Download 1. Solving Exponential Equations Method for solving equations of

CHAPTER 6
Mathematics of Finance
1. Solving Exponential Equations
Method for solving equations of the form y = abx.
y = abx
(1) Divide both sides by a.
y
= bx
a
(2) Take the log or ln of both sides.
y
y
log = log bx or ln = ln bx
a
a
(3) Bring the x down in front of log or ln.
y
y
log = x log b or ln = x ln b
a
a
(4) Divide both sides by log b or ln b, respectively.
log ay
ln ay
x=
or x =
log b
ln b
Problem (Page 389 #4).
5(0.5)x = 0.125
0.5x = 0.025
ln 0.5x = ln 0.025
x ln 0.5 = ln 0.025
ln 0.025
x=
⇡ 5.32
ln 0.5
119
120
6. MATHEMATICS OF FINANCE
Problem (Page 389 #14).
t = # of years since the end of 1980.
M = annual per capita milk beverage consumption in gallons.
M (t) = 27.76(0.9914)t
When is M (t) = 20? Solve
20 = 27.76(0.9914)t
20
= 0.9914t
27.76
20
ln
= ln 0.9914t
27.76
ln 20 ln 27.76 = t ln 0.9914
ln 20 ln 27.76
t=
⇡ 37.96
ln 0.9914
The average person will drink 20 gallons of milk in 2018.
1. SOLVING EXPONENTIAL EQUATIONS
x
x
Method for solving equations of the form ab = cd .
abx = cdx
(1) Divide both sides by a.
cdx
b =
a
x
(2) Divide both sides by dx.
bx
c
=
dx a
⇣ b ⌘x
bx
(3) Rewrite x as
.
d
d
⇣ b ⌘x
d
(4) Solve using previous rule.
ln
⇣ b ⌘x
d
⇣b⌘
=
c
a
= ln
⇣c⌘
a
⇣c⌘
x ln
= ln
d
a
c
ln a
x=
ln db
121
122
6. MATHEMATICS OF FINANCE
Example.
4 · 5x = 7 · 2x
7 · 2x
x
5 =
4
x
5
7
=
2x 4
⇣ 5 ⌘x 7
=
2
4
⇣ 5 ⌘x
⇣7⌘
ln
= ln
2
4
⇣5⌘
⇣7⌘
x ln
= ln
2
4
ln 74
ln 7 ln 4
x=
=
⇡ .61
5
ln
5
ln
2
ln 2
1. SOLVING EXPONENTIAL EQUATIONS
123
Problem (Page 389 #16).
t = # of years since the end of 1980.
M (t) = per capita gallons of milk consumed in year t
M (t) = 27.76(0.9914)t
W (t) = per capita gallons of bottled water consumed in year t
W (t) = 2.593(1.106)t
When is M (t) = W (t)?
27.76(0.9914)t = 2.593(1.106)t
0.9914t 2.593
=
1.106t
27.76
⇣ 0.9914 ⌘t 2.593
=
1.106
27.76
⇣ 0.9914 ⌘t
2.593
ln
= ln
1.106
27.76
⇣ 0.9914 ⌘
2.593
t ln
= ln
1.106
27.76
2.593
ln 27.76
t = 0.9914
⇡ 21.67
ln 1.106
Annual per capita consumption of water will surpass that of milk about twothirds of the way through 2002.
124
6. MATHEMATICS OF FINANCE
Problem (Page 391 #28).
t = time in years.
1st account: y = 500(1 + .092)t or y = 500(1.092)t.
2nd account: y = 1000(1 + .053)t or y = 1000(1.053)t.
On TI:
Y1 = 500 ⇤ 1.092 ^ X
Y2 = 1000 ⇤ 1.053 ^ X
Set a window [0, 25] ⇥ [0, 3500].
Use 2nd/CALC/5:intersect to find that the graphs meet at the point
(19.06, 2675.92).
