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58 Pages

### HW 2 solutions

Course: INDUSTRIAL IT 450, Fall 2011
School: Purdue
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Word Count: 1036

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45000 IT Production Cost Analysis HW 2 Page 2 of 5 1. What compound interest rate per year is equivalent to a 12% per year simple interest rate over a 15year period? F P(1 i) compound F P(1 ni) simple setting both equal to each other: n P(1 i)n P(1 ni) (1 i)15 1 (15)(.12) (1 i)15 2.80 1 i 15 2.80 1 i 1.0711 i 7.11% 2. How much could Tesla-Sino Inc., a maker of superconducting magnetic energy...

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45000 IT Production Cost Analysis HW 2 Page 2 of 5 1. What compound interest rate per year is equivalent to a 12% per year simple interest rate over a 15year period? F P(1 i) compound F P(1 ni) simple setting both equal to each other: n P(1 i)n P(1 ni) (1 i)15 1 (15)(.12) (1 i)15 2.80 1 i 15 2.80 1 i 1.0711 i 7.11% 2. How much could Tesla-Sino Inc., a maker of superconducting magnetic energy storage systems, afford to spend now on new equipment in lieu of spending \$85,000 five years from now? The companys rate of return is 18%. P F (P / F ,18%,5) 85,000(0.4371) P \$37,153.50 3. French car maker Renault signed a \$95 million contract with ABB of Zurich, Switzerland, for automated underbody assembly lines, body assembly workshops, and line control systems. If ABB will be paid in 3 years (when the systems are ready), what is the present worth of the contract at 12% per year interest? P F (P / F ,12%,3) 95,000,000(0.7118) P \$67,621,000 4. A pulp and paper company is planning to set aside \$150,000 now for possibly replacing its large synchronous refiner motors. If the replacement isnt needed for 8 years, how much will the company have in the account if it earns interest at a rate of 8% per year?7 F P(F / P ,8%,8) 150,000(1.8509) F \$277,635 IT 450 Production Cost Analysis HW 2 Page 3 of 5 5. A mechanical consulting company is examining its cash flow requirements for the next 7 years. The company expects to replace office machines and computer equipment at various times over the 7year planning period. Specifically, the company expects to spend \$7000 two years from now, \$9000 three years from now and \$5000 five years from now. What is the present worth of the planned expenditures at an interest rate of 10% per year? P 7000(P / F ,10%,2) 9000(P / F ,10%,3) 5000(P / F ,10%,5) 7000(0.8264) 9000(0.7513) 5000(0.6209) 5784.80 6761.70 3104.50 P \$15,651 6. A young college graduate wants to buy a first home but only can afford monthly payments of \$500. Thirty-year loans are available at 12% per year compounded monthly. With a \$5000 down payment what is the price of the most expensive house he can afford to purchase? P 5000 500(P / A,1%,360) 5000 500(97.2183) 5000 48,609.15 P \$53,609.15 7. HydroKlean, LLC, an environmental soil cleaning company, borrowed \$3.5 million to finance start-up costs for a site reclamation project. How much must the company receive each year in revenue to earn a rate of return of 20% per year for the 5-year project period? A P(A / P ,20%,5) 3,500,000(0.3344) A \$1,170,400 8. Southwestern Moving and Storage wants to have enough money to purchase a new tractor-trailer in 5 years at a cost of \$290,000. If the company sets aside \$100,000 in year 2 and \$75,000 in year 3, how much will the company have to save in year 4 in order to have the money it needs, if the money set aside earns 9% per year? 100,000(F / P ,9%,3) 75,000(F / P ,9%,2) x(F / P ,9%,1) 290 ,000 100,000(1.2950) 75,000(1.1881) x(1.09) 290,000 129,500 89,107.50 1.09 x 290,000 1.09 x 290,000 218,607.50 1.09 71,392.50 x x \$65,497.71 IT 450 Production Cost Analysis HW 2 Page 4 of 5 9. Allen Bradley claims that its electronic overspeed detection relay modules provide customers a costeffective monitoring and control system for turbo machinery. If the equipment provides more efficient turbine performance to the extent of \$20,000 in year 1, \$22,000 in year 2, and amounts increasing by \$2000 per year, how much could Mountain Power and Light afford to spend now at 10% per year interest, if it wanted to recover its investment in 5 years? P 20,000(P / A,10%,5) 2000(P / G ,10%,5) 20,000(3.7908) 2000(6.8618) 75,816 13,723.60 P \$89,539.60 10. A low-cost noncontact temperature measuring tool may be able to identify railroad car wheels that are in need of repair long before a costly structural failure occurs. If the BSNF railroad saves \$100,000 in year 1, \$110,000 in year 2, and amounts increasing by \$10,000 each year for five years, what is the equivalent annual worth of the savings, if interest is 12% per year? A A G(A / G ,12%,5) 100,000 10,000(1.7746) 100,000 17,746 P \$117,746 11. A concept car that will get 100 miles per gallon and can carry 4 persons would have a carbon-fiber and aluminum composite frame with a 900 cc three-cylinder turbo-diesel/electric hybrid power plant. The extra cost of these technologies is estimated to be \$11,000. If gasoline savings over a comparable conventional car would be \$900 in year 1, increasing by 10% each year, what is the present worth of the savings over a 10-year period at an interest rate of 8% per year? 1 1 g n 1 1.10 10 1 i 900 1.08 PA i g .08 .10 0.2014 900 0.02 900 10.0702 P \$9063.21 12. Find the future worth in year 10 of an investment that starts at \$8000 in year 1 and increases by 7% each year. The interest rate is 7% per year. n P0 A 1 i 10 8000 1.07 8000 9.3458 P0 \$74,766.36 F10 P0 (F / P ,7%,10) 74,766.36(1.9672) F10 \$147,080.37 IT 450 Production Cost Analysis HW 2 Page 5 of 5 13. An engineer plans to retire when he has \$1.6 million in his brokerage account. If he started with \$100,000 in the account, how long will it be (from the time he started) before he can retire, if the account makes a rate of return of 18% per year? P(F / P ,18%, n) F 100,000(F / P ,18%, n) 1,600,000 (F / P ,18%, n) 16.000 n 17 yrs Solve for the F/P factor, then look up the closest corresponding value in the 18% table. 14. The business plan for a start-up company that manufactures multi-gas portable detectors showed equivalent annual cash flows of \$400,000 for the first 5 years. If the cash flow in year 1 was \$320,000 and the increase thereafter was \$50,000 per year, what interest rate was used in the calculation? A G(A / G ,12%,5) AT 320,000 50,000(A / G , i %,5) 400,000 50,000(A / G , i %,5) 80,000 (A / G , i %,5) 1.6000 i % 22.58% Use a similar approach as above. Or, it's easier to use the Excel interest table from Blackboard. Look at the A/G column at row 5. Change the interest rate until the A/G value is 1.6000.
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