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Conversion for Arithmetic Gradient Series
Conversion for Geometric Gradient Series
Quiz Review
Project Review
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Series and Arithmetic Series
• A series is the sum of the terms of a
sequence.
• The sum of an arithmetic progression (an
arithmetic series, difference between one
and the previous term is a constant)
sn  a  (a  d )  (a  2d )  (a  3d )  ...  (a  (n  1)d )
• Can we find a formula so we don’t have to
add up every arithmetic series we come
across?
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Sum of terms of a finite AP
S n  a  (a  d )  (a  2d )  ...  [a  (n  2)d ]  [a  (n  1)d ]
S n  a  (a  d )  (a  2d )  ...  (a  nd  2d )  (a  nd  d )
S n  (a  nd  d )  (a  nd  2d )  ...  (a  2d )  (a  d )  a
2 S n  a  (2a  nd )  (2a  nd )  ...  (2a  nd )  (2a  nd )  a
There are (n) 2a terms  2a  n  2an;
There are (n - 1) nd terms  nd  (n - 1)  nd (n - 1) ; Therefore,
2Sn  2an  nd (n  1)
2 S n  n[2a  (n  1)d ]
n
S n  [2a  (n  1)d ]
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Arithmetic Gradient Series
• A series of N receipts or disbursements that increase
by a constant amount from period to period.
• Cash flows: 0G, 1G, 2G, ..., (N–1)G at the end of
periods 1, 2, ..., N
• Cash flows for arithmetic gradient with base annuity:
A', A’+G, A'+2G, ..., A'+(N–1)G at the end of
periods 1, 2, ..., N where A’ is the amount of the base
annuity
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Arithmetic Gradient to Uniform Series
• Finds A, given G, i and N
• The future amount can be “converted” to an
equivalent annuity. The factor is:
1
N
( A / G, i , N )  
i (1  i )N  1
• The annuity equivalent (not future value!)
to an arithmetic gradient series is A =
G(A/G, i, N)
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Arithmetic Gradient to Uniform Series
• The annuity equivalent to an arithmetic
gradient series is A = G(A/G, i, N)
• If there is a base cash flow A', the base
annuity A' must be included to give the
overall annuity:
Atotal = A' + G(A/G, i, N)
• Note that A' is the amount in the first year
and G is the uniform increment starting in
year 2.
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Arithmetic Gradient Series with
Base Annuity
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Example 3-8
• A lottery prize pays $1000 at the end
of the first year, $2000 the second,
$3000 the third, etc., for 20 years. If
there is only one prize in the lottery,
10 000 tickets are sold, and you can
invest your money elsewhere at 15%
interest, how much is each ticket
worth, on average?
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Example 3-8: Answer
• Method 1: First find annuity value of prize
and then find present value of annuity.
A' = 1000, G = 1000, i = 0.15, N = 20
A = A' + G(A/G, i, N) = 1000 + 1000(A/G,
15%, 20)
= 1000 + 1000(5.3651) = 6365.10
• Now find present value of annuity:
P = A (P/A, i, N) where A = 6365.10, i =
15%, N = 20
P = 6365.10(P/A, 15, 20)
