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Engineering Economics 4-1 - Valparaiso University

Professional Publications, EconomicsCash FlowCash flow is the sum of money recordedas receipts or disbursements in aproject s financial cash flow diagram presents the flow ofcash as arrows on a time line scaled tothe magnitude of the cash flow , whereexpenses are down arrows and receiptsare up convention ~ expenses occurring during the year are assumed to occur at the end of (FEIM):A mechanical device will cost $20,000when purchased. Maintenance will cost$1000 per year. The device will generaterevenues of $5000 per year for 5 salvage value is $ Publications, EconomicsDiscount Factors and EquivalencePresent Worth (P): present amount at t = 0 Future Worth (F): equivalent future amount at t = n of any presentamount at t = 0 Annual Amount (A): uniform amount that repeats at the end of eachyear for n yearsUniform Gradient Amount (G): uniform gradient amount that repeats

Engineering Economics 4-1 Cash Flow Cash flow is the sum of money recorded as receipts or disbursements in a project’s financial records. A cash flow diagram presents the flow of ... Cost-Benefit Analysis Project is considered acceptable if B – C ≥ 0 or B/C ≥ 1. Example (FEIM):

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Transcription of Engineering Economics 4-1 - Valparaiso University

1 Professional Publications, EconomicsCash FlowCash flow is the sum of money recordedas receipts or disbursements in aproject s financial cash flow diagram presents the flow ofcash as arrows on a time line scaled tothe magnitude of the cash flow , whereexpenses are down arrows and receiptsare up convention ~ expenses occurring during the year are assumed to occur at the end of (FEIM):A mechanical device will cost $20,000when purchased. Maintenance will cost$1000 per year. The device will generaterevenues of $5000 per year for 5 salvage value is $ Publications, EconomicsDiscount Factors and EquivalencePresent Worth (P): present amount at t = 0 Future Worth (F): equivalent future amount at t = n of any presentamount at t = 0 Annual Amount (A): uniform amount that repeats at the end of eachyear for n yearsUniform Gradient Amount (G): uniform gradient amount that repeats atthe end of each year, starting at the end of the second year andstopping at the end of year Publications, EconomicsDiscount Factors and EquivalenceNOTE.

2 To save time,use the calculatedfactor table providedin the NCEES Publications, EconomicsDiscount Factors and EquivalenceExample (FEIM):How much should be put in an investment with a 10% effective annualrate today to have $10,000 in five years?Using the formula in the factor conversion table,P = F(1 + i) n = ($10,000)(1 + ) 5 = $6209Or using the factor table for 10%,P = F(P/F, i%, n) = ($10,000)( ) = $6209 Professional Publications, EconomicsDiscount Factors and EquivalenceExample (FEIM):What factor will convert a gradient cash flow ending at t = 8 to a futurevalue? The effective interest rate is 10%.

3 The F/G conversion is not given in the factor table. However, there aredifferent ways to get the factor using the factors that are in the table. Forexample,NOTE: The answers arrived at using the formula versus the factor tableturn out to be slightly different. On Economics problems, one should notworry about getting the exact ( )( )= (F/G,i%,8) = (F/A,10%,8)(A/G,10%,8)(F/G,i%,8) = (P/G,10%,8)(F/P,10%,8)= ( )( )= Publications, EconomicsNonannual CompoundingEffective Annual Interest RateAn interest rate that is compounded more than once in a year isconverted from a compound nominal rate to an annual effective Interest Rate Per PeriodEffective Annual Interest RateExample (FEIM):A savings and loan offers a rate per annum compound daily over365 days per year.

4 What is the effective annual rate? ! ie=1+rm" # $ % & ' m(1=1+ " # $ % & ' 365(1= Publications, EconomicsDiscount Factors for Continuous CompoundingThe formulas for continuous compounding are the same formulas in thefactor conversion table with the limit taken as the number of periods, n,goes to Publications, EconomicsComparison of AlternativesPresent WorthWhen alternatives do the same job and have the same lifetimes,compare them by converting each to its cash value today. The superioralternative will have the highest present (EIT8):Professional Publications, EconomicsComparison of AlternativesCapitalized CostsUsed for a project with infinite life that has repeating expenses alternatives by calculating the capitalized costs ( , theamount of money needed to pay the start-up cost and to yield enoughinterest to pay the annual cost without touching the principal).))

