IE2308-006 Chapter 5 problem

apital Recovery and AW at must be realized at an interest rate of 5% per 6-month period? .o loro Company is expanding its U.S.-based plastic molding plant as it continues to transfer work from Juarez, Mexico contractors. The plant bought a S1.I million precision injection molding macninc to make plastic parts for Toro lawn mowers, trim- mers, and snow blowers. The plant also spent $275,000 for three smaller plastic injection mold- ing machines to make plastic parts for a new line of sprinkler systems.. The plant expects to hire 13 people, including some engineers for the ex- pansion. If the average loaded cost (i.e., including benefits) of each employee is $100,000 per year, determine the annual worth of the new systems 1 Heyden Motion Solutions ordered $7 milion worth of seamless tubes for manufacturing their high performance and precision linear motion products. If their annual operating costs are S860,000 per year, how much annual revenue is required over a 3-year planning period to recover the initial investment and operating costs at the company's MARR of 15% per year? 5.2 NRG Energy plans to construct a giant solar plant in Santa Teresa, NM to supply electricity to 30,000 southern NM and western TX homes. The plant will have 390,000 heliostats to concentrate sunlight onto 32 water towers to generate steam. NRG will spend $560 million in constructing the plant and $430,000 per year in operating it. If a salvage value of 20% of the initial cost is assumed, how much will the company have to make each year for 15 years in order to recover its investment at a MARR of 18% per year? 5.3 Environmental recovery company RexChem Part- ners plans to finance a site reclamation project that will require a 4-year cleanup period. The company will borrow $3.8 million now to finance the proj- ect. How much will the company have to receive in annual payments for 4 years, provided it will also receive a final lump sum payment after 4 years in the amount of $500,000? The MARR is 20% per year on this investment. 5.4 U.S. Steel is planning a plant expansion to pro- duce austenitic, precipitation-hardened, duplex and martensitic stainless steel round bar (as well as various nickels) that is expected to cost $13 mil- lion now and another $10 million 1 year from now. f total operating costs will be $1.2 million per year Starting 1 year from now, how much must the com- pany realize in annual revenue in years 1 through over a five-year planning period at an interest rate of 10% per year. Assume a 25% salvage value for the new equipment. 5.7 In an effort to retain troops who are proficient with weapons and who can speak the languages of Middle Eastern countries, the Pentagon offered bonuses of $150,000 to specialized personnel who were near or already eligible for retirement. If 400 enlisted personnel accepted the bonus in year one, 300 in year two, and 600 in year three, what was the equivalent annual cost of the program over the 3-year period at an interest rate of 6% per year? 5.8 A company that manufactures magnetic flow me- ters expects to undertake a project that will have the cash flows below. At an interest rate of 10% per year, what is the equivalent annual cost of the proj- ect? Find the AW value using (a) tabulated factors, (b) calculator functions, and (c) a spreadsheet. Which method did you find the easiest to use? First cost, $ -800,0000 Equipment replacement cost in year 2, $ -300,000 10 to recover its investment plus15% per year? S.5 A small metal plating company wants to become involved in electronic commerce. A modest Annual operating cost, S/year -950,000 Salvage value, $ 250,000 C-commerce package is available for $20,000. Semiannual updates and site maintenance will cost $300. The salvage value of the package is esti- mated to be $1500 after 3 years. If the company Life, years 4 5.9 A small commercial building contractor pur- chased a used crane 2 years ago for $60,000. Its operating cost was $2500 in month one, $2550 in month two, and amounts increasing by $50 per wants to recover its total cost in 3 years, what is the equivalent semiannual amount of new income

per month? 5.10 A 600-ton press used to produce Composite- material fuel cell components for automobiles using proton exchange membrane (PEM) technol- ogy can reduce the weight of enclosure parts up to 75%. At MARR = 12% per year, calculate (a) capital recovery and (b) annual revenue required. (c) Solve using a spreadsheet. Installed cost = $-3.8 million n = 12 years Salvage value = $250,000 Annual operating costs = $-350,000 in year 1, increasing by $25,000 per year Evaluating Alternatives Using AW 5.11 Two machines with the following cost estimates are under consideration for a dishwasher assembly process. Using an interest rate of l0% per year, de- termine which alternative should be selected on the basis of an annual worth analysis.

