Solve for “R” in classic annuity formula
$begingroup$
$$p=frac{c}r*(1-frac1{(r+1)^t} )$$
I'm stuck. I'm building an excel model where I will be able to put in inputs for p, c, and t... but it will need to solve for "r."
For some reason, I can't figure out the algebra.
Can you help me solve for "r"?
algebra-precalculus finance
$endgroup$
add a comment |
$begingroup$
$$p=frac{c}r*(1-frac1{(r+1)^t} )$$
I'm stuck. I'm building an excel model where I will be able to put in inputs for p, c, and t... but it will need to solve for "r."
For some reason, I can't figure out the algebra.
Can you help me solve for "r"?
algebra-precalculus finance
$endgroup$
$begingroup$
Excel has a solver add-in that is useful in situations like this.
$endgroup$
– spaceisdarkgreen
Jan 6 at 1:14
$begingroup$
I found the link for the "green" formula. Added a few lines at the bottom to show you what you could do for an incredible accuracy. Cheers.
$endgroup$
– Claude Leibovici
Jan 7 at 7:57
add a comment |
$begingroup$
$$p=frac{c}r*(1-frac1{(r+1)^t} )$$
I'm stuck. I'm building an excel model where I will be able to put in inputs for p, c, and t... but it will need to solve for "r."
For some reason, I can't figure out the algebra.
Can you help me solve for "r"?
algebra-precalculus finance
$endgroup$
$$p=frac{c}r*(1-frac1{(r+1)^t} )$$
I'm stuck. I'm building an excel model where I will be able to put in inputs for p, c, and t... but it will need to solve for "r."
For some reason, I can't figure out the algebra.
Can you help me solve for "r"?
algebra-precalculus finance
algebra-precalculus finance
edited Jan 6 at 1:13
Antoni Parellada
2,96521341
2,96521341
asked Jan 6 at 0:48
Sir Roland Baggybottom IVSir Roland Baggybottom IV
83
83
$begingroup$
Excel has a solver add-in that is useful in situations like this.
$endgroup$
– spaceisdarkgreen
Jan 6 at 1:14
$begingroup$
I found the link for the "green" formula. Added a few lines at the bottom to show you what you could do for an incredible accuracy. Cheers.
$endgroup$
– Claude Leibovici
Jan 7 at 7:57
add a comment |
$begingroup$
Excel has a solver add-in that is useful in situations like this.
$endgroup$
– spaceisdarkgreen
Jan 6 at 1:14
$begingroup$
I found the link for the "green" formula. Added a few lines at the bottom to show you what you could do for an incredible accuracy. Cheers.
$endgroup$
– Claude Leibovici
Jan 7 at 7:57
$begingroup$
Excel has a solver add-in that is useful in situations like this.
$endgroup$
– spaceisdarkgreen
Jan 6 at 1:14
$begingroup$
Excel has a solver add-in that is useful in situations like this.
$endgroup$
– spaceisdarkgreen
Jan 6 at 1:14
$begingroup$
I found the link for the "green" formula. Added a few lines at the bottom to show you what you could do for an incredible accuracy. Cheers.
$endgroup$
– Claude Leibovici
Jan 7 at 7:57
$begingroup$
I found the link for the "green" formula. Added a few lines at the bottom to show you what you could do for an incredible accuracy. Cheers.
$endgroup$
– Claude Leibovici
Jan 7 at 7:57
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
As Hello_World answered, the simplest way would be to use Newton method for finding the zero of function
$$f(r) = frac{c}r left(1-frac1{(r+1)^t}right)-p$$
If you start using $r_0=0$, the first iterate would be $r_1=frac{2 (c t-p)}{ct left(t+1right)}$ and you just need to continue until convergence to the desired precision.
The is another thing you can do since $r ll 1$. Build the Taylor series at $r=0$ to get
$$f(r)+p=c t-frac{1}{2} r (c t (t+1))+frac{1}{6} c r^2 t (t+1) (t+2)-frac{1}{24} r^3 (c t
(t+1) (t+2) (t+3))+frac{1}{120} c r^4 t (t+1) (t+2) (t+3) (t+4)-frac{1}{720}
r^5 (c t (t+1) (t+2) (t+3) (t+4) (t+5))+Oleft(r^6right)$$ and now use series reversion to get
$$color{blue}{r=x+frac{t+2}{3} x^2+frac{5 t^2+17 t+14}{36} x^3+frac{17 t^3+78 t^2+117 t+58}{270}
x^4+frac{193 t^4+1094 t^3+2301
t^2+2144 t+748 }{6480}x^5+Oleft(x^{6}right)}$$ where $color{blue}{x=frac{2 (c t-p)}{ct left(t+1right)}}$ (this is $r_1$)
Let us try using $p=100000$, $c=1000$ and $t=120$. This gives $x=frac 1 {363}$. Using the expansion above, this leads to $r=frac{31795409217001}{10210526568429660}approx 0.00311398$ while the "exact" solution obtained using Newton method would be obtained after the following iterates
$$left(
begin{array}{cc}
n & r_n \
0 & 0.000000000000 \
1 & 0.002754820937 \
2 & 0.003161445197 \
3 & 0.003114792911 \
4 & 0.003114182051 \
5 & 0.003114181946
end{array}
right)$$
As you can see, without any iterative procedure we can get a very close solution of the problem (for the working case, the relative error is about $0.0065$% !).
