3) Using benefit-cost ratio analysis, a 10-year useful life and a MARR of 25%, determine which of the following mutually exclusive models should be selected. Initial Cost Annual Benefits A $100 $37 B $200 $60 C $300 $83 D $400 $137 E $500 $150
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- Alternatives B and C are replaced at the end of their useful lives with identical replacements. Find the best alternative using MARR = 10%. Data Initial Cost Uniform Annual Benefit Useful Life a) Benefit to Cost ratio analysis b) Payback Period Analysis Alt. A $6,00 $150 20 Alt. B $900 $300 5 Alt. C $1,800 $450 10Following 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 6Evaluate the two alternatives A and B and decide the economic justified alternative using: Present worth method 3 Annual worth method, Future worth method E.R.R Method , E.R.R.R method I.R.R method M.A.R.R=15%, the details of alternatives are shown in the table below Alternatives A B Investments $120,000 $155,000 Useful life (years) 15 20 Annual disbursements $25,000 $35,000 Annual revenues $45,000 $60,000 Salvage values $25,000 $30,000
- Complete the following analysis of cost alternatives and select the preferred alternative. The s A D Сapital investment $15,000 S16,000 $13,000 $18,000 Annual costs Market value 250 300 500 100 1,000 1,300 1,750 2,000 at EOY 10 FW(12%) -$49,975 –$53,658 ??? –$55,660 tudy period is 10 years and the MARR = 12% per %3D year.Given the following pertinent data for four alternatives, what is the best alternative using the incremental ROR method given MARR=10% Initial Cost Annual CF, $ Life, years 30 O Alt. A Alt. B O Alt. C A O Alt. D B +22,000 +35,000 с -200,000 -275,000 -190,000 -331,350 30 D +19500 +42,000 30 301. The example in the previous chapter, National Homebuilders, Inc. evaluated cut-and-finish equipment from vendor A (6-year life) and vendor B (9-year life). The PW analysis used the LCM of 18 years. Consider only the vendor A option now. The cash diagram shows the cash flows for all three life cycles (first cost $-15,000; annual M&O costs $-3500; salvage value $1000). Demonstrate the equivalence at i= 15% of PW over three life cycles and AW over one cycle. In previous example, present worth for vendor A was calculated as PW = $-45,036. cost (A) income year net cash flow 0 -15000 0 -15000 1 -3500 0 -3500 2 -3500 0 -3500 3 -3500 0 -3500 4 -3500 0 -3500 5 -3500 0 -3500 6 -18500 1000 -17500 7 -3500 0 -3500 8 -3500 0 -3500 9 -3500 0 -3500 10 -3500 0 -3500 11 -3500 0 -3500 12 -18500 1000 -17500 13 -3500 0 -3500 14 -3500 0 -3500 15 -3500 0 -3500 16 -3500 0 -3500 17 -3500 0 -3500 18 -3500 1000 -2500 ($45,036.36) annual W ($7,349.32)
- Consider two air-conditioning systems with the cash flow estimates as given below. Use AW analysis to determine the sensitivity of the economic decision to MARR values of 4%, 6%, and 8% per year. Air-Conditioning System 1 System First cost, $ AOC. $ per year Salvage valué, $ New compressor and motor cost at midlife, $ Life, years -9,000 -850 -100 Proy -2,200 8 The annual worth of air-conditioning system 1 when the MARR is 4% is $ The annual worth of air-conditioning system 2 when the MARR is 4% is $ The annual worth of air-conditioning system 1 when the MARR is 6% is $ The annual worth of air-conditioning system 2 when the MARR is 6% is $ 10 www Air-Conditioning System 2 -13,000 -105 -300 -3,140 12Fiesta 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 A(b) Use the easiest technique to determine which one of the following alternatives should be selected. MARR = 10% and the chosen alternative is needed for 6 years. Write down any assumptions that you make. Alternative 1 2 Capital Investment Annual Revenues less Expenses Useful Life -$10,000 $6,000 -$15,000 $6,500 2 3
- If the MARR=10%, compute the value of X that makes the alternatives equally desirable. Do not use spreadsheets. Alternatives Machine A Machine B First cost $12,000 $20,000 Annual Operating cost $1,400/year X Salvage value $2,000 $3,000 Life 4 years 8 years:[A] { > Incremental analysis ([ B Alternative], [B wins ]): C A company considering 2 different machines at MARR at 12% Both life spans = 10 years Initial Cost Annual Operating lost Benefits per yin ar Salvage Value \table MM If you are and of frying investment to company decide More than 2 alternatives if the additional increment is worth while, compare Alternative: A Incremental analysis (Alternative) pairs then B C A MARR Company at Considering 2 different machines at 12% Both life spans = 10 years. M/C X м/с у Initial Cost 160000 285000 Annual Operating Cost Benefits per year Salvage Value 45 000 90000 45000 105000 20000 40000Given the following two alternatives and using the repeatability assumption and MARR=15%/year, the equation for computing the present worth of vendor B is Vendor A First Cost, $ -15,000 Annual cost, $ per year -3500 Salvage Value, $ 1000 Life, years 3 Vendor B -18000 -3100 2000 4 OPW=18000-18000(P/F,15%, 4)-18000(P/F,15%, 8)+2000(P/F, 15%, 4)+2000(P/F, 15%, 8) +2000 (P/F, 15%, 12)-3100(P/A, 15%, 12) PW=18000-18000(P/F, 15%, 12)+2000(P/F, 15%, 4)+2000 (P/F, 15%, 8)+2000(P/F, 15%, 12)-3100(P/A, 15%, 12) O PW=-18000-18000(P/F,15%, 4) - 18000(P/F,15%, 8) +2000(P/F, 15%, 4)+2000(P/F, 15%, 8) +2000 (P/F, 15%, 12)-3100(P/A, 15%, 12) O PW=-18000(P/F,15%, 4) -18000(P/F,15%, 8) +2000(P/F, 15%, 4)+2000(P/F, 15%, 8) +2000(P/F, 15%, 12)-3100(P/A, 15%, 12)