Thus it takes a little over 19 years for the two accounts to have the same amount
of money, $2675.92.
Problem (Page 391 #32).
Does
3x =
x2 + 4x
2
have a solution.
On TI, graph:
Y1 = 3 ^ X
Y2 = X ^ 2 + 4 ⇤ X
2
We see the graphs never meet, so there is no solution.
2. SIMPLE AND COMPOUND INTEREST
125
Problem (Page 391 #34).
5(2x+1) = 22x
1
22x 1
5 = x+1
2
5 = 2(2x 1) (x+1)
5 = 2x
2
ln 5 = ln 2x 2
ln 5 = (x 2) ln 2
ln 5
=x 2
ln 2
ln 5
x=2+
⇡ 4.32
ln 2
2. Simple and Compound Interest
What is interest?
It is a fee paid for using another’s money. It is a dollar amount.
Simple Interest — the basis for all interest.
I = P rt
I = interest (usually in dollars).
P = principal or present value of an investment.
r = interest rate (usually annual interst rate - we assume this unless told
otherwise).
t = time (usually in years).
Note. r and t must agree in time units, e.g., dollars per year and years,
dollars per month and months, etc.
126
6. MATHEMATICS OF FINANCE
Related formulas:
P =
I
,
rt
r=
I
,
Pt
t=
I
Pr
Simple Interest:
The future value A of an initial investment P earning a simple interest rate r
(or 100r%) is given by
A = P + P rt = P (1 + rt)
where t is the number of years after the initial investment is made. The rate r
is the decimal form of the percentage rate (e.g., .072 for 7.2%).
Time line:
P
A
0
t years
Problem (Page 402 #2).
r = 4.25% = .0425, t = 5, P = 1000
A = P (1 + rt) = 1000(1 + .0425(5)) = 1212.50
I = A P = 1212.50 1000 = 212.50
The CD will be worth $1212.50 at maturity with $212.50 coming from interest.
Problem (Page 402 #4).
Let A = 2P and t = 7.
A = P (1 + rt)
2P = P (1 + 7r)
2 = 1 + 7r
1 = 7r
1
r = ⇡ .1429 = 14.29%
7
The investor will need a 14.29% simple interest rate.
2. SIMPLE AND COMPOUND INTEREST
127
Simple interest investments earn interest only on the initial amount invested.
A=
P
+ |{z}
Pr t
|{z}
A-intercept
slope
Simple interest investments grow linearly with time.
Compound interest investments earn interest on the initial amount invested and
any previously earned interest.
Example. Suppose $100 is invested at 4% (annual rate) compounded monthly.
We use
A = P (1 + rt)
for one month.
100
0
Use
A
1 month
1
.04
r = .04 and t =
or r =
and t = 1 =)
12
12
|
{z
}
|
{z
}
yearly rate
monthly rate
⇣
.04 ⌘
Month 1: A = 100 1 +
12
h ⇣
⇣
.04 ⌘i⇣
.04 ⌘
.04 ⌘2
Month 2: A = 100 1 +
1+
= 100 1 +
12
12
12
h ⇣
⌘
i⇣
⌘
⇣
.04 2
.04
.04 ⌘3
Month 3: A = 100 1 +
1+
= 100 1 +
12
12
12
...
...
128
6. MATHEMATICS OF FINANCE
Compound Interest:
The future value A of an initial investment P earning a compound interest rate
100r percent is given by
⇣
r ⌘nt
A=P 1+
n
where n is the number of times the interest is paid annually and t is the number
of years after the initial investment is made. Again, r is the decimal form of
the percentage rate.
A
P
0
1
2
nt periods
Note that
I = A P.
The compounding frequency is the number of times the interest is paid each
year.
