# Inflation: How long does it take for prices to double?

A question that often comes up is how long it would take for prices to double if the rate of inflation remained constant. It also helps to turn an abstract percentage number into a value that is easier to grasp and interpret.

If we start at a certain value for the consumer price index CPI0 and apply a constant annual inflation factor f (which is just the annual inflation rate expressed in decimals plus one), the CPI would grow exponentially according to this formula:

CPIn = CPI0 · f n

where CPIn symbolizes the Consumer Price Index for year n. The prices have doubled when CPIn equals 2 · CPI0. So we get:

2 · CPI0 = CPI0 · f n

Or, after solving this equation for n:

n = ln(2) / ln(f)

with ln being the natural logarithm. Using this formula, we can calculate how many years it would take for prices to double given a constant inflation rate (and thus inflation factor). Let’s look at some examples.

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In 1918, the end of World War I and the beginning of the Spanish Flu, the inflation rate in the US rose to a frightening r = 0.204 = 20.4 %. The corresponding inflation factor is f = 1.204. How long would it take for prices to double if it remained constant?

Applying the formula, we get:

n = ln(2) / ln(1.204) = ca. 4 years

More typical values for the annual inflation rate are in the region several percent. Let’s see how long it takes for prices to double under normal circumstances. We will use r = 0.025 = 2.5 % for the constant inflation rate.

n = ln(2) / ln(1.025) = ca. 28 years

Which is approximately one generation.

One of the highest inflation rates ever measured occurred during the Hyperinflation in the Weimar Republic, a democratic ancestor of the Federal Republic of Germany. The monthly (!) inflation rate reached a fantastical value of r = 295 = 29500 %. To grasp this, it is certainly helpful to express it in form of the doubling time.

n = ln(2) / ln(296) = ca. 0.12 months = ca. 4 days

Note that since we used the monthly inflation rate as the input, we got the result in months as well. Even worse was the inflation at the beginning of the nineties in Yugoslavia, with a daily (!) inflation rate of r = 0.65 = 65 %, meaning prices doubled every 33 hours.

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This was an excerpt from “Business Math Basics – Practical and Simple”. I hope you enjoyed it. For more on inflation check out my post about the Time Value of Money.

# The Time Value of Money and Inflation

To make a point, I’ll start this blog entry in an unusual way, that is, by talking about vectors. A vector is basically an ordered row of numbers. Consider this expression for example:

(12, 3, 5)

This vector could represent a lot of things. For example a point in a three dimensional coordinate system, with the vector components being the x-, y- and z-values respectively. Or for a company offering three products, it could stand for the sales of these products in a certain year.

Why this talk about vectors? You were probably very surprised when you heard grandma say that she paid only 150 \$ for her first car. It seems so amazingly cheap. But it is not. Your dear grandma is talking about 1950’s money, while you are thinking of today’s money. These two have a very different value.

If you want to specify the costs of a good precisely, merely giving an amount of money will not be sufficient. The value of money changes over time and thus to be absolutely precise, you should always couple this amount with a certain year. For example, this is what grandma’s car really cost:

(150 \$, 1950)

This is far from (150 \$, 2012), which is what you were thinking of when grandma shared the story with you. Using an online inflation calculator, we can conclude that this is actually what the car would cost in today’s money:

(1410 \$, 2012)

Not an expensive car, but certainly more than 150 \$ in today’s money. Now you can see why I started this chapter using vectors. They allow us to easily and clearly couple an amount with a year. A true pedant would even ask for one more component since we are still missing the respective months. But let’s not get too pedantic.

How can we justify saying that 150 \$ in 1950’s money is the same as 1410 \$ in today’s money? We can look at how much of a certain good these amounts would buy in the given year. With 150 \$ in 1950 you could fill your basket with about as many apples as you can with 1410 \$ today. The same goes for most other common goods: oranges, potatoes, water, cinema tickets, and so on.

This is inflation, goods get more expensive each year. At a later point we will take a look at what reasons there are for inflation to occur. But before that, let’s define the rate of inflation and see how it is measured …

This was an excerpt from the ebook “Business Math Basics – Practical and Simple”, available for Kindle here: http://www.amazon.com/dp/B00FXB8QSO.