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# Investments

5. Consider the following four investments.
a) You invest \$ 3,000 annually in a mutual fund that earns 10 percent annually, and you reinvest all distributions. How much will you have in the account at the end of 20 years?
b) You invest \$ 3,000 annually in a mutual fund with a 5 percent load fee so that only \$ 2,850 is actually invested in the fund. The fund earns 10 percent annually, and you reinvest all distributions. How much will you have in the account at the end of 20 years? ( Assume that all distributions are not subject to the load fee.)
c) You invest \$ 3,000 annually in a no- load mutual fund that charges 12b- 1 fees of 1 percent. The fund earns 10 percent annually before fees, and you reinvest all dis-tributions. How much will you have in the account at the end of 20 years?
d) You invest \$ 3,000 annually in no- load mutual fund that has a 5 percent exit fee. The fund earns 10 percent annually before fees, and you reinvest all distributions. How much will you have in the account at the end of 20 years?
In each case you invest the same amount (\$ 3,000) every year; the fund earns the same return each year ( 10 percent), and you make each investment for the same time period ( 20 years). At the end of the 20 years, you withdraw the funds. Why is the final amount in each mutual fund different?

### 1 Answer by Expert Tutors

Keith M. | CMU graduate student tutoring Mathematics and Computer ScienceCMU graduate student tutoring Mathematic...
4.8 4.8 (62 lesson ratings) (62)
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This is a tough problem, and requires a lot of diligence in order to solve it.

Consider what this question is asking, in that all distributions are reinvested in the mutual fund.  In this case, we can treat the annual investments as earning interest independent of one another:

• Funds compounding at 10% interest for 20 years will be scaled by (1.10)20.
• Funds compounding at 10% interest for 19 years will be scaled by (1.10)19.
• Funds compounding at 10% interest for 18 years will be scaled by (1.10)18.
• Funds compounding at 10% interest for   1 year  will be scaled by (1.10)1.

Each of these scaling factors will affect one of the annual investment amounts (The initial investment will compound for 20 years, the next year for only 19, and so on), meaning that we need to sum each of the scale factors multiplied by the annual investment amount.

Depending on the fee incorporated, the formula will need to be adjusted accordingly.  For load fees, the amount of the principle that gets invested is lessened, so we adjust the principle.  For 12b- 1 fees, the distributions are affected, so you adjust the interest rate.  For exit fees, the final fund amount is the only thing affected, so we adjust the amount accordingly.

So, the amount in the accounts at the end of the 20 year investment period is as follows:

a) \$ 3,000 * [ (1.10)20 + (1.10)19 + (1.10)18 + … + (1.10)1 ] = \$ 189,007.50

b) (\$ 3,000 * 0.95) * [ (1.10)20 + (1.10)19 + (1.10)18 + … + (1.10)1 ] = \$ 179,557.12

c) \$ 3,000 * [ (1.10)20 * (0.99)20 + (1.10)19 * (0.99)19 + … + (1.10)1 * (0.99)1 ] = \$ 165,276.02

d) \$ 3,000 * [ (1.10)20 + (1.10)19 + (1.10)18 + … + (1.10)1 ] * 0.95 = \$ 179,557.12

Notice the differences in where the fees affect the formula!  A small fee that is compounded (like the 12b- 1 fee) will make a huge dent in your gains, whereas a larger fee that is not (like the front-end and back-end load fees) will have a smaller effect.  Another point to watch in this problem: since the interest gained is independent of the amount invested (which is not entirely accurate for a stock market investment), the front-end and back-end load fees result in the same final amounts for subproblems b) and d).

You can save yourself a lot of trouble if you write a simple computer program (e.g., using a graphing calculator or other computational tool) to compute the sums for you.

For reference, the Wolfram Alpha commands to compute the above sums are here:

[ (1.10)20 + (1.10)19 + (1.10)18 + … + (1.10)1 ]
• Sum (1.1)^j, j=1 to 20

[ (1.10)20 * (0.99)20 + (1.10)19 * (0.99)19 + … + (1.10)1 * (0.99)1 ]
• Sum (1.1*.99)^j, j=1 to 20