I think you will have to try various interest rates and graph that function
if i = .1 or 10%
6.25 = 5.76
if i = .08
6.25 = 6.24 close to 8%
an annuity has an initial balance of $5000. annual withdrawals are made in the amount of $800 for 9 years, at which point the account balance is zero. what annual rate of interest, compounded annually, was earned over the duration of this annuity?
PV = $5000
R = $800
n = 9
i = ?
PV = (R[1-(1+i)^-n])/i
5000 = (800[1-(1+i)^-9])/i
IDK how to isolate for i and solve for it. please help me!
4 answers
i know the answer is 0.08 but how can i isolate for i instead of using the method guess and check
5000 = (800[1-(1+i)^-9])/i
5000 = (800[1-(1+i)^-9])/i
5000 = (800[1-(1+i)^-9])/i
or
(1-(1+i)^-9)/i = 6.25
Unfortunaltely, there is no easy way to algebraically isolate the i in this type of equation.
The method that Damon has used is about the best way to go about it.
notice that if we use .08 exactly,
800(1 - 1.08^-9)/.08 = 4997.51 so we were everso-slightly off by 2.50
we could try something like .0801
800(1 - 1.0801^-9)/.0801 = 4995.43 off by 4.50 and more than before, so I actually "guessed" too high.
rate result
.08 -- 4997.51
r ----5000
.0801 -4995.43
so now could set up a ratio:
(r-.08)/(5000-4997.51) = (.0801-.08)/(4995.43-4997.51)
(r-.08)/2.49 = .0001/-2.07715
r-.08 = -.0001199
r = .07988
let's try that:
800(1 - 1.07988^-9)/.07988 = 5000.0025
This method is called interpolation, one finds two values very close to the unknown one, and then use a ratio similar to the one I used.
(Actually for my example it would be extrapolation)
There are other methods, such as Newton's Method, but it requires Calculus.
We could alway use the Magic of Wolfram:
http://www.wolframalpha.com/input/?i=solve+%281-%281%2Bx%29%5E-9%29%2Fx+%3D+6.25
Notice how close my answer of .07988 was to .0798802..
Also notice that I had to change the i to x, Wolfram knows that i = √-1
or
(1-(1+i)^-9)/i = 6.25
Unfortunaltely, there is no easy way to algebraically isolate the i in this type of equation.
The method that Damon has used is about the best way to go about it.
notice that if we use .08 exactly,
800(1 - 1.08^-9)/.08 = 4997.51 so we were everso-slightly off by 2.50
we could try something like .0801
800(1 - 1.0801^-9)/.0801 = 4995.43 off by 4.50 and more than before, so I actually "guessed" too high.
rate result
.08 -- 4997.51
r ----5000
.0801 -4995.43
so now could set up a ratio:
(r-.08)/(5000-4997.51) = (.0801-.08)/(4995.43-4997.51)
(r-.08)/2.49 = .0001/-2.07715
r-.08 = -.0001199
r = .07988
let's try that:
800(1 - 1.07988^-9)/.07988 = 5000.0025
This method is called interpolation, one finds two values very close to the unknown one, and then use a ratio similar to the one I used.
(Actually for my example it would be extrapolation)
There are other methods, such as Newton's Method, but it requires Calculus.
We could alway use the Magic of Wolfram:
http://www.wolframalpha.com/input/?i=solve+%281-%281%2Bx%29%5E-9%29%2Fx+%3D+6.25
Notice how close my answer of .07988 was to .0798802..
Also notice that I had to change the i to x, Wolfram knows that i = √-1
okay thank you ! :)
0 answers