!
"
#
$!
&
%
&
('' )
' (
&
%"
''
'
()
*
)
()
*
+
() *
) $)
$
(
&
&
() *
) , !&
()
(
*
'
('' )
&
'
#
)
'
&
&
.
)
)
#
&
.
&
)
+
)
x(n + 1) = Rx(n) − P
!
./ '
/#
0
12/
& '

→ x(n + 1) − x(n) 
→
rx ( n )
x(n + 1) − x(n) = rx(n) − p
x(n + 1) = (1 + r ) x(n) − P
R = 1+ r
x(n + 1) = Rx(n) − P
P
)
x(n + 1) = Rx(n) − P
x(0) = x(0)
x(1) = Rx(0) − P
x(2) = Rx(1) − P
= R[ Rx(0) − P ] − P
= R 2 x(0) − RP − P
= R x(0) − P ( R + 1)
2
)
x(2) = R 2 x(0) − P( R + 1)
x(3) = Rx(2) − P
= R[ R x(0) − P( R + 1)] − P
2
= R 3 x(0) − RP( R + 1) − P
= R x(0) − P[ R( R + 1) + 1]
3
= R x(0) − P( R + R + 1)
3
2
)
x(3) = R 3 x(0) − P( R 2 + R + 1)
x(4) = Rx(3) − P
= R[ R 3 x(0) − P( R 2 + R + 1)] − P
= R x(0) − RP( R + R + 1) − P
4
2
= R x(0) − P[ R( R + R + 1) + 1]
4
2
= R x(0) − P( R + R + R + 1)
4
3
2
)
,
'
x(1) = Rx(0) − P
x(2) = R x(0) − P( R + 1)
2
x(3) = R 3 x(0) − P( R 2 + R + 1)
x(4) = R x(0) − P( R + R + R + 1)
4
3
2
x(n) = R n x(0) − P( R n −1 + R n − 2 + ... + R + 1)
)
n −1
+R
n−2
+ ... + R + 1)
x(n) = R x(0) − P( R
n
n−2
+ ... + R + 1)
3)
S = (R
n −1
+R
'
S=R
n −1
+R
n−2
+ ... + R + 1
RS = R n + R n −1 + ... + R 2 + R
1 + RS = R n + ( R n −1 + ... + R 2 + R + 1)
1 + RS = R n + S
RS − S = R n − 1
S ( R − 1) = R n − 1
Rn −1
S=
,R ≠1
R −1
)
) 3)
-
x(n) = R n x(0) − P( R n −1 + R n − 2 + ... + R + 1)
!
n
−1
R
n
x(n) = R x(0) − P
,R ≠1
R −1
)
)
!)
'
!
)
!
Simulation 1
Balance =
$
yearly interest =
monthly payment = $
Months
0
1
2
3
4
5
10
20
30
40
41
$
$
$
$
$
$
$
$
$
$
$
3,000
18%
100
Balance
3,000.00
2,945.00
2,889.18
2,832.51
2,775.00
2,716.63
2,411.35
1,728.20
935.37
15.27
-
$
$
$
$
$
$
$
$
$
$
$
Interest
45.00
44.18
43.34
42.49
41.63
37.11
27.02
15.30
1.70
-
total payment
$ 4,015.27
total interest
$ 1,015.27
effective interest rate 25.29%
$
$
$
$
$
$
$
$
$
$
$
Payment
100.00
100.00
100.00
100.00
100.00
100.00
100.00
100.00
100.00
15.27
-
Simulation 2
Balance
$
yearly interest
monthly payment $
Months
0
1
2
3
4
5
10
20
30
40
50
60
61
62
$
$
$
$
$
$
$
$
$
$
$
$
$
$
4,000
18%
100
Balance
4,000.00
3,960.00
3,919.40
3,878.19
3,836.36
3,793.91
3,571.89
3,075.05
2,489.45
1,829.28
1,052.69
151.41
53.69
-
$
$
$
$
$
$
$
$
$
$
$
$
$
$
Interest
60.00
59.40
58.79
58.17
57.55
54.26
46.92
38.40
28.51
17.03
3.72
2.27
-
total payment $
6,153.69
total interest $
2,153.69
effective interest rate 35.00%
$
$
$
$
$
$
$
$
$
$
$
$
$
$
Payment
100.00
100.00
100.00
100.00
100.00
100.00
100.00
100.00
100.00
100.00
100.00
100.00
53.69
-
)' &
('
4
.
