Econ practice exam 2011

Engineering Economics Practice Problems
1. In a situation where the effective interest rate per year is 12%, based on monthly
compounding. What is the nominal interest rate per year?
2. If 10% nominal annual interest is compounded daily, what is the effective annual
interest rate?
3. An individual wishes to deposit a certain quantity of money now so that he will have
$500 at the end of five years. What is the amount of the deposit with an interest rate
of 4% per year, compounded semiannually?
4. A consulting engineer bought a fax machine. There will be no maintenance cost for
the first year as it was sold with one year’s free maintenance. In the second year the
maintenance is estimated at $20. In subsequent years the maintenance cost will
increase $20 per year (that is, 3rd year maintenance will be $40, 4th year will be $60
and so forth). What is the amount that must be set aside now at 6% interest to pay
the maintenance costs on the fax machine for the first six years?
5. A project has an initial cost of $10,000, uniform annual benefits of $2,400, and a
salvage value of $3,000 at the end of its 10-year useful life. At 12% interest, what is
the net present worth (NPW) of the project?
6. A person borrows $5,000 at an interest rate of 18%, compounded monthly. Monthly
payments of $167.10 were agreed upon. What is the length of the loan?
7. A machine costing $2,000 to buy and $300 per year to operate will save labor
expenses of $650 per year for 8 years. The machine will be purchased if its salvage
value at the end of 8 years is sufficiently large to make the investment economically
attractive. If the interest rate is 10%, what is the minimum salvage value?
8. An engineer deposited $200 quarterly in her savings account for three years at 6%
interest, compounded quarterly. Then for five years she made no deposits or
withdrawals. What is amount in her account after eight years?
9. An investor is considering the investment of $10,000 in a piece of land. The property
taxes are $100 per year. What is the lowest selling price the investor must receive if
she wishes to earn 10% interest rate after keeping the property for 10 years?
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Engineering Economics Practice Problems
10. Two alternatives are being considered:
Initial Cost
Uniform Annual Benefit
Useful Life
A
$500
$140
8
B
$800
$200
8
What is the Benefit Cost Ratio of the difference between the alternatives, based on
12% interest rate?
11. $20,000 is invested today. If the annual inflation rate is 6% and the effective annual
return on investment is 10%, what will be the approximate future value of the
investment, adjusted for inflation, in 5 years?
12. A factory that manufactures chairs costs $180,000 per year to operate, including
rent, depreciation charges on equipment, and salaries. Each chair costs $31 to
make, and sells for $59. What is the factory’s break-even sales volume?
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Engineering Economics Practice Problems
SOLUTIONS
1. In a situation where the effective interest rate per year is 12%, based on monthly
compounding. What is the nominal interest rate per year?
Solution: ie = (1+r/m)m -1
0.12 = (1+r/12)12 -1
(1.12)1/12 = (1+r/12)
1.00949 = (1+r/12)
r = 0.00949 x 12 = 0.1138 = 11.38%
2. If 10% nominal annual interest is compounded daily, what is the effective annual
interest rate?
Solution: ie = (1+r/m)m -1 = (1+0.10/365)365 -1 = 0.1052 = 10.52%
3. An individual wishes to deposit a certain quantity of money now so that he will have
$500 at the end of five years. What is the amount of the deposit with an interest rate
of 4% per year, compounded semiannually?
Solution: P = F (P/F, i, n) = 500 (0.8203) = $410
4. A consulting engineer bought a fax machine. There will be no maintenance cost for
the first year as it was sold with one year’s free maintenance. In the second year the
maintenance is estimated at $20. In subsequent years the maintenance cost will
increase $20 per year (that is, 3rd year maintenance will be $40, 4th year will be $60
and so forth). What is the amount that must be set aside now at 6% interest to pay
the maintenance costs on the fax machine for the first six years?
