## Review for the second midterm

... Sections 4B,4C,4D ...

## Section 4B

### Compound interest formulas

##### Compound interest formulas:

for interest payed once a year: $$A = P \times (1+APR)^Y$$

for interest paid $$n$$ times a year: $$A = P \times (1+\frac{APR}{n})^{(nY)}$$

for continuous compounding: $$A = P\times e^{(APR \times Y)}$$

$$A$$ = accumulated balance

$$P$$ = starting principal

$$APR$$ = annual percentage rate (in decimal)

$$Y$$ = number of years

$$n$$ = number of compounding periods per year

$$e \approx 2.71828$$ is a special number

##### Exercise 4 p.213

An account with an APR of 4% and quarterly compounding increases in value every three months by

(a) 1% (b) 1/4% (c) 4%

Solution. An APR of 4% means an increase by that 4% every year, which is four quarters. Which means 1% every quarter, that is 1% every three months (a quarter).

(a) is the right choice.

You deposit $3,200 in an account with an annual interest rate of 3.5%. Calculate the amount of money you will have after 5 years assuming that you earn simple interest. Solution. Simple interest means that you earn the same amount of $$3200 \times 0.035 = 112$$ yearly. Over a period of 5 years you earn $$112 \times 5 = 560$$ After 5 years, you have $$3200+560=\3,760$$ <=== Answer ### Compound interest. Exercise 59, p.214 Use the compound interest formula to compute the balance if $$\ 15,000$$ is invested for 25 years in an account with an APR of 3.2% compounded annually Solution. The formula $$A = P \times (1+APR)^Y$$ gives us $$A=15000 \times (1+0.032)^{25} = \ 32,967.32$$ ### Compound interest. Exercise 67, p.214 What is the accumulated balance if $$\2000$$ is invested for 15 years with an APR of 5% and monthly compounding? Solution. The compound interest formula for interest paid $$n$$ times a year is $$A = P \times (1+\frac{APR}{n})^{(nY)}$$ In this case, n=12, P=2000, Y=15, APR=0.05, and $$A = 2000 \times (1+\frac{0.05}{12})^{(12 \times 15)} = \ 4227.41$$ <=== answer ### Annual Percentage Yield (APY). Exercise 73, p.214 Find the APY if a bank offers an APR of 1.23% compounded monthly Solution. Invest $$\100$$, and after 1 year you will have $$A = P \times (1+\frac{APR}{n})^{(nY)} = 100 \times (1+ \frac{0.0123}{12})^{(12 \times 1)} = \ 101.237$$ The relative difference is the APY $$APY= \frac{101.237-100}{100} =0.01237 =1.237\% \approx 1.24\%$$ ### Continuous compounding. Exercise 79, p.215 Compute the balance after 20 years for a $$\3000$$ deposit in an account with APR of 6%, and continuous compounding Solution. Continuous compounding formula $$A = P\times e^{(APR \times Y)}$$ gives us $$A = 3000\times e^{(0.06 \times 20)} = \ 9,960.35$$ ### Planning ahead. Exercise 83, p.215 How much must you deposit today into an account with monthly compounding and an APR of 6% in order to have $$\ 25,000$$ in 8 years? Solution. Compound interest formula for interest paid $$n$$ times a year $$A = P \times (1+\frac{APR}{n})^{(nY)}$$ solved for the principal $$P= \frac{A}{(1+\frac{APR}{n})^{(nY)} }$$ gives us $$P= \frac{25000}{(1+\frac{0.06}{12})^{(12\times 8)} } \approx \15,488.10$$ Savings Plan Formula (regular payments) Section 4C $$A = PMT \times \frac{ \left[ \left( 1 + \frac{APR}{n} \right)^{(nY)} -1 \right] }{\left( \frac{APR}{n} \right)}$$ $$A$$ = accumulated savings plan balance (FV -- future value) $$PMT$$ = regular payment (deposit) amount $$APR$$ = annual percentage rate (in decimal) $$n$$ = number of payment periods per year $$Y$$ = number of years Savings Plan Formula solved for payments: $$PMT= A \times \frac {\left( \frac{APR}{n} \right)} { \left[ \left( 1 + \frac{APR}{n} \right)^{(nY)} -1 \right] }$$ ##### Saving plans Exercise 21, p. 233 You put$300 per month in an investment plan that pays an APR of 3.5%. 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 $$APR=0.035,\ \ \ PMT=300, \ \ \ n=12, \ \ \ Y=18$$: $$A = 300 \times \frac{ \left[ \left( 1 + \frac{0.035}{12} \right)^{(12 \times 18)} -1 \right] }{\left( \frac{0.035}{12} \right)} = \ 90,091.51$$

##### Saving plans

Exercise 21, p. 233 ... continued

You put $300 per month in an investment plan that pays an APR of 3.5%. What is the total deposit made over the period of 18 years? Solution. You deposit$300 every month over a period of 18 years.
Total deposit $$= 300 \times 12 \times 18 = \ 64,800$$

##### Saving plans

Exercise 21, p. 233 ... continued

Solution. We start with calculating the lump sum necessary for the indicated income. $200,000 should be 6% of this amount. The necessary amount is thus $$\frac{200000}{0.06} = \ 3,333,333.33$$ ##### Exercise 28, p. 233 ... continued The problem becomes: find the monthly payment for a saving plan with an APR of 6% which accumulates a balance of$3,333,333.33 over the period of 65-25=40 yeas.

