Notes: Compound Interest and Annual Yield I. Compound ...

[Pages:13]Notes: Compound Interest and Annual Yield

I. Compound Interest A. The Formula

? P dollars invested at an annual rate r, compounded n times per year, has a value of F dollars after t years.

F

P 1

r

nt

n

? Think of P as the present value, and F as the future value of

the deposit.

Number of Times Compounded

Period

Annual Semiannual

Interest Credited

year 6 months

Times Credited per year

1

2

Quarterly

quarter

4

Monthly

month

12

Ex 1: Suppose you invest $5000 in an IRA account that compounds quarterly at 5.5%. How much money will be in an

account after 1 year?

F

P 1

r

nt

n

F1

50001

0.055 4

41

$5280.72

After 10 years?

F

P 1

r

nt

n

F10

50001

0.055 4

410

$8633.85

Decrease in value: When something is decreasing in value (such as a new car) we can use the compound interest formula. However, we will need to use subtraction instead of addition.

Ex 2: Your parents bought a car two years ago for $32,000. They are going to give it to you when you graduate high school next year. If the value of the car decreases 15% each year, how much will it be worth by the time they give it to you?

F

32000 1

.15 1(3)

1

= $19652

If you need to solve for time, use the log or ln function. Use either

one, just be consistent.

Ex 3: Lets say we want to know how long it will take $32,000 to grow to $50,000 invested in an account that has 5.2% annual interest compounded quarterly.

F

P 1

r

nt

n

50000 32000 1 .052 4t 4

1.5625 1.0134t

ln1.5625 ln1.0134t

ln1.5625 (4)(t) ln1.013

ln1.5625 (4)(t) ln1.013

ln1.5625

4 ln1.013

(t)

8.64 (t)

B. Continuous Compounding

A Pert

? P = principal amount invested ?r = the interest rate ? t = the number of years interest is being compounded ? A = the compound amount, the balance after t years

Ex 1: Ten thousand dollars is invested at 6.5% interest compounded continuously. When will the investment be worth $41,787?

A Pert 41,787 10,000e0.065t

4.1787 e0.065t ln 4.1787 0.065t

Since the interest rate is 6.5%, r = 0.065. Since ten thousand dollars is being invested, P = 10,000. And since the investment is to grow to become $41,787, A = 41,787. We will make the appropriate substitutions and then solve for t.

Divide by 10,000.

Rewrite the equation in logarithmic form.

22 t

Divide by 0.065 and solve for t.

Therefore, the $10,000 investment will grow to $41,787, via 6.5% interest compounded continuously, in 22 years.

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