CE 561 Lecture Notes. Engineering Economic Analysis. Set 2. Time value of money. Cash Flow Diagram. Interest. Inflation Opportunity cost

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1 CE 56 Lecture otes Set 2 Engineering Economic Analysis Time value of money Inflation Opportunity cost Cash Flow Diagram P A A PInvestment AYearly Return 0 o. of Years Interest Profit Motive MARR Public Project Opportunity Cost Stable Economy 5-8% Developing Countries 0-5%

2 Types of Compounding and Interest Rates Interest Compound Simple Discreet Continuous Fixed Variable Example of Compounding (Int. Rate 0%) Year Principal Interest Amt. at the end of the year P000 F *Compounding vs. Discounting ominal and Effective Rates Compounded interest rate for periods less than a year If nominal rate 6% per year, compounded every 3 months 4 periods at.5% per period F P Effective Rate 00% P 2

3 P00 F? P00 00 ( ) (.05) (.05) (.05) 06.4 F Effective Rate % Effective Rate SPCAF i/m,m - m umber of Periods in a Year Interest Equations: Equivalence P i/period F? Compounding At the end of the st period P+P i P(+ At the end of the 2 nd period P(+ + P(+i P(+ 2 At the end of the th period P(+ Single Payment Compound Amount Factor (SPCAF i, ) (+ 3

4 Discounting P? i/period F 0 2 P F/(+ Single Payment Present Worth Factor SPPWF i, ( + Examples. How much will be accumulated in a fund at the end of 25 years if $2,000 is invested now with 6% compounded annually? F 2000 SPCAF 6%, x ( +.06) x $ How much money should be invested now at 6% compounded annually so that $8584 can be received 25 years hence? P 8584 SPPWF 6%, x $2000 4

5 Interest Equations Three sets Uniform Periodic Sums Uniformly Increasing Periodic Sums-Linear Gradient Exponentially Increasing Periodic Sums- Geometric Uniform Series Compound Amount Factor A A A A F? F A 2 ( + + A( A( + + A Multiplying by ( + ( + A( + + A( A( + F [ ] Subtracting, Fi A ( + USCAF i, [( + ]/ i Example. How much money will be accumulated in a fund at the end of 25 years from now if $2000 is invested at the end of each year (interest 6% year)? F A x USCAF 6%, x ( +.06) x $09,729 5

6 Sinking Fund Deposit Factor A A? A A A F The amount of uniform series of end-of-period deposits for periods to provide F at the end of periods. A F ( + SFDF i,n i i ( + Capital Recovery Factor P A A? A A A Given P, find A F P + A F i P CRF ( /[ + i] ( ) i( + ( + i, i( + ( + Examples. A county expects that a bridge reconstruction project 6 years from now will require $35 million. What annual payment should be made to a fund with 6% rate? A F x SFDF 6%,6 35 x 0.06 (.06) 6-35 x $.36 million/year (0.97 million/year if interest rate 0%) 6

7 . If an agency borrows $20 million to build a toll road, what must be the prospective annual toll income for 20 years to pay back with 7% interest? A P x CRF 7%,20 20 x 0 6 x 0.07(.07) 20 (.07) x 0 6 x $.9 million Uniform Series Present Worth Factor Given A, find P Reciprocal of CRF USPWF i,n ( + i( + For i 9% Year SPCAF SPPWF CRF USPWF SFDF USCAF

8 Example A county decided to raise its share of a construction project through local resources. The Bd. of Commissioners decided on Jan., 989 to invest an assured annual income in a special fund starting Jan., 990 until the start of the project. The last investment will be on Jan., The fund was expected to earn an interest of 0% per annum and the county would be able to withdraw $0 million each year for 4 years as from Jan., 2005, the start of the project. How much is being deposited each year? $0m Jan A Present Worth 0 million (USPWF 0%,4 )(SPCAF 0%, ) On Jan., million Annual Amt (A) so that million is available at 2005 is: A (SFDF 0%,6 ) $970,000/yr for 6 years ( ) Present Worth of Periodic Payments in Perpetuity R R R R R R R PW + + K 2 ( + ( + R + + K 2 ( + ( + R ( ) + i R ( + Kα 8

