Engineering economics and decision-making

Engineering is the art of doing for one dollar what anyone can do for two. Nearly every design decision has a cost side: whether to buy a more efficient motor, whether to repair a bridge now or replace it in ten years, whether a solar pump justifies its higher price, whether a new production line will pay for itself. Engineers are expected to compare such options in money terms, and to do it correctly, because money now and money later are not worth the same. This lesson introduces engineering economics: the time value of money, interest and equivalence, the standard formulas, present worth, annual worth, rate of return and payback, life-cycle cost, inflation, and how to handle uncertainty with sensitivity analysis. It finishes by placing economic analysis among the other factors, such as safety, environment and fairness, that money alone cannot capture.

The time value of money

A dollar today is worth more than a dollar a year from now, for three reasons: today's dollar can be invested to earn a return, future money is uncertain, and inflation erodes purchasing power. Interest is the price of using money over time. With simple interest, interest is paid only on the original amount; with compound interest, which is how almost all real finance works, interest is earned on accumulated interest too.

If an amount P is invested at an interest rate i per period, after n periods it grows to a future amount F:

Turning a future amount into its present equivalent is called discounting, and the rate used is the discount rate. For engineering decisions, the discount rate represents the organization's cost of capital or the return it could earn on other investments, often called the minimum attractive rate of return (MARR).

Equivalence and uniform series

Two cash flows are equivalent if they have the same present value at the given interest rate. Equivalence lets engineers compare options whose costs occur at different times by converting everything to a common basis. Many engineering cash flows are uniform series, equal amounts A every period for n periods: annual maintenance, energy costs, loan payments. Their present value is:

The second formula, the capital recovery factor, converts a present amount into an equivalent annual amount, exactly as a mortgage payment spreads the price of a house over its term.

Comparing alternatives: present worth and annual worth

To compare mutually exclusive options, such as two ways to supply power to the community pump of Lessons 2 and 9, engineers list each option's cash flows over the same study period, discount them at the MARR, and compare. The present worth method sums the present values; for options that provide the same service, the one with the lowest present cost wins. The annual worth method converts the present worth to an equivalent uniform annual amount, which is often easier to explain ("the solar option costs the equivalent of $4,700 a year") and convenient when options have different lives.

# Life-cycle cost of two ways to power the community pump (20 years)
N = 20
solar  = {"capital": 45000, "annual": 800,        "year10": 6000}   # pump replacement in year 10
diesel = {"capital": 18000, "annual": 5500 + 2000, "year10": 8000}  # fuel + maintenance; overhaul

def cashflows(opt):
    c = [opt["capital"]] + [opt["annual"]] * N
    c[10] += opt["year10"]
    return c                                   # costs by year 0..N

def npv(flows, i):
    return sum(f / (1 + i) ** t for t, f in enumerate(flows))

def annual_equiv(pv, i):                       # capital recovery factor A/P
    return pv * i * (1 + i) ** N / ((1 + i) ** N - 1)

i = 0.05
for name, opt in [("Solar", solar), ("Diesel", diesel)]:
    pv = npv(cashflows(opt), i)
    print(f"{name:6s}: present cost ${pv:9,.0f}, equivalent annual cost ${annual_equiv(pv, i):7,.0f}")

# Incremental investment: extra capital for solar, repaid by lower running costs
inc = [d - s for s, d in zip(cashflows(solar), cashflows(diesel))]   # savings by year
lo, hi = 0.0, 1.0
for _ in range(60):                            # internal rate of return by bisection
    mid = (lo + hi) / 2
    if npv(inc, mid) > 0:
        lo = mid
    else:
        hi = mid
print(f"Incremental IRR of solar over diesel: {lo:.1%}")
cum, year = inc[0], 0
while cum < 0:
    year += 1
    cum += inc[year] / (1 + i) ** year
print(f"Discounted payback at 5%: {year} years")
for r in [0.03, 0.08, 0.12]:
    diff = npv(cashflows(diesel), r) - npv(cashflows(solar), r)
    print(f"At {r:.0%}: solar saves ${diff:8,.0f} in present value")

Solar : present cost $   58,653, equivalent annual cost $  4,706
Diesel: present cost $  116,378, equivalent annual cost $  9,338
Incremental IRR of solar over diesel: 24.7%
Discounted payback at 5%: 5 years
At 3%: solar saves $  74,167 in present value
At 8%: solar saves $  39,708 in present value
At 12%: solar saves $  23,689 in present value

The solar option costs $27,000 more to build but saves $6,700 a year in fuel and maintenance, and its 20-year present cost is about half the diesel option's. The extra investment earns an internal rate of return of about 25 %, far above any reasonable MARR, and pays back, even after discounting, in five years. The conclusion holds across a wide range of discount rates. Notice that the diesel option looked cheaper on first cost alone; choosing on purchase price is the most common economic mistake in engineering, and the pump example in Lesson 11 makes the same point.

