Engineering Economics · Study · FE Other Disciplines · FE → PE Prep
Engineering Economics
6% of exam
Time value of money, cost analysis, break-even and benefit-cost, depreciation, replacement, and decision analysis.
5 concepts
A. Time value of money
Time Value of Money
The six discrete compound-interest factors, gradients, and nominal-vs-effective conversions that turn any cash-flow diagram into a single present, annual, or future value.
A dollar today is worth more than a dollar next year, because today's dollar can be invested to earn interest. Engineering Economics turns that single idea into machinery: every problem on this part of the FE is really a request to move money along a time line and compare it at one instant. The NCEES FE Reference Handbook (Engineering Economics chapter) hands you a one-page factor table — your job is fluency, knowing which factor moves which kind of cash flow and in which direction. Points are lost not on hard algebra but on reading the diagram wrong: discounting when you should compound, or sliding a present-worth value to the wrong year.
Simple vs compound interest
Simple interest is charged only on the original principal: after n periods you owe P(1+in)
. Compound interest charges interest on previously accrued interest, so growth is geometric,
P(1+i)n
. The FE — and essentially all of finance — uses compound interest unless a problem explicitly says 'simple.' The gap widens fast: at
i=10%
over 20 years, simple interest triples your money while compounding multiplies it by
6.7
.
Fsimple=P(1+in),Fcompound=P(1+i)n
The cash-flow diagram
Before any factor, draw the time line. A horizontal axis marks periods 0,1,2,…,n; upward arrows are receipts (positive), downward arrows are disbursements (negative). The single most important convention is end-of-period: a uniform series A and the period-n future value F both sit at the END of their periods, while a present value P sits at time 0 — which is the end of period 0, i.e. one period BEFORE the first A. Getting this offset wrong is the dominant source of error on the whole topic.
The two single-payment factors
These move one lump sum across time. The single-payment compound-amount factor (F/P,i,n)=(1+i)n pushes a present amount forward to a future value; its reciprocal, the present-worth factor (P/F,i,n)=(1+i)−n, discounts a future amount back. The notation reads like a fraction: (F/P,i,n) multiplied by P leaves F, because the P's 'cancel.' Memorize that pattern — it makes every factor self-checking.
(F/P,i,n)=(1+i)n,(P/F,i,n)=(1+i)−n
The four uniform-series factors
A uniform series is n equal end-of-period payments A (an annuity). Four factors connect A to either P or F. The present-worth (P/A) and capital-recovery (A/P) factors are reciprocals and place P one period before the first A. The sinking-fund (A/F) and compound-amount (F/A) factors are reciprocals and place F at the same instant as the last A. Capital recovery (A/P) is the one you use to turn a first cost into an equivalent annual cost — the backbone of annual-worth comparisons.
Real cash flows often trend. An arithmetic gradient adds a constant G each period: the series is 0,G,2G,…,(n−1)G — note the gradient is zero in year 1, so a flow that starts at base A1 and rises by G is the sum of a uniform series A1 plus the gradient. Use (P/G,i,n) to get its present worth or (A/G,i,n) to convert it to an equivalent uniform A. A geometric gradient grows by a constant rate g (e.g. inflation-indexed costs); it has its own closed form below, with a special case i=g where P=A1n/(1+i).
PG=G(P/G,i,n)=G[i2(1+i)n(1+i)n−1−i(1+i)nn]
Geometric gradient present worth
For a flow A1 in year 1 escalating at rate g per period, the present worth collapses to a single expression. When i=g use the ratio form; this is the standard model for costs that track an escalation rate. Always anchor A1 as the year-1 amount, not the year-0 amount.
P=A1i−g1−(1+i1+g)n(i=g),P=1+iA1n(i=g)
Nominal vs effective rate and compounding
The nominal annual rate r is just a quoted yearly figure; it ignores intra-year compounding. The effective annual rate ie is what you actually earn after compounding m times per year. They are equal only when m=1. The cardinal rule for the factors: i and n must use the SAME period. If interest compounds monthly and payments are monthly, use the monthly rate r/m and n in months — do not annualize. Convert to an effective rate only when periods differ (e.g. annual payments but monthly compounding). As m→∞, continuous compounding gives ie=er−1.
ie=(1+mr)m−1
Exam strategy
Draw the diagram, then ask three questions in order: what do I HAVE, what do I WANT, and over how many periods? That triple picks the factor — (want/have,i,n). Confirm i and n are in the same period before plugging in; this single check kills most errors. The handbook's factor tables give pre-computed values at common rates, so for clean rates you can read the number directly instead of evaluating the formula — but know the closed forms for off-table rates. When a flow has a base plus a trend, split it: uniform part with (P/A), trend with (P/G) or the geometric form, then add. Carry full precision and round only the final answer to 3 significant figures.
Key equations
Single payment, future from present (F/P)F=P(1+i)n
Compounds a single present amount P forward n periods at rate i to its future value F (dollars).
