ENGINEERING ECONOMY, Sixth Edition
LEARNING OBJECTIVES
by Blank and Tarquin
1. Dealing with
shifted series
2. Shifted series
and single
amounts
CHAPTER 3
COMBINING FACTORS
3. Shifted
gradients
4. Decreasing
gradients
Mc
Graw
Hill
5. Spreadsheet
applications
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Sct 3.1 Calculations for Uniform
Series that are Shifted
For a shifted series the present worth point
in time is NOT t = 0.
It is shifted either to the left of “0” or to the
right of “0”.
Remember, when dealing with a uniform
series:
The PW point is always one period to the left of the
first series value, no matter where the series falls
on the time line.
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3-3
Example of Shifted Series P and F
F6
0
1
2
3
4
5
6
7
8
A = $-500/year
P0
P2
• F for this series is at t = 6; F6 = A(F/A,i%,4)
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3-2
Shifted Uniform Series
Consider:
0
1
2
3
4
5
6
7
8
A = $-500/year
P0
P2
P of this series is at t = 2 (P2)
or F2
P2 = -500(P/A,i%,4)
or could refer to as F2
P0 = P2(P/F,i%,2)
or could be F2(P/F,i%,2)
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Sct 3.2 Calculations Involving
UniformUniform-Series and RandomlyRandomly-Placed
Single Amounts
Draw and correctly label the cash flow diagram that
defines the problem
Locate the present and future worth points for each
series
Write the time value of money equivalence
relationships
Substitute the correct factor values and solve
• P0 for this series at t = 0 is
P0 = -500(P/A,i%,4)(P/F,i%,2)
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1
Series with additional
single cash flows
3.2 Series with Other cash flows
It is common to find cash flows that are combinations of
series and single, randomly-placed cash flows
F4 = $300
♦ Consider:
A = $500
For present worth, P
Solve for the series present worth values then move to t = 0
Then solve for the P at t = 0 for the single cash flows using the
P/F factor for each cash flow
Add the equivalent P values at t = 0
0
1
2
3
4
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2
3.2 The PW Points are:
1
4
5
6
7
8
i = 10%
t = 1 is the PW point for the $500 annuity;
“n” = 3
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0
1
2
2
3
3
4
5
6
7
8
i = 10%
F5 = -$400
3-9
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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)
Back 5 Periods
F5 = -$400
t = 1 is the PW point for the two other
single cash flows
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3.2 Substitute the factors and
solve
P = $500( 2.4869 )( 0.8264 )
+
$300( 0.6830 )
400( 0.6209 )
=
Substituting the factor values into the
equivalence expression and solving….
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F4 = $300
A = $500
3
3
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Back 4 periods
A = $500
1
8
•Find the PW at t = 0 for this cash flow
F4 = $300
1
7
F5 = -$400
3.2 The PW Points are:
0
6
i = 10%
For future worth, F
Convert all cash flows to an equivalent F using the F/A and F/P
factors in year t = n
Add the equivalent F values at t = n
5
$1,027.58
$204.90
$248.36
$984.12
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2
Sct 3.3 Calculation for Shifted
Gradients
The Present Worth of an arithmetic gradient
(linear gradient) is always located:
One period to the left of the first cash flow
in the series ( “0” gradient cash flow) or,
Two periods to the left of the “1G” cash
flow
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Shifted Gradient
A Shifted Gradient
has its present value
point removed from
time t = 0
A Conventional
Gradient has its
present worth point at
t=0
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Example of a Shifted Gradient
Example of a Conventional Gradient
♦ Consider:
Gradient Series
Gradient Series
……..Base Series ……..
……..Base Series ……..
0
1
2
n-1
…
n
0
1
2
n-1
…
This represents a conventional gradient
n
The present worth point for the
base series and the gradient is
here!
The present worth point is t = 0
This represents a shifted gradient
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Shifted Gradient: Numerical Example
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Shifted Gradient: Numerical Example
PW of the base series
P0
nseries = 8 time periods
P2
G = $+100
Base Series = $500
0
1
2
3
4 ………..
………..
Base Series = $500
9
10
0
1
2
3
4 ………..
………..
