Every engineering decision is also a money decision. Four factor types and a cash-flow diagram handle nearly every economics question.
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A dollar today is worth more than a dollar later because it can earn interest.
What is the present worth of $10,000 received 5 years from now at 6%?
P = 10,000(1.06)−5 = 10,000/1.3382 = $7,473.
Common trap: (C) moves the money forward instead of back.
Engineering EconomicsSingle PaymentNominal and EffectiveThat was 1 of 3 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
An annuity is the same payment A every period.
You save $2,000 at the end of each year for 15 years at 5%. How much do you have at the end?
F = 2000(F/A, 5%, 15) = 2000(21.579) = $43,157.
Common trap: (C) ignores interest entirely.
Uniform Seriescapital recoverysinking fundThat was 1 of 3 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
To choose between projects, bring every cash flow to the same basis: present worth, annual worth, or a benefit-cost ratio.
A park needs $3,000 a year forever for maintenance. At 6%, what is the capitalized cost?
Capitalized cost = A/i = 3000/0.06 = $50,000.
Present WorthCapitalized CostsBenefit-CostThat was 1 of 3 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
Depreciation spreads an asset’s cost over its life for tax purposes.
A $50,000 asset is depreciated by 5-year MACRS (20%, 32%, 19.2%, …). What is the year-2 depreciation?
Year 2 rate = 32%: 0.32(50,000) = $16,000. MACRS ignores salvage.
Common trap: (D) is straight-line over 5 years.
DepreciationMACRSBreak-EvenThat was 1 of 3 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
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