Interest is Compounded Compounding Frequency n
Annually
1
Semiannually
2
Quarterly
4
Monthly
12
Daily
365
r
Compound interest investments grow exponentially. b = 1 + is the periodic
n
r
growth factor and is the periodic growth rate.
n
Problem (Page 402 #6).
t = 5, r = 4.69% = .0469, n = 12, P = 1000
⇣
⇣
r ⌘nt
.0469 ⌘12(5)
A=P 1+
= 1000 1 +
= $1263.70
n
12
I = A P = 1263.70 1000 = $263.70
2. SIMPLE AND COMPOUND INTEREST
129
Annual Percentage Yield (APY)
The annual percentage yield is the percentage rate at which interest would need
to be paid if interest were calculated and paid only once a year.
APY is used to compare investments.
Theorem. The annual percentage yield, APY, for an account earning
100r percent compounded n times per year is
⇣
r ⌘n
APY = 1 +
1.
n
For a simple interest account,
APY = r.
Note.
⇣
h⇣
r ⌘nt
r ⌘nit
A=P 1+
=P 1+
= P bt
n
n
if r and n are known. Recall that b is the annual growth factor, and
AP Y = b
1
is the annual growth rate of the investment.
Problem (Page 403 #12).
r = 1.25% = .0125, n = 365
⇣
.0125 ⌘365
APY = 1 +
1 ⇡ .01258 = 1.258%
365
130
6. MATHEMATICS OF FINANCE
Continuous Compound Interest
What happens as we make more and more compounding periods in a year?
Suppose r = .05.
Compounding
Number of times
⇣
r ⌘n
Frequency
compounded yearly (n) APY = 1 +
1
n
Annually
1
0.05000
Quarterly
4
0.05095
Monthly
12
0.05116
Daily
365
0.05127
Hourly
8760
0.05127
Every Minute
525,600
0.05127
Every Second
31,536,000
0.05127
A limit of sorts has been reached. Recall
⇣
1 ⌘n
1+
! e as n ! 1.
n
Similarly,
⇣
r ⌘n
1+
! er as n ! 1.
n
For instance,
⇣
⌘1,000,000
.05
1+
= 1.051271095
1, 000, 000
and
e.05 = 1.051271096.
2. SIMPLE AND COMPOUND INTEREST
131
Continuous Compounding
For infinitely large n, we use continuous compounding.
Theorem. The future value A of an initial investment P earning a
continuous compound interest rate 100r percent is given by
A = P ert
where t is the number of years after the initial investment is made. Then
APY = er
1.
Problem (Page 403 #22).
We have 2.98% compounded continuously.
r = APY = e0.0298
Problem (Page 403 #24).
1 ⇡ .0302 = 3.02%
We have a 4.75% annual rate.
er
1 = .0475 (AP Y )
er = 1.0475
ln er = ln 1.0475
r ln e = ln 1.0475
r = ln 1.0475 (since ln e = 1)
r ⇡ .0464 = 4.64% continuous rate.
132
6. MATHEMATICS OF FINANCE
Problem (Page 403 #18).
Account 1: P = 2000,
r = .04,
continuous
A = P ert = 2000e.04t
Account 2: P = 3000,
r = .05, n = 12, compounded monthly
⇣
⇣
r ⌘nt
.05 ⌘12t
A=P 1+
= 3000 1 +
n
12
Equate the right hand sides:
⇣ ⌘t
h⇣
.05 ⌘12it
.04
2000 e
= 3000 1 +
12
t
t
2000( 1.04081
| {z } ) = 3000( 1.05116
| {z } )
APY=4.081%
APY=5.116%
Account on the left (Account 1) starts lower and has a lower APY than the
account on the right (Account 2), so the accounts will never be equal in the
future. We continue to see how the algebra plays out.
1.04081t 3000
=
1.05116t 2000
⇣ 1.04081 ⌘t 3
=
1.05116
2
⇣ 1.04081 ⌘t
3
ln
= ln
1.05116
2
⇣ 1.04081 ⌘
3
t ln
= ln
1.05116
2
3
ln
t = ⇣ 2 ⌘ = 40.98
ln 1.04081
1.05116
This shows that the accounts could only have been equal in the past.