9
10
Cash flows start at t = 3
$500/year increasing by $100/year through
year 10; i = 10%; Find P at t = 0
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P2 = 500(P/A,10%,8) = 500(5.3349) = $2667.45
P0 = 2667.45(P/F,10%,2) = 2667.45(0.8264)
= $2204.38
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Shifted Gradient: Numerical Example
PW for the gradient component
P0
For the base series
P0 = $2204.38
For the arithmetic gradient
P0 = $1,324.61
P2
G = +$100
1
0
Example: Total Present Worth Value
2
3
4 ………..
………..
9
10
Total present worth
P2 = $100(P/G,10%,8) = $100(16.0287) = $1,602.87
P = $2204.38 + $1,324.61 = $3528.99
P0 = $1,602.87(P/F,10%,2) = $1,602.87(0.8264)
= $1,324.61
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To Find A for a Shifted Gradient
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Shifted Geometric Gradient
Conventional Geometric Gradient
1) Find the present worth of the gradient
at actual time 0
2) Then apply the (A/P,i,n) factor to
convert the present worth to an
equivalent annuity (series)
A1
0
1
2
3
…
…
…
n
Present worth point is at t = 0 for a conventional
geometric gradient
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Shifted Geometric Gradient
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Shifted Geometric Gradient Example
Shifted Geometric Gradient
i = 10%/year
0
1
2
3
4
5
6
7
8
A = $700/year
A1
0
1
2
3
…
…
…
n
A1 = $400 in year t = 5
12% increase/year
Present worth point is at t = 2 for this example
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Geometric Gradient Example
3.3 The Gradient Amounts
i = 10%/year
0
1
2
3
4
t
5
6
7
Base Amt
$400.00
51
62
73
84
8
$448.00
$501.76
$561.97
A = $700/year
Present Worth of the Gradient at t = 4
PW point for the A
series is t = 0
3.73674
P4 = $400{ P/A1,12%,10%,4 } = $
1,494.70
12% increase/year
P0 = $1,494.70( P/F,10%,4) = $1,494.70( 0. 6830 )
PW point for the
gradient is t = 4
P0 = $1,020,88
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3.3 The Annuity Present Worth
♦ PW of the Annuity
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3.3 Total Present Worth
♦ Geometric Gradient @ t =
i = 10%/year
♦
P0 = $1,020,88
♦ Annuity
0
1
2
3
4
5
6
7
8
♦
P0 = $2,218.94
♦ Total Present Worth”
A = $700/yr
P0 = $700(P/A,10%,4)
♦
$1,020.88 + $2,218.94
♦
= $3,239.82
= $700( 3.1699 ) = $2,218.94
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Sct 3.4 Shifted Decreasing Arithmetic
Gradients
PW for Shifted Decreasing Gradient
Given the following shifted, decreasing gradient
First, find PW at t = 2
F3 = $1,000; G = $-100
F3 = $1,000; G = -$100
i = 10%/year
i = 10%/year
0
1
2
3
4
5
6
7
8
Find the present worth at t = 0
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0
1
2
3
4
5
6
7
8
PW point at t = 2
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Shifted Decreasing Gradient Example
Shifted Decreasing Gradient Example
F3 = $1,000; G = -$100
Second, find the PW at t = 0
i = 10%/year
i = 10%/year
Base amount = $1,000
F3 = $1,000; G = $-100
0
1
2
3
4
5
6
7
8
0
1
2
P2
3
4
5
6
7
8
P2
The gradient amount is subtracted from the base
amount
P0 here
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P0 here
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3.4 Time Periods Involved
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3.4 Time Periods Involved
F3 = $1,000; G=-$100
F3 = $1,000; G=-$100
G = -$100/yr
$1,000
i = 10%/year
1
0
1
2
2
3
3
4
4
5
5
6
1
7
8
P2 or, F2: Take back to t = 0
P0 here
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Dealing with n = 5.
3-33
i = 10%/year
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0
1
2
2
3
3
4
4
5
5
6
7
8
P2 = $1,000( P/A,10%,5 ) – 100( P/G,10%.5 )
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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Summary
Chapter summarizes cash flow
patterns that are shifted away from time
t=0
Illustrations of using multiple factors
to perform PW or FW analysis for
shifted cash flows
Illustrations of shifted arithmetic and
geometric gradients
Illustrations of the power of Excel
financial functions3-35
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CHAPTER 3
End of Slide Set
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6