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Re: Math help

The equation for interest is an exponential. We can reverse the order--start at 265 and let it rise to 937. The length of time should be the same. The equation is 265*e^(0.015*t); where t is in years. We then set the equation to 937 and solve for t.

937 = 265*e^(0.015*t)
ln(937) = ln(265) + 0.015*t
t = (ln(937) - ln(265))/0.015 = 84.1969 years.

Re: Re: Math help

The posing of the problem is slightly ambiguous. There are three possible interpretations. Our accountant friends would adopt the discrete case. Assuming the annual change applies to the remainder rather than the principal value and that it is applied once a year (cf compound interest) the words translate to:
n = N (1+r)^y
And the solution for y is:
y = (log n – log N)/log(1+r)
Given the values n = 265, N = 937, r = -0.015, then y = 83 and a bit years (i.e. 84)

The alternative interpretation (cf simple interest) gives:
n = N(1+ry)
And:
y = (n-N)/Nr
Putting in the numbers yields y = 47.81 years.
For the continuous case, which we engineers would assume, the solution is as above. The solutions agree because the series for ln(1+r) is of the form r + higher terms and in this case r is small.

Re: Re: Re: Math help

John,

It's not simple interest. There is an annual compounding, so using the continuous formula for compound interest gives a value that's too low.

The correct equation is P*((1+r/n)^(t*n)); where P is the principal, r is the interest, n is number of times the interest is compounded during the year, and t is time in years. The limit of this equation as n goes to infinity is P*e^(t*r).

We can set this equation to 937 and solve for t:

937 = 265*(1+0.015/n)^(t*n)
ln(937) = ln(265) + t*n*ln(1+0.015/n)
t = (ln(937) - ln(265))/(n*ln(1+0.015/n)

n appears to be 1, so we have t = 84.8268.

Jim

Re: Re: Math help

I get a slightly different answer.

937 (.985)^t = 265
t ln .985 = ln 265 - ln 937
t = 83.6 years.

Not that anyone should get excited over half a year in 80.

I was perplexed momentarily at the difference, then realized as usual that I was probably being unduly digital.

ln 1.015 is 0.0149 and the ln 0.985 is -0.0151... My memory of logarhythms isn't 100% so I am not sure if its a digital thing (rounding issues for small numbers) or a approximation difference nor do I know which one is the approximation, and in the end who really gives a ****.

....

Re: Re: Re: Math help

When I plug the values into an Excel worksheet, yours and John's answer is correct. Reversing the order seems to add a year.

This is a trick question, since the value of 265 is never obtained. At 83 years, the value is slightly above 265, and at 84 years the value is slightly below 265. We're compounding at one-year intervals.

Jim

Re: Re: Re: Re: Math help

In 7th grade, they taught us the sales discount equation. I learned then that 40% off is not the same as 40% more than the discounted price. You are usually in the same ballpark, but you might be short of the wall by a few meters (or maybe many) meters.

Re: Re: Re: Math help

My mistake is falling for the continuity trap. If we put the non-swapped values into the continuous formula, we get the same answer as before:

265 = 937*e^(-0.015*t)
ln(265) = ln(937) -0.015*t
t = (ln(265 – ln(937))/(-0.015) = 84.1969 years

However, putting the correct values into the correct formula give us a slightly different answer. If we set A = 265, P = 937, r = -1.5%, and n = 1; we get:

265 = 937*(1 -0.015/1)^(1*t)
ln(265) = ln(937) +t*ln(0.985)
t = (ln(265) – ln(937))/ln(0.985) = 83.5638 years.

But we can’t get 265 exactly. We are slightly above it at 83 years and slightly below it at 84 years.

Jim

you can also use a spreadsheet

If you don't know math, you can also calculate it using excel:
type 937 in box A1
type =A1-0.015*A1 in box A2
drag down
in box A85 you'll see the number 263.2588
So after 84 years.

Re: you can also use a spreadsheet

If the rate of reduction was .5% instead of 1.5%, would the number of years be increased 3 fold to about 255 years?

Re: Re: you can also use a spreadsheet

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