Engineering Economics

5% of exam

Time value of money, cost estimation, risk, break-even and benefit-cost analysis.

3 concepts

Exam areas

A. Time value of money

Time Value of Money

Move money through time with the six compound-interest factors, convert nominal to effective rates, and read a cash-flow diagram the way the FE wants.

A dollar today is not a dollar next year: it can earn interest, so amounts at different dates are not directly comparable until you move them to a common point in time. Every engineering-economics question on the FE — present worth, annual cost, rate of return — is really one skill applied over and over: slide a cash flow forward or backward with a compound-interest factor. The NCEES handbook gives you the whole factor family in one table (Engineering Economics, 'Interest Rate Tables'); your job is to know which factor each problem calls for and to keep the rate and the period count consistent. Master this and the rest of the topic is bookkeeping.

The cash-flow diagram

Draw the timeline first — it prevents almost every setup error. A horizontal line marks periods

0,1,2,\dots,n

; upward arrows are receipts (positive), downward arrows are disbursements (negative). Period

0

is 'now', and by the end-of-period convention a payment in year

k

sits at the end of period

k

. The single most common slip is putting the first payment of a uniform series at time

0

instead of time

1

: the standard

P/A

factor assumes the series starts one period after

P

Single-payment factors: P/F and F/P

These move one lump sum across time. The compound-amount factor

(F/P,i,n)=(1+i)^n

grows a present amount

P

into a future amount

F

; its reciprocal, the present-worth factor

(P/F,i,n)=(1+i)^{-n}

, discounts a future

F

back to today. Read the notation as 'find the first letter, given the second':

(F/P,i,n)

finds

F

given

P

. Here

i

is the interest rate per period and

n

the number of periods.

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

Uniform-series factors: A/P, P/A, A/F, F/A

A uniform series is an equal end-of-period amount

A

repeated

n

times. Two pairs cover it. The capital-recovery factor

(A/P,i,n)

turns a present sum into the level payment that exactly repays it with interest (it is the loan-payment formula); its inverse

(P/A,i,n)

is the present worth of a level stream. The sinking-fund factor

(A/F,i,n)

is the deposit per period needed to accumulate a future target

F

; its inverse

(F/A,i,n)

is the future worth of a level stream. Memorize the shape:

A/P

and

A/F

have the bracketed quantity in the denominator.

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

Sinking fund and series compound amount

The future-oriented pair completes the set. The sinking-fund factor finds the deposit that grows to a goal; the series-compound-amount factor finds what a string of deposits is worth at the end. Note

(A/P)=(A/F)+i

— capital recovery is just the sinking-fund deposit plus interest on the principal, a useful check and a fast way to build an equivalent uniform annual cost.

\left(\frac{A}{F},i,n\right)=\frac{i}{(1+i)^n-1}, \qquad \left(\frac{F}{A},i,n\right)=\frac{(1+i)^n-1}{i}

Uniform gradient

When a cash flow grows by a constant amount

G

each period (the first cash flow being

0

at period 1, then

G

2G

\dots

(n-1)G

), the gradient factors convert it.

(P/G,i,n)

gives its present worth and

(A/G,i,n)

converts it to an equivalent uniform series so it can be added to a base annuity. An arithmetic-gradient stream that starts at a base

A_1

and rises by

G

is handled as a uniform series of

A_1

plus a gradient of

G

— never treat the base as part of the gradient.

\left(\frac{A}{G},i,n\right)=\frac{1}{i}-\frac{n}{(1+i)^n-1}

Nominal vs. effective rate and compounding

A nominal annual rate

r

ignores within-year compounding; the effective annual rate

i_e

accounts for it. If interest compounds

m

times per year, the rate per period is

r/m

and the effective annual rate is

(1+r/m)^m-1

. The non-negotiable rule for the factors: the period of

i

must match the period of

n

. For monthly payments use the monthly rate

i=r/12

and

n

in months; only convert to an effective annual rate when the cash flows themselves are annual. As

m\to\infty

, continuous compounding gives

i_e=e^{r}-1

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

Exam strategy

Work every problem in three moves: (1) sketch the cash-flow diagram and label time 0; (2) confirm the rate-per-period matches the period count — if payments are monthly,

i

is monthly; (3) pick the single factor whose 'find/given' letters match what you have and want. Decide direction by where you want the answer: if you want a value 'now', you are finding

P

; at the end, finding

F

; level over time, finding

A

. The handbook's factor table and interest tables are on the reference — use the tabulated factor when

i

and

n

are round, but be ready to evaluate the closed form for off-table rates. State present worth as a single number at time 0 and never add cash flows that sit at different dates.

Key equations

Single payment compound amount (F/P)

F = P(1+i)^n

Grows a present sum $P$ to future $F$ . $i$ = rate per period, $n$ = number of periods.

Single payment present worth (P/F)

P = F(1+i)^{-n}

Discounts a future sum $F$ to present $P$ . Reciprocal of $F/P$ .

Capital recovery (A/P)

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

Level payment $A$ that repays present $P$ over $n$ periods; the loan-payment factor.

Uniform series present worth (P/A)

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

Present worth of $n$ equal end-of-period amounts $A$ .

