lOMoARcPSD|2110249Sample/practice exam February 2016, questions and answers - This is a practice midterm from February 2016 Engineering Economics (Memorial University Of Newfoundland) Distributing prohibited | Downloaded by Seb Teg (
[email protected]) lOMoARcPSD|2110249 Student Name: _______________________________ Student Number: _______________________________ ENGINEERING ECONOMICS ENGINEERING 4102 Midterm February 11, 2016 7-8:30 PM General Instructions: Answer all questions in the answer booklet(s) provided. All questions have equal value. State all assumptions. Return examinations with answer books. Please note that there are four questions in this exam. Question 1 A metal company is examining two options – Machine A and Machine B - for a core- cutting machine. Machine A Machine B Initial Cost $ 300,000 $ 200,000 Annual Savings $25,000 $ 23,000 Annual M9590aintenance $12,000/year + $0.05/unit $22,000 + $0.01/widget Cost Labor 1.25/widget 0.50/widget Life 5years 10 years Salvage Value $20,000 18,000 MARR is 10 percent and the company’s sale projection is 20,000 widgets/year. Which machine should be purchased? Use the present worth comparison. LCM = 10 years Yearly Cost = 12000+20000*1.25+20000*0.05=38000 PW (Machine A) = -300,000 - 38000(P/A,10%,10) + 20000(P/F,10%,5) – 300000(P/F,10%,5) + 20000(P/F,10%,10)+ 25,000(P/A, 10%, 10) = -546026.6 PW – Machine B For Yearly Costs: 22,000 + (0. 5)* 20000 + 0.01*20000=32,200 PW = - 200,000 – 32200 (P/A, 10%,10) + 18,000 (P/F, 10%,10) +23000(P/A,10%,10) 1 Distributing prohibited | Downloaded by Seb Teg (
[email protected]) lOMoARcPSD|2110249 = -249590.6 So, Machine B must be purchased. Question 2 Sarah wants to buy a house for $325 000. She paid 20% as a down payment and obtained a bank loan for the rest. The loan has a nominal interest rate of 10% compounded monthly with a 10-year amortization period. The loan term is 10 years. What are the firm's monthly payments to the bank? Down-payment is $65 000 (20% of $325 000). Subtract down-payment from the total truck costs to get the amount that should be borrowed from a bank: $325 000 - $65 000 = $260 000. The amortization period is the duration over which the bank loan should be paid back to the bank. Therefore: A = $260 000 x (A/P, 10/12%, [10 x 12]) = $3435.9 Question 3 The maintenance cost of a machine is approximately $300 per year. With age the costs increase by $50 a year. What is the future worth of the maintenance costs in four years' time if the interest rate is 5% compounded monthly? Ie= (1+0.05/12)12-1=5.1% The maintenance costs are an arithmetic gradient series. The arithmetic gradient to annuity conversion factor is (A/G, 5.1%, 4). This means that the geometrically-increasing annuity of $300 per year is equivalent to a uniform annuity of $300 + $50 x 1.4378= $371.89 per year over the four-year period. So the future worth of the maintenance costs is $371.89 x (F/A, 5.1%,4) = $1605.28 Question 4 A metal company is considering buying one of two pieces of equipment. Data for the equipment is shown below: Piece 1 Piece 2 Life 15 years 10 years Initial Cost $ 250,000 $ 170,000 Annual Savings $28,000 $ 20,000 Annual Maintenance Cost $1500, increasing by 15% $1000 each year thereafter Salvage Value $32,500 23,000 2 Distributing prohibited | Downloaded by Seb Teg (
[email protected]) lOMoARcPSD|2110249 If MARR is 12 percent and the relevant information is as given above, which one should be purchased? Use the annual worth comparison. AW- Piece 1 g = 0.15 iϒ = (1 + i)/(1 + g) − 1 = 1.12/1.15 − 1 = −0.026 AW(hydraulic) = −250 000(A/P, 12%, 15) + 28 000 - 32,500(A/F, 12%, 15) -1500[(P/A, iϒ, 15)/(1 + g)](A/P, 12%, 15) = −250 000(0.14682) + 28 000 + 32500 (0.02682) − 3573.15 = -11406.5 AW- Piece 2 = −170 000(A/P, 12%, 10) + 20,000 − 1000 + 23000(A/F,12%,10) = −170 000(0.17698) + 20,000 – (1000 )+ 23,000(A/F,12%,10)= =-9776.06 So, Piece 2 must be purchased. 3 Distributing prohibited | Downloaded by Seb Teg (
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