2. SIMPLE AND COMPOUND INTEREST
Problem (Page 404 #30).
The lender receives $593.10 after 14 days.
14
P = 500, A = 593.10, t =
365
⇣
⌘
h⇣
⌘
i
r nt
r n t
A=P 1+
=P 1+
= P bt
n
n
14/365
593.10 = 500b
1.1862 = b14/365
365/14
1.1862365/14 = b14/365
=b
b = 85.78
AP Y = b 1 = 84.78 = 8478%
Problem (Page 404 #34).
P = 81.9,
(a)
A = 234.7, t = 20
⇣
r ⌘nt
A=P 1+
n
234.7 = 81.9(1 + r)20
234.7
= (1 + r)20
81.9
⇣ 234.7 ⌘1/20
1+r =
81.9
⇣ 234.7 ⌘1/20
AP Y = r =
1 = .05405 = 5.405%
81.9
Tomato prices were increasing by 5.405% per year.
(b)
A = P (1 + r)t = .86(1.05405)20 = 2.46
Tomatoes would be $2.46 per pound in 2010.
133
134
6. MATHEMATICS OF FINANCE
Problem (Page 405 #42).
P = 25,
A = 125.62,
n = 1 t = 23 +
1
277
=
12
12
125.62 = 25(1 + r)277/12
125.62
= (1 + r)277/12
25
⇣ 125.62 ⌘12/277
1+r =
= 1.07244
25
AP Y = r = .07244 = 7.244%
3. Future Value of an Increasing Annuity
An annuity is a financial instrument that requires a series of payments of set
size and frequency.
For an ordinary annuity, payments are made at the end of each time period.
For an annuity due, payments are made at the beginning of each time period.
These have slightly di↵erent formulas, but we will not study these in this course.
An increasing annuity is one where the balance increases with time.
3. FUTURE VALUE OF AN INCREASING ANNUITY
135
Example. If you deposit $100 per quarter at 8% compounded quarterly for
one year, how much money is in the account at the end of the year?
For instance, how much money does payment 2 result in at the end of the year?
100
0
1
2
A
3
4 quarters
We note that this payment receives two quarters of interest.
⇣
⇣
r ⌘nt
.08 ⌘4( 12 )
A=P 1+
= 100 1 +
= 100(1 + .02)2
n
4
Thus A = 100(1.02)2
Similarly,
0
1
2
3
4 quarters
100
100
100
100
100(1.02)
100(1.02)23
100(1.02)
The amount we are looking for is the sum of the numbers in the right hand
column. But we are going to add them a special way that helps us to generate
a formula.
S = 100 + 100(1.02) + 100(1.02)2 + 100(1.02)3
1.02S = 100(1.02) + 100(1.02)2 + 100(1.02)3 + 100(1.02)4
Subtract the top equation from the bottom one. Notice the cancelling.
1.02S
S = 100(1.02)4
.02S = 100(1.024
100
1)
1.024 1
S = 100
= 412.16
.02
Based on this work, we make the following definitions.
136
6. MATHEMATICS OF FINANCE
Definition. Future Value of an Increasing Annuity.
(1) With a Zero Present Value.
The future value F V of an increasing annuity with an initial balance of 0 dollars
is given by
(1 + i)m 1
F V = P MT
i
r
where i = is the periodic interest rate, m = nt is the number of payments,
n
and P M T is the payment amount.
(2) With a Nonzero Present Value.
The future value F V of an increasing annuity with a present value P V is given
by
(1 + i)m 1
m
F V = P V (1 + i) + P M T
.
i
Example. We want $1,000,000 after 40 years of monthly contributions at
12% compounded monthly. What is the monthly payment?
