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Saturday
Mar202010

We Don't Need No 37-cent Piece (But 30- and 45-cent Pieces Might be Nice)

Last night I was reading this Freakonomics blog explaining why a 37-cent piece would make for more efficient US coinage. In the article the question asked was what set of 4 different coin denominations could most efficiently be used to make up any amount between 1c and 99c. Two equally efficient answers were found: a set comprising a 1-cent, 3-cent, 11-cent and 37-cent piece, and one comprising a 1-cent, 3-cent, 11-cent and 38-cent piece. Either combination can be used to produce any total between 1c and 99c using, on average, just 4.1 coins.

Well Australia's different from the US in oh so many ways, and one of those ways is relevant for the present topic: we round all amounts to the nearest 5 cents, having disposed of the 1- and 2-cent pieces in 1991.

So, I wondered, what set of 4 coins would most efficiently meet our needs.

Just so you're clear what I'm on about, consider the 4 coin set comprising a 5 cent, 10 cent, 70 cent and 90 cent piece. A transaction totalling 5 cents can be met with just 1 coin, a transaction of 10 cents can also be met with just one coin, and a transaction of, say, 65 cents can be most efficiently met with 7 coins (1 5-cent piece and 6 10-cent pieces). To determine the overall efficiency of the (5,10,70,90) coin set we calculate how many coins would be needed to meet each transaction size from 5 cents to 95 cents (in 5 cent increments) and we average each of the 19 estimates so obtained. (The answer is 3.16 coins per transaction for this particular combination of denominations, which makes it a fairly inefficient combination, not surprising given that the 70- and 90-cent pieces are useful in so few of the 19 transaction sizes.)

For Australian conditions, it turns out, we'd need to substitute at least two of our current four sub-dollar denominations (viz the 5c, 10c, 20c and 50c pieces) to create an optimal set. Nine solutions are all equally efficient, each requiring an average of about 2.11 coins to meet every amount from 5 cents to 95 cents incrementing in 5 cent lots. The optimal coin sets are:
  • A 5,10,30 and 45 cent solution
  • A 5,15,20 and 45 cent solution
  • A 5,15,35 and 40 cent solution
  • A 5,15,35 and 45 cent solution
  • A 5,15,35 and 60 cent solution
  • A 5,15,40 and 45 cent solution
  • A 5,20,30 and 65 cent solution
  • A 5,20,35 and 45 cent solution
So, Aussies don't need to consider a 37-cent coin, we need to ponder 15-, 35-, 40- and 45-cent coins.

Maybe that's a little too much change (if you'll forgive the dreadful pun). As I noted earlier, each of the optimal solutions listed above necessitates our changing at least two of our existing coin denominations. If you'd prefer that we change only one, the best solution is the (5,10,20,45) set, which is only marginally less efficient than the optimum, requiring an average of 2.21 coins for each transaction in the 5 cent to 95 cent range.

Another sub-optimal but, I contend, attractive option is the (5,10,30,75) set, which needs an average of only 2.16 coins per transaction and which includes a 75-cent piece that would surely come in handy for purchases over a dollar.

Finally, you might be curious how inefficient our current (5,10,20,50) coin set is. It's not too bad, requiring 2.32 coins per transaction, which makes it about 10% less efficient than the optimum.

So the next time you're weighed down with a purse, wallet or pocket full of coins, just think how much more efficient it would be if some of those coins were 30 and 45 cent pieces (and think how much more fun it would be waiting for someone at the checkout to pause, look skyward, give up and then scan a reference sheet to find out how to provide you with the correct change).

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