STMN 7) The alternatives are mutually exclusive and the MARR is 10% per year. The annu worth (AW) of alternative X & Y is closest to First Cost, $ Annual Income, S/year Annual Operating Cost, S/year Maintenance Cost every years, S Life, years a. 45.75 & -29.50 b.-35.96 & 29.50 c. 35.96 & -36.87 d. 45.75 & 36.87 e. 35.96 & 29.50 -200 120 -50 -25 10 Y -550 110 -20 -50 S
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- Which alternative in the table below should be selected when the MRR = 6% per year? The life of each Kalternative is 10 years. Increment Considered A Investment cost A (Annual Revenues less. Costs) IRR on A Investment Cost A(A-DN) $800 $154 14.1% The IRR on A(C-B) is %. (Round to one decimal place.) A(B-A) $700 $114 10.0% A(C-B) $1,100 $170 ? A(D-C) $1,300 $130 2period of 5 years. Salvage value is estimated at 10% of its first cost. The interest rate is 12% per at $ 150,000. Labor and other operating costs are estimated to be $35,000 per year over the study high-use component can be purchased (bought) for S 25 per unit with a quick delivery. QWA Electronics is considering between two alternatives for their electronic components' needs. A Mestion #6 (20 points) cost of $ 5 per unit. For this option of making the product, QWA needs to purchase equipment today Alternatively, QWA can make the components in-house and have it already available at a variable year. Determine the breakeven quantity based on annual worth of cash flows. 0. Indicate whether to buy or make the components at the expected usage level of 5000 units per year. Explain your answer. a.Dexcon Technologles, Inc., Is evaluating two alternatives to produce its new plastic filament with tribological (I.e., low friction) properties for creating custom bearings for 3-D printers. The estimates associated with each alternative are shown below. Using a MARR of 17% per year, which alternative has the lower present worth? Method First Cost M&O Cost, per Year Salvage Value Life DDM LS $-140,000 $-30,000 $6,000 2 years $-480,000 $-35,000 $29,000 4 years The present worth for the DDM method is $ -217913 The present worth for the LS method is $ The method is selected.
- Advanced Technologies, Inc. is evaluating two alternatives to produce its new plastic filament with tribological (low friction) properties for creating custom bearing for 3-D printers. The estimates associated with its alternatives are shown below. Use LCM way and a MARR of 20% per year. Method DDM LS First Cost $ -165,000 -375,000 M&O cost, $/year -50,000 -25,000 Salvage value $ 30,000 Life, years 2 4 Which Method should be selected? O a. Select Method DDM since it generates higher revenue. b. Select Method LS since it generates higher revenue. c. Select Method DDM since it is less costly. O d. Select Method LS since it is less costly.Which alternative in the table below should be selected when the MARR =7% per year? The life of each alternative is 10 years. A(A - DN) A(B - A) $700 A(C - B) $1,100 Increment Considered A(D - C) $1.300 A Investment cost A (Annual Revenues less Costs) $800 $140 $108 $161 $130 IRR on A Investment Cost 11.7% 8.8% ? The IRR on A(C - B) is || %. (Round to one decimal place.)For the following table, if the MARR is 10% per year and a useful life for each alternative of six years that equals the study period. The rank order of alternatives from least capital investment to greatest capital investment is Do nothing Fill the table with the missing values. > A C в Do nothing > A ? (2) ? (4) A capital investment A annual revenues A annual costs A market value A IRR -15,000 -2000 -3000 4,000 900 460 -150 -1,000 6,000 -100 -2,220 3,350 ? (5) (6) 12.7% 10.9% Alternative chosen (1) (3) 1. write the alternative chosen 2. write the two alternatives compared 3. write the alternative chosen 4. write the two alternatives compared 5. what is IRR incremental (final answer) 6. write the alternative chosen
- Investment Cost Useful Life 15 years Market Value, aka salvage value (EOY 15) 3,000 Annual Operating Expenses Overhaul cost--EOY 5 Overhaul cost--EOY10 ('000s of $) 13,000 $19,553,380. NO. 1,000 $19,210,488. YES 200 Evaluate a combined cycle power plant on the basis of the PW method when the MARR is 12% per year. Pertinent cost data are shown in the table. Should the plant be built? 550 $19,553,380. Impossible to say without additional information. O $19,210,488. Impossible to say without additional information.A company that manufactures magnetic membrane switches is investigating two production options that have the estimated cash flows shown ($1 million units) in Table below. Which one should be selected on the basis of Present Worth (PW) analysis at 10% per year? In-house Contract First cost, $ Annual cost, $ Annual income, $ per year Salvage value, $ Life, years -30 per year -5 -2 14 3.1 10% TABLE 15 Discrete Cash Flow: Compound Interest Factors 10% Single Payments Uniform Series Payments Arithmetic Gradients Compound Present Worth Sinking Fund Compound Capital Recovery A/P Present Gradient Gradient Amount Amount Worth Present Worth Uniform Series F/P P/F A/F F/A P/A P/G A/G 1 1.1000 0.9091 1.00000 1.0000 1.10000 0.9091 1.2100 0.8264 0.47619 2.1000 0.57619 1.7355 0.8264 0.4762 3 1.3310 0.7513 0.30211 3.3100 0.40211 2.4869 2.3291 0.9366 4 1.4641 0.6830 0.21547 4.6410 0.31547 3.1699 4.3781 1.3812 0.6209 0.5645 0.5132 1.6105 0.16380 6.1051 0.26380 3.7908 6.8618 1.8101 6 1.7716 0.12961…An investment of P270,000 can be made in a project that will produce a uniform annual revenue of P185,400 for 5 years and then have a salvage value of 10% of the investment. Out-of-pocket cost for operation andmaintenance will be P81,000 per year. Taxes and insurance will be 4% of the first cost per year. The company expects capital to earn not less than 25% before income taxes. (a) Future Worth Method(b) Present Worth Method What is the future worth of net cash flows
- K Dexcon Technologies, Inc., is evaluating two alternatives to produce its new plastic filament with tribological (ie, low friction) properties for creating custom bearings for 3-D printers. The estimates associated with each alternative are shown below. Using a MARR of 15% per year, which alternative has the lower present worth? Method First Cost M&O Cost, per Year Salvage Value Life DDM $-200,000 $-40,000 $2,000 2 years The present worth for the DDM method is $2635271 The present worth for the LS method is $ The [(Click to select) method is selected LS $-500,000 $-40,000 $21,000 4 yearsMARR = 30% B. -12,000 5,000 2,500 Initial Investment -18,000 Estimated net income per year: 7.000 3,000 Salvage value Estimated competitive life, years (a) Perform incremental ROR analysis (b) Develop the PW vs. i graphs for each alternative and for the increment. (c) Create Choice Table.An environmental engineer is considering three methods for disposing of a non-hazardous chemical sludge: land application, fluidized-bed incineration, and private disposal contract. The estimates for each method are below. Determine the equivalent present worth value of the LAND APPLICATION using its AW value based on 10% per year. Refer to table 2 for additional information. Table 2 Land Application Inciner- ation Contract Firg cost, S -150.000 -a00 000 Annual cost, Syear -95.000 -60,000 -170,000 Salvage value, S Life, years 25.000 300,000 4. 6. O $1,158,329 O $-1,158,329 O $933,027 O $-933,027 O No correct answer