.12
F V = 1, 000, 000, i =
= .01, m = 12(40) = 480
12⇥
⇤
1, 000, 000(.01)
F V (i)
⇥
⇤ = $85.00
P MT =
=
(1 + i)m 1
(1 + .01)480 1
Problem (Page 415 #12).
.0811
i=
, P V = 3200,
12
m = 12(25) = 300,
P M T = 125
(1 + i)m 1
F V = P V (1 + i) + P M T
i i
h
300
.0811
⇣
⌘
1
+
1
300
12
.0811
F V = 3200 1 +
+ 125
= $145, 164.60
.0811
12
12
m
Note. Using i = .00676 gives an answer of $145,216.04.
3. FUTURE VALUE OF AN INCREASING ANNUITY
137
Problem (Page 415 #10 adjusted).
Suppose also that the account has a $5000 present value.
.08
P V = 5000, i =
⇡ .00676, F V = 250, 000, P M T = 300
12
(1 + i)m 1
m
F V = P V (1 + i) + P M T
i
⇥
⇤
m
(1
+
.0067)
1
250, 000 = 5000(1 + .0067)m + 300
.0067
m
250, 000 = 5000(1.0067) + 44776.1194 (1.0067m 1)
250, 000 = 5000(1.0067)m + 44776.1194 (1.0067m) 44776.1194
294776.1194 = 5000(1.0067)m + 44776.1194 (1.0067m)
294776.1194 = (5000 + 44776.1194)(1.0067)m
294776.1194 = 49776.1194(1.0067)m
5.922 = 1.0067m
ln 5.922 = ln 1.0067m
ln 5.922 = m ln 1.0067
ln 5.922
m=
= 266.36 or 267 months
ln 1.0067
Note. Without rounding o↵ i, we get m = 283 months.
A sinking fund is an account established for the purpose of accumulating money
to pay o↵ future debts or obligations. It is always an increasing annuity.
138
6. MATHEMATICS OF FINANCE
The time value of money (TVM) refers to the fact that a given amount now
is worth more than the same amount in the future since money now can gain
interest in the interim.
Problem (Page 416 #22).
We use the TVM Solver as described on page 411. Note that for increasing
annuities, P V and P M T are entered as negatives.
N = number of payments = 21 ⇤ 12
P V = negative of P V =
10, 185.22
P M T = negative of P M T =
650
F V = future value = 2, 900, 000
P/Y = payments per year = 12
C/Y = compound periods per year = 12
P M T = beginning or end of period for interest calculation = EN D
I% = annual interes rate
I = 19.55%
Problem (Page 417 #28).
Find N, I% = 4.69, P V = 24020, P M T = 1000,
F V = .20 ⇤ 525000 20000, P/Y = 12, C/Y = 365, EN D
It will take 51 months (round up).
4. Present Value of an Decreasing Annuity
Amortization is the process of paying o↵ a loan, usually with equal periodic
payments.
The present value of an annuity is the amount of money that would need to
be invested today in order to provide the desired numbers of equal periodic
payments.
4. PRESENT VALUE OF AN DECREASING ANNUITY
139
Example. In order to withdraw $100 for 4 quarters at 8% compounded
quarterly, how much money must one start with?
For instance, how much money is needed at the beginning for payment 3 — a
compound interest problem?
P
100
1
0
2
3
4 quarters
⇣
⇣
r ⌘nt
.08 ⌘4( 34 )
A=P 1+
=) 100 = P 1 +
n
4
100
3
P =
=)
P
=
100(1.02)
(1 + .02)3
=)
Similarly,
0
100(1.02)
100(1.02)-2
100(1.02)-3
100(1.02)-4
-1
1
2
3
4 quarters
100
100
100
100
The amount we are looking for is the sum of the numbers in the left hand
column. But, again, we are going to add them a special way thats helps us to
generate a formula.
