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Engineering Economy IEN255 Chapter 5 - Annual Equivalent Worth Analysis Another common method Equal payments on an annual basis Convert NPW to AE
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Engineering Economy Annual Equivalent Worth Criterion AE(I) = PW(I)(A/P,I,N) (5.1)
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Engineering Economy Good or bad? (for revenue) If AW(i) > 0, accept If AW(i) = 0, indifferent If AW(i) < 0, reject
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Engineering Economy Cyclic flow pattern figure 5.2
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Engineering Economy Why do AE? Consistency of report formats Need for unit costs/profits Unequal project lives indefinite service period
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Engineering Economy Capital vs Operating Costs Operating Costs - incurred during operation (labor, matls, etc.) recurring costs Capital Costs - purchasing assets (non-recurring)
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Engineering Economy A Formula CR(i) = I(A/P,i,N) - S(A/F,i,N) (5.2) If I = initial cost and S = salvage value we know that, (A/F,i,N) = (A/P,i,N) - I so rewriting, CR(i) = I(A/P,i,N) - S[(A/P,i,N) - I] and rearranging CR(i) = (I-S)(A/P,I,N) +iS
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Engineering Economy Example 5.3 figure 5.4
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Engineering Economy Applications - Unit Profit/Cost Calculation Determine the number of units to be produced each year over the life of the asset Identify the cash flow series over life calculate NPW Divide equivalent AE by the number of units to be produced (if # varies each year, you need to convert them to equivalent annual units).
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Engineering Economy Example 5.4
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Engineering Economy Make or Buy Determine the time span determine the annual quantity of the part (or product) Obtain the unit cost of purchasing the part from outside determine (extra) equipment, labor, etc. required to make the part. Estimate the cash flows with the make option compute AE of producing the part over the time span compute the unit cost choose minimum cost
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Engineering Economy Example 5.6
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Engineering Economy Break even point (cost reimbursement) unit of use calculation depreciation - loss in value figure 5.6
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Engineering Economy Mutually exclusive projects Analysis period equals project lives Example 5.8 operating cost per kWh per unit determine total input power for both motor types determine total kWh per year for both motor types determine annual energy costs for both motor types determine capital costs for both motor types determine total AE (= capital cost + annual energy cost) then calculate unit cost per kWh based on output power determine savings per operating hour for switching
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Engineering Economy Break even
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Engineering Economy Analysis periods differs from lives Service of the selected alternative is required on a continuous basis each alternative will be replace by an identical alternative Iff these two criteria are true we can solve for only initial life span. (saves time from doing LCM)
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Engineering Economy Example 5.9 Quick way for example 4.14 figure 5.8
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Engineering Economy Design economics AE(i) = a +bx +c/x (5.5) minimize and (5.6)
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Engineering Economy IEN255 Summer’99 Chapter 3, 4 & 5 Test #2 Test#2 Over Chapters 3 (commercial loans (3.6) and bonds (3.7)), 4, and 5. Don’t forget chapter 5!!!! 1 hour and 30 minutes for the test done at 3:30!
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