Interest Formulas – Equal
Payment Series
Engineering Economy
Equal Payment Series
F
Using equal payment
series you can find
present value or future
value
0
A
A
1
2
N
A
P
0
1
2
N
0
N
Compound Amount Factor
F
A(1+i)N-2
A
A
A
A(1+i)N-1
0
1
N
2
F  A(1  i)
N 1
0
1
 A(1  i)
2
N 2
N

A
This is a geometric series with the first term as A and the constant r = (1+i)
The formula for a geometric series = A (1- r^n)/1-r
Equal Payment Series Compound Amount Factor
(Future Value of an annuity)
F
0
1
2
3
N
A
(1  i ) N  1
FA
i
 A( F / A, i , N )
Example:
 Given: A = $5,000, N = 5 years, and i = 6%
 Find: F
 Solution: F = $5,000(F/A,6%,5) = $28,185.46
Finding an Annuity Value
F
0
1
2
3
N
A=?
i
A F
N
(1  i )  1
 F ( A / F ,i, N )
Example:
• Given: F = $5,000, N = 5 years, and i = 7%
• Find: A
• Solution: A = $5,000(A/F,7%,5) = $869.50
Example: Handling Time Shifts in a Uniform
Series
F=?
First deposit occurs at n = 0
i = 6%
0
1
2
3
4
$5,000 $5,000 $5,000 $5,000 $5,000
5
 Annuity
Due
F5  $5,000( F / A,6%,5)(1.06)
 $29,876.59
 Excel
Solution
Beginning period
=FV(6%,5,5000,0,1)
Sinking Fund Factor
The term between the brackets is called the equal-payment-series sinkingfund factor.
F
0
1
2
3
N


i
A F 

N
 (1  i )  1
A
Example: College Savings Plan
• Given: F = $100,000, N = 8 years, and i = 7%
• Find: A
• Solution:
A = $100,000(A/F,7%,8) = $9,746.78
Excel Solution
 Given:
 F = $100,000
 i = 7%
 N = 8 years
• Find: A
Using the equation:


i
A F 

N
(
1

i
)

1


Using built in Function:
=PMT(i,N,pv,fv,type)
=PMT(7%,8,0,100000,0)
=$9,746.78
$100,000
Current age: 10 years old
0
1
2
3
4
A=?
i = 8%
5
6
7
8
Capital Recovery Factor
If we need to find A, given P,I, and N


i
A F 

N
 (1  i )  1
Remember that:
F  P(1  i)
N
Replacing F with its value


i
A  P(1  i) 

N
(
1

i
)

