dollar amount of "profit" over the minimum amount required by the investors (owners). For P(i%) > 0, the following holds true:

18. 18 / 68 CE533 - PW Analysis 4.2 THE PRESENT WORTH METHOD – Depends upon the Discount Rate Used The present worth is purely a function of the MARR (the discount rate one uses). If one changes the discount rate, a different present worth will result.

19. 19 / 68 CE533 - PW Analysis 4.2 THE PRESENT WORTH METHOD For P(i%) > 0, the following holds true: Acceptance or rejection of a project is a function of the timing and magnitude of the project's cash flows, and the choice of the discount rate.

20. 20 / 68 CE533 - PW Analysis 4.2 PRESENT WORTH: Special Applications Present Worth of Equal Lived Alternatives Alternatives with unequal lives: Beware Capitalized Cost Analysis Require knowledge of the discount rate before we conduct the analysis

21. 21 / 68 CE533 - PW Analysis 4.2 PRESENT WORTH: Equal Lives Present Worth of Equal Lived Alternatives – straightforward Compute the Present Worth of each alternative and select the best, i.e., smallest if cost and largest if profit.

22. 22 / 68 CE533 - PW Analysis 4.2 Equal Lives – Straightforward! Given two or more alternatives with equal lives…. Alt. 1 Alt. 2 Alt. N N = for all alternatives Find PW(i%) for each alternative then compare

23. 23 / 68 CE533 - PW Analysis 4.2 PRESENT WORTH: Example Consider:Machine AMachine B First Cost $2,500 $3,500 Annual Operating Cost 900 700 Salvage Value 200 350 Life5 years5 years i = 10% per year Which alternative should we select?

24. 24 / 68 CE533 - PW Analysis 4.2 PRESENT WORTH: Cash Flow Diagram Which alternative should we select? 0 1 2 3 4 5 $2,500 A = $900 F 5 =$200 MAMA 0 1 2 3 4 5 $3,500 F 5 =$350 A = $700 MBMB

25. 25 / 68 CE533 - PW Analysis 4.2 PRESENT WORTH: Solving P A = 2,500 + 900 (P|A,.10, 5) – 200 (P|F,.01, 5) = 2,500 + 900 (3.7908) - 200 (.6209) = 2,500 + 3,411.72 - 124.18 = $5,788 P B = 3,500 + 700 (P|A,.10, 5) – 350 (P|F,.10, 5) = 3,500 + 2,653.56 - 217.31 = $5,936 SELECT MACHINE A: Lower PW cost!

26. 26 / 68 CE533 - PW Analysis 4.2 Present Worth of Bonds Often corporations or government obtain investment capital for projects by selling bonds. A good application of PW method is the evaluation of a bond purchase alternative. If PW < 0 at MARR, do-nothing alternative is selected. A bond is similar to an IOU for time periods such as 5, 10, 20 or more years.

27. 27 / 68 CE533 - PW Analysis 4.2 Present Worth of Bonds Each bond has a face value V of $100, $1000, $5000 or more that is fully returned to the purchaser when the bond maturity is reached. Additionaly, bond provide the purchaser with periodic interest payments I (bond dividends) using the bond coupon (interest) b, and c, the number of payment periods per year.

28. 28 / 68 CE533 - PW Analysis 4.2 Bonds – Notation and Example (Bond face value)(bond coupon rate) V.b I = ------------------------------------------ = ---- number of payments per year c At the time of purchase, the bond may sell for more or less than face value. Example: V = $5,000 (face value) b = 4.5% per year paid semiannually c = 10 years

29. 29 / 68 CE533 - PW Analysis 4.2 PW Bonds – Example – Continued The interest the firm would pay to the current bondholder is calculated as: The bondholder, buys the bond and will receive $112.50 every 6 months for the life of the bond

30. 30 / 68 CE533 - PW Analysis 4.2 Example 4.2 Ayşe has some extra money, requires safe investment. Her employer is offering to employees a generous 5% discount for 10- year 5000 YTL bonds tat carry a coupon rate of 6% paid semiannually. The expectation is to match her return on other safe investments, which have averaged 6.7% per year compounded semiannually. (This is an effective rate of 6.81% per year). Should Ayşe buy the bond?