9 Continuous Compounding infinite number of periods i nominal rate, m no. of periods F P + P + Lim α m ( i/ m) m i m + i F Pe m i i m i m i e Continuous Compound Amount Factor CCAF e i Effective Annual Interest Rate F P P e P P i ( ) P i e Computation of PW of Continuous Benefit/Cost Flows R R R 2 R 3 R n 0 n r rate of growth i interest Amounts compounded continuously Interest compounded continuously 9

10 Bring all future streams to the present R PW of R R R Re r R Re r /e i R 2 R e r R 2 Re 2r /e 2i r e e Take summation, Σ R + + i e e 2r 2i n( r i ) e R r i + + e e nr ni Computation of PW of Benefits/Costs logα Annual growth rate r Y α Future Annual Value Estimate (F) First Year Value Estimate (P) Y Period of the estimate logf logp Y r Assuming continuous compounding, ry F Pe e ry F / P α ry logα logα r Y logf logp logf /P r Y Y 0

11 Example If the current user benefit from a road improvement project is $5 million ( st yr) and the total project cost is $80 million, what must be the minimum rate of growth of benefit (continuously increasing)? Assume the project life is 5 years and interest rate of 5% per year. PW of Benefit must be at least equal to PW of cost PW of Benefit $80 million PW of Benefit st Yr Benefit *Present Worth Factor (PWF) ( r n e PWF r i ( r.05) 5 e 80 6 r.05 5 r > i Try r 6% (.06.05) 5 e minimum rate of growth 6%

12 Methods of Economic Analysis Equivalent Uniform AC Method Present Worth of Costs Equivalent Uniform Annual et Return et Present Value Benefit-Cost Ratio Internal Rate of Return Formulas Related SPCAF USCAF SPPWF SFDF CRF USPWF CRF SFDF + i Annual Cost Method AC P * CRFi,n T * SFDFi, n + K ( P T) * CRF + T * i K i, n + P Initial Cost T Salvage K Annual Maint. & Op. Cost 2

13 Example A 00,000 50,000 50,000 50,000 60,000 60,000 60, T40,000 i 5% B 80,00055,000 55,000 55,000 65,000 65,000 65, T30,000 AC A 00,000 CRF(5%,6) 40,000 SFDF(5%,6) + 50,000 USPWF(5%,3) + 60,000 USPWF(5%,3) SPPWF(5%,3) CRF(5%,6) 75,850 AC B 76,680 System A is economically more desirable Present Worth of Cost A B 000 K800/yr T00 4 yrs 800 K900/yr T0 4 yrs Present Worth of A (8% interest) SPPWF + 8%,4 800USPWF8%, ,576 3

14 PW of B USPWF 8%, Advantage of A ( ) $204 Choose A Equivalent Uniform et Annual Return A B C EAR Annual Income (Benefit) Annual Cost Initial Inv. (P) 25,000 70,000 85,000 Terminal Value (T) 30,000 50,000 70,000 Annual Maint. (K) 0,000 3,000 4,000 Annual Benefit (R) 60,000 75,000 85,000 i 0%, Proj. Pd 2 yrs EAR A R [ P * CRF( 0%,2) + K T * SFDF( 0%,2) ] [( 25, ) + 0,000 30,000 * ] 60,000 23,058 EAR B 39,389 EAR C 47,22 Highest Annual et Return 4

15 Internal Rate of Return IRR is the interest rate which makes the costs equivalent to the benefits. P K/yr T 0 R/yr n R Benefit/yr i unknown ( ) T * SFDF( i,n) + K R P * CRF i,n Find i by trial and error Example P30,000 R5,000/yr T5,000 0 K2,000/yr 0 Find the internal rate of return 30 0,000 CRFi 5,000 SFDFi, + 2,000 5,000,0 First trial, i 0% Selecting zero shows whether the income is actually sufficient to recover the costs 30,000 5, ,000 5, ,000 0,500 At 5%, Annual Cost > Annual Benefit by,500 An approximate rate of return, ,000 5% 5

