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Savings Plan Formula. Exercise 16, p. 233

Find the savings plan balance after 5 years with an APR of 2.5% and monthly payments of $ 100

Solution. Use Savings Plan Formula \( A = PMT \times \frac{ \left[ \left( 1 + \frac{APR}{n} \right)^{(nY)} -1 \right] }{\left( \frac{APR}{n} \right)} \) with \(APR=0.025,\ \ \ PMT=100, \ \ \ n=12, \ \ \ Y=5 \): \( A = 100 \times \frac{ \left[ \left( 1 + \frac{0.025}{12} \right)^{(12 \times 5)} -1 \right] }{\left( \frac{0.025}{12} \right)} = \$ 6,384.05 \)

Remark. The pure investment is \( 5 \times 12 \times 100 = \$ 6,000 \), and one expects some extra from the interest.

Savings Plan Formula. Exercise 18, p. 233

Find the savings plan balance after 24 months with an APR of 5% and monthly payments of $ 250

Solution. Use Savings Plan Formula \( A = PMT \times \frac{ \left[ \left( 1 + \frac{APR}{n} \right)^{(nY)} -1 \right] }{\left( \frac{APR}{n} \right)} \) with \(APR=0.05,\ \ \ PMT=250, \ \ \ n=12, \ \ \ Y=2 \): \( A = 250 \times \frac{ \left[ \left( 1 + \frac{0.05}{12} \right)^{(12 \times 2)} -1 \right] }{\left( \frac{0.05}{12} \right)} = \$ 6296.48 \)

Remark. The pure investment is \( 24 \times 250 = \$ 6,000 \), and one expects some extra from the interest.

Exercise 20, p. 233

A friend has an IRA with an APR of 6.25%. She started the IRA at age 25 and deposits $50 per month. How much will her IRA contain when she retires at age 65? Compare that amount to the total deposits made over the time period.

Solution. Use Savings Plan Formula \( A = PMT \times \frac{ \left[ \left( 1 + \frac{APR}{n} \right)^{(nY)} -1 \right] }{\left( \frac{APR}{n} \right)} \) with \(APR=0.0625,\ \ \ PMT=50, \ \ \ n=12, \ \ \ Y=65-25=40 \): \( A = 50 \times \frac{ \left[ \left( 1 + \frac{0.0625}{12} \right)^{(12 \times 40)} -1 \right] }{\left( \frac{0.625}{12} \right)} = \$ 106,595.63 \)

Exercise 20, p. 233 ... Continued

Calculate the total deposits made over the time period.

She deposits $50 per month over a period of 40 years

\( \text{total deposit} = 50 \times 12 \times 40 = \$ 24,000 \)

Comparison. While the total deposits over the period of time is $24,000, the accumulated balance is $106,595.63.

This time the interest makes a big difference. That is both because the higher interest rate and the much lengthier period of time.

Exercise 22, p. 233

You put $200 per month in an investment plan that pays an APR of 4.5%. How much money will you have in 18 years?

Solution. Use Savings Plan Formula \( A = PMT \times \frac{ \left[ \left( 1 + \frac{APR}{n} \right)^{(nY)} -1 \right] }{\left( \frac{APR}{n} \right)} \) with \(APR=0.045,\ \ \ PMT=200, \ \ \ n=12, \ \ \ Y=18 \): \( A = 200 \times \frac{ \left[ \left( 1 + \frac{0.045}{12} \right)^{(12 \times 18)} -1 \right] }{\left( \frac{0.045}{12} \right)} = \$ 66,373.60 \)

Exercise 22, p. 233 ... Continued

Compare that amount to the total deposits made over that time period.

Calculate the total deposits made over the time period.

You deposit $200 per month over a period of 18 years

\( \text{total deposit} = 200 \times 12 \times 18 = \$ 43,200 \)

Comparison. While the total deposits over the period of time is $43,200, the accumulated balance is $66,373.60.

Exercise 24, p. 233

At age 35, you start saving plan for retirement. If your investment plan pays an APR of 6%, and you want to have $2 million when you retire in 30 years, how much should you deposit monthly?

