Engineering Economics

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

.

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

The cash-flow diagram

Before any factor, draw the time line. A horizontal axis marks periods

0,1,2,\dots,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, \qquad (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.

(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 and geometric gradients

Real cash flows often trend. An arithmetic gradient adds a constant

G

each period: the series is

0, G, 2G, \dots, (n-1)G

— note the gradient is zero in year 1, so a flow that starts at base

A_1

and rises by

G

is the sum of a uniform series

A_1

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=A_1 n/(1+i)

.

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

Geometric gradient present worth

For a flow

A_1

in year 1 escalating at rate

g

per period, the present worth collapses to a single expression. When

i\neq g

use the ratio form; this is the standard model for costs that track an escalation rate. Always anchor

A_1

as the year-1 amount, not the year-0 amount.

P = A_1\left[\dfrac{1-\left(\dfrac{1+g}{1+i}\right)^n}{i-g}\right]\quad (i\neq g), \qquad P=\dfrac{A_1\,n}{1+i}\ (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

i_e

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\to\infty

, continuous compounding gives

i_e=e^r-1

.

i_e=\left(1+\dfrac{r}{m}\right)^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 —

(\text{want}/\text{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 $P$ (dollars).

Uniform series present worth (P/A)

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

Present worth $P$ of $n$ equal end-of-period payments $A$ ; $P$ falls one period before the first $A$ .

Capital recovery (A/P)

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

Equal end-of-period payment $A$ that recovers a present amount $P$ over $n$ periods at $i$ ; reciprocal of (P/A).

Uniform series compound amount (F/A)

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

Future value $F$ of $n$ equal payments $A$ ; $F$ occurs at the same time as the last payment.

Sinking fund (A/F)

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

Equal deposit $A$ needed each period to accumulate a future amount $F$ ; reciprocal of (F/A).

Arithmetic 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 a flow $0,G,2G,\dots,(n-1)G$ ; gradient $G$ in dollars per period, starting in year 2.

Arithmetic gradient uniform series (A/G)

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

Converts an arithmetic gradient $G$ into an equivalent uniform annual amount $A$ .

Geometric gradient present worth

P = A_1\dfrac{1-\left(\frac{1+g}{1+i}\right)^n}{i-g}\ (i\neq g)

Present worth of a flow growing at rate $g$ from year-1 base $A_1$ ; for $i=g$ , $P=A_1 n/(1+i)$ .

Effective annual rate

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

Effective annual rate from nominal annual $r$ with $m$ compoundings per year; continuous limit (not printed in the handbook) $i_e=e^r-1$ .

Simple interest

F = P(1+in)

Interest on principal only; used only when a problem explicitly states simple interest.

Inflation-adjusted (combined) rate

d = i + f + if

Combines real rate $i$ and inflation $f$ into the market rate $d$ used to discount actual (then-current) dollars.

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)=\dfrac{(1.08)^{10}-1}{0.08(1.08)^{10}}=6.7101

.

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 = 1200\,\dfrac{(1.08)^{10}-1}{0.08(1.08)^{10}} = \$8{,}052

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

A_1=\$1{,}000

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\times(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

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?

Solution.

i_e=\left(1+\dfrac{0.09}{12}\right)^{12}-1=(1.0075)^{12}-1=0.09381=9.38\%

.

F = 2000(1.09381)=\$2{,}188

. Sanity: monthly compounding must give an effective rate above the nominal

9\%

, and

9.38\%>9\%

as expected. Answer:

i_e=9.38\%

,

F=\$2{,}190

(3 s.f.).

i_e=\left(1+\dfrac{0.09}{12}\right)^{12}-1 = 9.38\%

Common pitfalls

•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 $A_1$ rising by $G$ is $A_1\,(P/A)+G\,(P/G)$ — not $(A_1)\,(P/G)$ .
•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 $P=A_1 n/(1+i)$ ; 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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