)
&
!
'
&
&
&
#
>#
%
+)
;:::
6=8
<
9:::
678
5
;::
+)
How do we pay off both cards?
3)
b1 , b 2 = the balances on both cards
where : b1 = 3000, b 2 = 5000
r1 , r2 = the interest rates on both cards
where : r1 = .18, r2 = .12
F = the available funds in your budget
where : F = 300
P = interest charged next month
P = r1 ( b1 − x ) + r 2 ( b2 − y )
4
#
Question:
What is the most effective way to pay of the two
credit card balances?
Answer:
Pay the card with the highest interest rate.
!
'
&
&
&
#
>#
%
+)
6
6
<
7
7
5
+
+)
)'
5 )
'
"
!
)
)
"
!
) #
3)
y=F − x
f ( x) = r1 ( b1 − x) + r 2 ( b2 − ( F − x))
f ( x ) = r 1 b1 − r 1 x + r 2 b 2 − r 2 F + r 2 x
f ( x) = ( r 2 − r1 ) x + r1 b1 + r 2 ( b2 − F )
+
) #
f ( x) = ( r 2 − r1 ) x + r1 b1 + r 2 ( b2 − F )
f (x)
x
Do not attempt this in the real world.
Why?
Your credit card company will charge you late
fees in the real world.
Alternative:
Consolidate your credit cards with a home
equity loan or low interest credit card.
Good Debt
Examples of Good Debt
Education
House
Land
Example: We will use is taking a loan out for an
Applied Math degree.
Assumptions
After 108 months ( 5 years after you graduate)
Till retirement at age of 65
Your Math Degree Pays
•And 758,648 ahead of the
associate degree grad.
•As you can see you
come out 911,616 of the
high school grad
•Well worth your 26,000
in loans.
Conclusions
There is a right time to go into debt.
Just think before you act.
Do the Math.
And make you good investments.
)
('' )
&
) )
) )
"
?
&
*
&.
)
!
"
#
$!
&
%
&
('' )
' (
&
%"
''
'
()
*
)
()
*
+
() *
) $)
$
(
&
&
() *
) , !&
()
(
*
'
('' )
&
'
#
)
'
&
&
.
)
)
#
&
.
&
)
+
)
x(n + 1) = Rx(n) − P
!
./ '
/#
0
12/
& '

→ x(n + 1) − x(n) 
→
rx ( n )
x(n + 1) − x(n) = rx(n) − p
x(n + 1) = (1 + r ) x(n) − P
R = 1+ r
x(n + 1) = Rx(n) − P
P
)
x(n + 1) = Rx(n) − P
x(0) = x(0)
x(1) = Rx(0) − P
x(2) = Rx(1) − P
= R[ Rx(0) − P ] − P
= R 2 x(0) − RP − P
= R x(0) − P ( R + 1)
2
)
x(2) = R 2 x(0) − P( R + 1)
x(3) = Rx(2) − P
= R[ R x(0) − P( R + 1)] − P
2
= R 3 x(0) − RP( R + 1) − P
= R x(0) − P[ R( R + 1) + 1]
3
= R x(0) − P( R + R + 1)
3
2
)
x(3) = R 3 x(0) − P( R 2 + R + 1)
x(4) = Rx(3) − P
= R[ R 3 x(0) − P( R 2 + R + 1)] − P
= R x(0) − RP( R + R + 1) − P
4
2
= R x(0) − P[ R( R + R + 1) + 1]
4
2
= R x(0) − P( R + R + R + 1)
4
3
2
)
,
'
x(1) = Rx(0) − P
x(2) = R x(0) − P( R + 1)
2
x(3) = R 3 x(0) − P( R 2 + R + 1)
x(4) = R x(0) − P( R + R + R + 1)
4
3
2
x(n) = R n x(0) − P( R n −1 + R n − 2 + ... + R + 1)
)
n −1
+R
n−2
+ ... + R + 1)
x(n) = R x(0) − P( R
n
n−2
+ ... + R + 1)
3)
S = (R
n −1
+R
'
S=R
n −1
+R
n−2
+ ... + R + 1
RS = R n + R n −1 + ... + R 2 + R
1 + RS = R n + ( R n −1 + ... + R 2 + R + 1)
1 + RS = R n + S
RS − S = R n − 1
S ( R − 1) = R n − 1
Rn −1
S=
,R ≠1
R −1
)
) 3)
-
x(n) = R n x(0) − P( R n −1 + R n − 2 + ... + R + 1)
!
n
−1
R
n
x(n) = R x(0) − P
,R ≠1
R −1
)
)
!)
