This document contains solutions to homework problems from an engineering economics course. It addresses topics like the time value of money, simple vs compound interest, calculating rates of return, and cash flows. Example calculations include determining the rate of increase in company profits, the rate of return on an investment, and the amount originally borrowed given the loan payoff amount. The document also covers the time it would take investments to double or quadruple at given compound interest rates using the Rule of 72.

1. Zarqa University
Engineering Economics
HW # 1
Due: After 1 week
Solution
1.1 What is the definition of Time Value of Money?
Time value of money means that there is a certain worth in having money and the
worth changes as a function of time.
1.6 What is the difference between simple and compound interest?
In simple interest, the interest rate applies only to the principal, while compound
interest generates interest on the principal and all accumulated interest.
1.10 If Ford Motor Company's profits increased from 22 cents per share to
29 cents per share in the April-June quarter compared to the previous
quarter, what was the rate of increase in profits for that quarter?
Rate of increase = [(29 – 22)/22](100) = 31.8%
1.12 A design-build engineering firm completed a pipeline project wherein
the company realized a profit of $2.3 million in 1 year. If the amount of
money the company had invested was $6 million, what was the rate of return
on the investment?
Rate of return = (2.3/6)(100)= 38.3%
1.14 A publicly traded construction company reported that it just paid off a
loan that it received 1 year earlier. If the total amount of money the company
paid was $1.6 million and the interest rate on the loan was 10% per year,
how much money did the company borrow 1 year ago?
P + P(0.10) = 1,600,000
1.1P = 1,600,000
P = $1,454,545
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2. 1.16 At an interest rate of 8% per year, $10,000 today is equivalent to how
much (a) 1 year from now and (b) 1 year ago?
(a) Equivalent future amount = 10,000 + 10,000(0.08)
= 10,000(1 + 0.08)
= $10,800
(b) Equivalent past amount: P + 0.08P = 10,000
1.08P = 10,000
P = $9259.26
1.19 At what interest rate would $100,000 now be equivalent to $80,000 one
year ago?
80,000 + 80,000(i) = 100,000
i = 25%
1.20 Certain certificates of deposit accumulate interest at 10% per year
simple interest. If a company invests $240,000 now in these certificates for
the purchase of a new machine 3 years from now, how much will the
company have at the end of the 3-year period?
F = 240,000 + 240,000(0.10)(3) = $312,000
1.21 A local bank is offering to pay compound interest of 7% per year on
new savings accounts. An e-bank is offering 7.5% per year simple interest
on a 5-year certificate of deposit. Which offer is more attractive to a
company that wants to set aside $1,000,000 now for a plant expansion 5
years from now?
Compound amount in 5 years = 1,000,000(1 + 0.07)5
= $1,402,552
Simple amount in 5 years = 1,000,000 + 1,000,000(0.075)(5)
= $1,375,000
Compound interest is better by $27,552
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3. 1.22 Badger Pump Company invested $500,000 five years ago in a new
product line that is now worth $1,000,000. What rate of return did the
company earn (a) on a simple interest basis and (b) on a compound interest
basis?
Simple: 1,000,000 = 500,000 + 500,000(i)(5)
i = 20% per year simple
Compound: 1,000,000 = 500,000(1 + i)5
(1 + i)5
= 2.0000
(1 + i) = (2.0000)0.2
i = 14.87%
1.23 How long will it take for an investment to double at 5% per year (a)
simple interest and (b) compound interest?
Simple: 2P = P + P(0.05)(n)
P = P(0.05)(n)
n = 20 years
Compound: 2P = P(1 + 0.05)n
(1 + 0.05)n
= 2.0000
n = 14.2 years
1.30 Write the engineering economy symbol that corresponds to each of the
following
Excel functions.
(a) PV
(b) PMT
(c) NPER
(d) IRR
(e) FV
(a) PV = P (b) PMT = A (c) NPER = n (d) IRR = i (e) FV = F
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4. 1.39 Construct a cash flow diagram for the following cash flows: $10,000
outflow at time zero, $3000 per year outflow in years I through 3 and $9000
inflow in years 4 through 8 at an interest rate of 10% per year, and an
unknown future amount in year 8.
1.41 Use the rule of 72 to estimate the time it would take for an initial
investment of $10,000 to accumulate to $20,000 at a compound rate of 8%
per year.
Note: Read page 35 in the text book for information about rule of 72
Time to double = 72/8 = 9 years
1.42 Estimate the time it would take (according to the rule of 72) for money
to quadruple in value at a compound interest rate of 9% per year.
Note: Read page 35 in the text book for information about rule of 72
Time to double = 72/9 = 8 years
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