Two pumps are being considered for purchase for a service life of 10 years. Interest rate is 5%.AlternativesPump 1 Pump 2Initial Cost$60,000 $75,000Estimated salvage value at end of usefullife$10,000 $12,000Useful Life6 years12 yearsEstimated market value, end of 10-year $12,000 $15,000Apply the present worth technique, which pump should be considered?Solution:1. Neither input nor output is the same, calculate the net present worth (NPW)2. NPW;:o Initial cost: PW=o Equivalent PW for salvage and replacement together: PW=k;o Cash flow at the end of 10 years, PW=k;o Thus NPW =k.3. NPW2:o Initial cost: PW=o Equivalent PW for salvage and replacement together: PW=k;o Cash flow at the end of 10 years, PW=k;o Thus NPW2=k.4. Hence use

Initial Cost = $- 60000 Equivalent PVV for salvage & replacement cost = 10000/(1.05)6 -…

Two pumps are being considered for purchase for a serVice life of 10 years. Interest rate is 5%. Alternatives Pump 1 Pump 2 Initial Cost $60,000 $75,000 Estimated salvage value at end of useful life $10,000 $12,000 Useful Life 6 years 12 years Estimated market value, end of 10-year $12,000 $15,000 Apply the present worth technique, which pump should be considered?

Two pumps are being considered for purchase for a serVice life of 10 years. Interest rate is 5%. Alternatives Pump 1 Pump 2 Initial Cost $60,000 $75,000 Estimated salvage value at end of useful life $10,000 $12,000 Useful Life 6 years 12 years Estimated market value, end of 10-year $12,000 $15,000 Apply the present worth technique, which pump should be considered?

FINANCIAL ACCOUNTING

10th Edition

ISBN:9781259964947

Author:Libby

Publisher:Libby

Chapter1: Financial Statements And Business Decisions

Section: Chapter Questions

Problem 1Q

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Compare the alternatives C and D on the basis of a present worth analysis using an interest rate of 15.00% per year and a study period of 10 years. (Include a minus sign if necessary.) Alternative First Cost AOC, per Year Annual Increase in Operating Cost, per Year Salvage Value Life, Years C $-50000 $-9000 $-700 $6000 10 The present worth of alternative C is $ (Click to select) offers the lower present worth. D $-28000 $-5000 $-900 $700 5 and that of alternative D is $
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A company wishes to evaluate two alternative machines. Use the AW method to select the better alternative, considering the interest rateas 29% Machine R Machine S First cost, $ —250,000 —370,500 Annual operating cost, —40,000 -50,000 $ per year Salvage value, $ 20,000 30,000 Life, years 3 5 [
determine the most economic alternative using AW analysis at 12% per year. Give an explanation and draw the results.
Assume that a company is choosing between two alternatives-keep an existing machine or replace it with a new machine. The costs associated with the two alternatives are summarized as follows: Multiple Choice Purchase cost (new) Remaining book value Overhaul needed now Annual cash operating costs Salvage value (now) Salvage value (eight years from now) $ 6,000 Click here to view Exhibit 14B-1 and Exhibit 14B-2, to determine the appropriate discount factor(s) using the tables provided. If the company overhauls its existing machine, it will be usable for eight more years. If it buys the new machine, it will be used for eight years. Assuming a discount rate of 12%, what is the net present value of the cash flows associated with keeping the existing machine? $(50,760) $(56,760) Existing Machine $ 15,000 $ 6,000 $ 5,000 $ 10,500 $ 2,000 $ 1,000 $(71,760) New Machine $ 22,000 $ 7,000
An engineer must choose between two technically equivalent alternatives, A and B. Based upon the data in the table below over the 12year analysis period and an interest rate of 9%, the PW of machine A is: A B First Cost $3,000 $6,000 Annual Maintenance $500 $300 End of Useful Life Salvage Value $700 $1,000 Useful Life 5 years |15 уears O PWA=-3000-3000(P/F,9%,5)-3000(P/F,9%,10)-500(P/A,9%,15)+700(P/F,9%,5)+700(P/F,9%,10)+700(P/F,9%,15) O PWA=-3000-3000(P/F,9%,10)-500(P/A,9%,15)+700(P/F,9%,5)+700(P/F,9%,10)+700(P/F,9%,15) O PWA=-3000-3000(P/F,9%,5)-3000(P/F,9%,10)-500(P/F,9%,15)+700(P/F,9%,5)+700(P/F,9%,10)+ 700(P/F,9%,15) O PWA=-3000-3000(P/F,9%,5)-3000(P/F,9%,10)-500(P/A,9%,10)+700(P/F,9%,5)+700(P/F,9%,10)+ 700(P/F,9%,15)
Required Information [The following information applies to the questions displayed below.] Project A requires a $365,000 initial investment for new machinery with a five-year life and a salvage value of $42,000. The company uses straight-line depreciation. Project A is expected to yield annual net income of $25,300 per year for the next five years. Compute Project A's accounting rate of return. Choose Numerator: Accounting Rate of Return I Choose Denominator: = Accounting Rate of Return Accounting rate of return
NEW PROJECT ANALYSIS You must evaluate the purchase of a proposed spectrometer for the R&D department. The base price is $100,000, and it would cost another $15,000 to modify the equipment for special use by the firm. The equipment falls into the MACRS 3-year class and would be sold after 3 years for $45,000. The applicable depreciation rates are 33%, 45%, 15%, and 7%. The equipment would require a $14,000 increase in net operating working capital (spare parts inventory). The project would have no effect on revenues, but it should save the firm $67,000 per year in before-tax labor costs. The firm's marginal federal-plus-state tax rate is 40%. a. What is the initial investment outlay for the spectrometer, that is, what is the Year 0 project cash flow? Round your answer to the nearest cent. Negative amount should be indicated by a minus sign. $ b. What are the project's annual cash flows in Years 1, 2, and 3? Round your answers to the nearest cent. In Year 1 $ In Year 2 $ In Year 3 $ c.…
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…
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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
Compare the alternatives C and D on the basis of a present worth analysis using an interest rate of 10% per year and a study period of 10 years. Alternative C First Cost $-40,000 $-11,000 $-24,000 AOC, per Year Annual Increase in Operating Cost, per Year Salvage Value Life, Years $-2,000 $-200 $-500 $6,000 $200 10 The present worth of alternative C is $ and that of alternative D is $ (Click to select) voffers the lower present worth.