Algebra 1

Estimating a huge quantity from small facts

How many piano tuners work in a city of four million people? Nobody has counted, so how can anyone give an answer worth trusting?

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Estimating a huge quantity from small facts

Why this exists

Some questions ask for a number nobody has written down. How many times does a heart beat in a lifetime? How many grains of rice would fill a classroom? No table holds the answer, and counting is out of the question.

What you do have is small knowledge, each piece guessable to within a factor of two or so: roughly how fast a heart beats, how big a grain of rice is. A chain of such facts builds the huge number, and the result is good enough to say whether a plan is possible, or whether a claim is off by a factor of a thousand.

The method is named after the physicist Enrico Fermi, who was known for quick estimates of this kind. It rests on conversion factors, which cancel units along a chain, friendly numbers, which keep the arithmetic in your head, and Scientific notation, which keeps track of the zeros.

Definition

A Fermi estimate is an approximate value for a quantity, built by writing it as a product (sometimes with a quotient) of smaller quantities, estimating each from everyday knowledge, and multiplying them with their units. Four habits go with it.

Every assumption is stated, so anyone who disagrees can change that one number and redo the chain.
Numbers keep one or two Significant figures, because the inputs are known no better than that.
The answer is judged by its order of magnitude.
Where possible it is bounded and checked: a low and a high estimate fence it in, and a known figure, if one exists, tests it.

The order of magnitude of a positive number is the power of ten in its scientific notation: written as a \times 10^{n} with 1 \le a < 10, it has order of magnitude 10^{n}. So 2\,900 = 2.9 \times 10^{3} has order of magnitude 10^{3}, and 80 = 8 \times 10^{1} has 10^{1}. Two estimates agree to an order of magnitude when the larger is less than ten times the smaller, which can happen across a change of power: 8 \times 10^{4} and 1.2 \times 10^{5} agree.

Intuition

Picture a ruler on which each step to the right multiplies by ten: 1, 10, 100, 1000, equally spaced. On this power-of-ten scale, distance measures how many times bigger one number is than another. A guess twice the truth and a guess half the truth sit equally far from it; an estimate of 3 \times 10^{6} is a short step from a true 5 \times 10^{6} and three whole steps from 3 \times 10^{9}. Digits after the first one or two barely move you along the ruler, so they are not worth keeping.

Figure 1The words-a-year estimate of Example 4 on a power-of-ten scale. Each step multiplies by ten, so the estimate, four times the low bound and a quarter of the high one, sits in the middle.

The middle of a range moves too. Between 10 and 1000 the ordinary average, 505, sits near the top end; the point halfway along is \displaystyle 100 = \sqrt{10 \times 1000}, ten times the low end and a tenth of the high one. And multiplying two factors adds their steps, so a factor guessed twice too big and one guessed half too small cancel. That is why long chains still work.

Results

The chain. The wanted quantity Q is a product of estimated factors, each carrying its unit:

Q \approx q_1 \times q_2 \times \cdots \times q_k

The units must cancel to leave exactly the unit of Q; if they do not, a factor is missing or upside down.

Bounds. If every factor is positive and L_i \le q_i \le H_i, then

L_1 L_2 \cdots L_k \;\le\; q_1 q_2 \cdots q_k \;\le\; H_1 H_2 \cdots H_k

For a factor that divides, the low bound of the answer uses its high guess, and the high bound its low guess.

The middle of a range. Between positive bounds L and H, the value \displaystyle M = \sqrt{LH} is the same number of times above L as it is below H:

\frac{M}{L} = \frac{H}{M} = \sqrt{\frac{H}{L}}

How errors grow. If each of k factors is within a factor f of its true value (at most f times too big or too small), the product is within a factor f^{k} of the true product.

Why it is true

Units cancel because a factor such as \dfrac{70\ \text{beats}}{1\ \text{min}} is a conversion factor, and units multiply and divide the way numbers do.

The bounds. Multiplying an inequality by a positive number keeps its direction. So L_1 \le q_1 gives L_1 L_2 \le q_1 L_2, and L_2 \le q_2 gives q_1 L_2 \le q_1 q_2; together L_1 L_2 \le q_1 q_2, and each further factor comes in the same way. The upper bound is the same argument turned round. For a divisor with 0 < L \le d \le H, dividing L \le d by Ld and d \le H by dH gives \dfrac{1}{H} \le \dfrac{1}{d} \le \dfrac{1}{L}.