Adding more terms will give a better result (if you want them, just ask).
Edit
Working a bit more, the blue formula may be written in a more compact form (for an even better accuracy) using a Padé approximant
$$r=x ,frac{1+a_1x+a_2x^2}{1+b_1x+b_2x^2}$$ where
$$a_1=-frac{2 left(2 t^2+3 t+1right)}{11 t+13}qquad a_2=frac{4 t^3-3 t-1}{45 (11 t+13)}$$
$$b_1=-frac{23 t^2+53 t+32}{3 (11 t+13)}qquad b_2=frac{67 t^3+240 t^2+291 t+122}{60 (11 t+13)}$$ For the work example, this would give $r=frac{13916770468}{4468838772519}approx 0.0031141805$
Update
Asking a banker friend of mine, he mentioned an approximation he saw somewhere in the past (he does not remember when and/or where. I just found where). I is
$$color{green}{rsimeq left(left(1+frac{c}{p}right)^{frac{1}{q}}-1right)^q-1}qquad text{where} qquad color{green}{q=log_2left(1+frac 1t right)}$$ Applied to the worked example, this gives $0.00310743$.
Using this estimate as $r_0$, Newton iterates would be
$$left(
begin{array}{cc}
n & r_n \
0 & 0.003107429977 \
1 & 0.003114180156 \
2 & 0.003114181946
end{array}
right)$$ which is real fast. I suppose that one single iteration of Newton method will be more than sufficient.
You could use as a very safe solution
$$r=r_0+frac{2,f(r_0) ,f'(r_0)}{f(r_0), f''(r_0)-2, f'(r_0)^2}$$ where $r_0$ is the result of the green formula.
Applied to the worked example, this would lead to
$r=0.00311418194589$ while th exact solution would be
$r=0.00311418194600$
$endgroup$
add a comment |
$begingroup$
You could try Newton's method to solve for $r.$ Take,
$$F(r) = p-frac{c}rcdot left(1-frac1{(r+1)^t}right)$$
then your goal is to find $r$ such that $F(r)=0.$ For this you start with some choice $r_0$ and then use the following recursive definition:
$$r_{n+1}= r_n - frac{F(r_n)}{F'(r_n)}.$$
This will converge to the root of $F.$
$endgroup$
add a comment |
$begingroup$
I prefer to add a second answer.
Being just fascinated by David W. Cantrell's approximation
$$color{green}{rsimeq left(left(1+frac{c}{p}right)^{frac{1}{q}}-1right)^q-1}qquad text{where} qquad color{green}{q=log_2left(1+frac 1t right)}$$ totally inspired by it, I tried something in the same spirit
$$frac{c}r left(1-frac1{(r+1)^t}right)=pimplies 1+frac{c}{p}=1+frac{r}{1-(r+1)^{-t}}$$ Taking logarithms of both sides
$$log left(1+frac{c}{p}right)=log left(1+frac{r}{1-(r+1)^{-t}}right)$$ Now, expanding the rhs as a Taylor series at $r=0$
$$log left(1+frac{c}{p}right)=log left(1+frac{1}{t}right)+frac{r}{2}+frac{2t-5}{24} r^2 +Oleft(r^3right)$$
Neglecting the second order term, we the obtain the first approximation
$$color{blue}{r_1 =2 log left(frac{t ,(c+p)}{p, (t+1)}right)}$$
Using the complete expansion to $Oleft(r^3right)$, we then have the second approximation
$$color{blue}{r_2=frac{sqrt{1+4, alpha, r_1}-1}{2 alpha }}qquad text{where}qquad color{blue}{alpha=frac {2t-5}{12}}$$
We could even avoid quadratic equations building the simplest Padé approximant instead of the Taylor series. This gives
$$log left(1+frac{c}{p}right)=frac{log left(1+frac{1}{t}right)+frac{1}{12} left((5-2 t) log
left(1+frac{1}{t}right)+6right) r} {1+ frac{5-2t}{12} r }implies
color{blue}{r_3=frac{12,r_1}{12+(2t-5),r_1}}$$
Applied to the worked example, this would give
$$r_1=2 log left(frac{606}{605}right)approx 0.00330306$$
$$r_2=frac{2}{235} left(sqrt{9+1410 log left(frac{606}{605}right)}-3right)approx 0.00311325$$
$$r_3=frac{12 log left(frac{606}{605}right)}{6+235 log
left(frac{606}{605}right)}approx 0.00310238$$
while the exact solution is $$r=0.00311418$$
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063370%2fsolve-for-r-in-classic-annuity-formula%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As Hello_World answered, the simplest way would be to use Newton method for finding the zero of function
$$f(r) = frac{c}r left(1-frac1{(r+1)^t}right)-p$$
If you start using $r_0=0$, the first iterate would be $r_1=frac{2 (c t-p)}{ct left(t+1right)}$ and you just need to continue until convergence to the desired precision.
