Chapter 1 Engineering Economic Decisions
There are no end-of-chapter questions in this introductory chapter. However, the following
questions could be added as a part of instruction:
1. Ask students to review the contents of The Wall Street Journal for the past 3 months.
Then, identify and categorize the types of investment decisions appeared in the
journal according to the types of strategic economics decisions discussed in the text.
2. Work in small groups and brainstorm ideas about how a common appliance, device
or tool could be redesigned to improve it in some way. Identify the steps involved
and the economic factors, which you would need to consider prior to making a
decision to manufacture the redesigned product. A detailed design and actual cost
estimates are not required. Some items, which could be considered for this redesign
exercise, are: a shopping cart, telephone, can opener, screwdriver, etc.
3. Many oil price forecasts in the early 2000 indicated that the price of oil in the year
2007 would not exceed $50 per barrel. What is the price of today? Why are these
prices so difficult to predict? Imagine what the consequences would be if you used
these optimistic estimates in your economic analysis in your early project
undertaking. What would be some practical ways to consider this type of variation
in economic analysis?
,Chapter 2: Time Value of Money
2.1) I = iPN = (0.09)($3,000)(5) = $1,350
2.2)
• Simple interest:
F = P(1 + iN )
$4, 000 = $2, 000(1 + 0.08 N )
N = 12.5 years (or 13 years)
• Compound interest:
$4, 000 = $2, 000(1 + 0.07) N
2 = 1.07 N
log 2 = N log 1.07
N = 10.24 years (or 11 years)
2.3)
• Simple interest:
I = iPN = (0.07)($10, 000)(20)
= $14, 000
• Compound interest:
, I = P (1 + i) N − 1 = $10,000 (1.07)20 − 1
= $28,696.84
2.4)
• Compound interest:
F = $1, 000(1 + 0.06)5
= $1,338.23
• Simple interest:
F = $1, 000(1 + 0.07(5))
= $1,350
The simple interest option is better.
2.5)
• Loan balance calculation:
Principal Interest Remaining
End of period Payment Payment Balance
0 $0.00 $0.00 $5,000.00
1 $835.46 $450.00 $4,164.54
2 $910.65 $374.81 $3,253.89
3 $992.61 $292.85 $2,261.28
4 $1,081.94 $203.52 $1,179.33
5 $1,179.32 $106.14 $0.00
, 2.6) P = $8, 000( P / F ,8%,5) = $8, 000(0.6806) = $5, 444.8
2.7) F = $20,000(F / P,10%,2) = $20,000(1.21) = $24,200
2.8)
• Alternative 1
P = $100
• Alternative 2
P = $120(P / F,8%,2) = $120(0.8573) = $102.88
• Alternative 2 is preferred
2.9) (a) F = $7,000(F / P,9%,8) = $7,000(1.9926) = $13,948.2
(b) F = $1,250(F / P,4%,12) = $1,250(1.6010) = $2,001.25
(c) F = $5,000(F / P,7%,31) = $5,000(8.1451) = $40,725.5
(d) F = $20,000(F / P,6%,7) = $20,000(1.5036) = $30,072
There are no end-of-chapter questions in this introductory chapter. However, the following
questions could be added as a part of instruction:
1. Ask students to review the contents of The Wall Street Journal for the past 3 months.
Then, identify and categorize the types of investment decisions appeared in the
journal according to the types of strategic economics decisions discussed in the text.
2. Work in small groups and brainstorm ideas about how a common appliance, device
or tool could be redesigned to improve it in some way. Identify the steps involved
and the economic factors, which you would need to consider prior to making a
decision to manufacture the redesigned product. A detailed design and actual cost
estimates are not required. Some items, which could be considered for this redesign
exercise, are: a shopping cart, telephone, can opener, screwdriver, etc.
3. Many oil price forecasts in the early 2000 indicated that the price of oil in the year
2007 would not exceed $50 per barrel. What is the price of today? Why are these
prices so difficult to predict? Imagine what the consequences would be if you used
these optimistic estimates in your economic analysis in your early project
undertaking. What would be some practical ways to consider this type of variation
in economic analysis?
,Chapter 2: Time Value of Money
2.1) I = iPN = (0.09)($3,000)(5) = $1,350
2.2)
• Simple interest:
F = P(1 + iN )
$4, 000 = $2, 000(1 + 0.08 N )
N = 12.5 years (or 13 years)
• Compound interest:
$4, 000 = $2, 000(1 + 0.07) N
2 = 1.07 N
log 2 = N log 1.07
N = 10.24 years (or 11 years)
2.3)
• Simple interest:
I = iPN = (0.07)($10, 000)(20)
= $14, 000
• Compound interest:
, I = P (1 + i) N − 1 = $10,000 (1.07)20 − 1
= $28,696.84
2.4)
• Compound interest:
F = $1, 000(1 + 0.06)5
= $1,338.23
• Simple interest:
F = $1, 000(1 + 0.07(5))
= $1,350
The simple interest option is better.
2.5)
• Loan balance calculation:
Principal Interest Remaining
End of period Payment Payment Balance
0 $0.00 $0.00 $5,000.00
1 $835.46 $450.00 $4,164.54
2 $910.65 $374.81 $3,253.89
3 $992.61 $292.85 $2,261.28
4 $1,081.94 $203.52 $1,179.33
5 $1,179.32 $106.14 $0.00
, 2.6) P = $8, 000( P / F ,8%,5) = $8, 000(0.6806) = $5, 444.8
2.7) F = $20,000(F / P,10%,2) = $20,000(1.21) = $24,200
2.8)
• Alternative 1
P = $100
• Alternative 2
P = $120(P / F,8%,2) = $120(0.8573) = $102.88
• Alternative 2 is preferred
2.9) (a) F = $7,000(F / P,9%,8) = $7,000(1.9926) = $13,948.2
(b) F = $1,250(F / P,4%,12) = $1,250(1.6010) = $2,001.25
(c) F = $5,000(F / P,7%,31) = $5,000(8.1451) = $40,725.5
(d) F = $20,000(F / P,6%,7) = $20,000(1.5036) = $30,072