5 NOTE: The factor conversion for a project with no end is the limit of theP/A factor as the number of periods, n, goes to Publications, EconomicsComparison of AlternativesExample (EIT8):Professional Publications, EconomicsComparison of AlternativesAnnual CostWhen alternatives do the same job but have different lives, compare thecost per year of each alternatives are assumed to be replaced at the end of their lives byidentical alternatives. The initial costs are assumed to be borrowed atthe start and repaid evenly during the life of the (EIT8):Professional Publications, EconomicsComparison of AlternativesCost-Benefit AnalysisProject is considered acceptable if B C 0 or B/C (FEIM):The initial cost of a proposed project is $40M, the capitalized perpetualannual cost is $12M, the capitalized benefit is $49M, and the residualvalue is $0.

6 Should the project be undertaken?B = $49M, C = $40M + $12M + $0B C = $49M $52M = $3M < 0 The project should not be Publications, EconomicsComparison of AlternativesRate of Return on an Investment (ROI)The ROI must exceed the minimum attractive rate of return (MARR).The rate of return is calculated by finding an interest rate that makes thepresent worth zero. Often this must be done by trial and Publications, EconomicsDepreciationStraight Line DepreciationThe depreciation per year is the cost minus the salvage value divided bythe years of Publications, EconomicsDepreciationAccelerated Cost Recovery System (ACRS)The depreciation per year is the cost times the ACRS factor (see thetable in the NCEES Handbook).

7 Salvage value is not Publications, EconomicsDepreciationExample (FEIM):An asset is purchased that costs $9000. It has a 10-year life and a salvagevalue of $200. Find the straight-line depreciation and ACRS depreciationfor 3 depreciation/yearACRS depreciationFirst year($9000)( ) = $ 900 Second year($9000)( ) = $1620 Third year($9000)( ) = $1296 ! =$9000"$20010=$880/yrProfessional Publications, EconomicsDepreciationBook ValueThe assumed value of the asset after j years. The book value (BVj) is theinitial cost minus the sum of the depreciations out to the j th (FEIM):What is the book value of the asset in the previous example after 3 yearsusing straight-line depreciation?

8 Using ACRS depreciation?Straight-line depreciation$9000 (3)($800) = $6360 ACRS depreciation$9000 $900 $1620 $1296 = $5184 Professional Publications, EconomicsTax ConsiderationsExpenses and depreciation are deductible, revenues are (EIT8):Professional Publications, EconomicsTax ConsiderationsTax CreditA one-time benefit from a purchase that is subtracted from income (EIT8):Professional Publications, EconomicsTax ConsiderationsGain or loss on the sale of an asset:If an asset has been depreciated and then is sold for more than the bookvalue, the difference is Publications, EconomicsBondsBond value is the present worth of payments over the life of the yield is the equivalent interest rate of the bond compared to thebond (EIT8):Professional Publications, EconomicsBreak-Even AnalysisCalculating when revenue is equal to cost, or when one alternative is equal toanother if both depend on some (FEIM):How many kilometers must a car be driven per year for leasing and buying to costthe same?

9 Use 10% interest and year-end : $ per kilometerBuying: $5000 purchase cost, 3-year life, salvage $1200,$ per kilometer for gas and oil, $500 per year for insuranceEUAC (leasing)= $ , where x is kilometers drivenEUAC (buying)= $ + $500 + ($5k)(A/P,10%,3) ($ )(A/F,10%,3)= $ + $500 + ($5k)( ) ($ )( )= $ + $2148 Setting EUAC (leasing) = EUAC (buying) and solving for x$ $ + $2148x= 19,527 km must be driven to break evenProfessional Publications, EconomicsInflationInflation-Adjusted Interest RateProfessional Publications, EconomicsAdditional ExamplesExample 1 (FEIM):What is the uninflated present worth of $2000 in 2 years if the averageinflation rate is 6% and i is 10%?

10 D = i + f + if = + + ( )( ) = ($2000)(P/F, , 2) = ($2000)(1 + d) n= ($2000)(1 + ) 2 = $1471 Professional Publications, EconomicsAdditional ExamplesExample 2 (FEIM):It costs $75 per year to maintain a cemetery plot. If the interest rate , how much must be set aside to pay for maintenance on each plotwithout touching the principal?(A) $1150(B) $1200(C) $1250(D) $1300P = ($75)(P/A,6%, ) = ($75)(1 ) = $1250 Therefore, (C) is Publications, EconomicsAdditional ExamplesExample 3 (FEIM):It costs $1000 for hand tools and $ labor per unit to manufacture aproduct. Another alternative is to manufacture the product by anautomated process that costs $15,000, with a $ per-unit cost.


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