Section 4C

exercises

...

Stocks

Price-to-Earning Ratio. Exercise 42, p. 234

General Mills close at $52.65 per share with a P/E ratio of 16.14

(a) How much were earnings per share?

Solution. Recall that
\( \text{P/E ratio} = \frac{\text{share price}}{\text{earnings per share }} \)
Thus in this case \( 16.14= \frac{52.65}{\text{earnings per share }}
\)
and
\( {\text{earnings per share }} = \frac{52.65}{16.14} \approx \$ 3.26 \)

Price-to-Earning Ratio. Exercise 42, p. 234 ... continued

(b) Does the stock seem to be overpriced, underpriced, or about right given that the historical P/E ratio is 12 - 14?

Solution. The P/E ratio of 16.14 is higher than the historical numbers 12-14. Recall that
\( \text{P/E ratio} = \frac{\text{share price}}{\text{earnings per share }} \)
With given earning per share, the price must be lower in order to have the P/E ratio within the historical interval. The stock is overpriced. For example, if the P/E ratio was 13, the share price would be
\( \text{share price} = \text{P/E ratio} \times \text{earnings per share } = 13 \times 3.26 = 42.38 \)

Stocks

Price-to-Earning Ratio. Exercise 44, p. 234

Google closed at $393.50 with P/E ratio of 28.78

(a) How much were earnings per share?

Solution. Recall that
\( \text{P/E ratio} = \frac{\text{share price}}{\text{earnings per share }} \)
Thus in this case \( 28.78= \frac{393.50}{\text{earnings per share }}
\)
and
\( {\text{earnings per share }} = \frac{393.50}{28.78} \approx \$ 13.67 \)

Price-to-Earning Ratio. Exercise 44, p. 234 ... continued

(b) Does the stock seem to be overpriced, underpriced, or about right given that the historical P/E ratio is 12 - 14?

Solution. The P/E ratio of 28.78 is higher than the historical numbers 12-14. Recall that
\( \text{P/E ratio} = \frac{\text{share price}}{\text{earnings per share }} \)
With given earning per share, the price must be lower in order to have the P/E ratio within the historical interval. The stock is overpriced. For example, if the P/E ratio was 13, the share price would be
\( \text{share price} = \text{P/E ratio} \times \text{earnings per share } = 13 \times 13.67 = 177.71 \)

Bond yields

Exercise 48, p. 234

Compute the current yield of a \( \$ 1000 \) Treasury bond with coupon rate of 2.5% that has a market value of \( \$ 1050 \).

Solution. Recall that
\( \text{current yield} = \frac{\text{annual interest payment}}{\text{current price of bond}} \)
While the current price of the bond is $ 1050,
the annual interest payment = face value \(\times \) coupon rate = \( 1000 \times 0.025 = 25 \)
Thus
\( \text{current yield} = \frac{25}{1050} \approx 0.0238 = 2.38\%. \)

Bond yields. Exercise 50, p. 234

Compute the current yield of a \( \$ 10,000 \) Treasury bond with coupon rate of 3% that has a market value of \( \$ 9,500 \).

Solution. Recall that
\( \text{current yield} = \frac{\text{annual interest payment}}{\text{current price of bond}} \)
While the current price of the bond is $ 9500,
the annual interest payment = face value \(\times \) coupon rate = \( 10000 \times 0.03 = 300 \)
Thus
\( \text{current yield} = \frac{300}{9500} \approx 0.0316 = 3.16\%. \)

Bond interest

Exercise 52, p. 234

Compute the annual interest that you would earn with a $1000 Treasury bond with a current yield of 1.5% that is quoted at 98 points

Solution. Recall that
\( \text{current yield} = \frac{\text{annual interest }}{\text{current price }} \)
Thus \( \text{annual interest } = \text{current yield} \times \text{current price } \)
While the current yield is 1.5% = 0.015,
current price = 98% of the face value = 0.98 \(\times \)1000 = $980
Thus \( \text{annual interest } = 0.015 \times 980 = \$ 14.7 \)

Bond interest

Exercise 54, p. 234

Compute the annual interest that you would earn with a $10,000 Treasury bond with a current yield of 3.6% that is quoted at 102.5 points

Recall that
\( \text{current yield} = \frac{\text{annual interest }}{\text{current price }} \)
Thus \( \text{annual interest } = \text{current yield} \times \text{current price } \)
While the current yield is 3.6% = 0.036,
current price = 102.5% of the face value = 1.025 \(\times \)10000 = $10250
Thus \( \text{annual interest } = 0.036 \times 10250 = \$ 369 \)

Further applications. Savings plans

Exercise 58, p. 235

Polly deposits \( \$ 50 \) per month in an account with an APR of 6%.
Quint deposits deposits \( \$ 40 \) per month in an account with an APR of 6.5%.

