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By Assoc. Prof. Dr. Ahmet ÖZTAŞ

CE 533 - ECONOMIC DECISION ANALYSIS IN CONSTRUCTION. By Assoc. Prof. Dr. Ahmet ÖZTAŞ. CHP I V - PRESENT WORTH ANALYSIS. Gaziantep University Department of Civil Engineering. CHP I V - PRESENT WORTH ANALYSIS. TOPICS. Formulating Alternatives PW of Equal-Life Alternatives

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By Assoc. Prof. Dr. Ahmet ÖZTAŞ

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

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

  3. 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 CE533 - PW Analysis

  4. 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. CE533 - PW Analysis

  5. 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 CE533 - PW Analysis

  6. 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. CE533 - PW Analysis

  7. 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) CE533 - PW Analysis

  8. 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! CE533 - PW Analysis

  9. 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 CE533 - PW Analysis

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

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

  12. 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! CE533 - PW Analysis

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

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

  15. 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! CE533 - PW Analysis

  16. 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! CE533 - PW Analysis

  17. 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: CE533 - PW Analysis

  18. 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. CE533 - PW Analysis

  19. 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. CE533 - PW Analysis

  20. 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 CE533 - PW Analysis

  21. 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. CE533 - PW Analysis

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

  23. 4.2 PRESENT WORTH: Example Consider: Machine AMachine B First Cost $2,500 $3,500 Annual Operating Cost 900 700 Salvage Value 200 350 Life 5 years 5 years i = 10% per year Which alternative should we select? CE533 - PW Analysis

  24. F5=$200 F5=$350 MA 0 1 2 3 4 5 0 1 2 3 4 5 A = $900 $2,500 MB A = $700 $3,500 4.2 PRESENT WORTH: Cash Flow Diagram Which alternative should we select? CE533 - PW Analysis

  25. 4.2 PRESENT WORTH: Solving • PA = 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 • PB = 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! CE533 - PW Analysis

  26. 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. CE533 - PW Analysis

  27. 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. CE533 - PW Analysis

  28. 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 CE533 - PW Analysis

  29. 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 CE533 - PW Analysis

  30. 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? CE533 - PW Analysis

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

  32. 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. CE533 - PW Analysis

  33. 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! CE533 - PW Analysis

  34. 4.3 PRESENT WORTH: UnequalLives • 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) CE533 - PW Analysis

  35. 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. CE533 - PW Analysis

  36. 4.3 PRESENT WORTH: Example Unequal Lives • EXAMPLE Machine A Machine B First Cost $11,000 $18,000 Annual Operating Cost 3,500 3,100 Salvage Value 1,000 2,000 Life 6 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 CE533 - PW Analysis

  37. 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! CE533 - PW Analysis

  38. Machine A F6=$1,000 0 1 2 3 4 5 6 A 1-6 =$3,500 F6=$2,000 $11,000 0 1 2 3 4 5 6 7 8 9 A 1-9 =$3,100 Machine B $18,000 4.3 PRESENT WORTH: Unequal Lives i = 15% per year LCM(6,9) = 18 year study period will apply for present worth CE533 - PW Analysis

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

  40. 4.3 Example: Unequal Lives Solving • LCM = 18 years • Calculate the present worth of a 6-year cycle for A PA = 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 CE533 - PW Analysis

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

  42. F6=$2,000 0 1 2 3 4 5 6 7 8 9 A 1-9 =$3,100 $18,000 4.3 Unequal Lives Example: Machine B • Calculate the Present Worth of a 9-year cycle for B CE533 - PW Analysis

  43. 4.3 9-Year Cycle for B Calculate the Present Worth of a 9-year cycle for B PB = 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 CE533 - PW Analysis

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

  45. 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 LCM = 35 yrs Alt-2: N= 7 yrs Could assume a study period of, say, 5 years. CE533 - PW Analysis

  46. 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 Now, evaluate both over 5 years using the PW method! Alt-2: N= 7 yrs CE533 - PW Analysis

  47. 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 CE533 - PW Analysis

  48. 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 CE533 - PW Analysis

  49. 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 CE533 - PW Analysis

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

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