Fermi Problems — Definition & Examples

Often, you might find it useful to estimate the size of something without very much information. Rather than to get a precise answer, you might aim to get an order of magnitude estimate. The goal is to get a sense of approximate size. This type of problem is known as a Fermi Problem, named for physicist Enrico Fermi. Enrico Fermi was known for his contribution to the Manhattan Project during World War II. He was a professor and research scientist at Columbia University and the University of Chicago.

Let's do an example of a Fermi Problem. Here's a classic example typically attributed to Enrico Fermi:

How many piano tuners are there in Chicago?

Take a minute to think about how you might approach this problem. How do you estimate this quantity without looking up anything online?

At first glance, this might seem like a really hard question to answer without additional information. However, we can produce a reasonable estimate by making a few sensible assumptions.

Let's say that Chicago has around $5$ million people. Pianos are generally owned by families rather than by individuals, and perhaps each family in Chicago has $4$ people. Then there are around $1.25$ million families in Chicago.

Let's assume that one in every ten families owns a piano that is tuned regularly. That would mean that there are around $125{,}000$ pianos in Chicago.

How many piano tuners do we need to maintain these pianos? We can assume that each piano should be tuned once per year, so there are $125{,}000$ piano tunings in Chicago each year.

If each piano tuner works $8$ hours per day on each weekday, and if each piano takes around $2$ hours to tune, then a piano tuner can tune around $4$ pianos per day for $250$ days per year. That's around $1{,}000$ piano tunings per piano tuner each year.

So we can estimate that we need around $125$ piano tuners to perform the $125{,}000$ piano tunings in Chicago each year.

While we can't claim this answer is exactly accurate, we can claim that it is a fair estimate of the actual value. That's the goal of a Fermi estimation problem.

This TED-Ed video demonstrates even more shortcuts we can take to calculate the number of piano tuners even faster.

Tips and Tricks for Fermi Problems: Use Nice Numbers

One of the highlights of Fermi Problems is that you are asked to estimate a given quantity rather than find its exact value. This gives you some slack in your estimations to save a lot of work when you actually go to calculate something. In other words, it's helpful to estimate smartly. This usually means using numbers that are easier to do calculations with.

But what types of numbers qualify? Here's a general pool of numbers you often want to use when possible:

• $1$
• Numbers ending in $5$ or $0$
• Numbers with lots of $0$s
• Numbers without a lot of digits past the decimal point (when applicable)

To emphasize that we want an estimate instead of an exact value, Fermi problems often ask to give an answer to the nearest order of magnitude, which means writing the number in scientific notation and then just using the factor that is a power of $10.$ In other words, you know how many digits are in the number (or how many $0$s there are after the decimal point for really small numbers).

Now let's talk about why we use the types of numbers I listed before. It takes less time to add, multiply, divide, and subtract these numbers from each other. For instance, would you rather calculate $123{,}456 \times 654{,}321$ or $100{,}000 \times 600{,}000?$ The second one is much faster while the first pretty much requires a calculator or multiplying out by hand. Let's wrap up with a full example that shows how smart estimation can lead to very quick calculations.

Example: You have giant ice cubes that have side lengths of $3.121485 \ \text{cm}.$ Estimate the total volume of $9$ of these ice cubes.

Solution: Alright let's get to work! We want to replace the numbers in the problem with "cleaner-looking" numbers that have almost the same value. The side length of an ice cube is $3.12148\,5 \ \text{cm},$ which is just a tiny bit bigger than $3 \ \text{cm},$ so let's round the side length down to $3 \ \text{cm}.$

In a similar vein, let's round up the total number of ice cubes to $10,$ which is only slightly bigger than $9,$ and has the advantage of ending in the digit $0.$ Now we can do the calculation.

The ice cubes are all the same so they have the same volume. This means if we find the volume of one ice cube, we can multiply that volume by $10$ to get the final answer. As one can see from our volume formulas, the volume of one ice cube is the side length raised to the third power, which is

$$(3 \ \text{cm})^3 = 27 \ \text{cm}^3.$$

Like we said, multiply this by $10$ to get our estimate:

$$27 \ \text{cm}^3 \times 10 = 270 \ \text{cm}^3.$$

That didn't take long, did it? If we had tried to directly cube the number $3.12148\,5,$ we'd be up all night making sure we didn't make a mistake in the multiplication. And since we only changed the numbers in the problem by small amounts, we know that the final answer is close to $270 \ \text{cm}^3.$

(Video) A Clever Way to Estimate Enormous Numbers

by TED-Ed

Here's a great video by TED-Ed that explains Fermi Problems and their uses.

$\large{\text{Summary}}$

The first part of the video goes how you can estimate numbers as an order of magnitude. For example, the speed of light is $299{,}792{,}458\text{ m/s}$. We can quickly express this as $3\times10^{8}$.

This is the key concept behind Fermi problems. Instead of being exact, we round numbers to their closest magnitude, or power of $10$, and estimate the answer using that.

They next go over one of the most classic Fermi problems. How many piano tuners are in the city of Chicago? This sounds difficult to figure out exactly! Luckily, this is the perfect question for a Fermi problem.

Start by estimating how many people are in the city of Chicago. Remember that we only want a power of $10$ as a guess. We can estimate the population of Chicago as $10^6$. No matter if the population is $1{,}000{,}000$, $5{,}000{,}000$, or $9{,}000{,}000$, the magnitude will still be $10^6$. And, if we were to check, we would see the population about $3$ million.

Next, we want to find the number of pianos. Or, in other words, how many pianos there are per person. We could say there is $1$ piano for every $10$ people. This would be a magnitude of $10^{-1}$. We could also say there's $1$ piano for every $100$ people. This would be a magnitude of $10^{-2}$. Since our population is also including children, let's choose $\frac{1}{100}$ as our estimate. Thus, we have a population of $10^6$ people, and $10^{-2}$ pianos per person. By using scientific notation we avoid having to multiply decimals together. $$\begin{align} &10^6\text{ people}\times10^{-2}\,\frac{\text{piano}}{\text{people}}\\ &=10^6\times10^{-2}\text{ pianos}\\ &=10^{6+(-2)}\text{ pianos}\\ &=10^4\text{ pianos}\\ &=10{,}000\text{ pianos} \end{align}$$

With this many pianos, we now need to find how many piano tuners there are. We could focus on a bunch of variables like how often pianos are tuned, how many pianos are tuned in one day, or how many days a piano tuner works. But that's not the point of quick estimation. Just think in orders of magnitude. Let's assume a piano tuner tunes $10^2$ pianos each year. This means on average they tune a few hundred pianos a year.

Given that there are $10^4$ pianos and $10^2$ pianos tuned per year, we can guess that there are: $$\begin{align} &\frac{10^4\text{ pianos}}{10^2\,\frac{\text{pianos}}{\text{tuners}}}\\ &=10^4\text{ pianos}\times10^{-2}\,\frac{\text{tuners}}{\text{pianos}}\\ &=10^{4+(-2)}\text{ tuners}\\ &=10^2\text{ tuners}\\ &=100\text{ tuners} \end{align}$$ Looking through a phonebook, we see there are actually $81$ piano tuners. Not bad at all for a guess!