# Exponential Decay with Replenishment

I am working on a problem where 10% of a group of 10500 objects is diminished per year. However, 1000 new objects are added to the group each year, and I have to find the "long-term viability" of the group.

The issue is that I don't know how to integrate the replenishing factor into the exponential decay function. Is there some way to do this?

Thanks!

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Let $a_n$ be the number of objects after $n$ years. Then you have $$a_0=10500,\qquad a_n=(.9)a_{n-1}+1000$$ Have you studied such linear recurrences?

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Well my only issue is that using this model would require that I calculate the number of objects recursively, wouldn't I? Because I am adding 1000 to .1 of the population (.9 left), I would have to know the value after each decay, right? –  nmagerko Sep 8 '12 at 3:05
@nmagerko: If we define $b_n=a_n-10000$, the recurrence becomes $b_n=0.9b_{n-1}$ with the solution $b_n=b_0\cdot 0.9^n$, so $a_n=10000+500\cdot 0.9^n$ –  Ross Millikan Sep 8 '12 at 3:11
You could use this model to calculate the number of objects recursively, but if you have studied linear recurrences you may have learned that you can solve them and get a formula for $a_n$ that depends only on $n$ (in particular, not on previous values). Even if you haven't studied the topic, you mau be able to figure out the solution on your own, by doing a few iterations, guessing the pattern, and then proving it. Also, if all that interest you is the long-term viability, just think about solving $a_{n+1}=a_n$, and the implications of a solution. –  Gerry Myerson Sep 8 '12 at 3:15

For a simple approach, if there is a stable population, it must satisfy $n=0.9n+1000$. This is easily solved, giving $n=10000$ You can prove that if the population is greater than 10000 in any given year, it will be greater next year, but closer. Similarly, (not important for your problem), if the population is less than 10000 one year, it will be less than 10000 next year, but closer. So the population converges on 10000, regardless of the start.

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Well just doing the first two years by hand, I am taking .9 of 10500 and adding 1000, taking .9 of that result and adding 1000, seeing a significant decrease in the number of objects. Am I understanding your response incorrectly? –  nmagerko Sep 8 '12 at 3:08
@nmagerko: I get 10450 for the first year and 10405 for the second. Is that what you get? –  Ross Millikan Sep 8 '12 at 3:13
Yes, that's what I get. I think your comment on the other answer helped me figure out what you were saying! –  nmagerko Sep 8 '12 at 3:17