Interest Formulas – Equal
Payment Series
Lecture No.5
Professor C. S. Park
Fundamentals of Engineering Economics
Copyright © 2005
Equal Payment Series
F
0
A
A
1
2
N
A
P
0
1
2
N
0
N
Equal Payment Series – Compound Amount
Factor
F
A
A
A
0
1
2
N
F
0
1
2
N
0
1
2
N
A
A
A
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
Equal Payment Series Compound Amount Factor (Future Value of
an annuity)
F
(1  i )  1
FA
i
 A( F / A, i , N )
N
0
1
2
3
N
A
Example 2.9:
 Given: A = $5,000, N = 5 years, and i = 6%
 Find: F
 Solution: F = $5,000(F/A,6%,5) = $28,185.46
Validation
$5,000(1  0.06)  $6,312.38
F =?
4
$5,000(1  0.06)3  $5,955.08
$5,000(1  0.06)  $5,618.00
i = 6%
2
0
1
2
3
4
5
$5,000(1  0.06)1  $5,300.00
$5,000(1  0.06)  $5,000.00
0
$28.185.46
$5,000 $5,000 $5,000 $5,000 $5,000
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 2.10 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
F
i
L
O
A  FM
P
(
1

i
)

1
N Q
N
0
1
2
3
N
A
 F( A / F, i, N )
Example 2.11 – 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
$100,000
Current age: 10 years old
• Find:
0
1
=PMT(i,N,pv,fv,type)
=PMT(7%,8,0,100000,0)
=$9,746.78
2
3
4
A=?
i = 8%
5
6
7
8
Capital Recovery Factor
P
i (1  i )
A P
N
(1  i )  1
 P( A / P, i , N )
N
1
2
3
0
N
A=?
Example 2.12: 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 2.14 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
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 2.14:Powerball Lottery
 Given: A = $7.92M, N = 25 years, and i = 8%
 Find: P
 Solution: P = $7.92M(P/A,8%,25) = $84.54M
Excel Solution

Given:




A = $7.92M
i = 8%
N = 25
Find: P
=PV(8%,25,7.92,0)
= $84.54M
A = $7.92 million
0
1
2
25
i = 8%
P=?
Example 2.15 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