Future Value of an Annuity with Monthly Deposits and Annual Growth

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


# 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)

## Answer by Chris Degnen (score 1, accepted)

https://quant.stackexchange.com/a/11236

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
```

## Answer by plockc (score 0)

https://quant.stackexchange.com/a/11307

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:

$futureValue=\frac{annualPayment}{r-g}((1+r)^n-(1+g)^n) = 75693$