Matakuliah : D0762 – Ekonomi Teknik Tahun : 2009

1. Matakuliah : D0762 – Ekonomi Teknik Tahun : 2009 Annual Worth AnalysisCourse Outline 5

2. Outline next next next next next next next next - http://www.cs.wright.edu/~snarayan/hfeeconomics/ise481_07_files/v3_document.htm • Principle and Benefit • Equivalent Annual Worth • Capital Ownership Cost • AW by salvage sinking-fund method • AW by Salvage present-worth method • AW by Capital recovery plus interest method • Spreadsheet Refererences • Engineering Economy – Leland T. Blank, Anthoy J. Tarquin p.180-199 • Engineering Economic Analysis, Donald G. Newman, p. 141-163 • Engineering Economy, William G. Sulivan, p.137-194, p. 212-284 2

3. Annual Worth Analysis Principle: Measure investment worth on annual basis Benefit:  By knowing annual equivalent worth, we can:• Seek consistency of report format• Determine unit cost (or unit profit)• Facilitate unequal project life comparison 3

4. Computing Equivalent Annual Worth $120 $80 $70 0 1 2 3 4 5 6 $50 $189.43 $100 A = $46.07 2 3 4 5 6 0 1 PW(12%) = $189.43 AE(12%) = $189.43(A/P, 12%, 6) = $46.07 http://www.cs.wright.edu/~snarayan/hfeeconomics/ise481_07_files/v3_document.htm 4

5. Annual Equivalent Worth $800 $800 $700 $700 $500 $500 $400 $400 $400 $400 $1,000 $1,000 Repeating cycle RepeatingCash Flow Cycles http://www.cs.wright.edu/~snarayan/hfeeconomics/ise481_07_files/v3_document.htm 5

6. Annual Equivalent Worth First Cycle:PW(10%) = -$1,000 + $500 (P/F, 10%, 1)       + . . . + $400 (P/F, 10%, 5)     = $1,155.68AE(10%) = $1,155.68 (A/P, 10%, 5) = $304.87 Both Cycles:PW(10%) = $1,155.68 + $1,155.68 (P/F, 10%, 5)       + . . . + $400 (P/F, 10%, 5)     = $1,873.27AE(10%) = $1,873.27 (A/P, 10%,10) = $304.87 http://www.cs.wright.edu/~snarayan/hfeeconomics/ise481_07_files/v3_document.htm 6

7. Annual Equivalent Cost Capital costs + Annual Worth Costs Operating costs When only costs areinvolved, the AEmethod is called theannual equivalent cost. Revenues must covertwo kinds of costs:Operating costs andcapital costs. http://www.cs.wright.edu/~snarayan/hfeeconomics/ise481_07_files/v3_document.htm 7

8. Capital (Ownership) Costs S 0 N I 0 1 2 3 N CR(i) • Definition: The cost of owning an equipment is associatedwith two transactions— (1)its initial cost (I) and (2) its salvage value (S). Capital costs: Taking intothese sums, we calculatethe capital costs as: http://www.cs.wright.edu/~snarayan/hfeeconomics/ise481_07_files/v3_document.htm 8

9. Example - Capital Cost Calculation $50,000 0 5 $ 200,000 Given:I = $200,000N =  5 years      S = $50,000i = 20% Find: CR(20%) http://www.cs.wright.edu/~snarayan/hfeeconomics/ise481_07_files/v3_document.htm 9

10. Justifying an investment based on AE Method Given: I = $20,000, S =$4,000, N = 5 years, i =10% Find: see if an annualrevenue of $4,400 isenough to cover thecapital costs. Solution:CR(10%) = $4,620.76 Conclusion: Need anadditional annual revenuein the amount of $220.76. http://www.cs.wright.edu/~snarayan/hfeeconomics/ise481_07_files/v3_document.htm 10

11. Salvage -Sinking Fund Method • General Equation • Example 6.1 Calculate the AW of a tractor attachment that has an initial cost of $8000 and a salvage value of $500 after 8 years. Annual operating cost for the machine are estimated to be $900 and an interest rate of 20% per year is applicable. Solution The problem indicates there are 2 cashflow AW = A1 + A2 Where A1 = annual cost of initial investment with salvage value considered Equation above = -8000(A/P,20%8) + 500(A/F,20%,8) = $2055 A2 = annual operating cist = $-900 The Annual worth for the attachment is : AW = -2055 – 900 = $-2955 11

12. Salvage Present-Worth Method • General Equation • The Steps to determine the complete asset AW are • Calculate the present worth of the salvage value via the P/F factor • Combine the value obtained in step 1 with the investment cost P • Annualize the resulting difference over the life of the asset using the A/P factor • Combine any uniform annual worth with the value from step 3 • Convert any other cash flows into an equivalent uniform annual worth and combine with the value obtained in step 4 12

13. Example 6.2 • Compute the AW of the attachment detailed in Example 6.1 using the salvage present worth method Solution Using the steps outline and equation before AW = [-8000+500(P/F,20%,8)(A/P,20%,8) – 900 = $-2955 13

14. Capital Recovery Plus Interest Method • General equation • The steps to be followed for this method are : • Reduce the initial cost by the amount of the salvage value • Annualize the value in step 1 using A/P factor • Multiply the salvage value by the interest rate • Combine the values obtained in steps 2 and 3 • Combine any uniform annual amounts • Convert all other cash flows into equivalent uniform amount and combine them with the value from step 5 Step 1 through 4 are accomplished bya pplying equation before 14

15. Example 6.1 • Use the value of Example 6.1 to compute the AW using the capital recovery plus interest Solution From equation and steps before : AW =-(8000-500)(A/P,20%,8)-500(0,20) – 900 = $-2955 15

16. Comparing Alternatives • Following costs are estimated for two equal service tomato peeling machines to evaluated by a canning plant manager’ if the minimum required rate of return is 15% per year, help the manager decide which machine to select ! • Solution AW A = -26,000(A/P,15%,6) + 2000(A/F,15%,6)-11800 = $-18,442 AW B = -36,000(A/P,15%,10)+3000(A/F,15%,10) = $-16,925 16

17. Spreadsheet Example 6.10 If Ms.Kaw Deposits $10,000 now at an interest rate of 7% per year, hom many year s must the money accumulate before she can withdraw $1400 per year forever 17