NCEES Engineering Economics 4

Question 1

What is the future worth of $2,000 invested today at 7% per year compounded annually for 8 years? (F/P, 7%, 8) = 1.7182

Accepted Answer

$3,436.40

Answer

F = P × (F/P, i, n) = $2,000 × 1.7182 = $3,436.40.

Question 2

Which of the following best describes the minimum attractive rate of return (MARR)?

Accepted Answer

The minimum interest rate a company will accept for any investment

Answer

MARR is the minimum return a firm requires before it will invest; projects below this rate are rejected.

Question 3

A geometric gradient series has a first payment of $1,000 and increases by 5% each year for 4 years at i = 10%. What is the present worth? (P/G, geom formula: P = A₁[(1−(1+g)ⁿ(1+i)⁻ⁿ)/(i−g)])

Accepted Answer

$3,397

Answer

Using the geometric gradient PW formula with A₁=$1,000, g=5%, i=10%, n=4, PW ≈ $3,397.

Question 4

A project requires $200,000 now and yields $50,000/year for 6 years. At MARR = 10%, (P/A,10%,6)=4.3553, the net present value is:

Accepted Answer

$17,765

Answer

NPV = −200,000 + 50,000(4.3553) = −200,000 + 217,765 = +$17,765.

Question 5

Book value under straight-line depreciation after year k is calculated as:

Accepted Answer

Cost − k × (Cost − Salvage) / n

Answer

BV_k = Cost − k × annual depreciation = Cost − k × (Cost − Salvage)/n.

NCEES Practice Test

NCEES Practice Test

NCEES Engineering Economics 4