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- 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
- Given the following: Alt. A Alt. B Alt. C Initial Cost $25,000 $40,000 $32,000 Benefits Year 1 5000 2000 6000 Year 2 5500 9000 6000 Year 3 6000 8000 6000 Year 4 6500 7000 6000 Year 5 7200 6000 7000 Salvage Value 4000 8000 6500 MARR 8% Find the best alternative using uniform annual cash flow analysis.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 4If 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 firm is considering the “make vs. buy” question for a subcomponent. If the part is made in-house, the production data would be: first cost = $350, 000; annual costs for operation = $45, 000; salvage value = $15, 000; project life = 5 years; interest = 10%; and material cost per unit = $8.50. If annual production is 10,000 units, the maximum amount that the firm should be willing to pay to an outside vendor for the subcomponent is nearest? (a) $10 per unit (b) $16 per unit (c) $22 per unit (d) $28 per unit?The company will invest for a 2-year project. The first year, return of the investment is 240.000 USD and the second year, return is 432.000 USD. If the annual interest rate is 20%, how much money should be invested at present value for this investment to be profitable?Please no written by hand and no emage Solve in excel Carp, Inc. wants to evaluate two machines for packaging their products.Machine A:Initial cost is $700,001st year O&M cost is 18,000; this cost increases $900 each year.The annual benefits are $154,000It can be sold at the end of 10 years useful life for $145,000 Machine B:Initial cost is $1,600,001st year O&M cost is 28,000; this cost increases $650 each year.The annual benefits are $300,000It can be sold at the end of 20 years useful life for $210,000The companies uses an interest rate of 15% Use annual cash flow analysis to decide which is the most desirable alternative.
- As an engineer, you are responsible to find the best alternative that is economically best. The cash flows shown in the table and the MARR is 11 % per year. Then please answer the following questions. Cash Flow Project 01 Project 02 Project 03 First Cost, $ -90000 220000 230000 Annual Cost, $ Per Year -60000 -21000 -17000 Overhaul Every 10 Years, $ _ _ -60000 Salvage Value, $ 8000 29000 11000 Life, Years 3 10 a) Find the Aw for Project 01. b) calculate the AW for Project 02 c) Calculate Aw for Project 03. What is your decision?7. Two machines are considered for purchase. Assume 10% interest, Use Benefit Cost analysis. Machine X Machine Y Initial Cost $200 $700 Uniform annual benefit $95 $120 Salvage value $50 $150 Useful life in years 6 12 a. Which machine should be bought? b. List the decision guideline for single project c. List the selection rule for incremental analysis d. List the procedures for the incremental analysis. Solve all this question......you will not solve all questions then I will give you down?? upvote.....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) − 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)