Basics Of Engineering Economy
Basics Of Engineering Economy
2nd Edition
ISBN: 9780073376356
Author: Leland Blank, Anthony Tarquin
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 6, Problem 37P
To determine

Selection of independent alternative on the basis of overall rate of return.

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Street lighting fixtures and their sodium vapor bulbs for a two-block area of a large city need to be installed at a first cost (investment cost) of $120,000. Annual maintenance expenses are expected to be $6,650 for the first 10 years and $10,000 each year thereafter upto 25 years. With an interest rate of 10% per year, what is the present worth cost of this project? Choose the closest answer below. OA. The present worth cost of the project is $251,632. OB. The present worth cost of the project is $271,133. OC. The present worth cost of the project is $236,922. OD. The present worth cost of the project is $190,186.
Choose the formula that you will use for the following case: A major repair on the suspension system of a 5-year-old car cost $2000. The cost of periodic maintenance has been $800 every 2 years. If the owner donates the car to charity after 8 years of ownership, what is the quivalent annual cost of the repair and maintenance in the 8-year period of ownership? Use an interest rate of 8% per year, and assume that the owner paid the $800 maintenance cost immediately before donating the car in year 8. AW = -2000 (A/P, 8%, 8) - 800 (A/P, 8%, 2) - 800 (A/P,8%, 4) - 800 (A/P, 8%, 6) - 800 (A/P,8%, 8) PW = -2000 (P/A, 8%, 8) - 800 (P/A, 8%, 2) - 800 (PIA, 8%, 4) - 800 (P/A, 8%, 6) - 800 (P/A, 8%, 8) PW = -2000 (P/F, 8%, 8) - 800 (P/F, 8%, 2) - 800 (P/F, 8%, 4) - 800 (P/F, 8%, 6) - 800 (P/F, 8%, 8) AW = -2000 (A/F, 8%, 8) - 800 (A/F, 8%, 2) - 800 (A/F,8%, 4) - 800 (A/F, 8%, 6) - 800 (A/F,8%, 8)
Identify the engineering economy symbols and their values. Use a question mark with the symbol whose value is to be determined. Fifteen years ago your grandfather invested $18,000 in a stock fund to pay for your college education. Each year thereafter, he deposited $2,600 into the fund. If the investments grew at a rate of 8% per year over the 15 years, how much is in the fund now?   $18,000 =                  (Click to select) n A i F P $2,600 =                  (Click to select) F i A n P 0.08 =                  (Click to select) P n i A F 15 =                  (Click to select) n P i A F ? =                  (Click to select) P n i A F

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Basics Of Engineering Economy

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