Engineering Economics

A. Time value of money

Time Value of Money and Equivalence

How interest moves money across time: compound vs simple interest, the six discrete factors, gradients, and converting between nominal and effective rates.

A dollar today is worth more than a dollar next year, because the dollar you hold can earn interest in the meantime. Every engineering-economics problem on the FE is, at bottom, an exercise in moving cash flows along the time line until they sit at the same instant, where they can finally be compared or added. That operation is called finding an equivalent value, and the machinery for it — the interest-factor table in the Engineering Economics chapter of the FE Reference Handbook — is the single most-used tool in this whole topic. Master the six factors and the rest of the chapter is bookkeeping.

Simple versus compound interest

Under simple interest, only the original principal earns interest, so growth is linear:

F = P(1 + in)

. Under compound interest, interest itself earns interest each period, so growth is geometric:

F = P(1+i)^n

. The FE — and essentially all of finance — uses compound interest unless a problem explicitly says 'simple.' The gap widens fast:

\$25{,}000

at

6\%

for

8

years grows to

\$37{,}000

simple but

\$39{,}846

compound. Always read

i

as the rate per compounding period and

n

as the number of those periods, and make sure they refer to the same period length.

F = P(1+i)^n \quad \text{(compound)}, \qquad F = P(1+in) \quad \text{(simple)}

The single-payment factors (P/F and F/P)

Two factors move a single lump sum across time. The single-payment compound-amount factor

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

pushes a present amount

P

forward

n

periods to its future worth

F

. Its reciprocal, the single-payment present-worth factor

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

, discounts a future amount back to today. The Handbook writes factors in the functional form '(to find / given,

i

,

n

)' — read

(P/F,i,n)

as 'find

P

given

F

.' On exam day you can either pull the value from the printed interest-rate table or evaluate the closed form on your calculator; the closed form is faster once you trust it.

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

The uniform-series factors (P/A, A/P, F/A, A/F)

A uniform series

A

is an equal end-of-period cash flow repeated for

n

periods — a loan payment, an annual maintenance cost, a recurring saving. Four factors connect

A

to a present worth

P

or a future worth

F

. The present-worth factor

(P/A,i,n)

collapses the whole stream to a single value one period before the first payment; the capital-recovery factor

(A/P,i,n)

does the reverse and is exactly how a loan payment is computed. The sinking-fund factor

(A/F,i,n)

finds the deposit needed to accumulate a target

F

, and

(F/A,i,n)

accumulates a series to a future lump sum. Note the convention baked into these formulas:

A

occurs at the END of each period, and

P

sits at time zero, one full period BEFORE the first

A

.

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

Arithmetic gradients (P/G and A/G)

When cash flows rise by a constant amount

G

each period —

0

in year 1,

G

in year 2,

2G

in year 3, and so on — they form a uniform (arithmetic) gradient. The Handbook supplies

(P/G,i,n)

to find the present worth of the gradient part and

(A/G,i,n)

to convert it into an equivalent uniform series. The key convention trips people up: the gradient contributes ZERO in the first period, so a stream of

\$5{,}000

growing by

\$500

per year is modeled as a base annuity of

A = \$5{,}000

PLUS a gradient with first increment in year 2. For flows that grow by a constant PERCENTAGE instead of a constant dollar amount, you have a geometric gradient — but be aware the FE Reference Handbook provides NO geometric-gradient factor or rate adjustment; that method is standard-textbook material (Newnan), so on the exam any percentage-growth handling must be built from the basic factors or given in the problem.

P = A\,(P/A,i,n) + G\,(P/G,i,n), \qquad (P/G,i,n)=\dfrac{(1+i)^n-1}{i^2(1+i)^n}-\dfrac{n}{i(1+i)^n}

Nominal versus effective interest rates

A nominal annual rate

r

stated with

m

compounding periods per year is not the rate you actually earn — compounding within the year boosts it. The effective annual rate is

i_e = (1 + r/m)^m - 1

. A loan quoted at

9\%

compounded monthly carries

i_e = (1+0.09/12)^{12}-1 = 9.38\%

. The non-negotiable rule when using the factors: the period interest rate and the period count must match the compounding period. For monthly payments at

9\%

nominal, use

i = 0.09/12 = 0.75\%

per month and

n

in months — never the annual rate with annual periods. As

m \to \infty

you approach continuous compounding,

i_e = e^r - 1

— a useful limit, but note the FE Reference Handbook lists only the discrete effective-rate formula, so do not go hunting for continuous-compounding factors on exam day.

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

Cash-flow diagrams and equivalence

Before substituting into any factor, draw the time line. Put time zero at the analysis 'present,' mark each period, and draw arrows up for receipts and down for disbursements (or pick your own sign convention and hold it). Two cash-flow sets are equivalent at rate

i

if they have the same value at any single point in time — so you can move everything to

t=0

(present worth), to

t=n

(future worth), or spread it as an annuity, and the comparison is valid as long as you used the same

i

throughout. Getting the diagram right — especially the off-by-one placement of

P

relative to the first

A

— prevents the most common errors on this entire topic.

Exam strategy

Identify what you are given and what you want, then name the single factor that connects them — '

F

given

P

' is

(F/P)

, '

A

given

P

' is

(A/P)

, and so on. Confirm

i

and

n

share the same period before you touch the keypad; mismatched periods are the number-one error. The printed interest-rate tables only cover common rates at whole-number

n

, so for odd rates evaluate the closed forms directly. Decompose any messy stream into pieces you have factors for: a base annuity plus a gradient, a lump sum plus a series, a deferred annuity discounted one extra step. And always sketch the diagram first — ten seconds there saves a sign or timing blunder worth the whole question.

