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We're trying to figure out something, and things aren't adding up. The senario is that you make $\$ 50,000$ a year. Every year you get a $15\%$ bonus of that income, which then gets added to your next year's income. So, the first year, you get $\$ 7,500$ which then makes your base $\$ 57,500$ the next year. When we try to figure out what the bonus will be past $6$ or $7$ years the numbers just sort of plateau at around $\$ 8,823$. All we were doing is adding the amount of the bonus to $\$ 50,000$ and then finding $15\%$ of the resulting number, and repeating. Is this correct? Why are the numbers plateauing? What would the bonus actually be after $6$ or $7$ years? I'm sorry if this is a simple question for this forum, it was recommended to me. Thanks so much in advance.

callculus42
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Nicholas Compton
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  • Welcome to MSE! Could you show us some more of that data so we can try to understand what might be going on better? It seems to me like one way to model the scenario you're talking about is with the recurrence relation $x_{n+1}=1.15x_n$, where $x_0=50000$ and $x_n$ is the income you make in the $n$th year after the bonuses start. Is that accurate? – Robert Howard Aug 13 '18 at 06:09
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    If your values are not growing, you're probably making a typo in the calculator. As far as I know, the function $$ f(x) = 50000 \times 1.15^x $$ is an increasing function ... – Matti P. Aug 13 '18 at 06:10
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    The base would be $57,500$ for the second year, not $57,000$ – Ross Millikan Aug 13 '18 at 16:36

3 Answers3

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You start with a bonus of 7500. Then every year the bonus decreases about $85\%$. But you carry over the bonuses of the previous years. That means that at the first year we have $b_1=7500$. And in the second year $b_2=7500+0.15\cdot 7500=8625$ and so on:

$b_3=7500+0.15\cdot 7500+0.15^2\cdot 7500$

...

$b_n=7500\sum_{i=1}^{n} 0.15^{i-1}$

We can use the formula for the partial sum of a geometric series to calculate $b_n$

$$b_n=\frac{7500}{0.15}\sum_{i=1}^{n} 0.15^{i}=\frac{7500}{0.15}\cdot 0.15\cdot \frac{1-0.15^n}{1-0.15}=7500\cdot \frac{1-0.15^n}{1-0.15}$$

And for $$ n \to \infty, b_n=\frac{7500}{1-0.15}\approx 8823.53$$

This is the upper bound of b$_n$

The graph below shows that the increase of the bonus is large in the first $3$ years and it is very close to the upper bound in year 4: $b_4=8819.06$

enter image description here

callculus42
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I am not sure this is what you want, but here is the calculation for the bonuses in the first $3$ years which should get you going:

1st Year:

$$ \text{Base} = 50,000 \quad \Rightarrow \quad \text{Bonus}=0.15*50,000 = 7,500. $$

2nd Year:

$$ \text{Base} = 50,000+7,500=57,500 \quad \Rightarrow \quad \text{Bonus}=0.15*57,500 = 8,625. $$

3rd Year:

$$ \text{Base} = 57,500+8,625=66,125 \quad \Rightarrow \quad \text{Bonus}=0.15*57,500 \approx 9,919. $$

mzp
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  • The third year base would be 50,000 + 8,625 the amount made in the year is always 50,000 and the bonus is 15% of the amount made, plus the bonus from the year before. – Nicholas Compton Aug 14 '18 at 14:02
  • If that is the case, then the bonus would indeed reach a limit. If you know that should not be the case, I suspect that is probably the mistake you are making. – mzp Aug 14 '18 at 14:35
  • @NicholasCompton Could you comprehend what I´ve written in my answer? – callculus42 Dec 03 '20 at 15:32
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The general rule is that in year $n$ the base is $50,000\cdot 1.15^n$ and the bonus is $7,500\cdot 1.15^n$ This increases without bound. enter image description here

Ross Millikan
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