exponential growth saving function
$begingroup$
I'm figuring out some kind of money saving function where the amount, in the end, should be 100.000 value with a time period of 10 years.
I thought of an exponential growth function since people start with jobs and with work experience, you will earn more with results into a higher amount that you can save.
With some search, I found this formula: $P(t) = P0 e^r*e^t$ (I can't format this propper) Where $P(t) = 100.000$ , $r=0.5298$, $P0=500$ and $t=10$
This will lead to an insane amount of money in year 9 and 10.
I'm looking for a more reasonable function where you start with the least amount in year 0 and the most amount in year 10 so the amount that you saved is $100.000$. But it should be reasonable like you can't save 20k in year 10. Also, $P0$ may differ for each person, but the time and the end goal is fixed. I do not take interest into account.
exponential-function
asked Dec 12 '18 at 20:11
HelperHelper
101
$endgroup$
add a comment |
$begingroup$
I'm figuring out some kind of money saving function where the amount, in the end, should be 100.000 value with a time period of 10 years.
I thought of an exponential growth function since people start with jobs and with work experience, you will earn more with results into a higher amount that you can save.
With some search, I found this formula: $P(t) = P0 e^r*e^t$ (I can't format this propper) Where $P(t) = 100.000$ , $r=0.5298$, $P0=500$ and $t=10$
This will lead to an insane amount of money in year 9 and 10.
I'm looking for a more reasonable function where you start with the least amount in year 0 and the most amount in year 10 so the amount that you saved is $100.000$. But it should be reasonable like you can't save 20k in year 10. Also, $P0$ may differ for each person, but the time and the end goal is fixed. I do not take interest into account.
exponential-function
asked Dec 12 '18 at 20:11
HelperHelper
101
$endgroup$
add a comment |
$begingroup$
I'm figuring out some kind of money saving function where the amount, in the end, should be 100.000 value with a time period of 10 years.
I thought of an exponential growth function since people start with jobs and with work experience, you will earn more with results into a higher amount that you can save.
With some search, I found this formula: $P(t) = P0 e^r*e^t$ (I can't format this propper) Where $P(t) = 100.000$ , $r=0.5298$, $P0=500$ and $t=10$
This will lead to an insane amount of money in year 9 and 10.
I'm looking for a more reasonable function where you start with the least amount in year 0 and the most amount in year 10 so the amount that you saved is $100.000$. But it should be reasonable like you can't save 20k in year 10. Also, $P0$ may differ for each person, but the time and the end goal is fixed. I do not take interest into account.
exponential-function
asked Dec 12 '18 at 20:11
HelperHelper
101
$endgroup$
I'm figuring out some kind of money saving function where the amount, in the end, should be 100.000 value with a time period of 10 years.
I thought of an exponential growth function since people start with jobs and with work experience, you will earn more with results into a higher amount that you can save.
With some search, I found this formula: $P(t) = P0 e^r*e^t$ (I can't format this propper) Where $P(t) = 100.000$ , $r=0.5298$, $P0=500$ and $t=10$
This will lead to an insane amount of money in year 9 and 10.
I'm looking for a more reasonable function where you start with the least amount in year 0 and the most amount in year 10 so the amount that you saved is $100.000$. But it should be reasonable like you can't save 20k in year 10. Also, $P0$ may differ for each person, but the time and the end goal is fixed. I do not take interest into account.
exponential-function
exponential-function
asked Dec 12 '18 at 20:11
HelperHelper
101
asked Dec 12 '18 at 20:11
HelperHelper
101
asked Dec 12 '18 at 20:11
HelperHelper
101
asked Dec 12 '18 at 20:11
HelperHelper
101
asked Dec 12 '18 at 20:11
HelperHelper
101
101
add a comment |
add a comment |
1 Answer
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$begingroup$
You can impose the following model:
Save $P(0)$ in year 0, $P(1) = alpha P(0)$ in year 1, $P(2) = alpha P(1) = alpha^2 P(0)$ in year 2, etc. for some $alpha>1$. So you are geometrically/exponentially increasing the amount you save each year. After 10 years you will have $P(0) + P(1) + ; ... ; + P(10)$ = $P(0)[1+ alpha + alpha^2 + ; ... ; + alpha^{10}] = P(0) frac{alpha^{10}-1}{alpha-1}$.
