Engineering Economy Problems



Engineering Economy Problems 1What is meant by the term time value of money?List three intangible factorsWhat is meant by evaluation criterion?What is the primary evaluation criterion used in economic analysis?List three evaluation criteria besides the economic one for selecting the best restaurant.Discuss the importance of identifying alternatives in the engineering economic process.What is the difference between simple and compound interest?Trucking giant Yellow Corp agreed to purchase rival Roadway for $966 million in order to reduce so-called back-office costs (e.g., payroll and insurance) by $45 million per year. If the savings were realized as planned, what would be the rate of return on investment?If Astra International’s profits increased from 22 rupiahs per share to 29 rupiahs per share in the April-June quarter compared to the previous quarter, what was the rate of increase in profits for that quarter? At an interest rate of 8% per year, $10,000 today is equivalent to how much (a) 1 year from now (b) 1 year ago?A medium-size consulting engineering firm is trying to decide whether it should replace its office furniture now or wait and do it 1 year from now. If it waits 1 year, the cost is expected to be $16,000. At an interest rate of 10% per year, what would be the equivalent cost now?Certain certificates of deposit accumulate interest at 10% per year simple interest. If a company invests $240,000 now in these certificates for the purchase of a new machine 3 years from now, how much will the company have at the end of the 3-year period? A local bank is offering to pay compound interest of 7% per year on new savings accounts. An e-bank is offering 7.5% per year simple interest on a 5-year certificate of deposit. Which offer is more attractive to a company that wants to set aside $1,000,000 now for a plant expansion 5 years from now? A company that manufactures in-line mixers for bulk manufacturing is company borrow now or 1 year from now? Assume the total amount due will be considering borrowing $1.75 million to update a production line. If it borrows the money now, it can do so at an interest rate of 7.5% per year simple interest for 5 years. If it borrows next year, the interest rate will be 8% per year compound interest, but it will be for only 4 years. (a) How much interest (total) will be paid under each scenario, and (b) Should the paid when the loan is due in either case. Engineering Economy Problems 2Define the symbols involved when a construction company wants to know how much money it can spend 3 years from now in lieu of spending $50,000 now to purchase a new truck, when the compound interest rate is 15% per year?State the purpose for each of the following built-in Excel functions:FV(i%,n,A,P)IRR(first_cell:last_cell)PMT(i%,n,P,F)PV(i%,n,A,F)Identify the following as cash inflows or cash outflows to Daimler-Chrysler: income taxes, loan interest, salvage value, rebates to dealers, sales revenues, accounting services, cost reductions.Construct a cash flow diagram for the following cash flows: $10,000 outflow at time zero, $3000 per year outflow in years 1 through 3 and $9000 inflow in years 4 through 8 at an interest rate of 10% per year, and an unknown future amount in year 8. Use the rule of 72 to estimate the time it would take for an initial investment of $10,000 to accumulate to 20,000 at a compound rate of 8% per year.If you now have $62,500 in your retirement when the account is worth $2 million, estimate the rate of return that the account must earn if you want retire in 20 years without adding any more money to the account. From the Case StudyProblem Formulation. The president of Innovations Plastics wants a recommendation on whether the company should plan on offering the new technology to the major manufacturers and an estimate of the necessary capital investment to enter this market rmation gathering.The technology and equipment are expected to last about 10 years.The inflation and income taxes are ignored for simplicityThe expected ROI were compound rates of 15%, 5% and 18%. And 5% rate for enhancing an employee-safety Equity capital financing beyond $5 millions is not possible. The amount of debt financing and its cost are unknownAnnual operating cost 8% of first cost for major equipmentIncreased annual training cost and salary range from $800,000 to $1.2 millionAnother economic and non economic factors that may influence to each alternative.Seeking possible alternativesAlternatives AAlternative BAlternative C Alternatives Analysis Nominal and Effective Interest RatesNominal interest rate, r, is an interest rate that does not include any consideration of compounding. By definition:r = interest rate per period x number per periodEffective interest rate is actual rate that applies for a stated period of time. The compounding of interest during the time period of the corresponding nominal rate is accounted for by the effective interest rate. It is commonly expressed on an annual basis as the effective annual rate ia, but any time basis can be used.Examples: 4% per year compounded monthly.ieff = ( 1 + r/m)m – 1where: ieff = effective rate for specified time period r = nominal interest rate for same time period m = number of times interest is compounded per state time period Equivalence Relations: Single Amounts with PP ≥ CPA present sum of $ 5000 at an interest rate of 8% per year, compounded semi-annually, is equivalent to how much money 8 years ago?The optical products division of Panasonic is planning a $3.5 million building expansion for manufacturing its powerful Lumix DMC digital zoom camera. If the company uses an interest rate of 20% per year compounded quarterly, for all new investments, what is the uniform amount per quarter the company must make to recover its investment in 3 years? Northwest Iron and Steel is considering getting involved in electronic commerce. A modest e-commerce package is available for $20,000. If the company wants to recover the cost in 2 years, what is the equivalent amount of new income that must be realized every 6 months, if the interest rate is 3% per quarter?Equivalence When PP < CPAn engineer deposit $300 per month into a savings account that pays interest at a rate of 6% per year, compounded semi-annually. How much will be in the account at the end of 15 years? Assume no inter-period compounding.For the transactions shown below, determine the amount of money in the account at the end of year 3 if the interest rate is 8% per year, compounded semi-annually. Assume no inter-period compounding. End of Quarter Amount of deposit, Amount of Withdrawal, $/Quarter $/Quarter 1 900 2-4 700 7 1000 2600 11 - 1000 Continuous CompoundingEffective Interest rate for Continues Compounding;i% = er – 1where : r is nominal interest rate e = 2.71828What effective interest rate per year, compounding continuously, is equivalent to a nominal rate of 13% per year?Corrosion problems and manufacturing defects rendered a gasoline pipeline between El Paso and Phoenix subjects to longitudinal weld seam failures. Therefore, pressure was reduced to 80% of the design value. If the reduced pressure results in delivery of $100,000 per month less product, what will be the value of the lost revenue after a 2-year period at an interest rate of 15% per year, compounded continuously?Because of a chronic water shortage in Santa Fee, new athletic fields must use artificial turf or xeriscape landscaping. If the value of water saved each month is $6000, how much can a private developer afford to spend on artificial turf if he wants to recover his investment in 5 years at an interest rate of 18% per year, compounded continuously?An interest rate of 21% per year, compounded every 4 months, is equivalent to what effective rate per year?In ‘N Out Payday Loans advertises that for a fee of only $10, you can immediately borrow up to $200 for one month. If a person accepts the offer, what are (a) the nominal interest rate per year and (b) the effective rate per year?How much will be in a high-yield account at the National Bank of Arizona 12 years from now if you deposit $5000 now and $7000 five years from now? The account earns interest at rate of 18% per year, compounded quarterly. Varying Interest RateHow much money could the maker of fluidized bed scrubbers afford