Single payment, present from future (P/F)P=F(1+i)−n
Discounts a single future amount F back n periods to present worth P
Uniform series present worth (P/A)P=Ai(1+i)n(1+i)n−1
Interest on principal only; used only when a problem explicitly states simple interest.
Inflation-adjusted (combined) rated=i+f+if
Combines real rate i and inflation f into the market rate d
Worked examples
Single sum forward
Problem. You deposit $5,000 today in an account earning 6% compounded annually. What is the balance after 8 years?
Solution. Have P, want F, so use (F/P,6%,8)=(1.06)8=1.5938.
F=5000(1.5938)=$7,969.
Sanity: 6% over 8 years should roughly multiply by (1.06)8, and money grew about 59% — consistent with the rule-of-72 doubling time of 72/6=12 years, so we should be well under double. Answer: $7,970 (3 s.f.).
F=5000(1+0.06)8=$7,969
Annuity present worth
Problem. A maintenance contract pays out $1,200 at the end of each year for 10 years. At a MARR of 8%, what lump sum today is equivalent?
Solution. Have A, want P: (P/A,8%,10)=0.08(1.08)10(1.08)10−1=6.7101
Base plus arithmetic gradient
Problem. An operating cost is $1,000 in year 1 and rises by $150 each year through year 6. At i=7%, find the present worth of the 6-year cost stream.
Solution. Split into a uniform A1=$1,000
Nominal to effective rate
Problem. A credit line quotes a nominal 9% per year compounded monthly. What is the effective annual rate, and what is $2,000 worth after one year?
•Placing P at the same instant as the first A. The (P/A) factor puts P exactly one period BEFORE the first payment — if the first payment is at year 0, you must handle it separately.
•Mixing the rate and the count: using an annual i with n in months (or vice versa). Always make i and n share the same period before applying any factor.
•Confusing nominal and effective rates. With m>1 compoundings, the effective rate exceeds the nominal rate; only annualize when payment and compounding periods differ.
•Forgetting that an arithmetic gradient is zero in year 1. A flow of A1 rising by G is A1(P/A)+G(P/G)
•Inverting a factor: multiplying by (A/P) when you meant (P/A). Use the 'want/have' notation as a built-in check — the wanted quantity is the numerator.
•Using simple interest by habit. The FE and finance use compound interest unless a problem explicitly says 'simple.'
•Dividing instead of subtracting for geometric gradients when i=g. That case has its own formula P=A1n/(1+i); the general ratio form divides by zero.
References
NCEES FE Reference Handbook — Engineering Economics
Newnan, Eschenbach & Lavelle, Engineering Economic Analysis — factor derivations and gradient series
B. Cost analysis
Cost Analysis
How to classify, estimate, and escalate costs — fixed vs variable, direct vs indirect, marginal vs average, and why sunk costs never belong in a decision.
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C. Economic analyses
Economic Analysis Methods
Present worth, annual worth, future worth, rate of return with the incremental rule, benefit-cost ratio, break-even, and payback — and when each gives the right answer.
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D. Uncertainty
Uncertainty and Expected Value in Economics
Expected monetary value, decision trees, sensitivity analysis, and the value of perfect information — how to choose when outcomes are probabilistic.
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E. Project selection
Project Selection and Depreciation
Compare unequal-life alternatives, run lease/buy/make decisions, depreciate with straight-line, declining-balance, and MACRS, and time replacements by minimizing annual cost.
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(dollars).
Present worth P of n equal end-of-period payments A; P falls one period before the first A.
Equal end-of-period payment A that recovers a present amount P over n periods at i; reciprocal of (P/A).
of
n
equal payments
A
;
F
occurs at the same time as the last payment.
needed each period to accumulate a future amount
F
; reciprocal of (F/A).
Present worth of a flow 0,G,2G,…,(n−1)G; gradient G in dollars per period, starting in year 2.
Converts an arithmetic gradient G into an equivalent uniform annual amount A.
Present worth of a flow growing at rate g from year-1 base A1; for i=g, P=A1n/(1+i).
Effective annual rate from nominal annual r with m compoundings per year; continuous limit ie=er−1.
used to discount actual (then-current) dollars.
.
P=1200(6.7101)=$8,052
.
Sanity: undiscounted total is
$12,000
; the present worth must be less, and
$8,052
is a reasonable
67%
of that for a 10-year,
8%
stream. Answer:
$8,050
(3 s.f.).
P=12000.08(1.08)10(1.08)10−1=$8,052
plus a gradient
G=$150
.
(P/A,7%,6)=4.7665
and
(P/G,7%,6)=10.9784
.
P=1000(4.7665)+150(10.9784)=4766.5+1646.8=$6,413
.
Sanity: the average annual cost is about
$1,375
;
1375×(P/A)=$6,554
, close to our split result and slightly higher because more weight sits in later (more-discounted) years. Answer:
$6,410
(3 s.f.).
P=1000(P/A,7%,6)+150(P/G,7%,6)=$6,413
.
F=2000(1.09381)=$2,188
.
Sanity: monthly compounding must give an effective rate above the nominal