Compound Interest

Compound Interest By Ms. Karen Overman using Tan’s 5th edition Applied Calculus for the managerial , life, and social sciences text

Simple Interest You may recall from previous math courses the formula for simple interest, I = Prt, where I is the interest earned P is the principal or the amount invested, r is the interest rate and t is the time in years. The amount after the interest in added, the accumulated amount is the sum of the principal and the interest or A = P+I = P+Prt = P(1+rt).

Getting to Compound Interest Suppose we have an account where the simple interest is added in each year and then that money also earns interest. This is called compound interest. Let’s look at the pattern for several years. 1st year: A(1) = P(1+rt) The A(1) denotes the amount after 1 year, etc… 2nd year: A(2) = A(1) (1+rt) But A(1) = P(1+rt), so … A(2) = P(1+rt) (1+rt) So, A(2) = P(1+rt)² 3rd year: A(3) = A(2) (1+rt) A(3) = P(1+rt)² (1+rt) A(3) = P(1+rt)³

Are you starting to see a pattern? A(1) = P(1+rt) A(2) = P(1+rt)² A(3) = P(1+rt)³ So, what would be the accumulated amount after n years? It is also possible to have the interest compound more than once a year, say m times a year. If the interest rate for the year is r and the interest is compounded m times a year, then we replace r with i, where i = r/m. And for n to keep the meaning of how many times it was compounded, n = mt. This leads us to …

Compound Interest Formula A = the accumulated amount P = the principal (amount invested) r = the yearly interest rate m = the number of times it is compounded in a year t = the time in years

Let’s look at an example: Find the amount in an account after 4 years if $5000 is invested at 6% per year and is compounded a.) quarterly and b.) monthly. It helps to identify what is given and what you need to find. P = 5000, r = 0.06, t = 4 and in a.) m = 4 in b.) m = 12 We are trying to find A in each case. Now, just substitute into the compound interest formula. a.)

Example continued… b.) So after 4 years, if the $5000 is compounded quarterly (every 3 months) or 4 times a year the accumulated amount is $6344.93. If the $5000 is compounded monthly or 12 times a year the accumulated amount is $6352.45. You can see from this example if the yearly interest rate is constant, the accumulated amount is larger for larger numbers of compoundings in a single year.

You can see from the previous example that the amount of money earned depends on the number of times it is compounded. To serve as a method of comparing interest rates, the effective rate is the simple interest rate that would give you the same accumulated amount. Recall the accumulated amount of simple interest is A=P(1+rt). Thus the accumulated amount with the effective rate , in 1 year would be (assuming t = 1). The accumulated amount if it were compound interest over 1 year would be (assuming t = 1).

Set the accumulated amount with the effective rate (or simple interest) equal to the accumulated amount from compounded interest. Now, solve for the effective rate.

The Effective Rate is given by where r is the yearly interest rate and m is the number of compoundings per year. Let’s look at an example to make this a little more clear.

Let’s find the effective rates for the compound interest problem we looked at before: $5000 invested for 4 years at 6% per year compounded a.)quarterly and b.)monthly. a.) r = 0.06, m = 4 So, having an interest rate of 6% yearly and compounding it quarterly is equivalent to having a simple interest rate of 6.13%. This is what the effective rate does. It serves as a method of comparing rates to one standard. b.) r = 0.06, m = 12

Let’s look at another problem. Suppose you want to invest enough now so that you’ll have $60,000 for your child’s education in 18 years. What would you need to invest if the interest rate is 4 ½ % compounded monthly? Think about the compound interest formula. What parts do you know and what are you trying to find? You know: A = $60,000 – that’s how much you want to have in 18 years, the accumulated amount r = 4 ½% = 0.045 t = 18 years m = 12 – compounded monthly You want to find: P – the amount to invest now

Example continued… Now simply substitute into the compound interest formula and solve for P. Wow, you would need to invest about $26,732 now! Fortunately most of us will invest a little each year to reach the same goal instead of all at one time.

Present Value The previous problem was an example of present value. The equation for present value is simply the compound interest formula solved for P. So the equation for Present Value is P is called the present value and A is called the future value.

Continuously Compounded Interest There is also a formula for interest compounded continuously. To show the derivation of this formula requires a limit we have not studied so the derivation is omitted here. Continuously Compounded Interest

Example Let’s consider that first problem again only this time we’ll calculate the accumulated amount after investing $5000 for 4 years at a yearly interest rate of 6% compounded continuously. P = 5000, r = 0.06 and t = 4 So, the accumulated amount will be $6356.25. You should note this amount is more than both of our previous calculations where it was compounded quarterly and monthly.

Another example Using those same numbers, let’s see how long it would take to double our investment of $5000. Think about what you know and what you are trying to find! You know P = 5000, r = 0.06 and if you know you are trying to double your investment then A = 10,000. You are trying to find the time to double or t. Substitute the values into the continuously compounded interest formula and solve for t.

Example continued… Substituting into the formula. Divide both sides by 5000. Take the ln of both sides and solve for t. So it would take just over 11 ½ years to double the initial investment.

Compound Interest