Sinking fund (A/F)

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

Deposit per period needed to accumulate future target $F$ .

Series compound amount (F/A)

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

Future worth of $n$ equal deposits $A$ .

Capital recovery identity

\left(\tfrac{A}{P},i,n\right)=\left(\tfrac{A}{F},i,n\right)+i

Capital recovery equals sinking-fund deposit plus interest on principal; a fast EUAC builder and an algebra check.

Uniform gradient to series (A/G)

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

Converts an arithmetic gradient $G$ (0, G, 2G, ...) into an equivalent uniform series.

Effective annual rate

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

$r$ = nominal annual rate, $m$ = compounding periods per year. Rate per period is $r/m$ .

Continuous compounding

i_e=e^{r}-1

Limit of effective rate as $m\to\infty$ ; use when interest compounds continuously. Not printed in the FE Reference Handbook 10.6, which gives only non-annual compounding.

Capitalized cost (perpetuity)

P=\dfrac{A}{i}

Present worth of a perpetual uniform series $A$ at rate $i$ ; used for endless maintenance/replacement.

Worked examples

Single deposit grows (F/P)

Problem. You deposit $\$5{,}000$ in an account paying $6\%$ compounded annually. What is the balance after $8$ years?

Solution. Single future amount, so use

F=P(1+i)^n

with

P=\$5{,}000

i=0.06

n=8

F=5000(1.06)^8 = 5000(1.59385)=\$7{,}969

. Sanity:

(1.06)^8\approx 1.59

, so the balance should be about

1.6\times

the deposit — it is. Final:

\$7{,}970

(3 s.f.).

F = 5000(1+0.06)^{8} = \$7{,}969

Monthly loan payment (A/P)

Problem. A $\$25{,}000$ equipment loan carries a nominal $6\%$ annual rate compounded monthly over $5$ years. Find the monthly payment.

Solution. Payments are monthly, so use a monthly rate

i=0.06/12=0.005

and

n=5\times 12=60

(A/P,0.5\%,60)=\dfrac{0.005(1.005)^{60}}{(1.005)^{60}-1}=0.019333

A = 25000(0.019333)=\$483.32

. Sanity:

60

payments of

\$483

total

\$28{,}999

, i.e. about

\$4{,}000

interest on

\$25{,}000

over 5 years — reasonable. Final:

\$483

per month.

A = 25000\cdot\frac{0.005(1.005)^{60}}{(1.005)^{60}-1}=\$483

Effective rate then accumulate (F/A)

Problem. An account pays a nominal $8\%$ compounded quarterly. You deposit $\$2{,}000$ at the end of each year for $10$ years. What is the value just after the last deposit?

Solution. Cash flows are annual but compounding is quarterly, so convert to an effective annual rate:

i_e=(1+0.08/4)^4-1=(1.02)^4-1=0.08243

(i.e.

8.24\%

). Future worth of an annuity:

(F/A,8.243\%,10)=\dfrac{(1.08243)^{10}-1}{0.08243}=14.655

F=2000(14.655)=\$29{,}310

. Sanity:

10

deposits of

\$2{,}000

contribute

\$20{,}000

of principal; the rest is interest — plausible at

8\%

. Final:

\$29{,}300

(3 s.f.).

i_e=(1.02)^4-1=8.24\%,\qquad F=2000\cdot\frac{(1.08243)^{10}-1}{0.08243}

Common pitfalls

•Mismatching $i$ and $n$ : for monthly payments use the monthly rate $r/12$ and $n$ in months. Plugging an annual rate with $n$ in months (or vice-versa) is the top FE error here.
•Confusing nominal and effective rates: $8\%$ compounded quarterly is an effective $8.24\%$ per year. Convert before applying annual factors.
•Inverting a factor: $(A/P)$ has the bracket in the denominator (payment from a present sum); $(P/A)$ has it in the numerator. Reading 'find/given' backwards flips the answer.
•Placing the first annuity payment at time 0. The standard $P/A$ and $F/A$ factors assume the series begins at the end of period 1; an annuity-due needs an extra adjustment.
•Including the base amount in a gradient. An arithmetic gradient runs 0, G, 2G, ...; the constant base is a separate uniform series added on top.
•Adding cash flows at different dates. You must move every amount to one common time point with the factors before summing.
•Sign sloppiness: keep receipts positive and disbursements negative throughout, or a present-worth comparison silently changes sign.

References

NCEES FE Reference Handbook — Engineering Economics: Factor Table and Nomenclature
NCEES FE Reference Handbook — Engineering Economics: Non-Annual Compounding — Effective interest rate formula.
NCEES FE Reference Handbook — Engineering Economics: Interest Rate Tables — Tabulated P/F, P/A, A/P, A/F, F/A, A/G factors.

B. Cost estimation

Cost Estimation

Classify fixed, variable, direct, and indirect costs, update historical prices with cost indexes, scale equipment with six-tenths, and apply learning curves.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders

D. Analysis

Economic Analysis and Decision Methods

Choose between alternatives with present worth, annual cost, rate of return, and benefit-cost ratio; find break-even and payback; depreciate with SL and MACRS.

Unlocks with an access pass — one-time payment, no auto-renew. View passes

Pass holders