Engineering Economics
Time value of money, NPV, IRR, payback, depreciation and choosing between projects.
The time value of money
F = P (1 + i)n P = F ÷ (1 + i)n annuity: P = A [1 − (1 + i)−n] ÷ i A = P i (1 + i)n ÷ [(1 + i)n − 1]
Worked example: interest and loans
Worked example: growth and a mortgage
- $10 000 at 6 percent a year for 10 years becomes F = 10 000 × 1.0610 = $17 908. Discounted back, $17 908 in ten years is worth $10 000 today. The rule of 72 gives the doubling time: 72 ÷ 6 = 12 years (exact: 11.9).
- A $200 000 loan at 6 percent (0.5 percent a month) over 30 years (360 payments): A = $1 199.10 a month, a total of $431 676, of which $231 676 is interest.
Net present value and IRR
NPV = −I + Σ Ct ÷ (1 + r)t IRR: the r that makes NPV = 0 simple payback = I ÷ annual saving profitability index = PV of returns ÷ I
Worked example: an investment decision
Worked example: a $50 000 machine saving $12 000 a year for 6 years
- Simple payback = 50 000 ÷ 12 000 = 4.17 years.
- At a MARR of 8 percent the annuity factor is [1 − 1.08−6] ÷ 0.08 = 4.6229, so the savings are worth $55 475 today and NPV = $5 475 (profitability index 1.11). Annual worth = $1 184 a year.
- IRR = 11.53 percent. At a MARR of 15 percent the NPV is −$4 586: the same project is rejected if the required return is higher. The decision depends on the rate as well as the cash flows.
- The machine has no salvage value here; tax, maintenance and downtime would change the cash flows and are best handled in a spreadsheet.
Comparing alternatives
Alternatives with different lives and cost patterns are compared on the same basis, usually the equivalent uniform annual cost (EUAC): capital recovery (P × the capital-recovery factor) plus annual costs minus the salvage annuity.
Worked example: an expensive durable machine or a cheap one
| Machine A | Machine B | |
|---|---|---|
| First cost | $80 000 | $50 000 |
| Running cost per year | $6 000 | $11 000 |
| Salvage after 10 years | $10 000 | $5 000 |
| EUAC at 8 percent | $17 232 a year | $18 106 a year |
The machine with the higher first cost is cheaper per year here (17 232 against 18 106) because its running costs are lower. At a much higher discount rate the cheaper machine would win: capital is expensive when the rate is high.
Depreciation
Depreciation spreads the cost of an asset over its life for accounting and tax. Straight line: (cost − salvage) ÷ life. Declining balance writes off a fixed fraction of the remaining book value each year (double-declining uses 2 ÷ life). Tax systems often prescribe their own schedules (MACRS in the US), so check the rules.
| Year | Straight line | Double-declining | Book value after (DDB) |
|---|---|---|---|
| 1 | $18 000 | $40 000 | $60 000 |
| 2 | $18 000 | $24 000 | $36 000 |
| 3 | $18 000 | $14 400 | $21 600 |
| 4 | $18 000 | $8 640 | $12 960 |
| 5 | $18 000 | $2 960 | $10 000 |
On a $100 000 asset with a $10 000 salvage over 5 years: straight line is $18 000 a year; double-declining starts at $40 000 and is capped so the book value never falls below the salvage.
Inflation, tax and risk
- Real and nominal rates: (1 + nominal) = (1 + real)(1 + inflation). With a 7 percent nominal return and 3 percent inflation the real return is 3.88 percent. Do not mix constant-dollar cash flows with nominal discount rates.
- Taxes reduce the savings and give a depreciation shield; always work with after-tax cash flows when a company’s tax rate is material.
- Risk: test how the answer changes when the savings, life or rate change (sensitivity analysis). If the decision flips with a small change, the project is marginal.
- Life-cycle cost counts purchase, installation, energy, maintenance and disposal; for motors and pumps the energy over the life is often many times the purchase price (see Induction Motors and Drives).
Related
- Previous: Statistics and Probability for Engineers. Back to the overview: Engineering Fundamentals.
- Hydraulics and Pneumatics and Electrical Power for energy-saving projects.
- Prototyping Techniques for tooling break-even.
FAQ
Which is better: NPV or IRR?
NPV is more reliable, especially when comparing projects of different size or with changing signs of cash flow. IRR is intuitive but can mislead.
What discount rate should I use?
The minimum return your organization requires (its cost of capital plus a risk margin). For personal projects, the return you could earn elsewhere.
Why does a long payback still make sense?
Because payback ignores everything after the payback date: a project that pays back in 6 years but lasts 30 can be far better than one that pays back in 3 and lasts 4.
References
- Sullivan, W. G., Wicks, E. M., Koelling, C. P., Engineering Economy, Pearson.
- Blank, L., Tarquin, A., Engineering Economy, McGraw-Hill.
- Park, C. S., Contemporary Engineering Economics, Pearson.
Last reviewed: October 2026. This is educational information, not financial or tax advice.