Exercise 24, p. 233 ... continued

Solution.
Use Savings Plan Formula solved for payments
\(
PMT= A \times \frac {\left( \frac{APR}{n} \right)} { \left[ \left( 1 + \frac{APR}{n} \right)^{(nY)} -1 \right] }
\)

with \(A= \$2,000,000, \ \ \ APR=0.06, \ \ \ n=12, \ \ \ Y=30 \):
\(
PMT= 2000000 \times \frac {\left( \frac{0.06}{12} \right)} { \left[ \left( 1 + \frac{0.06}{12} \right)^{(12 \times 30)} -1 \right] } = \$ 1,991.01
\)

Your total deposit is \(1991.01 \times 12 \times 30 = \$716,763.6 \) which is much smaller than the accumulated $ 2 million

Exercise 26, p. 233

At age 20 when you graduate, you start saving for retirement. If your investment plan pays an APR of 4.5%, and you want to have $ 5 million when you retire in 45 years, how much should you deposit monthly?

Exercise 26, p. 233 ... continued

Solution.
Use Savings Plan Formula solved for payments
\(
PMT= A \times \frac {\left( \frac{APR}{n} \right)} { \left[ \left( 1 + \frac{APR}{n} \right)^{(nY)} -1 \right] }
\)

with \(A= \$5,000,000, \ \ \ APR=0.045, \ \ \ n=12, \ \ \ Y=45 \):
\(
PMT= 5000000 \times \frac {\left( \frac{0.045}{12} \right)} { \left[ \left( 1 + \frac{0.045}{12} \right)^{(12 \times 45)} -1 \right] } = \$ 2,863.70
\)

Your total deposit over this period of time is much smaller:

\( \$ 2,863.7 \times 12 \times 45 = \$1,546,398 \)

Exercise 30, p. 233

You pay \(\$ 8,000\) for a municipal bond. When it matures after 20 years, you receive \(\$ 12,500\). Compute the total and annual return.

Solution.
For the total return, we use the formula

\(
\text{total return} = \frac{A-P}{P} \times 100 \%
\)

with \(A=12500, \ \ \ \text{and} \ \ \ P=8000 \):

\(
\text{total return} = \frac{12500-8000}{8000} \times 100 \% =56.25\%
\)

Exercise 30, p. 233 ... continued

For the annual return, we use the formula\( \text{annual return} = \left( \frac{A}{P} \right) ^{(1/Y)} -1 \)

with \(A=12500, \ \ \ P=8000, \text{and} \ \ \ Y=20\):

\( \text{annual return} = \left( \frac{12500}{8000} \right) ^{(1/20)} -1 \approx 0.0226 = 2.26 \% \)

Exercise 32, p. 233

Three years after buying 200 shares of XYZ stock for \( \$ 25 \) per share, you sell the stock for \(\$ 8,500\). Compute the total and annual return.

Solution.
For the total return, we use the formula

\(
\text{total return} = \frac{A-P}{P} \times 100 \%
\)

with \(A=8500, \ \ \ \text{and} \ \ \ P=25 \times 200= 5000 \):

\(
\text{total return} = \frac{8500-5000}{5000} \times 100 \% =70 \%
\)

Exercise 32, p. 233 ... continued

For the annual return, we use the formula\( \text{annual return} = \left( \frac{A}{P} \right) ^{(1/Y)} -1 \)

with \(A=8500, \ \ \ P=5000, \text{and} \ \ \ Y=3\):

\( \text{annual return} = \left( \frac{8500}{5000} \right) ^{(1/3)} -1 \approx 0.1935 = 19.35 \% \)

Exercise 34, p. 233

Five years after paying \(\$ 5,000 \) for shares in a new company, you sell the shares for \(\$ 3,000 \) (at a loss). Compute the total and annual return.

Solution.
For the total return, we use the formula

\(
\text{total return} = \frac{A-P}{P} \times 100 \%
\)

with \(A=3000, \ \ \ \text{and} \ \ \ P=5000 \):

\(
\text{total return} = \frac{3000-5000}{5000} \times 100 \% = -40 \%
\)

Pay attention at the negative sign which indicates a loss.

Exercise 34, p. 233 ... continued

For the annual return, we use the formula\( \text{annual return} = \left( \frac{A}{P} \right) ^{(1/Y)} -1 \)

with \(A=3000, \ \ \ P=5000, \text{and} \ \ \ Y=3\):

\( \text{annual return} = \left( \frac{3000}{5000} \right) ^{(1/3)} -1 \approx -0.1566 = -15.66 \% \)

The negative sign again indicates a loss.