'
!
)
!
Simulation 1
Balance =
$
yearly interest =
monthly payment = $
Months
0
1
2
3
4
5
10
20
30
40
41
$
$
$
$
$
$
$
$
$
$
$
3,000
18%
100
Balance
3,000.00
2,945.00
2,889.18
2,832.51
2,775.00
2,716.63
2,411.35
1,728.20
935.37
15.27
-
$
$
$
$
$
$
$
$
$
$
$
Interest
45.00
44.18
43.34
42.49
41.63
37.11
27.02
15.30
1.70
-
total payment
$ 4,015.27
total interest
$ 1,015.27
effective interest rate 25.29%
$
$
$
$
$
$
$
$
$
$
$
Payment
100.00
100.00
100.00
100.00
100.00
100.00
100.00
100.00
100.00
15.27
-
Simulation 2
Balance
$
yearly interest
monthly payment $
Months
0
1
2
3
4
5
10
20
30
40
50
60
61
62
$
$
$
$
$
$
$
$
$
$
$
$
$
$
4,000
18%
100
Balance
4,000.00
3,960.00
3,919.40
3,878.19
3,836.36
3,793.91
3,571.89
3,075.05
2,489.45
1,829.28
1,052.69
151.41
53.69
-
$
$
$
$
$
$
$
$
$
$
$
$
$
$
Interest
60.00
59.40
58.79
58.17
57.55
54.26
46.92
38.40
28.51
17.03
3.72
2.27
-
total payment $
6,153.69
total interest $
2,153.69
effective interest rate 35.00%
$
$
$
$
$
$
$
$
$
$
$
$
$
$
Payment
100.00
100.00
100.00
100.00
100.00
100.00
100.00
100.00
100.00
100.00
100.00
100.00
53.69
-
)' &
('
4
.
)
&
!
'
&
&
&
#
>#
%
+)
;:::
6=8
<
9:::
678
5
;::
+)
How do we pay off both cards?
3)
b1 , b 2 = the balances on both cards
where : b1 = 3000, b 2 = 5000
r1 , r2 = the interest rates on both cards
where : r1 = .18, r2 = .12
F = the available funds in your budget
where : F = 300
P = interest charged next month
P = r1 ( b1 − x ) + r 2 ( b2 − y )
4
#
Question:
What is the most effective way to pay of the two
credit card balances?
Answer:
Pay the card with the highest interest rate.
!
'
&
&
&
#
>#
%
+)
6
6
<
7
7
5
+
+)
)'
5 )
'
"
!
)
)
"
!
) #
3)
y=F − x
f ( x) = r1 ( b1 − x) + r 2 ( b2 − ( F − x))
f ( x ) = r 1 b1 − r 1 x + r 2 b 2 − r 2 F + r 2 x
f ( x) = ( r 2 − r1 ) x + r1 b1 + r 2 ( b2 − F )
+
) #
f ( x) = ( r 2 − r1 ) x + r1 b1 + r 2 ( b2 − F )
f (x)
x
Do not attempt this in the real world.
Why?
Your credit card company will charge you late
fees in the real world.
Alternative:
Consolidate your credit cards with a home
equity loan or low interest credit card.
Good Debt
Examples of Good Debt
Education
House
Land
Example: We will use is taking a loan out for an
Applied Math degree.
Assumptions
After 108 months ( 5 years after you graduate)
Till retirement at age of 65
Your Math Degree Pays
•And 758,648 ahead of the
associate degree grad.
•As you can see you
come out 911,616 of the
high school grad
•Well worth your 26,000
in loans.
Conclusions
There is a right time to go into debt.
Just think before you act.
Do the Math.
And make you good investments.
)
('' )
&
) )
) )
"
?
&
*
&.
)