to spend now instead of spending $150,000 in year 5 if the interest rate is 10% per year in years 1 through 4 and 1% per month in years 5 through 8?For the cash flows shown below, determine (a) the future worth in year 5 and (b) the equivalent A value for years 0 through 5.Year Cash Flow, $/year Interest rate per year. % 0 5000 12 1 – 4 6000 12 5 9000 20 PRESENT WORTH ANALYSIS (PW ANALYSIS) A FUTURE AMOUNT OF MONEY CONVERTED INTO ITS EQUIVALENT VALUE NOW HAS A PW THAT ALWAYS LESS THAN ACTUAL CASH FLOW, FOR ANY INTEREST RATE > 0.P/F factor < 1PW values are often referred to as DCF (Discounted Cash Flow)Interest Rate = Discounted RatePW = PV = NPV Purpose: To compare mutually exclusive alternatives on a PW basis. PW ANALYSIS PW ANALYSIS will help us:Identify mutually exclusive and independent projectsDefine a service and a revenue alternativeSelect the best equal-life alternatives using PWASelect the best different-life alternatives using PWA Select the best alternative using FWASelect the best alternative using CC calculationsDetermine the payback period at i = 0% and i > 0%Perform a life-cycle cost analysis for the acquisition & operation phases of the systemCalculate the PW of a bond investmentDevelop spreadsheet that use PWA, including payback period Mutually exclusive project: only one of the viable projects can be selected Independent project: more than one viable project may be selected Example. Perform a PWA of equal-service machines, if MARR is 10%/year Revenues for all three alternatives are expected to be the same Electric Powered(E)Gas Powered (G)Solar Powered (S)First Cost, $AOC, $Salvage Value, $Life, Years-2500-9002005-3500-7503505-6000-501005 Answer:PWE = -2500 – 900(P/A,10%,5) + 200(P/F, 10%, 5) = 4 -5788PWG = -3500 – 750(P/A, 10%, 5) + 350(P/F, 10%, 5) = $ -5936PWS = -6000 – 50(P/A, 10%, 5) + 100(P/F, 10%,5) = $ -6127PWE is selected since PW of its cost is lowest, or numerically largest PW value. If there are different life alternativesCompare the alternatives over a period of time equal to the LCM (Least Common Multiple) of their livesThe assumptions of a PWA of different-life alternatives:The service provided by alternatives will be needed for the LCM of years or moreThe selected alternative will be repeated over each life cycle of the LCM in exactly the same mannerThe cash flow estimates will be the same in every life-cycleExample.A project engineer with EnvironCare is assigned to start up a new office in a city where a 6-year contract has been finalized to take and analyze ozone-level readings. There are two options: Location ALocation BFirst Cost, $Annual Cost, $Deposit Return, $Lease Time, years-15,000-3,50010006-18,000-3,10020009 Determine which lease option should be selected on the basis of PW, if MARR is 15%/yearIf a study period of 5 years is used, which location should be selected ?If a study period of 6 years is used, and the deposit return of alternative B is estimated to be $ 6000 after 6 years, which location should be selected?Answer.Since the lease have different term, the LCM(6,9) is 18 PWA = -15,000 – 15,000(P/F, 15%,6) + 1000(P/F, 15%, 6) – 15,000(P/F, 15%,12) + 1000(P/F, 15%, 12) + 1000(P/F, 15%, 18) – 3,500(P/A, 15%,18) = $ - 45,036 PWB = -18,000 – 18,000(P/F, 15%,9) + 2000(P/F, 15%, 9) + 2,000(P/F, 15%,18) - 3100(P/A, 15%, 18) = $ -41,384Location B is selected, because the PWB value is numerically larger than PWA(b) For 5 year study period, no cycle repeats are necessary PWA = -15,000 – 3,500(P/F, 15%,5) + 1000(P/F, 15%, 5) = $ – 26,236PWB = -18,000(P/F, 15%, 12)–3,100(P/A,15%,5)+2,000(P/F,15%,5) = $ - 27,397Location A is the better choice(c) For 6 year study period, PWA = -15,000 – 3,500(P/F, 15%,6) + 1000(P/F, 15%, 6) = $ – 27,813PWB = -18,000(P/F, 15%, 6)–3,100(P/A,15%,6)+6,000(P/A,15%,6) = $ - 27,397Location B is numerically larger than AFuture Worth AnalysisFW of an alternative may be determined directly from the cash-flows by determining the FW value, or multiplying the PW by the F/P factor, at the established MARR.FW analysis is applicable to large capital investment decision when a prime goal is to maximize the future wealth of a corporation’s stockholders.FW