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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      :

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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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