A company with a MARR of 10% must install one of two production machines. The economic parameters of each machine are asfollows:MachineInitial cost (5)Service life (years)Salvage value endof life (S)$15,0004 years$22,0006 years52,000$1,000The Future worth (FW) for the machine Y over the 12 years analysis period is:

Project Y MARR = 10℅ Initial cost = $22,000 Salvage value = $1,000 Service Life = 6 Years Time…

A company with a MARR of 10% must install one of two production machines. The economic parameters of each machine are as follows: Machine Inital cost (5) Service e yean) $15,000 4 years $22.000 6 years Salvage value end of fe (5) 52,000 S1,000 The Future worth (Fw) for the machine Y over the 12 years analysis period is:

A company with a MARR of 10% must install one of two production machines. The economic parameters of each machine are as follows: Machine Inital cost (5) Service e yean) $15,000 4 years $22.000 6 years Salvage value end of fe (5) 52,000 S1,000 The Future worth (Fw) for the machine Y over the 12 years analysis period is:

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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Alternative X has a first cost of $30,000, an operating cost of $9,000 per year, and a $5,000 salvage value after 5 years. Alternative Y will cost $45,000 at nO with an operating cost of $4,000 per year after that and a salvage value of $7,000 after 5 years. At an MARR of 12% per year, what is the present worth far alternative X2 $49,606 $58,777 $59,606 -$32,443
Project XYZ has a first cost of $75,000, operating and maintenance costs of $10,000 during each year of its 10 years life, and a $15,000 salvage value. What is its equivalent uniform annual cost (EUAC) if the interest rate is 10%/year? Most nearly number for answer. O No correct answer 21546 O 21262 21622 26223 22126
Compare two alternatives, A and B, on the basis of a present worth evaluation using i = 10% per year and a study period of 8 years. Alternative A B First cost, $ −15,000 −28,000 Annual operating cost, $/year −6,000 −9,000 Overhaul in year 4, $ — −2000 Salvage value, $ 3,000 5,000 Life, years 4 8
If produced by Plan A, a producer’s initial capital cost will be $120,000, its annual operating cost will be $30,000, and its salvage value after 4 years will be 20,000. With Plan B, there is a first cost of $240,000, an annual operating cost of $10,000, and $40,000 salvage value after 4-year life. Based on a present worth analysis at 15% interest rate, which plan should be chosen?
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 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
Compare the machines below using present worth analysis at i10% per year and find which one should be selected Machine Y First Cost Operating Cost Savage Lfe cycle Machine X 20.000 3.000 3.000 3years Firat Cost 30.000 .000 Annuel Operating Cost Salvage Life cycle 3.000 6 yoars
Alternative X has a first cost of 36000 an annual operating cost of 6900, and a salvage value of 11025 after 13 year. Alternative Y has a first cost of 37000 an annual operating cost of 7000, and a salvage value of 16280 after 13 year. If MARR of 13% per year, approximately what is the PW of each alternative? Select one: O a. PW of X= -77518 and PW of Y= -78067 O b. PW of X= -75990 and PW of Y= -76529 O c. PW of X= -76750 and PW of Y= -77294 O d. PW of X= -78293 and PW of Y= -78848
Use Service-Output Method. A Machine costs ₱80,000 and an estimated life of 10 years with a salvage value of ₱5,000. Assuming that the total service of the machine is 1,497,600 hours and the number of working days per year is 260 days. What is the book value after 4 years if the machine production time a is 24 hours? Answer: BV4 = P 78,750.00Show Solution
$500,000 is invested in equipment having a negligible salvage value regardless of the number of years used. O&M costs equal $125,000 the first year and increase $25,000 per year. Based on a MARR of 10%, what are the optimum replacement interval and minimum EUAC?
please as soon as possible . thank you
How long will it take to recover an investment of $245,000 in enhanced CNC controls that include axis control to 8 axes (on the milling model) if theassociated income is $92,000 per year, expenses are $38,000 per year, and the salvage value is assumed to be 15% of the first cost? Use a MARR of 15% per year.

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