ECE 192: Engineering Economics and Impact on Society
Dario Peralta Moarry
Estimated study time: 1 hr 16 min
Table of contents
Sources and References
Primary textbook — N. M. Fraser, E. M. Jewkes, and M. Pirnia, Engineering Economics: Financial Decision Making for Engineers, 6th ed., Pearson Canada, 2016. (eTextbook edition required for Winter 2026.)
Supplementary textbook — D. G. Newnan, T. G. Eschenbach, and J. P. Lavelle, Engineering Economic Analysis, 4th ed., Oxford University Press, 2018.
Open educational resources — T. Eschenbach, Engineering Economy: Applying Theory to Practice, 3rd ed. (public domain excerpts); MIT OpenCourseWare 1.011 “Project Evaluation and Finance” (open access); T. Ward & J. Uysal, Introduction to Engineering Economics (open textbook, BCcampus); Canadian Revenue Agency, Capital Cost Allowance regulations (public government document).
Chapter 1: Economic Analysis and Cost Estimation
1.1 The Role of Economic Analysis in Engineering Decision-Making
Engineers are routinely asked to make choices: which design to build, whether a piece of capital equipment is worth replacing, how to compare two technically equivalent power-generation technologies with different cost profiles. These questions are not answered by Kirchhoff’s laws or structural mechanics alone. They require a systematic framework for comparing costs and benefits that occur at different points in time and that carry different degrees of uncertainty. That framework is engineering economics.
The central insight of engineering economics is that money has a time value: a dollar received today is worth more than a dollar received a year from now, because the dollar in hand can be invested and will earn a return. Conversely, a future cash obligation is less burdensome in today’s terms than an equivalent present obligation. This single principle — the time value of money — underlies virtually every technique covered in this course, from simple compound-interest calculations to sophisticated benefit-cost ratios and decision-tree analysis.
Engineering economics also insists on a lifecycle perspective. A piece of equipment is not merely purchased; it must be maintained, eventually overhauled, and ultimately disposed of. A power plant does not merely generate electricity; it also produces emissions that impose costs on society. The full accounting of costs and benefits across the entire life of a project is what distinguishes sound engineering analysis from naive comparisons of first cost alone.
1.2 Classification of Engineering Costs
Before any economic analysis can proceed, costs must be identified, classified, and estimated. Engineers distinguish several fundamental cost categories.
Fixed costs do not vary with the level of output or activity. A factory lease payment, annual software licence fees, and insurance premiums are fixed for the duration of the contract period. Variable costs scale with activity: raw materials, direct labour tied to production volume, and electricity consumed per unit manufactured are all variable. The total cost of an operation is the sum of its fixed and variable components, and the relationship between them defines the break-even point (discussed in Chapter 8).
Sunk costs are past expenditures that cannot be recovered regardless of the decision being made. A fundamental principle of rational economic analysis is that sunk costs must be ignored when evaluating future alternatives. If your firm has already spent $500,000 on a software system that is not working, that half-million dollars is gone; it should not influence whether you spend an additional $200,000 attempting to fix it or $150,000 replacing it with a competitor’s product. Only the prospective costs and benefits of each alternative are relevant.
Opportunity costs capture what is given up by choosing one option over the next best alternative. If capital is tied up in a project earning 6% per year, the opportunity cost is the 8% the same capital could have earned in an alternative investment of similar risk. Opportunity cost is the economic rationale for the minimum attractive rate of return (MARR), introduced in Section 1.4.
1.3 Cost Estimation Methods
Accurate cost estimation is both the starting point and one of the most difficult tasks in economic analysis. The three principal estimation approaches, in decreasing order of effort and precision, are as follows.
The detailed estimate (also called bottom-up or engineering estimate) involves identifying every component, activity, and resource required by the project and pricing each one individually. This approach yields the most accurate result but requires a fully developed design and substantial effort. It is appropriate for final investment decisions on major capital projects.
The analogous estimate (also called parametric or top-down estimation) extrapolates from historical data on similar completed projects. If a 50 MW natural-gas peaking plant cost $85 million in 2020, an engineer might estimate that a 75 MW plant in 2026 will cost roughly
\[ \text{Cost}_{2026} = \$85\text{M} \times \left(\frac{75}{50}\right)^{0.6} \times \frac{\text{CPI}_{2026}}{\text{CPI}_{2020}}, \]where the exponent 0.6 reflects the economy-of-scale (six-tenths) rule and the CPI ratio accounts for inflation. The six-tenths rule holds approximately across many types of chemical and electrical processing equipment.
The order-of-magnitude estimate (also called a conceptual estimate) provides a rough sense of project cost, typically accurate to within −30% to +50%, and is used early in project screening before significant design effort is committed.
1.4 Cash Flow Diagrams
A cash flow diagram is the fundamental bookkeeping tool of engineering economics. It represents a project’s revenues and expenditures as arrows on a timeline, with time measured in discrete periods (usually years or months). The convention is:
- Downward arrows represent costs (cash outflows, expenditures).
- Upward arrows represent revenues or savings (cash inflows, benefits).
- The height of each arrow is proportional to the magnitude of the cash flow.
- Period 0 is “now” (the present); the arrow at period \( n \) represents a cash flow occurring at the end of period \( n \).
The cash flow diagram shows: a downward arrow of $120,000 at period 0; downward arrows of $8,000 at the end of each of years 1–8; an additional downward arrow of $25,000 at year 4; and an upward arrow of $15,000 at year 8 (the salvage value is a positive inflow when the asset is sold). The net outflow at year 4 is $8,000 + $25,000 = $33,000. The net inflow at year 8 is −$8,000 + $15,000 = +$7,000.
1.5 The Minimum Attractive Rate of Return
The minimum attractive rate of return (MARR), sometimes called the hurdle rate, is the lowest rate of return an organization considers acceptable for a new investment. It reflects the opportunity cost of capital: if the firm can earn MARR on other investments of comparable risk, then any proposed project must clear this threshold to deserve funding.
The MARR is set by management, not calculated from the project data. It accounts for the cost of capital (the weighted average of equity return and debt interest), a risk premium for the type of project, and sometimes a premium for capital rationing (when investment opportunities exceed available funds). In this course, problems will specify the MARR (often labelled \( i^* \) or \( i_{MARR} \)) and you will use it as the discount rate in all equivalence calculations.
Chapter 2: Time Value of Money and the Power of Compounding
2.1 Simple vs. Compound Interest
Interest is the price paid for the use of borrowed money, or equivalently, the return earned on invested money. Two conventions exist for computing interest.
Simple interest applies the interest rate only to the original principal \( P \), regardless of how long the money has been invested. After \( n \) periods at rate \( i \) per period:
\[ F = P(1 + ni). \]Simple interest is rarely used for multi-period investments in practice, though it does appear in some short-term lending instruments such as Canadian Treasury Bills and certain mortgages.
Compound interest applies the interest rate to the accumulated balance at the beginning of each period, meaning that interest itself earns interest. After \( n \) periods at rate \( i \) per period:
\[ F = P(1 + i)^n. \]The quantity \( (1 + i)^n \) is called the single-payment compound-amount factor, and this relation is the cornerstone of all engineering economic analysis.
Simple interest:
\[ F = \$10{,}000 \times (1 + 5 \times 0.08) = \$10{,}000 \times 1.40 = \$14{,}000. \]Compound interest:
\[ F = \$10{,}000 \times (1.08)^5 = \$10{,}000 \times 1.4693 = \$14{,}693. \]The difference of $693 illustrates the benefit of compounding: the interest earned in years 1 through 4 itself earns interest in subsequent years.
2.2 Nominal vs. Effective Interest Rates
Interest rates are often quoted as annual rates, but compounding may occur more frequently — monthly for mortgages and credit cards, daily for many savings accounts, continuously for theoretical analysis.
Let \( r \) be the nominal (stated) annual interest rate and \( m \) the number of compounding periods per year. Then the interest rate per compounding period is \( r/m \), and the effective annual interest rate is
\[ i_e = \left(1 + \frac{r}{m}\right)^m - 1. \]As \( m \to \infty \), compounding becomes continuous and \( i_e = e^r - 1 \), where \( e \) is Euler’s number.
The effective cost to the cardholder is nearly 22% per year, despite the stated rate of “just under 20%.” This discrepancy is why Canadian law (the Bank Act) requires financial institutions to disclose the effective annual rate (EAR) for consumer lending products.
2.3 Present Worth and Future Worth Factors
Engineering economics uses a compact factor notation to represent time-value calculations. The six standard factors appear throughout the course:
| Factor name | Symbol | Formula | Meaning |
|---|---|---|---|
| Single-payment compound-amount | \( (F/P,\, i,\, n) \) | \( (1+i)^n \) | Given P, find F |
| Single-payment present-worth | \( (P/F,\, i,\, n) \) | \( (1+i)^{-n} \) | Given F, find P |
| Uniform-series compound-amount | \( (F/A,\, i,\, n) \) | \( \frac{(1+i)^n - 1}{i} \) | Given A, find F |
| Uniform-series sinking-fund | \( (A/F,\, i,\, n) \) | \( \frac{i}{(1+i)^n - 1} \) | Given F, find A |
| Uniform-series capital-recovery | \( (A/P,\, i,\, n) \) | \( \frac{i(1+i)^n}{(1+i)^n - 1} \) | Given P, find A |
| Uniform-series present-worth | \( (P/A,\, i,\, n) \) | \( \frac{(1+i)^n - 1}{i(1+i)^n} \) | Given A, find P |
These factors are tabulated in most textbooks and can also be computed directly. The functional notation reads naturally: \( (F/P,\, 8\%,\, 5) \) means “the factor that converts a present amount P into an equivalent future amount F, at 8% per period, over 5 periods.”
Chapter 3: Interest, Equivalence, and Cash Flow Analysis
3.1 Economic Equivalence
Two cash flow sequences are economically equivalent at a given interest rate if they have the same value when expressed at any common point in time. Equivalence is the foundation of all comparison methods: present worth, annual worth, and future worth analyses all work by converting a set of cash flows to a single equivalent amount at a reference point.
The concept of equivalence is subtler than it first appears. Consider $1,000 today versus $1,080 one year from now at an interest rate of 8% per year. These two amounts are equivalent: an investor who receives $1,000 today and invests it for one year at 8% will have exactly $1,080 at year-end. Neither amount is “more” than the other in an economic sense — they have identical value to an investor facing an 8% market rate.
3.2 Uniform (Ordinary Annuity) Cash Flows
An ordinary annuity (or uniform series) is a sequence of equal cash flows \( A \) occurring at the end of each period for \( n \) periods. This pattern arises frequently in engineering: equal annual loan payments, uniform annual maintenance contracts, and constant annual revenues from a stable product line.
The present worth of a uniform series is
\[ P = A \cdot \frac{(1+i)^n - 1}{i(1+i)^n} = A \cdot (P/A,\, i,\, n). \]The inverse relation gives the capital recovery formula — the equal annual payment required to repay a present amount \( P \) with interest:
\[ A = P \cdot \frac{i(1+i)^n}{(1+i)^n - 1} = P \cdot (A/P,\, i,\, n). \]First, convert the nominal rate to an effective monthly rate. With semi-annual compounding at 5% nominal:
\[ i_{\text{semi}} = 0.05/2 = 0.025 \text{ per half-year.} \]Effective monthly rate:
\[ i_{\text{monthly}} = (1 + 0.025)^{1/6} - 1 = (1.025)^{1/6} - 1 \approx 0.004124 = 0.4124\% \text{ per month.} \]Number of payments: \( n = 25 \times 12 = 300 \).
\[ A = \$400{,}000 \times (A/P,\, 0.4124\%,\, 300) = \$400{,}000 \times \frac{0.004124 \times (1.004124)^{300}}{(1.004124)^{300} - 1}. \]Computing \( (1.004124)^{300} \approx 3.437 \):
\[ A = \$400{,}000 \times \frac{0.004124 \times 3.437}{3.437 - 1} = \$400{,}000 \times \frac{0.01418}{2.437} \approx \$400{,}000 \times 0.005818 \approx \$2{,}327/\text{month.} \]3.3 Arithmetic Gradient Series
In many engineering problems, costs or revenues do not remain constant but increase (or decrease) by a fixed dollar amount each period. An arithmetic gradient adds (or subtracts) an amount \( G \) to the cash flow in each successive period beyond the first. The cash flow at period \( t \) is \( A_1 + (t-1)G \) for \( t = 1, 2, \ldots, n \).
The present worth of the gradient component alone (the part that grows at \( G \) per period) is
\[ P_G = G \cdot \frac{(1+i)^n - 1 - ni}{i^2(1+i)^n} = G \cdot (P/G,\, i,\, n). \]The full present worth of an arithmetic gradient series starting at \( A_1 \) in period 1 and increasing by \( G \) per period is \( P = A_1 (P/A,\, i,\, n) + G (P/G,\, i,\, n) \).
From tables: \( (P/A,\, 10\%,\, 6) = 4.3553 \), \( (P/G,\, 10\%,\, 6) = 9.6842 \).
\[ P = \$5{,}000 \times 4.3553 + \$1{,}000 \times 9.6842 = \$21{,}777 + \$9{,}684 = \$31{,}461. \]3.4 Geometric Gradient Series
A geometric gradient grows each period by a fixed percentage \( g \) rather than a fixed dollar amount. This models inflation-escalating costs, revenue from a product with market growth, or energy prices that rise at a steady annual rate. If the first-period cash flow is \( A_1 \), the cash flow at period \( t \) is \( A_1(1+g)^{t-1} \).
When \( g \neq i \), the present worth of the geometric gradient series is
\[ P = A_1 \cdot \frac{1 - (1+g)^n (1+i)^{-n}}{i - g}. \]When \( g = i \), this formula is indeterminate (0/0), and the present worth reduces to \( P = nA_1/(1+i) \).
3.5 Non-Standard Cash Flows and Superposition
Real projects often combine multiple types of cash flows. The superposition principle allows engineers to decompose a complex cash flow series into standard components, compute the present worth (or annual worth, or future worth) of each component separately using the factor formulas, and sum the results.
The key skill is drawing a clear cash flow diagram before any algebra begins, labelling each arrow with its amount and timing. Once the diagram is correct, the calculation almost writes itself.
Chapter 4: Depreciation
4.1 Why Depreciation Matters
Capital equipment — machinery, vehicles, buildings, computer hardware — loses value over time through physical wear, technological obsolescence, and market changes. Depreciation is the systematic allocation of an asset’s cost over its useful life, representing the decline in the asset’s book value.
Depreciation has two important economic roles. First, it provides a more accurate accounting of a firm’s true cost of producing goods or services: the production cost of a widget should include not only labour and materials but also the fraction of the machine’s value consumed in making that widget. Second, and critically for after-tax analysis, depreciation is a non-cash expense that reduces taxable income. A firm that depreciates a $100,000 machine over 5 years reports $20,000 per year in depreciation expense, which reduces its taxable income by $20,000 per year — generating a tax shield even though no cash actually leaves the firm.
Three parameters define any depreciation calculation: the initial cost \( P \) (the purchase price plus installation), the salvage value \( S \) (the estimated market value at end of life), and the useful life \( n \) (the number of periods over which the asset is depreciated).
4.2 Straight-Line Depreciation
Straight-line (SL) depreciation allocates an equal fraction of the depreciable amount \( (P - S) \) to each period:
\[ d_t = \frac{P - S}{n} \quad \text{for } t = 1, 2, \ldots, n. \]The book value at the end of period \( t \) is
\[ BV_t = P - t \cdot \frac{P-S}{n}. \]Straight-line depreciation is the simplest method and is commonly used for accounting purposes in Canada (and required for certain asset classes under the Income Tax Act).
4.3 Declining Balance Depreciation
Declining balance (DB) depreciation applies a fixed percentage \( d \) of the current book value each period. Because the book value falls each year, the depreciation charge also falls — an accelerated front-loading of expenses. The depreciation in period \( t \) is
\[ D_t = P \cdot d \cdot (1 - d)^{t-1}, \]and the book value after \( t \) periods is
\[ BV_t = P (1-d)^t. \]A common choice is the double-declining balance (DDB) method, which sets \( d = 2/n \), twice the straight-line rate. Since DDB does not inherently reach the salvage value \( S \) at year \( n \), many implementations switch to straight-line for the remaining periods once the straight-line charge would exceed the declining-balance charge.
4.4 Capital Cost Allowance (CCA) — Canadian Tax Depreciation
For Canadian income-tax purposes, the depreciation system is called Capital Cost Allowance (CCA) and is governed by the Income Tax Act. Unlike accounting depreciation (which the firm controls), CCA rates and rules are set by the Canada Revenue Agency (CRA) and apply uniformly across all taxpayers.
Key CCA features:
- Assets are grouped into CCA classes. For example, computers and systems software (Class 50) have a 55% declining-balance CCA rate; most manufacturing equipment (Class 43) uses 30%; most buildings (Class 1) use 4%.
- The half-year rule (half-year convention): in the year an asset is acquired, only half the usual CCA may be claimed, regardless of when during the year the purchase occurred. This prevents bunching of purchases at year-end to maximize deductions.
- CCA is claimed on the undepreciated capital cost (UCC) of the class, not on individual assets. When an asset is sold, the proceeds reduce the UCC (to a minimum of zero); the remaining UCC continues to attract CCA.
The CCA claimed in year \( t \geq 2 \) is \( d \times \text{UCC}_{t-1} \); in year 1 it is \( (d/2) \times P \).
| Year | Opening UCC | CCA (55%) | Closing UCC |
|---|---|---|---|
| 1 | $80,000 | $80,000 × 0.55/2 = $22,000 | $58,000 |
| 2 | $58,000 | $58,000 × 0.55 = $31,900 | $26,100 |
| 3 | $26,100 | $26,100 × 0.55 = $14,355 | $11,745 |
| 4 | $11,745 | $11,745 × 0.55 = $6,460 | $5,285 |
Note the half-year rule in year 1: only $22,000 is claimed rather than $44,000. By the end of year 4, the server’s UCC has fallen to $5,285, even though it may still be perfectly functional.
4.5 Sum-of-the-Years’-Digits (SOYD) Depreciation
SOYD depreciation is another accelerated method. The depreciation in year \( t \) is
\[ D_t = \frac{(n - t + 1)}{\text{SOYD}} \cdot (P - S), \quad \text{where } \text{SOYD} = \frac{n(n+1)}{2}. \]For a 5-year life, SOYD = 1 + 2 + 3 + 4 + 5 = 15. Year 1 claims 5/15, year 2 claims 4/15, …, year 5 claims 1/15 of the depreciable amount. SOYD is less common in Canadian practice than declining balance but appears in the optional textbook (Newnan et al.) and is occasionally used for specialized asset valuation.
Chapter 5: Present Worth, Annual Worth, and Future Worth Analysis
5.1 The Logic of Equivalence-Based Comparison
When an engineer must choose among two or more alternative projects, the comparison must be made on an equivalent basis — that is, all alternatives must be evaluated at the same point in time and over the same time horizon. The three most widely used equivalence-based methods are present worth (PW) analysis, annual worth (AW) analysis, and future worth (FW) analysis. They always give consistent rankings because they differ only in the reference point chosen (now, per period, or a future date).
A crucial preliminary step is establishing whether the alternatives have equal lives or unequal lives. If the service period differs, the comparison must either use the least common multiple (LCM) of the lifetimes (assuming repeat purchase) or a specified study period.
5.2 Present Worth Analysis
The present worth (PW) of an alternative is the sum of all its cash flows discounted to time 0 at the MARR:
\[ PW = \sum_{t=0}^{n} C_t (1 + i)^{-t}, \]where \( C_t \) is the net cash flow in period \( t \) (positive for inflows, negative for outflows) and \( i \) is the MARR. An alternative is acceptable if its PW is non-negative (for revenue projects) or if it is the lower cost option (for cost-only projects). Among mutually exclusive alternatives, choose the one with the highest PW.
Pump A:
\[ PW_A = -\$45{,}000 - \$6{,}000 (P/A,\, 8\%,\, 10) + \$5{,}000 (P/F,\, 8\%,\, 10). \]\[ PW_A = -\$45{,}000 - \$6{,}000 \times 6.7101 + \$5{,}000 \times 0.4632 = -\$45{,}000 - \$40{,}261 + \$2{,}316 = -\$82{,}945. \]Pump B:
\[ PW_B = -\$30{,}000 - \$9{,}000 \times 6.7101 + \$2{,}000 \times 0.4632 = -\$30{,}000 - \$60{,}391 + \$926 = -\$89{,}465. \]Since both PWs are negative (cost-only alternatives), choose the less negative: Pump A at a total cost of $82,945 in today’s dollars.
5.3 Annual Worth Analysis
The annual worth (AW) of an alternative converts all cash flows to a uniform annual equivalent over the study period:
\[ AW = PW \times (A/P,\, i,\, n). \]Annual worth is particularly useful when comparing alternatives with unequal lives, because the AW of a repeating sequence of identical projects is just the AW of one cycle — provided the LCM assumption holds. It is also intuitive in a business context: an AW of −$15,000/year can be compared directly to a contract offer or a budget line.
5.4 Future Worth Analysis
Future worth (FW) discounts all cash flows to a future point \( N \) rather than to the present. It is most useful when the decision-maker is specifically interested in accumulated value at a future date (e.g., a retirement savings calculation or a project that delivers all benefits at completion).
\[ FW = PW \times (F/P,\, i,\, N). \]Because PW, AW, and FW are all proportional to one another (through the compound interest factors), rankings among alternatives are always identical regardless of which method is used.
5.5 Payback Period Analysis
The payback period is the time \( T \) required to recover the initial investment from net cash inflows, without discounting. For a project with initial cost \( P \) and uniform annual net inflow \( A \):
\[ T = \frac{P}{A}. \]The discounted payback period incorporates the time value of money by finding \( T \) such that
\[ \sum_{t=1}^{T} C_t (1+i)^{-t} = P. \]Chapter 6: Rate of Return Analysis
6.1 Internal Rate of Return (IRR)
The internal rate of return (IRR) of a project is the interest rate \( i^* \) that makes the project’s present worth exactly equal to zero:
\[ PW(i^*) = \sum_{t=0}^{n} C_t (1 + i^*)^{-t} = 0. \]Equivalently, \( i^* \) is the rate at which the present worth of all inflows equals the present worth of all outflows. The decision rule is straightforward: a project is acceptable if \( i^* \geq \text{MARR} \), and among mutually exclusive alternatives, choose the one with the highest IRR — but only if IRR analysis is applied correctly (see Section 6.3).
Finding \( i^* \) analytically is rarely possible except for the simplest cash flows. In practice, engineers use linear interpolation between two trial interest rates or a spreadsheet solver (Goal Seek / IRR function in Excel). The interpolation formula is
\[ i^* \approx i_1 + (i_2 - i_1) \cdot \frac{|PW_1|}{|PW_1| + |PW_2|}, \]where \( i_1 < i^* < i_2 \) are two trial rates with \( PW_1 > 0 \) and \( PW_2 < 0 \).
Try \( i = 10\% \):
\[ PW = -\$20{,}000 + \$3{,}500 \times (P/A,\, 10\%,\, 10) = -\$20{,}000 + \$3{,}500 \times 6.1446 = -\$20{,}000 + \$21{,}506 = +\$1{,}506. \]Try \( i = 12\% \):
\[ PW = -\$20{,}000 + \$3{,}500 \times (P/A,\, 12\%,\, 10) = -\$20{,}000 + \$3{,}500 \times 5.6502 = -\$20{,}000 + \$19{,}776 = -\$224. \]Interpolating:
\[ i^* \approx 10\% + (12\% - 10\%) \times \frac{1{,}506}{1{,}506 + 224} = 10\% + 2\% \times 0.870 = 11.74\%. \]Since 11.74% < 12% MARR, the project is marginally unacceptable at this MARR. (The true IRR is approximately 11.72%.)
6.2 Incremental IRR Analysis (ΔIRR)
When comparing mutually exclusive alternatives, selecting the one with the highest IRR is correct only if the alternatives have identical initial investments. In general, a lower-cost alternative may have a higher IRR but the incremental investment (the extra dollars spent to choose the more expensive option) might also yield a return above the MARR — making the more expensive alternative the better choice.
Incremental analysis proceeds as follows:
- Order the alternatives from lowest to highest initial investment. The cheapest alternative is the baseline.
- Form the incremental cash flows: \( \Delta C_t = C_t^{\text{high}} - C_t^{\text{low}} \) for each period.