Say $alpha=1.5$. At the end you will have $113.33 P(0)$, which means $P(0) = 882.38$ if you want the final amount to be 100,000. With this model, you will have to save $$882.38$ in year 0, $$1323.60$ in year 1, ..., and $$5088.2$ in year 10 (which is still reasonable).
answered Dec 12 '18 at 20:32
Aditya DuaAditya Dua
1,11418
$endgroup$
$begingroup$
care to explain how you go from $113.33P(0)$ to $P(0)=882.38$ as I people start with saving 500 euro in P(0) so I get $113.33 *500 = 56665$. I know you calculate to get 100.000 in the end. But I don't quite get it
$endgroup$
– Helper
Dec 13 '18 at 20:00
$begingroup$
I simply set $113.33P(0) = 100,000$, which gives $P(0) = 883.28$. I assumed that $alpha = 1.5$ (just as an example). You can fix $P(0) = 500$, and then find an $alpha$ such that $P(0)frac{alpha^10-1}{alpha-1} = 100,000$. Is that clearer?
$endgroup$
– Aditya Dua
Dec 14 '18 at 3:33
1
$begingroup$
Yes, now I understand the formula. My thanks to you for the explanation.
$endgroup$
– Helper
Dec 15 '18 at 19:00
$begingroup$
Although If I calculate year 10 by multiplying all years before with 1.5 I get 50882.48 value. Which I call insane value.
$endgroup$
– Helper
Dec 15 '18 at 19:13
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You can impose the following model:
Save $P(0)$ in year 0, $P(1) = alpha P(0)$ in year 1, $P(2) = alpha P(1) = alpha^2 P(0)$ in year 2, etc. for some $alpha>1$. So you are geometrically/exponentially increasing the amount you save each year. After 10 years you will have $P(0) + P(1) + ; ... ; + P(10)$ = $P(0)[1+ alpha + alpha^2 + ; ... ; + alpha^{10}] = P(0) frac{alpha^{10}-1}{alpha-1}$.
Say $alpha=1.5$. At the end you will have $113.33 P(0)$, which means $P(0) = 882.38$ if you want the final amount to be 100,000. With this model, you will have to save $$882.38$ in year 0, $$1323.60$ in year 1, ..., and $$5088.2$ in year 10 (which is still reasonable).
answered Dec 12 '18 at 20:32
Aditya DuaAditya Dua
1,11418
$endgroup$
$begingroup$
care to explain how you go from $113.33P(0)$ to $P(0)=882.38$ as I people start with saving 500 euro in P(0) so I get $113.33 *500 = 56665$. I know you calculate to get 100.000 in the end. But I don't quite get it
$endgroup$
– Helper
Dec 13 '18 at 20:00
$begingroup$
I simply set $113.33P(0) = 100,000$, which gives $P(0) = 883.28$. I assumed that $alpha = 1.5$ (just as an example). You can fix $P(0) = 500$, and then find an $alpha$ such that $P(0)frac{alpha^10-1}{alpha-1} = 100,000$. Is that clearer?
$endgroup$
– Aditya Dua
Dec 14 '18 at 3:33
1
$begingroup$
Yes, now I understand the formula. My thanks to you for the explanation.
$endgroup$
– Helper
Dec 15 '18 at 19:00
$begingroup$
Although If I calculate year 10 by multiplying all years before with 1.5 I get 50882.48 value. Which I call insane value.
$endgroup$
– Helper
Dec 15 '18 at 19:13
add a comment |
$begingroup$
You can impose the following model:
Save $P(0)$ in year 0, $P(1) = alpha P(0)$ in year 1, $P(2) = alpha P(1) = alpha^2 P(0)$ in year 2, etc. for some $alpha>1$. So you are geometrically/exponentially increasing the amount you save each year. After 10 years you will have $P(0) + P(1) + ; ... ; + P(10)$ = $P(0)[1+ alpha + alpha^2 + ; ... ; + alpha^{10}] = P(0) frac{alpha^{10}-1}{alpha-1}$.
