1 / 68

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

gloria-mendoza
Download Presentation

By Assoc. Prof. Dr. Ahmet ÖZTAŞ

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related