The is another thing you can do since $r ll 1$. Build the Taylor series at $r=0$ to get
$$f(r)+p=c t-frac{1}{2} r (c t (t+1))+frac{1}{6} c r^2 t (t+1) (t+2)-frac{1}{24} r^3 (c t
(t+1) (t+2) (t+3))+frac{1}{120} c r^4 t (t+1) (t+2) (t+3) (t+4)-frac{1}{720}
r^5 (c t (t+1) (t+2) (t+3) (t+4) (t+5))+Oleft(r^6right)$$ and now use series reversion to get
$$color{blue}{r=x+frac{t+2}{3} x^2+frac{5 t^2+17 t+14}{36} x^3+frac{17 t^3+78 t^2+117 t+58}{270}
x^4+frac{193 t^4+1094 t^3+2301
t^2+2144 t+748 }{6480}x^5+Oleft(x^{6}right)}$$ where $color{blue}{x=frac{2 (c t-p)}{ct left(t+1right)}}$ (this is $r_1$)
Let us try using $p=100000$, $c=1000$ and $t=120$. This gives $x=frac 1 {363}$. Using the expansion above, this leads to $r=frac{31795409217001}{10210526568429660}approx 0.00311398$ while the "exact" solution obtained using Newton method would be obtained after the following iterates
$$left(
begin{array}{cc}
n & r_n \
0 & 0.000000000000 \
1 & 0.002754820937 \
2 & 0.003161445197 \
3 & 0.003114792911 \
4 & 0.003114182051 \
5 & 0.003114181946
end{array}
right)$$
As you can see, without any iterative procedure we can get a very close solution of the problem (for the working case, the relative error is about $0.0065$% !).
Adding more terms will give a better result (if you want them, just ask).
Edit
Working a bit more, the blue formula may be written in a more compact form (for an even better accuracy) using a Padé approximant
$$r=x ,frac{1+a_1x+a_2x^2}{1+b_1x+b_2x^2}$$ where
$$a_1=-frac{2 left(2 t^2+3 t+1right)}{11 t+13}qquad a_2=frac{4 t^3-3 t-1}{45 (11 t+13)}$$
$$b_1=-frac{23 t^2+53 t+32}{3 (11 t+13)}qquad b_2=frac{67 t^3+240 t^2+291 t+122}{60 (11 t+13)}$$ For the work example, this would give $r=frac{13916770468}{4468838772519}approx 0.0031141805$
Update
Asking a banker friend of mine, he mentioned an approximation he saw somewhere in the past (he does not remember when and/or where. I just found where). I is
$$color{green}{rsimeq left(left(1+frac{c}{p}right)^{frac{1}{q}}-1right)^q-1}qquad text{where} qquad color{green}{q=log_2left(1+frac 1t right)}$$ Applied to the worked example, this gives $0.00310743$.
Using this estimate as $r_0$, Newton iterates would be
$$left(
begin{array}{cc}
n & r_n \
0 & 0.003107429977 \
1 & 0.003114180156 \
2 & 0.003114181946
end{array}
right)$$ which is real fast. I suppose that one single iteration of Newton method will be more than sufficient.
You could use as a very safe solution
$$r=r_0+frac{2,f(r_0) ,f'(r_0)}{f(r_0), f''(r_0)-2, f'(r_0)^2}$$ where $r_0$ is the result of the green formula.