Compare the balances, and the amounts deposited after 10 years.

Exercise 58, p. 235 ... continued

Solution. We firstly calculate the amounts deposited.

Polly deposits \( \$ 50 \) per month during 10 years:

\( 50 \times 12 \times 10 = \$ 6,000 \)

Quint deposits \( \$ 40 \) per month during 10 years:

\( 40 \times 12 \times 10 = \$ 4,800 \)

Polly deposits bigger amount of money then Quint.

Exercise 58, p. 235 ... continued

We now calculate their accumulated balances using Savings Plan Formula

\( A = PMT \times \frac{ \left[ \left( 1 + \frac{APR}{n} \right)^{(nY)} -1 \right] }{\left( \frac{APR}{n} \right)} \)

Polly: \(PMT=50, \ \ \ APR=0.06 \ \ \ n=12, \ \ \ Y=10 \):

\( A = 50 \times \frac{ \left[ \left( 1 + \frac{0.06}{12} \right)^{(12 \times 10)} -1 \right] }{\left( \frac{0.06}{12} \right)}=\$ 8,193.97 \)

Quint: \(PMT=40, \ \ \ APR=0.065 \ \ \ n=12, \ \ \ Y=10 \):

\( A = 40 \times \frac{ \left[ \left( 1 + \frac{0.065}{12} \right)^{(12 \times 10)} -1 \right] }{\left( \frac{0.065}{12} \right)}=\$ 6,736.13 \)

Exercise 58, p. 235 ... continued

Summary

Over the period of 10 years, Polly deposited \( \$ 6,000 \) and accumulated \( \$ 8,193.97 \)

Over the period of 10 years, Quint deposited \( \$ 4,800 \), and accumulated \( \$ 6,736.13 \)

It is interesting to calculate the total return in both cases

Polly's total return = \( \frac{8193.97-6000}{6000} \times 100\% \approx 36.6 \% \)

Quint's total return = \( \frac{6736.13 -4800}{4800} \times 100 \% \approx 40.3 \% \)

That is because Quint's APR was higher!

Further applications. Savings plans

Exercise 60, p. 235

George deposits \( \$ 40 \) per month in an account with an APR of 7%.
Harvey deposits deposits \( \$ 150 \) per quarter in an account with an APR of 7.5%.

Compare the balances, and the amounts deposited after 10 years.

Exercise 60, p. 235 ... continued

Solution. We firstly calculate the amounts deposited.

George deposits \( \$ 40 \) per month during 10 years:

\( 40 \times 12 \times 10 = \$ 4,800 \)

Harvey deposits \( \$ 150 \) per quarter during 10 years:

\( 150 \times 4 \times 10 = \$ 6,000 \)

Harvey deposits bigger amount of money then George.

Exercise 60, p. 235 ... continued

We now calculate their accumulated balances using Savings Plan Formula

\( A = PMT \times \frac{ \left[ \left( 1 + \frac{APR}{n} \right)^{(nY)} -1 \right] }{\left( \frac{APR}{n} \right)} \)

George: \(PMT=40, \ \ \ APR=0.07 \ \ \ n=12, \ \ \ Y=10 \):

\( A = 40 \times \frac{ \left[ \left( 1 + \frac{0.07}{12} \right)^{(12 \times 10)} -1 \right] }{\left( \frac{0.07}{12} \right)}=\$ 6,923.39 \)

Harvey: \(PMT=150, \ \ \ APR=0.075 \ \ \ n=4, \ \ \ Y=10 \):

\( A = 150 \times \frac{ \left[ \left( 1 + \frac{0.075}{4} \right)^{(4 \times 10)} -1 \right] }{\left( \frac{0.075}{4} \right)}=\$ 8,818.79 \)

Exercise 60, p. 235 ... continued

Summary

Over the period of 10 years, George deposited \( \$ 4,800 \) and accumulated \( \$ 6,923.39 \)

Over the period of 10 years, Harvey deposited \( \$ 6,000\), and accumulated \( \$ 8,818.7 \)

It is interesting to calculate the total return in both cases

George's total return = \( \frac{6923.39-4800}{4800} \times 100\% \approx 44.2\% \)

Harvey's total return = \( \frac{8818.7 -6000}{6000} \times 100 \% \approx 46.9 \% \)

That is bcause Harvey's APR was higher.