Engineering Economics · Study · FE Chemical · FE → PE Prep
Engineering Economics
4% of exam
Time value of money (present, annual, and future worth, rate of return), economic analyses (break-even, benefit-cost, optimal economic life), uncertainty with expected value and risk, and project selection with unequal lives, depreciation, and discounted cash flow.
3 concepts
A. Time value of money
Time Value of Money and Interest Factors
The six interest factors, gradients, nominal-vs-effective rate, and rate of return — the toolkit that converts cash flows across time on the FE.
Engineering economics is the most formulaic block on the FE Chemical exam, and the handful of questions here are among the most reliably bankable points on the whole CBT — every one of them reduces to moving money along a timeline. A dollar today is not a dollar next year: it can earn interest, so cash flows at different times are not comparable until you bring them to a common point. The entire subject is built on one idea — equivalence — and a small family of multipliers called interest factors that the NCEES handbook tabulates for you. The 'Engineering Economics' chapter gives you the factor table, the gradient and effective-rate formulas, and a numeric factor table; your job is to know which factor to reach for, not to re-derive any of them. Treat the timeline as a bookkeeping diagram and almost every problem solves itself.
Single-payment factors: moving one lump sum
Start with a single amount. If you have a present amount P and let it compound at rate
i
per period for
n
periods, it grows to a future amount
F
by the single-payment compound-amount factor
(F/P,i,n)=(1+i)n
. Run it backwards — what is a future
F
worth today? — and you divide instead, the single-payment present-worth factor
(P/F,i,n)=(1+i)−n
. The notation is deliberately readable:
(F/P,i,n)
means 'find
F
given
P
,' and the two named quantities cancel like a fraction, so
F=P(F/P,i,n)
. Every other factor is just an annuity built out of these.
F=P(1+i)n,P=F(1+i)−n
Uniform-series factors: the annuity family
A uniform series A is an equal end-of-period payment running for n periods (the FE assumes the ordinary-annuity, end-of-period convention). Four factors connect A to P and F. To find the present worth of a series, use (P/A,i,n); to recover capital as equal payments — the loan-payment or capital-recovery factor — use its reciprocal (A/P,i,n). To accumulate a future sum from deposits, use (F/A,i,n); to find the deposits that build a target fund — the sinking-fund factor — use (A/F,i,n). The single most common error is an off-by-one in n: the first A lands one period after the P, and the last A coincides with F.
P=Ai(1+i)n(1+i)n−1,A=P(1+i)n−1i(1+i)n
Arithmetic gradients
Many cash-flow streams are not flat — maintenance that climbs by a fixed amount each year, output that ramps up. An arithmetic gradient G adds 0 in year 1, G in year 2, 2G in year 3, and so on, layered on top of a base annuity. The gradient present-worth factor (P/G,i,n) converts the G-stream to a present value, and (A/G,i,n) converts it to an equivalent level annuity, so a ramping series becomes Aeq=Abase+G(A/G,i,n). Watch the convention: the gradient is referenced so that the first nonzero increment G occurs at the end of period 2, not period 1.
(A/G,i,n)=i1−(1+i)n−1n
Nominal versus effective interest
Rates are quoted nominally — '6% compounded monthly' — but money actually grows at the effective rate, which is larger whenever compounding happens more than once per period. With nominal annual rate r and m compounding periods per year, the annual effective rate ie follows the handbook's non-annual-compounding formula. The discipline is simple but unforgiving: the interest factors only work when i and n use the same period. If payments are monthly, use the monthly rate r/m and count n in months; if you keep an annual timeline, first convert r to the annual effective ie and use years. Continuous compounding is the limit ie=er−1.
ie=(1+mr)m−1
Present, annual, and future worth
Any cash-flow diagram can be summarized at three reference points, and the FE will ask for whichever is convenient. Present worth PW discounts everything to time zero; future worth FW pushes everything to time n; and annual worth (equivalent uniform annual cost/benefit) AW spreads the net into an equal yearly amount. They are equivalent statements — AW=PW(A/P,i,n) and FW=PW(F/P,i,n) — so a project that is attractive on one measure is attractive on all three. One special case is worth memorizing: a perpetual uniform series (capitalized cost) has present worth P=A/i, because (P/A,i,n)→1/i as n→∞.
AW=PW(A/P,i,n),P∞=iA
Rate of return
The rate of return (ROR, or internal rate of return IRR) of a cash-flow stream is the interest rate that makes its present worth exactly zero — the breakeven discount rate at which benefits and costs are equivalent. You rarely solve for it in closed form; instead you write PW(i∗)=0, then interpolate between two trial rates or read it off the factor table. Compare i∗ to the minimum acceptable rate of return (MARR): accept the project when i∗≥MARR. Because PW(i) can change sign more than once when cash flows alternate, a stream with multiple sign changes can have multiple IRRs — when in doubt, fall back on present worth at the stated MARR.
PW(i∗)=t=0∑n(1+i∗)tCFt=0
Exam strategy
Draw the cash-flow diagram first, every time — it tells you which factor you need and exposes off-by-one errors in n. Confirm that i and n share a period before touching any factor; if compounding is sub-annual, either work in the sub-annual period with r/m or convert to ie and work in years. Read whether the answer wanted is P, A, or F, and pick the single factor that gets there directly rather than chaining three. The handbook's factor table (and the P/F,P/A,F/P,… numeric tables) saves time — interpolate linearly between tabulated rates. For ROR, set PW=0 and interpolate between two rates that bracket a sign change; for an attractive/reject decision, compare i∗ with MARR or simply check the sign of PW at the MARR.
