Percentage Plateau
Clash Royale CLAN TAG#URR8PPP
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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 $$ 50,000$ and then finding $15%$ of the resulting number, and repeating. Is this correct? Why are the numbers plateauing? What would the bonus actually be after $6$ or $7$ years? I'm sorry if this is a simple question for this forum, it was recommended to me. Thanks so much in advance.
economics
edited Aug 14 at 14:51
callculus
16.5k31427
asked Aug 13 at 5:17
Nicholas Compton
111
add a comment |Â
up vote
1
down vote
favorite
We're trying to figure out something, and things aren't adding up. The senario is that you make $$ 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.
economics
edited Aug 14 at 14:51
callculus
16.5k31427
asked Aug 13 at 5:17
Nicholas Compton
111
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 at 6:09
1
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 at 6:10
1
The base would be $57,500$ for the second year, not $57,000$
– Ross Millikan
Aug 13 at 16:36
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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 $$ 50,000$ and then finding $15%$ of the resulting number, and repeating. Is this correct? Why are the numbers plateauing? What would the bonus actually be after $6$ or $7$ years? I'm sorry if this is a simple question for this forum, it was recommended to me. Thanks so much in advance.
economics
edited Aug 14 at 14:51
callculus
16.5k31427
asked Aug 13 at 5:17
Nicholas Compton
111
We're trying to figure out something, and things aren't adding up. The senario is that you make $$ 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.
economics
edited Aug 14 at 14:51
callculus
16.5k31427
asked Aug 13 at 5:17
Nicholas Compton
111
edited Aug 14 at 14:51
callculus
16.5k31427
edited Aug 14 at 14:51
callculus
16.5k31427
edited Aug 14 at 14:51
callculus
16.5k31427
16.5k31427
asked Aug 13 at 5:17
Nicholas Compton
111
asked Aug 13 at 5:17
Nicholas Compton
111
asked Aug 13 at 5:17
Nicholas Compton
111
111
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 at 6:09
1
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 at 6:10
1
The base would be $57,500$ for the second year, not $57,000$
– Ross Millikan
Aug 13 at 16:36
add a comment |Â
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 at 6:09
1
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 at 6:10
1
The base would be $57,500$ for the second year, not $57,000$
– Ross Millikan
Aug 13 at 16:36
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 at 6:09
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 at 6:09
1
1
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 at 6:10
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 at 6:10
1
1
The base would be $57,500$ for the second year, not $57,000$
– Ross Millikan
Aug 13 at 16:36
The base would be $57,500$ for the second year, not $57,000$
– Ross Millikan
Aug 13 at 16:36
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
0
down vote
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:
$$ textBase = 50,000 quad Rightarrow quad textBonus=0.15*50,000 = 7,500. $$
2nd Year:
$$ textBase = 50,000+7,500=57,500 quad Rightarrow quad textBonus=0.15*57,500 = 8,625. $$
3rd Year:
$$ textBase = 57,500+8,625=66,125 quad Rightarrow quad textBonus=0.15*57,500 approx 9,919. $$
answered Aug 13 at 16:35
mzp
1,15611134
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 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 at 14:35
add a comment |Â
up vote
0
down vote
The general rule is that in year $n$ the base is $50,000cdot 1.15^n$ and the bonus is $7,500cdot 1.15^n$ This increases without bound.
answered Aug 13 at 16:38
Ross Millikan
277k21187352
add a comment |Â
up vote
0
down vote
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 $b_2=7500+0.15cdot 7500=8625$ and so on:
$b_3=7500+0.15cdot 7500+0.15^2cdot 7500$
...
$b_n=7500sum_i=1^n 0.15^i-1$
We can use the formula for the partial sum of a geometric series to calculate $b_n$
$$b_n=frac75000.15sum_i=1^n 0.15^i=frac75000.15cdot 0.15cdot frac1-0.15^n1-0.15=7500cdot frac1-0.15^n1-0.15$$
And for $$ n to infty, b_n=frac75001-0.15approx 8823.53$$
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: $b_4=8819.06$
answered Aug 14 at 15:58
callculus
16.5k31427
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
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:
$$ textBase = 50,000 quad Rightarrow quad textBonus=0.15*50,000 = 7,500. $$
2nd Year:
$$ textBase = 50,000+7,500=57,500 quad Rightarrow quad textBonus=0.15*57,500 = 8,625. $$
3rd Year:
$$ textBase = 57,500+8,625=66,125 quad Rightarrow quad textBonus=0.15*57,500 approx 9,919. $$
answered Aug 13 at 16:35
mzp
1,15611134
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 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 at 14:35
add a comment |Â
up vote
0
down vote
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:
$$ textBase = 50,000 quad Rightarrow quad textBonus=0.15*50,000 = 7,500. $$
2nd Year:
$$ textBase = 50,000+7,500=57,500 quad Rightarrow quad textBonus=0.15*57,500 = 8,625. $$
3rd Year:
$$ textBase = 57,500+8,625=66,125 quad Rightarrow quad textBonus=0.15*57,500 approx 9,919. $$
answered Aug 13 at 16:35
mzp
1,15611134
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 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 at 14:35
add a comment |Â
up vote
0
down vote
up vote
0
down vote
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:
$$ textBase = 50,000 quad Rightarrow quad textBonus=0.15*50,000 = 7,500. $$
2nd Year:
$$ textBase = 50,000+7,500=57,500 quad Rightarrow quad textBonus=0.15*57,500 = 8,625. $$
3rd Year:
$$ textBase = 57,500+8,625=66,125 quad Rightarrow quad textBonus=0.15*57,500 approx 9,919. $$
answered Aug 13 at 16:35
mzp
1,15611134
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:
$$ textBase = 50,000 quad Rightarrow quad textBonus=0.15*50,000 = 7,500. $$
2nd Year:
$$ textBase = 50,000+7,500=57,500 quad Rightarrow quad textBonus=0.15*57,500 = 8,625. $$
3rd Year:
$$ textBase = 57,500+8,625=66,125 quad Rightarrow quad textBonus=0.15*57,500 approx 9,919. $$
answered Aug 13 at 16:35
mzp
1,15611134
answered Aug 13 at 16:35
mzp
1,15611134
answered Aug 13 at 16:35
mzp
1,15611134
answered Aug 13 at 16:35
mzp
1,15611134
1,15611134
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 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 at 14:35
add a comment |Â
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 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 at 14:35
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 at 14:02
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 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 at 14:35
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 at 14:35
add a comment |Â
up vote
0
down vote
The general rule is that in year $n$ the base is $50,000cdot 1.15^n$ and the bonus is $7,500cdot 1.15^n$ This increases without bound.