Applied to the worked example, this would lead to
$r=0.00311418194589$ while th exact solution would be
$r=0.00311418194600$
$endgroup$
add a comment |
$begingroup$
As Hello_World answered, the simplest way would be to use Newton method for finding the zero of function
$$f(r) = frac{c}r left(1-frac1{(r+1)^t}right)-p$$
If you start using $r_0=0$, the first iterate would be $r_1=frac{2 (c t-p)}{ct left(t+1right)}$ and you just need to continue until convergence to the desired precision.
The is another thing you can do since $r ll 1$. Build the Taylor series at $r=0$ to get
$$f(r)+p=c t-frac{1}{2} r (c t (t+1))+frac{1}{6} c r^2 t (t+1) (t+2)-frac{1}{24} r^3 (c t
(t+1) (t+2) (t+3))+frac{1}{120} c r^4 t (t+1) (t+2) (t+3) (t+4)-frac{1}{720}
r^5 (c t (t+1) (t+2) (t+3) (t+4) (t+5))+Oleft(r^6right)$$ and now use series reversion to get
$$color{blue}{r=x+frac{t+2}{3} x^2+frac{5 t^2+17 t+14}{36} x^3+frac{17 t^3+78 t^2+117 t+58}{270}
x^4+frac{193 t^4+1094 t^3+2301
t^2+2144 t+748 }{6480}x^5+Oleft(x^{6}right)}$$ where $color{blue}{x=frac{2 (c t-p)}{ct left(t+1right)}}$ (this is $r_1$)
Let us try using $p=100000$, $c=1000$ and $t=120$. This gives $x=frac 1 {363}$. Using the expansion above, this leads to $r=frac{31795409217001}{10210526568429660}approx 0.00311398$ while the "exact" solution obtained using Newton method would be obtained after the following iterates
$$left(
begin{array}{cc}
n & r_n \
0 & 0.000000000000 \
1 & 0.002754820937 \
2 & 0.003161445197 \
3 & 0.003114792911 \
4 & 0.003114182051 \
5 & 0.003114181946
end{array}
right)$$
As you can see, without any iterative procedure we can get a very close solution of the problem (for the working case, the relative error is about $0.0065$% !).
Adding more terms will give a better result (if you want them, just ask).
Edit
Working a bit more, the blue formula may be written in a more compact form (for an even better accuracy) using a Padé approximant
$$r=x ,frac{1+a_1x+a_2x^2}{1+b_1x+b_2x^2}$$ where
$$a_1=-frac{2 left(2 t^2+3 t+1right)}{11 t+13}qquad a_2=frac{4 t^3-3 t-1}{45 (11 t+13)}$$
$$b_1=-frac{23 t^2+53 t+32}{3 (11 t+13)}qquad b_2=frac{67 t^3+240 t^2+291 t+122}{60 (11 t+13)}$$ For the work example, this would give $r=frac{13916770468}{4468838772519}approx 0.0031141805$
Update
Asking a banker friend of mine, he mentioned an approximation he saw somewhere in the past (he does not remember when and/or where. I just found where). I is
$$color{green}{rsimeq left(left(1+frac{c}{p}right)^{frac{1}{q}}-1right)^q-1}qquad text{where} qquad color{green}{q=log_2left(1+frac 1t right)}$$ Applied to the worked example, this gives $0.00310743$.
Using this estimate as $r_0$, Newton iterates would be
$$left(
begin{array}{cc}
n & r_n \
0 & 0.003107429977 \
1 & 0.003114180156 \
2 & 0.003114181946
end{array}
right)$$ which is real fast. I suppose that one single iteration of Newton method will be more than sufficient.
You could use as a very safe solution
$$r=r_0+frac{2,f(r_0) ,f'(r_0)}{f(r_0), f''(r_0)-2, f'(r_0)^2}$$ where $r_0$ is the result of the green formula.
Applied to the worked example, this would lead to
$r=0.00311418194589$ while th exact solution would be
$r=0.00311418194600$
$endgroup$
add a comment |
$begingroup$
As Hello_World answered, the simplest way would be to use Newton method for finding the zero of function
$$f(r) = frac{c}r left(1-frac1{(r+1)^t}right)-p$$
If you start using $r_0=0$, the first iterate would be $r_1=frac{2 (c t-p)}{ct left(t+1right)}$ and you just need to continue until convergence to the desired precision.
