Warm-Up Review
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Rule of 72
Single Sum Compounding
Annuities
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The Rule of 72
• Estimates how many years an investment
will take to double in value
• Number of years to double =
72 / annual compound interest rate
• Example -- 72 / 8 = 9 therefore, it will
take 9 years for an investment to double
in value if it earns 8% annually
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Example: Double Your Money!!!
Quick! How long does it take to double $5,000 at
a compound rate of 12% per year?
Approx. Years to Double = 72 / i%
72 / 12% = 6 Years
[Actual Time is 6.12 Years]
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Single Sum Problems: Future Value
Given:
•
Amount of deposit today (PV):$50,000
•
Interest rate: 11%
•
Frequency of compounding: Annual
•
Number of periods (5 years): 5 periods
What is the future value of this single sum?
FVn = PV(1 + i)n
$50,000 x (1.68506) = $84,253
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Single Sum Problems: Present Value
Given:
• Amount of deposit end of 5 years: $84,253
•
Interest rate (discount) rate:
11%
•
Frequency of compounding:
Annual
•
Number of periods (5 years):
5 periods
What is the present value of this single sum?
• FVn = PV(1 + i)n
$84,253 x (0.59345) = $50,000
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Annuity Computations
An annuity requires that:
• the periodic payments or receipts
(rents) always be of the same
amount,
• the interval between such payments
or receipts be the same, and
• the interest be compounded once
each interval.
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Example of Annuity
End of Year
0
1
2
3
$1,000
$1,000
$1,070
4
7%
$1,000
$1,145
If one saves $1,000 a year at the end
$3,215 = FVA3
of every year for three years in an
2
FVA3
=
$1,000(1.07)
+
account earning 7% interest,
$1,000(1.07)1 +
compounded annually, how much
$1,000(1.07)0 = $3,215
will one have at the end of the
third year?
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Annuities: Future Value
Given:
• Deposit made at the end of each period:
$5,000
• Compounding:
Annual
• Number of periods:
Five
• Interest rate:
12%
What is future value of these deposits?
F = A[(1+i)n - 1] / i
$5,000 x (6.35285) = $ 31,764.25
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Annuities: Present Value
Given:
• Rental receipts at the end of each period:
$6,000
• Compounding:
Annual
• Number of periods (years):
5
• Interest rate:
12%
What is the present value of these receipts?
F = A[(1+i)n - 1] / i
$6,000 x (3.60478) = $ 21,628.68
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Key of Annuity Calculation
Fv =
n
Pv[(1+i)
- 1] / i
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Topics Today
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•
•
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Single Payment, Present/Future Value Factor
Sinking Factor, Capital Recovery Factor
Conversion for Arithmetic Gradient Series
Conversion for Geometric Gradient Series
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Compound Amount Factor
(Single Payment)
• This factor finds the equivalent
future worth, F, of a present
investment, P, held for n periods at i
rate of interest.
• Example: What is the value in 9
years of $1,200 invested now at
10% interest
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Compound Amount Factor
(Single Payment)
F
1
2
3
4
9
P = $1,200
F  P(1  i)
n
 $1,200(1  0.10)
 $2,829
9
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Present Worth Factor
(Single Payment)
• This factor finds the equivalent present
value, P, of a single future cash flow, F,
occurring at n periods in the future when
the interest rate is i per period.
• Example: What amount would you have to
invest now to yield $2,829 in 9 years if the
interest rate per year is 10%?
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Present Worth Factor
(Single Payment)
F = $2,829
1
2
3
4
9
P
P  F (1  i)
n
 $2,829(1  0.10)
 $1,200
9
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Compound Amount Factor
(Uniform Series)
• This factor finds the equivalent future
value, F, of the accumulation of a uniform
series of equal annual payments, A,
occurring over n periods at i rate of
interest per period.
• Example: What would be the future worth
of an annual year-end cash flow of $800
for 6 years at 12% interest per year?
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Compound Amount Factor
(Uniform Series)
F
1
$800
2
$800
3
4
$800
5
$800
6
$800
$800
(1  i )  1
FA
i
(1  0.12) 6  1
 $800
0.12
 $6,492
n
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Sinking Fund Factor
• This factor determines how much must be
deposited each period in a uniform series,
A, for n periods at i interest per period to
yield a specified future sum.
• Example: If a $1.2 million bond issue is to
be retired at the end of 20 years, how
much must be deposited annually into a
sinking fund at 7% interest per year?
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Sinking Fund Factor
F = $1,200,000
1
2
A
3
A
4
A
20
A
A
A
i
A F
(1  i ) n  1
0.07
 $1,200,000
(1  0.07) 20  1
 $29,272
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Capital Recovery Factor
• This factor finds an annuity, or uniform
series of payments, over n periods at i
interest per period that is equivalent to a
present value, P.
• Example: What savings in annual
manufacturing costs over an 8 year period
would justify the purchase of a $120,000
machine if the firm’s minimum attractive
rate of return (MARR) were 25%?
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Capital Recovery Factor
A
1
$120,000
A
2
A
A
3
A
8
i (1  i ) n
A P
(1  i ) n  1
0.25(1  0.25)8
 $120,000
(1  0.25)8  1
 $36,048
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Present Worth Factor
(Uniform Series)
• This factor finds the equivalent
present value, P, of a series of endof-period payments, A, for n periods
at i interest per period.
• Example: What lump sum payment
would be required to provide
$50,000 per year for 30 years at an
annual interest rate of 9%?
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Present Worth Factor
(Uniform Series)
$50,000 $50,000 $50,000
1
P
2
$50,000 $50,000
3
30
(1  i ) n  1
PA
i (1  i ) n
(1  0.07)30  1
 $50,000
0.07(1  0.07)30
 $620,452
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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 ]
2
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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
1
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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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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Summary
•
•
•
•
Single Sum Compounding
Annuities
Conversion for Arithmetic Gradient Series
Conversion for Geometric Gradient Series
• Key: Compound Interests Calculation
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