Compound Interest Formulas II | EME 460: Geo-Resources Evaluation and Investment Analysis (2024)

3. Uniform Series Compound-Amount Factor

The third category of problems in Table 1-5 demonstrates the situation that equal amounts of money, A, are invested at each time period for n number of time periods at interest rate of i (given information are A, n, and i) and the future worth (value) of those amounts needs to be calculated. This set of problems can be noted as $F/A_{i,n}$ . The following graph shows the amount occurred. Think of it as this example: you are able to deposit A dollars every year (at the end of the year, starting from year 1) in an imaginary bank account that gives you i percent interest and you can repeat this for n years (depositing A dollars at the end of the year). You want to know how much you will have at the end of year n^th.

Figure 1-4: Uniform Series Compound-Amount Factor, $F/A_{i,n}$

In this case, utilizing Equation 1-2 can help us calculate the future value of each single investment and then the cumulative future worth of these equal investments.

Future value of first investment occurred at time period 1 equals $A{\left(1+i\right)}^{n-1}$
Note that first investment occurred in time period 1 (one period after present time) so it is n-1 periods before the n^th period and then the power is n-1.

And similarly:
Future value of second investment occurred at time period 2: $A{\left(1+i\right)}^{n-2}$
Future value of third investment occurred at time period 3: $A{\left(1+i\right)}^{n-3}$
Future value of last investment occurred at time period n: $A{\left(1+i\right)}^{n-n}=A$
Note that the last payment occurs at the same time as F.

So, the summation of all future values is
$$F=A{\left(1+i\right)}^{n-1}+A{\left(1+i\right)}^{n-2}+A{\left(1+i\right)}^{n-3}+\dots +A$$

By multiplying both sides by (1+i), we will have
$$F\left(1+i\right)=A{\left(1+i\right)}^n+A{\left(1+i\right)}^{n-1}+A{\left(1+i\right)}^{n-2}+\dots +A\left(1+i\right)$$

By subtracting first equation from second one, we will have
$$\begin{array}{l}F\left(1+i\right)–F=A{\left(1+i\right)}^n+A{\left(1+i\right)}^{n-1}+A{\left(1+i\right)}^{n-2}+\dots \\ +A\left(1+i\right)–\left[A{\left(1+i\right)}^{n-1}+A{\left(1+i\right)}^{n-2}+A{\left(1+i\right)}^{n-3}+\dots +A\right]\\ F+Fi–F=A{\left(1+i\right)}^n+A{\left(1+i\right)}^{n-1}+A{\left(1+i\right)}^{n-2}+\dots \\ +A\left(1+i\right)–A{\left(1+i\right)}^{n-1}-A{\left(1+i\right)}^{n-2}-A{\left(1+i\right)}^{n-3}-\dots -A\end{array}$$

which becomes:
$$Fi=A{\left(1+i\right)}^n–A$$

then

$$F=A\left[{\left(1+i\right)}^n-1\right]/i$$

Equation 1-3

Therefore, Equation 1-3 can determine the future value of uniform series of equal investments as $F=A\left[{\left(1+i\right)}^n-1\right]/i$ . Which can also be written regarding Table 1-5 notation as: $F=A*F/A_{i,n}$. Then $F/A_{i,n}=\left[{\left(1+i\right)}^n-1\right]/i$.
The factor $\left[\left(1+i\right)n-1\right]/i$ is called “Uniform Series Compound-Amount Factor” and is designated by F/A_i,n. This factor is used to calculate a future single sum, “F”, that is equivalent to a uniform series of equal end of period payments, “A”.

Note that n is the number of time periods that equal series of payments occur.

Please review the following video, Uniform Series Compound-Amount Factor (3:42).

Example 1-3:

Assume you save 4000 dollars per year and deposit it at the end of the year in an imaginary saving account (or some other investment) that gives you 6% interest rate (per year compounded annually), for 20 years. How much money will you have at the end of the 20th year?

So
A =$4000
n =20
i =6%
F=?

Please note that nis the number of equal payments.

Using Equation 1-3, we will have
$$\begin{array}{l}F=A*F/A_{i,n}=A\left[{\left(1+i\right)}^n-1\right]/i\\ F=A*F/A_{6\%,20}=4000*\left[{\left(1+0.06\right)}^{20}-1\right]/0.06\\ F=4000*36.78559=147142.4\end{array}$$

So, you will have 147,142.4 dollars at 20th year.

4. Sinking-Fund Deposit Factor

The fourth group in Table 1-5 is similar to the third group but instead of A as given and F as unknown parameters, F is given and A needs to be calculated. This group illustrates the set of problems that ask you to calculate uniform series of equal payments (or investment), A, to be invested for n number of time periods at interest rate of i and accumulated future value of all payments equal to F. Such problems can be noted as $A/F_{i,n}$ and are displayed in the following graph. Think of it as this example: you are planning to have F dollars in n years and there is a saving account that can give you i percent interest. You want to know how much you have to deposit every year (at the end of the year, starting from year 1) to be able to have F dollars after n years.

Figure 1-5: Sinking-Fund Deposit Factor, $A/F_{i,n}$

Equation 1-3 can be rewritten for A (as unknown) to solve these problems:

$$A=F\left\{i/\left[{\left(1+i\right)}^n-1\right]\right\}$$

Equation 1-4

Equation 1-4 can determine uniform series of equal investments, A, given the cumulated future value, F, the number of the investment period, n, and interest rate i. Table 1-5 notes these problems as: $A=F*A/F_{i,n}$. Then $A/F_{i,n}=i/[{\left(1+i\right)}^n-1]$. The factor $i/[{\left(1+i\right)}^n-1]$ is called the “sinking-fund deposit factor”, and is designated by $A/F_{i,n}$ . The factor is used to calculate a uniform series of equal end-of-period payments, A, that are equivalent to a future sum F.

Note that n is the number of time periods that equal series of payments occur.

Please watch the following video, Sinking Fund Deposit Factor (4:42).

Example 1-4:

Referring to Example 1-3, assume you plan to have 200,000 dollars after 20 years, and you are offered an investment (imaginary saving account) that gives you 6% per year compound interest rate. How much money (equal payments) do you need to save each year and invest (deposit it to your account) in the end of each year?

So
F=$200,000
n=20
i=6%
A=?

Using Equation 1-4, we will have
$$\begin{array}{l}A=F*A/F_{i,n}=F\left\{i/\left[{\left(1+i\right)}^n-1\right]\right\}\\ A=F*A/F_{6\%,20}=200,000*0.06/[{\left(1+0.06\right)}^{20}-1]\\ A=200,000*0.027185=5436.912\end{array}$$

So, in order to have 200,000 dollars at 20th year, you have to invest 5,436.9 dollars in the end of each year for 20 years at annual compound interest rate of 6%.

Note that $i/\left[{\left(1+i\right)}^n-1\right]$