27107689 engineering economy chapter 3x
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CHAPTER 3
COMBINING FACTORS
Mc GrawHill
ENGINEERING ECONOMY, Sixth Edition
by Blank and Tarquin
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Learning Objectives
Shifted Series
Shifted Series and Single Amounts
Shifted Gradients
Decreasing Gradients
Spreadsheets
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Section 3.1
Calculations For UniformSeries That Are Shifted
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1. Uniform Series that are SHIFTED
A shifted series is one whose presentworth point in time is NOT t = 0.
Shifted either to the left of “0” or to the
right of t = “0”. Dealing with a uniform series:
The PW point is always one period to theleft of the first series value
No matter where the series falls on the time line.
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Shifted Series
0 1 2 3 4 5 6 7 8
A = -$500/year
Consider:
P of this series is at t = 2 (P3 or F3)P3 = $500(P/A,i%,4) or, could refer to as F3
P0 = P3(P/F,i%,2) or, F3(P/F,i%,2)
P3 P0
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Shifted Series: P and F
A = -$500/year
Consider:
F for this series is at t = 6
F6 = A(F/A,i%,4)
0 1 2 3 4 5 6 7 8
P3 P0
F at t = 6
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3.2 Series with other single cash flows
It is common to find cash flows that arecombinations of series and other singlecash flows.
Solve for the series present worthvalues then move to t = 0
Solve for the PW at t = 0 for the singlecash flows
Add the equivalent PW’s at t = 0
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3.2 Series with Other cash flows
Consider:
0 1 2 3 4 5 6 7 8
A = $500
F5 = -$400
F4 = $300
•Find the PW at t = 0 and FW at t = 8 for thiscash flow
i = 10%
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3.2 The PW Points are:
F5 = -$400
F4 = $300
A = $500
0 1 2 3 4 5 6 7 8
i = 10%
t = 1 is the PW point for the $500 annuity;“n” = 3
1 2 3
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3.2 The PW Points are:
F5 = -$400
F4 = $300
A = $500
0 1 2 3 4 5 6 7 8
i = 10%
t = 1 is the PW point for the two othersingle cash flows
1 2 3
Back 4 periods
Back 5 Periods
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3.2 Write the Equivalence Statement
P = $500(P/A,10%,3)(P/F,10%,2)
+
$300(P/F,10%,4)
-
400(P/F,10%,5)
Substituting the factor values into theequivalence expression and solving….
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3.2 Substitute the factors and solve
P = $500( 2.4869 )( 0.8264 )
+
$300( 0.6830 )
-
400( 0.6209 )
=
$831.06
$1,027.58
$204.90
$248.36
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Section 3.3
Calculations for ShiftedGradients
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3.3 The Linear Gradient - revisited
The Present Worth of an arithmeticgradient (linear gradient) is alwayslocated:
One period to the left of the first cash flowin the series ( “0” gradient cash flow) or,
Two periods to the left of the “1G” cash flow
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3.3 Shifted Gradient
A Shifted Gradient is one whose presentvalue point is removed from time t = 0.
A Conventional Gradient is one whose
present worth point is t = 0.
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3.3 Example of a Conventional Gradient
Consider:
……..Base Annuity ……..
Gradient Series
0 1 2 … n-1 n
This Represents a Conventional Gradient.
The present worth point is t = 0.
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3.3 Example of a Shifted Gradient
Consider:
……..Base Annuity ……..
Gradient Series
0 1 2 … n-1 n
This Represents a Shifted Gradient.
The Present Worth Point for the
Base Annuity and the Gradientwould be here!
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3.3 Shifted Gradient: Example
Given:
Base Annuity = $100
G = +$100
0 1 2 3 4 ……….. ……….. 9 10
Let C.F start at t = 3:
$500/ yr increasing by $100/yearthrough year 10; i = 10%; Find thePW at t = 0
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3.3 Shifted Gradient: Example
PW of the Base Annuity
Base Annuity = $100
0 1 2 3 4 ……….. ……….. 9 10
P2 = $100( P/A,10%,8 ) = $100( 5.3349 ) = $533.49
Nannuity = 8 time periods
P0 = $533.49( P/F,10%,2 ) = $533.49( 0.8264 )
= $440.88
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3.3 Gradient Present Worth
For the gradient component
0 1 2 3 4 ……….. ……….. 9 10
G = +$100
• PW of gradient is at t = 2:
•P2 = $100( P/G,10%,8 ) = $100( 16.0287 ) = $1,602.87
•P0 = $1,602.87( P/F,10%,2 ) = $1,602.87( 0.8264 )
• = $1,324.61
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py g p , q p p y
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3.3 Example: Final Solution
For the Base Annuity
P0 = $440.88
For the Linear Gradient
P0 = $1,324.61
Total Present Worth:
$440.88 + $1,324.61 = $1,765.49
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py g p , q p p y
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3.3 Shifted Geometric Gradient
Conventional Geometric Gradient
0 1 2 3 … … … n
A1
Present worth point is at t = 0 for a conventionalgeometric gradient!