The is another thing you can do since $r ll 1$. Build the Taylor series at $r=0$ to get
$$f(r)+p=c t-frac{1}{2} r (c t (t+1))+frac{1}{6} c r^2 t (t+1) (t+2)-frac{1}{24} r^3 (c t
(t+1) (t+2) (t+3))+frac{1}{120} c r^4 t (t+1) (t+2) (t+3) (t+4)-frac{1}{720}
r^5 (c t (t+1) (t+2) (t+3) (t+4) (t+5))+Oleft(r^6right)$$ and now use series reversion to get
$$color{blue}{r=x+frac{t+2}{3} x^2+frac{5 t^2+17 t+14}{36} x^3+frac{17 t^3+78 t^2+117 t+58}{270}
x^4+frac{193 t^4+1094 t^3+2301
t^2+2144 t+748 }{6480}x^5+Oleft(x^{6}right)}$$ where $color{blue}{x=frac{2 (c t-p)}{ct left(t+1right)}}$ (this is $r_1$)
Let us try using $p=100000$, $c=1000$ and $t=120$. This gives $x=frac 1 {363}$. Using the expansion above, this leads to $r=frac{31795409217001}{10210526568429660}approx 0.00311398$ while the "exact" solution obtained using Newton method would be obtained after the following iterates
$$left(
begin{array}{cc}
n & r_n \
0 & 0.000000000000 \
1 & 0.002754820937 \
2 & 0.003161445197 \
3 & 0.003114792911 \
4 & 0.003114182051 \
5 & 0.003114181946
end{array}
right)$$
As you can see, without any iterative procedure we can get a very close solution of the problem (for the working case, the relative error is about $0.0065$% !).
Adding more terms will give a better result (if you want them, just ask).
Edit
Working a bit more, the blue formula may be written in a more compact form (for an even better accuracy) using a Padé approximant
$$r=x ,frac{1+a_1x+a_2x^2}{1+b_1x+b_2x^2}$$ where
$$a_1=-frac{2 left(2 t^2+3 t+1right)}{11 t+13}qquad a_2=frac{4 t^3-3 t-1}{45 (11 t+13)}$$
$$b_1=-frac{23 t^2+53 t+32}{3 (11 t+13)}qquad b_2=frac{67 t^3+240 t^2+291 t+122}{60 (11 t+13)}$$ For the work example, this would give $r=frac{13916770468}{4468838772519}approx 0.0031141805$
Update
Asking a banker friend of mine, he mentioned an approximation he saw somewhere in the past (he does not remember when and/or where. I just found where). I is
$$color{green}{rsimeq left(left(1+frac{c}{p}right)^{frac{1}{q}}-1right)^q-1}qquad text{where} qquad color{green}{q=log_2left(1+frac 1t right)}$$ Applied to the worked example, this gives $0.00310743$.
Using this estimate as $r_0$, Newton iterates would be
$$left(
begin{array}{cc}
n & r_n \
0 & 0.003107429977 \
1 & 0.003114180156 \
2 & 0.003114181946
end{array}
right)$$ which is real fast. I suppose that one single iteration of Newton method will be more than sufficient.
You could use as a very safe solution
$$r=r_0+frac{2,f(r_0) ,f'(r_0)}{f(r_0), f''(r_0)-2, f'(r_0)^2}$$ where $r_0$ is the result of the green formula.
Applied to the worked example, this would lead to
$r=0.00311418194589$ while th exact solution would be
$r=0.00311418194600$
$endgroup$
As Hello_World answered, the simplest way would be to use Newton method for finding the zero of function
$$f(r) = frac{c}r left(1-frac1{(r+1)^t}right)-p$$
If you start using $r_0=0$, the first iterate would be $r_1=frac{2 (c t-p)}{ct left(t+1right)}$ and you just need to continue until convergence to the desired precision.
The is another thing you can do since $r ll 1$. Build the Taylor series at $r=0$ to get
$$f(r)+p=c t-frac{1}{2} r (c t (t+1))+frac{1}{6} c r^2 t (t+1) (t+2)-frac{1}{24} r^3 (c t
(t+1) (t+2) (t+3))+frac{1}{120} c r^4 t (t+1) (t+2) (t+3) (t+4)-frac{1}{720}
r^5 (c t (t+1) (t+2) (t+3) (t+4) (t+5))+Oleft(r^6right)$$ and now use series reversion to get
$$color{blue}{r=x+frac{t+2}{3} x^2+frac{5 t^2+17 t+14}{36} x^3+frac{17 t^3+78 t^2+117 t+58}{270}
x^4+frac{193 t^4+1094 t^3+2301
t^2+2144 t+748 }{6480}x^5+Oleft(x^{6}right)}$$ where $color{blue}{x=frac{2 (c t-p)}{ct left(t+1right)}}$ (this is $r_1$)
Let us try using $p=100000$, $c=1000$ and $t=120$. This gives $x=frac 1 {363}$. Using the expansion above, this leads to $r=frac{31795409217001}{10210526568429660}approx 0.00311398$ while the "exact" solution obtained using Newton method would be obtained after the following iterates
$$left(
begin{array}{cc}
n & r_n \
0 & 0.000000000000 \
1 & 0.002754820937 \
2 & 0.003161445197 \
3 & 0.003114792911 \
4 & 0.003114182051 \
5 & 0.003114181946
end{array}
right)$$
As you can see, without any iterative procedure we can get a very close solution of the problem (for the working case, the relative error is about $0.0065$% !).