answered Aug 13 at 16:38
Ross Millikan
277k21187352
add a comment |Â
up vote
0
down vote
The general rule is that in year $n$ the base is $50,000cdot 1.15^n$ and the bonus is $7,500cdot 1.15^n$ This increases without bound.
answered Aug 13 at 16:38
Ross Millikan
277k21187352
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The general rule is that in year $n$ the base is $50,000cdot 1.15^n$ and the bonus is $7,500cdot 1.15^n$ This increases without bound.
answered Aug 13 at 16:38
Ross Millikan
277k21187352
The general rule is that in year $n$ the base is $50,000cdot 1.15^n$ and the bonus is $7,500cdot 1.15^n$ This increases without bound.
answered Aug 13 at 16:38
Ross Millikan
277k21187352
answered Aug 13 at 16:38
Ross Millikan
277k21187352
answered Aug 13 at 16:38
Ross Millikan
277k21187352
answered Aug 13 at 16:38
Ross Millikan
277k21187352
277k21187352
add a comment |Â
add a comment |Â
up vote
0
down vote
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 $b_2=7500+0.15cdot 7500=8625$ and so on:
$b_3=7500+0.15cdot 7500+0.15^2cdot 7500$
...
$b_n=7500sum_i=1^n 0.15^i-1$
We can use the formula for the partial sum of a geometric series to calculate $b_n$
$$b_n=frac75000.15sum_i=1^n 0.15^i=frac75000.15cdot 0.15cdot frac1-0.15^n1-0.15=7500cdot frac1-0.15^n1-0.15$$
And for $$ n to infty, b_n=frac75001-0.15approx 8823.53$$
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: $b_4=8819.06$
answered Aug 14 at 15:58
callculus
16.5k31427
add a comment |Â
up vote
0
down vote
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 $b_2=7500+0.15cdot 7500=8625$ and so on:
$b_3=7500+0.15cdot 7500+0.15^2cdot 7500$
...
$b_n=7500sum_i=1^n 0.15^i-1$
We can use the formula for the partial sum of a geometric series to calculate $b_n$
$$b_n=frac75000.15sum_i=1^n 0.15^i=frac75000.15cdot 0.15cdot frac1-0.15^n1-0.15=7500cdot frac1-0.15^n1-0.15$$
And for $$ n to infty, b_n=frac75001-0.15approx 8823.53$$
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: $b_4=8819.06$
answered Aug 14 at 15:58
callculus
16.5k31427
add a comment |Â
up vote
0
down vote
up vote
0
down vote
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 $b_2=7500+0.15cdot 7500=8625$ and so on:
$b_3=7500+0.15cdot 7500+0.15^2cdot 7500$
...
$b_n=7500sum_i=1^n 0.15^i-1$
We can use the formula for the partial sum of a geometric series to calculate $b_n$
$$b_n=frac75000.15sum_i=1^n 0.15^i=frac75000.15cdot 0.15cdot frac1-0.15^n1-0.15=7500cdot frac1-0.15^n1-0.15$$
And for $$ n to infty, b_n=frac75001-0.15approx 8823.53$$
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: $b_4=8819.06$
answered Aug 14 at 15:58
callculus
16.5k31427
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 $b_2=7500+0.15cdot 7500=8625$ and so on:
$b_3=7500+0.15cdot 7500+0.15^2cdot 7500$
...
$b_n=7500sum_i=1^n 0.15^i-1$
We can use the formula for the partial sum of a geometric series to calculate $b_n$
$$b_n=frac75000.15sum_i=1^n 0.15^i=frac75000.15cdot 0.15cdot frac1-0.15^n1-0.15=7500cdot frac1-0.15^n1-0.15$$
And for $$ n to infty, b_n=frac75001-0.15approx 8823.53$$
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: $b_4=8819.06$
answered Aug 14 at 15:58
callculus
16.5k31427
edited Aug 14 at 16:04
answered Aug 14 at 15:58
callculus
16.5k31427
answered Aug 14 at 15:58
callculus
16.5k31427
answered Aug 14 at 15:58
callculus
16.5k31427
16.5k31427
add a comment |Â
add a comment |Â
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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 at 6:09
1
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 at 6:10
1
The base would be $57,500$ for the second year, not $57,000$
– Ross Millikan
Aug 13 at 16:36