7 - Engineering Economic Analysis

Transcription

1 Construction Project Management (CE ) 7 - Engineering Economic Analysis Dr. Khaled Hyari Department of Civil Engineering Hashemite University Introduction Is any individual project worthwhile? Determine whether a proposed solution is financially viable (profitable or not)? Given a list of feasible projects, which one is the best? What is the best option between competing solutions to a specific problem? How does each project rank compared to the others on the list? 7-2 1

2 Engineering Economic Decision Analysis Every decision made costs money (an investment). If these decisions are not made wisely, money will be lost with prospects of bankruptcy Economic Analysis considers the economic viability of each and every investment project such that money is made, not lost Objective: Strive to secure the highest net dollar return on capital investments which is compatible with the risks incurred 7-3 Economic Decision Steps Problem (Need or Opportunity) Recognition and Definition Generation of Alternative Solutions Development of Cash Flows for Feasible Alternative Solutions and Information Gathering Economic Evaluation of Alternatives Selection and Implementation of Best Alternative Post-implementation Analysis and Evaluation 7-4 2

3 Example 1: Project Evaluation (Examples) Housing Project A Construction=10 months Cost = /month Sale Value=$150,000 Total Cost? Profit? Housing Project B Construction=20 months Cost=/month Sale Value=$280,000 Total Cost? Profit? Example 2: Project requires an investment of $1,000 today and returns $1,100 in one year Is it profitable to make such investment? 7-5 Time Value of Money Money has a time value A dollar received tomorrow in not equivalent to a dollar received today (The money have the same nominal quantity but the dollar does not have the same usefulness or buying power) Because of this value differential we cannot estimate benefits or costs simply by adding dollars amount that are realized in different periods 7-6 3

4 Time Value of Money II Not all projects can be described completely in terms of monetary costs and benefits. Social Benefits (e.g. Hospital, School) Intangible Benefits (e.g. new cafeteria to workers ) 7-7 Discount Rate, Interest Rate, and Minimum Attractive Rate of Return Discount Rate: Represents the way money now is worth more than the money later (Investment opportunities + Inflation). Specified as a rate - given as some percentage per year Assumed to be constant over the time 7-8 4

5 Discount Rate, Interest Rate, and Minimum Attractive Rate of Return II Interest Rate: Contractual arrangement between a borrower and a lender discount rate > interest rate Minimum Attractive Rate of Return (MARR) Minimum discount rate accepted by a person or organization to accept the risks of a determined project or product 7-9 Interest Formulas i = Effective interest rate per interest period (discount rate or MARR) n = Number of compounding periods P = Present sum of money F = Future sum of money A = End-of-period cash flow in a uniform series counting for n periods

6 Interest Formulas II Future Value FV 1 = 1 x (1+ i ) FV 1 = Future Value of $1 at the end of year 1 If i = 10% FV 1 = 1 x ( ) = 1.10 FV 2 = 1.10 x ( ) = 1.21 FV 2 = Future Value of $1 at the end of year 2 FV 2 = 1 (1 + i ) (1 + i ) = 1 (1 + i ) 2 In general the future value of $1 at the end of n years will be: FV n = 1 (1 + i ) n 7-11 Interest Formulas III Inflation of 1% day: Not compounding: 1 X 365 = 365%/year? FV = 1 X ( ) = 4.65 Compounding: FV = 1 X ( ) 365 FV = ,680% >>>>> 365 %

7 Single Payment Compound Amount Factor (F/P, i%, n) = (1 + i ) n F = P (1 + i) n Single Payment Present Worth Factor (P/F, i%, n) = 1/ [(1 + i ) n ] = 1/ (F/P, i%, n) P = F [1/(1 + i) n ] Interest Formulas IV F P 7-13 Interest Formulas V Uniform Series Sinking Fund Factor (A/F, i%, n) = i / [(1 + i ) n 1] A = F { i / [(1 + i) n 1] } Uniform Series Compound Amount Factor (F/A, i%, n) = [(1 + i ) n 1] / i = 1 / (A/F, i%, n) F = A { [(1 + i ) n 1] / i} F A A A A A

