# What is the correct Formula for calculating fund growth with fees

I am considering two different options for a roth IRA. A managed fund that has a higher historical growth rate but 1% fee and a self directed vanguard account, which would have a lower growth rate but lower fees.

What I want is to understand are the higher fees worth it. Which as I understand it comes down to will the higher growth rate of the managed fund cover the fees (I understand past performance is not a guarantee of future results). To do this objectively I want to do the math. I am using the formula for compound interest

A = P(1 + r/n)^(nt) * (1 - f)

where

• A is the future value or ending balance.
• P is the initial principal amount
• r is the annual growth rate
• n is the number of times that growth is compounded per year. If it's compounded annually, n would be 1, compounded semi-annually, n would be 2, and so on.
• t is the number of years the money is invested for
• f is the fee expressed as a decimal

Based on some articles it doesn't look like my equation is correct. Because conclusions from the above equation yield only a 1% difference in the ending balance for a 1% fee. The article concludes a much higher difference.

What is the proper equation for incorporating fees with fund growth rates?

Any info is greatly appreciated.

• Different kinds of fees: Purchase load and sale load would be handled the way you did, because they are independent of holding time. But expense ratio applies periodically just like annual growth. Nov 6, 2023 at 15:02

In simplest terms, if the fee is a percentage, you can subtract it directly from the percentage yield. That is, a fund with a 1% fee and 8% yield will be no better for you as an investor than a fund with 7.01% yield but 0.01% fee.

(Which is why Warren Buffet won his bet. It is not at all easy for a fund manager to reliably produce results sufficiently better than index funds to justify the much higher fee the actively managed funds charge. Unless you have good reason for buying into a specific fund, hunting for low fees with decent return and acceptable risk seems to be a net win.)

The results in the article quoted can be calculated as follows.

With

``````s = future value
a = periodic deposit
n = number of periods
r = periodic rate
``````

Setting the future value `s` equal to the sum of the appreciated payments `a`. Formula is by induction.

$s=\sum_{k=1}^{n}a(1+r)^{k}=\frac{a(1+r)((1+r)^{n}-1)}{r}$

``````  r = 0.097/12
a = 1000
n = 40*12
∴ s = (a (1 + r) ((1 + r)^n - 1))/r = \$5,820,873.14
``````

Changing `r` to include a 1% fee

``````  r = (0.097 - 0.01)/12
∴ s = (a (1 + r) ((1 + r)^n - 1))/r = \$4,314,469.72
``````
• Sorry, I'm not following. What does the 0.097 represent in r=0.097/12 ? Is that like a 9.7% growth rate? And I'm assuming dividing by 12 is assuming that growth is annual Nov 10, 2023 at 1:39
• Hi. 9.7% is the rate mentioned in the article. They don't say so I assumed 9.7% nominal annual interest compounded monthly. If it were an effective rate I would have set the monthly periodic rate `r = (1 + 0.097)^(1/12) - 1`. Or could have just left it all annual with `r = 0.097 and n = 40`. Nov 10, 2023 at 9:26
• As a nominal annual interest compounded monthly 9.7% has a periodic monthly rate of `r = 0.097/12 = 0.00808333`. It is done that way to make calculation easier, i.e. "the true calculation [being] not readily available" – Fed 2008. 9.7% nominal compounded monthly is actually 10.14307% annually: `(1 + 0.00808333)^12 - 1 = 0.1014307` in which case the monthly periodic rate would again be `r = (1 + 0.1014307)^(1/12) - 1 = 0.00808333`. If a rate is quoted as nominal the compounding period should also be stated. Nov 10, 2023 at 12:21