A company that manufactures magnetic flow meters expects to undertake a project that will have the cash flowsestimated.First cost, $-810,000Equipment replacement cost inyear 2, $-300,000Annual operating cost, $/yearSalvage value, $Life, years-860,000250.0004an 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 $-820000

Equivalent annual cost is that cost which includes the all cost including the cost of operating and…

Answered: ted. First cost, $ Equipment replaceme…

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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Find the Annual Recovery for Year 1. Method Purchased Cost Recovery Class Annual Recovery for Year 1 MACRS Jul 08 $6,000 5
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Question: A piece of machinery has a cost basis of $45,000. Its salvage value will be $5000 after 10,000 hours of operation. With units-of-production depreciation, what is the allowable depreciation rate per hour? What is the book value after 4,000 hours of operation? Suggestion: Use straight-line on an hourly basis for the rate. Use that rate to determine the BV @ 4,000 hours.
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Problems 43 and 44 are based on the following statement: Used New machine machine Initial Cost 15,000 40,000 ($) Annual 8,000 2,000 operating cost ($/year) Salvage 5,000 10,000 value ($) Life (years) 3 6 Use an interest rate of 10% per year 43. The present worth of the new machine is closest to: (A) $43,630 (B) $43,066 (C) $45,210 (D) $49,660 44. To compare the machines on the basis of a present worth analysis, the present worth values to use are: (A) PWused $31,139, PW new = $43,066 (B) PWused $31,139, PW new = $37,461 (C) PWused $54,533, PW new = $43,066 (D) PWused $54,533, PW new = $37,461
1. Maintenance costs for a small bridge with an expected 50-yr life are estimated to be P 2,000.00 each year for the first 5 years, followed by a P10.000 expenditure in the 15th year and a P 20,000 expenditure in the 25th year. IfI 12% per year, what is the equivalent uniform annual maintenance costs over the entire 50-yr period? a. P 1,230 b. P 852 C. P 936 d. P 1,159
Soru 1 Henüz cevaplanmadi 1,00 0zerinden işaretlenmiş P Soruyu işaretle New digital scanning graphics equipment is expected to cost 550023, to be used for 6 years, and to have an annual operating cost (AOC) of $10071. Determine the AW values for one and three life cycles at i 10% per year.
profile-image Time remaining: 00 : 09 : 41 Accounting We have the following data for a power plant: 1 MW CENTRAL ECONOMIC BALANCE Power: 1 Mwa Maintenance cost: 3% per year Initial cost of operation: 5% per year Regulated Rate: $0.015/kWh (20 years) Project cost = $750,000.00 Plant factor= 60% Energy: 5256000 KWh Income from regulated tariff: 78840 Annual revenue increase: 6% Initial maintenance cost/year of annual maintenance (2% installation): $ 22,500.00 Annual maintenance increase: 5% Initial cost of operation/year (1% Installation) $ 37,500.00 Annual operation increase: 3% It is assumed that in year 10, an equipment update will be carried out whose cost will be 25% of the value of the investment = $187,500.00 SOLVE: In what year will the investment be recovered?
Correct answer please fast
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Question 5 A machine that have the following cost is under consideration for a new manufacturing process. The interest rate is 6% compounded semiannually. First cost = $70,000 Semiannual operating cost = 6,000 Semiannual income = 18,000 Salvage value = 9,000 Life in years = 5 years a. PW = $29,059 b. PW = $39,059 O c. PW = -$39,059 d. PW = $19,059
At i = 10% per year, a machine has a first cost of $9,000 and estimated operating costs and year-end salvage values as shown. The annual equivalent cost if you operate the machine for 3 years is closest to which value? Operating Salvage Cost, $ Value, $ 7000 Year -1000 -1200 5000 -1500 4200 -2000 3000 -3000 2000 O-$-3,568 = $-3,770 -1500 4200 -2000 3000 -3000 2000 O =$-3,568 O = $-3,770 = 2-3,636 %3D O =$-3,454 O = $-3,832 123 n 345

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ted. First cost, $ Equipment replaceme year 2, $ Annual operating cos Salvage value, $ Life, years interest rate of 10% pe equivalent annual cost