R is the rate of interest and n is the number of years. The n is an exponential value of 1+r. It can also be written as (r^-1)(1-(1+r)^n).
As written the formula for the Present value of annuity of $1 in Arrears; (1/r)(1-(1/(1+r)^n)). For each dollar invested per n years, the principal will grow by the ammount given. I am trying to solve for r. So it could be A=(P)(1/r)(1-(1/(1+r)^n)) where A is the Amount (or total), P is the Principal, r is the rate, and n is the number of years.
How would you solve for for R in the Present Value of Annuity (1/r)(1-(1/(1+r)^n))?
19
May
starman2718
May 19, 2010 at 6:41 am
Please rewrite this problem. You’re missing some information. What is the above expression equal to?
Dan D
May 19, 2010 at 7:02 am
At t = 0, you put in P so A[0] = P
At t = 1, you have P(1 + r) and you add in another P for a total of A[1] = P(1 + (1 + r))
At t = 2, you have P(1 + (1 + r)) (1 + r) and you add in another P for a total of A[2] = P(1 + (1 + r) + (1 + r)^2).
Etc.
At t = n-1, just after putting in another P, your total is
A[n-1] = P(1 + (1 + r) + (1 + r)^2 + … + (1 + r)^(n-1))
(1 + r)A[n-1] = P((1 + r) + (1 + r)^2 + … + (1 + r)^n)
((1 + r)A[n-1] – A[n-1]) = r A[n-1] = P((1 + r)^n – 1)
So A[n-1] = P (1/r) ((1 + r)^n – 1)
Likewise a year later after adding another P you have
A[n] = P (1/r) ((1 + r)^(n+1) – 1)
A[n] is n years after your first contribution, but is immediately after your n+1st contribution. A[n-1] is immediately after your nth contribution but is only n-1 years after your first contribution.
Assume it is A[n-1] = P (1/r) ((1 + r)^n – 1) that you want to be equal to A. Then you need to solve iteratively. For example, if you want n = 30 then you can check this following table of r vs. A/P = A[n-1]/P = A[29]/P
1% 34.78
2% 40.57
3% 47.58
4% 56.08
5% 66.44
6% 79.06
7% 94.46
8% 113.28
9% 136.31
10% 164.49
11% 199.02
12% 241.33
If you want your 30 contributions to P to add up to 100P immediately after that last contribution, then you need r to be between 7% and 8%. Zooming in on the table near there:
7.30% 99.72
7.31% 99.90
7.32% 100.08
7.33% 100.26
and then zooming in further to
7.3156% 99.997946
7.3157% 99.999756
7.3158% 100.001567
7.3159% 100.003378
Basically you use an iterative root finder (the most basic is the bisection method but there are better methods) to find the desired r to more and more precision.
Dan