Question

6-7 the average age of engineering students at graduation is a little over 23 years. this means that the working career of most engineers is almost exactly 500 months. how much would an engineer need to save each month to accrue $5 million by the end of her working career? assume a 9% interest rate, compounded monthly.

Explanation:

Step1: Identify the formula for future - value of an ordinary annuity

The formula for the future - value of an ordinary annuity is $FVA = A\times\frac{(1 + r)^{n}-1}{r}$, where $FVA$ is the future value of the annuity, $A$ is the amount of each payment (the monthly savings we want to find), $r$ is the interest rate per period, and $n$ is the number of periods.

Step2: Calculate the monthly interest rate

The annual interest rate is $i = 9\%=0.09$. Since the interest is compounded monthly, the monthly interest rate $r=\frac{0.09}{12}=0.0075$.

Step3: Identify the number of periods

The working career is $n = 500$ months.

Step4: Rearrange the formula to solve for $A$

We know that $FVA = 5000000$, and from $FVA = A\times\frac{(1 + r)^{n}-1}{r}$, we can solve for $A$:
\[A=\frac{FVA\times r}{(1 + r)^{n}-1}\]

Step5: Substitute the values into the formula

\[A=\frac{5000000\times0.0075}{(1 + 0.0075)^{500}-1}\]
First, calculate $(1 + 0.0075)^{500}$. Let $x=(1 + 0.0075)^{500}$. Using the formula $a^{b}=e^{b\ln(a)}$, we have $\ln(x)=500\times\ln(1.0075)\approx500\times0.007472\approx3.736$. Then $x = e^{3.736}\approx41.977$.
\[A=\frac{5000000\times0.0075}{41.977 - 1}=\frac{37500}{40.977}\approx915.1\]

Answer:

$\$915.1$