Derivation of Compound Interest Formulas
Derivation of Compound Interest Formulas
There is some logic to the creation of the compound interest formulas (shown in Illustration 10-1 of the text). The next few pages offer an explanation. The explanations require a basic understanding of exponents, compound fractions, factoring, roots, and equation-solving skills. We will derive each of the formulas except Formula 5; the derivation of Formula 5 is beyond the scope of the material you are covering. The derivations will be done in the following order: Formula 1A, 2A, 1B, 2B, 3, 4A, and 4B.
Derivation of Formula 1A
Suppose you deposit $100 in a savings account earning 3% compounded annually. At the end of the first year you will earn 3% (or $3), ending up with a balance of $103. Stated another way, your ending balance will be 1.03 times as great as what you started with ($100 ? 1.03 = $103). At the end of the second year you will earn 3% on the $103 beginning balance (or $3.09), ending up with a balance of $106.09. This is compound interest because you are earning interest on interest. Each year the balance increases by a multiple of 1.03:
(Yr 1) (Yr 2) (Yr 3) (Yr 4)
$100 (1.03) $100 (1.03) (1.03) = $100 (1.03)2 $100 (1.03) (1.03) (1.03) = $100 (1.03)3 $100 (1.03) (1.03) (1.03) (1.03) = $100 (1.03)4
Let's use symbols to represent the process. The ending balance is a future value (FV), the initial deposit is a present value (PV), the 1.03 is "1 + i" (1 + .03 = 1.03), and the exponent is the n-value. We end up with Formula 1A:
FV = PV(1 + i) n
Derivation of Formula 2A
The Golden Rule of Equation Solving (see Unit 3.2 of the text) states that we can "Do unto one side of an equation as we do unto the other" and the equation will still be in balance. Let's start with Formula 1A:
FV = PV(1 + i) n
If we want a formula that is designed to solve for PV, we can manipulate Formula 1A. By dividing both sides of Formula 1A by "(1 + i)n " we get:
FV ' PV (1 % i)n
If the left side of the above equation equals the right side, then it follows that the right side equals the left side, so it is okay to put the variable PV on the left side, giving us Formula 2A:
PV ' FV (1 % i)n
Derivation of Formula 1B
Suppose you deposit $100 at the end of each year and earn 3% compounded annually. At the end of 1 year, your balance will be $100 (you will earn no interest the first year). Here is one way to calculate the balance at the end of year 2:
Deposit at the end of year 2 Deposit at the end of year 1, plus interest for next year: $100.00 (1.03) Balance, year 2
$100.00 103.00
$203.00
Let's use the same procedure to figure the balance at the end of year 4:
Deposit at the end of year 4 Deposit at the end of year 3, plus interest for the next year: $100 (1.03) Deposit at the end of year 2, plus interest for the next 2 years: $100 (1.03) (1.03) Deposit at the end of year 1, plus interest for the next 3 years: $100 (1.03) (1.03) (1.03) Balance, year 4
$100.00 103.00 106.09 109.27
$418.36
You might notice a pattern developing. The ending balance that results from making periodic deposits of $100 at the end of each period for 4 years, earning 3% compounded annually can be written as:
Ending Balance = $100 + $100 (1.03) + $100 (1.03)2 + $100 (1.03)3
Let's use symbols to represent the process. The ending balance is a future value (FV), the annual deposits are periodic payments (PMT), and the 1.03 is "1 + i" (1 + .03 = 1.03). Each exponent represents the number of periods each deposit will earn interest; notice, the first deposit earns interest for one period less than the total n-value. We end up with:
FV = PMT + PMT (1 + i) + PMT (1 + i)2 + PMT (1 + i)3 + ..... PMT (1 + i)n - 2 + PMT (1 + i)n - 1
The process of deriving Formula 1B gets a bit tricky from here. The next step is to divide each side of the above equation by (1 + i):
FV ' PMT % PMT % PMT (1 % i) % PMT (1 % i)2 % ..... PMT (1%i)n & 2 (1 % i) (1 % i)
Now we will subtract the second equation from the first equation. That is, we will subtract the left side of the second equation from the left side of the first equation, and subtract the right side of the second equation from the right side of the first equation. You may wonder if doing so is mathematically sound. Remember, since the left and right sides of the second equation are equal we are, in effect, subtracting equal values from both sides of the first equation. By following this procedure most of the terms on the right side are canceled out. We are left with:
FV & FV ' PMT (1 % i)n&1 & PMT
(1 % i)
(1 % i)
Now let's factor out FV on the left and PMT on the right:
FV 1 & 1 ' PMT (1 % i)n&1 & 1
(1 % i)
(1 % i)
Next let's get common denominators inside each set of brackets:
FV (1 % i) & 1 ' PMT (1 % i) n & 1
(1 % i) (1 % i)
(1 % i) (1 % i)
Let's combine terms inside of each set of brackets: FV i ' PMT (1 % i) n & 1
(1 % i)
(1 % i)
Finally, we will multiply each side by (1 % i) . We get Formula 2B: i
FV ' PMT (1 % i)n & 1 i
Derivation of Formula 2B
Let's start out with Formula 1B: FV ' PMT (1 % i)n & 1 i
In Formula 1A, FV = PV(1 + i)n . Let's substitute "PV (1 + i)n " for FV in the above formula:
PV (1 + i)n = PMT (1 % i)n & 1 i
Now, let's divide each side of the equation by (1 + i) n : PV = PMT (1 % i)n & 1 i (1 % i)n
Move the "i" from the denominator inside of the brackets to create a denominator for PMT outside of the brackets. Doing so does not change the value:
PV = PMT (1 % i)n & 1
i
(1 % i)n
The numerator inside of the brackets is a binomial (two terms) and the denominator is a monomial (one term). Let's create two separate fractions inside of the brackets (this is the opposite of combining fractions with the same denominator):
PV = PMT (1 % i)n & 1 i (1 % i)n (1 % i)n
Now we will simplify: PV = PMT 1 & 1
i
(1 % i)n
Finally, divide the value inside of the brackets by the "i" outside the brackets (doing so does not change the value):
1& 1
PV ' PMT
(1 % i)n
i
Derivation of Formula 3 Let's start with Formula 1A: FV = PV(1 + i)n Now, let's divide both sides by PV: FV ' (1 % i)n PV
Next, we want to take the "nth " root of both sides. Before we do that, let's consider what taking a root
means. To illustrate, taking the "third root of 64" is the same as finding "what value to the third power is 64?" The answer is 4, because 43 = 64. Mathematically, we can write the "third root of 64" as either:
3
64 OR
1
(64) 3
1
By taking the "nth " root of both sides of the equation shown above, we end up with: FV n ' 1 % i PV
1
Subtract "1" from both sides of the equation: FV n & 1 ' i PV
Finally, rewrite with "i" on the left:
1
i ' FV n & 1 PV
Derivation of Formula 4A
Let's start with Formula 1B: FV ' PMT (1 % i)n & 1 i
Multiply both sides by
i
. We end up with:
FV ( i ) ' PMT
(1 % i)n & 1
(1 % i)n & 1
Rewrite the equation with PMT on the left:
PMT ' FV ( i ) (1 % i)n & 1
Derivation of Formula 4B
1& 1
Let's start with Formula 2B: PV ' PMT
(1 % i)n
i
Next, multiply both sides of the equation by
i
. We get PV ( i ) ' PMT
1& 1
1& 1
(1 % i)n
(1 % i)n
Rewrite the equation with PMT on the left:
PMT ' PV ( i ) 1& 1 (1 % i)n
................
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