31. 31 / 68 CE533 - PW Analysis 4.2 Example 4.2 – Cash-Flow Diagram A = 150 0 1 2 3 4 …. ….. 19 20 P=?? $5,000 i=3.35% Find the PW(3.35%) of the future cash flows to the potential bond buyer

32. 32 / 68 CE533 - PW Analysis 4.2 Example 4.2 – Solving I = (5000)(0.06)/2 = 150 YTL every 6 months for a total of n=20 dividend payments. The semiannual MARR is 6.7/2 = 3.35%, and the purchase price now is – 5000(0.95)= -4750 YTL. Using PW evaluation: PW = -4750 + 150(P/A, 3.35%, 20) + 5000(P/F, 3.35%, 20) = - 2.13 YTL <0 Effective rate is slightly less than 6.81% per year since PW<0. She sould not buy.

33. 33 / 68 CE533 - PW Analysis 4.3 Present Worth Analysis of Different-Life Alternatives In an analysis one cannot effectively compare the PW of one alternative with a study period different from another alternative that does not have the same study period. This is a basic rule!

34. 34 / 68 CE533 - PW Analysis 4.3 PRESENT WORTH: Unequal Lives If the alternatives have different lifes, there are 2 ways to use PW analysis to compare alternatives: A) The lowest common Multiple (LCM) B) Study period (planning horizon)

35. 35 / 68 CE533 - PW Analysis 4.3 PRESENT WORTH: Lowest Common Multiple (LCM) of Lives If the alternatives have different study periods, you find the lowest common life for all of the alternatives in question. Example: {3,4, and 6} years. The lowest common life (LCM) is 12 years. Evaluate all over 12 years for a PW analysis.

36. 36 / 68 CE533 - PW Analysis 4.3 PRESENT WORTH: Example Unequal Lives EXAMPLE Machine AMachine B First Cost$11,000$18,000 Annual Operating Cost 3,500 3,100 Salvage Value 1,000 2,000 Life6 years9 years i = 15% per year Note: Where costs dominate a problem it is customary to assign a positive value to cost and negative to inflows

37. 37 / 68 CE533 - PW Analysis 4.3 PRESENT WORTH: Example Unequal Lives A common mistake is to compute the present worth of the 6-year project and compare it to the present worth of the 9-year project. NO! NO! NO!

38. 38 / 68 CE533 - PW Analysis 4.3 PRESENT WORTH: Unequal Lives i = 15% per year 0 1 2 3 4 5 6 $11,000 F 6 =$1,000 A 1-6 =$3,500 Machine A 0 1 2 3 4 5 6 7 8 9 F 6 =$2,000 A 1-9 =$3,100 $18,000 Machine B LCM(6,9) = 18 year study period will apply for present worth

39. 39 / 68 CE533 - PW Analysis 4.3 Unequal Lives: 2 Alternatives i = 15% per year Machine A LCM(6,9) = 18 year study period will apply for present worth Cycle 1 for ACycle 2 for ACycle 3 for A 6 years Cycle 1 for B Cycle 2 for B 18 years 9 years Machine B

40. 40 / 68 CE533 - PW Analysis 4.3 Example: Unequal Lives Solving LCM = 18 years Calculate the present worth of a 6-year cycle for A P A = 11,000 + 3,500 (P|A,.15, 6) – 1,000 (P|F,.15, 6) = 11,000 + 3,500 (3.7845) – 1,000 (.4323) = $23,813, which occurs at time 0, 6 and 12