16 Try 6%, 30,000 CRF6 %,0 5,000 SFDF6%, 0 + 2,000 5,000 4,938 5, Try 7%,,000 CRF 5,000 SFDF + 2,000 5, %,0 7%, 0 5,86 5,000 By interpolation, 62 i 6% + % 6.25% Direction finder 62 i 6% % 30,000 approximate rate of return after 2 nd trial et Present Value PW of Benefit > PW of Cost P30,000 K/yr2,000 R/yr5,000 T5,000 0 ( ) [ P T SPPWF( i,n) + K USPWF( i,n) ] R * USPWF i,n 0 [ 30,000 5,000SPPWF 2,000USPWF ] 6 6%,0 %,0 6%,0 [ 30,000 5, , ] PW 5,000USPWF + 5, [ 30,000 8,376 4,720] 36,805 + i 6% 46< 0 Therefore, the project is not feasible at 6%. 6

17 Benefit-Cost Ratio Method US Flood Control Act of 936 benefit to whomsoever it may accrue should exceed estimated costs. B/C ratio expresses the ratio of equivalent uniform annual benefit (or its present worth) to the equivalent uniform annual cost (or its present worth). Same proposed project may indicate different B/C ratios depending upon whether certain items are considered disbenefits or costs Example Alt 00 K 54 K 58 K Alt 2 05 K 6 K 65 K Alt 3 K 62 K 7 K Alt PW of Cost 00 K i 0% PW of Benefit ( +.) ( +.) B / C / Similarly, B / C B / C Alt 2 is best 7

18 Benefit-Cost Ratio Method- Mutually Exclusive Projects B/C Benefit Additional Cost R A R B AC B AC A Ann. Rd. User Cost in A Ann. Rd. User Cost in B Ann. Cost of B Ann. Cost of A Choosing from several mutually exclusive projects A, B, C Comparing B with A, Ben/Cost.2 So, B is better than A C with A, Ben/Cost. C is better than A C with B, Ben/Cost 0.6 C is not better than B Most economic solution is B Example: Incremental Analysis Loc H J K Construction Cost $0, ,000,58,000 Ann. Maint. 35,000 2,000 7,000 Ann. Road User Cost 266,500 99,900 73,200 i 7%, 20 yrs 8

19 Rate of Return Computations Balance the prospective savings in road user and maintenance costs to recover the extra construction outlay. MARR 7% J compared to H [( 266,500 99,900) + ( 35,000 2,000) ] USPWF( i%,20) 622,000 0,000 i * 2.8% Rate of Return Computations J compared to H will yield 2.8% K compared to J will yield 3% location J is economically superior Benefit-Cost Ratio Benefit savings in road user costs Costs Initial outlay plus maintenance Compare J with H: B (266,000 99,900)USPWF(7%, 20) C (622,000 0,000) (35,000 2,000)USPWF(7%, 20) 364,000 PV B-C 342,000 > 0 B/C 706,000 or 66, J is better than H 364,000 34,400 9

20 Compare K with J: PV B-C -2,000 < 0 B/C 282,000 or 26, ,000 46,500 favorable location is J 0.57 Discussion of Various Methods AC Method Benefits are not considered Can be used in project formulation phase ot useful in evaluation Alternatives must have the same level of service et Present Value Profitability expressed as a lump sum, not a rate Unambiguous, direct index Can be used in project evaluation as well as in project formulation Standard procedure-aashto Rate of Return Benefit or Income is considered Solution is not unique Wrong assumptions when terminal dates are different-reinvestment of funds Used by World Bank Benefit-Cost Ratio Concern over significance in relative values of B/C ratio Definition of benefits and costs Policy-maker s lack of understanding 20

21 Other Methods Capitalized Cost--PW of costs with the analysis period of infinity Payback Period--period required to accumulate savings to equal investment Breakeven Analysis--example 2

Engineering Economy Chapter 4 More Interest Formulas

Engineering Economy Chapter 4 More Interest Formulas 1. Uniform Series Factors Used to Move Money Find F, Given A (i.e., F/A) Find A, Given F (i.e., A/F) Find P, Given A (i.e., P/A) Find A, Given P (i.e.,