8 Interest Formulas VI Uniform Series Capital Recovery Factor (A/P, i%, n) = [ i (1 + i ) n ] / [(1 + i ) n 1] A = P { [ i (1 + i ) n ] / [(1 + i ) n 1] } Uniform Series Present Worth Factor (P/A, i%, n) = [(1 + i ) n 1] / [i (1 + i ) n ]= 1 / (A/P, i%, n) P = A { [(1 + i ) n 1] / [i (1 + i ) n ] } A A A A A P 7-15 (Example 3) $ 20,000 equipment expected to last 5 years $ 4,000 salvage value Minimum attractive rate of return = 15% What are the? A - Annual equivalent B - Present Equivalent

9 (Example 3) II 7-17 (Example 3) III P = -$ 20,000 + $ 4,000 (P/F, 15%, 5) = - $ 20,000 + $ 4,000 [1/ (1.15) 5 ] = -$18,011 A = -$ 20,000 (A/P, 15%, 5) + $ 4,000 (A/F, 15%, 5) = -$ 20,000 { [ 0.15 (1.15) 5 ] / [(1.15) 5 1] } + $ 4,000 { 0.15 / [(1.15) 5 1] } =-$5,

10 Comparison of Alternatives The first step is to lay out the estimated cash flows Sequence of benefits (returns) and costs (payments) over time Several Evaluation Criteria: Equivalent Worth Methods: Present Worth Method Annual Worth Method Future Worth Method Ratio Methods Internal Rate-of-Return 7-19 Comparison of Alternatives II Correct selection of the discount rate is fundamental. Its choice can easily change the ranking of projects Example: 2 equipment one needs human operator (initial cost, annual $4,200 for labor) the second is fully automated (initial cost $18,000, annual $3,000 for power). n=10 years. Is the $8,000 more in the initial investment worth the $1,200 annual savings?

11 Comparison of Alternatives III Net Present Value (i=5%) = -8, ,200 x [(1.05) 10-1) / (0.05 (1.05) 10 )] = 1,264 There is a critical value of i that change the equipment choice (around 8.5%) 7-21 Equivalent Worth Method Coverts all relevant cash flows into equivalent (comparable) amounts at the MARR

12 Present Worth Method 7-23 Machine A Present Worth Method $20,000 $4,000 $20,000 $4,

13 Machine A Present Worth Method $4, $4,000 $20,000 $20,000 NPV = -$20, $20,000 (P/F, 15%,5) + $4,000 (P/F, 15%,5) + $4,000 (P/F, 15%,10) + (P/A, 15%,10) = $1, Machine B Present Worth Method $30,

14 Machine B Present Worth Method $30,000 NPV = -$30,000 + (P/A, 15%,10) = -$2, Equivalent Worth Methods Only provides a good comparison between projects when they are strictly comparable in terms of investment or total budget Focus attention on quantity of money Net present value is different than profit (discount rate X interest rate) Give no indication of the scale of effort required to achieve the result

15 Internal Rate of Return Method (IRR) Involves the calculation of the interest rate that is compared against a minimum standard of desirability (MARR). It is the interest rate for which the Net Present Value of the project is ZERO (PV of inflow = PV of outflow) Eliminates the need to find a proper MARR Rankings cannot be manipulated by the choice of MARR 7-29 Internal Rate of Return Method (IRR) II 0(i%) = -20, ,600 (P/A, i%, 5) + 4,000 (P/F, i%, 5) I = % > 15% then the project is justified

16 Internal Rate of Return Method (IRR) III 7-31 Internal Rate of Return Method (IRR) III 0(i%) = -30, ,400 (P/A, i%, 10) = % < 15% then the project is not justified

17 Continuous Compounding and Discounting Variety of interest rates offered on deposits, banking, saving and loan associations Semi-annually Quarterly Daily The greater the frequency with which the interest is compounded, the higher the future value of the deposit 7-33 Continuous Compounding and Discounting II Example: Deposit $100 in a bank for one year at an annual rate of 6% 100 (1+0.06) = $ Compounded twice a year 100 (1+0.06/2)(1+0.06/2) = $ The higher the interest rate and the longer the duration, the greater is the impact of the compounding method

18 Nominal Interest Rate of Return r = the nominal interest rate per year M = the compounding frequency or the number of interest periods per year r/m = interest rate per compounding period Effective interest rate = the rate that truly represents the amount of interest earned in a year or some other time period i effective = (1 + r/m) M 1 i effective = effective annual interest rate 7-35 Example If a savings bank pays 1 ½% interest every three months, what are the nominal and effective interest rates per year? Nominal %/year, r = 1 1/2% x 4 = 6% Effective interest rate per year, i effective = ( /4) 4 1 = = 6.1% Notice that when M=1, i effective = r