Chapter 24
Personal Money, Investment
and Pension Examples
Volume 2, Three Skills for Algebra
Geometric sums appear in calculations involving (i) loan
and mortgage repayments, and (ii) the purchase cost for a pension plan,
here an annuity with a constant monthly payments This chapter indicates
how and why geometric sums appear in money matters.
1. Two Calculations
A Multiple Investment Calculation
Initial Example: John's investments earn 6 percent compounded
monthly. His investment began with the initial amount of 5000 dollars.
Then
- 10 months after his initial investment, he invests another 3000
dollars,
- 2 years after his initial investment, he invests another 2000
dollars,
- 30 months after his initial investment, he invests another 6000
dollars,
- 40 months after his initial investment, he invests another 3500
dollars,
Find the total value of his investments after five years.
Solution: Five years equal sixty months. The monthly interest
rate is i = [6%/12] = 0.5% = 0.005.
- His first 5000 dollars stays invested for five years = sixty months.
It grows to 5000(1+.005)60 dollars.
- His second investment of 3000 dollars is invested for 60-10=50
months. Thus at the end of five years, it has grown to
3000(1+.005)50 dollars.
- His third investment of 2000 dollars is invested for sixty months
minus two years, that is, 60-24 months = 36 months. Thus at the end of
five years, it has grown to 2000(1+.005)36 dollars.
- His fourth investment of 6000 dollars is invested for 60-30=30
months. Thus at the end of five years, it has grown to
6000(1+.005)30 dollars.
- His fifth investment of 3500 dollars is invested for 60-40=20 months.
Thus at the end of five years, it has grown to 3500(1+.005)20
dollars
At the end of five years his investment is worth
S = $5000(1.005)60 + $3000(1.005)50
+ $2000(1.005)36 + $6000(1.005)30
+ $3500(1.0005)20
This gives
S = $5000·(1.34885) + $3000·(1.283226)
+ $2000·(1.19668) + $6000·(1.1614)
+ $3500·(1.1049)
Therefore his investments grows to
S = $6744.25+$3849.68+$2393.36+$6968.40+$3867.13
Addition now implies
S = 23822.82
dollars. This gives the value of his investments after five years.
1.1 A Debt Calculation - Many
Amounts Borrowed
Second Example: Fred can borrow money at 6 percent compounded
monthly. He initially borrows 5000 dollars. Then
- 10 months after his initial loan, he borrows another 3000 dollars,
- 2 years (24 months) after his initial loan, he borrows another 2000
dollars,
- 30 months after his initial loan, he borrows another 6000 dollars,
- 40 months after his initial loan, he borrows another 3500 dollars,
Find the total amount of the debt after five years.
The solution of this second example involves exactly the same numbers and
computations as the solution first example. For instance, the initial
amount of 5000 dollars grows to a debt of
5000(1.005)60 = 6744.25
dollars after five years. Likewise, the second debt of 3000 dollars grows
to a debt of
3000(1.005)60-10 =
3000(1.005)50 = 3849.68
dollars. This second example shows that the same arithmetic or algebra
may give the solution of two different problems - recycled mathematics.
2 Credit and Debt
Accounts (I)
Example: Elizabeth takes a mortgage (loan, debt)
of 50000 dollars. The initial term of the mortgage is 5 years. The
interest rate is 9 percent compounded monthly. At the end of each month,
she repays 800 dollars. The mortgage will be renewed or renegotiated at
the end of the initial 5 year term. Find the outstanding balance, that
is, how much is still owing, after five years.
Solution: Here [9%÷12] = 0.75% = 0.0075 is the monthly interest
rate.
There are several ways to calculate the outstanding
balance. We will use the simplest one, namely the two account method
(otherwise known as the merchant method). That is, we imagine that her
initial debt accumulates in the DEBT account at 9 percent compounded
monthly while her deposits accumulate in the CREDIT account, also at 9
percent compounded monthly. At any time, the difference between the two
accounts gives the total debt or credit. (How or why a single account can
be treated account as the difference of two accounts will be explained in
the next section.)
At five years or sixty months, the initial debt of 50000
dollars in the DEBT account has compounded to the amount
The calculation of the amount in the CREDIT account is
slightly more complicated.
1. The first payment of 800 made one months after
the loan begins, stays in the account for 60-1=59 months. At the end of this stay with compound interest, this credit
grows to 800(1.0075)59dollars.