analysis is often utilized if the asset might be sold or traded at some time after its start-up, but before the expected life is reached.Capitalized Cost Calculation and AnalysisCC is present worth of an alternative that will last “forever” Projects fall into this category are:Bridges (Suramadu)Dams (Asahan)Irrigation Systems (Jatiluhur)Rail Road (Jabodetabek) and Toll Road (Cipularang)CC is derived from the relation P = (A/P, i, n), where n = P = A[{(1+i)n – 1}/{i(1+i)n}] or P = A[{1-1/(1+i)n}/i}] = A/i so, CC = A/i = AW/iProcedure to calculate CC for an infinite sequence of cash-flows.Step 1. Draw cash flow diagram showing all non-recurring (one-time) cash-flows and at least two cycles of all recurring cash-flows.Step 2. Find the PW of all non-recurring amounts . This is their CC valueStep 3. Find the equivalent uniform AW (annual worth/value) through one life cycle of all recurring amounts.Step 4. Divide the AW in step 3 by i to obtain a CC value.Step 5. Add CC values obtained in step 2 and 4. Payback Period AnalysisThe payback period np is the estimated time (in years), it will take for the estimated revenues and other economic benefits to recover the initial investment and a stated rate of return. npIf the stated rate i>0%, then 0 = -P + NCFt (P/F, i, t) t=1Where: P : is the initial investment (first cost) NCFt : Net Cash Flow for each year t = receipt – disbursementIf NCF values are expected to be equal each year: 0 = - P + NCF (P/F, i, np)If the NCF series are uniform: np = P/NCFLife Cycle CostTypical applications for LCC areBuildingsNew product linesManufacturing plantsCommercial AircraftNew Automobile modelsDefense Equipment SystemsGenerally, LCC estimates may be categorized into simplified format for the major phases of acquisition and operation as following:Acquisition PhaseRequirement definition Stage:Determination of user needAnticipated systemPreparation and documentationPreliminary designFeasibility StudyConceptualEarly Stage PlansDetail Design. Detail Plan & ResourcesCapital, human, facilities, information system, marketing(b) Some acquisition of assetsOperation PhaseAll activities are functioning, products & services are availableConstruction & implementation, testing and preparationUsage Stage: to generate products and servicesPhase Out & Disposal StagePresent Worth of BondsEquity Financing: The corporation uses its own funds from cash on hand, stock sales, or retained earning. The individuals can use their own cash, savings or investment.Debt Financing: The corporation borrows from outside sources and repays the principal and interest. Sources of dept capital may be bonds, loans, mortgages, venture capita pools. Individuals can utilize debt sources, such as credit card or from a credit union. A time-tested method of raising capital is through the issuance of IOU (I owe you), which is financing through debt. One form of IOU is bond.A bond: is a long term note issued by a corporation (financing institution) or government to finance major project.Bonds are usually issued in face value amounts of $100, $1000, $5000 or $10,000.Bond Interest I (also called bond dividend) is paid periodicallyThe bond interest is paid c times per year.The stated interest rate is called the bond coupon rate bI = (face value)(bond coupon rate)/number of payment periods per year = Vb/cThe steps to calculate the PW of a bond are as follow:Determine I, the interest per payment periodConstruct the cash flow diagramEstablish the required MARR or rate of returnCalculate the PW value of the bond interest payments and the face value at i=MARRExample.Determine the purchase price you should be willing to pay now for a 4.5% $5000 10-year bond with interest paid semiannually. Assume your MARR is 8% per year compounded quarterly.Solution..I = 5000(0.045)/2 = $112.50 every 6 monthThe nominal semiannual MARR is r = 8%/2 = 4%Effective I = ( 1 + 0.04/2)2 – 1 = 4.04% per 6 monthsThe PW of the bond is determined for n = 2(10) = 20 semiannual periodPW = $112.50(P/A,4.04%,20) + 5000(P/F,4.04%,20) = $3788Problems5.9, 5.11, 5.14, 5.16, 5.26, 5.28, 5.38, 5.39, 5.42, 5.49, 5.50 ................
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