Engineering Economics

5% of exam

Time value of money and equivalence, cost types and breakdowns, benefit-cost, break-even, and life-cycle analyses, and project selection with depreciation and unequal lives.

3 concepts

Exam areas

A. Time value of money

Time Value of Money and Life-Cycle Cost

The interest-factor family (P/F, F/P, P/A, A/P, A/F, F/A), gradients, nominal vs effective rates, and putting capital and annual O&M on one basis for life-cycle cost.

A dollar today is worth more than a dollar next year because today's dollar can earn interest, and almost every environmental decision — lining a landfill, sizing a pump, choosing a membrane over media filtration — hinges on comparing money that arrives at different times. The whole of engineering economics is one idea applied six ways: move a cash flow forward or backward along a time line by multiplying it by an interest factor. The FE Reference Handbook gives you the full factor table in its Engineering Economics chapter; the points are lost not in the formulas but in choosing the right factor, lining up

n

with the compounding period, and matching the interest rate to that period. Master the factor family and life-cycle costing collapses into bookkeeping you can do under exam pressure.

The cash-flow diagram and sign convention

Before any factor, draw a horizontal time line marked

0,1,2,\dots,n

, with up arrows for receipts and down arrows for disbursements. Present worth

P

sits at time

0

; future worth

F

sits at time

n

; a uniform series

A

is an equal end-of-period amount at each of

1

through

n

(the "end-of-year" convention the FE uses unless told otherwise). Getting the picture right — especially whether the first

A

lands at year

1

or year

0

, and where

P

sits relative to the series — eliminates most factor errors before you touch a number.

Single-payment factors: moving one lump sum

A single amount carried forward

n

periods at rate

i

grows by the single-payment compound-amount factor

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

; carrying it backward uses its reciprocal, the present-worth factor

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

. Read the notation literally as a fraction:

(F/P)

means "

F

given

P

," so

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

. Every other factor is just a packaged combination of these, so if you ever forget one you can rebuild it from

(1+i)^n

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

Uniform-series factors: lump sum to a stream and back

Most real cash flows are streams — annual O&M, debt service, recurring chemical costs — so the uniform-series factors do the heavy lifting. The present-worth factor

(P/A,i,n)

converts an end-of-year annuity

A

into a single present value, and its reciprocal, the capital-recovery factor

(A/P,i,n)

, spreads a present amount into equal annual payments — exactly how you annualize a capital cost. The sinking-fund factor

(A/F,i,n)

and compound-amount factor

(F/A,i,n)

play the same roles against a future value. Note the tidy identity

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

, which says a capital-recovery payment is just the sinking-fund deposit plus interest on the principal.

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

Gradients: when the stream changes each year

Maintenance and energy costs rarely stay flat — they climb by a roughly constant amount each year. A cash flow that starts at zero in year

1

and increases by a constant

G

each year (so

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

) is a uniform arithmetic gradient. Convert it to a present worth with

(P/G,i,n)

or, more usefully, to an equivalent flat annuity with

(A/G,i,n)

. A series that starts at a base amount

A_1

in year

1

and rises by

G

is handled by superposition: treat the flat base

A_1

with

(P/A)

and the rising part with

(P/G)

, then add. Do not apply

(P/G)

to the whole first-year amount — the gradient part by definition contributes nothing in year

1

P = A_1\,(P/A,i,n) + G\,(P/G,i,n), \qquad A_{\text{eq}} = A_1 + G\,(A/G,i,n)

Nominal vs effective interest and non-annual compounding

A quoted "6% compounded monthly" is a nominal annual rate

r

; the rate that actually applies over a year, accounting for interest-on-interest within the year, is the effective annual rate

i_e

. With

m

compounding periods per year,

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

, which always exceeds

r

when

m>1

— at

r=6\%

monthly,

i_e=6.17\%

. The cardinal rule is consistency: the interest rate and the number of periods

n

must reference the same period. Either work in monthly steps with the periodic rate

r/m

and

n

months, or convert to an annual effective rate and use

n

in years — never mix a yearly

n

with a monthly rate.

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

Putting capital and O&M on one basis: life-cycle cost

A treatment process is never just its purchase price. Life-cycle cost (LCC) gathers the first cost, recurring operation and maintenance, periodic replacements, and end-of-life salvage onto a single comparable number — usually a present worth (the net cost in today's dollars) or an equivalent uniform annual cost (EUAC, the levelized annual charge). Annualizing is often the cleaner route: multiply the capital by

(A/P)

, add the annual O&M directly, and subtract the salvage spread over the life with

(A/F)

. Choosing the alternative with the lowest LCC — not the lowest sticker price — is the entire point of the exercise, and it routinely reverses the ranking a first-cost comparison would give.

