FINANCIAL ACCOUNTING
10th Edition
ISBN: 9781259964947
Author: Libby
Publisher: MCG
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what is the NPV of the new construction equipment? Initial cost = $100k, Salvage Value in 6 yrs = $25K, Increase Yearly Net Sales = $25k, Bank Rate = 10%
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- Osler Company is considering an investment with the following data: Initial cost $200,000 Annual net cash inflows $25,000 Expected life 10 years Salvage value none Depreciation will be taken on a straight-line basis over the expected life of the investment.What is the accounting rate of return for the investment? a.12.5% b.2.5% c.20% d.25% e.10%arrow_forwardGiven the following, calculate the project's cash flows, NPV, and IRR. Initial investment $325,000 Expected life is 5 years First Year Revenues: 145,000 First-Year Expenses: $65,000 • Growth for revenue and expenses: 4.5 percent per year Straight Line Depreciation over 5 years Salvage Value: $50,000 One-time net working capital investment of $10,000 is required at the start of the project and will be recovered at project end The tax rate is 34 percent The risk-free rate is 4 percent Beta is 1.1 • The expected market return is 8 percent Answer the following: ● ● ● ● ● ● ● ● ● What are the cash flows for each year? What is the NPV? What is the IRR?arrow_forwardConsidering an interest rate of 8%, that the cost of the equipment is $300,000 and you will replace the equipment after 7 years, what is the annual cost if the expected salvage value is $75,000?arrow_forward
- What must the annual earnings be from a new piece of equipment that costs $10 million and has a projected operating life of 10 years? Assume the salvage value is $800,000 at the end of year 10. Use an interest rate of 10% compounded annually.arrow_forwardRequired information A company that manufactures magnetic flow meters expects to undertake a project that will have the cash flows estimated. First cost, $. Equipment replacement cost in year 2, $ Annual operating cost, $/year Salvage value, $ Life, years -870,000 -300,000 -920,000 250,000 4 At an interest rate of 10% per year, what is the equivalent annual cost of the project? Find the AW value using tabulated factors. The equivalent annual cost of the project is $-1arrow_forwardThe cash flow of an energy management opportunity is estimated as follows: Initial cost:$12,000 Energy saving:$2,700/year for 12 years Maintenance cost:$1,200/year for 12 years Salvage value:$2,500@the end of 12 years If the interest rate is 10%, 1) What is the simple payback period (SPP) (in years)? (a)5.2 (b)4.2 (c)4.6 (d)8.02) With an annual discount rate is 10%, what is the discounted payback period (in years)? (a)9.5 b) 15.1 (c)8.1 (d) 16.9 (e) 6.53)With an annual discount rate is 10%, what is the benefit-cost ratio (BCR)? (Hint: Benefit = Annual saving-Maintenance; Cost= Initial investment - Salvage)(a) 1.04(b) 0.80(c) 1.25(d) 1.12(e) 1.43arrow_forward
- I will rate. Solve correctly and show all works.arrow_forwardCalculate Net Present Value of minitor that costs $ 35,000.00 Amortization period 5 years with savings of 8000 per year with a hurdle rate of 12%, 5%. Which investment is more attractive?arrow_forwardQ1: For the machines indicated below. Consider i= 10% per year First cost Annual cost Salvage value Life duration Machine A 20,000 $ 5,000 $ 7,500 $ 3 Machine B 25,000 4,000 6,000 4 A- Draw cash flow diagram for each machine for one cycle of each project B- Draw cash flow diagram for each project considering the LCM life cycle (Hint: different project duration, need to have the LCM life cycle) C- Compare the machines to select best alternative one based on Present worth analysis method D- Repeat part B considering Future worth analysisarrow_forward
- What is the annual equivalent cost of purchasing a lift truck that has an initial cost of $85,000, an annual operating cost of $13,500, and an estimated salvage value of $23,000 after six years of use at an annual interest rate of 6%? X More Info Equal Payment Series Single Payment Compound Present Amount Worth Compound Amount Factor (F/A, I, N) Sinking Present Fund Worth Factor Factor Capital Recovery Factor (A/P, i, N) Factor Factor (F/P, i, N) (P/F, i, N) (A/F, i, N) (P/A, i, N) 1.0800 0.9434 1.0000 1.0000 0.9434 1.0800 1.1238 0.8900 2.0800 0.4854 1.8334 0.5454 1.1910 0.8396 3.1836 0.3141 2.6730 0.3741 1.2625 0.7921 4.3746 0.2288 3.4861 0.2886 1.3382 0.7473 5.6371 0.1774 4.2124 0.2374 1.4185 0.7050 6.9753 0.1434 4.9173 0.2034 0.6851 8.3938 0.1191 5.5824 0.1791 1.5038 1.5938 1.6895 0.6274 9.8975 0.1010 6.2098 0.1610 0.5919 11.4913 0.0870 6.8017 0.1470 1.7908 0.5584 13.1808 0.0759 7.3801 0.1359 SAWNIN 1 2 3 4 5 67899 10arrow_forward4(25). Some data is given about a project below: Life = 3 years Cost of Project now = 100 Cash inflows for next 2 years is 80 tl every year. However, at the end of the third year he has to spend 10 tl for the repairs. a. (15). What is the profitability of the Project? b. (10) . How will he decide on the Project?arrow_forwardA company that manufactures magnetic flow meters expects to undertake a project that will have the cash flows estimated. First cost, $ -810,000 Equipment replacement cost in year 2, $ -300,000 Annual operating cost, $/year Salvage value, $ Life, years -860,000 250.000 4 an interest rate of 10% per year, what is the equivalent annual cost of the project? Find the AW value using tabulated factors. e equivalent annual cost of the project is $-820000arrow_forward
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