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Mechanical Engineering

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Feb 20, 2024

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ME 489 Engineering Economics Homework Spring 2024 Due: February 3, 2024 Name Tyler Betts CWID 11962268 You are encouraged to use the Casio fx-260 Solar II calculator on this homework. The problems contain digits from your CWID. Consider your CWID as abcdefgh. So if your CWID is 12345678, a = 1, b = 2, c = 3, d = 4, e = 5, f = 6, g = 7, h = 8 If your CWID is 11912345, a = 1, b = 1, c = 9, d = 1, e = 2, f = 3, g = 4, h = 5, etc. 1. Alan receives a graduation present of $10,000 from his rich uncle. If he invests all of the money at 4% interest, how much will he have in (5 + 2d) years? D=6 10k 17 years F/P=(1+i)^n i=.04 n=17 P=10,000 F=19,479$ 2. A company is considering the purchase of equipment that will have an initial cost of $25,000, operating and maintenance costs of ($1,000 + 100e) per year, and an estimated salvage value of $2,000 at the end of 30 years. Assume an interest rate of 8%. What is the present equivalent cost of the investment for the 30 year project? Draw a cashflow diagram. E=2 P=-25,000$ P/A=((1+i)^n-1)/(i*(1+i)^n)=(1.08^30-1)/(.08*1.08^30) P=-13509.34$ P/F=1/(1+i)^n=1/1.08^30 P=198.75$ Equivalent cost=38,310.60$ S
3. Compute the life cycle cost (or present cost) of a reciprocating compressor with first cost of $120,000. The annual maintenance cost is $5000, but at the end of four years, it increases by $500*(1+0.1f) per year. The salvage value is $25,000, and life of ten years. The rate of return is 10%. F=2 4. Machine A costs $11,500 and has an annual operating cost of ($3000+$100g). At the end of its 10-yr life, it has a salvage value of $1500. Machine B costs $15,000 and has an annual operating cost of ($1200+200g). At the end of its 10-yr life, it has a salvage value of $4000. At 8% interest, what is the annual net equivalent advantage of the additional investment in machine B? G=6 5. A carbon-steel reactor costs $20,000 and has an estimated life of three years in a corrosive environment. A stainless-steel reactor costs $35,000 and has an estimated life of 10 years in the same corrosive environment. The salvage value of the carbon-steel reactor is
$2500(1+0.1h), and the salvage value of the stainless steel reactor is $5000. The reactor will be needed indefinitely. Using 10% interest which reactor should be purchased? Why? H=8 To better evaluate these, as they will be needed indefinitely, I expanded the scope to look at 30 years. So over the first thirty years, the carbon steel reactor would cost 68,688.90$ in current value, whereas the stainless steel reactor would cost 50,739.10$ in current value. With that in mind the stainless steel should be purchased.
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