The middle. With \displaystyle M = \sqrt{LH}:

\begin{aligned} \frac{M}{L} &= \frac{\sqrt{LH}}{L} = \sqrt{\frac{H}{L}} \\ \frac{H}{M} &= \frac{H}{\sqrt{LH}} = \sqrt{\frac{H}{L}} \end{aligned}

How errors grow. An estimate e_i within a factor f of q_i means \dfrac{q_i}{f} \le e_i \le f q_i. Multiplying these bounds, as above, puts the product of the estimates between \dfrac{1}{f^{k}} and f^{k} times the true product.

Worked examples

Example 1

Heartbeats in a lifetime

Estimate how many times a human heart beats in a lifetime.

Solution

Two assumptions, both estimates: a resting heart beats about 70 times a minute, and a lifetime is about 80 years. A year has 60 \times 24 \times 365 = 525\,600 minutes, about 5.3 \times 10^{5}.

\begin{aligned} \text{beats} &\approx \frac{70\ \text{beats}}{1\ \text{min}} \times \frac{5.3 \times 10^{5}\ \text{min}}{1\ \text{year}} \\ &\qquad \times 80\ \text{years} \\ &= (7 \times 5.3 \times 8) \times 10^{1+5+1} \\ &= 296.8 \times 10^{7} \approx 3 \times 10^{9} \end{aligned}

Minutes and years cancel, leaving beats; the leading parts and the powers of ten are multiplied separately, as in any product in scientific notation.

About three billion beats. The unrounded 70 \times 525\,600 \times 80 = 2\,943\,360\,000 rounds to the same 3 \times 10^{9}. The assumptions matter more: 60 beats instead of 70 lowers the answer by a seventh.

Example 2

Piano tuners in a large city

Estimate how many piano tuners make a living in a city of 4 million people.

Solution

The chain runs from people to pianos to tunings, then divides by the work one tuner does (Figure 2). The assumptions, each an estimate:

a household is about 2.5 people, and about one household in 20 has a piano (schools and concert halls are left out);
a piano is tuned about once a year;
a tuning takes about 2 hours with travel, so a tuner does about 4 a day, 5 days a week, 50 weeks a year: 4 \times 5 \times 50 = 1000 tunings a year.

Figure 2The piano-tuner chain. Each arrow is one stated assumption, and anyone can change one without redoing the rest.

\begin{aligned} \text{pianos} &\approx 4 \times 10^{6}\ \text{people} \times \frac{1\ \text{household}}{2.5\ \text{people}} \\ &\qquad \times \frac{1\ \text{piano}}{20\ \text{households}} \\ &= \frac{4 \times 10^{6}}{50}\ \text{pianos} = 8 \times 10^{4}\ \text{pianos} \end{aligned}

At one tuning each, that is 8 \times 10^{4} tunings a year, and one tuner supplies 1000:

\text{tuners} \approx \frac{8 \times 10^{4}\ \text{tunings a year}}{10^{3}\ \text{tunings a year per tuner}} = 80

So 80 = 8 \times 10^{1}, order of magnitude 10^{1} and near the top of it: tens of tuners, not a handful and not a thousand. Anyone who thinks one household in 10 has a piano changes one line and gets 160.

Example 3

Water a town uses in a day, with bounds

A town has 20\,000 people. Estimate how many liters of water its homes use in a day, with a low and a high estimate.

Solution

The chain is people times liters per person. Build the second factor from estimated pieces:

Use	Liters per person per day
shower, 8 liters a minute for 8 minutes	64 \approx 60
toilet, 5 flushes of 6 liters	30
washing clothes and dishes	30
drinking, cooking and sinks	30
total	150

Bound the total too: at least 100 liters (short showers), at most 300 (baths, gardens). Each day,

2 \times 10^{4}\ \text{people} \times 1.5 \times 10^{2}\ \frac{\text{L}}{\text{person}} = 3 \times 10^{6}\ \text{L}

By the bounds result the low estimate is 2 \times 10^{4} \times 100 = 2 \times 10^{6} liters and the high one 2 \times 10^{4} \times 300 = 6 \times 10^{6} liters: millions of liters, not thousands and not billions. “Homes only” leaves out shops, factories and leaking pipes, so the whole town uses more.

Example 4

Words spoken in a year, choosing the middle

Estimate how many words a person speaks in a year.

Solution

Speed times time. Assume people say about 150 words a minute in conversation. Time is far less certain: perhaps 15 minutes a day for a quiet person, 240 minutes for a talkative one with a talking job.

