Engineering Economics

HW2 – 23/10/2012

Q 3.3

Amalgamated Iron and Steel purchased a new machine for ram cambering large I-beams. The company expects to bend 80 beams at $2000 per beam in each of the first 3 years, after which the company expects to bend 100 beams per year at $2500 per beam through year 8. If the company's minimum attractive rate of return is 18% per year, what is the present worth of the expected income?

P = 80(2000)(P/A,18%,3) + 100(2500)(P/A,18%,5)(P/F,18%,3) = 160,000(2.1743) + 250,000(3.1272)(0.6086) = $823,691

Q3.22

Use the cash flow diagram below to calculate the amount of money in year 5 that is equivalent to all the cash flows shown, if the interest rate is 12% per year.

Amt, year 5 = 1000(F/A,12%,4)(F/P,12%,2) + 2000(P/A,12%,7)(P/F,12%,1) = 1000(4.7793)(1.2544) + 2000(4.5638)(0.8929) = $14,145

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Q 3.30

Find the value of x in the diagram below that will make the equivalent present worth of the cash flow equal to $15,000, if the interest rate is 15% per year.

15,000 = 2000 + 2000(P/A,15%,3) + 1000(P/A,15%,3)(P/F,15%,3) + x(P/F,15%,7) 15,000 = 2000 + 2000(2.2832) + 1000(2.2832)(0.6575) + x(0.3759) x = $18,442

Q3.35

Exxon-Mobil is planning to sell a number of producing oil wells. The wells are expected to produce 100,000 barrels of oil per year for 8 more years at a selling price of $28 per barrel for the next 2 years, increasing by $1 per banel through year 8. How much should an independent refiner be willing to pay for the wells now, if the interest rate is 12% per year?

P = [2,800,000(P/A,12%,7) + 100,000(P/G,12%,7) + 2,800,000](P/F,12%,1) = [2,800,000(4.5638) + 100,000(11.6443) + 2,800,000](0.8929) = $14,949,887

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Q 3.40

A start-up company selling color-keyed carnuba car wax borrows $40,000 at an interest rate of 10% per year and wishes to repay the loan over a 5-year period with annual payments such that the third through fifth payments are $2000 greater than the first two. Determine the size of the first two payments.

40,000 = x(P/A,10%,2) + (x + 2000)(P/A,10%,3)(P/F,10%,2) 40,000 = x(1.7355) + (x + 2000)(2.4869)(0.8264) 3.79067x = 35,889.65 x = $9467.89 (size of first two payments)

Q3.45

Calculate the present worth for a machine that has an initial cost of $29,000, a life of 10 years, and an annual operating cost of $13,000 for the first 4 years, increasing by 10% per year thereafter. Use an interestrate of l0% per year.

P = 29,000 + 13,000(P/A,10%,3) + 13,000[7/(1 + 0.10)](P/F,10%,3) = 29,000 + 13,000(2.4869) + 82,727(0.7513) = $123,483

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Q3.49

Compute the present worth (year 0) of the following cash flows at i = 12% per year.

P = 5000 + 1000(P/A,12%,4) + [1000(P/A,12%,7) – 100(P/G,12%,7)](P/F,12%,4) = 5000 + 1000(3.0373) + [1000(4.5638) – 100(11.6443)](0.6355) = $10,198

Q3.50

For the cash flow tabulation, calculate the equivalent uniform annual worth in periods I through 10, if the interest rate is 10% per year.

Find P in year 0 and then convert to A. P = 2000 + 2000(P/A,10%,4) + [2500(P/A,10%,6) – 100(P/G,10%,6)](P/F,10%,4) = 2000 + 2000(3.1699) + [2500(4.3553) – 100(9.6842)](0.6830) = $15,115 A = 15,115(A/P,10%,10) = 15,115(0.16275) = $2459.97

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