Engineering Economics Flashcards

A firm's after-tax MARR is 12%. Its effective tax rate is 35%. What is the before-tax MARR?

Answer: 18.5%

Before-tax MARR = After-tax MARR / (1 − tax rate) = 12% / (1 − 0.35) ≈ 18.5%.

Which of the following best describes 'sunk cost' in engineering economics?

Answer: Past expenditure that cannot be recovered

A sunk cost is a past expenditure that is irrelevant to future decision-making because it cannot be recovered.

Two alternatives have the following annual costs: Alt A = $15,000/yr, Alt B = $12,000/yr with an incremental first cost of $20,000 over Alt A's life of 5 years at 10%. Should Alt B be selected?

Answer: Yes, if the PV of savings exceeds $20,000

PV of annual savings = 3,000·(P/A,10%,5) = 3,000·3.791 = $11,373 < $20,000, so Alt B is NOT preferred — select Alt A.

What is the present value of $5,000 received at the end of each year for 8 years at 6% interest?

Answer: $31,047

PV = 5,000·(P/A,6%,8) = 5,000·6.210 = $31,050 ≈ $31,047.

In a replacement analysis, the 'defender' is:

Answer: The currently owned asset

The defender is the existing asset under consideration for replacement; the challenger is the new alternative.

A geometric gradient series has a first-year payment of $1,000 growing at 5% per year for 10 years. At i = 8%, the present value is approximately:

Answer: $8,108

PV = A₁·[1−(1+g)^n·(1+i)^(-n)]/(i−g) = 1000·[1−(1.05/1.08)^10]/0.03 ≈ $8,108.

If inflation is 4% and the market interest rate is 9%, the real interest rate is approximately:

Answer: 4.81%

Real rate = (1 + market)/(1 + inflation) − 1 = (1.09/1.04) − 1 ≈ 4.81%.