Say $alpha=1.5$. At the end you will have $113.33 P(0)$, which means $P(0) = 882.38$ if you want the final amount to be 100,000. With this model, you will have to save $$882.38$ in year 0, $$1323.60$ in year 1, ..., and $$5088.2$ in year 10 (which is still reasonable).
answered Dec 12 '18 at 20:32
Aditya DuaAditya Dua
1,11418
$endgroup$
$begingroup$
care to explain how you go from $113.33P(0)$ to $P(0)=882.38$ as I people start with saving 500 euro in P(0) so I get $113.33 *500 = 56665$. I know you calculate to get 100.000 in the end. But I don't quite get it
$endgroup$
– Helper
Dec 13 '18 at 20:00
$begingroup$
I simply set $113.33P(0) = 100,000$, which gives $P(0) = 883.28$. I assumed that $alpha = 1.5$ (just as an example). You can fix $P(0) = 500$, and then find an $alpha$ such that $P(0)frac{alpha^10-1}{alpha-1} = 100,000$. Is that clearer?
$endgroup$
– Aditya Dua
Dec 14 '18 at 3:33
1
$begingroup$
Yes, now I understand the formula. My thanks to you for the explanation.
$endgroup$
– Helper
Dec 15 '18 at 19:00
$begingroup$
Although If I calculate year 10 by multiplying all years before with 1.5 I get 50882.48 value. Which I call insane value.
$endgroup$
– Helper
Dec 15 '18 at 19:13
add a comment |
$begingroup$
You can impose the following model:
Save $P(0)$ in year 0, $P(1) = alpha P(0)$ in year 1, $P(2) = alpha P(1) = alpha^2 P(0)$ in year 2, etc. for some $alpha>1$. So you are geometrically/exponentially increasing the amount you save each year. After 10 years you will have $P(0) + P(1) + ; ... ; + P(10)$ = $P(0)[1+ alpha + alpha^2 + ; ... ; + alpha^{10}] = P(0) frac{alpha^{10}-1}{alpha-1}$.
Say $alpha=1.5$. At the end you will have $113.33 P(0)$, which means $P(0) = 882.38$ if you want the final amount to be 100,000. With this model, you will have to save $$882.38$ in year 0, $$1323.60$ in year 1, ..., and $$5088.2$ in year 10 (which is still reasonable).
answered Dec 12 '18 at 20:32
Aditya DuaAditya Dua
1,11418
$endgroup$
You can impose the following model:
Save $P(0)$ in year 0, $P(1) = alpha P(0)$ in year 1, $P(2) = alpha P(1) = alpha^2 P(0)$ in year 2, etc. for some $alpha>1$. So you are geometrically/exponentially increasing the amount you save each year. After 10 years you will have $P(0) + P(1) + ; ... ; + P(10)$ = $P(0)[1+ alpha + alpha^2 + ; ... ; + alpha^{10}] = P(0) frac{alpha^{10}-1}{alpha-1}$.
Say $alpha=1.5$. At the end you will have $113.33 P(0)$, which means $P(0) = 882.38$ if you want the final amount to be 100,000. With this model, you will have to save $$882.38$ in year 0, $$1323.60$ in year 1, ..., and $$5088.2$ in year 10 (which is still reasonable).
answered Dec 12 '18 at 20:32
Aditya DuaAditya Dua
1,11418
answered Dec 12 '18 at 20:32
Aditya DuaAditya Dua
1,11418
answered Dec 12 '18 at 20:32
Aditya DuaAditya Dua
1,11418
answered Dec 12 '18 at 20:32
Aditya DuaAditya Dua
1,11418
1,11418
$begingroup$
care to explain how you go from $113.33P(0)$ to $P(0)=882.38$ as I people start with saving 500 euro in P(0) so I get $113.33 *500 = 56665$. I know you calculate to get 100.000 in the end. But I don't quite get it
$endgroup$
– Helper
Dec 13 '18 at 20:00
$begingroup$
I simply set $113.33P(0) = 100,000$, which gives $P(0) = 883.28$. I assumed that $alpha = 1.5$ (just as an example). You can fix $P(0) = 500$, and then find an $alpha$ such that $P(0)frac{alpha^10-1}{alpha-1} = 100,000$. Is that clearer?
