- We are going to look at investments for a this section.
- In this case, you put the money into an account, and withdraw it
some time later.
- One method would be just simple interest.
- Example: I have $1000, I deposit for two years at 5% interest.
A = p + prt (or A = p(1+rt))
= 1000 (1 + 0.05 × 2)
= 1100
- What if I decided to take the money out once a year and redeposit it
- Example 2:
Year 1:
A = 1000(1+0.05 × 1)
= 1050
Year 2:
A = 1050(1+0.05 × 1)
= 1102.50
- What happened in the previous example? Compound interest
- There is a formula, which we will investigate in a moment
A = p(1+ r/n)nt
r = interest rate
n = number of compoundings per year
t = number of years
p = principal
- Example 3:
In the previous example, n = 1, t = 2
A = 1000(1+.05/1)(1× 2)
= 1102.5
- Example 4:
p = 5000
r = 5%
t = 9
n = 1 (annually)
A = 5000(1+.05/1)^(1*9)
= 7756.64
i = A-p
= 5000 -7756.64 = 2756.64
- Other compounding periods
Name | n |
annually | 1 |
semiannually | 2 |
quarterly | 4 |
monthly | 12 |
daily | 360 |
- Example: $2500 is deposited at 4 1/4 percent interest compounded daily for three years. How much interest is earned?
p = 2500
r = 4.24
n = 360
t = 3
A = 2500(1+.0425/360)(3× 360)
= 2839.94
i = A -p
= 2839.94 - 2500 = 339.94
- APY or annual percentage yield, or effective annual yield.
- APY is the simple interest rate that gives the same amount of interested as the compound rate over the same period of time.
- This can be solved in general:
r = 3.5%, n=2
Assume p = 1 and t = 1
A = 1(1+0.035/2)(1× 2)
= 1.0353
or 3.53%
- Sometimes, we have a goal.
- We would like to have 100,000 in 20 years, a bank offers 9.5% interest compounded quarterly. How much must we deposit now?
A = 100,000
p = ?
r = 9.5%
n = 4
t = 20
A = p(1+r/n)(nt)
p = A/(1+r/n)nt
p = 100000/(1+0.05/4)(4*20)
= 15,292.79
- This is called present value.
- To derive the Compound interest formula:
use A=p(1+rt)
assume n = 2 and t = 1
The first 1/2 year The second 1/2 yeaer
A = p+pr 1/2 A = p(1+r/2)(1+r/2)
= p(1+r/2) = p(1+r/2)2
Assume n = 3 and t = 1
The first 1/3 The second 1/3 The last 1/3
A = p(1+r 1/3) A = p(1+r/3)(1+r/3) A = p(1+r/3)2(1+r/3)
= p(1+r/3) = p(1+r/3)2 = p(1+r/3)3
Assume n=3, t = 2
Each time we will have 2/6 years or 1/3 year
The first 1/6 The second 1/6 The third 1/6
A = p(1+r 1/3) A = p(1+r/3)(1+r/3) A = p(1+r/3)2(1+r/3)
= p(1+r/3) = p(1+r/3)2 = p(1+r/3)3
The fourth 1/6 The fifth 1/6 The last 1/6
A = p(1+r/3)3(1+r/3) A = p(1+r/3)4(1+r/3) A = p(1+r/3)5(1+r/3)
= p(1+r/3)4 = p(1+r/3)5 = p(1+r/3)6