= 6365.10(6.2593) = 39 841.07
• Since 10 000 tickets are to be sold, on
average each ticket is worth (39
841.07)/10,000 = $3.98.
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Arithmetic Gradient Conversion Factor
(to Uniform Series)
• The arithmetic gradient conversion factor (to
uniform series) is used when it is necessary
to convert a gradient series into a uniform
series of equal payments.
• Example: What would be the equal annual
series, A, that would have the same net
present value at 20% interest per year to a
five year gradient series that started at $1000
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and increased $150 every year thereafter?
Arithmetic Gradient Conversion Factor
(to Uniform Series)
1
2
3
4
5
1
2
3
4
5
$1000
$1150
A
A
A
A
A
$1300
$1450
(1  i ) n  (1  ni )
A  Ag  G
i[(1  i ) n  1]
$1600
(1  0.20)5  (1  5 * 0.20)
 $1,000  $150
0.20[(1  0.20)5  1]
 $1,246
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Arithmetic Gradient Conversion Factor
(to Present Value)
• This factor converts a series of cash
amounts increasing by a gradient value,
G, each period to an equivalent present
value at i interest per period.
• Example: A machine will require $1000
in maintenance the first year of its 5
year operating life, and the cost will
increase by $150 each year. What is the
present worth of this series of
maintenance costs if the firm’s minimum
attractive rate of return is 20%?
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Arithmetic Gradient Conversion Factor
(to Present Value)
$1600
$1450
$1300
$1150
$1000
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2
3
4
5
P
(1  i ) n  1
1  (1  ni )(1  i )  n
PA
G
n
i (1  i )
i2
(1  0.20)5  1
1  (1  5 * 0.20)(1  0.20) 5
 $1,000
 $150
5
0.20(1  0.20)
(0.20) 2
 $3,727
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Geometric Gradient Series
• A series of cash flows that increase or decrease
by a constant proportion each period
• Cash flows: A, A(1+g), A(1+g)2, …, A(1+g)N–1
at the end of periods 1, 2, 3, ..., N
• g is the growth rate, positive or negative
percentage change
• Can model inflation and deflation using
geometric series
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Geometric Series
• The sum of the consecutive terms of a
geometric sequence or progression is
called a geometric series.
• For example:
Sn  a  ak  ak 2  ak 3  ....  ak n 2  ak n 1
Is a finite geometric series with quotient
k.
• What is the sum of the n terms of a finite
geometric series
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Sum of terms of a finite GP
Sn  a  ak  ak 2  ....  ak n  2  ak n 1
kSn  ak  ak 2  ....  ak n  2  ak n 1  ak n
Sn  kSn  a  0  0  .....  0  0  ak n
Sn (1  k )  a (1  k n )
(1  k n )
Sn  a
(1  k )
• Where a is the first term of the geometric progression,
k is the geometric ratio, and n is the number of terms
in the progression.
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Geometric Gradient to
Present Worth
• The present worth of a geometric series is:
A
A(1  g )
A(1  g )N 1
P


