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- 5-27image If produced by Method A, a product’s initial capital cost will be $100,000, its annual operating cost will be $20,000, and its salvage value after 3 years will be $20,000. With Method B there is a first cost of $150,000, an annual operating cost of $10,000, and a $50,000 salvage value after its 3-year life. Based on a present worth analysis at a 15% interest rate, which method should be used?A purchase of a new machine is being evaluated. If we assume 12% annual interest, what is the Equivalent Uniform Annual Cost? Choose the nearest answer. Initial Cost $30,000 Annual Maintenance $1,500 Salvage Value $5,000 Life 6 yearsA manufacturer is considering replacing a production machine tool. The new machine, costing $3700, would have a life of 4 years and no salvage value, but would save the firm $500 per year in direct labor costs and $200 per year indirect labor costs. The existing machine tool was purchased 4 years ago at a cost of $4000. It will last 4 more years and will have no salvage value at the end of that time. It could be sold now for $1000 cash.Assume that money is worth 8% and that the difference in taxes, insurance, and so forth, for the two alternatives is negligible. Use an annual cash flow analysis to determine whether the new machine should be purchased.
- A city has developed a plan to provide for future municipal water needs. The plan proposes an aqueduct that passes through 150 meters of tunnel in a nearby mountain. Two alternatives are being considered. The first proposes to build a full-capacity tunnel now for $556 000. The second proposes to build a half capacity tunnel now and a second identical half-capacity tunnel in 20 years. Each of half capacity tunnel costs $402 000. The maintenance cost of the tunnel lining for the full-capacity tunnel is $40 000 every 10 years, and for each half-capacity tunnel it is $32 000 every 10 years. The friction losses in the half-capacity tunnel will be greater than if the full-capacity tunnel were built. The estimated additional pumping costs for each half-capacity tunnel will be $2 000 per year. Using present worth method and a 7% interest rate, which alternative should be selected? Give typing answer with explanation and conclusionConsider your self as a businessman, you owned 5 storey building with a total of 35-unit apartment near at the downtown area of Davao City. You felt that because the location of the apartment will be occupied 95% at all time. You desires a rate of return 30%. Other pertinent data are the following: Land investment - 8,000,000.00 Building investment - 20,000,000.00 Study period - 30 yrs Cost of the land after 30 yrs - 25,000,000.00 Cost of the building after 30 yrs - 5,000,000.00 Rent per unit per month - 7,500.00 Upkeep per unit per year - 1,500.00 Property Taxes - 1% Insurance - 0.5% Is this a good investment? And what is the Payback period of investment? Note: use all the method.3) The cost of painting the public bridge is $15000. If the bridge is painted now and every 7 years, what is the capitalized cost (CC) of painting at an interest rate of 10% per year. a.-16650 b.-155000 c.-35750 d. -55990 e.-31500
- Calculate the future worth (FW) at 10% of a project that will save $25K per year for 20 years. The first cost is $120K, and the salvage value is $20K. Compare this with the PW and the EAW. (Please show the process and solution ty.)there are three machines in the mechanicle angineering lab A ,B , and C and need to be evaluted economically, machine A has first cost of $4500,an annual operating cost(AOE) of $900, a salvage value of $200 and a service life 4 year. machine B has first cost of $3500, an annual operating cost(AOE) of $700, a salvage value of $350, and a service life 4 year. machine C has first cost of $6000, an annual operating cost(AOE) of $50, a salvage value of $100 and a service life 8 year. which machine should be selected ? the MARR is 10% per year. (a) machine A (B) machine B (c) machine C (d) noneA 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) − 2000(F/P, 10%, 3) (b) PW = − 12,500 + (5000 − 3000) (P/A, 10%, 3 ) − 2000(P/F, 10%, 3) (c) PW = 12,500(F/P, 10%, 3) + (5000 − 3000) (F/A, 10%, 3) + 2000 (d) PW = − 12, 500 + 5000(P/A, 10%, 3) − 3000 (P/A, 10%, 3) + 2000(P/F, 10%, 3)
- The product development group of a high-tech electronics company developed five proposals for new products. The company wants to expand its product offerings, so it will undertake all projects that are economically attractive at the company’s MARR of 20% per year. The cash flows (in $1000 units) associated with each project are estimated. Which projects, if any, should the company accept on the basis of a present worth analysis? Project A B C D E Initial investment, $ −400 −510 −660 −820 −900 Operating cost, $/year −100 −140 −280 −315 −450 Revenue, $/year 360 235 400 605 790 Salvage value, $ — 22 — 80 95 Life, years 3 10 5 8 4Municipal Engineer wants to evaluate three alternatives for supplementing the water supply. 1st alternative – continue deep well pumping at an annual cost of $10,500 2nd alternative – install a 10” pipeline from a surface reservoir. First cost is $25,000 and annual pumping cost is $7,000 3rd alternative – install a 20” pipeline from the reservoir. First cost of $34,000 and annual pumping cost of $5,000. Life of all alternatives is 20 years. For the second and third alternatives, salvage value is 10% of first cost. With interest at 8%, which alternative should the engineer recommend? Use present worth analysis PW (deepwell) = ? PW (10”pipeline) = ? PW (20”pipeline) = ?A company has annual fixed costs of $2,500,000 and variable costs of 0.15¢ per unit produced. For the firm to break even if they charge $1.85 for their product, the level of annual production is nearest to what value? (a) 375,000 units (b) 1,315,789 units (c) 1,351,351 units (d) 1,562,500 units