$endgroup$
– Aditya Dua
Dec 14 '18 at 3:33
1
$begingroup$
Yes, now I understand the formula. My thanks to you for the explanation.
$endgroup$
– Helper
Dec 15 '18 at 19:00
$begingroup$
Although If I calculate year 10 by multiplying all years before with 1.5 I get 50882.48 value. Which I call insane value.
$endgroup$
– Helper
Dec 15 '18 at 19:13
add a comment |
$begingroup$
care to explain how you go from $113.33P(0)$ to $P(0)=882.38$ as I people start with saving 500 euro in P(0) so I get $113.33 *500 = 56665$. I know you calculate to get 100.000 in the end. But I don't quite get it
$endgroup$
– Helper
Dec 13 '18 at 20:00
$begingroup$
I simply set $113.33P(0) = 100,000$, which gives $P(0) = 883.28$. I assumed that $alpha = 1.5$ (just as an example). You can fix $P(0) = 500$, and then find an $alpha$ such that $P(0)frac{alpha^10-1}{alpha-1} = 100,000$. Is that clearer?
$endgroup$
– Aditya Dua
Dec 14 '18 at 3:33
1
$begingroup$
Yes, now I understand the formula. My thanks to you for the explanation.
$endgroup$
– Helper
Dec 15 '18 at 19:00
$begingroup$
Although If I calculate year 10 by multiplying all years before with 1.5 I get 50882.48 value. Which I call insane value.
$endgroup$
– Helper
Dec 15 '18 at 19:13
$begingroup$
care to explain how you go from $113.33P(0)$ to $P(0)=882.38$ as I people start with saving 500 euro in P(0) so I get $113.33 *500 = 56665$. I know you calculate to get 100.000 in the end. But I don't quite get it
$endgroup$
– Helper
Dec 13 '18 at 20:00
$begingroup$
care to explain how you go from $113.33P(0)$ to $P(0)=882.38$ as I people start with saving 500 euro in P(0) so I get $113.33 *500 = 56665$. I know you calculate to get 100.000 in the end. But I don't quite get it
$endgroup$
– Helper
Dec 13 '18 at 20:00
$begingroup$
I simply set $113.33P(0) = 100,000$, which gives $P(0) = 883.28$. I assumed that $alpha = 1.5$ (just as an example). You can fix $P(0) = 500$, and then find an $alpha$ such that $P(0)frac{alpha^10-1}{alpha-1} = 100,000$. Is that clearer?
$endgroup$
– Aditya Dua
Dec 14 '18 at 3:33
$begingroup$
I simply set $113.33P(0) = 100,000$, which gives $P(0) = 883.28$. I assumed that $alpha = 1.5$ (just as an example). You can fix $P(0) = 500$, and then find an $alpha$ such that $P(0)frac{alpha^10-1}{alpha-1} = 100,000$. Is that clearer?
$endgroup$
– Aditya Dua
Dec 14 '18 at 3:33
1
1
$begingroup$
Yes, now I understand the formula. My thanks to you for the explanation.
$endgroup$
– Helper
Dec 15 '18 at 19:00
$begingroup$
Yes, now I understand the formula. My thanks to you for the explanation.
$endgroup$
– Helper
Dec 15 '18 at 19:00
$begingroup$
Although If I calculate year 10 by multiplying all years before with 1.5 I get 50882.48 value. Which I call insane value.
$endgroup$
– Helper
Dec 15 '18 at 19:13
$begingroup$
Although If I calculate year 10 by multiplying all years before with 1.5 I get 50882.48 value. Which I call insane value.
$endgroup$
– Helper
Dec 15 '18 at 19:13
add a comment |
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