2. The second payment of 800 made two months after the loan begins, stays in the account for
60-2=58 months. At the end of this stay with compound interest, this
credit grows to 800(1.0075)58 dollars.
3. The third payment of 800 made three months after the loan begins, stays in the account for
60-3=57 months. At the end of this stay with compound interest, this
credit grows to 800(1.0075)57
dollars.
.
.
.
58. The fifty-eight payment of 800 made 58 months after
the loan begins, stays in the account for 60-58=2 months. At the end of
this stay with compound interest, this credit grows to
800(1.0075)2 dollars.
59. The fifty-ninth payment of 800 made 59 months after the loan
begins, stays in the account for 60-59=1 months. At the end of this
stay with compound interest, this credit grows to
800(1.0075)1 dollars.
60. The last payment of 800 made 60 months after the loan begins, stays
in the account for 60-60=0 months. At the start and end of this stay
with compound interest, this credit is 800(1.0075)0 = 800
dollars.
The total amount in the CREDIT account is
given by the geometric sum \begin{eqnarray*} A_{credit}&=&
\$[800(1.0075)^{59}+800(1.0075)^{58}+800(1.0075)^{57}+{} \\
&&\quad\cdots+800(1.0075)^{2}+800(1.0075)^{1}+800(1.0075)^{0}] \\
&=&\$\sum_{j=0}^{59}800(1.0075)^j \end{eqnarray*} The value of this
geometric sum \[ A_{credit}=\$800 \frac{(1.0075)^{60}-1}{1.0075-1} \]
Therefore \[ A_{credit}=\$800\cdot\frac{1.565681027-1}{.0075}=\$60339.31\]
Thus her credit is 60339.31 dollars while her debt is 78284.05 dollars. The
debt account exceeds the credit by 17944.74 dollars. Thus 17944.74 dollars
is the amount still owed at the end of the five years.
Exercises
-
What would her debt-credit be at the end of the five
year term if she made an additional monthly payment of
[17944.75/75.424137] dollars? How much is this additional
payment?
-
Solve the equation
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($800+x)·75.424137 = $78284.05
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How is the solution x
related to that in the previous question?
-
Solve the equation
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p·(75.424137) = $78284.05
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How is the solution p
related to that in the previous questions?
-
Suppose the 12th payment is missed. How
does this affect the calculations? Hint: At the end of the five year
term, the 12th payment can be regarded as a loss of
$800(1.0075)60-12 to the credit
account or an additional debt of the same amount. See the next
section.
3 Credit and Debt Accounts (II)
Suppose you have an account with a
business (bank, firm, insurance company, etc). Assume all your payments
to the business are recorded as credits. Further assume all the payments
from the firm to you are recorded as debts or debits. The business may be
content to continuing dealing with you if the value of credits is greater
than the value of your debts, or if you have an arrangement to pay your
debts in regular fashion. For the calculations that follow, we assume
that a dollar credit grows with compound interest precisely at the same
rate as dollar of debt. Otherwise, the following calculations and the
conclusions drawn from them will be false.
The difference between the accumulating
total credit and the accumulating total debt or debits gives what is owed
by you to or by you to the business in question. Now if you
withdraw a dollar from the firm and then immediately return it, there is
an extra dollar of debt and also an extra dollar of credit recorded. The
difference does not immediately change. Moreover, because the extra
dollar of debt grows at the same rate as the extra dollar of credit, the
future difference is not changed by the simultaneous addition of a dollar
debt and a dollar credit. This implies we can add or subtract the same
amount from credit and debt record, or changing the difference now and in
the future because they both grow or compound at the same interest
rate.1 Only the difference matters.
Because of this when you make a payment
to the business in question you have the option of recording it as an
credit and then subtracting the same amount from both the credit and debt
balance. The latter is equivalent to subtracting the payment from the
outstanding debt and leaving the credit balance unchanged. This implies
choice or flexibility in tracking the amount owed to or by you. Three
options are described.
-
-
Record all the credits and debits, and
how they accumulate and compound, separately. The difference gives your
account balance. This was the option described above. This is the
two-account method used in the previous section.
-
Record the credit and debits together
in a single account. This single account balance will equal the
previous difference (provided the interest rate here is always the same
as that for the credits and debits recorded separately.)
-
Mix the previous two approaches as
convenient in your calculations of the difference: That is, you may use
some payments to the business to either increase your credit balance or
lessen your debt balance; and use payments from the business to reduce
your credit balance or increase in your debt balance.
