Chapter 10A, Problem 2P

To determine

(a)

The annual investment.

Answer to Problem 2P

The annual investment is $\$12133.84$.

Explanation of Solution

Given:

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

The time period is $35\text{years}$.

The earning of engineer is $\$68000$ 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:

As per the table of return data for capital allocation line, the interest rate is $7.8\%$ for portfolio of $20\%$ of T-bills, $40\%$ stocks and $40\%$ bonds.

Calculate the annual investment.

Substitute $\$2000000$ for $F$, $7.8\%$ for $i$ and $35$ for $n$ in Equation (I).

$\begin{array}{c}A=\$2000000\left[\frac{0.078}{{\left(1+0.078\right)}^{35}-1}\right]\\ =\$2000000\left[\frac{0.078}{{\left(1.078\right)}^{35}-1}\right]\\ =\$2000000\left(0.00606\right)\\ =\$12133.84\end{array}$.

Conclusion:

Thus, the annual investment is $\$12133.84$.

To determine

(b)

The annual investment.

Answer to Problem 2P

The annual investment is $\$8136.83$.

Explanation of Solution

Given:

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

The time period is $40\text{years}$.

The earning of engineer is $\$68000$ per year.

Calculations:

Calculate the annual investment.

Substitute $\$2000000$ for $F$, $7.8\%$ for $i$ and $40$ for $n$ in Equation (I).

$\begin{array}{c}A=\$2000000\left[\frac{0.078}{{\left(1+0.078\right)}^{40}-1}\right]\\ =\$2000000\left[\frac{0.078}{{\left(1.078\right)}^{40}-1}\right]\\ =\$2000000\left(0.004068\right)\\ =\$8136.83\end{array}$.

Conclusion:

Thus, the annual investment is $\$8136.83$.

To determine

(c)

The investment per year by the employee and the employer.

Answer to Problem 2P

The investment by employer is $\$2720$ per year.

The investment by employee is $\$9413.84$ per year.

Explanation of Solution

Given:

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

The time period is $35\text{years}$.

The earning of engineer is $\$68000$ per year.

$4\%$ of the salary is contributed each year to the retirement savings.

Calculations:

Calculate the annual investment.

Substitute $\$2000000$ for $F$, $7.8\%$ for $i$ and $35$ for $n$ in Equation (I).

$\begin{array}{c}A=\$2000000\left[\frac{0.078}{{\left(1+0.078\right)}^{35}-1}\right]\\ =\$2000000\left[\frac{0.078}{{\left(1.078\right)}^{35}-1}\right]\\ =\$2000000\left(0.00606\right)\\ =\$12133.84\end{array}$

The annual investment is $\$12133.84$.

Calculate amount to be investment by the employer.

$\begin{array}{c}{A}^{\prime }=\$68000\times 4\%\\ =\$68000\times \frac{4}{100}\\ =\$2720\end{array}$

The amount to be investment by employer is $\$2720$ per year.

Calculate the amount to be investment by employee.

$\begin{array}{c}{A}_i=\$12133.84-\$2720\\ =\$9413.84\end{array}$

The amount to be investment by employee is $\$9413.84$ per year.

Conclusion:

Thus, the amount to be invested by the employer is $\$2720$ per year.

The amount to be invested by the employee is $\$9413.84$ per year.