41. 41 / 68 CE533 - PW Analysis 4.3 Example: Unequal Lives P A = 23,813+23,813 (P|F,.15, 6)+ 23,813 (P|F,.15, 12) = 23,813 + 10,294 + 4,451 = 38,558 0 6 12 18 $23,813 Machine A

42. 42 / 68 CE533 - PW Analysis 4.3 Unequal Lives Example: Machine B Calculate the Present Worth of a 9-year cycle for B 0 1 2 3 4 5 6 7 8 9 F 6 =$2,000 A 1-9 =$3,100 $18,000

43. 43 / 68 CE533 - PW Analysis 4.3 9-Year Cycle for B Calculate the Present Worth of a 9-year cycle for B P B = 18,000+3,100(P|A,.15, 9) – 1,000(P|F,.15, 9) = 18,000 + 3,100(4.7716) - 1,000(.2843) = $32,508 which occurs at time 0 and 9

44. 44 / 68 CE533 - PW Analysis 4.3 Alternative B – 2 Cycles P B = 32,508 + 32,508 (P|F,.15, 9) = 32,508 + 32,508(.2843) P B = $41,750 Choose Machine A 0 9 18 $32,508 Machine A: PW =$38,558

45. 45 / 68 CE533 - PW Analysis 4.3 Unequal Lives – Assumed Study Period Study Period Approach Assume alternative: 1 with a 5-year life Alternative: 2 with a 7-year life Alt-1: N = 5 yrs Alt-2: N= 7 yrs LCM = 35 yrs Could assume a study period of, say, 5 years.

46. 46 / 68 CE533 - PW Analysis 4.3 Unequal Lives – Assumed Study Period Assume a 5-yr. Study period Estimate a salvage value for the 7-year project at the end of t = 5 Truncate the 7-yr project to 5 years Alt-1: N = 5 yrs Alt-2: N= 7 yrs Now, evaluate both over 5 years using the PW method!

47. 47 / 68 CE533 - PW Analysis FUTURE WORTH APPROACH FW(i%) is an extension of the present worth method Compound all cash flows forward in time to some specified time period using (F/P), (F/A),… factors or, Given P, the F = P(1+i) N

48. 48 / 68 CE533 - PW Analysis Applications of Future Worth Projects that do not come on line until the end of the investment period Commercial Buildings Marine Vessels Power Generation Facilities Public Works Projects Key – long time periods involving construction activities

49. 49 / 68 CE533 - PW Analysis Life Cycle Costs (LCC) Extension of the Present Worth method Used for projects over their entire life span where cost estimates are employed Used for: Buildings (new construction or purchase) New Product Lines Commercial aircraft New automobile models Defense systems

50. 50 / 68 CE533 - PW Analysis Life Cycle: Two General Phases TIME Cost-$ Acquisition PhaseOperation Phase Cumulative Life Cycle Costs

51. 51 / 68 CE533 - PW Analysis 4.4 CAPITALIZED COST ANALYSIS CAPITALIZED COST- the present worth of a project that lasts forever. Government Projects Roads, Dams, Bridges (projects that possess perpetual life) Infinite analysis period

52. 52 / 68 CE533 - PW Analysis 4.4 Derivation for Capitalized Cost Start with the closed form for the P/A factor Next, let N approach infinity and divide the numerator and denominator by (1+i) N

53. 53 / 68 CE533 - PW Analysis 4.4 Derivation - Continued Dividing by (1+i) N yields Now, let n approach infinity and the right hand side reduces to….

54. 54 / 68 CE533 - PW Analysis 4.4 Derivation - Continued Or, CC(i%) = A/i

55. 55 / 68 CE533 - PW Analysis 4.4 CAPITALIZED COST Assume you are called on to maintain a cemetery site forever if the interest rate = 4% and $50/year is required to maintain the site. Find the PW of an infinite annuity flow 1 2 3 4 5.. N=inf. A=$50/yr P = ? …………………..