The ordinary average, \dfrac{15 + 240}{2} = 127.5 minutes, is 8.5 times the low guess but less than 2 times below the high one. Halfway on a scale where each step multiplies by ten is the value that is the same number of times above the low guess as below the high one, and that value is the square root of their product:

\sqrt{15 \times 240} = \sqrt{3600} = 60\ \text{minutes}

which is 4 times 15 and a quarter of 240. Use an hour a day:

\begin{aligned} \text{words} &\approx \frac{150\ \text{words}}{1\ \text{min}} \times \frac{60\ \text{min}}{1\ \text{day}} \\ &\qquad \times \frac{365\ \text{days}}{1\ \text{year}} \\ &= 9000 \times 365 \\ &= 3\,285\,000 \approx 3.3 \times 10^{6} \end{aligned}

The bounds come from the two ends of the time range:

\begin{aligned} \text{low: } 150 \times 15 \times 365 &= 821\,250 \approx 8 \times 10^{5} \\ \text{high: } 150 \times 240 \times 365 &= 13\,140\,000 \approx 1.3 \times 10^{7} \end{aligned}

Figure 1 shows all three, the estimate exactly halfway because the one uncertain factor took its middle value: a few million words a year.

Example 5

Rice to fill a classroom

Estimate how many grains of rice would fill a classroom.

Solution

Divide the space the rice fills by the space one grain takes. Three assumptions, all estimates: the room is 8 m long, 7 m wide and 3 m high; a grain is roughly a block 5 mm by 2 mm by 2 mm; and poured grains leave gaps, filling about half the space.

The room holds 8 \times 7 \times 3 = 168 \approx 1.7 \times 10^{2} cubic meters, so the rice itself takes up about \tfrac{1}{2} \times 1.7 \times 10^{2} = 85 cubic meters. A grain is 5 \times 2 \times 2 = 20 cubic millimeters. The units must match, so convert with \dfrac{1\ \text{m}}{10^{3}\ \text{mm}}, which equals one, used three times because a cubic millimeter has three lengths in it:

20\ \text{mm}^{3} \times \left(\frac{1\ \text{m}}{10^{3}\ \text{mm}}\right)^{3} = 2 \times 10^{-8}\ \text{m}^{3}

\frac{85\ \text{m}^{3}}{2 \times 10^{-8}\ \text{m}^{3}\ \text{per grain}} = 42.5 \times 10^{8} \approx 4 \times 10^{9}

About four billion grains, order of magnitude 10^{9}: about half as many grains as there are people on Earth (about 8 billion).

Example 6

Checking an estimate against a known figure

Estimate how many people live on Earth, starting from the distance around the Earth, about 40\,000 km. Then check the result against the known world population.

Solution

People equals land area times people per square kilometer. Treat the Earth as a sphere, which it very nearly is. A circle with circumference C has radius r = \dfrac{C}{2\pi}, and a sphere of radius r has surface area 4\pi r^{2}, so the Earth’s surface is about

\begin{aligned} 4\pi r^{2} &= 4\pi \cdot \frac{C^{2}}{4\pi^{2}} = \frac{C^{2}}{\pi} \\ &= \frac{(4 \times 10^{4})^{2}}{\pi} = \frac{1.6 \times 10^{9}}{\pi} \\ &\approx 5.1 \times 10^{8}\ \text{km}^{2} \end{aligned}

Two assumptions, both estimates. On a world map land looks like about \tfrac{3}{10} of the surface, so land is about 0.3 \times 5.1 \times 10^{8} \approx 1.5 \times 10^{8} square kilometers. Averaged over all land, deserts and ice included, there are surely more than 10 and fewer than 100 people per square kilometer; the middle is \displaystyle \sqrt{10 \times 100} = \sqrt{1000} \approx 32, so use 30. Then

1.5 \times 10^{8}\ \text{km}^{2} \times \frac{30\ \text{people}}{1\ \text{km}^{2}} = 4.5 \times 10^{9}\ \text{people}

with bounds 1.5 \times 10^{9} (at 10) and 1.5 \times 10^{10} (at 100).

Check. The world’s population is about 8 billion: inside the bounds, and \dfrac{8 \times 10^{9}}{4.5 \times 10^{9}} \approx 1.8, so the estimate is low by less than a factor of 2, and both have order of magnitude 10^{9}. The check also finds the weak link: \dfrac{8 \times 10^{9}}{1.5 \times 10^{8}} \approx 53 people per square kilometer, so the density guess came out low.

