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Determine the present worth, future worth, and annual worth of the following engineering project when the MARR is 15% per year. Is the project acceptable?
Investment cost $10,000
Expected life 5 years
Market (salvage) value $1,000
Annual receipts $8,000
Annual expenses $4,000
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- Problem 2: Select one of the following two plans using the present worth method. Given i= 20% per year. 1st Cost ($) Life (years) Salvage value ($) Annual Cost ($) Plan 1 25,000 10 5,000 6,000 Plan 2 50,000 20 0 2,500A 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.Given 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)
- :[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 40000Ch 5 economics Nancy’s Notions pays a delivery firm to distribute its products in the metro area. Delivery costs are $31,500 per year. Nancy can buy a used truck for $8,000 that will be adequate for the next 3 years. Operating and maintenance costs are estimated to be $24,000 per year. At the end of 3 years, the used truck will have an estimated salvage value of $3,200. Nancy’s MARR is 25%/year. What is the present worth of this investment? $Q#1. Complete the analysis based on IRR and ERR methods separately to identify if the given project is a good investment or not. The MARR is 10% potential cost and revenue data are as follow. Also draw the cash flow diagram. Project A $19000 $500 Investment Cost Annual Operation Cost Annual Revenue $3500 Salvage Value Useful Life $3000 10 Years (Also draw the cash flow diagram).
- An electronics firm is planning to manufacture a new handheld gaming device for the preteen market. The data have been estimated for the product. Assuming a negligible market (salvage) value for the equipment at the end of five years, determine the breakeven annual sales volume for this product.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 years1. 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)
- Evaluate 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 $10,000Alternatives 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 10A 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 %3D
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