Key equations
Single-payment compound amount (F/P)F=P(1+i)n
Future worth of a present lump sum. i = rate per period, n = number of periods (same period as i).
Single-payment present worth (P/F)P=F(1+i)−n
Discounts a single future amount F to the present. The base building block of all discounting.
Gradient to uniform series (A/G)(A/G,i,n)=i1−(1+i)n−1n
Effective interest rateie=(1+mr)m−1
Equivalence among PW, AW, FWAW=PW(A/P,i,n),FW=PW(F/P,i,n)
Capitalized cost (perpetuity)P∞=iA
Present worth of an infinite uniform series — capitalized cost of perpetual service or maintenance.
Rate of return conditionPW(i∗)=∑t=0n(1+i∗)tCFt=0
Worked examples
Lump-sum growth, annual vs monthly compounding
Problem. You deposit $5,000 in an account paying a nominal 6% per year. Find the balance after 8 years if interest is compounded (a) annually and (b) monthly, and report the effective annual rate in case (b).
Solution. Case (a): annual compounding, i=6%, n=8.
F=5000(1+0.06)8=5000(1.59385)=$7,969.
Case (b): monthly rate r/m=0.06/12=0.005, n=12×8=96 months.
F=5000(1.005)96=5000(1.61414)=$8,071.
Effective annual rate: ie=(1+0.06/12)12−1=0.06168=6.17%.
Sanity check: monthly compounding earns more than annual ($8,071>$7,969) and the effective rate 6.17% exceeds the nominal 6%, exactly as expected. Final answers: $7,970 and $8,070 (3 sig figs).
Loan payment and the present-worth check
Problem. A process-control upgrade costs $25,000, financed at 8% per year over 5 years with equal end-of-year payments. Find the annual payment, then verify by discounting the payments back to the present.
Problem. A pump's maintenance is $1,000 at the end of year 1 and rises by $200 each year through year 5. At i=10%, find the present worth of the maintenance stream and its equivalent uniform annual cost.
Solution. Split into a $1,000 base annuity (n=5
Internal rate of return by interpolation
Problem. A heat-recovery project costs $10,000 now and saves $3,000 per year for 5 years. Find the rate of return and decide whether to accept it if MARR=12%.
Solution. Set PW=0
Common pitfalls
•Mixing the period of i and n: monthly payments with an annual rate. Either use r/m with n in months, or convert to the effective annual rate ie and use years — never blend them.
•Confusing nominal and effective rates. '6% compounded monthly' is a nominal rate; the money grows at ie=6.17%. The interest factors take the per-period rate, so for monthly work use 0.5%, not 6%.
•Off-by-one in n for annuities: the first A falls one period AFTER the present P, and the last A coincides with F. Drawing the cash-flow diagram catches this.
•Mis-timing the gradient: (P/G) assumes the first nonzero increment G occurs at the end of period 2, with the base amount handled separately by (P/A). Do not double-count year 1.
•Using (A/P) when you want (P/A) (or (A/F) vs (F/A)) — they are reciprocals, so a wrong pick gives the inverse of the right answer. Confirm whether you are solving for A
•Assuming a single IRR always exists. Cash flows with more than one sign change can yield multiple IRRs; when signs alternate, decide with PW at the MARR instead.
•Forgetting the perpetuity shortcut: capitalized cost is P=A/i, not A(P/A,i,n) with a large n — use the closed form for infinite service.
References
NCEES FE Reference Handbook — Engineering Economics
Newnan, Eschenbach & Lavelle, Engineering Economic Analysis — Standard treatment of interest factors, gradients, and rate-of-return analysis.
Blank & Tarquin, Engineering Economy — Factor notation and effective-rate conventions matching the handbook.
B. Economic analyses
Economic Analyses, Depreciation and Project Selection
Present worth, equivalent annual cost, benefit-cost, break-even, depreciation, and the unequal-lives rule that decide which alternative wins on the FE.
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C. Uncertainty
Decision-Making Under Uncertainty (Expected Value and Risk)
Expected monetary value, decision trees, sensitivity analysis, and the value of information — folding probability into an economic project comparison.
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Present worth of n equal end-of-period payments A. Reciprocal is capital recovery (A/P).
Equal payment that repays a present amount P (loan payment); also used to annualize a present cost.
equal deposits
A
. Reciprocal is the sinking-fund factor
(A/F)
.
in
n
periods.
Present worth of an arithmetic gradient G (first increment at end of period 2). Add to base-annuity P.
Converts a gradient G to an equivalent level annuity: Aeq=Abase+G(A/G,i,n).
Annual effective rate from nominal annual r with m compoundings/yr. Continuous limit: ie=er−1.
The three worth measures are interconvertible; sign and accept/reject conclusion are identical.
i∗ = internal rate of return. Accept if i∗≥MARR. Beware multiple roots with sign changes.
A=25,000(0.25046)=$6,261per year.
Check:
(P/A,8%,5)=0.08(1.08)5(1.08)5−1=3.9927
, so
P=6261(3.9927)=$25,000.
The present worth of the five payments returns the original principal, confirming the factor. Final answer: