Water Resources HubFE section 5 of 16 · free theory
Engineering Economics
Time value of money, the interest factors, NPV, IRR, benefit-cost, breakeven, depreciation, and inflation — the section where fluent factor use turns 4–6 questions into free marks.
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The time value of money
A dollar today is worth more than a dollar next year, because today's dollar can earn interest. Every question in this section is a way of moving cash flows between points in time — and the interest factors are the machinery for doing it:
F = P(1 + i)n
| P | present worth (value at time 0) |
| F | future worth (value at time n) |
| i | effective interest rate per period |
| n | number of periods |
The standard interest factors
Learn to read the factor notation fluently: (X/Y, i%, n) means "find X given Y at interest rate i over n periods." There are six factors built from single payments and uniform series:
(F/P, i, n) = (1 + i)n find F given P
(P/F, i, n) = (1 + i)−n find P given F
(F/A, i, n) = [(1 + i)n − 1] / i find F given A
(A/F, i, n) = i / [(1 + i)n − 1] find A given F (sinking fund)
(P/A, i, n) = [(1 + i)n − 1] / [i(1 + i)n] find P given A
(A/P, i, n) = [i(1 + i)n] / [(1 + i)n − 1] find A given P (capital recovery)
Each factor is the reciprocal of its partner: (P/F) = 1/(F/P), (A/F) = 1/(F/A), (A/P) = 1/(P/A). Also note (A/P, i, n) = (A/F, i, n) + i — capital recovery is the sinking-fund payment plus interest on the principal. The first payment of a uniform series A occurs at the end of period 1 (ordinary annuity); if the question says payments begin immediately, shift the series by one period.
Arithmetic gradient series
Some cash flows grow by a constant amount G each period (maintenance costs are the classic example). The gradient starts at zero in period 1 and reaches G in period 2:
(P/G, i, n) = (1/i)[(1 + i)n − 1] / [i(1 + i)n] − n/(i(1 + i)n) find P given G
(A/G, i, n) = (P/G, i, n) × (A/P, i, n) find A given G
Treat a gradient series as a base uniform series plus a gradient: P = Abase(P/A, i, n) + G(P/G, i, n). If the series decreases instead of growing, G is negative — the formulas do not care.
Nominal vs effective interest
The quoted rate is rarely the rate you use. When a nominal rate r is compounded m times per year, the effective annual rate is higher — and it is the effective rate that goes into the factors:
ieff = (1 + r/m)m − 1
| r | nominal annual rate (quoted) |
| m | compounding periods per year |
| ieff | effective annual rate (the one you use) |
Compounding continuously is the limit: ieff = er − 1. The golden rule: the interest period and the payment period must match. If payments are monthly, convert the annual rate to a monthly effective rate first.
Comparing alternatives: PW, AW, FW
To compare projects, reduce every alternative to a single equivalent number using the same rate (the minimum attractive rate of return, MARR). Present worth, annual worth, and future worth all rank alternatives identically — pick whichever is easiest:
PW = Σ receipts(P/F or P/A …) − Σ disbursements(P/F or P/A …)
AW = PW × (A/P, i, n)
Choose the alternative with the highest (least negative) PW / AW / FW
Alternatives must be compared over a common study period. For unequal lives, use the least common multiple of the lives (repeatability assumed) or convert everything to annual worth — AW needs no common multiple, which is why it is the workhorse on the FE.
NPV, IRR, and benefit-cost ratio
These are the three decision metrics the exam tests. NPV is the workhorse; IRR is the rate; benefit-cost is the ratio:
NPV = −C0 + Σt=1..n Ct/(1 + i)t accept if NPV > 0
IRR: the rate i* such that NPV(i*) = 0 accept if IRR > MARR
B/C = PW(benefits) / PW(costs) accept if B/C > 1
| C0 | initial investment (at time 0) |
| Ct | net cash flow in period t |
| i* | internal rate of return — the break-even discount rate |
NPV and IRR usually agree on accept/reject, but they can rank mutually exclusive projects differently when cash-flow timing differs — when in doubt on the exam, NPV (incremental analysis) is the tiebreaker. Watch for multiple IRRs when cash flows change sign more than once.