Keeping track of a single account
means that credits are added to the account balance while debits are
subtracted. In contrast, keeping track of the credits and debits separately
implies that credits are added to credits and debits are added to the
accumulating debits. Thus subtraction is avoided, except when computing the
difference between the accumulating credit and debits or debts. And only
the difference is important. An example was given earlier. An example
involving a pension plan follows next.
4 Pension Plan Example
Example: John is seventy years
old. The A1Z26 Pension Provider Company offers (sells) a 15 year annuity
pension plan consisting of constant monthly payments. The size of the
payment of the payment depends on the cost of the plan and present
interest rate assumptions. John has up to $350000 dollars to buy a plan.
The A1Z26 Pension Provider company announces an interest rate of 4
percent, compounded monthly. What is the maximum monthly payment2 John can
obtain?
First Solution: First imagine that
the $350000 dollars is placed in a credit account with the company, also
at 4 percent, compounded monthly. Then this credit grows to the
amount
|
Acredit = $350000·(1+
|
.04
12
|
)180
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Therefore
|
Acredit = $350000·(1.82030163) =
$637105.57
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This amount could be paid to the
company at the end of the fifteen years.
Second, imagine that John has asked the
A1Z26 Pension Provider Company to let him debit $1.00 from his pension
plan the end of each month for the next 15 years, at the interest rate of
4 percent compounded monthly. We will calculate the amount of debt that
the payments represent at the end of fifteen years, and then see what
amount of credit is then required to cancel this debt.
1. At the end of the first month, the
first payment of \$1.00 to John represents a debt which grows at 4
percent, compounded monthly for 180-1 =179 months. This contributes a
debt of 1.00(1+[.04/12])179 = (1+[.04/12])179
dollars.
2. At the end of the second month, the second payment of $1.00 to John
represents a debt which grows at 4 percent, compounded monthly for
180-2 =178 months. This contributes a debt of
1.00(1+[.04/12])178 = (1+[.04/12])178
dollars.
3. At the end of the third month, the third payment of \$1.00 to John
represents a debt which grows at 4 percent, compounded monthly for
180-3 =177 months. This contributes a debt of
1.00(1+[.04/12])177 = (1+[.04/12])177
dollars.
.
.
.
178. At the end of the 178th month,
the 178th payment of \$1.00 to John represents a debt which grows at 4
percent, compounded monthly for 180-178 =2 months. This contributes a
debt of $1.00(1+[.04/12])2 = (1+[.04/12])2
dollars.
179. At the end of the 179th month, the 179 payment of \$1.00 to John
represents a debt which grows at 4 percent, compounded monthly for
180-179 =1 month. This contributes a debt of
\$1.00(1+[.04/12])1 = (1+[.04/12])1 dollars.
180. At the end of the 180th month, the last payment of $1.00 to John or
his estate represents a debt which grows at 4 percent, compounded monthly
for 180-180 =0 months. This contributes a debt of
1.00(1+[.04/12])0 = 1 dollar.
The total amount of debt at the end of fifteen years is
given by the geometric sum
\begin{eqnarray*} A_{debt} = &\quad\$\left[(1+\frac{.04}{12})^{179}+
(1+\frac{.04}{12})^{178}+ (1+\frac{.04}{12})^{177}+\right. \\
&\quad+\cdots+ \left.(1+\frac{.04}{12})^{2}+ (1+\frac{.04}{12})^{1} +
(1+\frac{.04}{12})^{0}\right] \\ =&\$\sum_{j=0}^{179}
(1+\frac{.04}{12})^j \end{eqnarray*}
The value of this geometric sum is \[ A_{debt}= \$
\frac{(1+\frac{.04}{12})^{180}-1}{1+\frac{.04}{12}-1} \]
Therefore \[
A_{debt}=\$\frac{1.820301518-1}{\frac{.04}{12}}=\$246.09\]
Thus 180 end of month payments of \$1.00 from the A1Z26
Pension Provider Company will result in a debt of 246.09 dollars to the
A1Z26 Pension Provider Company at the end of the fifteen years.
Now at the end of fifteen years, a deposit of 3,500,000
dollars today will leave John with a credit of $637105.57$ dollars.
Observe $637,105.57\div246.09=2588.91$. Therefore if John receives
2588.91 dollars at the end of each month for the next 15 years = 180
months, he will need to give 637,105.57 = 2588.91×246.09 dollars to the
A1Z26 Pension Provider Company at the end of the fifteen years. But that
is precisely the accumulated amount in his credit account.