Adding more terms will give a better result (if you want them, just ask).
Edit
Working a bit more, the blue formula may be written in a more compact form (for an even better accuracy) using a Padé approximant
$$r=x ,frac{1+a_1x+a_2x^2}{1+b_1x+b_2x^2}$$ where
$$a_1=-frac{2 left(2 t^2+3 t+1right)}{11 t+13}qquad a_2=frac{4 t^3-3 t-1}{45 (11 t+13)}$$
$$b_1=-frac{23 t^2+53 t+32}{3 (11 t+13)}qquad b_2=frac{67 t^3+240 t^2+291 t+122}{60 (11 t+13)}$$ For the work example, this would give $r=frac{13916770468}{4468838772519}approx 0.0031141805$
Update
Asking a banker friend of mine, he mentioned an approximation he saw somewhere in the past (he does not remember when and/or where. I just found where). I is
$$color{green}{rsimeq left(left(1+frac{c}{p}right)^{frac{1}{q}}-1right)^q-1}qquad text{where} qquad color{green}{q=log_2left(1+frac 1t right)}$$ Applied to the worked example, this gives $0.00310743$.
Using this estimate as $r_0$, Newton iterates would be
$$left(
begin{array}{cc}
n & r_n \
0 & 0.003107429977 \
1 & 0.003114180156 \
2 & 0.003114181946
end{array}
right)$$ which is real fast. I suppose that one single iteration of Newton method will be more than sufficient.
You could use as a very safe solution
$$r=r_0+frac{2,f(r_0) ,f'(r_0)}{f(r_0), f''(r_0)-2, f'(r_0)^2}$$ where $r_0$ is the result of the green formula.
Applied to the worked example, this would lead to
$r=0.00311418194589$ while th exact solution would be
$r=0.00311418194600$
edited Jan 7 at 7:22
answered Jan 6 at 5:47
Claude LeiboviciClaude Leibovici
120k1157132
120k1157132
add a comment |
add a comment |
$begingroup$
You could try Newton's method to solve for $r.$ Take,
$$F(r) = p-frac{c}rcdot left(1-frac1{(r+1)^t}right)$$
then your goal is to find $r$ such that $F(r)=0.$ For this you start with some choice $r_0$ and then use the following recursive definition:
$$r_{n+1}= r_n - frac{F(r_n)}{F'(r_n)}.$$
This will converge to the root of $F.$
$endgroup$
add a comment |
$begingroup$
You could try Newton's method to solve for $r.$ Take,
$$F(r) = p-frac{c}rcdot left(1-frac1{(r+1)^t}right)$$
then your goal is to find $r$ such that $F(r)=0.$ For this you start with some choice $r_0$ and then use the following recursive definition:
$$r_{n+1}= r_n - frac{F(r_n)}{F'(r_n)}.$$
This will converge to the root of $F.$
$endgroup$
add a comment |
$begingroup$
You could try Newton's method to solve for $r.$ Take,
$$F(r) = p-frac{c}rcdot left(1-frac1{(r+1)^t}right)$$
then your goal is to find $r$ such that $F(r)=0.$ For this you start with some choice $r_0$ and then use the following recursive definition:
$$r_{n+1}= r_n - frac{F(r_n)}{F'(r_n)}.$$
This will converge to the root of $F.$
$endgroup$
You could try Newton's method to solve for $r.$ Take,
$$F(r) = p-frac{c}rcdot left(1-frac1{(r+1)^t}right)$$
then your goal is to find $r$ such that $F(r)=0.$ For this you start with some choice $r_0$ and then use the following recursive definition:
$$r_{n+1}= r_n - frac{F(r_n)}{F'(r_n)}.$$
This will converge to the root of $F.$
answered Jan 6 at 1:20
Hello_WorldHello_World
4,16421631
4,16421631
add a comment |
add a comment |
$begingroup$
I prefer to add a second answer.
