ENGR.ECONOMIC ANALYSIS
ENGR.ECONOMIC ANALYSIS
14th Edition
ISBN: 9780190931919
Author: NEWNAN
Publisher: Oxford University Press
Question
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Chapter 10A, Problem 1P
To determine

(a)

The annual investment that should be made.

Expert Solution
Check Mark

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 30years.

The earning of engineer is $65000 per year.

Concept used:

Write the expression for annual investment.

A=F[i(1+i)n1] ...... (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).

A=$2000000[0.1(1+0.1)301]=$2000000[0.1(1.1)301]=$2000000(0.00607)=$12158.5.

Conclusion:

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

To determine

(b)

The investment per year by employee and employer.

Expert Solution
Check Mark

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 30years.

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).

A=$2000000[0.1(1+0.1)301]=$2000000[0.1(1.1)301]=$2000000(0.00607)=$12158.5

Calculate the investment by the employer.

A=$65000×3%=$65000×3100=$1950

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

Calculate the investment by employee.

Ai=$12158.5$1950=$10208.5

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.

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