Chapter 10A, Problem 1P

To determine

(a)

The annual investment that should be made.

Answer to Problem 1P

The annual investment is $\$12158.5$.

Explanation of Solution

Given:

The compound amount in future is $\$2000000$.

The time period is $30\text{years}$.

The earning of engineer is $\$65000$ per year.

Concept used:

Write the expression for annual investment.

$A=F\left[\frac{i}{{\left(1+i\right)}^n-1}\right]$ ...... (I)

Here, the compound amount in future is $F$, the interest rate is $i$ and time period is $n$.

Calculations:

Assume the interest rate is $10\%$.

Calculate the annual investment.

Substitute $\$2000000$ for $F$, $10\%$ for $i$ and $30$ for $n$ in Equation (I).

$\begin{array}{c}A=\$2000000\left[\frac{0.1}{{\left(1+0.1\right)}^{30}-1}\right]\\ =\$2000000\left[\frac{0.1}{{\left(1.1\right)}^{30}-1}\right]\\ =\$2000000\left(0.00607\right)\\ =\$12158.5\end{array}$.

Conclusion:

Thus, the annual investment to be made is $\$12158.5$.

To determine

(b)

The investment per year by employee and employer.

Answer to Problem 1P

The amount to be invested by the employer is $\$1950$ per year.

The amount to be invested by the employee is $\$10208.5$ per year.

Explanation of Solution

Given:

The compound amount in future is $\$2000000$.

The time period is $30\text{years}$.

The earning of engineer is $\$65000$ per year.

$3\%$ of the salary is contributed each year to retirement savings.

Calculations:

Assume the interest rate is $10\%$.

Calculate the annual investment.

Substitute $\$2000000$ for $F$, $10\%$ for $i$ and $30$ for $n$ in Equation (I).

$\begin{array}{c}A=\$2000000\left[\frac{0.1}{{\left(1+0.1\right)}^{30}-1}\right]\\ =\$2000000\left[\frac{0.1}{{\left(1.1\right)}^{30}-1}\right]\\ =\$2000000\left(0.00607\right)\\ =\$12158.5\end{array}$

Calculate the investment by the employer.

$\begin{array}{c}{A}^{\prime }=\$65000\times 3\%\\ =\$65000\times \frac{3}{100}\\ =\$1950\end{array}$

The amount to be invested by employer is $\$1950$ per year.

Calculate the investment by employee.

$\begin{array}{c}{A}_i=\$12158.5-\$1950\\ =\$10208.5\end{array}$

The amount to be invested by employee is $\$10208.5$ per year.

Conclusion:

Thus, the amount to be invested by the employer is $\$1950$ per year.

The amount to be invested by the employee is $\$10208.5$ per year.