\text{EUAC} = P\,(A/P,i,n) + A_{\text{O\&M}} - S\,(A/F,i,n)

Exam strategy

Sketch the cash-flow diagram first, then pick the factor by reading its name as the fraction you want: you have

P

and want

A

, so reach for

(A/P)

. Confirm

i

and

n

share a period before substituting — convert nominal-to-effective or drop to a monthly rate up front. The FE supplies the factor tables, so you can either read tabulated values or evaluate the closed form on your calculator; the closed form is safer for odd rates the table skips. For life-cycle problems, annualize everything with

(A/P)

and

(A/F)

and compare EUACs, or discount everything to present worth — just never compare a present worth against an annual amount. A quick sanity check: present worths are always smaller in magnitude than the future sums they represent, and EUAC must exceed the bare annual O&M because it also carries the capital.

Key equations

Single-payment compound amount (F/P)

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

Grows a present lump sum $P$ to its future value $F$ after $n$ periods at periodic rate $i$ . Factor symbol $(F/P,i\%,n)$ .

Single-payment present worth (P/F)

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

Discounts a future amount $F$ back to the present. Reciprocal of $(F/P)$ ; the building block of all present-worth analysis.

Uniform-series present worth (P/A)

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

Present value of an end-of-period annuity $A$ for $n$ periods. Used to capitalize annual O&M or benefits.

Capital recovery (A/P)

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

Spreads a present cost $P$ into $n$ equal payments $A$ ; this annualizes capital cost. Reciprocal of $(P/A)$ .

Sinking fund (A/F)

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

Equal deposits $A$ that accumulate to a future amount $F$ (e.g., a replacement reserve). Note $(A/P)=(A/F)+i$ .

Uniform-series compound amount (F/A)

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

Future value of an annuity $A$ after $n$ periods. Reciprocal of $(A/F)$ .

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 $0,G,2G,\dots,(n-1)G$ . $G$ = constant yearly increase.

Gradient to uniform series (A/G)

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

Equivalent flat annuity of an arithmetic gradient. Add to a base $A_1$ to levelize a rising cost.

Effective annual interest rate

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

Converts nominal annual rate $r$ with $m$ compoundings/yr to the effective annual rate $i_e$ . $m\to\infty$ gives $i_e=e^{r}-1$ .

Equivalent uniform annual cost (EUAC)

\text{EUAC} = P\,(A/P,i,n) + A_{\text{O\&M}} - S\,(A/F,i,n)

Levelized annual life-cycle cost: capital recovery plus annual O&M less salvage credit. $S$ = salvage at year $n$ .

Capitalized cost (perpetual)

P = \dfrac{A}{i}

Present worth of a uniform cost continuing forever ( $n\to\infty$ ); used for permanent works and replacement-in-perpetuity comparisons.

Worked examples

Single sum plus effective-rate growth

Problem. A utility sets aside $5,000 today in an account earning a nominal $6\%$ compounded monthly. What is the balance after $8$ years?

Solution. Match the rate and period. Working in years requires the effective annual rate:

i_e=(1+0.06/12)^{12}-1 = 0.061678

(i.e.

6.168\%

). Apply the single-payment compound-amount factor over

n=8

years:

F = P\,(1+i_e)^{8} = 5000\,(1.061678)^{8}

(1.061678)^{8}=1.61414

, so

F = 5000(1.61414)=\$8{,}070.7

. (Equivalently, work in months:

F=5000(1+0.005)^{96}=\$8{,}070.7

— identical, as it must be.) Sanity check: at

6\%

effective for

8

years the factor would be

1.5938

, giving $7,969; monthly compounding earns slightly more, so the larger $8,071 is consistent.

\boxed{F \approx \$8{,}070}

F = 5000\left(1+\dfrac{0.06}{12}\right)^{96} = \$8{,}070.7

Life-cycle cost of a pump station (PW and EUAC)

Problem. A pump station costs $120,000 installed, has annual operation and maintenance of $18,000, and a salvage value of $15,000 at the end of its $15$ -year life. At a discount rate of $6\%$ , find the present worth of total cost and the equivalent uniform annual cost.

