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

Exam areas

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, \qquad 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

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 = A\,\dfrac{(1+i)^n-1}{i(1+i)^n}, \qquad A = P\,\dfrac{i(1+i)^n}{(1+i)^n-1}

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

A_\text{eq}=A_\text{base}+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)=\dfrac{1}{i}-\dfrac{n}{(1+i)^n-1}

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

i_e

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

i_e

and use years. Continuous compounding is the limit

i_e=e^r-1

i_e=\left(1+\dfrac{r}{m}\right)^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)\to 1/i

n\to\infty

AW = PW\,(A/P,i,n), \qquad P_\infty = \dfrac{A}{i}

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^*\ge 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^*)=\sum_{t=0}^{n}\dfrac{CF_t}{(1+i^*)^t}=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

i_e

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,\dots

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.

Uniform-series present worth (P/A)

P = A\,\dfrac{(1+i)^n-1}{i(1+i)^n}

Present worth of $n$ equal end-of-period payments $A$ . Reciprocal is capital recovery $(A/P)$ .

Capital recovery (A/P)

A = P\,\dfrac{i(1+i)^n}{(1+i)^n-1}

Equal payment that repays a present amount $P$ (loan payment); also used to annualize a present cost.

Uniform-series compound amount (F/A)

F = A\,\dfrac{(1+i)^n-1}{i}

Future worth of $n$ equal deposits $A$ . Reciprocal is the sinking-fund factor $(A/F)$ .

Sinking fund (A/F)

A = F\,\dfrac{i}{(1+i)^n-1}

Equal deposit needed to accumulate a target future fund $F$ in $n$ periods.

Gradient present worth (P/G)

P = G\left[\dfrac{(1+i)^n-1}{i^2(1+i)^n}-\dfrac{n}{i(1+i)^n}\right]

Present worth of an arithmetic gradient $G$ (first increment at end of period 2). Add to base-annuity $P$ .

Gradient to uniform series (A/G)

(A/G,i,n)=\dfrac{1}{i}-\dfrac{n}{(1+i)^n-1}

Converts a gradient $G$ to an equivalent level annuity: $A_\text{eq}=A_\text{base}+G\,(A/G,i,n)$ .

Effective interest rate

i_e=\left(1+\dfrac{r}{m}\right)^m-1

Annual effective rate from nominal annual $r$ with $m$ compoundings/yr. Continuous limit (not printed in the handbook): $i_e=e^r-1$ .

Equivalence among PW, AW, FW

AW = PW\,(A/P,i,n),\quad FW = PW\,(F/P,i,n)

The three worth measures are interconvertible; sign and accept/reject conclusion are identical.

Capitalized cost (perpetuity)

P_\infty=\dfrac{A}{i}

Present worth of an infinite uniform series — capitalized cost of perpetual service or maintenance.

Rate of return condition

PW(i^*)=\sum_{t=0}^{n}\dfrac{CF_t}{(1+i^*)^t}=0

$i^*$ = internal rate of return. Accept if $i^*\ge MARR$ . Beware multiple roots with sign changes.

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\times 8 = 96

months.

F = 5000\,(1.005)^{96} = 5000(1.61414)=\$8{,}071.

Effective annual rate:

i_e=(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.

Solution. Capital-recovery factor:

(A/P,8\%,5)=\dfrac{0.08(1.08)^5}{(1.08)^5-1}=\dfrac{0.08(1.46933)}{0.46933}=0.25046.

A = 25{,}000(0.25046)=\$6{,}261\ \text{per year}.

Check:

(P/A,8\%,5)=\dfrac{(1.08)^5-1}{0.08(1.08)^5}=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:

\$6{,}260

per year (3 sig figs).

Base annuity plus arithmetic gradient

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

) plus a gradient

G=\$200

(P/A,10\%,5)=3.7908

(P/G,10\%,5)=6.8618

PW = 1000(3.7908)+200(6.8618)=3790.8+1372.4=\$5{,}163.

Equivalent uniform:

(A/G,10\%,5)=\dfrac{1}{0.10}-\dfrac{5}{(1.10)^5-1}=10-8.1899=1.8101.

A_\text{eq}=1000+200(1.8101)=\$1{,}362\ \text{per year}.

Sanity check:

A_\text{eq}(P/A)=1362(3.7908)=\$5{,}163

, matching the

PW

. Final answers:

PW=\$5{,}160

A_\text{eq}=\$1{,}360

per year (3 sig figs).

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

-10{,}000 + 3000\,(P/A,i^*,5)=0\Rightarrow (P/A,i^*,5)=3.3333.

Bracket with trials: at

i=15\%

PW = -10{,}000+3000(3.3522)=+\$56.5

; at

i=16\%

PW=-10{,}000+3000(3.2743)=-\$177.

Linear interpolation:

i^*\approx 15\% + 1\%\times\dfrac{56.5}{56.5+177}=15.24\%.

Since

i^*=15.2\% > MARR=12\%

, accept the project. Sanity check:

PW

changes sign once between

15\%

and

16\%

, so a single unique IRR exists. Final answer:

i^*\approx 15.2\%

, accept (3 sig figs).

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 $i_e$ and use years — never blend them.
•Confusing nominal and effective rates. ' $6\%$ compounded monthly' is a nominal rate; the money grows at $i_e=6.17\%$ . 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 $A$ or for $P/F$ .
•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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