Growth Arithmetic

Here is some basic algebra/arithmetic that can be used for calculations involving growth. This could be growth of GDP, growth of money in a savings account, or growth in the population of gerbils. The algebra is the same.

There are four variables involved.

Y0 is the initial value of the variable. For example, we could start out with a population of 1000 gerbils in period 0.

t is the number of periods over which we are calculating growth. The periods could be days, months, or years. In the example, let us think in terms of years, so that if t=4 that means that we are observing growth in the gerbil population over 4 years.

Yt is the value of the variable at the end of the observation interval. If t=4, then Y is the size of the gerbil population after 4 years.

r is the average periodic rate of growth, expressed as a decimal. For example, if the gerbil population increases at an average rate of 24 percent per year, then r=.24

The fundamental equation for growth is given by

[1] Yt = Y0(1+r)t

For example, if we start with 1000 gerbils and the population grows at an average rate of 24 percent for 4 years, we have:

Y4 = 1000(1+.24)4 = 2364 gerbils.

Calculating cumulative growth

Cumulative growth is the total growth from the beginning to the end of the observation interval, expressed in percent. That is,

Cumulative growth = [(Yt - Y0)/Y0]*100

I find it easier to work with the equivalent formula

Cumulative growth = (Yt/Y0 - 1)*100

Using the equivalent formula, we have in the gerbil example

Cumulative growth = (2364/1000 - 1)*100 = 136.4 percent.

Solving for r

Suppose that we know the value we started with and the value at the end of the observation interval, and we want to calculate the average growth rate, r. Solving equation [1] for r gives

r = (Yt/Y0)(1/t) - 1

For example, if we start with 1000 gerbils and after five years we have 1800 gerbils, then the average annual growth rate is

r = (1800/1000)(1/5) - 1 = .125, or 12.5 percent per year

Solving for t

Suppose we ask, how long will it take for the gerbil population to reach 4000, starting from a population of 1000 and growing at an average annual rate of 24 percent? This is a problem where we are given Y0, Yt, and r, and we need to solve for t.

To do this, we use something I'll bet you never thought you would use: logs! If you take the log of both sides of equation [1], to get

logYt = logY0 + t[log(1+r)]

Solving for t gives

t = (logYt - logY0)/log(1+r)

In our example where we want to know when the gerbil population will reach 4000, we have

t = (log[4000] - log[1000])/log[1+.24] = 6.44 years