# 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 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.

``````  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 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 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 at 12:21