Four mutually exclusive investment alternatives have a 5-year useful life and no salvage value. The MARR is 6%. Which investment should be selected? Initial Cost Uniform Annual Benefit 0 000 U A B A B C D $400 $125 $225 $500 $100 $35 $50 $135
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- build a model Each of the three alternatives shown has a 5-yearuseful life. MARR is 10%, a) Using FW, which alternative should be selected? B) Using benefit–cost ratio analysis, which alternative should be selected? C) what is the discounted payback period of each alternative? A B C Cost $ 650.00 $ 500.00 $ 250.00 Uniform annual benefit $ 180.00 $ 140.00 $ 66.00 Useful life, years 3 6 3When assessing mutually exclusive alternatives using Present Worth Analysis, which alternative/alternatives should be selected? a. Select none unless all PW are positive b. Select all with positive PW c. Select the alternative with the most positive PW d. Discard the negative PW and select all with positive PW In assessing an alternative, the capital recovery is found to be $ -2.47 (millions). What is this proper interpretation of this number? a. annually the alternative must have net revenues of at least $2.47 to recover the initial cost at a required MARR b. the alternative will lose $ 2.47 annually c. The AW is $ -2.47 d. The alternative will recover $ -2.47 annually once in operationTwo hazardous environment facilities are being evaluated, with the projected life of each facility being ten years. The company uses a minimum attractive rate of return (MARR) of 15%. Using incremental net present value, which alternative should be selected? B $300,000 $25,000 $92,000 -$5,000 A $615,000 $10,000 $158,000 $65,000 First Cost O&M Cost Annual Benefit Salvage Value
- Following cash flows for alternatives X and Y at an interest rate of 10% per year. Initial cost, $ AOC, $/year Annual revenue, $/year Salvage value, $ Life, years Multiple Choice In comparing the alternatives on a present worth basis, the PW of Machine X is closest to O O $23.160. $40,560 $58,950. Machine X Machine Y -146,000 -220,000 $72,432 -15,000 -10,000 80,000 75,000 10,000 25,000 3 6A Cost $4000 $2000 $6000 $1000 $9000 Annual benefit 639 410 761 117 750 Useful life, in years 20 20 20 20 20 i* 15% 20% 11.1% 9.9% 5.45% Perform incremental ROR analysis to select best one. MARR=6%; (P/A, 6%, 20) = 11.47 %3DEvaluate the two alternatives A and B and decide the economic justified alternative using: Present worth method, Annual worth method, Future worth method I.R.R method , E.R.R Method .E.R.R.R method M.A.R.R=15%, the details of alternatives are shown in the table below Alternatives A B Investments $60,000 $75,000 Useful life (years) 5 15 Annual disbursements $25,000 $35,000 Annual revenues $45,000 $60,000 Salvage values $5,000 $10,000
- QUESTION 7 for th below two machines and based on AW analysis which machine we should select? MARR-10%. Machine A Machine B First cost, $ 100,000 Annual cost, S/year 7,000 Salvage value, $ Life, years 3 infinite Answer the below question: A-the AW for machine A QUESTION 8 For th below two machines and based on AW analysis which machine we should select? MARR-10% Machine A Machine B First cost, $ 135,982 Annual cost, Siyear 8,740 12.308 Salvage value, $ Life, years 3 infinite Answer the below question: B- the AW for machine B QUESTION 9 C-Based on the AW value you got in the previous 2 questions, which machine we should select? type you explanation below For the toolbar, press ALT+F10 (PC) or ALT-FN-F10 (Mac) B IUS Paragraph M Arial Y 10pt 田饼印园 BE 0) 20,000 9,000 4,000 23,720 4.805- !!! 4A project is being planned that has an initial investment at time 0, annual revenuesand expenses, and a salvage value at the end of the project lifespan (20 years). The financialvalues are summarized below:Initial investment amount at time 0 $150,000Estimated annual revenue $34,500 per yearEstimated annual expenses $8,700 per yearEstimated salvage value at end of lifespan $10,000Minimum attractive rate of return (MARR) 15%a. Calculate the capital recovery amount CR(i%).b. Using the annual worth (AW) method, determine whether purchasing the equipmentis economically justified.c. Repeat part (a) using the internal rate of return (IRR) method based on annual worth(AW).d. Using the present worth (PW) method, determine the break-even time period afterwhich purchase of the equipment generates a profit. (Find N when PW = 0) year period.A B C $9,000 $12,000 $1,800 Initial cost $15,000 $8,000 $5,000 2 years Annual benefit Salvage value Life in years $2,000 $9,000 3 Years Infinity MARR 10% i. The NPW of alt. A is A) $13,420 B) $17,380 C) $11,000 D) $6,00 ii. The NPW of alt. B is A) $13,420 B) $17,380 C) $11,000 D) $6,000 iii. The NPW of alt. C is A) $13,420 B) $17,380 C) $11,000 D) $6,000 iv. The best alternative among the three alternatives using the PW analysis is A) Alt. A B) Alt. B C) Alt. C D) Either alt. B or alt.
- Fiesta Foundry is considering a new furnace that will allow them to be more productive. Three alternative furnaces are under consideration. Perform an incremental analysis of these alternatives using the IRR method for each increment of cash flows. The MARR is 10% per year. E Click the icon to view the description of the alternatives. Click the icon to view the interest and annuity table for discrete compounding when the MARR is 10% per year. Perform the incremental PW Analysis. Fill-in the table below (Hint: Order alternatives by increasing capital investment). (Round to the nearest dollar.) Incremental Investment Inc. PW Alternative to be selected A V A3) A and B are mutually exclusive projects.. What MARR has to be for A to be chosen? A B Initial cost Useful life Annual benefit Salvage value Rate of return $30,000 6 years $8,577 $0 18%lyr $50,000 6 years $8,577 $29,098 12%/yr a) MARR<8.4% b) MARR< 6.4% c) 8.4%Xanadu Mining is considering three mutually exclusive alternatives, as shown in the table below. MARR is 10.0%/year. EOY A001 B002 C003 0 -$210 - $110 -$160 1 $100 $60 $76 2 $110 $60 $76 3 $120 $60 $76 4 $130 $70 $76 Click here to access the TVM Factor Table Calculator art a What is the present worth of each alternative? Alternative A001: $ Alternative B002: $ Alternative C003: $SEE MORE QUESTIONS