Mathematics and Personal Finance



Geometry Interest and Rates of Return Budgeting Compound Interest Video

Interest and Rate of Return

Interest

     So many aspects of one's personal finance involve interest: loans, credit cards, savings, house payments, car payments, etc. Richeson's textbook provides the formulas for calculating these. For simple interest we have:

\[I=Pit\]
and
\[A=P+I\]
     These equations are given with P (the principal), I (the interest), i (the rate), A (the amount), and t (the time) (Richeson 13). Now for compound interest we have,
\[s=(1+i)^n\]
with "s being the compound amount of 1 for n periods at the rate i per period," (Richeson 14). The difference between compound interest and simple interest is explained, "Simple interest is based on the principal amount of a loan or deposit. In contrast, compound interest is based on the principal amount and the interest that accumulates on it in every period," (Nickolas). These formulas are a perfect example of how knowledge of algebra is connected to personal finance.


Rate of Return

     In Suresh Chandra's textbook Financial Mathematics: An Introduction there is a chapter dedicated to the basics of financial mathematics. One of the topics that is covered a lot in this chapter is investing. Investing tends to be a topic that many find intimidating and risky, however if you understand the mathematics behind it, you can evaluate the risk and make choices that are safest and most beneficial for you. Chandra starts by sharing the formula for the rate of return on an asset. This formula is,

\[r_s= {S(1)-S(0)\over S(0)}\] and \[r_b={B(1)-B(0) \over B(0)}\]
\(r_s\) representing a stock and \(r_b\) representing a bond (Chandra 2). These formulas are easy for any secondary mathematics students to use and understand. They are later used to calculate the expected return for a stock. The example given is as follows:
\[Let B(0)=100, B(1)=110,S(0)=80 and\]

\(S(1)=\{100,\) with probability \(0.8\}\)

\(\{60,\)with probability \(0.2\}\)

Solution:

\(r_S={100-80 \over 80}\) with probability of \(0.8\)

and

\(r_S={60-80 \over 80}\) with probability of \(0.2\)

Therefore \(E(r_S)=(0.25)(0.8)+(-0.25)(0.2)=0.15\)

(Chandra 4)

Now to calculate it for a bond we have:

\(r_B={B(1)-B(0) \over B(0)}\)

Thus, \(r_B={110-100 \over 100}={1/10}=.1\)

This is the rate of return on the bond.

(Chandra 2)

This is just one of many examples of the mathematical formulas that most secondary education students already have experience solving presented in this textbook.