Engineering Economics

4% of exam

Time value of money and annuities, cost types and breakdowns, and economic analyses including cost-benefit, break-even, and life cycle.

3 concepts

Exam areas

A. Time value of money

Time Value of Money and Annuities

Interest factors (P/F, F/P, P/A, A/P, A/F, F/A), gradients, and nominal-versus-effective rates that turn any cash-flow diagram into a single comparable number.

A dollar today is worth more than a dollar next year, because today's dollar can earn interest in the meantime. Engineering economics is the discipline of moving money across time correctly so that alternatives evaluated at different dates can be compared on equal footing. On the FE Mechanical exam this single idea — the time value of money — is the engine behind every economics question, and the points are won or lost on two things: picking the right interest factor and reading the cash-flow diagram correctly. Master the six basic factors plus gradients and you can solve the whole topic; the FE Reference Handbook gives them all in one table (FE Reference Handbook — Engineering Economics), so the exam is testing whether you can ·apply· them, not recall them.

The cash-flow diagram comes first

Before any factor, draw a horizontal time line with the present at

t=0

and integer periods marching to the right. Money received (inflow) points up; money spent (outflow) points down. A single present amount is

P

t=0

; a single future amount is

F

t=n

; a level series of equal end-of-period payments is the annuity

A

, drawn as equal arrows at

t=1,2,\dots,n

. The convention that every

A

lands at the ·end· of its period — and that the first

A

of a present-worth series occurs one period ·after·

P

— is the source of more lost points than any formula. When in doubt, sketch it: the diagram tells you which factor you need and how many periods separate the two amounts you are relating.

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

The most fundamental move is shifting one lump sum across

n

periods at rate

i

per period. Growing a present amount forward uses the single-payment compound-amount factor

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

; discounting a future amount back to the present uses its reciprocal, the present-worth factor

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

. Read the handbook's factor notation literally:

(F/P,i\%,n)

means 'find

F

given

P

,' so

F=P\,(F/P,i,n)

. These two factors underlie all the others, because an annuity or gradient is just a bundle of single payments summed.

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

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

An annuity

A

relates to a present lump sum through the uniform-series present-worth factor

(P/A)

and its inverse the capital-recovery factor

(A/P)

— the workhorse for loan payments and equivalent annual cost. It relates to a future lump sum through the uniform-series compound-amount factor

(F/A)

and its inverse the sinking-fund factor

(A/F)

— the workhorse for retirement and replacement funds. All four collapse to the same geometric-series sum; memorize

(A/P)

and

(P/A)

and you can derive the rest, but knowing the pattern is faster than re-deriving on the clock.

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

The arithmetic gradient: P/G and A/G

Many real cash flows are not level but ramp by a constant amount

G

each period — maintenance that rises, production that grows. The handbook models this as a ·base· annuity plus a ·gradient· whose first nonzero term (

G

) occurs at the end of period 2, with

2G

at period 3, and so on up to

(n-1)G

at period

n

. Convert the gradient to a present worth with

(P/G,i,n)

or to an equivalent level annuity with

(A/G,i,n)

, then add the base annuity separately. The classic mistake is putting

G

at period 1; the gradient timeline always starts one period late.

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

Nominal versus effective interest

A rate quoted 'per year compounded monthly' is a ·nominal· annual rate

r

; the ·effective· annual rate

i_e

is what actually accrues once you account for compounding

m

times per year. The two are equal only when

m=1

. Whenever the compounding period and the cash-flow period differ — monthly loan payments under an annual quote, or quarterly compounding — convert first: either work in the compounding period with

i=r/m

, or compute

i_e

and work in years. Mixing an annual rate with monthly periods is the single most common numerical error on this topic. The continuous-compounding limit is

i_e=e^{r}-1

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

Equivalence: the unifying idea

Two cash-flow patterns are ·equivalent· at rate

i

if they have the same value at any common reference date. This is why a loan principal

P

and its stream of payments

A

are interchangeable, and why you can compare a one-time purchase against a leasing annuity by bringing both to present worth. Every factor is just a tool for enforcing equivalence; once both alternatives are expressed at the same point in time (present, future, or annual), you simply compare the numbers. Choose whichever reference point makes the algebra shortest — usually present worth, sometimes annual cost when lives differ.