P = 100(1.02)
1
+ 100(1.02)
1.02P = 100 + 100(1.02)
1
2
+ 100(1.02)
+ 100(1.02)
2
3
+ 100(1.02)
+ 100(1.02)
4
3
Subtract the top equation from the bottom one. Notice the cancelling.
1.02P
P = 100
.02P = 100(1
P = 100
1
1.02
.02
100(1.02)
1.02 4)
4
= 380.77
4
140
6. MATHEMATICS OF FINANCE
Based on this work, we make the following definitions.
Definition. Present Value of a Decreasing Annuity.
(1) With a Future Value of Zero.
The present value P V of an decreasing annuity with a future value of 0 dollars
is given by
P V = P MT
1
(1 + i)
i
m
r
where i = is the periodic interest rate, m = nt is the number of payments,
n
and P M T is the payment amount.
(2) With a Nonzero Future Value.
The present value P V of an decreasing annuity with a future value F V is given
by
P V = F V (1 + i)
m
+ P MT
(1 + i)
i
1
m
(3) The payment P M T required to amortize (pay o↵) a debt of P V dollars is
given by
P MT =
i(P V )
1 (1 + i)
m
Problem (Page 433 #6).
P V = 49330
.0699
18500 = 30830, i =
= .005825,
12
⇥
⇤
.005825(30830)
⇤ = 738.12
P MT = ⇥
48
1 (1.005825)
See the associated amortization schedule:
m = 48
4. PRESENT VALUE OF AN DECREASING ANNUITY
141
Home Amortization at 9.6%
48-Month Auto Loan of $30830
at 6.99% Compounded Monthly
Payment
Number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
Principal
(Present
Value)
$30,830.00
$30,271.46
$29,709.67
$29,144.61
$28,576.26
$28,004.60
$27,429.61
$26,851.27
$26,269.56
$25,684.46
$25,095.95
$24,504.01
$23,908.63
$23,309.78
$22,707.44
$22,101.59
$21,492.21
$20,879.28
$20,262.78
$19,642.69
$19,018.99
$18,391.66
$17,760.67
$17,126.01
$16,487.65
$15,845.57
$15,199.75
$14,550.17
$13,896.80
$13,239.63
$12,578.63
$11,913.78
$11,245.06
$10,572.44
$9,895.90
$9,215.42
$8,530.98
$7,842.55
$7,150.11
$6,453.64
$5,753.11
$5,048.50
$4,339.79
$3,626.95
$2,909.96
$2,188.79
$1,463.42
$733.82
Interest
(0.005825 x
Principal)
Payment
$179.58
$176.33
$173.06
$169.77
$166.46
$163.13
$159.78
$156.41
$153.02
$149.61
$146.18
$142.74
$139.27
$135.78
$132.27
$128.74
$125.19
$121.62
$118.03
$114.42
$110.79
$107.13
$103.46
$99.76
$96.04
$92.30
$88.54
$84.75
$80.95
$77.12
$73.27
$69.40
$65.50
$61.58
$57.64
$53.68
$49.69
$45.68
$41.65
$37.59
$33.51
$29.41
$25.28
$21.13
$16.95
$12.75
$8.52
$4.27
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.12
$738.09
$4,599.73
$35,429.73
New Balance
(Principal +
Interest Payment)
$30,271.46
$29,709.67
$29,144.61
$28,576.26
$28,004.60
$27,429.61
$26,851.27
$26,269.56
$25,684.46
$25,095.95
$24,504.01
$23,908.63
$23,309.78
$22,707.44
$22,101.59
$21,492.21
$20,879.28
$20,262.78
$19,642.69
$19,018.99
$18,391.66
$17,760.67
$17,126.01
$16,487.65
$15,845.57
$15,199.75
$14,550.17
$13,896.80
$13,239.63
$12,578.63
$11,913.78
$11,245.06
$10,572.44
$9,895.90
$9,215.42
$8,530.98
$7,842.55
$7,150.11
$6,453.64
$5,753.11
$5,048.50
$4,339.79
$3,626.95
$2,909.96
$2,188.79
$1,463.42
$733.82
$0.00
142
6. MATHEMATICS OF FINANCE
Problem (Page 433 #14).
