Machine A
Equivalent uniform annual worth = Capital recovery cost + Annual operating cost
Capital recovery cost = - First cost
Capital recovery factor + Salvage value
Sinking fund factor
Capital recovery cost machine A = - 15,000
(A/P, 10%, 2 years ) + 3000
(A/F, 10% , 2 years)
Capital recovery cost machine A = - 15,000
0.5762 + $ 3000
0.4762 = - 7214.4
Equivalent uniform annual worth = - 7214.4 - 1500
Equivalent uniform annual worth machine A = $ - 8,714.4
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Machine B
Equivalent uniform annual worth = Capital recovery cost + Annual operating cost
Capital recovery cost machine B = - 25,000
(A/P, 10%, 10 years ) + 6000
(A/F, 10% , 10 years)
Capital recovery cost machine B = - 25000
0.1627 + 6000
0.0627
Capital recovery cost machine B = - 3691.3
Equivalent uniform annual worth = - 3691.3 - 400 - 50
( P/A, 10% , 10 years)
( P/A, 10% , 10 years) = Uniform series present worth
Equivalent uniform annual worth machine B = - 3691.3 - 400 - 50
6.145
Equivalent uniform annual worth machine B = $- 4398.6
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Thus machine B must be selected because it has a lower equivalent uniform annual worth of cost when compared to machine A.
Ismael Nofal كان نشظا منذ 48 دقيقة 4 125 points Machines that have the following costs...
1. Machines that have the following costs are under consideration for a continuous production process. Using an interest rate of 8% per year, determine which alternative should be selected on the basis of an annual-worth analysis. First Cost Annual Operation Cost Salvage Value Life, Years Machine A 50.000 10,000 8,000 Machine B 60,000 15,000 10,000
please dont use excel, show me the formula used
5. Machines that have the following costs are under consideration for a robotized welding process. Using an interest rate of 10% per year, determine which alternative should be selected on the basis of a present worth analysis. Machine X Machine Y First cost, $ Annual operating cost, $ per year Salvage value, $ Life, years - 250,000 --60.000 70.000 -430.000 -40,000 95.000
Machines that have the following costs are under consideration for a robotized welding process. Using an interest rate of 10% per year, determine which alternative should be selected on the basis of a present worth analysis. Machine X Machine Y Initial cost ($) 300,000 machine x, 400,000 machine y Annual operation cost ($ per year) 45,000 machine x, 50,000 machine y Annual operation cost increased by 8% machine x, 300 machine y Salavage value ($) 70,000 machine x, 95,000 machine...
You have two machines under consideration for an improved automated wrapping process for Snickers Fun Size candy bars as detailed below. Compare them on the basis of annual worths at i = 12% Machine First Cost Annual Cost per Year Salvage Value $-45,000 $-10,000 $10.000 3 years $-65.000 $-15.000 $16,000 6 vears Life Machine (Click to select) is selected for an improved automated wrapping process.
Two machines are under consideration and only one can be bought. MARR is 10%. Use the following information and find out which option should be purchased. Use an annual worth comparison. Initial cost Annual savings Annual maintenance cost Life Salvage value Machine A $280,000 $40,000 $2000 for year 1, increasing by 5% each year thereafter 15 years $19,250 Machine B $185,000 $32,000 $1000 for year 1, increasing by $350 each year thereafter 10 years $14,800 3(a) Advertisements suggest that a...
Required information The two machines shown are being considered for a chip manufacturing operation. Assume the MARR is a real return of 13% per year and that the inflation rate is 4.9% per year. Machine First Cost, $ M&0, $ per year Salvage Value, $ Life, years А -150,000 -70,000 40,000 B -800,000 -5,000 200,000 00 Which machine should be selected on the basis of an annual worth analysis if the estimates are in future dollars? What is the annual...
Exercise 2 A firm is considering which of two machines to install to reduce costs. Both machines have useful life of 5 years and no salvage value. Machine A costs 820.5 QAR and can be expected to result in 150 QAR savings first year increasing by 50 annually. Machine B costs 1,389 QAR and will provide savings of 300 QAR the first year increasing 50 QAR annually, making the second-year savings 350 QAR, the third-year savings 400 QAR and so...
Required information Problem 14.056 The two machines shown are being considered for a chip manufacturing operation. Assume the MARR is a real return of 14% per year and that the inflation rate is 5.2% per year. 0.000 Machine First Cost, $ M&0. $ per year Salvage Value, $ Life, years -145.000 -70.000 40,000 5.000 00.000 Problem 14.056.a: Compare two alternatives based on their AW values without inflation consideration Which machine should be selected on the basis of an annual worth...
Required information Problem 14.056 The two machines shown are being considered for a chip manufacturing operation. Assume the MARR is a real return of 14% per year and that the inflation rate is 5.2% per year. -780.000 Machine First Cost. $ M&O. $ per year Salvage Value, $ Life, years -145,000 - 70.000 40,000 -5,000 200,000 Problem 14.056.b: Compare two alternatives based on their AW values with inflation consideration Which machine should be selected on the basis of an annual...
4. 125 points] For the cash flows below, use a capitalised cost method to alternative is best at MARR value of 10% per year. to determine which Alternative 1 Alternative 2 First Cost, S -250000 -160000 Maintenance and Operation Cost S/year 3000 -15000 Cost Every 6 years,S -1000 Cost Every 15 yearss -12000 Salvage Value,S 10000 50000 ife, years 12