Exam strategy

Draw the diagram, label

P

F

A

G

, then match each transfer to a factor by its 'find/given' notation. Confirm the period: if compounding and payments are both monthly, use

i=r/m

and

n

in months — do not annualize. The handbook prints interest-factor ·tables· for common rates; if the problem's

i

is in the table, read the factor directly rather than evaluating

(1+i)^n

by hand. Keep the base annuity and the gradient as two separate terms and add at the end. Finally, sanity-check magnitude:

(P/A)

is always less than

n

(F/A)

is always greater than

n

, and

(A/P)>i

for finite

n

— if your factor violates these, you grabbed the wrong one.

Key equations

Single-payment compound amount (F/P)

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

Grows a present amount $P$ forward $n$ periods at rate $i$ to a future amount $F$ . $i$ is the rate per period; $n$ counts periods, not necessarily years.

Single-payment present worth (P/F)

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

Discounts a future amount $F$ back $n$ periods to its present worth $P$ . Reciprocal of the (F/P) factor.

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 $A$ running $n$ periods. First $A$ occurs at $t=1$ , one period after $P$ .

Capital recovery (A/P)

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

Level payment $A$ that repays (recovers) a present amount $P$ over $n$ periods at $i$ ; the standard loan-payment factor. Inverse of (P/A).

Uniform-series compound amount (F/A)

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

Future worth at $t=n$ of an end-of-period annuity $A$ . Used for savings/replacement funds.

Sinking fund (A/F)

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

Level deposit $A$ each period needed to accumulate a future target $F$ in $n$ periods. Inverse of (F/A).

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 that adds $G$ per period; first increment ( $G$ ) lands at $t=2$ . Add a separate base annuity for the period-1 amount.

Gradient to uniform series (A/G)

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

Equivalent level annuity of an arithmetic gradient $G$ over $n$ periods. Lets a ramping cash flow be treated as a constant $A$ .

Effective annual rate

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

Converts nominal annual rate $r$ compounded $m$ times per year to the effective annual rate $i_e$ . Continuous limit (not printed in the handbook): $i_e=e^{r}-1$ .

Worked examples

Loan payment via capital recovery

Problem. You finance a $\$25{,}000$ machine at $8\%$ effective annual interest, repaid in $10$ equal end-of-year payments. What is the annual payment?

Solution. This is a present amount converted to an annuity, so use the capital-recovery factor

(A/P,8\%,10)

(A/P,8\%,10)=\dfrac{0.08(1.08)^{10}}{(1.08)^{10}-1}=0.149029

A = P\,(A/P) = 25{,}000\times 0.149029 = \$3{,}725.74

. Sanity check:

10

payments total

\$37{,}257

, comfortably above the

\$25{,}000

principal because of interest, and

(A/P)=0.149>i=0.08

as required. Final answer:

A \approx \$3{,}730\,/\text{yr}

(3 sig figs).

Sinking fund for replacement

Problem. A plant must accumulate $\$50{,}000$ in $20$ years to replace a compressor. With a fund earning $6\%$ per year, what equal annual end-of-year deposit is required?

Solution. A future target from an annuity uses the sinking-fund factor

(A/F,6\%,20)

(A/F,6\%,20)=\dfrac{0.06}{(1.06)^{20}-1}=\dfrac{0.06}{2.2071}=0.027185

A = F\,(A/F)=50{,}000\times 0.027185 = \$1{,}359.23

. Sanity check:

20

deposits sum to

\$27{,}185

, and interest supplies the remaining

\$22{,}815

to reach

\$50{,}000

— reasonable for

6\%

over

20

years. Final answer:

A \approx \$1{,}360\,/\text{yr}

Base annuity plus arithmetic gradient

Problem. Maintenance is $\$2{,}000$ at the end of year 1 and rises $\$500$ each year thereafter through year 5. At $i=10\%$ , find the present worth of the maintenance stream.