Breakeven analysis
Breakeven is the point where total revenue equals total cost — the minimum volume that avoids a loss. In economic terms it is also the point where NPV = 0:
Breakeven volume Q* = Fixed costs / (unit price − unit variable cost)
| Q* | units to produce and sell to break even |
| unit price − unit variable cost | contribution margin per unit |
The denominator is the contribution margin — each unit sold contributes that much toward the fixed costs. The exam loves the distractor of dividing by price alone and ignoring variable costs.
Depreciation
Depreciation spreads an asset's cost over its useful life. The FE tests straight-line and declining-balance computation, plus MACRS concepts:
Straight-line: Dt = (Cost − Salvage) / n BVt = Cost − t × D
Declining balance (rate R): Dt = R × BVt−1 BVt = Cost × (1 − R)t
| Dt | depreciation charge in year t |
| BVt | book value at end of year t |
| R | declining-balance rate (double-declining uses R = 2/n) |
MACRS (the US tax system) uses prescribed recovery periods and switches to straight-line when that gives larger deductions — conceptually, know that it accelerates depreciation relative to straight-line; the exam rarely asks you to look up MACRS tables. Note: land is not depreciable, and book value can never drop below salvage under straight-line.
Inflation-adjusted analysis
When cash flows are in today's dollars (real) but the discount rate is quoted in nominal terms, adjust one to match the other:
if = (1 + i)(1 + f) − 1 = i + f + if
| if | inflated (nominal/market) interest rate |
| i | real interest rate |
| f | inflation rate |
Match like with like: discount real cash flows with the real rate, actual (inflated) cash flows with the inflated rate. Mixing them is the trap. For small rates, if ≈ i + f is a fine approximation.
Worked example Sinking into the future — the (F/A) factor
Given: You deposit $3,000 at the end of each year for 8 years into an account earning 5% effective annual interest. What is the accumulated amount just after the last deposit?
Solution:
- Identify the pattern: a uniform series A = $3,000, find F — the (F/A, 5%, 8) factor.
- (F/A, 5%, 8) = [(1.05)8 − 1] / 0.05 = (1.47746 − 1) / 0.05 = 9.54911.
- F = 3,000 × 9.54911 = $28,647.
- Sanity check: 8 × $3,000 = $24,000 deposited, so $28,647 after interest is reasonable.
Answer: F ≈ $28,647.
Worked example Growing maintenance costs — the gradient factor
Given: A pump's maintenance costs $2,000 in year 1 and increases by $300 each year through year 6. At 7% interest, what is the present worth of all maintenance costs?
Solution:
- Split into a base uniform series A = $2,000 plus an arithmetic gradient G = $300.
- Base: (P/A, 7%, 6) = [(1.07)6 − 1] / [0.07(1.07)6] = 4.76654. PWbase = 2,000 × 4.76654 = $9,533.
- Gradient: (P/G, 7%, 6) = (1/0.07)[4.76654 − 6/(1.07)6] = 10.97838. PWgrad = 300 × 10.97838 = $3,294.
- P = 9,533 + 3,294 = $12,827.
Answer: P ≈ $12,827.
Worked example NPV and benefit-cost for a project decision
Given: A drainage improvement costs $50,000 now and delivers $9,000 per year in flood-damage reduction for 8 years. The agency's MARR is 6%. Should the project be accepted?
Solution:
- Present worth of benefits: PWb = 9,000 × (P/A, 6%, 8). (P/A, 6%, 8) = [(1.06)8 − 1] / [0.06(1.06)8] = 6.20979. PWb = $55,888.
- NPV = 55,888 − 50,000 = $5,888. Positive — the project earns more than the 6% MARR.
- B/C = 55,888 / 50,000 = 1.12. Greater than 1 — same verdict.
- (IRR would be the rate making (P/A, i*, 8) = 50,000/9,000 = 5.556, i.e. about 8.8% — above the 6% MARR, confirming acceptance.)
Answer: Accept — NPV ≈ +$5,888 and B/C ≈ 1.12.