Engineering Economics & Investment Decisions

Deciding whether projects are worth it. Learn the time value of money, simple and compound interest, cash-flow diagrams, present/future/annual worth, NPV and discount rates (MARR), IRR, payback and discounted payback, benefit–cost ratio, comparing alternatives with unequal lives, depreciation (straight-line, declining balance), taxes, inflation, break-even and sensitivity analysis — with Python evaluating an automation investment.

Every engineering improvement competes for money: a new robot, a better layout, an energy-saving motor. Engineering economics converts future costs and benefits into comparable values so decisions are made on evidence rather than intuition.

The central idea is that money has time value: ₹100 today is worth more than ₹100 next year because it can earn a return. Discounting future cash flows with an appropriate rate makes projects comparable.

An analogy

Discounting is like judging fruit that will arrive in the future: the later it arrives, the less it's worth today, because you could have planted the money and grown your own in the meantime.

Time value of money

Interest compensates for delayed consumption, risk and inflation. Compound interest grows exponentially: at 10%, money doubles in about 7.2 years (rule of 72). Cash-flow diagrams show receipts (up arrows) and payments (down arrows) over time — draw one for every problem.

Evaluation methods

Consistent methods give the same accept/reject decision for single projects:

Comparing alternatives

For mutually exclusive options, choose the highest NPV (not highest IRR — use incremental IRR if comparing by rates). For unequal lives, compare annual worth or use a common study period / least common multiple of lives. Include all relevant costs: installation, training, maintenance, energy, downtime, salvage; ignore sunk costs.

Taxes, depreciation, inflation and risk

Depreciation (straight-line, declining balance, MACRS or country-specific rules) is not a cash flow but reduces taxes — after-tax cash flow = (revenue − costs − depreciation)(1 − t) + depreciation. Use real rates with real cash flows or nominal rates with inflated cash flows, never mixed. Sensitivity analysis (tornado diagrams), scenario analysis and Monte Carlo quantify uncertainty; real options value flexibility to expand, defer or abandon.

Formulas

Worked example: Compound growth

Given: ₹1 lakh at 8% for 10 years

Find: F.

Answer: ₹2.16 lakh

Worked example: Annuity present worth

Given: ₹70 000/yr for 6 years at 10%

Find: P.

Answer: ₹3.05 lakh

Worked example: Break-even

Given: Fixed ₹1 lakh/month, price ₹250, variable cost ₹150

Find: Q_BE.

Answer: 1000 units/month

NPV, IRR and payback for an automation project

# Engineering economics: NPV, IRR, payback for an automation project
invest = 250000.0
savings = [70000, 70000, 70000, 70000, 70000, 70000]
salvage = 20000.0
r = 0.10
flows = [-invest] + savings[:-1] + [savings[-1] + salvage]
def npv(rate):
    return sum(cf / (1 + rate) ** t for t, cf in enumerate(flows))
lo, hi = 0.0, 1.0
for _ in range(200):
    mid = (lo + hi) / 2
    if npv(mid) > 0: lo = mid
    else: hi = mid
cum, payback = -invest, None
for t, s in enumerate(savings, 1):
    if cum + s >= 0 and payback is None:
        payback = t - 1 + (-cum) / s
    cum += s
print(f"NPV at 10% = {npv(r):,.0f}, IRR = {mid*100:.1f}%, simple payback = {payback:.2f} years")
# Straight-line vs double-declining-balance depreciation, year 1
print(f"depreciation year 1: SL {(invest-salvage)/6:,.0f}, DDB {invest*2/6:,.0f}")
# NPV at 10% = 66,158, IRR = 18.3%, simple payback = 3.57 years
# depreciation year 1: SL 38,333, DDB 83,333

A positive NPV at the company's MARR (10%) means the project creates value; the IRR of 18.3% exceeds MARR, confirming it. Payback (3.6 years) is easy to understand but ignores the time value of money and cash flows after payback — use it as a risk indicator, not the decision rule.

Common mistakes

Applications

Key takeaways

Going further