- Find the IRR of the incremental cash flow stream, \( \Delta i^* \).
- If \( \Delta i^* \geq \text{MARR} \), the extra investment is justified; keep the higher-cost alternative as the new baseline.
- Repeat until all alternatives have been compared.
IRR of Design X: solve \( -50{,}000 + 12{,}000 (P/A, i^*, 7) = 0 \), giving \( (P/A, i^*, 7) = 4.167 \). From tables: IRR ≈ 16.3% > 10%, so X is acceptable.
Incremental cash flows (Y − X): \( \Delta C_0 = -\$30{,}000 \), \( \Delta A = +\$6{,}000 \)/year.
Solve \( -30{,}000 + 6{,}000 (P/A, \Delta i^*, 7) = 0 \): \( (P/A, \Delta i^*, 7) = 5.000 \). From tables: \( \Delta i^* \approx 9.2\% \).
Since \( \Delta i^* = 9.2\% < \text{MARR} = 10\% \), the extra $30,000 invested in Y does not earn the MARR. Choose Design X.
6.3 Modified Internal Rate of Return (MIRR)
The standard IRR has a known pitfall: when a project’s cash flows change sign more than once (non-conventional cash flow), there may be multiple positive IRR values, none of which equals the “true” rate of return in an economically meaningful sense. Descartes’ rule of signs tells us the number of sign changes equals an upper bound on the number of positive real roots of the polynomial PW equation.
The Modified Internal Rate of Return (MIRR) resolves this ambiguity by explicitly specifying a reinvestment rate for positive cash flows. The procedure is:
- Find the future worth of all positive (inflow) cash flows at the reinvestment rate \( \epsilon \): \( FW^+ = \sum_{t: C_t>0} C_t (1+\epsilon)^{n-t} \).
- Find the present worth of all negative (outflow) cash flows at the MARR: \( PW^- = \sum_{t: C_t<0} |C_t| (1+i_{\text{MARR}})^{-t} \).
- Solve for MIRR: \( PW^- (1 + \text{MIRR})^n = FW^+ \), so
Accept the project if MIRR ≥ MARR.
6.4 External Rate of Return (ERR)
The External Rate of Return (ERR) is a related concept that also handles non-conventional cash flows by bringing all negative (net cost) cash flows to the present at the MARR and all positive (net benefit) cash flows to the future at an explicitly specified external reinvestment rate. In many formulations, ERR and MIRR are identical when the same rates are used for financing and reinvestment.
Chapter 7: Cost-Benefit Analysis
7.1 The Benefit-Cost Framework
Cost-benefit analysis (CBA) is the standard framework for evaluating public-sector engineering projects — dams, highways, transit systems, power grids, environmental remediation programs — where the project’s outputs are not sold in a market and the “revenue” must be estimated as willingness-to-pay or monetized social value.
The benefit-cost ratio (BCR) is defined as
\[ BCR = \frac{PW(\text{Benefits})}{PW(\text{Costs})}. \]A project is justified if BCR ≥ 1.0 (i.e., the present worth of benefits at least equals the present worth of costs). Among competing projects, the highest BCR does not necessarily indicate the best project — incremental analysis is again required.
7.2 Identifying Benefits, Disbenefits, and Costs
Clear categorization of cash flows is essential in CBA:
- Benefits are positive consequences to the public, expressed in dollar terms: reduced travel time, accident prevention, flood damage avoided, health outcomes improved.
- Disbenefits are negative consequences to the public: noise, displacement, environmental damage, loss of existing tax base.
- Costs are expenditures by the sponsoring agency: capital costs, operating and maintenance costs.
The BCR can be reformulated to include disbenefits:
\[ BCR = \frac{PW(B) - PW(D)}{PW(C)}, \]or equivalently written as \( BCR = PW(B) / [PW(C) + PW(D)] \), depending on the convention. Both conventions yield BCR ≥ 1 for the same projects, but the numerical value differs — always specify which convention is in use.
7.3 Incremental Benefit-Cost Analysis
When comparing mutually exclusive public alternatives, choose the alternative with the highest total net benefit (i.e., largest PW of benefits minus PW of costs). Equivalently, proceed through incremental B/C analysis:
- Order alternatives from lowest to highest cost.
- Compute the incremental BCR for each successive pair: \( \Delta BCR = \Delta PW(B) / \Delta PW(C) \).
- If \( \Delta BCR \geq 1.0 \), the higher-cost alternative is preferred; otherwise, retain the lower-cost alternative.
| Alternative | PW Costs (M) | PW Benefits (M) | BCR |
|---|---|---|---|
| Do nothing | $0 | $0 | — |
| Alt A (grade improvements) | $15 | $22 | 1.47 |
| Alt B (bypass highway) | $40 | $55 | 1.375 |
| Alt C (full freeway) | $70 | $88 | 1.257 |
Incremental A vs. Do Nothing: \( \Delta BCR = 22/15 = 1.47 \geq 1.0 \). Accept A.
Incremental B vs. A: \( \Delta BCR = (55-22)/(40-15) = 33/25 = 1.32 \geq 1.0 \). Accept B.
Incremental C vs. B: \( \Delta BCR = (88-55)/(70-40) = 33/30 = 1.10 \geq 1.0 \). Accept C.
Select Alternative C: despite its lower BCR compared to A, the incremental investment in the full freeway still earns more than a dollar of benefit per dollar of cost above what Alt B would achieve.
7.4 Sensitivity Analysis
Real projects are built on estimated quantities — costs, demand, interest rates, equipment lifetimes — that are uncertain. Sensitivity analysis systematically varies one or more input parameters across a plausible range to determine how much the conclusion (PW, IRR, BCR) changes and which parameters the decision is most sensitive to.
A spider plot (or tornado chart) displays the change in the output metric as each input varies by ±10%, ±20%, etc. from its base value. Parameters that produce steep slopes — large changes in PW per unit change in input — deserve the most attention in risk management.
Break-even analysis (Chapter 8) is a special case of sensitivity analysis in which one asks: at what value of the uncertain parameter does the project’s PW exactly equal zero (or two alternatives become equivalent)?