56. 56 / 68 CE533 - PW Analysis 4.4 CAPITALIZED COST P 0 = $50[1/0.04] P 0 = $50[25] = $1,250.00 Invest $1,250 into an account that earns 4% per year will yield $50 of interest forever if the fund is not touched and the i-rate stays constant.

57. 57 / 68 CE533 - PW Analysis 4.4 CAPITALIZED COST: Endowments Assume a wealthy donor wants to endow a chair in an engineering department. The fund should supply the department with $200,000 per year for a deserving faculty member. How much will the donor have to come up with to fund this chair if the interest rate = 8%/yr.

58. 58 / 68 CE533 - PW Analysis 4.4 CAPITALIZED COST: Endowed Chair The department needs $200,000 per year. P = $200,000/0.08 = $2,500,000 If $2,500,000 is invested at 8% then the interest per year = $200,000 The $200,000 is transferred to the department, but the principal sum stays in the investment to continue to generate the required $200,000

59. 59 / 68 CE533 - PW Analysis EXAMPLE Calculate the Capitalized Cost of a project that has an initial cost of $150,000. The annual operating cost is $8,000 for the first 4 years and $5000 thereafter. There is an recurring $15,000 maintenance cost each 15 years. Interest is 15% per year. 4.4 Capitalized Cost Example

60. 60 / 68 CE533 - PW Analysis $4,000 0 1 2 3 4 5 6 7 15 30 ……… $150,000 $8,000 $15,000 “i”=15%/YR N=  How much $$ at t = 0 is required to fund this project? The capitalized cost is the total amount of $ at t = 0, when invested at the interest rate, will provide annual interest that covers the future needs of the project. 4.4 Cash Flow Diagram

61. 61 / 68 CE533 - PW Analysis 4.4 CAPITALIZED COST - Example Continued 1. Consider $4,000 of the $8,000 cost for the first four years to be a one-time cost, leaving a $4,000 annual operating cost forever. P 0 = 150,000 + 4,000 (P|A,.15, 4) = $161,420 2.855

62. 62 / 68 CE533 - PW Analysis 4.4 CAPITALIZED COST - Continued Recurring annual cost is $4,000 plus the equivalent annual of the 15,000 end-of- cycle cost. ……. 0 15 30 45 60 …….. Take any 15-year period and find the equivalent annuity for that period using the F/A factor.

63. 63 / 68 CE533 - PW Analysis 4.4 CAPITALIZED COST: One Cycle Take any 15-year period and find the equivalent annuity for that period using the F/A factor $15,000 A for a 15-year period 0 15 30 45 60 …….. …….

64. 64 / 68 CE533 - PW Analysis 4.4 CAPITALIZED COST 2. Recurring annual cost is $4,000 plus the equivalent annual of the 15,000 end-of-cycle cost. A= 4,000 + 15,000 (A|F,.15, 15) = 4,000 + 15000 (.0210) = $5,315 Recurring costs = $5,315/i = 5,315/0.15 =$3,443/yr

65. 65 / 68 CE533 - PW Analysis 4.4 CAPITALIZED COST Capitalized Cost = 161,420 + 5315/.15 = $196,853 Thus, if one invests $196,853 at time t = 0, then the interest at 15% will supply the end-of-year cash flow to fund the project so long as the principal sum is not reduced or the interest rate changes (drops).

66. 66 / 68 CE533 - PW Analysis 4.6 USING EXCEL FOR PW ANALYSIS General format to determine the PW is; PW = P – PV (i%,n,A,F) When different-life alternatives are evaluated using LCM; develop NPV function PW = P + NPV(i%,year_1_CF_cell:last_year_CF_cell)

67. 67 / 68 CE533 - PW Analysis Summary: Present Worth PW represents a family of methods Annual worth Future Worth Capitalized Cost Life-cycle cost analysis – application Bond Problems – application

68. 68 / 68 CE533 - PW Analysis End of Chapter 4