HW#1 Solution
4-18
(a) Cash flow in year k = ($14/ton)(10,000 tons)(1.05)k-1
$187,613
$178,679
i = 10% / yr
$170,171
$162,068
$154,350
$140,000
0
1
$147,000
2
3
4
5
6
7
End of Year
(b)
Solution based on Section 4.8:
F7 = $140,000(F/P,15%,6) + $147,000(F/P
Chapter 3
INTEREST AND EQUIVALENCE
1
Where economic decisions are immediate we need
to consider:
amount of expenditure
Taxes
Where economic decisions occur over a
considerable period of time we also need to
consider:
interest
inflation
2
Money has
Chapter 4
MORE INTEREST FORMULAS
1
Calculation of P and F are fundamental.
Some problems are more complex and require an
understanding of added components:
Uniform series.
Arithmetic or geometric gradients.
Nominal and effective interest rates
Conti
170 El Chapter 5 Present Worth Analysis
Solution: The analysis period can conveniently be selected as the useful life of the devices,
ce both devices cost $1000, we have a situation where, in choosing either
or ﬁve years. Sin
he appropriate decision crite
Alex bought $1000 worth of home furniture. If
the items are expected to last 10 years, what
will the equivalent uniform annual cost be if
interest is 7%
Equivalent Uniform Annual Cost = EUAC = PWC(A/P, i, n)
Equivalent Uniform Annual Benefit = EUAB = PWB(
ISE 405
ENGINEERING ECONOMIC ANALYSIS
JAN. 2016
Chapter 2
Outline:
Engineering Costs
Cost Estimating and Estimating Models
Objective:
Understand various cost concepts
Understand various cost estimation models
Be able to estimate engineering costs w