For a steam power plant the repair costs are estimated to be 1.5$ per short ton coal burned. The Fuel will be coal at $38.6, a short ton having a high heat value of 14200 Btu/lb. If the fuel and repair costs are 278950 $ find the repair cost Select one: a. 10943 $ b. 10558 $ c. 9190 $ d. 10435 $ e. 9942 $
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- ENGINEERING ECONOMY RATE WILL BE GIVEN. WRITE THE COMPLETE SOLUTIONS/EXPLANATION LEGIBLY OR TYPEWRITTEN. GIVE STRAIGHT TO THE POINT EXPLANATION. A tire manufacturer produces tires at a variable cost of $25 per unit. The plant has annual sales of $480,000 at a price of $40 per unit. Find the breakeven point in units of production. The annual fixed costs for the plant are $150,000.Nissan is advertising a 24-month lease on the all-electric Leaf for $190 payable at the beginning of each month. The lease requires a $2000 down payment plus the first months payment on signing. No security deposit is required. The lessee pays for maintenance, excess wear and tear, and $0.18 per mile over 12,000 miles per year. Lease end purchase option is $17500 and lease payments total $4560. A disposition fee of $350 is due at the end of the lease period. Assuming an interest rate of 6% compounded monthly, what is the total cost of leasing if you put 40,000 miles at the end of lease?Data for two 50-hp motors are as follows: Alpha motor Beta motor Original cost P37 500 P48 000 Annual maintenance 1500 750 Life, years Efficiency 10 10 87% 92 % Taxes and insurance 3% 3% Power cost is P2 per kWh. If money is worth 20%, how many hours per year would the motors have to be operated at full load for them to be equally economical? If the expected number of hours of operation per year exceeds the break-even point, which motor is more economical?
- Required information Assume you started a sideline business in commercial photography last year using your then-owned equipment. Due to excellent success, you plan to purchase new equipment and upgrade your studio facility. First cost of equipment, $ -160,000 Annual expenses, $ per year -45,000 Annual revenue, $ per year 75,000 Determine the no-return payback period. The no-return payback period is years.12.9 A three-year maintenance contract for a local computer network costs $4000. The network is expected to be needed for fifteen years and the maintenance contract will be purchased for the same price at the beginning of every three year period. If the interest rate is 6%, the present equivalent cost of the maintenance contract is nearest: (A) $12,950 (B) $14,500 (C) $15,400 (D) $16,000You plan to start saving money by depositing $5000 into your savings account now. Also, you estimate that you will make a deposit of $200 each year for the next 9 ears. What would be the future worth if the investment interest rate is 7% per year? (solve using both tables and rules)
- B) Calculate the AEC (Average Equivalent Cost) of the hospital building using the cost requirements given in Table 3 ) Table 3 Construction Cost OMR 1 million OMR 150,000 Annual Running Cost OMR 30,000 Repainting Costs @ every 2 Years Repair Costs @every 2 years OMR 25,000 Major replacement cost @ 5 years OMR 75,000 (Including the service costs) The demolition cost is estimated at OMR 200,000 (excluding the Salvaged material valued at OMR 60,000) at the end of 15th year. The interest rate (r) is 5% and the Annual Sinking Fund (ASF) is 1.5%. Note: The PV and SF tables is provided as attachment to this question paper.Problem 13 The Keiser Mining Company must deposit money in an account TODAY (n=0) to cover the anticipated cleanup costs that will begin when the mine closes. These cleanup costs of $6000 will begin in year 5 and repeat every 2 years, as shown below. The account will pay 5% annual interest. $6 $6 $6 $in thousands of 5 - ST 10 11 = The amount Keiser Mining Company should deposit TODAY is closest to: a. $45,900 b. $48,200 c. $50,600 d. $58,500Q1:A person wants to have (5000 CU) pay for his education at the end of each year for (6 years). If the bank rate of interest is (11.5%), how much money has to be deposit at present. Q2:A company invests in one of three alternatives, the life of all alternatives is estimated to be (5 years) with the following investments, annual returns, and salvage value: Alternatives 3 200000 CU 70000 CU 50000 CU Details Alternatives 1 Alternatives 2 Investment 150000 CU 175000 CU Annual equal return 60000 CU 70000 CU 15000 CU 25000 CU Salvage value Determine the best alternative based on Net Present Worth Comparison (NPW) by assuming i=10%. نوں
- A large textile company is trying to decide which sludge dewatering process it should use ahead of its sludge drying operation. The costs associated with centrifuge and belt press systems are shown. Compare them on the basis of their annual worths using an interest rate of 10% per year. System First cost, $ Centrifuge -250,000 -31,000 Belt Press -170,000 -35,000 -26,000 10,000 4 AOC, $/year Overhaul in year 2, $ Salvage value, $ Life, years The annual worth of the centrifuge system is $- The system selected on the basis of the annual worth analysis is the (Click to select) ✓ system. 40,000 6 and the annual worth of the belt press system is $-A piece of asset was purchased 4 years ago. Current market value is now $13,000. Estimated Future Market Values and AOC's: (Assume the interest rate is 10% per year) are as follows. Market EOY value, $ AOC 1 9000 2500 8000 2700 6000 3000 4 2000 3500 4500 What is the ESL? 2 years 3 years 4 years 45yearsAn engineering firm has identified five ways to cut costs in its main office. Only one of the options can be implemented, however, since each involves significant training time for staff engineers. Data are provided in the table. Each option has a lifetime of seven years, and the firm sets a MARR at 15%. Option A B C D E Capital cost ($ million) 2.713 0.375 1.650 0.088 0.950 Annual cost ($ million/yr 0.093 0.275 0.132 0.147 0.228 Annual benefit ($ million/yr) 0.890 0.288 0.841 0.312 0.505 a) Solve by present worth analysis. b) Solve by annual cash flow analysis. c) Solve by incremental benefit-cost ratio analysis. d) Solve by an incremental rate of return analysis using the full detailed procedure