# HW-Chap-5-2 - (9.818 PW of Fuel Oil Cost = \$103,635 PW of...

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Engineering Economy ENGR 300 Chapter 5 5-45 PW of Cost of 30 years of Westinghome = \$45,000 + \$2,700 (A/P, 10%, 30) + \$42,000 (P/F, 10%, 10) + \$42,000 (P/F, 10%, 20) − \$3,000 (P/F, 10%, 30)= \$45,000 + \$2,700 (9.427) + \$42,000 (0.3855) + \$42,000 (0.1486) − \$3,000 (0.0573) = \$92,713 PW of Cost of 30 years of Itis = \$54,000 + \$2,850 (P/A, 10%, 30) + \$49,500 (P/F, 10%, 15) − \$4,500 (P/F, 10%, 30) = \$54,000 + \$2,850 (9.427) + \$49,500 (0.2394) − \$4,500 (0.0573) = \$92,459 The Itis bid has a slightly lower cost. 5-53 Capitalized Cost = PW of an infinite analysis period When n = ∞ or P = A/ i PW = \$5,000/0.08 + \$150,000 (A/P, 8%, 40)/0.08 = \$62,500 + \$150,000 (0.0839)/0.08 = \$219,800 5-64 For fixed output, minimize PW of Cost: Natural Gas PW of Cost = \$30,000 + \$7,500 (P/A, 8%, 20) + PW of Fuel Oil Cost = \$30,000 + \$7,500
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Unformatted text preview: (9.818) + PW of Fuel Oil Cost = \$103,635 + PW of Fuel Oil Cost Fuel Oil PW of Cost = \$55,000 + PW of Fuel Oil Cost Coal PW of Cost = \$180,000 − \$15,000 (P/A, 8%, 20) + PW of Fuel Oil Cost = \$180,000 − \$15,000 (9.818) + PW of Fuel Oil Cost = \$32,730 + PW of Fuel Oil Cost. Install coal-fired steam boiler. 5-73 In this problem, choosing to Maximize NPW per share leads to Western House. But the student should recognize that this is a faulty criterion. An investment of some lump sum of money (like \$1,000) will purchase different numbers of shares of the various stock. It would buy 83 shares of Spartan Products, but only 42 shares of Western House. The criterion, therefore, is to Maximize NPW for the amount invested. This could be stated as Maximize NPW per \$1 invested. Buy Spartan Products....
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