We apply saving 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 $$A=\ 3,333,333.33$$, $APR=0.06$, $Y=40$, and $n=12$ to find

$$PMT= 3,333,333.33 \times \frac {\left( \frac{0.06}{12} \right)} { \left[ \left( 1 + \frac{0.06}{12} \right)^{(12*40)} -1 \right]} = \1,673.79$$

#### Total and Annual Returns

Exercise 31, p. 233

Twenty years after purchasing shares in a mutual fund for $$\6,500$$, you sell them for $$\11,300$$. Compute total and annual return.

Solution. For the total return, we use the formula
$$\text{total return} = \frac{A-P}{P} \times 100 \%$$
with $$A=11,300 \ \ \ \text{and} \ \ \ P=6500$$:
$$\text{total return} = \frac{11300-6500}{6500} \times 100 \% =73.8\%$$

#### Total and Annual Returns

Exercise 31, p. 233 ... continued

For the annual return, we use the formula
$$\text{annual return} = \left( \frac{A}{P} \right) ^{(1/Y)} -1$$
with $$A=11300, \ \ \ P=6500, \text{and} \ \ \ n=20$$:
$$\text{annual return} = \left( \frac{11300}{6500} \right) ^{(1/20)} -1 \approx 0.0280 = 2.8 \%$$

Loan Payment formula (Installment Loans) Section 4D

$$PMT= \frac{ P \times \left( \frac{APR}{n} \right) } {1 - \left( 1 + \frac{APR}{n} \right)^{ \left( -n Y \right) } }$$

$$PMT =$$ regular payment amount

$$P=$$ staring loan principal

$$APR =$$ annual percentage rate (in decimal)

$$n=$$ number of payment periods per year

$$Y=$$ loan term in years

##### Exercise 18 p 250

Consider a home mortgage of $$\ 150,000$$ at a fixed APR of 4% for 15 years.

(a) Calculate the monthly payment

Solution. We use the Loan Payment formula $$PMT= \frac{ P \times \left( \frac{APR}{n} \right) } {1 - \left( 1 + \frac{APR}{n} \right)^{ \left( -n Y \right) } }$$
with $$P=150000$$, $$APR =0.04$$, $$n=12$$, $$Y=15$$:
$$PMT= \frac{ 150000 \times \left( \frac{0.04}{12} \right) } {1 - \left( 1 + \frac{0.04}{12} \right)^{ \left( -12 \times 15 \right) } } \approx \ 1109.53$$

##### Exercise 18 p 250 ... continued

Consider a home mortgage of $$\ 150,000$$ at a fixed APR of 4% for 15 years.

(b) Determine the total amount paid over the term of the loan.

Solution. Over a period of 15 years, every month, an amount of $1,109.53 was paid. All together, an amount of $$n \times Y \times PMT = 12 \times 15 \times 1109.53 = \ 199,715$$ was paid. ##### Exercise 18 p 250 ... continued Consider a home mortgage of $$\ 150,000$$ at a fixed APR of 4% for 15 years. (c) Of the total amount paid, what percentage is paid toward the principal, and what percentage is paid for interest? Solution. The total amount paid is $$\ 199,715$$, while the loan principal was $$\ 150,000$$. Thus the percentage paid towards the principal is $$\frac{150000}{199,715} \approx 0.751 = 75.1 \%$$. The rest $$100\% - 75.1\%=24.9\%$$ is paid for interest. ##### Exercise 18 p 250 ... continued ... modification Consider a home mortgage of $$\ 150,000$$ at a fixed APR of 4% for 15 years. (d) Of the first monthly payment, what percentage is paid toward the principal, and what percentage is paid for interest? Solution. The first payment (in fact, every payment) is$1109.53.

By the end of the first month, the new principal is
$$150,000 + \frac{0.04}{12} \times 150,000 = 150,500$$

The payment of $$\1109.53$$ splits into $$\500$$ for interest and

$$1109.53 - 500 = \600.53$$ toward the principal.

##### Exercise 18 p 250 ... continued ... modification

Consider a home mortgage of $$\ 150,000$$ at a fixed APR of 4% for 15 years.

(d) Of the first monthly payment, what percentage is paid toward the principal, and what percentage is paid for interest?

Solution (continued): We have just found that out of $1109.53$500 is for interest.

That is $$\frac{500}{1109.53} \times 100\% = 45.06\%$$ <==== answer

Observation. Of the total amount paid over 15 years, 24.9% is for interest, while of the first payment, 45.06% is for interest.