Solution: Using single payment present worth factors:
P = 20(P/F, 6%, 2) + 40 (P/F, 6%, 3) + 60 (P/F, 6%, 4) + 80 (P/F, 6%, 5) + 100 (P/F, 6%, 6)
= 20 (0.8900) + 40 (0.8396) + 60 (0.7921) + 80 (0.7473) + 100 (0.7050) = $229.19
An alternative solution is to use the gradient present worth factor.
P = 20 (P/G, 6%, 6) = 20 (11.459) = $229.18
5. A project has an initial cost of $10,000, uniform annual benefits of $2,400, and a
salvage value of $3,000 at the end of its 10-year useful life. At 12% interest, what is
the net present worth (NPW) of the project?
Solution: NPW = PW of Benefits – PW of Cost
= 2,400(P/A,12%,10) + 3,000(P/F,12%,10) – 10,000 =
= 2,400(5.650) + 3,000(.3220) – 10,000 = $4,526
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Engineering Economics Practice Problems
6. A person borrows $5,000 at an interest rate of 18%, compounded monthly. Monthly
payments of $167.10 were agreed upon. What is the length of the loan?
Solution: PW of Benefits = PW of Cost
$5,000 = $167.10(P,A,1.5%,n)
(P/A,1.5%,n) = 5,000/167.10 = 29.92 From 1½% Table n= 40
7. A machine costing $2,000 to buy and $300 per year to operate will save labor
expenses of $650 per year for 8 years. The machine will be purchased if its salvage
value at the end of 8 years is sufficiently large to make the investment economically
attractive. If the interest rate is 10%, what is the minimum salvage value?
Solution: NPW = PW of Benefits –PW of Cost = 0
= (650-300)(P/A,10%,8) + S8(P/F,10%,8) –2,000 = 0
= 350(5.335) + S8(0.4665) – 2,000 = 0
S8 = $284.57
8. An engineer deposited $200 quarterly in her savings account for three years at 6%
interest, compounded quarterly. Then for five years she made no deposits or
withdrawals. What is amount in her account after eight years?
Solution: FW 200(F/A,1.5%,12)(F/P,1.5%,20) = 200(13.041)(1.347) = $3,513.25
9. An investor is considering the investment of $10,000 in a piece of land. The property
taxes are $100 per year. What is the lowest selling price the investor must receive if
she wishes to earn 10% interest rate after keeping the property for 10 years?
Solution: Min Sale Price = 10,000(F/P,10%,10) + 100(F/A,10%,10) =
= 10,000(2.594) + 100(15.937) = $27,533.70
10. Two alternatives are being considered:
Initial Cost
Uniform Annual Benefit
Useful Life
A
$500
$140
8
B
$800
$200
8
What is the Benefit Cost Ratio of the incremental cash flow of the two alternatives,
based on 12% interest rate?
Solution:
B/C = PW of Benefits/PW of Cost = 60(P/A,12%,8)/300 = 60(4.968)/300 = 0.99
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Engineering Economics Practice Problems
Alternative
B/C = EUAB/EUAC = 60/300(A/P,12%,8) = 60/300(0.2013) = 0.99
11. $20,000 is invested today. If the annual inflation rate is 6% and the effective annual
return on investment is 10%, what will be the approximate future value of the
investment, adjusted for inflation, in 5 years?
Solution:
The interest rate adjusted for inflation is d = I +f +i*f = .10+.06+.1*.06 = 0.166
So F = P(i+d)n = $20000(1+0.166)5 = $43105
12. A factory that manufactures chairs costs $180,000 per year to operate, including
rent, depreciation charges on equipment, and salaries. Each chair costs $31 to
make, and sells for $59. What is the factory’s break-even sales volume?
Solution:
The break-even point occurs when costs and sales total zero. Let x be the number of
chars per year at break-ven.
Costs = $180000+$31*x
Revenues = $59*x
So, $180,000+$31*x = $59*x
X = 6429 chairs/year
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