Engineering Economy

Engineering Economy

Annual Worth Analysis An alternative to Present Worth (PW) and Future Worth (FW) analysis is Annual Worth (AW) analysis. Annual worth analysis will select the same projects that would be selected by PW or FW analysis. The relationship between AW, PW and FW is: AW = PW(A/P,i,n) = FW(A/F,i,n) where n is the number of years in the planning horizon or LCM used to obtain PW or FW. EGR 312 - 13

1 Annual Worth Analysis Annual Worth (AW) analysis is often desirable since an annual worth is often more intuitive to individuals who think in terms of annual cash flows. AW analysis is also desirable since AW has to be calculated for only one life cycle. It is not necessary to use LCM as we did for the PW and FW analyses. EGR 312 - 13 2 Annual Worth Analysis

When alternative have different lives, the AW method make the following assumptions: 1. The services provided are needed for the indefinite future (forever) 2. The selected alternative will be repeated for succeeding life cycles in exactly the same manner as for the first life cycle. 3. All cash flows will have the same estimated values in every life cycle. EGR 312 - 13 3 Annual Worth Analysis Earlier PW analysis for different-life alternatives: Service Example: You are evaluating the purchase of a new or used cars that needs to last you for 10 years. i = 8%. New Car Used Car

Purchase Price $20,000 $10,000 Annual Maint. $ 750 $ 1,000 Resale Value $ 4,000 $ 4,000 Life, years 10 5 PW (LCM 10) -$23,180 -$18,941 AW(New) = _________________________________________________

AW(Used) = ________________________________________________ What is AW when it is calculated using the PW (LCM 10) values? EGR 312 - 13 4 Capital Recovery Capital Recovery (CR) is the equivalent annual cost of owning an asset plus the return on the initial investment. Consider an alternative where P is the total of all initial investments, A is the equivalent of all annual cash flows (cost only in a service project, receipts and costs in revenue projects), and S is the salvage value. Then the relationship between CR and AW is: AW = -CR A

So that CR = [P(A/P,i,n) - S(A/F,i,n)]. EGR 312 - 13 5 Capital Recovery Example: Recall the evaluation of two income properties. You expect a MARR of 15%. What is the Annual Worth of the larger home? Purchase Price Annual Maint. Annual Income Resale (after Expenses) Life, years $75K Home $15,000 $ 6,000 $ 7,500 $90,000

15 CR = _____________________________________ A = -$1,500 (Note, A represents a cost, but because a profit is being made, A in this case is negative) AW = ___________________________________ EGR 312 - 13 6 Evaluation of Alternatives Using AW To evaluate alternatives: 1. First, calculate the AW at the MARR. Then, for A. One alternative: If AW > 0, then the MARR is met or exceeded. B. Two or more alternatives: Choose the project with lowest cost AW value (service projects) or highest income AW value (revenue projects.) EGR 312 - 13

7 Evaluation of Alternatives Using AW Service Example: You are evaluating the purchase of a new or used cars that needs to last you for 10 years. MARR = 8%. Purchase Price Annual Maint. Resale Value Life, years New Car $20,000 $ 750 $ 4,000 10 Used Car $10,000 $ 1,000

$ 4,000 5 AW(new) = -$20,000(A/P,8%,10) -$750 + $4000(A/F,8%,10) = -$3454.48 AW(used) = -$10,000(A/P,8%,5) - $1000 + $4000(A/F,8%,5) AW(used) = -$2822.76 Decision? EGR 312 - 13 8 AW of a Permanent Investment For a permanent investment with an initial investment of P, the capital recovery is simply the annual interest earned, in other words: CR = Pi In addition, to find the annual worth, all cash flows must be converted to equivalent annual amounts, A. Then,

AW = -CR - A EGR 312 - 13 9 AW of a Permanent Investment Example: A toll-road was just completed at a cost of $1.5 billion, with major maintenance expenditures of $500 million forecast every 10 years. Annual receipts minus maintenance results in a positive cash flow of $150 million. What is the annual worth, assuming i = 5%? CR = ________________________ AW = ______________________________ EGR 312 - 13 10

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