Chapter 8: Break-Even Analysis and Accounting for Uncertainty
8.1 Break-Even Analysis
Break-even analysis finds the value of an input variable at which a financial outcome (profit, present worth, BCR) transitions from unfavorable to favorable. In cost-volume-profit analysis, the break-even output quantity \( Q^* \) is where total revenue equals total cost:
\[ Q^* = \frac{FC}{P_{\text{unit}} - VC_{\text{unit}}}, \]where \( FC \) is total fixed cost per period, \( P_{\text{unit}} \) is the selling price per unit, and \( VC_{\text{unit}} \) is the variable cost per unit. The denominator is the contribution margin — each unit sold contributes this amount toward covering fixed costs.
At 5,000 units the firm neither profits nor loses. For every unit produced beyond 5,000, the firm earns $40 in contribution margin toward profit.
Break-even analysis also extends to comparing two alternatives: the break-even value of an input parameter is the value at which both alternatives have equal present worth. Below the break-even, one alternative is preferred; above it, the other.
8.2 Basic Probability and Expected Value
When project outcomes are not deterministic, probability theory provides a framework for quantifying uncertainty. If a project can result in several discrete outcomes with associated probabilities, the expected monetary value (EMV) is
\[ EMV = \sum_j p_j \cdot V_j, \]where \( p_j \) is the probability of outcome \( j \) and \( V_j \) is its monetary value (PW, NPV, or profit). The decision rule is to choose the alternative with the highest EMV.
EMV of drilling: \( 0.30 \times (\$2{,}000{,}000 - \$500{,}000) + 0.70 \times (-\$500{,}000) \) \( = 0.30 \times \$1{,}500{,}000 + 0.70 \times (-\$500{,}000) \) \( = \$450{,}000 - \$350{,}000 = +\$100{,}000. \)
EMV of not drilling: $0.
Since EMV(drill) = $100,000 > $0, drilling is preferred on an expected-value basis.
8.3 Decision-Tree Analysis
A decision tree is a graphical tool for structuring multi-stage decisions under uncertainty. It contains:
- Decision nodes (squares): points at which the decision-maker chooses among alternatives.
- Chance nodes (circles): points at which an uncertain event occurs, with associated probabilities summing to 1.0.
- Terminal nodes (triangles or leaves): the final outcomes with their monetary values.
The tree is solved by backward induction (rollback): starting from the terminal nodes, compute the EMV at each chance node, then select the maximum-EMV branch at each decision node, working from right to left.
EMV of immediate launch:
\[ EMV_{\text{immediate}} = 0.40 \times \$800{,}000 + 0.60 \times \$100{,}000 = \$320{,}000 + \$60{,}000 = \$380{,}000. \]EMV of study → favorable result → launch:
\[ EMV_{\text{fav}} = 0.70 \times \$800{,}000 + 0.30 \times \$100{,}000 = \$560{,}000 + \$30{,}000 = \$590{,}000. \]EMV of study → unfavorable result → don’t launch: $0.
EMV of study branch (before subtracting study cost):
\[ EMV_{\text{study}} = 0.50 \times \$590{,}000 + 0.50 \times \$0 - \$50{,}000 = \$295{,}000 - \$50{,}000 = \$245{,}000. \]Since $380,000 > $245,000, the firm should launch immediately without conducting the market study.
Chapter 9: Inflation and After-Tax Analysis
9.1 Inflation and Purchasing Power
Inflation is a sustained rise in the general price level, which erodes the purchasing power of money over time. The general inflation rate \( f \) (often proxied by the Consumer Price Index, CPI, or the GDP deflator) measures the average annual percentage increase in prices across the economy.
Two types of dollar values appear in economic analysis:
- Nominal (then-current, actual) dollars represent cash flows expressed at the price level of the period in which they occur. A salary of $75,000 in 2030 is a nominal dollar amount.
- Real (constant, today’s) dollars express all cash flows at the purchasing power of a chosen base year. To convert a nominal amount \( F_n \) in year \( n \) to its real equivalent in base year 0:
The relationship between the nominal interest rate \( i_n \), the real interest rate \( i_r \), and the inflation rate \( f \) is given by the Fisher equation:
\[ 1 + i_n = (1 + i_r)(1 + f), \quad \Rightarrow \quad i_r = \frac{i_n - f}{1 + f} \approx i_n - f. \]Consistency rule: When performing a present-worth calculation, you must be consistent — discount nominal cash flows at the nominal rate, or discount real cash flows at the real rate. Mixing the two (nominal flows at the real rate, or vice versa) will give wrong answers.
Energy cost is escalating at 4% per year in real terms. In today’s dollars, the cost in year \( t \) is \( \$50{,}000 \times (1.04)^t \). This is a geometric gradient in real dollars, so use the real MARR:
\[ i_r = \frac{0.12 - 0.03}{1.03} \approx 8.74\%. \]Apply the geometric gradient PW formula with \( g = 4\% \), \( i = 8.74\% \), \( n = 10 \), \( A_1 = \$52{,}000 \) (year-1 real cost):
\[ PW = \$52{,}000 \times \frac{1 - (1.04/1.0874)^{10}}{0.0874 - 0.04} \approx \$52{,}000 \times \frac{1 - (0.9569)^{10}}{0.0474}. \]\[ PW \approx \$52{,}000 \times \frac{1 - 0.6469}{0.0474} \approx \$52{,}000 \times 7.446 \approx \$387{,}200. \]9.2 After-Tax Analysis Framework
Canadian corporations pay income tax on net taxable income. Because depreciation (or CCA) reduces taxable income, the after-tax cash flows of a capital investment differ substantially from the before-tax cash flows, and ignoring taxes can significantly overstate a project’s attractiveness.
The after-tax cash flow (ATCF) in year \( t \) is
\[ ATCF_t = \text{BTCF}_t - \text{Tax}_t, \]where \( \text{BTCF}_t \) is the before-tax cash flow, and
\[ \text{Tax}_t = \tau \times (\text{BTCF}_t - D_t), \]with \( \tau \) the marginal corporate tax rate and \( D_t \) the depreciation (or CCA) claimed in year \( t \). Combining:
\[ ATCF_t = (1-\tau) \cdot \text{BTCF}_t + \tau \cdot D_t. \]The second term \( \tau \cdot D_t \) is the tax shield provided by depreciation: depreciation is a non-cash expense that reduces taxes paid, effectively returning \( \tau \cdot D_t \) in cash to the firm each period.
