IE343_HW5_Solutions

# IE343_HW5_Solutions - HomeworkAssignment#5Solutions...

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Unformatted text preview: HomeworkAssignment#5Solutions  PU R D U E U N I V E RSI T Y I E 343 : E n gi n e e r i n g E con om i cs F a l l 2010 I nst r u c to r : A . C a p pon i H om e w o r k A ssign m e n t #5 A ssign e d : 24 Se p t 2010 D u e : 01 O c t 2010 Missing name on the top of each page submitted Not stapled together 2 Not completed on the correct template 10 2 P r ob l e m 1 (15 points) Consider a machine that cost \$15,000 and has a 5 year useful life. At the end of 5 years, it can be sold for \$3,500. The machine increases the profit for the firm by \$4,000 per year. (a) Draw cash flow diagram 3,500 \$4,000 0 \$4,000 \$4,000 \$4,000 1 2 3 4 \$4,000 5 \$15,000 (b) Should the machine be purchased if the MARR is 10% per year and the annual worth method is used? We have CR(0.1) = \$15,000 (A/P, 0.1,5) - \$3,500 (A/F, 0.1, 5) = \$3,957-\$573.3 = \$3,384. Thus, we get that the annual worth is AW(0.1) = \$4,000 - \$3,384 = \$616 > 0. Since AW(0.1) >0, we can conclude that buying the machine is a worthwhile investment. IE343 HomeworkAssignment#5Solutions Page1 HomeworkAssignment#5Solutions  P r ob l e m 2 (15 points) The heat loss through the exterior walls of a processing plant is estimated to cost the owner \$3,000, starting from next year. A salesman from Superfiber, Inc. claims he can reduce the heat loss by 80% with the installation of \$15,000 of Superfiber now. The cost of heat loss rises by \$200 per year after next year (gradient), and the owner plans to keep the building ten more years. (a) Draw the cash flow diagram The savings due to heat loss in the next year is 0.8 * 3000 + 0.8 * 200. The savings due to heat loss in two years from now is 0.8 * 3000 + 0.8 * 400. The savings due to heat loss in three years from now is 0.8 * 3000 + 0.8 * 600. And so on. Hence, the cash flow diagram can be decomposed as \$ 0.8 *3,000 \$0.8* 3,000 \$0.8*600 \$0.8 * 3,000 \$0.8*400 + 0 1 2 \$0.8*200 10 0 \$ ­15,000 10 1 2 3 4 (b) What is the yearly internal rate of return? Find the IRR via linear interpolation of 12% and 15%. The expression for the present worth is: PW(i) = -\$15,000 + \$0.8 * 3,000 * (P/A, i, 10) + \$0.8 * 200 * (P/G, i, 10) We have PW(12%) = \$1,800.64 and PW(15%) =- \$237.76 If we interpolate, we find If you complete this problem in Excel, the equation is IRR(-15,000, 24000, 2560, 2720, 2880, 3040, 3200. 3360, 3520, 3680) = 14.91% (c) Assume the MARR = 10%. Is the installation of Superfiber a worthwhile investment, when using the IRR an evaluation criteria? Since the internal rate of return i is higher than MARR = 10%, then installing Superfiber is a worthwhile investment. IE343 HomeworkAssignment#5Solutions Page2 HomeworkAssignment#5Solutions  (10 points) P r ob l e m 3 Textbook Problem 5-21   This project is acceptable as it results in a positive future value. P r ob l e m 4 Textbook Problem 5-22 \$12,000 Revenue estimated annual expenses = 1,800 EOY1 = 1,300 (10 points) 500 = 1,300 for the first year. EOYi+1= EOYi +100 FW(15%) = -10,000(F|P,15%,6) + 1,300(F|A,15%,6) + 100(P|G,15%,6)(F|P,15%,6) + 12,000 FW(15%)= -10,000(2.3131) + 1,300(8.7537) + 100(7.937)(2.3131) + 12,000=\$2,085.72 Since \$2,085.72 > 0, the investment result in a profit, and thus is worth pursuing it. IE343 HomeworkAssignment#5Solutions Page3 HomeworkAssignment#5Solutions  (10 points) P r ob l e m 5 Textbook Problem 5-25 Use the AW method to evaluate this investment. The annual worth of the investment is negative, thus this is not a good investment. (15 points) P r ob l e m 6 Textbook Problem 5-27 C a sh O u t f low s Investment: C a sh I n f low s \$300,000 + \$600,000 + \$250,000 + \$100,000 = \$1,250,000 (year 0) Annual revenue: Annual expenses: \$750,000 (annually) \$475,000 (annually) Market (salvage value) + working capital: \$400,000 + \$350,000 + \$50,000 + \$100,000 = \$900,000 (year 10)  The capital recovery cost is    Since the Annual Worth at MARR = 15% is equal to \$70,245 which is greater than zero, then the company should invest a plant to manufacture the proposed new product. IE343 HomeworkAssignment#5Solutions Page4 ...
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## This note was uploaded on 01/20/2012 for the course INDUSTRIAL 343 taught by Professor Capponi during the Spring '11 term at Purdue.

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