Question

An engineer must choose between two materials for an underground storage tank. A steel tank costs $207,000 and is expected to last 15 years. A fiberglass tank costs $313,000 and is expected to last 30 years. Neither tank has a salvage value at the end of its useful life. Determine the MARR below which the FRP tank is preferred. Express your answer in \% (not a decimal) to the nearest 0.1%.

Answer 1

To determine the MARR below which the FRP tank is preferred, we need to compare the present worth of costs for both tanks and solve an inequality to find the interest rate (MARR). The MARR is approximately 8.1%.

Step-by-step explanation:

To determine the minimum attractive rate of return (MARR) below which the fiberglass tank is preferred, we need to compare the present worth of costs for both tanks.

The present worth of the steel tank is $207,000.

The present worth of the fiberglass tank is calculated by multiplying the cost of the tank by the present worth factor (PWF) for its useful life. The PWF can be calculated using the formula PWF = (1+i)^(-n), where i is the interest rate and n is the number of years. For the fiberglass tank, the PWF is (1+i)^(-30).

Since the fiberglass tank is preferred only when its present worth is less than the present worth of the steel tank, we can set up the inequality 313,000 * (1+i)^(-30) < 207,000 and solve for i. By trial and error or using a financial calculator, we find that the MARR below which the fiberglass tank is preferred is approximately 8.1%.