.0685
i=
, P V = .80(220, 000), m = 360
12
h
i
.0685
12 (176, 000)
i = 1153.26
P MT = h
360
.0685
1
1 + 12
Problem (Page 435 #30).
.06
= .005, P M T = 4000
12
1 (1 + i) m
m
P V = F V (1 + i) + P M T
i
h 1 (1.005) m i
300000 = 60000(1.005) m + 4000
.005
⇥
⇤
m
300000 = 60000(1.005) + 800000 1 (1.005) m
P V = 300, 000,
F V = 60, 000,
300000 = 60000(1.005)
m
i=
+ 800000
500000 = 60000(1.005)
m
800000(1.005)
800000(1.005)
500000 = 740000(1.005) m
50
= 1.005 m
74
⇣ 50 ⌘
ln
= ln 1.005 m
74
⇣ 50 ⌘
ln
= m ln 1.005
74
ln 50
74
m=
= 78.6 or 79 payments
ln 1.005
m
m
4. PRESENT VALUE OF AN DECREASING ANNUITY
143
Estimate of the Future Balance of a Credit Card.
The balance on a credit card, Bn, after n minimum payments have been made
may be estimated by
Bn = B(1 + i r)n
where B is the initial balance, i is the monthly periodic interest rate, and r is
the percentage of the balance that is the minimum amount required to be paid.
Problem (Page 435 #32).
.219
i=
, r = .0225, n = 24
12
⇣
⌘24
.219
B24 = 1890.25 1 +
.0225
= 1706.58
12
What about after 10 years?
⇣
⌘120
.219
B120 = 1890.25 1 +
.0225
= 1133.85
12
An underestimate of how much we have paid:
B = 1890.25,
monthly payment = .0225(1133.85) = 25.51
120 monthly payments = 120(25.51) = 3061.20
Problem (Page 435 #36).
There are two parts to this problem:
saving
age
year
21
0
spending
65
44
(1) Find out how much money you need for the withdrawals.
(2) Find the monthly payment to get that amount.
90
25
144
6. MATHEMATICS OF FINANCE
(1) Decreasing annuity:
P M T = 3500,
P V = P MT
1
(1 + i)
i
m
.08
, m = 25(12) = 300
12h
i
300
.08
1
1 + 12
= 3500
= $453, 475.83
.08
i=
12
(2) Increasing annuity:
.08
, m = 44(12) = 528
12
(1 + i)m 1
F V = P MT
i
h
i
.08 528
1 + 12
1
453, 475.83 = P M T
.08
F V = 453, 475.83,
i=
12
453, 475.83 = P M T 4858.81
P M T = $93.34
Note.
I = 300(3500)
528(93.94) = 1, 050, 000
49, 283.52 = $1, 000, 716.48
4. PRESENT VALUE OF AN DECREASING ANNUITY
What about?
age
year
letting it sit
saving
30
9
21
0
145
spending
65
35
90
25
(1) Same as before.
(2) Money just gains compound interest:
.08
A = 453, 475.83, i =
, m = 35(12) = 420
12
A = P (1 + i)m
⇣
.08 ⌘420
453, 475.83 = P 1 +
12
453, 475.83 = P (16.29)
P = $27, 837.69
(3) Increasing annuity:
.08
, m = 9(12) = 108
12
(1 + i)m 1
F V = P MT
i
h
i
.08 108
1 + 12
1
27, 837.69 = P M T
.08
F V = 27, 837.69,
i=
12
27, 837.69 = P M T 157.42
P M T = $176.84
Note.
I = 1, 050, 000
108(176.84) = 1, 050, 000
19, 098.72 = $1, 030, 901.28