1


N
Capital Recovery Factor
This factor is called capital
recovery factor
P
1
2
3
0
N
A=?
i (1  i )
A P
N
(1  i )  1
 P( A / P, i , N )
N
Example: Paying Off Education Loan
• Given: P = $21,061.82, N = 5 years, and i =
6%
• Find: A
• Solution: A = $21,061.82(A/P,6%,5) = $5,000
Example: Deferred Loan Repayment Plan
P =$21,061.82
i = 6%
0
1
Grace period
2
3
4
A
A
A
5
A
6
A
P’ = $21,061.82(F/P, 6%, 1)
i = 6%
0
1
2
3
4
A’
A’
A’
5
A’
6
A’
Two-Step Procedure
Adding the first year interest to the principal then calculating the annuity
payment
P '  $21, 061.82( F / P, 6%,1)
 $22,325.53
A  $22,325.53( A / P, 6%,5)
 $5,300
Present Worth of Annuity Series
P=?
1
2
3
0
N
A
(1  i ) N  1
P A
i (1  i ) N
 A( P / A, i , N )
Example:Powerball Lottery
• Given: A = $7.92M, N = 25 years, and i = 8%
• Find: P
• Solution: P = $7.92M(P/A,8%,25) = $84.54M
Example: Early Savings Plan – 8% interest
?
Option 1: Early Savings Plan
0
1
2
3
4
5
6
7
8
9
10
44
$2,000
?
Option 2: Deferred Savings Plan
0
1
2
3
4
5
6
7
8
9
10 11 12
44
$2,000
Option 1 – Early Savings Plan
?
F10  $2, 000( F / A,8%,10)
 $28,973
Option 1: Early Savings Plan
0 1 2 3 4 5 6 7 8 9 10
F44  $28,973( F / P,8%,34)
 $396, 645
44
$2,000
Age
31
65
Option 2: Deferred Savings Plan
?
F44  $2, 000( F / A,8%,34)
 $317, 233
Option 2: Deferred Savings Plan
0
11 12
44
$2,000
At What Interest Rate These Two
Options Would be Equivalent?
Option 1:
F44  $2, 000( F / A, i,10)( F / P, i,34)
Option 2:
F44  $2, 000( F / A.i,34)
Option 1 = Option 2
$2, 000( F / A, i,10)( F / P, i,34)  $2, 000( F / A.i,34)
Solve for i
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
40
41
42
43
44
45
46
47
A
B
C
Year
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
37
38
39
40
41
42
43
44
Option 1
Option 2
$
$
$
$
$
$
$
$
$
$
(2,000)
(2,000)
(2,000)
(2,000)
(2,000)
(2,000)
(2,000)
(2,000)
(2,000)
(2,000)
D
E
Interest rate
$
$
$
$
$
$
$
$
$
$
$
$
$
$
$
$
$
(2,000)
(2,000)
(2,000)
(2,000)
(2,000)
(2,000)
(2,000)
(2,000)
(2,000)
(2,000)
(2,000)
(2,000)
(2,000)
(2,000)
(2,000)
(2,000)
(2,000)
F
0.08
FV of Option 1
$ 396,645.95
FV of Option 2
$ 317,253.34
Target cell
$ 79,392.61
Using Excel’s Goal Seek Function
Result
$396,644
Option 1: Early Savings Plan
0
1
2
3
4
5
6
7
8
9
10
44
$2,000
$317,253
Option 2: Deferred Savings Plan
0
1
2
3
4
5
6
7
8
9
10 11 12
44
$2,000
Interest Formulas
(Gradient Series)
Linear Gradient Series
Gradient-series present –
worth factor
 1  i   iN  1
P G  2

N
 i 1  i 

P  G ( P / G , i, N )
N
P
Gradient Series as a Composite Series
We view the cash flows as composites of two series a uniform with a
payment amount of A1 and a gradient with a constant amount of G
Example:
$2,000
$1,250 $1,500
$1,750
$1,000
0
1
P =?
2
3
4
5
How much do you have to deposit
now in a savings account that
earns a 12% annual interest, if
you want to withdraw the annual
series as shown in the figure?
Method 1:
$2,000
$1,250 $1,500
$1,750
$1,000
0
1
P =?
2
3
4
5
$1,000(P/F, 12%, 1) = $892.86
$1,250(P/F, 12%, 2) = $996.49
$1,500(P/F, 12%, 3) = $1,067.67
$1,750(P/F, 12%, 4) = $1,112.16
$2,000(P/F, 12%, 5) = $1,134.85
$5,204.03
Method 2:
P1  $1,000( P / A,12%,5)
 $3,604.80
P2  $250( P / G,12%,5)
 $1,599.20
P  $3,604.08  $1,599.20
 $5,204
Example: Supper Lottery
$3.44 million
Cash Option
0
1
2
3
4
5
6
7
25
26
Annual Payment Option
G = $7,000
$189,000
$357,000
$196,000
$175,000
0
1
2
3
4
5
6
7
25
26
Equivalent Present Value of Annual Payment
Option at 4.5%
The
gradient
series is
delayed by
one period
P  [$175, 000  $189, 000( P / A, 4.5%, 25)
$7, 000( P / G, 4.5%, 25)]( P / F , 4.5%,1)
 $3,818,363
To return the
calculations to
year zero
Geometric Gradient Series
Geometric Gradient is a gradient series that is been determined by a fixed rate
expressed as a percentage instead of a fixed dollar amount
For example the economic problems related to construction cost which involves
cash flows that increase or decrease by a constant percentage
Present Worth Factor
Geometric-gradient-series
present-worth factor
 1  (1  g ) N (1  i )  N 
 A1 
 , if i  g
ig
 

