Imagine riding your motor scooter one sunny afternoon to your auntie’s home who lives near the sea, 20 miles from your home. On your trip there, the wind is at your back and the terrain slopes downward, resulting in an average speed of 25 mph. On the way home however, you’re motoring on a slight upward slope and into the wind, resulting in an average speed of 15 mph.

What is your average speed for the entire roundtrip?

Now before you shout, “20 mph!”, consider this: If 20 mph were the correct answer, why would I write an article about that?

Let’s to some quick math:

On first 20-mile leg of the trip, you averaged 25 mph, which means it took .8 hours to get there (20 miles / 25 mph = .8 hours). On the second 20-mile leg of your trip, you averaged 15 mph, which means it took you 1.33 hours to get home (20 miles / 15 mph = 1.33 hours). In total, you travelled 40 miles in 2.13 hours, for a total trip average of 18.75 mph.

So, what happened here?? The “average” we learned in the 5^{th} grade – and for most people, the only average they ever use — is more precisely termed the *arithmetic average *or arithmetic mean. It’s found by adding up the numbers of interest then dividing by their count. However, certain types of data, including many ratios, don’t average using the arithmetic method.

Way back in the fifth century BC, Pythagoras identified three different averages: arithmetic, geometric, and harmonic. Each intended to average a different type of number. And it’s the harmonic average that’s best suited to deal with our trip to auntie’s house. Let’s look at the formula:

$$ \displaystyle A_{H}=n/(\frac{1}{x1}+\frac{1}{x2}+\ldots\ \frac{1}{xn}) $$

The harmonic average, when each number is weighted equally, is the reciprocal of the arithmetic average of the reciprocals of the numbers of interest. Back to our trip to auntie’s house:

$$ \displaystyle A_{H}=\frac{2}{\frac{1}{25}+\frac{1}{15}}=18.75 $$

Consider another application: Production line A manufactures 1,000 parts per hour and Production Line B manufactures 500 parts per hour. You plan to manufacture 10,000 parts from each, for a total of 20,000 parts. What is your average production rate?

Using our harmonic average formula,

$$ \displaystyle A_{H}=\frac{2}{\frac{1}{1,000}+\frac{1}{500}}=667\ parts\ per\ hour $$

This harmonic average formula works in any problem of averaging ratios when the units of output are fixed in the same units as found in the numerators of the ratios. In our first problem, the total trip length was fixed (40 miles) in the same units as in the numerator of the ratios (* miles* per hour). In our second problem, the total parts needed, 20,000, was fixed and used the same unit as the ratios we were averaging (

*per hour).*

__parts__Once you understand the harmonic average formula for cases when the weights of the numbers you’re averaging are equal, you’re a short leap from the more general, weighted harmonic average formula:

$$ \displaystyle A_{H}=\ \sum_{i=1}^{n}w_{i}\ /\sum_{i=1}^{n}\frac{Wi}{Xi} $$

If in our production rate problem, we had manufactured 10,000 parts on Line A at 1,000 parts per hour, and 20,000 parts on Line B at 500 parts per hour on Line B, our average production rate across both lines is:

$$ \displaystyle A_{H}=\frac{30,000}{\frac{10,000}{1,000}+\frac{20,000}{500}}=600\ parts\ per\ hour $$

Of the three Pythagorean Means, the harmonic mean is always the lowest and the arithmetic mean is always the highest when averaging positive numbers Using the incorrect mean therefore will always bias your results.

Ray Harkins is a manufacturing professional and online educator. He teaches a variety of low-cost, high-quality manufacturing and business-related courses at Udemy.com. Click on the following coupon codes to receive substantial discounts on courses such as **Root Cause Analysis and the 8D Corrective Action Process**.

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