Acme Molding is examining 5 alternatives for a piece of material handling equipment. Each has an expected life of 8 years with no salvage value, and Acme's MARR is 12%. Using an incremental analysis, which material handling alternative should be chosen? The table below includes initial investment, net annual income, and IRR for each alternative. Alternatives A B D E Capital investment $12,000 $2,500 $12,500 $2,520 $14,400 $3,050 $16,250 S3,620 $20,000 $4,400 Net annual income IRR 12.99% 12.04% 13.48% 14.99% 14.61%

Acme Molding is examining 5 alternatives for a piece of material handling equipment. Each has an expected life of 8 years with no salvage value, and Acme's MARR is 12%. Using an incremental analysis, which material handling alternative should be chosen? The table below includes initial investment, net annual income, and IRR for each alternative. Alternatives A B D E Capital investment $12,000 $2,500 $12,500 $2,520 $14,400 $3,050 $16,250 S3,620 $20,000 $4,400 Net annual income IRR 12.99% 12.04% 13.48% 14.99% 14.61%

Essentials Of Investments

11th Edition

ISBN:9781260013924

Author:Bodie, Zvi, Kane, Alex, MARCUS, Alan J.

Publisher:Bodie, Zvi, Kane, Alex, MARCUS, Alan J.

Chapter1: Investments: Background And Issues

Section: Chapter Questions

Problem 1PS

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Saved Help Save & Prairie Corporation has provided the following information for a proposed investment project: Discount rate 10% Life of the project 4 years Initial investment cost $23,775 Annual cost savings 7,500 Salvage value 1,500 What is the net present value of the proposed investment project? (Select the answer that is closest to your calculations.) Present value tables are provided below. Present Value of $1 Table (Exhibit 11B-1) (Partial table) Periods 4% 5% 6% 7% 8% 9% 10% 11% 12% 13% 14% 0.962 0.952 0.943 0.935 0.926 0.917 0.909 0.901 0.893 0.885 0.877 0.925 0.907 0.890 0.873 0.857 0.842 0.826 0.812 0.797 0.783 0.769 0.889 0.864 0.840 0.816 0.794 0.772 0.751 0.731 0.712 0.693 0.675 0.855 0.823 0.792 0.763 0.735 0.708 0.683 0.659 0.636 0.613 0.592 0.822 0.784 0.747 0.713 0.681 0.650 0.621 0.593 0.567 0.543 0.519 4. 0.790 0.746 0.705 0.666 0.630 0.596 0.564 0.535 0.507 0.480 0.456 0.760 0.711 0.665 0.623 0.583 0.547 0.513 0.482 0.452 0.425 0.400 0.731 0.677 0.627 0.582 0.540…
A company with a MARR of 10% must install one of two production machines. The economic parameters of each machine are as follows: Machine Initial cost (5) Service life (years) Salvage value end of life (S) $15,000 4 years $22,000 6 years 52,000 $1,000 The Future worth (FW) for the machine Y over the 12 years analysis period is:
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 value, $ New compressor and motor cost at midlife, $ Life, years -10,000 -500 -100 -1,850 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 $ The annual worth of air-conditioning system 1 when the MARR is 8% is $ The annual worth of air-conditioning system 2 when the MARR is 8% is $ Air-Conditioning System 2 -14,000 -85 -300 -2,540 12
Two machines with the following cost estimates are under consideration for a dishwasher assembly process. Using an interest rate of 10% per year, determine which alternative should be selected on the basis of an annual worth analysis.
What is the present worth equivalent of the machine considered to be the most preferred choice? (i=10%) show detailed solution
Pitt Company is considering two alternative Investments. The company requires a 12% return from its Investments. Neither option has a salvage value. Project X Project Y $243,046 $175,883 Initial Investment Net cash flows anticipated: Year 1 Year 2 Year 3 Year 4 Year 5 82,000 60,000 91,000 82,000 75,000 A. Compute the IRR for both projects using the IRR spreadsheet function. Project X Project Y B. Which project should be recommended. Project X ✓ % % 35,000 54,000 73,000 69,000 26,000
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onsider the following financial informationabout a retooling project at a computer manufacturing company:• The project costs $2.5 million and has a five-yearservice life.• The retooling project can be classified as sevenyear property under the MACRS rule.• At the end of the fifth year, any assets held for theproject will be sold. The expected salvage valuewill be about 10% of the initial project cost.• The firm will finance 40% of the project moneyfrom an outside financial institution at an interestrate of 10%. The firm is required to repay the loanwith five equal annual payments.• The firm’s incremental (marginal) tax rate on theinvestment is 35%.• The firm’s MARR is 18%.With the preceding financial information,(a) Determine the after-tax cash flows.(b) Compute the annual equivalent worth for thisproject.
The Fence Company is setting up a new production line to create top rails. The relevant data for two alternatives are shown below. Flow Line Manufacturing Cell Installed Cost Expected Life Salvage Value Variable Cost per Top Rail Click here to access the TVM Factor Table Calculator $15,000 5 years $0 $6.00 $10,000 5 years $0 $7.00
expense of $500. Consider a yearly Two machines are evaluated for a new project, consider a yearly interest rate of 6% and the information described in the following table. Determine EUAC for both alternatives and determine the best option. Purchase cost Salvage value Duration Machine A Machine B $10,000 $15,000 $1,250 $1,950 5 years 10 years 11 13 L
Compare three alternatives on the basis of their capitalized costs at /= 8% per year and select the best alternative. Alternative E F G $-460,000 $-30,000 $90,000 4 First Cost AOC, per Year Salvage Value Life, Years $-80,000 $-65,000 $40,000 2 The capitalized cost of alternative E is $ The best alternative is (Click to select) ▼ alternative F is $ $-960,000 $-3,000 $450,000 ∞0 and alternative G is $
Compare the Annual Worth of the two systems and identify the better option at MARR = 10% per year: Solar: First cost $1,500,000; AOC $-700,000; Salvage value $100,000; Life of 8 years Geothermal: First cost $2,250,000; AOC $-600,000; Salvage value $50,000; Life of 8 years Select the Geothermal System with a calculated AW of $-1,017,368 Select the Geothermal System with a calculated AW of $-727,591 Select the Solar System with a calculated AW of $-1,238, Select the Solar System with a calculated AW of $-972,416

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