Example
Suppose we will deposit $100 each month into an account paying 6% interest. We assume that the account is compounded with the same frequency as we make deposits unless stated otherwise. Write an explicit formula that represents this scenario.
Answer:
In this example:
- r = 0.06 (6%)
- k = 12 (12 compounds/deposits per year)
- d = $100 (our deposit per month)
Writing out the recursive equation gives
Pm=(1+120.06)Pm−1+100=(1.005)Pm−1+100
Assuming we start with an empty account, we can begin using this relationship:
P0=0
P1=(1.005)P0+100=100
P2=(1.005)P1+100=(1.005)(100)+100=100(1.005)+100
P3=(1.005)P2+100=(1.005)(100(1.005)+100)+100=100(1.005)2+100(1.005)+100
Continuing this pattern, after
m deposits, we’d have saved:
Pm=100(1.005)m−1+100(1.005)m−2+L+100(1.005)+100
In other words, after
m months, the first deposit will have earned compound interest for
m-1 months. The second deposit will have earned interest for
m-2 months. The last month's deposit (L) would have earned only one month's worth of interest. The most recent deposit will have earned no interest yet.
This equation leaves a lot to be desired, though – it doesn’t make calculating the ending balance any easier! To simplify things, multiply both sides of the equation by 1.005:
1.005Pm=1.005(100(1.005)m−1+100(1.005)m−2+⋯+100(1.005)+100)
Distributing on the right side of the equation gives
1.005Pm=100(1.005)m+100(1.005)m−1+⋯+100(1.005)2+100(1.005)
Now we’ll line this up with like terms from our original equation, and subtract each side
1.005PmPm==100(1.005)m+100(1.005)m−1+⋯+100(1.005)m−1+⋯+100(1.005)100(1.005)+100
Almost all the terms cancel on the right hand side when we subtract, leaving
1.005Pm−Pm=100(1.005)m−100
Factor
Pm out of the terms on the left side.
Pm(1.005−1)=100(1.005)m−100(0.005)Pm=100(1.005)m−100
Solve for
Pm
0.005Pm=100((1.005)m−1)Pm=0.005100((1.005)m−1)
Replacing
m months with 12
N, where
N is measured in years, gives
PN=0.005100((1.005)12N−1)
Recall 0.005 was
r/k and 100 was the deposit
d. 12 was
k, the number of deposit each year.
Generalizing this result, we get the savings annuity formula.
Annuity Formula
PN=(kr)d((1+kr)Nk−1)
- PN is the balance in the account after N years.
- d is the regular deposit (the amount you deposit each year, each month, etc.)
- r is the annual interest rate in decimal form.
- k is the number of compounding periods in one year.
If the compounding frequency is not explicitly stated, assume there are the same number of compounds in a year as there are deposits made in a year.
For example, if the compounding frequency isn’t stated:
When do you use this?
Annuities assume that you put money in the account
on a regular schedule (every month, year, quarter, etc.) and let it sit there earning interest.
Compound interest assumes that you put money in the account
once and let it sit there earning interest.
- Compound interest: One deposit
- Annuity: Many deposits.
Examples
A traditional individual retirement account (IRA) is a special type of retirement account in which the money you invest is exempt from income taxes until you withdraw it. If you deposit $100 each month into an IRA earning 6% interest, how much will you have in the account after 20 years?
Answer:
In this example,
d = $100 |
the monthly deposit |
r = 0.06 |
6% annual rate |
k = 12 |
since we’re doing monthly deposits, we’ll compound monthly |
N = 20 |
we want the amount after 20 years |
Putting this into the equation:
P20=(120.06)100((1+120.06)20(12)−1)P20=(0.005)100((1.005)240−1)P20=(0.005)100(3.310−1)P20=(0.005)100(2.310)=$46200
The account will grow to $46,200 after 20 years.
Notice that you deposited into the account a total of $24,000 ($100 a month for 240 months). The difference between what you end up with and how much you put in is the interest earned. In this case it is $46,200 - $24,000 = $22,200.
This example is explained in detail here.
https://youtu.be/quLg4bRpxPA
Try It
A conservative investment account pays 3% interest. If you deposit $5 a day into this account, how much will you have after 10 years? How much is from interest?
Answer:
d = $5 the daily deposit
r = 0.03 3% annual rate
k = 365 since we’re doing daily deposits, we’ll compound daily
N = 10 we want the amount after 10 years
P10=3650.035((1+3650.03)365∗10−1)=21,282.07
Financial planners typically recommend that you have a certain amount of savings upon retirement. If you know the future value of the account, you can solve for the monthly contribution amount that will give you the desired result. In the next example, we will show you how this works.
Example
You want to have $200,000 in your account when you retire in 30 years. Your retirement account earns 8% interest. How much do you need to deposit each month to meet your retirement goal?
Answer:
In this example, we’re looking for d.
r = 0.08 |
8% annual rate |
k = 12 |
since we’re depositing monthly |
N = 30 |
30 years |
P30 = $200,000 |
The amount we want to have in 30 years |
In this case, we’re going to have to set up the equation, and solve for
d.
200,000=(120.08)d((1+120.08)30(12)−1)200,000=(0.00667)d((1.00667)360−1)200,000=d(1491.57)d=1491.57200,000=$134.09
So you would need to deposit $134.09 each month to have $200,000 in 30 years if your account earns 8% interest.
View the solving of this problem in the following video.
https://youtu.be/LB6pl7o0REc
Example
If you invest $100 each month into an account earning 3% compounded monthly, how long will it take the account to grow to $10,000?
Answer:
This is a savings annuity problem since we are making regular deposits into the account.
d = $100 |
the monthly deposit |
r = 0.03 |
3% annual rate |
k = 12 |
since we’re doing monthly deposits, we’ll compound monthly |
We don’t know
N, but we want
PN to be $10,000.
Putting this into the equation:
10,000=(120.03)100((1+120.03)N(12)−1) Simplifying the fractions a bit
10,000=0.0025100((1.0025)12N−1)
We want to isolate the exponential term, 1.002512
N, so multiply both sides by 0.0025
25=100((1.0025)12N−1) Divide both sides by 100
0.25=(1.0025)12N−1 Add 1 to both sides
1.25=(1.0025)12N Now take the log of both sides
log(1.25)=log((1.0025)12N) Use the exponent property of logs
log(1.25)=12Nlog(1.0025) Divide by 12log(1.0025)
12log(1.0025)log(1.25)=N Approximating to a decimal
N = 7.447 years
It will take about 7.447 years to grow the account to $10,000.
This example is demonstrated here:
https://youtu.be/F3QVyswCzRo