This formula consolidates the infinite geometric series of CCA claims into a closed-form expression, making after-tax analysis computationally tractable without tabulating every year’s CCA.
Chapter 10: Engineering Ethics and Societal Impact
10.1 The Engineer’s Social Responsibility
The preceding chapters have treated economic analysis primarily as a tool for optimizing resource allocation within a firm or public agency. But engineering projects do not occur in a vacuum. A power plant that is economically optimal for its owner may impose costs on neighboring communities through air pollution, water use, or disrupted land. A highway that maximizes traffic throughput may fragment ecosystems and displace residents. A semiconductor fabrication facility that is profitable for its shareholders may consume water in a drought-stressed region.
Professional engineers in Canada are licensed by provincial associations (Professional Engineers Ontario, Engineers Nova Scotia, etc.) and are bound by codes of ethics that explicitly require consideration of public safety, health, welfare, and the environment. The preamble to PEO’s Code of Ethics states that engineers must “hold paramount the safety, health and welfare of the public.” This paramount duty extends beyond the immediate client to society at large and to future generations.
10.2 Externalities and Social Cost
An externality is a cost or benefit of a transaction that falls on parties not directly involved in the transaction — what economists call third parties or “society at large.” When a steel mill emits sulphur dioxide that causes acid rain and damages forests and lakes hundreds of kilometres away, the downstream damage is an externality. The mill-owner bears no direct cost for this damage unless regulation, litigation, or carbon/pollution pricing forces them to internalize it.
The most important externality in contemporary energy engineering is the social cost of carbon (SCC) — the present-value economic damage associated with emitting one additional tonne of CO₂ (or equivalent greenhouse gas). As of the early 2020s, estimates from major government agencies range from approximately US$50 per tonne (U.S. Environmental Protection Agency lower bound, 2021 estimate) to over US$200 per tonne under high-damage scenarios. Canada’s federal backstop carbon price reached $65/tonne CO₂e in 2023 and is scheduled to increase to $170/tonne by 2030 under current policy.
10.3 Environmental Economics and Lifecycle Cost Analysis
Lifecycle cost analysis (LCCA) extends the time horizon of economic analysis to encompass all costs and benefits from initial design and construction through operation, maintenance, decommissioning, and disposal (including environmental remediation). For infrastructure with 30–100 year lifetimes — bridges, power plants, water treatment facilities — LCCA is essential for rational comparison.
Key lifecycle stages for infrastructure:
- Design and procurement: capital cost, procurement overhead, permitting, environmental impact assessment.
- Construction: labour, materials, site preparation, temporary works, contractor overhead.
- Operation: energy, staffing, consumables, insurance.
- Maintenance: routine (preventive) maintenance and periodic major overhauls.
- End-of-life: decommissioning, asset removal, site remediation (potentially very large for nuclear or mining facilities), and any residual asset value.
The engineering economic framework — discounting all lifecycle costs and benefits to present worth — provides the quantitative backbone of LCCA. The key additional challenge is the long time horizon: at a 5% real discount rate, a cost of $1 million 50 years from now has a present value of only $87,200. Long-horizon analysis is therefore sensitive to the choice of discount rate, and low discount rates (reflecting concern for future generations) give more weight to end-of-life and environmental costs.
10.4 Global Energy Systems and Engineering’s Role
Electricity generation is one of the largest contributors to global greenhouse gas emissions. The International Energy Agency (IEA) reported in 2023 that the energy sector accounts for approximately three-quarters of global GHG emissions. Engineering decisions about power generation infrastructure — which technologies to build, which fuels to use, what energy efficiency standards to set — have enormous societal consequences that persist for decades.
The levelized cost of energy (LCOE) is the per-kilowatt-hour cost of building and operating a power plant over its lifetime, expressed in today’s dollars:
\[ LCOE = \frac{PW(\text{all lifetime costs})}{PW(\text{all lifetime energy produced})}. \]This is an annual-worth-per-unit-output measure — the price per kWh at which the plant would exactly break even at the chosen discount rate. LCOE comparisons (with caution) inform the economic competition between coal, natural gas, nuclear, wind, solar photovoltaic, and hydroelectric power.
10.5 Engineering Ethics in Practice
Several canonical case studies illustrate the intersection of economic pressure and engineering ethics:
The Space Shuttle Challenger (1986): Under severe schedule and cost pressure, managers overrode the objections of Thiokol engineers who warned that O-ring performance at low temperatures was not characterized. The catastrophic failure killed all seven crew members. The Rogers Commission found that the decision process had systematically underweighted engineering safety concerns relative to schedule economics.
The Lac-Mégantic Rail Disaster (2013): Cost-cutting decisions — single-person crew operation, reduced brake testing, deferred maintenance — contributed to a runaway oil train derailing in downtown Lac-Mégantic, Quebec, killing 47 people and devastating the town centre. The disaster demonstrated that apparently “rational” cost minimization can externalize catastrophic risks onto communities and the environment.
The Athabasca Oil Sands: The economic viability of oil sands extraction depends heavily on the oil price, the cost of natural gas (used for steam generation), and the carbon price. Environmental externalities — tailings pond contamination, peat bog destruction, water use, and GHG emissions roughly 15–20% higher per barrel than conventional crude — are partly internalized by Canada’s carbon pricing system and regulatory requirements, but the full social cost remains contested in environmental economics literature.
These cases are not exceptional failures — they reflect structural pressures that arise whenever economic incentives misalign with public safety or environmental protection. The engineer’s professional and ethical obligation is to raise concerns through appropriate channels, document objections, and, if necessary, refuse to sign off on work that poses unacceptable risk to the public.
Chapter 11: Synthesis — Putting It All Together
11.1 A Complete After-Tax Project Evaluation
The chapters above have developed the building blocks of engineering economics in sequence. In practice, a complete project evaluation integrates all of them: identify and estimate costs, draw the cash flow diagram, apply CCA to compute the tax shield, compute the after-tax present worth, check IRR against the MARR, assess sensitivity to key uncertain parameters, and explicitly account for societal externalities.