P 
 A N , if i  g
 1 (1  i )
Alternate Way of Calculating P
ig
Let g ' 
1 g
A1
P
( P / A, g ', N )
(1  g )
Unconventional
Equivalence Calculations
EGN3613
Ch2 Part IV
$200
Composite
Cash Flows
$150 $150 $150 $150
$100 $100 $100
$50
0
1
2
3
4
5
6
PGroup 1  $50( P / F ,15%,1)
 $43.48
PGroup 2  $100( P / A,15%,3)( P / F ,15%,1)
 $198.54
PGroup 3  $150( P / A,15%, 4)( P / F ,15%, 4)
 $244.85
PGroup 4  $200( P / F ,15%, 9)
 $56.85
P  $43.48  $198.54  $244.85  $56.85
 $543.72
7
8
9
Unconventional Equivalence Calculations
Situation:
What value of A
would make the
two cash flow
transactions
equivalent if i =
10%?
Multiple Interest Rates
F=?
Find the balance at the end of year 5.
6%
6%
5%
4%
4%
0
1
2
3
4
$400
$300
$500
5
Solution
n  1:
$300( F / P, 5%,1)  $315
n  2:
$315( F / P, 6%,1)  $500  $833.90
n  3:
$833.90( F / P, 6%,1)  $883.93
n  4:
$883.93( F / P, 4%,1)  $400  $1, 319.29
n  5:
$1, 319.29( F / P, 4%,1)  $1, 372.06
Cash Flows with Missing Payments
P=?
1
2
3
4
5
6
7
8
9 10
11 12 13
14 15
0
$100
i = 10%
Missing payment
Solution
P=?
1
2
$100
3
4
5
6
7
8
9 10
Add this cash flow to
offset the change
11 12 13
14 15
0
$100
i = 10%
Pretend that we have the 10th
payment
Approach
P=?
1
2
$100
3
4
5
6
7
8
9 10
11 12 13
0
$100
i = 10%
Equivalent Cash Inflow = Equivalent Cash Outflow
14 15
Equivalence Relationship
P  $100( P / F ,10%,10)  $100( P / A,10%,15)
P  $38.55  $760.61
P  $722.05
Unconventional Regularity in Cash Flow
Pattern
$10,000
i = 10%
1
2
3
4
5
6
7
8
9
10 11 12 13 14
0
C
C
C
C
Payment is made every other year
C
C
C
Approach 1: Modify the Original Cash
Flows
$10,000
i = 10%
1
2
3
A
A
A
4
5
6
7
8
9
10 11 12 13 14
0
A
A
A
A
A
A
A
A  $10, 000( A / P,10%,14)
 $1,357.46
A
A
A
A
Relationship Between A and C
$10,000
i = 10%
1
2
3
4
5
6
7
8
9
10 11 12 13 14
0
C
C
C
C
C
C
C
$10,000
i = 10%
1
2
3
A
A
A
4
5
6
7
8
9
10 11 12 13 14
0
A
A
A
A
A
A
A
A
A
A
A
Solution

C
A
A
i = 10%
A =$1,357.46
A  $10,000( A / P,10%,14)
 $1,357.46

C  A( F / P,10%,1)  A
 1.1A  A
 2.1A
 2.1($1,357.46)
 $2,850.67
Approach 2: Modify the Interest Rate
 Idea: Since cash flows occur every other year, let's find out the equivalent
compound interest rate that covers the two-year period.
 How: If interest is compounded 10% annually, the equivalent interest rate
for two-year period is 21%.
(1+0.10)(1+0.10) = 1.21
Solution
$10,000
i = 21%
1
1
2
2
3
4
3
5
6
4
7
8
5
9
6
7
10 11 12 13 14
0
C
C
C
C
C
C
C  $10,000( A / P, 21%,7)
 $2,850.67
C