Rate of return and payback

The internal rate of return (IRR) is the discount rate at which an investment's present worth is zero: the effective interest the investment earns. An investment is attractive if its IRR exceeds the MARR. When comparing options, the correct approach is to compute the IRR of the incremental investment, the extra money spent on the more expensive option, as the code does, because the option with the higher IRR on its own is not necessarily the better choice.

Payback period, the time for savings to repay the initial investment, is popular because it is easy to understand, and a short payback indicates low risk. But simple payback ignores the time value of money and everything after the payback date, so it can favour quick, small savings over larger long-term ones. Use it as a screening tool alongside present worth or IRR, not alone.

Life-cycle cost

Life-cycle cost analysis adds up all the costs of owning an asset over its life: design and construction, operation and energy, maintenance and repair, replacement of components, and disposal or salvage, all discounted to a common basis. For many engineered assets, such as pumps, motors, lighting, buildings and pavements, operating and maintenance costs over the life far exceed the purchase price. Public owners increasingly require life-cycle cost analysis for major decisions, and Lesson 11 links it to life-cycle environmental assessment, which counts carbon and other impacts in the same way.

Life-cycle thinking also reveals the cost of deferring maintenance. A road resurfaced at the right time costs a fraction of one allowed to deteriorate until it must be rebuilt; a bridge whose drains are cleaned and joints sealed lasts decades longer. Asset management, the systematic planning of maintenance, renewal and replacement for a whole network of assets, is a growing field for civil, mechanical and electrical engineers.

Inflation, taxes and depreciation

Inflation, the general rise in prices, means that a dollar in the future buys less. Analyses can be done in nominal (actual, inflated) dollars with a nominal discount rate, or in real (constant, today's) dollars with a real discount rate; mixing the two is a common error. The real rate is approximately the nominal rate minus the inflation rate: a 7 % nominal rate with 2 % inflation is about a 5 % real rate. Costs that rise faster than general inflation, such as energy in some periods, can be modelled with their own escalation rate.

For businesses, taxes reduce both the income from an investment and, through depreciation deductions, its effective cost. Depreciation spreads the cost of an asset over its useful life for accounting and tax purposes, and tax rules specify how quickly each class of asset may be written off.

In Canada, depreciation for tax purposes is claimed as capital cost allowance (CCA), with rates set for each asset class by the Income Tax Act; some clean energy equipment qualifies for accelerated rates.

Detailed after-tax analysis is the subject of a full engineering economics course; for now, recognize that taxes can change which option is best and that accountants and engineers need to work together on major investment decisions.

Uncertainty and sensitivity analysis

Every economic analysis rests on estimates: costs, savings, lives, interest rates, future prices. Rather than pretending to precision, good analyses test how sensitive the conclusion is to the inputs. A sensitivity analysis changes one input at a time over a plausible range and sees whether the preferred option changes; the code above does this for the discount rate. A break-even analysis finds the value of an input at which two options are equally attractive: for example, the diesel price below which diesel would be preferred. Scenario analysis combines several inputs into optimistic, likely and pessimistic cases. If the decision is the same across all plausible values, it is robust; if it flips, the uncertain input deserves more study before committing.

Beyond the numbers

Economic analysis is a tool for decision-making, not the decision itself. Some important factors are hard to express in money: safety, reliability, environmental impact, effects on communities, fairness between groups and between generations, and the preferences of the people who will live with the result. A diesel generator brings noise, fumes, fuel deliveries over winter roads and spill risk; a solar system brings independence and fewer moving parts. Benefit-cost analysis, used for public projects, attempts to value some of these effects, such as travel time saved or injuries avoided, but always involves judgements that should be stated openly. Discounting itself raises questions: at high discount rates, costs that fall on future generations, such as climate impacts or the cost of replacing infrastructure that has been neglected, almost vanish from the analysis.

Engineers should therefore present economic results alongside the other criteria, often in a decision matrix (Lesson 1), state their assumptions, and recognize that the code of ethics places the safety and welfare of the public above economic gain.