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Let's say I have a utility rate of 0.15 which inflates by 0.5% per year. I then want to convert that annual rate to a monthly rate and determine what the rate would be for month $n$.

The way I'm doing it now (which just feels incorrect) is: $$\mathrm{utilityRate}\Bigl( (1.0 + \text{rate of inflation})^{n/12}\Bigr)$$

Is this correct, or am I completely off base?

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A rate of $n/12$ compounded monthly gives you a larger rate than $n$ per year. –  Arturo Magidin Jul 2 '12 at 22:26
    
The number of months could be anything... in this particular case, I need to know the rate of month n where n = 1...300 –  Chris Cashwell Jul 2 '12 at 22:29
    
@Arturo: I though $n$ was the number of months rather than the rate. –  Henry Jul 2 '12 at 22:33
    
@Hnery; I started the comment before cleaning up the LaTeX, and misinterpreted; then I didn't correct the comment before sending. –  Arturo Magidin Jul 2 '12 at 22:34

1 Answer 1

up vote 2 down vote accepted

If the rate of inflation is constant every month then $(1 + \text{annual rate of inflation})^{n/12}$ is indeed what happens to the price level starting at $1$. So your formula looks sensible.

So in your example with $0.5\%$ annual inflation, this becomes $1.005^{n/12}$. The utilities I know would only change rates once a year if inflation was so low.

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I believe the rate of inflation (being yearly initially) should be divided by 12 in order to get the monthly inflation rate. Does that sound about right? –  Chris Cashwell Jul 2 '12 at 23:38
    
@ChrisCashwell: it will be close, particularly if the inflation rate is low. The correct monthly rate is as you implied in your question: $(1.0 + \text{annual rate of inflation})^{1/12}-1$. To the extent that $(1+\text{monthly rate})^{12}\approx 1+12*\text{monthly rate}$ they will agree.The next term is $6(\text{monthly rate})^2$, so you can see if you think that is big enough to care about. –  Ross Millikan Jul 2 '12 at 23:54

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