Solution. Split into a base annuity

A=\$2{,}000

and a gradient

G=\$500

(first $500 increment at year 2).

(P/A,10\%,5)=\dfrac{(1.10)^{5}-1}{0.10(1.10)^{5}}=3.7908

, so base

PW = 2{,}000\times 3.7908 = \$7{,}581.6

(P/G,10\%,5)=\dfrac{(1.10)^{5}-1}{0.10^{2}(1.10)^{5}}-\dfrac{5}{0.10(1.10)^{5}}=6.8618

, so gradient

PW = 500\times 6.8618 = \$3{,}430.9

P = 7{,}581.6 + 3{,}430.9 = \$11{,}012

. Sanity check: undiscounted payments are

\$2{,}000 + \$2{,}500 + \dots + \$4{,}000 = \$15{,}000

; a present worth of about

\$11{,}000

is the expected discount. Final answer:

P \approx \$11{,}000

P = 2000\,(P/A,10\%,5) + 500\,(P/G,10\%,5)

Nominal rate with monthly compounding

Problem. You deposit $200 at the end of each month into an account quoted at $6\%$ nominal annual interest compounded monthly. (a) What is the effective annual rate? (b) What is the balance after $5$ years?

Solution. (a)

i_e=\left(1+\dfrac{0.06}{12}\right)^{12}-1 = (1.005)^{12}-1 = 0.06168 \Rightarrow 6.17\%

. (b) Work in the compounding period:

i=0.06/12=0.005

per month,

n=5\times 12=60

months.

(F/A,0.5\%,60)=\dfrac{(1.005)^{60}-1}{0.005}=69.770

F = 200\times 69.770 = \$13{,}954

. Sanity check:

60

deposits total

\$12{,}000

; interest adds about

\$1{,}950

, consistent with

\sim 6\%

over a few years. Final answers:

i_e \approx 6.17\%

F \approx \$13{,}950

Common pitfalls

•Annualizing a nominal rate instead of matching periods: with monthly payments under a '6% compounded monthly' quote, use $i=r/m=0.5\%$ and $n$ in months — do NOT use $6\%$ with $n$ in years.
•Confusing nominal and effective rates: $r=12\%$ compounded monthly gives $i_e=12.68\%$ , not $12\%$ . Only when $m=1$ are they equal.
•Placing the first gradient term at period 1. The arithmetic-gradient series starts its increment $G$ at the END of period 2 ( $0,\,G,\,2G,\dots$ ); add the period-1 amount as a separate base annuity.
•Timing offset of $P$ and the annuity: in $(P/A)$ the present worth sits one period BEFORE the first $A$ . If the first payment is at $t=0$ (annuity-due), shift by one period or multiply by $(1+i)$ .
•Grabbing the reciprocal factor: $(A/P)$ (find payment) versus $(P/A)$ (find present worth), or $(A/F)$ versus $(F/A)$ . Read the 'find/given' notation literally and check that $(P/A)<n<(F/A)$ .
•Using $i$ as a percent in the formula. The factor equations take $i$ as a decimal (0.08), not 8; only the table label is in percent.
•Forgetting that interest factors assume END-of-period cash flows. Beginning-of-period (annuity-due) or mid-period conventions require an explicit shift.

References

NCEES FE Reference Handbook — Engineering Economics
Blank & Tarquin, Engineering Economy — factor derivations and interest tables
Newnan, Eschenbach & Lavelle, Engineering Economic Analysis — cash-flow diagrams and equivalence

B. Cost types and breakdowns

Cost Types, Breakdowns and Depreciation

Fixed/variable, direct/indirect, sunk and incremental costs, plus straight-line and MACRS depreciation with book value — the cost vocabulary every economic decision rests on.

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C. Economic analyses

Economic Analyses: Present Worth, B/C, Break-Even and Life-Cycle

Present worth, equivalent annual cost, rate of return, benefit-cost ratio, break-even and payback — and how to compare alternatives with unequal lives correctly.

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