3 Arch 449 Chapter 3-1
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Transcript of 3 Arch 449 Chapter 3-1
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1
20102011
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..
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A
A :
; ( )
( )
/
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Why Is Engineering Economics Important?
• Engineers , Architects DESIGN things and performPROJECTS
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• Therefore, engineers and Architects must be concerned with the
economic aspects of designs that they recommend, andprojects that they perform
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: A:
$18,000 ,
$600 3
?
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• It will help you make good decisions:
• In your professional life
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•And in your personal life!
• Knowledge of engineering economics will
have a significant impact on you personally!
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ENGINEERING ECONOMICS INVOLVES:
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FORMULATING, ESTIMATING, ANDEVALUATING ECONOMIC OUTCOMES
WHEN CHOICES OR ALTERNATIVES AREAVAILABLE
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BY USING SPECIFIC
MATHEMATICAL RELATIONSHIPS
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TO COMPARE THE CASH FLOWS OF THEDIFFERENT ALTERNATIVES
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: $100 ,
$100 ?
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( )
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( ).
().
.
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$$
Investment
Inflation
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. , % , $
( ) $
. A, $ $. .
, $. , $ , $
6%
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:
: ( )
:
Definitions
. ( )
: , (
)
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Interest : used to move money through time forcomparisons. The rent for loaned money (cost of
using money)
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, .
.
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: $, 1, $, .
A.
B. .
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Type of Interest
Simple interest: the practice of chargingan interest rate only to an initial sum(principal amount).
Compound interest: the practice ofcharging an interest rate to an initial sumand to any previously accumulated interest
that has not been withdrawn.
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Simple Interest
P = Principal amount
i = Interest rate
N = Number of
End ofYear
Beginning
Balance
Interestearned
EndingBalance
interest periods Example:
P = $1,000
i = 10%
N = 3 years
0 $1,000
1 $1,000 $100 $1,100
2 $1,100 $100 $1,200
3 $1,200 $100 $1,300
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(.. ),
.
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/
.
Simple Interest
I = P * i * n
Ex. $ 1000 loan for 2 years at 10 % per year – nocompounding
I = P * i * n = 1000 * .10 * 2 = $200
Payback = F = P + I
= 1000 + 200 = $1200
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Compound Interest
Compound interest: the practice ofcharging an interest rate to an initial sumand to any previously accumulated interest
that has not been withdrawn.
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Compound Interest
P = Principal amount
i = Interest rate
N = Number of
Endof
Year
BeginningBalance
Interestearned
EndingBalance
interest periods Example:
P = $1,000
i = 10%
N = 3 years
0 $1,000
1 $1,000 $100 $1,100
2 $1,100 $110 $1,210
3 $1,210 $121 $1,331
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.
, .
: $1,000 6%
, ?
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Interest for Year 1 = $1,000 X 0.06 = $60
Total amount due after year 1 = $1,000 + $60 = $1,060
Interest for year 2 = $1,060 X 0.06 = $63.60
Total amount due after year 2 = $1,060 + $63,60 = $1,123.60
Interest for year 3 = $1,123.60 X 0.06 = $67.42Total amount due after year 3 = $1,123.60 + $67.42 = $1,191.02
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, , $1,000 $1,191.02 $1,180 = $11.02 .
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,
=0, ()
:
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A ,
(2)
1
1 2.
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, 3
(3) 2 2 3.
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,
() ()
, () :
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To avoid the trouble of writing out each formula, a
standard notation is used:
(X/Y, i%, n)
The first letter in the parentheses (X) represent whatyou “Want to find”, while the second letter (Y)
represents what is “Given”.
For example, F/P means “find F when given P”. The iis the interest rate in percent and n represents the
number of periods involved.
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