Step 1: Before-tax annual cash flow (uniform portion)
\[ BTCF = \$55{,}000 + \$12{,}000 - \$8{,}000 = \$59{,}000 \text{ per year.} \]Step 2: After-tax uniform cash flow (excluding depreciation tax shield)
\[ ATCF_{\text{operating}} = (1 - 0.265) \times \$59{,}000 = 0.735 \times \$59{,}000 = \$43{,}365 \text{ per year.} \]Step 3: PW of after-tax operating flows
\[ PW_{\text{op}} = \$43{,}365 \times (P/A,\, 10\%,\, 8) = \$43{,}365 \times 5.3349 = \$231{,}353. \]Step 4: PW of CCA tax shield (using the Canadian formula with \( d = 0.20 \), \( i = 0.10 \), \( \tau = 0.265 \), \( P = \$250{,}000 \), \( S = \$20{,}000 \), \( n = 8 \)):
\[ PW_{\text{shield}} = \frac{0.265 \times 0.20 \times 250{,}000}{0.20 + 0.10} \cdot \frac{2 + 0.10}{2 \times 1.10} - \frac{0.265 \times 0.20 \times 20{,}000}{0.20 + 0.10} \cdot \frac{1}{(1.10)^8}. \]First term: \( \frac{13{,}250}{0.30} \times \frac{2.10}{2.20} = 44{,}167 \times 0.9545 = \$42{,}164. \)
Second term: \( \frac{1{,}060}{0.30} \times \frac{1}{2.1436} = 3{,}533 \times 0.4665 = \$1{,}648. \)
\[ PW_{\text{shield}} = \$42{,}164 - \$1{,}648 = \$40{,}516. \]Step 5: PW of after-tax salvage value
When an asset is sold, the proceeds reduce the UCC of the CCA class. If the class remains open (other assets in the class), the after-tax salvage is approximately \( S(1 - \tau d/(d+i) \cdot 1/(1+i)^n) \). A simplified (and common exam) approach for open classes is:
\[ PW_{\text{salvage}} \approx S(1-\tau) \times (P/F,\, 10\%,\, 8) = \$20{,}000 \times 0.735 \times 0.4665 = \$6{,}857. \](A more precise treatment requires adding back the lost tax shield on the salvage amount; the formula in Step 4 already incorporates the salvage correction in its second term.)
Step 6: Total after-tax PW
\[ PW_{\text{total}} = -\$250{,}000 + \$231{,}353 + \$40{,}516 + \$6{,}857 = +\$28{,}726. \]Since PW > 0 at the after-tax MARR of 10%, the automated inspection system is economically justified.
11.2 Sensitivity Analysis on the Complete Example
The result PW = $28,726 is a single-point estimate. Several inputs are uncertain:
- Annual savings (currently $67,000/year = $55,000 labour + $12,000 scrap): if actual savings are 15% lower, \( BTCF = 0.85 \times 67{,}000 - 8{,}000 \approx \$49{,}000 \), giving \( ATCF_{\text{op}} = 0.735 \times 49{,}000 = \$36{,}015 \) and \( PW_{\text{op}} = 36{,}015 \times 5.3349 = \$192{,}158 \) — a drop of $39,000 in PW.
- MARR sensitivity: at MARR = 12%, \( (P/A, 12\%, 8) = 4.9676 \), so \( PW_{\text{op}} = 43{,}365 \times 4.9676 = \$215{,}446 \). Combined with a reduced tax-shield PW (since the discount rate is higher), the total PW drops toward zero.
- Salvage value: modest influence given heavy discounting over 8 years at 10%.
The savings estimate is the most sensitive parameter and should be verified carefully through time-motion studies and waste audits before the investment decision is finalized.
11.3 Replacement Analysis
A recurring engineering economics question is whether to keep an existing asset (the defender) or replace it with a new asset (the challenger). The key principle is that the defender must be evaluated at its current market value (opportunity cost), not its book value or original cost (both sunk).
The economic life of an asset is the number of years that minimizes its annual worth of costs (capital recovery plus operating costs). Operating costs typically increase over time due to aging; capital recovery (the annualized first cost less salvage) decreases as the asset age increases and the annual cost is spread over more years. The sum is U-shaped, with a minimum at the economic life.
For a decision to replace now versus wait one more year, the one-year “cost of keeping the defender” is the sum of: (1) the forgone return on the defender’s market value (opportunity cost); (2) the expected change in market value (loss in value due to aging); and (3) the next year’s operating cost. If this one-year cost exceeds the AW cost of the challenger, replace now.
Summary: Key Formulas and Factor Reference
The following consolidated reference covers all factors used throughout this course. Let \( i \) be the interest rate per period and \( n \) the number of periods.
Single payment:
\[ F = P(1+i)^n, \quad P = F(1+i)^{-n}. \]Uniform series (ordinary annuity):
\[ F = A \cdot \frac{(1+i)^n-1}{i}, \quad P = A \cdot \frac{(1+i)^n-1}{i(1+i)^n}, \quad A = P \cdot \frac{i(1+i)^n}{(1+i)^n-1}. \]Arithmetic gradient (gradient component only):
\[ P_G = G \cdot \frac{(1+i)^n - 1 - ni}{i^2(1+i)^n}, \quad A_G = G \cdot \left[\frac{1}{i} - \frac{n}{(1+i)^n-1}\right]. \]Geometric gradient (\( g \neq i \)):
\[ P = A_1 \cdot \frac{1-(1+g)^n(1+i)^{-n}}{i-g}. \]Effective annual rate:
\[ i_e = \left(1 + \frac{r}{m}\right)^m - 1. \]Fisher equation (inflation):
\[ 1 + i_n = (1+i_r)(1+f). \]After-tax cash flow:
\[ ATCF = (1-\tau) \cdot BTCF + \tau \cdot D_t. \]Straight-line depreciation:
\[ d_t = \frac{P-S}{n}, \quad BV_t = P - t \cdot \frac{P-S}{n}. \]Declining balance:
\[ D_t = P \cdot d(1-d)^{t-1}, \quad BV_t = P(1-d)^t. \]Break-even quantity:
\[ Q^* = \frac{FC}{P_{\text{unit}} - VC_{\text{unit}}}. \]Benefit-cost ratio:
\[ BCR = \frac{PW(B) - PW(D)}{PW(C)}. \]MIRR:
\[ MIRR = \left(\frac{FW^+(\epsilon)}{PW^-(i_{\text{MARR}})}\right)^{1/n} - 1. \]