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Half Life of Money


Date: 04/13/2001 at 00:39:30
From: Grant Davis
Subject: Half Life of money

Dear Dr. Math:

I have come across two formulas for determining the amount of time it 
takes for money to lose half of its value due to inflation, etc. The 
first one is the common decay formula:

     t = half life
     r = interest rate (e.g. 9% = .09)

The formula is:

     t = -log.5/-r

This formula is also written as:

     t = log2/r

The second formula I found in a financial mathematics book written in 
the 1979 _Handbook of Financial Mathematics, Formulas and Tables_ by 
Vichas. He posts the formula for financial half-life as follows:

     t = log2/log(1+r)

The formulas give different answers. (They differ by about one year.) 
My question is, which formula is the standard accepted practice and 
why?


Date: 04/13/2001 at 04:27:13
From: Doctor Mitteldorf
Subject: Re: Half Life of money

Dear Grant,

I like to think concretely. Start with the idea that the money is 
worth $1 the 0th year, $0.91 after 1 year, (0.91)^2 = $0.828 after 2 
years, then $0.753, $0.685, $0.624, $0.568, and $0.517 after the 7th 
year. So the answer comes out a little more than 7 years.

We've assumed here that the money loses 9% of its value each year, 
that is, its value each year is 91% of what it was the year before, 
where 91% is 1-r. The question we're asking is, 0.91 raised to what 
power equals 0.5? The answer is log(.5) divided by log(0.91) = 7.35 
years.

On the other hand, there's enough ambiguity in the meaning of 
inflation to say that each year the value of the money is not 
multiplied by 0.91 but divided instead by 1.09. The difference is 
small but not too small, since 1/1.09 = 0.9174.  Using this figure, we 
get:

     log(0.5) / log(0.9174) = 8.04 years

Since log(1/(1+r)) is the same as -log(1+r), this is equivalent to 
your second formula. Your first formula, I believe, is my first method 
mistyped, using (1-r) for the factor each year.

So the difference amounts to the question, what do we mean by the 
inflation rate? If we mean that at the end of the year it takes $1 to 
buy what we could have bought for $0.91 at the beginning, then we use 
the first formula. If we mean that at the end of the year it takes 
$1.09 to buy what we could have bought for $1 at the beginning, then 
we use the second formula.

I believe the second is the more conventional of the two. Go with 
log(2)/log(1+r).

- Doctor Mitteldorf, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Exponents
High School Interest

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