February 22, 2003

100 Interesting Mathematical Calculations, Puzzles, and Amusements, Number 13: Introduction to Compound Growth

One Hundred Interesting... Index Page

Introduction to Compound Growth

"The problem is: 'Suppose there is a town with a population of 10,000. Its population grows at one percent per year. How large a population will it have in five years?' I guess that since .01 x 5 = 0.05, and since 0.05 x 10,000 = 500, that the answer is 10,500."

"Ah. They've tricked you. Do it year by year. What's one percent of 10,000? And how large will its population be after one year?"

"100, and 10,100."

"Right. Now what's one percent of 10,100?"

"101. Oh!"

"Yes. The population grows by 100 during the first year. And the population grows by 101 during the second year. What's the population after two years?"


"And how much does the population grow during the third year? Round the population to the nearest whole person--because that's what it has to be."

"102. So there are 10,303 people after three years?"

"Yes. And 10,406 people after four years."

"And 10,510 people after five years. Still, ten extra people doesn't seem all that many."

"Ah. But as growth compounds the differences add up. How long would you say it would take the population to double to 20,000--if it grows at a steady one percent per year?"

"Well, if you were only adding 100 people a year, it would take 100 years. But it's clearly less than that because of this compounding--the fact that by the time you've reached 15,000 people you're adding not 100 but 150 people a year. I don't know. 90 years?"

"70. After 70 years you have 20,067 people. How many do you think you'll have after 100?"

"This is what computers are good for: (1.01)^(100) = 2.7048. So after 100 years you'll have 27,048 people... Very close to the original 10,000 x e, that number 2.718281828... we have seen before..."

Posted by DeLong at February 22, 2003 01:45 PM | TrackBack


If you can teach compound growth to everybody and have
it stick, you'll deserve the medal of your choice. :-)

If that is your strategy here, though, I think I would
take out the "e" stuff unless you explicitly mention
before or after that what you just did with (1.01)^100
was calculate (1+1/x)^x for a medium size x, and that
that as x grows without bound...etc.

Posted by: Jonathan King on February 22, 2003 08:58 PM


James Grant used a cute example of compound interest in one of his recent Fortune columns (on whether or not to buy gold). He figures out how much the ancient investor, who bought and hid gold under his floor, would have earned had he invested the money instead and compounded the interest.

Posted by: Andrew Boucher on February 22, 2003 10:55 PM


Rule of 70. Because ln(2) is .693, and interest rates are multiples of .01, we can take 69.3 (or 70) divided by the rate and get the doubling period. That makes it easy to estimate.

You can create a young Malthusian by asking "You have a colony of bacteria filling half of a petrie dish. The population doubles every minute. How long 'til they completely fill the dish?"

Posted by: Brennan Peterson on February 23, 2003 10:14 PM


An important application concerns the current claims for restitution by early inhabitants whose land was taken over by invading Europeans. Suppose an early settler in California was granted 100,000 acres of the San Francisco Bay Area shoreline by the Spanish Crown, for services rendered. Suppose that we now decide to pay the original Native American tribes back for this 'stolen' land. Suppose that a fair price at the time of the takeover would have been 10c/acre, i.e., $10,000 in the currency of that time. Assume that was 150 years ago, and a rate of inflation of 2%. What should we pay them in current dollars? Answer:........

Comment; you think taxpayers can't afford that? Right; but there are volunteers who will contribute that amount without charge to taxpayers. Just give the tribes the rights to build casinos.

Posted by: michael scriven on December 8, 2003 10:10 AM


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