ADDITIONAL PROBLEMS AND FE EXAM REVIEW QUESTIONS 5.32 In comparing alternatives that have different lives by the annual worth method, a. the annual worth value of both alternatives must be calculated over a time period equal to $10 million first cost and the required rate of return on this investment. C. each year of its expected life, a net revenue of S1,985,000 must be realized to recover the $10 million first cost. the life of the shorter-lived one. b. the annual worth value of both alternatives must be calculated over a time period equal to the life of the longer-lived asset. d. the services provided by the asset will stop if less than $1,985,000 in net revenue is reported in any year C. the annual worth values must be calculated 5.35 The AWs of three cost alternatives are S-23.000 for Alternative A, $-21,600 for B. and S-27.300 for C. On the basis of AW values, the best over a time period equal to the least common multiple of the lives. d. the annual worth values can be compared over one life cycle of each alternative. 5.33 If you have the present worth of an alternative with a 5-year life, you can obtain its annual worth by: economic choice is: a. select alternative A. b. select alternative B. c. select alternative C. d. select the do nothing alternative. 5.36 The initial cost of a packed-bed degassing reactor for removing trihalomethanes from potable water is $84,000. The annual operating cost for power, a. multiplying the PW by i. b. multiplying the PW by (A/F,i,5). C. multiplying the PW by (PIA,i,5). d. multiplying the PW by (A/P,i,5). 5.34 An automation asset with a high first cost of S10 million has a capital recovery (CR) of $1,985,000 per year. The correct interpretation of this CR value is that: site maintenance, etc. is $13,000. If the salvage value of the pumps, blowers, and control systems is expected to be $9000 at the end of 10 years, the AW of the packed-bed reactor at an interest rate of 8% per year is closest to: a. $-26,140 a. the owner must pay an additional $1,985,000 each year to retain the asset. D. each year of its expected life, a net revenue of $1,985,000 must be realized to recover the b. $-25,520 c. $-24,900 d. $-13,140

140 Chapter 5 Annual Worth Analysis 5.39 The equivalent annual worth of alternat: of alternative B is closest to: 5.37 The AW values of three revenue alternatives are a. $-25,130 b. $-28,190 S-23.000 for A, $-21.600 lor B. and $ 27.300 for C. On the basis of these AW values, the correct c. $-37,080 d. $-39, 100 540 'The equivalent annual worth of alternative A an infinite time period is clOsest to: decision is to: a. select alternative A b. select alternative B. c. select altemative C d. select the do nothing alternative Over a. $-25,000 b. $-27,200 Problems 5.38 through 5.40 are hased on the following estimates. C. S-31,600 d. $-37, 100 se an interest rate of 10 per year. 5.41 If you have the capitalized cost of an alternative that has an infinite life, you can get its annual cost Alternative A B over a very long number of years by: a. multiplying the capitalized cost by i. b. multiplying the capitalized cost by (A/F.i.n). c. dividing the capitalized cost by (P/A.i.n). d. dividing the capitalized cost by i. 5.42 If you have the annual worth of an alternative that has a 5-year life, you can obtain its perpetual annual worth by: First cost. S 50,000 -80,000 Annual cost. S/year -20.000 - 10.000 Salvage value. S 10.000 25,000 Life. years 3 5.38 The equivalent annual worth of alternative A is closest t0: a. S-25.130 a. doing no calculations, since perpetual annu b. S-37.100 worth equals the annual worth. b. multiplying the annual worth by (AP.. C. dividing the annual worth by i. d. multiplying the annual worth by i C. S-41.500 d.-42.900