Key equations

Single-payment compound amount (F/P)

F = P(1+i)^n

Future worth of a present sum. $i$ = rate per period, $n$ = number of periods (must match compounding).

Single-payment present worth (P/F)

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

Discounts a future lump sum to time zero.

Uniform-series present worth (P/A)

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

Present worth of an end-of-period annuity $A$ ; $P$ sits one period before the first payment.

Capital recovery (A/P)

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

Equal payment that recovers a present sum over $n$ periods — the loan-payment formula.

Uniform-series compound amount (F/A)

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

Future worth accumulated by an end-of-period series.

Sinking fund (A/F)

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

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

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; first increment $G$ occurs in period 2.

Gradient to uniform series (A/G)

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

Converts an arithmetic gradient into an equivalent level annuity.

Effective annual rate

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

$r$ = nominal annual rate, $m$ = compounding periods per year. The continuous-compounding limit $i_e=e^r-1$ is beyond the handbook.

Worked examples

Single sum — compound versus simple

Problem. You invest $\$25{,}000$ in an account paying $6\%$ per year for $8$ years. Find the future worth under annual compounding, and compare with simple interest.

Solution. Compound:

F = P(1+i)^n = 25{,}000(1.06)^8

. Since

(1.06)^8 = 1.5938

,

F = 25{,}000(1.5938) = \$39{,}846

. Simple:

F = P(1+in) = 25{,}000[1 + (0.06)(8)] = 25{,}000(1.48) = \$37{,}000

. Final answer:

F \approx \$39{,}800

compound versus

\$37{,}000

simple. Sanity check: compounding must exceed simple interest for

n>1

, and the

\$2{,}846

surplus is exactly the interest-on-interest, confirming the result.

F = 25{,}000(1.06)^8 = \$39{,}846

Annuity plus salvage — present worth

Problem. A pump upgrade saves $\$12{,}000$ per year for $10$ years and has a salvage value of $\$15{,}000$ at the end of year 10. At a MARR of $8\%$ , what is the present worth of these benefits?

Solution. Discount the annuity with

(P/A,8\%,10)

and the salvage with

(P/F,8\%,10)

.

(P/A,8\%,10) = \dfrac{(1.08)^{10}-1}{0.08(1.08)^{10}} = 6.7101

, so the savings are worth

12{,}000(6.7101) = \$80{,}521

.

(P/F,8\%,10) = (1.08)^{-10} = 0.4632

, so the salvage is worth

15{,}000(0.4632) = \$6{,}948

.

PW = 80{,}521 + 6{,}948 = \$87{,}469 \approx \$87{,}500

. Sanity check: the salvage is one distant lump and should be small next to a decade of savings — it is under

9\%

of the total, as expected.

PW = 12{,}000\,(P/A,8\%,10) + 15{,}000\,(P/F,8\%,10)

Arithmetic gradient and an effective rate

Problem. Maintenance costs are $\$5{,}000$ at the end of year 1 and rise by $\$500$ each year through year 6. (a) Find the present worth at $7\%$ per year. (b) Separately, what effective annual rate corresponds to $9\%$ nominal compounded monthly?

Solution. (a) Split into a base annuity

A=\$5{,}000

plus a gradient

G=\$500

(first increment in year 2).

(P/A,7\%,6)=4.7665

and

(P/G,7\%,6)=10.9784

.

PW = 5{,}000(4.7665) + 500(10.9784) = 23{,}833 + 5{,}489 = \$29{,}322 \approx \$29{,}300

. (b)

i_e = (1+0.09/12)^{12}-1 = (1.0075)^{12}-1 = 0.0938 = 9.38\%

. Sanity check: the gradient adds only about

\$5{,}500

to a roughly

\$24{,}000

base, consistent with small

\$500

steps; and

i_e

exceeds the

9\%

nominal, as monthly compounding requires.

PW = A\,(P/A,7\%,6) + G\,(P/G,7\%,6)

Common pitfalls

•Mixing rate and period units: with monthly payments at $9\%$ nominal, use $i=0.75\%$ per month and $n$ in months — never $9\%$ with $n$ in years. Convert to a consistent period first.
•Treating a nominal rate as effective. $(1+r/m)^m-1$ is the rate actually earned; using $r$ directly understates interest whenever $m>1$ .
•Misplacing the gradient: the arithmetic-gradient series contributes $0$ in period 1, not $G$ . Model real growing flows as a base annuity PLUS the gradient.
•Forgetting that $(P/A)$ lands the present worth ONE period before the first $A$ . For a deferred annuity you must then discount that $P$ back the extra periods with $(P/F)$ .
•Defaulting to simple interest. Unless a problem says 'simple,' interest compounds — and the gap grows with $n$ .
•Adding cash flows that sit at different times. Two amounts can only be summed or compared after both are moved to the same instant using the same $i$ .
•Reading the factor table at the wrong rate or interpolating carelessly; for odd rates or non-tabulated $n$ , evaluate the closed-form factor instead of guessing between rows.

References

NCEES FE Reference Handbook — Engineering Economics
Newnan, Eschenbach & Lavelle, Engineering Economic Analysis — factor derivations and conventions

C. Analyses

Comparing Alternatives: Present Worth, B/C, Break-Even, Depreciation

Choosing among projects with present worth, annual cost, rate of return, and benefit-cost ratio, plus break-even, depreciation, and the cost types behind every analysis.

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D. Uncertainty

Uncertainty and Expected Value in Economics

Weighing economic alternatives when outcomes are uncertain: expected value and expected monetary value, decision trees, risk, and sensitivity analysis.

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