# How can I determine the interest rate, given increasing monthly deposits and a final amount?

I need help to determine the interest rate of this investment.

• The initial value is nothing
• The monthly investment is R500.
• It stays R500 a month until the year is over and increases by R50. Thus, R550 per month the 2nd year, and R600 the 3rd year, etc.
• Total period is 10 Years
• Total invested: R87000 over the 10 years.
• Total returned (deposits plus interest): R149028

• Assuming interest is paid monthly, this works out to 11% (Did it in Excel by creating an Amortization and using Goal Seek to make the final balance what you said.) Keep in mind that you last year's payments will be 950 a month, and there might be severe penalties if you miss a payment. I'm looking at this with my "Too good to be true" goggles. It sounds a bit ponzi-ish to me. – Chris Cudmore Aug 20 '12 at 17:52
• Thank you very much, if you make it an answer I can mark it. – Johan Aug 20 '12 at 19:24
• Is this homework? – JohnFx Aug 21 '12 at 20:47
• No, it is an actual offer I was given, still doing some research on the offer. Something seems off, I just can't seem to find it! – Johan Aug 22 '12 at 6:34

I'm assuming that compound interest is paid monthly. In that case, the interest rate is ~0.9162% per month (i.e. an APR of 11.57%).

The following formula is the one you are interested in:

sum_{y=0}^9 (sum_{m=0}^11 (500+50*y)*(1+T)^(120-12*y-m))=149028

I didn't attempt to solve this, but instead used trial and error. The value T=0.0091624 works nicely; you can verify this in Wolfram Alpha.

Getting at APR/EAR is simply (1+T)^12 - 1.

• +1 I agree with your result. Do you have a reference to your formula? – Chris Degnen Dec 30 '15 at 12:26

A method to work this out can be found by using a simpler quarterly example over two years. Using an example rate, `r = 0.01`, this is the example calculation for the first year

``````y1q1 = 0 + 500
y1q2 = y1q1 (1 + r) + 500
y1q3 = y1q2 (1 + r) + 500
y1q4 = y1q3 (1 + r) + 500
y1q4 (1 + r) =  2050.502505
``````

Equivalent to the summation

Continuing, this is the calculation for two years

``````y2q1 = y1q4 (1 + r) + 500 + 50
y2q2 = y2q1 (1 + r) + 500 + 50
y2q3 = y2q2 (1 + r) + 500 + 50
y2q4 = y2q3 (1 + r) + 500 + 50
y2q4 (1 + r) = 4389.313885
``````

Equivalent to this summation

To create a general formula this needs to be re-expressed as a double summation, where n is the total number of periods, `n = 8`

This can be generalised, where

``````y is the number of years
m is the number of months or quarters (or days)
p is the initial regular deposit
d is the annual deposit increase
``````

By induction, this can be reduced to a formula

Checking

``````r = 0.01
p = 500
d = 50
y = 2
m = 4
n = 8

((1 + r)^(1 + n) (d + p (-1 + (1 + r)^m) +
(1 + r)^(-m y) (-d + p + d y -
(1 + r)^m (p + d y))))/(r (-1 + (1 + r)^m)) = 4389.313885
``````

This can be used to solve for the OP's values

``````fv = 149028
p  = 500
d  = 50
y  = 10
m  = 12
n = 120
``````

Plot of future value for a range of `r` also showing the target `fv`

Solving exactly yields `r = 0.009162396432`

Giving an annual effective rate of

``````(1 + 0.009162396432)^12 - 1 = 0.115662 = 11.5662 %
``````