Being just fascinated by David W. Cantrell's approximation
$$color{green}{rsimeq left(left(1+frac{c}{p}right)^{frac{1}{q}}-1right)^q-1}qquad text{where} qquad color{green}{q=log_2left(1+frac 1t right)}$$ totally inspired by it, I tried something in the same spirit
$$frac{c}r left(1-frac1{(r+1)^t}right)=pimplies 1+frac{c}{p}=1+frac{r}{1-(r+1)^{-t}}$$ Taking logarithms of both sides
$$log left(1+frac{c}{p}right)=log left(1+frac{r}{1-(r+1)^{-t}}right)$$ Now, expanding the rhs as a Taylor series at $r=0$
$$log left(1+frac{c}{p}right)=log left(1+frac{1}{t}right)+frac{r}{2}+frac{2t-5}{24} r^2 +Oleft(r^3right)$$
Neglecting the second order term, we the obtain the first approximation
$$color{blue}{r_1 =2 log left(frac{t ,(c+p)}{p, (t+1)}right)}$$
Using the complete expansion to $Oleft(r^3right)$, we then have the second approximation
$$color{blue}{r_2=frac{sqrt{1+4, alpha, r_1}-1}{2 alpha }}qquad text{where}qquad color{blue}{alpha=frac {2t-5}{12}}$$
We could even avoid quadratic equations building the simplest Padé approximant instead of the Taylor series. This gives
$$log left(1+frac{c}{p}right)=frac{log left(1+frac{1}{t}right)+frac{1}{12} left((5-2 t) log
left(1+frac{1}{t}right)+6right) r} {1+ frac{5-2t}{12} r }implies
color{blue}{r_3=frac{12,r_1}{12+(2t-5),r_1}}$$
Applied to the worked example, this would give
$$r_1=2 log left(frac{606}{605}right)approx 0.00330306$$
$$r_2=frac{2}{235} left(sqrt{9+1410 log left(frac{606}{605}right)}-3right)approx 0.00311325$$
$$r_3=frac{12 log left(frac{606}{605}right)}{6+235 log
left(frac{606}{605}right)}approx 0.00310238$$
while the exact solution is $$r=0.00311418$$
$endgroup$
add a comment |
$begingroup$
I prefer to add a second answer.
Being just fascinated by David W. Cantrell's approximation
$$color{green}{rsimeq left(left(1+frac{c}{p}right)^{frac{1}{q}}-1right)^q-1}qquad text{where} qquad color{green}{q=log_2left(1+frac 1t right)}$$ totally inspired by it, I tried something in the same spirit
$$frac{c}r left(1-frac1{(r+1)^t}right)=pimplies 1+frac{c}{p}=1+frac{r}{1-(r+1)^{-t}}$$ Taking logarithms of both sides
$$log left(1+frac{c}{p}right)=log left(1+frac{r}{1-(r+1)^{-t}}right)$$ Now, expanding the rhs as a Taylor series at $r=0$
$$log left(1+frac{c}{p}right)=log left(1+frac{1}{t}right)+frac{r}{2}+frac{2t-5}{24} r^2 +Oleft(r^3right)$$
Neglecting the second order term, we the obtain the first approximation
$$color{blue}{r_1 =2 log left(frac{t ,(c+p)}{p, (t+1)}right)}$$
Using the complete expansion to $Oleft(r^3right)$, we then have the second approximation
$$color{blue}{r_2=frac{sqrt{1+4, alpha, r_1}-1}{2 alpha }}qquad text{where}qquad color{blue}{alpha=frac {2t-5}{12}}$$
We could even avoid quadratic equations building the simplest Padé approximant instead of the Taylor series. This gives
$$log left(1+frac{c}{p}right)=frac{log left(1+frac{1}{t}right)+frac{1}{12} left((5-2 t) log
left(1+frac{1}{t}right)+6right) r} {1+ frac{5-2t}{12} r }implies
color{blue}{r_3=frac{12,r_1}{12+(2t-5),r_1}}$$
Applied to the worked example, this would give
$$r_1=2 log left(frac{606}{605}right)approx 0.00330306$$
$$r_2=frac{2}{235} left(sqrt{9+1410 log left(frac{606}{605}right)}-3right)approx 0.00311325$$
$$r_3=frac{12 log left(frac{606}{605}right)}{6+235 log
left(frac{606}{605}right)}approx 0.00310238$$
while the exact solution is $$r=0.00311418$$
$endgroup$
add a comment |
$begingroup$
I prefer to add a second answer.
