Engineering Economics practice questions

Answer: A) $27,920

A single future amount is brought back to the present with the single-payment present-worth factor: P = F(1 + i)^−n = 50,000 × 0.5584 ≈ $27,920.

• ACorrect. P = F/(1 + i)^n = 50,000/(1.06)^10 = 50,000 × 0.5584 ≈ $27,920.
• BIncorrect. This discounts with simple interest, 50,000/(1 + 0.06 × 10) = 50,000/1.6. Present worth calls for compound discounting.
• CIncorrect. This takes 6% of the face value for each of 10 years off the amount, 50,000 × (1 − 0.60). That is not a valid discounting method.
• DIncorrect. This uses the sinking-fund factor (A/F, 6%, 10) = 0.0759. That gives the equal annual deposit needed to build up $50,000, not a single present amount.

Answer: C) $613,900

• AIncorrect. That's the future worth.
• BIncorrect. That's simple discounting at 5% for 10 years.
• CCorrect. P = 1,000,000/(1.05)¹⁰ = 1,000,000/1.6289.
• DIncorrect. That's one year of discounting.

Answer: C) $101,900

• AIncorrect. That's interest only.
• BIncorrect. That ignores interest.
• CCorrect. 1,000,000 × 0.1019.
• DIncorrect. Multiply by A/P, don't divide.

Answer: B) $13,439

Inflated future cost uses the compound-amount relationship F = P(1 + f)^n: 10,000 × (1.03)^10 ≈ $13,439.

• AIncorrect. This applies simple inflation, 10,000 × (1 + 0.03 × 10). Inflation compounds each year.
• BCorrect. F = P(1 + f)^n = 10,000 × (1.03)^10 = 10,000 × 1.3439 ≈ $13,439.
• CIncorrect. This applies only one year of inflation instead of ten.
• DIncorrect. This divides by (1.03)^10, which discounts the cost back in time instead of inflating it forward.

Answer: C) $90,000

• AIncorrect. Subtract, don't add, salvage.
• BIncorrect. Subtract salvage first.
• CCorrect. (2,000,000 − 200,000)/20.
• DIncorrect. Recheck the division.

Answer: A) $13,730

• ACorrect. 30,000(3.791) − 100,000.
• BIncorrect. That ignores interest.
• CIncorrect. Subtract the initial cost.
• DIncorrect. Recheck the multiplication.

Answer: B) Lowest equivalent annual cost over the study period

Compare alternatives on a life-cycle basis, such as equivalent annual cost or present worth, which discounts capital and all O&M costs to a common time basis. Then choose the alternative with the lowest life-cycle cost.

• AIncorrect. Looking only at first cost ignores O&M costs over the whole service life, and those can outweigh differences in capital cost.
• BCorrect. Equivalent annual cost (or present worth) converts capital cost and every future O&M cost to a common time basis using the interest rate, so the alternative with the lowest life-cycle cost is the most economical.
• CIncorrect. One year of O&M leaves out the capital cost and the O&M costs in all later years.
• DIncorrect. Simple payback ignores the time value of money and any costs after the payback point, so it doesn't measure total life-cycle cost.

Answer: D) $2.5 million

Cost indexes scale a historical cost by the ratio of the current index to the original index: $2.0M × 1,000/800 = $2.5M.

• AIncorrect. This flips the index ratio (2.0 × 800/1,000). The index went up, so the cost should go up too.
• BIncorrect. This ignores the change in the cost index. Historical costs must be brought up to the current index.
• CIncorrect. This divides the 200-point rise by the new index (20%) instead of the original index (25%). The correct escalation is 2.0 × (1 + 200/800).
• DCorrect. Current cost = original cost × (current index / original index) = 2.0 × 1,000/800 = $2.5 million.

Answer: A) Capital, O&M and replacement costs, less salvage value, discounted to present worth

A life-cycle cost analysis takes capital, O&M, energy and replacement costs over the study period, subtracts salvage value, and discounts everything to present worth (or an equivalent annual cost).

• ACorrect. Life-cycle cost includes investment cost, operation and maintenance, energy, and replacement costs over the study period, minus salvage value. All of these are discounted to present value so alternatives can be compared on the same basis.
• BIncorrect. Adding future costs without discounting ignores the time value of money. Life-cycle cost analysis converts future costs to present values.
• CIncorrect. Salvage (residual) value at the end of the study period is a credit that is subtracted from investment and replacement costs. Leaving it out misstates the life-cycle cost.
• DIncorrect. Using only the first year of O&M leaves out operating, maintenance and replacement costs for the rest of the facility's life.

Answer: A) 1.5

• ACorrect. 3/2; above 1 is justified.
• BIncorrect. That's inverted.
• CIncorrect. That's B − C in millions.
• DIncorrect. Divide, don't add.