Boundaries

A chain is only as good as its weakest factor. One factor known only to within a factor of 100 leaves the answer at least that uncertain.
Errors cancel only when the guesses are made separately. If one leaning pushes every factor the same way, the worst case f^{k} comes close to true.
Bounds and the middle need positive quantities. Multiplying inequalities keeps their direction only for positive numbers, and a range starting at 0 has no middle on the power-of-ten scale.
“Order of magnitude” has a second meaning. Some books round to the nearest power of ten on the power-of-ten scale, with the cut-off at \displaystyle \sqrt{10} \approx 3.16, the middle of 1 and 10; in that sense 80 has order of magnitude 10^{2}.
A check proves less than it seems. Agreement with a known figure shows the overall error was small, not that each assumption was right.

Where this goes next

With measured data in place of guesses, the same chains become reasoning with units and a sensible level of accuracy, and Measurement error asks how far a measurement can be trusted. Comparing results by their powers of ten is comparing numbers in scientific notation, and turning a messy question into stated assumptions plus a calculation is the start of modeling.

Exercises

Building a chain with units

Write a chain of conversion factors that turns one year into seconds, show the units canceling, and give the result to two significant figures.
Assume a person at rest takes about 16 breaths a minute. Estimate the breaths in one day, to two significant figures.
A town of 4000 people uses about 150 liters of water per person per day. Write the chain for one week with every unit shown, and give the total in scientific notation.
Assume a hamster’s heart beats about 400 times a minute and a hamster lives about 2 years. Using 525\,600 minutes in a year, estimate its heartbeats in a lifetime to two significant figures, and compare with Example 1.
Orrin reads 250 words a minute for 30 minutes every day. Estimate the words he reads in a year, to two significant figures.

Orders of magnitude and the middle of a range

Give the order of magnitude of (a) 2.9 \times 10^{9}, (b) 80, (c) 4200, (d) 0.002.
For the same quantity Lachlan estimates 6 \times 10^{4} and Persephone 2 \times 10^{5}. Give the order of magnitude of each, and say whether they agree to an order of magnitude.
A quantity is surely between 200 and 20\,000. Find the middle of the range on the power-of-ten scale and the ordinary average, and say how many times the low end each is.

Bounds

A town has between 1 \times 10^{4} and 2 \times 10^{4} people, each using between 100 and 300 liters of water a day. Give low and high bounds for the daily use, and the middle of that range on the power-of-ten scale.
A town needs between 5 \times 10^{4} and 2 \times 10^{5} haircuts a year, and one barber gives between 1500 and 5000 haircuts a year. What are the fewest and the most barbers the town could keep busy?
Niamh-Kate’s chain has three factors, each within a factor of 2 of the truth. Her estimate is 4 \times 10^{6}. Between what values must the true quantity lie?

Making your own estimates

Each answer shows one reasonable chain; other stated assumptions give other numbers.

How many liters of water does a person drink in a lifetime?
How many times does a person blink in a year?
How many grains of rice are in a 1 kg bag?
How many words are printed in all the books of a school library?
Rafferty-Jo wonders how long it would take to walk once around the Earth at the equator, about 40\,000 km, if the oceans were not in the way. Estimate the time in days and in years.
Using Example 4’s 3 \times 10^{6} words a person a year and a world population of about 8 billion, estimate the words spoken by everyone on Earth in a year. Someone claims the total is more than 10^{16}. Is that believable?

Reasoning

Mirabel’s estimate for the town in Example 3 was 3 \times 10^{6} liters a day. The water company later reports 4.1 million liters. Was her estimate wrong? Explain, and name an assumption that could account for the gap.
Using Example 1, explain why an estimate of this kind keeps only one or two significant figures.
Tamsin’s chain has five factors, each within a factor of 2 of the truth. Explain why her product is within a factor of 32 of the true value, and why it is usually much closer.

