A project is being considered that has a first cost of $12,500, creates $5000 in annual cost savings, requires $3000 in annual operating costs, and has a salvage value of $2000 after a project life of 3 years. If interest is 10% per year, which formula calculates the project's present worth? (a) PW = 12,500 (P/F, 10%, 1) + (-5000 + 3000) (P/A, 10%, 3) (b) PW = -12,500 + (5000-3000) (P/A, 10 %, 3 (d) PW=-12, 500+ - 2000 (F/P, 10%, 3) )-2000 (P/F, 10%, 3) (c) PW = 12,500 (F/P, 10%, 3) + (5000-3000) (F/A, 10%, 3) +2000 5000 (P/A, 10%, 3) - 3000 (P/A, 10%, 3) + 2000 (P/F, 10%, 3)

(engineering economic) Four alternatives with cash flows (million rupiah) as follows: Determine the best alternative based on a 12% MARR using the following methods: (a) Rate of Return; (b) Payback Period

How much will Sonja have in a savings account 12 years from now if she deposits $3000 now and $5000 four years from now? The account earns interest at a rate of 10% per year. (a) $10,720 (b) $9,415 (c) $20,133 (d) $15,630 (e) Greater than $22,000

Mechanical engineer at Company B is considering five equivalent projects, some of which have different life expectations. Salvage value is nil for all alternatives. Assuming that the company’s MARR is 13% per year, determine which should be selected (a) if they are independent, and (b) if they are mutually (c) Explain why your selection in part (b) is correct.     First Cost, $ Net Annual Income, $/Year Life, Years A -20,000 +5,500 4 B −10,000 +2,000 6 C −15,000 3,800 6 D −60,000 +11,000 12 E −80,000 +9,000 12