Determine the FW of the following engineering project when the MARR is 16% per year. Is the project acceptable? Proposal A $9,500 5 years Investment cost Expected life Market (salvage) value Annual receipts Annual expenses A negative market value means that there is a net cost to dispose of an asset -$1.200 $8.000 $4 500 Click the icon to view the interest and annuity table for discrete compounding when the MARR is 16% per vear The FW of the following engineering project is S (Round to the nearest dollar)

Determine the FW of the following engineering project when the MARR is 16% per year. Is the project acceptable? Proposal A $9,500 5 years Investment cost Expected life Market (salvage) value Annual receipts Annual expenses A negative market value means that there is a net cost to dispose of an asset -$1.200 $8.000 $4 500 Click the icon to view the interest and annuity table for discrete compounding when the MARR is 16% per vear The FW of the following engineering project is S (Round to the nearest dollar)

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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Two alternatives are being considered for installation. Which should be selected based on an interest rate of 5.6% per year ? Alternative A Alternative B First Cost, $ 135,000 462,000 Annual Operating Cost, $/yr 76,000 34,600 Salvage value, $ 71,200 n/a Life, years 9 ∞
Given the following table for two alternatives. consider that the interest rate = 10% Alternative A B First cost, $ -90,000 -750,000 AOC, $/year -50,000 -10,000 Salvage value, $ Life, years 8,000 2,000,000 00 All of the following equations for calculating the capitalized cost of alternative B are correct, except: Select one: = -750,000 – 10,000/0.10 a. + 2,000,000/(1+0.10) b. = [-750,000(0.10) – 10,000]/0.10 c. capitalized cost = -750,000 – 10,000/0.10 + 2000000 (0.10) d. capitalized cost = -750,000 – 10,000/0.10
Determine the FW of the following engineering project when the MARR is 14% per year. Is the project acceptable? Proposal A $10.000 4 years - $1,000 Investment cost Expected life Market (salvage) value Annual receipts $9,000 Annual expenses $4,500 A negative market value means that there is a net cost to dispose of an asset A Click the icon to view the interest and annuity table for discrete compounding when the MARR is 14% per vear The FW of the following engineering project is S. (Round to the nearest dollar.) According to the FW Decision Rule the project acceptable. - X More Info is is not Discrete Compounding; i-14% Single Payment Compound Amount Factor Compound Amount Factor Uniform Series Sinking Capital Recovery Factor To Find A Given P AIP Present Worth Factor Present Worth Factor Fund Factor To Find F Given P FIP To Find F Given A To Find P Given F PIF 0.8772 0.7695 0.6750 0.5921 To Find A To Find P Given A Given F AIF FIA PIA 1.1400 1.2996 1.4815 1.6890 1.0000 0.8772 1.6467…
Two mutually exclusive alternatives have the estimates shown below. Use annual worth analysis to determine which should be selected at an interest rate of 10% per year. Q $-44,000 R $-84,000 First Cost AOC per Year Salvage Value $-10,000 $-5,000 in year 1, increasing by $1,000 per year thereafter $4,000 $5,000 Life 2 years 4 years Alternative (Click to select) should be selected.
Two mutually exclusive alternatives have the estimates shown below. Use annual worth analysis to determine which should be selected at an interest rate of 15% per year. Q R $-44,000 $-11,000 $3,000 $-84,000 $-6,000 in year 1, increasing by $1,000 per year thereafter $10,000 First Cost AOC per Year Salvage Value Life 2 years 4 years Alternative (Click to select) vshould be selected. (Click to select) Q
An equipment cost P60,000 today and its salvage value is P20,000 after 6 years . Calculate its depreciation at year 1 with 3.83% interest rate.Use Sinking Fund Method NOTE: Please Solve and explain! and indicate your final answer!
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 $
The following table summarizes the financial data for a proposed new asset. The issue is how long the asset should be kept. Assume an interest rate of 12% and delermine the economic service life of the anset. End of YearN) Investment cost O&M Cost Salvage value $14,000 $3.400 $8.000 $4600 $6.000 $5.800 $4,000 01 year O2 years 3 years O Not enough information
Questions which require graphical solutions, please use Excel and submit Tables and Graphs with your 1.) Three mutually exclusive alternatives are being considered: A Initial investment Cost $43,000 $24,000 $17,000 Annual net income $4,150 $2,500 $1,700 Rate of Return 7.3% 8.3% 6.0% Each alternative has a 20 -year useful life and no salvage value a) Construct a choice table for interest rates from 1 to 60%. b) Calculate the PW of each alternative and draw the graph of PW vs Interest rate for each alternative on the same chart. c) If the minimum attractive rate of return is 7% which alternative should be selected? Submit the Excel graph and choice table with your solution. I
Compare the alternatives shown on the basis of their capitalized costs using an interest rate of 10% per yearquarterly. Alternative M Alternative N First cost, P Annual operating cost, P per year 50,000 Salvage value, Life, years 150,000 800,000 10,000 1,000,000 8,000 5 25
The following data pertain to an investment proposal (Ignore income taxes.): Cost of the investment Annual cost savings $ 35,000 $ 12,000 $ 6,000 Estimated salvage value Life of the project Discount rate. Click here to view Exhibit 14B-1 and Exhibit 14B-2, to determine the appropriate discount factor(s) using the tables provided. The net present value of the proposed investment is closest to: 5 years 18%
Two mutually exclusive alternatives have the estimates shown below. Use annual worth analysis to determine which should be selected at an interest rate of 10% per year. First cost, $ AOC, $ per year Salvage value, $ Life, years Q -42,000 -6,000 0 2 R -80,000 -7,000 in year 1, increasing by $1,000 per year thereafter 4,000 4

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