Posted by John on .
Need help finding a formula for:
Question:
Suppose a bank offers you the following deal:
You pay to the bank an annuity amount of $A per year over the next 10 years and the bank will in turn pay you $40,000 per year starting at the end of year 11 and ending the payments by the end of year 30.
Interest rate=10%/year throughout the 30 year period.
Find the annuity amount of $A you will be willing to pay over the next 10 yrs.

Finance 
economyst,
First, and excel spreadsheet is very helpful for solving these kinds of problems.
I believe you need to have a personal discount rate. How much would you pay to receive $40,000 thirty years from now. For this problem, I believe you are to assume the interest rate, r, is your personal discount rate. (While its not stated, I will also assume you pay $A at the end of the year)
So, you the value of the amount you pay at time zero will be:
A/(1.1) + A/(1.1)^2 + A/(1.1)^3 + ... A/(1.1)^10
= A * sum(i) of 1/(1.1)^i as i goes from 1 to 10
= A * 17.53117
The value of the amount you receive at time zero is 40000/(1.1)^11 + 40000/(1.1)^12 + ... 40000/(1.1)^30
= 40000 * sum(j) of 1/(1.1)^j as j goes from 11 to 30
= 40000 * 163.4123
Set these two equal and solve for A. 
Finance 
John,
Thanks.
What formula did you use to calculate 17.53117? I found a formula that I thought would work to get that sum, but I got a different number. It is:
1  1.1^(10)/ 0.1
and also: (40000* 163)/17.53 = $371933 is the answer? 
Finance 
economyst,
My bad. I did'nt properly apply my own formula.
17.53 is the sum of (1.1)^i as i goes from 1 to 10. What I really want, as my original formula says, is the sum of (1/(1.1)^i) as i goes from 1 to 10. This turns out to be 6.1446. So the present value of 10 payments of A over 10 years is A*6.1446.
Likewise, my 163.4123 is the sum of (1.1)^i as i goes from 11 to 30. What I really want is the sum of 1/(1.1)^i as i goes from 11 to 30. This new sum is 3.2823.
So, set A*6.1446 = 40000*3.2823.
A = 21367.
This makes much more sense.
Sorry for the confusion.