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py g p q p p y
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3.3 Shifted Geometric Gradient
Conventional Geometric Gradient
0 1 2 3 … … … n
A1
Present worth point is at t = 2 for this example
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3.3 Geometric Gradient Example
0 1 2 3 4 5 6 7 8
A = $700/yr
12% Increase/yr
i = 10%/year
A1 = $400 @ t = 5
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3.3 Geometric Gradient Example
0 1 2 3 4 5 6 7 8
A = $700/yr
12% Increase/yr
i = 10%/year
PW point for thegradient
PW point for theannuity
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3.3 The Gradient Amounts
t Base Amt
1 $400.00
2 $448.00
3 $501.76
4 $561.97
5
6
7
8
Present Worth of the Gradient at t = 4
P4
= $400{ P/A1,12%,10%,4 } = 1,494.70$
3.73674
P0 = $1,494.70( P/F,10%,4) = $1,494.70( 0. 6830 )
P0 = $1,020,88
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3.3 The Annuity Present Worth
PW of the Annuity
0 1 2 3 4 5 6 7 8
i = 10%/year
P0 = $700(P/A,10%,4)
= $700( 3.1699 ) = $2,218.94
A = $700/yr
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3.3 Total Present Worth
Geometric Gradient @ t =
P0 = $1,020,88
Annuity
P0 = $2,218.94
Total Present Worth”
$1,020.88 + $2,218.94
= $3,239.82
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3.4 Shifted Decreasing Linear Gradients
Given the following shifted, decreasinggradient:
0 1 2 3 4 5 6 7 8
F3 = $1,000; G=-$100
i = 10%/year
Find the Present Worth @ t = 0
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3.4 Shifted Decreasing Linear Gradients
0 1 2 3 4 5 6 7 8
F3 = $1,000; G=-$100
i = 10%/year
P2 or, F2: Take back to t = 0
P0 here
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3.4 Shifted Decreasing Linear Gradients
0 1 2 3 4 5 6 7 8
F3 = $1,000; G=-$100
i = 10%/year
P2 or, F2: Take back to t = 0
P0 here
Base Annuity = $1,000
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3.4 Time Periods Involved
0 1 2 3 4 5 6 7 8
F3 = $1,000; G=-$100
i = 10%/year
P2 or, F2: Take back to t = 0
P0 here
1 2 3 4 5
Dealing with n = 5.
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3.4 Time Periods Involved
0 1 2 3 4 5 6 7 8
F3 = $1,000; G=-$100
i = 10%/year
1 2 3 4 5
P2 = $1,000( P/A,10%,5 ) – 100( P/G,10%.5 )
$1,000 G = -$100/yr
P2= $1,000( 3.7908 ) - $100( 6.8618 ) = $3,104.62
P0 = $3,104.62( P/F,10%,2 ) = $3104.62( 0 .8264 ) = $2,565.65
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Section 3.5
Spreadsheet Applications
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3.5 Spreadsheet Applications
Assume Excel is the spreadsheet of choice
Instructors may vary on the degree of
emphasis placed on spreadsheet useStudent’s Goal:
Learn the Excel Financial Functions
Create your own spreadsheets to solve avariety of problems
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3.5 NPV Function in Excel
NPV function is basic
Requires that all cell in the range sodefined have an entry.
The entry can be $0…but not blank! Incorrect results can be generated if one or more cells in the defined range isleft blank .
A “0” value must be entered.
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3.5 Spreadsheets
It is assumed if an instructor desires toapply spreadsheets, he or she willprovide examples and go over each
example and the associated cellformulas.
See Appendix A for further details onExcel applications
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End of Slide Set
Mc GrawHill
ENGINEERING ECONOMY, Sixth Edition
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