This suggests that the A1Z26 Pension Plan Company would
be willing to sell John fifteen years of monthly payments of 2588.91
dollars (approximately) in exchange for the amount of 350000 dollars at
their announced 4% cent interest rate.
Second Solution: Suppose John want to receives a
payment of 1.00 dollar at the end of each month for the next fifteen
years = 180 months. Recall that a deposit of $P=\frac{A}{(1+i)^n}$ will
grow the amount $A$ at the end of $n$ periods in a compound interest
account. Suppose John want to receives a payment of 1.00 dollar at the
end of each month for the next fifteen years = 180 months from a compound
interest account, paying 4 percent compounded monthly. What does he need
today in the account today to be able to make all the withdrawals?
-
1. To withdraw 1.00 dollar from the account one
month after his initial deposit, he needs
$\frac{\$1.00}{(1+\frac{.04}{12})^{1}}$ in the initial
deposit.
-
2. To withdraw 1.00 dollar from the account two
months after his initial deposit, he needs
$\frac{\$1.00}{(1+\frac{.04}{12})^{2}}$ in the initial
deposit.
-
3. To withdraw 1.00 dollar from the account three
months after his initial deposit, he needs
$\frac{\$1.00}{(1+\frac{.04}{12})^{3}}$ in the initial
deposit.
-
\[ \vdots\]
-
177. To withdraw 1.00 dollar from the account 178
months after his initial deposit, he needs
$\frac{\$1.00}{(1+\frac{.04}{12})^{178}}$ in the initial
deposit.
-
179. To withdraw 1.00 dollar from the account 179
months after his initial deposit, he needs
$\frac{\$1.00}{(1+\frac{.04}{12})^{179}}$ in the initial
deposit.
-
180. To withdraw 1.00 dollar from the account 180
months after his initial deposit, he needs
$\frac{\$1.00}{(1+\frac{.04}{12})^{180}}$ in the initial
deposit.
To make all the withdrawals the total amount in the
initial deposit has to be
\begin{eqnarray*}
A_{deposit}=&
\frac{\$1.00}
{ (1+\frac{.04}{12})^{1}}
+ \frac{\$1.00}
{(1+\frac{.04}{12})^{2}}
+ \frac{\$1.00}
{(1+\frac{.04}{12})^{3}}+\cdots \\
&\quad+\frac{\$1.00}{(1+\frac{.04}{12})^{178}}+\frac{\$1.00}
{(1+\frac{.04}{12})^{179}}+\frac{\$1.00}{(1+\frac{.04}{12})^{180}}
\end{eqnarray*} Therefore \begin{eqnarray*}
A_{deposit}&=\frac{\$1.00}
{(1+\frac{.04}{12})^{180}}\left[(1+\frac{.04}{12})^{179}
+{(1+\frac{.04}{12})^{178}}\right. \\
&\left.\quad+{(1+\frac{.04}{12})^{177}}+\cdots+(1+\frac{.04}{12})^{2}+
{(1+\frac{.04}{12})^{1}}+1\right) \end{eqnarray*} From the value of the
geometric sum in the last expression, we conclude \[ A_{deposit}=
\frac{\$1.00}
{(1+\frac{.04}{12})^{180}}\frac{(1+\frac{.04}{12})^{180}-1}{1+\frac{.04}{12}-1}
\] Therefore \[ A_{deposit}=\frac{\$1.00}
{1.820301518}\cdot\frac{1.820301518-1}{\frac{.04}{12}}=\$135.1921388\]
Therefore, for each dollar John receives at the end
of each month for the next 15 years = 180 months, he will need to place an
initial deposit of 135.1921388 dollars with the A1Z26 Pension Provider
Company. Now 350000¸138.1921388 = 2588.91.
Therefore for John to receive 2588.94 dollars at the end of each month for
the next 15 years = 180 months, he will need to make an initial deposit of
350000 = 2588.91×135.19 dollars to the A1Z26 Pension Provider Company. This
assumes the 4% yearly interest rate, compounded monthly.
Note in computing, rounding should be done last and
avoided in intermediate calculations for the sake of preserving accuracy.
(The above calculation would be slightly less accurate if the
intermediate result 135.192188 was rounded to 135.19 dollars.
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Calculus Starter Lessons
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