2
(1  i ) (1  i )
(1  i )N
• Where A is the base amount and g is the
growth rate.
• Before we may get the factor, we need what
is called a growth adjusted interest rate:
i 
1 i
1
1 g
 1 so that

1 g
1 i  1 i
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Geometric Gradient to Present Worth
Factor: (P/A, g, i, N)
(1  i  )N  1 1 
(P / A, g, i , N ) 



 N 1 g

i (1  i ) 
(P/A,i ,N)

( 1  g)
Four cases:
(1) i > g > 0:
i° is positive  use tables or formula
(2) g < 0:
i° is positive  use tables or formula
(3) g > i > 0: i° is negative  Must use formula
(4) g = i > 0: i° = 0

 A 
P  N

1 g 
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Compound Interest Factors
Discrete Cash Flow, Discrete Compounding
To Find
F
P
Given
P
F
Name of Factor
Compound Amount
Factor (single payment)
Present Worth Factor
(single payment)
F
A
Compound Amount
Factor (uniform series)
A
F
Sinking Fund Factor
Factor
(1  i) n
(1  i )  n
(1  i ) n  1
i
i
(1  i ) n  1
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Compound Interest Factors
Discrete Cash Flow, Discrete Compounding
To Find
A
P
A
P
Given
P
A
G
G
Name of Factor
Factor
Capital Recovery Factor
i (1  i ) n
(1  i ) n  1
Present Worth Factor
(uniform series)
Arithmetic Gradient
Conversion Factor (to
uniform series)
Arithmetic Gradient
Conversion Factor (to
present value)
(1  i) n  1
i (1  i) n
(1  i ) n  (1  ni )
i[(1  i ) n  1]
1  (1  ni)(1  i )  n
i2
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Compound Interest Factors
Discrete Cash Flow, Continuous Compounding
To Find
F
P
F
A
Given
P
F
A
F
Name of Factor
Compound Amount
Factor (single payment)
Present Worth Factor
(single payment)
Factor
e  rn
Compound Amount
Factor (uniform series)
e rn  1
er 1
Sinking Fund Factor
er 1
e rn  1
e rn
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Compound Interest Factors
Discrete Cash Flow, Continuous Compounding
To Find
A
P
A
P
Given
P
A
G
G
Name of Factor
Factor
Capital Recovery Factor
e rn (e r  1)
e rn  1
Present Worth Factor
(uniform series)
Arithmetic Gradient
Conversion Factor (to
uniform series)
Arithmetic Gradient
Conversion Factor (to
present value)
e rn  1
e rn (e r  1)
1
n
 rn
r
e 1 e 1
e rn  1  n(e r  1)
e rn (e r  1) 2
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Compound Interest Factors
Continuous Uniform Cash Flow, Continuous Compounding
To Find
C
C
F
P
Given
F
P
C
C
Name of Factor
Sinking Fund Factor
(continuous, uniform
payments)
Capital Recovery Factor
(continuous, uniform
payments)
Compound Amount
Factor (continuous,
uniform payments)
Present Worth Factor
(continuous, uniform
payments)
Factor
r
e rn  1
re rn
e rn  1
e rn  1
r
e rn  1
re rn
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Quiz---When and Where
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Quiz: Tuesday, Sept. 27, 2005
11:30 - 12:20 (Quiz: 30 minutes)
Tutorial: Wednesday, Sept. 28, 2005
ELL 168 Group 1
(Students with Last Name A-M)
ELL 061 Group 2
(Students with Last Name N-Z)
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Quiz---Who will be there
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U, U, U, U, and U!!!!
CraigTipping ctipping@uvic.ca
Group 1 (Last NameA-M) ELL 168
LeYang
yangle@ece.uvic.ca
Group 2 (Last Name N-Z) ELL 061
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Quiz---Problems, Solutions
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Do not argue with your TA!
Question? Problems? TAWei
Solutions will be given on Tutorial
Bring: Blank Letter Paper, Pen, Formula
Sheet, Calculator, Student Card
• Write: Name, Student No. and Email
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Quiz---Based on Chapter 1.2.3.
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Important: Wei’s Slides
Even More Important: Examples in Slides
1 Formula Sheet is a good idea
5 Questions for 1800 seconds.
Wei used 180 seconds (relax)
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Quiz---Important Points
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Simple Interests
Compound Interests
Future Value
Present Value
Key: Compound Interest
Key: Understand the Question
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Quiz---Books in Library!!!
Engineering Economics in Canada, 3/E
Niall M. Fraser, University of Waterloo
Elizabeth M. Jewkes, University of Waterloo
Irwin Bernhardt, University of Waterloo
May Tajima, University of Waterloo
Economics: Canada in the Global Environment
by Michael Parkin and Robin Bade.
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Calculator Talk
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No programmable
No economic function
Simple the best
Trust your ability
Trust your teaching group
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• Questions?
• (Sorry I forget the problems)
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Project----Time Table
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Find your group: Mid-October
Select Topic: End of October
Survey finished: End of October
Project: November (3 Weeks)
Project Report Due: Final Quiz
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Project----Requirements
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Group: 3-6 Students
Topic: Practical, Small
Report: On Time, Original
Marks: 1 make to 1 report
Report: 25 marks out of 100
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Project Topic----What to do
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You Find it
Practical
Example: Run a Pizza Shop
Example: Run a Store for computer renting
Example: Survey on the Tuition Increase
Example: Why ??? Company failed…..
Team Work
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Project----Recourse
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Not your teaching group
No spoon feed: Independent work
Example: Government Web
Example: Library, Database, Google
Example: Economics Faculty
Example: Newspaper, TV
Example: Friends
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Summary
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Conversion for Arithmetic Gradient Series
Conversion for Geometric Gradient Series
Quiz: My slides and the examples in slides
Project: Good Idea, be open, independent
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