Being just fascinated by David W. Cantrell's approximation
$$color{green}{rsimeq left(left(1+frac{c}{p}right)^{frac{1}{q}}-1right)^q-1}qquad text{where} qquad color{green}{q=log_2left(1+frac 1t right)}$$ totally inspired by it, I tried something in the same spirit
$$frac{c}r left(1-frac1{(r+1)^t}right)=pimplies 1+frac{c}{p}=1+frac{r}{1-(r+1)^{-t}}$$ Taking logarithms of both sides
$$log left(1+frac{c}{p}right)=log left(1+frac{r}{1-(r+1)^{-t}}right)$$ Now, expanding the rhs as a Taylor series at $r=0$
$$log left(1+frac{c}{p}right)=log left(1+frac{1}{t}right)+frac{r}{2}+frac{2t-5}{24} r^2 +Oleft(r^3right)$$
Neglecting the second order term, we the obtain the first approximation
$$color{blue}{r_1 =2 log left(frac{t ,(c+p)}{p, (t+1)}right)}$$
Using the complete expansion to $Oleft(r^3right)$, we then have the second approximation
$$color{blue}{r_2=frac{sqrt{1+4, alpha, r_1}-1}{2 alpha }}qquad text{where}qquad color{blue}{alpha=frac {2t-5}{12}}$$
We could even avoid quadratic equations building the simplest Padé approximant instead of the Taylor series. This gives
$$log left(1+frac{c}{p}right)=frac{log left(1+frac{1}{t}right)+frac{1}{12} left((5-2 t) log
left(1+frac{1}{t}right)+6right) r} {1+ frac{5-2t}{12} r }implies
color{blue}{r_3=frac{12,r_1}{12+(2t-5),r_1}}$$
Applied to the worked example, this would give
$$r_1=2 log left(frac{606}{605}right)approx 0.00330306$$
$$r_2=frac{2}{235} left(sqrt{9+1410 log left(frac{606}{605}right)}-3right)approx 0.00311325$$
$$r_3=frac{12 log left(frac{606}{605}right)}{6+235 log
left(frac{606}{605}right)}approx 0.00310238$$
while the exact solution is $$r=0.00311418$$
$endgroup$
I prefer to add a second answer.
Being just fascinated by David W. Cantrell's approximation
$$color{green}{rsimeq left(left(1+frac{c}{p}right)^{frac{1}{q}}-1right)^q-1}qquad text{where} qquad color{green}{q=log_2left(1+frac 1t right)}$$ totally inspired by it, I tried something in the same spirit
$$frac{c}r left(1-frac1{(r+1)^t}right)=pimplies 1+frac{c}{p}=1+frac{r}{1-(r+1)^{-t}}$$ Taking logarithms of both sides
$$log left(1+frac{c}{p}right)=log left(1+frac{r}{1-(r+1)^{-t}}right)$$ Now, expanding the rhs as a Taylor series at $r=0$
$$log left(1+frac{c}{p}right)=log left(1+frac{1}{t}right)+frac{r}{2}+frac{2t-5}{24} r^2 +Oleft(r^3right)$$
Neglecting the second order term, we the obtain the first approximation
$$color{blue}{r_1 =2 log left(frac{t ,(c+p)}{p, (t+1)}right)}$$
Using the complete expansion to $Oleft(r^3right)$, we then have the second approximation
$$color{blue}{r_2=frac{sqrt{1+4, alpha, r_1}-1}{2 alpha }}qquad text{where}qquad color{blue}{alpha=frac {2t-5}{12}}$$
We could even avoid quadratic equations building the simplest Padé approximant instead of the Taylor series. This gives
$$log left(1+frac{c}{p}right)=frac{log left(1+frac{1}{t}right)+frac{1}{12} left((5-2 t) log
left(1+frac{1}{t}right)+6right) r} {1+ frac{5-2t}{12} r }implies
color{blue}{r_3=frac{12,r_1}{12+(2t-5),r_1}}$$
Applied to the worked example, this would give
$$r_1=2 log left(frac{606}{605}right)approx 0.00330306$$
$$r_2=frac{2}{235} left(sqrt{9+1410 log left(frac{606}{605}right)}-3right)approx 0.00311325$$
$$r_3=frac{12 log left(frac{606}{605}right)}{6+235 log
left(frac{606}{605}right)}approx 0.00310238$$
while the exact solution is $$r=0.00311418$$
answered Jan 8 at 9:20
Claude LeiboviciClaude Leibovici
120k1157132
120k1157132
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063370%2fsolve-for-r-in-classic-annuity-formula%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Excel has a solver add-in that is useful in situations like this.
$endgroup$
– spaceisdarkgreen
Jan 6 at 1:14
$begingroup$
I found the link for the "green" formula. Added a few lines at the bottom to show you what you could do for an incredible accuracy. Cheers.
$endgroup$
– Claude Leibovici
Jan 7 at 7:57