Solution. Gather the factors at

i=6\%

n=15

(P/A)=9.7122

(P/F)=0.41727

(A/P)=0.10296

(A/F)=0.04296

. Present worth (cost positive):

PW = 120{,}000 + 18{,}000\,(P/A) - 15{,}000\,(P/F)

PW = 120{,}000 + 18{,}000(9.7122) - 15{,}000(0.41727) = 120{,}000 + 174{,}820 - 6{,}259 = \$288{,}561

. EUAC by annualizing each piece:

\text{EUAC} = 120{,}000\,(A/P) + 18{,}000 - 15{,}000\,(A/F) = 120{,}000(0.10296) + 18{,}000 - 15{,}000(0.04296)

\text{EUAC} = 12{,}356 + 18{,}000 - 644 = \$29{,}711/\text{yr}

. Cross-check:

\text{EUAC} = PW\,(A/P) = 288{,}561(0.10296) = \$29{,}711/\text{yr}

— the two routes agree.

\boxed{PW = \$289{,}000,\quad \text{EUAC} = \$29{,}700/\text{yr}}

\text{EUAC} = 120{,}000\,(A/P,6\%,15) + 18{,}000 - 15{,}000\,(A/F,6\%,15)

Rising maintenance cost as a gradient

Problem. A blower's maintenance is expected to be $5,000 at the end of year $1$ and to increase by $1,200 each year thereafter, over a $10$ -year life. At $8\%$ , find the present worth of maintenance and the equivalent uniform annual maintenance cost.

Solution. Split into a flat base of $5,000 and an arithmetic gradient

G=\$1{,}200

(the gradient contributes $0 in year 1, $1,200 in year 2, and so on). Factors at

i=8\%

n=10

(P/A)=6.7101

(P/G)=25.977

(A/G)=3.8713

. Present worth:

PW = 5{,}000\,(P/A) + 1{,}200\,(P/G) = 5{,}000(6.7101) + 1{,}200(25.977) = 33{,}550 + 31{,}172 = \$64{,}723

. Equivalent uniform cost:

A_{\text{eq}} = 5{,}000 + 1{,}200\,(A/G) = 5{,}000 + 1{,}200(3.8713) = \$9{,}646/\text{yr}

. Sanity check:

A_{\text{eq}}(P/A) = 9{,}646(6.7101) = \$64{,}723

, matching the present worth, and the levelized $9,646 sensibly sits between the first-year $5,000 and the last-year $15,800.

\boxed{PW = \$64{,}700,\quad A_{\text{eq}} = \$9{,}650/\text{yr}}

PW = A_1\,(P/A,8\%,10) + G\,(P/G,8\%,10)

Common pitfalls

•Mismatching rate and period: using a yearly $n$ with a monthly rate (or vice versa). Convert nominal to effective with $i_e=(1+r/m)^m-1$ , or drop to the periodic rate $r/m$ with $n$ in months — but never mix.
•Confusing nominal and effective rates. " $6\%$ compounded monthly" is nominal; the effective annual rate is $6.17\%$ . Only when compounding is annual are the two equal.
•Applying the gradient factor to the whole first-year cash flow. The gradient is zero in year 1; use $(P/A)$ on the base amount $A_1$ and $(P/G)$ only on the increment $G$ .
•Reading the factor backwards: $(A/P)$ converts $P$ into $A$ , while $(P/A)$ does the reverse. They are reciprocals — using the wrong one inverts your answer.
•Comparing a present worth against an annual amount. Put every alternative on the same basis (all PW or all EUAC) before ranking.
•Forgetting the salvage credit, or adding it instead of subtracting it. Salvage is a future receipt that reduces life-cycle cost via $-S(A/F)$ or $-S(P/F)$ .
•Treating the lowest first cost as the cheapest option. Annual O&M discounted over a long life routinely dwarfs capital and reverses the ranking — that is exactly why life-cycle costing exists.

References

NCEES FE Reference Handbook — Engineering Economics
Newnan, Lavelle & Eschenbach, Engineering Economic Analysis — factor derivations and gradient series
Davis & Cornwell, Introduction to Environmental Engineering — life-cycle costing of treatment processes

B. Cost types and breakdowns

Cost Types, Breakdowns, and Depreciation

Fixed vs variable, direct vs indirect, sunk and incremental costs, capital vs O&M, and depreciation by straight-line and MACRS with book value.

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C. Economic analyses

Economic Analyses and Project Selection

Present worth, equivalent uniform annual cost, rate of return, benefit-cost ratio, break-even and payback, and comparing alternatives with unequal lives.

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