Quiz 6 sec 2 sol

# Quiz 6 sec 2 sol - 2 There are two other robotic arms that...

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IE 343 Spring 2010 Name: Section 2, Quiz 6 (last) (first) Closed book, closed notes. No calculators. 1. GM is thinking of buying a new robotic arm to be used in their assembly line. The cost of this arm is \$600,000 and has a useful life of 8 years. At the end of the 8 th year, it will be sold for \$50,000. This arm is estimated to increase revenues by \$75,000 per year. Expenses will amount to \$20,000 a year. GM has an MARR of 8% and a bank account that pays 5% interest per year (this is ε ). Write the equation you would use to determine the ERR. Total expenses at EOY 8 = [600k+20k(P/A,5%,8)](F/P,i'%,8) Total revenues at EOY 8 = 75k(F/A,5%,8)+50k By utilizing eq. (5-8) in textbook, we have the following relationship to solve for i': [600k+20k(P/A,5%,8)](F/P,i'%,8) = 75k(F/A,5%,8)+50k (F/P,i'%,8) = (1+i') 8 = [75k(F/A,5%,8)+50k] / [600k+20k(P/A,5%,8)]
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Unformatted text preview: 2. There are two other robotic arms that might fulfill GM’s needs. Their details are summarized in the following table: Arm A Arm B Capital investment \$100,000 \$80,000 Annual revenues \$75,000 \$50,000 Annual expenses \$24,000 \$18,000 Salvage value \$12,000 \$10,000 Useful life 3 years 2 years The MARR is still 8%. GM wants to decide which of these two arms is a better investment. Assuming repeatability, write expressions for the present worth of both these arms. Setting study period = 6 years (common multiple periods), PW A = –100k – 100k(P/F,8%,3) + 75k(P/A,8%,6) – 24k(P/A,8%,6) + 12k(P/F,8%,3) + 12k(P/F,8%,6) PW B = –80k – 80k(P/F,8%,2) – 80k(P/F,8%,4) + 50k(P/A,8%,6) – 18k(P/A,8%,6) + 10k(P/F,8%,2) + 10k(P/F,8%,4) + 10k(P/F,8%,6) Please write any concerns/comments on the back....
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