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

2. Outline 2 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 next - http://www.cs.wright.edu/~snarayan/hfeeconomics/ise481_07_files/v3_document.htm next

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

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

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

7. Annual Equivalent Cost When only costs are involved, the AE method is called the annual equivalent cost. Revenues must cover two kinds of costs: Operating costs and capital costs. 7 Capital costs Operating costs + Annual Worth Costs http://www.cs.wright.edu/~snarayan/hfeeconomics/ise481_07_files/v3_document.htm

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

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

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

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 1.Calculate the present worth of the salvage value via the P/F factor 2.Combine the value obtained in step 1 with the investment cost P 3.Annualize the resulting difference over the life of the asset using the A/P factor 4.Combine any uniform annual worth with the value from step 3 5.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 : 1.Reduce the initial cost by the amount of the salvage value 2.Annualize the value in step 1 using A/P factor 3.Multiply the salvage value by the interest rate 4.Combine the values obtained in steps 2 and 3 5.Combine any uniform annual amounts 6.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 Machine AMachine B First Cost26,00036,000 Annual maintenance cost,$800300 Annual Labor cost, $11,0007000 Extra annual income taxes, $-2,600 Salvage Value2,0003,000 Life, years610

17. Spreadsheet 17 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