# Is there a formula for future value of a growing annuity with yearly payment growth and monthly payments?

My example is saving for college:

• assume a start of 0 balance
• deposits of 200 made monthly, every year they increase by (g) 2% to account for salary increases, first deposit made at the end of the first month
• Interest Rate (r) is constant at 8% (effective rate)
• Goes for (n=15) years

What is the future value?

Even though I can convert the yearly rate into a compounded monthly rate to match the yearly rate, I can't use the "future value of a growing annuity" formula, that assumes timing of growth and payment are the same.

It is acceptable to make it a two or three steps (like use equation 1 to solve for a new value for payment to plug that into equation 2), I am just trying to avoid making calculations for each and every year as I'm doing now.

n(1) = 2486
n(2) = 5222.23
n(15)= 75693

Update
I found my own answer as well below that combines well known formulas to get to the same answer (and I presume, with substitution, would be equivalent to the accepted answer)

You can calculate it with the formula below, which is produced from a double sum.

P. S. The initial examples are for an annuity due (savings type annuity).

Future value = (r*(-1 + r^y)*(-b^(1 + a) + r^((1 + a)*y))*z)/((-1 + r)*(-b + r^y))


where

r = 1 + monthly rate = 1.08^(1/12) = 1.00643
y = months per year = 12
a = years - 1 = 14
b = deposit increase rate + 1 = 1.02
z = initial deposit amount = 200

(r*(-1 + r^y)*(-b^(1 + a) + r^((1 + a)*y))*z)/((-1 + r)*(-b + r^y)) = 76180.4


Mathematica was used to produce the formula from the double sum:

The double sum is produced from the workings below.

Edit

To illustrate the robustness of the formula here is another example with different period parameters: a twice-yearly deposit of 200 for three years, again incrementing annually by 2%, with 8% interest rate.

Running the calculation in four forms produces the same result. This proves the formula's robustness.

r = 1 + six-monthly rate = 1.08^(1/2) = 1.03923
y = periods per year = 2
a = years - 1 = 2
b = deposit increase rate + 1 = 1.02
z = initial deposit amount = 200


(r*(-1 + r^y)*(-b^(1 + a) + r^((1 + a)*y))*z)/((-1 + r)*(-b + r^y)) = 1402.25


2nd Edit

Recalculation for ordinary annuity (loan type), rather than annuity due (savings). - ref. Calculating The Present And Future Value Of Annuities

((-1 + r^y)*(-b^(1 + a) + r^((1 + a)*y))*z)/((-1 + r)*(-b + r^y)) = 1349.32

• I think that only solves for when a=14, if you try different values for a (changing the number of years), you get numbers all over the place – plockc May 10 '14 at 18:25
• I also have a different result for 15 years of 75963, not sure yet where the difference lies. – plockc May 10 '14 at 18:36
• So the first term has 1.02^0 which is 1 making it the last year of interest being applied, but it shouldn't be r^k, should be r^180-k so that first payment of the last year gets r^11 interest compounding? – plockc May 10 '14 at 18:59
• @plockc - Hi, the first term has 1.02^0 = 1 being the first year of deposits (at 200 * 1), which year's 12 deposits each accumulate 180 to 169 months of interest by the end of 15 years. Hope that clarifies things. I have added another, simpler example too. – Chris Degnen May 10 '14 at 22:05
• Hi @ChrisDegnen, maybe we're applying stuff at different times? I am stepping through fv = oldfv*r+200*1.02^year, with rate at 1.03923, year being 0 twice, then 1 twice, then 2 twice, I get 200, 408, 628, 856, 1098, then 1349 as my answer. From the summation, I think I'm not sure why there is an interest applied at both the first (k=1) and last (k=180) month (so interest applied 180 times where I think it should be 179 times), is what you have an annuity due solution (first $200 deposit immediately)? – plockc May 10 '14 at 22:56 I later figured out you can calculate a special payment and fit it into the normal future value of a growing annuity function that is set up in terms of years. First, solve for the get the monthly rate, this compounded by 12 will bring us back to r$monthlyRate=(1+r)^{1/n}-1 = .006434$Then figure out the effective annual payment (basically accounts for the different lengths of time of interest for each payment) by using the monthly payment in an ordinary annuity for a single year (n=12 months)$annualPayment= \frac{pmt}{monthlyRate}((1+monthlyRate)^n-1)=2486.77$Now since growth and rate are already defined in terms of a year, we get to use a standard growing annuity formula for annual periods:$futureValue=\frac{annualPayment}{r-g}((1+r)^n-(1+g)^n) = 75693\$