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Here's the problem:

The problem

Now, here is what I did:

Since we invested $\mathbb{$}1000$ at the beginning of $1989,1990$, and $1991$, we find the balance of each contribution over the specified intervals of time and sum them together. So, denoting $C_{y}$ to be the year in which our contribution was made, we obtain: $$C_{1989}=1000(1.06)(1.055)(1.05)(1.045)\approx 1227.05$$ $$C_{1990}=1000(1.065)(1.06)(1.055)(1.05)\approx 1250.54$$ $$C_{1991}=1000(1.06)(1.0555)(1.05)(1.05)\approx 1232.93$$ $$\implies B= C_{1989}+C_{1990}+C_{1991}=3710.52$$ So our balance in $1994$ is $B=\mathbb{$}3710.52$.

Apparently, my answer is off by $3$ dollars, as the answer is claimed to be $3713.16$.To my question: was my approach to the problem incorrect? There wasn't much to go off of in the chapter, so I wasn't quite sure how to play with this one other than one example.

user146925
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1 Answers1

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Your calculation is not right in total.

The value of the first investment in year ´94 is $1000*1.06*1.055*1.05*1.045*1.05$

After you have compounded the investment the first 3 years (until 1992), the investment has to be compounded with the portfolio rate for the next two years (until 1994).

The second investment has only be additionally compounded with the portfolio rate of the year 1993:

$1000*1.065*1.06*1.055*\color{blue}{1.05^1}$

In total the calculation is

$1000*(1.06*1.055*1.05*1.045*1.05+1.065*1.06*1.055*1.05^1+1.06*1.055*1.05)\approx 3,713.16$

callculus42
  • 30,550