Answers

About 3.2 \times 10^{7} seconds. 1\ \text{year} \times \dfrac{365\ \text{days}}{1\ \text{year}} \times \dfrac{24\ \text{h}}{1\ \text{day}} \times \dfrac{60\ \text{min}}{1\ \text{h}} \times \dfrac{60\ \text{s}}{1\ \text{min}}: each unit cancels against the next factor, leaving seconds, and 365 \times 24 \times 60 \times 60 = 31\,536\,000.
About 2.3 \times 10^{4} breaths. 16 \times 60 \times 24 = 23\,040.
4.2 \times 10^{6} liters. 4000\ \text{people} \times 150\ \dfrac{\text{L}}{\text{person} \cdot \text{day}} \times 7\ \text{days} leaves liters, and 4000 \times 150 \times 7 = 4\,200\,000.
About 4.2 \times 10^{8} beats. 400 \times 525\,600 \times 2 = 420\,480\,000. A human’s beats are \dfrac{70 \times 80}{400 \times 2} = 7 times as many: a slower heart, but a far longer life.
About 2.7 \times 10^{6} words. 250 \times 30 \times 365 = 2\,737\,500.
(a) 10^{9}. (b) 10^{1}, since 80 = 8 \times 10^{1}. (c) 10^{3}, since 4200 = 4.2 \times 10^{3}. (d) 10^{-3}, since 0.002 = 2 \times 10^{-3}.
Lachlan’s is 10^{4} and Persephone’s 10^{5}, yet they agree: \dfrac{2 \times 10^{5}}{6 \times 10^{4}} \approx 3.3, less than 10. The two sit close together on either side of 10^{5}.
Middle \displaystyle \sqrt{200 \times 20\,000} = 2000, which is 10 times the low end and a tenth of the high end. Average \dfrac{200 + 20\,000}{2} = 10\,100, which is 50.5 times the low end but only about half the high end.
Low 10^{4} \times 100 = 10^{6} liters; high 2 \times 10^{4} \times 300 = 6 \times 10^{6} liters. Middle \displaystyle \sqrt{10^{6} \times 6 \times 10^{6}} = \sqrt{6} \times 10^{6} \approx 2.4 \times 10^{6} liters.
Between about 10 and 130. Fewest: low demand over the busiest barbers, \dfrac{5 \times 10^{4}}{5000} = 10. Most: high demand over the least busy, \dfrac{2 \times 10^{5}}{1500} \approx 133, about 1.3 \times 10^{2}.
Between 5 \times 10^{5} and 3.2 \times 10^{7}. The product is within a factor of 2^{3} = 8, and \dfrac{4 \times 10^{6}}{8} = 5 \times 10^{5}, 4 \times 10^{6} \times 8 = 3.2 \times 10^{7}.
About 6 \times 10^{4} liters. Assuming 2 liters a day for 80 years: 2 \times 365 \times 80 = 58\,400.
About 5 \times 10^{6} blinks. Assuming 15 blinks a minute for 16 waking hours a day: 15 \times 60 \times 16 \times 365 = 5\,256\,000.
About 5 \times 10^{4} grains. Assuming a grain weighs about 20 milligrams: 1\ \text{kg} = 10^{6}\ \text{mg}, and \dfrac{10^{6}\ \text{mg}}{20\ \text{mg per grain}} = 5 \times 10^{4}.
About 9 \times 10^{8} words, close to a billion, order of magnitude 10^{8}. Assuming 10^{4} books of 300 pages with 300 words a page: 10^{4} \times 300 \times 300 = 9 \times 10^{8}.
About 1000 days, a little under 3 years. Walking 5 km an hour for 8 hours a day gives 5 \times 8 = 40 km a day; \dfrac{40\,000}{40} = 1000 days, and \dfrac{1000}{365} \approx 2.7 years.
About 2.4 \times 10^{16}, so the claim is believable. 3 \times 10^{6} \times 8 \times 10^{9} = 2.4 \times 10^{16}, and the total stays above 10^{16} unless the true average per person is below \dfrac{10^{16}}{8 \times 10^{9}} = 1.25 \times 10^{6} words, less than half the estimate.
Not wrong. \dfrac{4.1 \times 10^{6}}{3 \times 10^{6}} \approx 1.4, and 4.1 million lies inside her bounds of 2 to 6 million. “Homes only” could account for the gap: shops, schools, factories and leaks also use water.
The inputs, about 70 beats a minute and about 80 years, are good to one or two figures at most; an equally fair 60 beats changes the answer by a seventh. So 2\,943\,360\,000 claims a precision the assumptions do not have, while 3 \times 10^{9} says exactly what is known.
Each estimate lies between half and twice its true factor, so multiplying the five sets of bounds puts the product between \dfrac{1}{32} and 32 times the true product, since 2^{5} = 32. That needs all five errors to go the same way; guesses made separately usually split high and low and partly cancel.