Equivalent percentages

Exponential and logarithmic growth: Future value

Equivalent percentages

We want to compare two savings accounts on the basis of compound interest, but one has an interest rate per quarter and the other per year. We can compare these by calculating the interest rates for the same time unit (for example, both for a year or both for a quarter). More generally, we are concerned with conversions from rates for short periods to rates for longer periods, and vice versa.

Conversion formula for growth factors

Suppose that the growth factor per period #t_1# is equal to #g_1# and the growth factor per period #t_2# is equal to #g_2#, where #t_1# and #t_2# are expressed in the same units of time. Then \[ g_2=g_1^{\frac{t_2}{t_1}}\]

In order to convert the interest rate of one unit to another unit of time, we first transform the interest rate to a growth factor. Then we adjust the growth factor to the other unit of time. Next we transform this new growth factor to the interest rate. This procedure is illustrated by the following diagram:

We can see this by looking at the familiar formula for future value #S(n)=S(0) \cdot (1+i)^n# where we choose #S(0)# equal to #1# and #n# equal to the number of periods of length #t_1# that fit into the new unit of time: #n=\frac{t_2}{t_1}#.

\[\begin{array}{rcl} S(n)&=& S_0 \cdot (1+i)^n\\ &&\phantom{xxxxx}\color{blue}{\text{the formula}}\\ S(n) &=& (g_1)^{\frac{t_2}{t_1}}\\
&&\phantom{xxxxx}\color{blue}{g_1=1+i ,\ S_0=1 ,\ n=\frac{t_2}{t_1} \text{ used}}\\
\end{array}\]

Now #g_2=S(n)# is the growth factor for the new unit of time #t_2#.

The transformation of growth percentage from a period of length #t_1# to a period of length #t_2# is expressed by the formula below, where #p_1# and #p_2# represent the growth percentages pertaining to #t_1# and #t_2#, respectively.

\[ p_2 = 100\cdot \left( \left(\dfrac{p_1}{100}+1\right)^{\frac{t_2}{t_1}}-1\right)\]

This formula is derived from the conversion formula for growth factors by use of the equalities #g_k = \frac{p_k}{100}+1# for #k=1# and #k=2#.

Here are some examples.

Has your bank account ever been overdrawn? If so, then you probably received a statement from the bank that read something like:

\(\phantom{xxx}\) "Your balance shows a deficit. The borrowing rate is #1.39\%# on a monthly basis."

What is the annual interest rate in this case?

Give your answer as a percentage with a precision to #1# decimal point.

Percentage per year: #18.0# #\%#

After all,
\[\begin{array}{rcl}
i&=&\dfrac{1.39}{100}=0.0139\\
&&\color{blue}{\text{the growth rate of the monthly percentage}}\\
g_{\text{month}}&=&0.0139+1=1.0139\\
&&\color{blue}{\text{the corresponding growth factor}}\\
g_{\text{year}}&=&1.0139^{12}\approx 1.1802\\
&&\color{blue}{\text{there are }12\text{ months in a year}}\\
i_{\text{year}}&=&1.1802-1=0.1802\\
&&\color{blue}{\text{the growth rate corresponding to the growth factor}}\\
\text{percentage per year}&=&0.1802 \cdot 100 \approx 18.0\%\\
&&\color{blue}{\text{the corresponding percentage rounded to }1\text{ decimal}}
\end{array}\]

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