Project Management

9% of exam

Quantity take-off and cost estimating, CPM schedules with activity sequencing and float, and engineering-economic comparison of alternatives (present worth, EUAC, lifecycle, benefit-cost).

4 concepts

Exam areas

A. Quantity and cost estimating

Quantity Take-Off & Cost Estimating

Take earthwork and material quantities off the plans, price them with unit costs, escalate dated prices with a cost index, and build a bid with overhead, contingency, and profit.

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B. Project schedules

CPM Scheduling, Float & Crashing

Build an activity-on-node network, run the forward and backward passes for ES/EF/LS/LF, read total and free float off the critical path, and crash the schedule at least cost.

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C. Economic analysis

Engineering Economics & Benefit-Cost

Move cash flows to one date with the interest-factor family, compare alternatives by present worth and EUAC, and justify a project with the benefit-cost ratio.

Almost every transportation decision is a money-over-time decision: a roundabout that costs more to build but slashes crash costs, a longer-lasting pavement, an ITS upgrade that buys back travel-time delay year after year. Engineering economics is the language that puts those alternatives on equal footing, and the PE Civil exam tests it directly under economic analysis. The whole subject reduces to one idea — you cannot add cash flows that occur at different times until you move them to a common date at the interest rate

i

— and the handbook hands you the factors that do the moving in

\S1.7

Engineering Economics. This concept is the deliberate, full walkthrough, because the method here underwrites the life-cycle and alternatives comparisons every other planning question rests on.

Why a dollar has a date

A dollar today is worth more than a dollar a year from now because today's dollar can be invested and earn interest. That single fact forbids adding a first cost to an annual maintenance figure to a far-off salvage value as if they were the same currency — they are dollars at different dates, and dates carry value. The fix is mechanical: pick one reference date (usually time zero, 'present'), discount or grow every cash flow to that date at the rate

i

, and only then add. Master that habit and the arithmetic never lies to you.

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

The interest-factor family

The handbook's factor notation is a compact verb for 'move this cash flow.'

(F/P,i,n)

grows a present lump

P

forward

n

periods;

(P/F,i,n)

discounts a future lump back.

(P/A,i,n)

and its inverse the capital-recovery factor

(A/P,i,n)

trade a level annual series

A

for a present sum — the workhorse of annual-cost work.

(F/A,i,n)

and the sinking-fund factor

(A/F,i,n)

link a series to a future sum, the natural home of salvage. The gradient factors

(P/G,i,n)

and

(A/G,i,n)

handle a series that grows by a constant amount

G

each period — exactly the shape of benefits that rise as traffic grows. Read the factor name as 'what you want / what you have,' and the algebra writes itself.

(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}

Present worth — the master comparison

Present worth (PW) discounts every cash flow of an alternative to time zero and sums them. For a cost stream the better alternative is the one with the smallest (least-negative) PW; for net benefits, the largest. PW is the cleanest method when the alternatives share the same study period — line them up over the same

n

and total them. A single positive net present worth means the project earns more than the discount rate; a negative one means it does not. When lives differ, PW needs the least-common-multiple of lives or a switch to an annual basis, which is the next tool.

\text{PW} = -P_0 + \sum_{t=1}^{n}\dfrac{A_t}{(1+i)^t} + \dfrac{S}{(1+i)^n}

Equivalent uniform annual cost

Equivalent uniform annual cost (EUAC) spreads every cash flow into a single level annual figure using the capital-recovery factor

(A/P,i,n)

, with salvage credited back through the sinking-fund factor

(A/F,i,n)

. Because the result is already per-year, EUAC compares alternatives of different lives without any least-common-multiple gymnastics — pick the lowest EUAC. It is the same information as present worth, rotated onto an annual axis;

\text{PW} = \text{EUAC}\times(P/A,i,n)

converts one to the other.

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

Benefit-cost ratio and rate of return

Public transportation projects are usually justified with a benefit-cost ratio: the present worth (or annual worth) of benefits divided by that of costs, with the project economically justified when

B/C \ge 1

. Benefits are user goods — travel-time savings, crash-cost reductions, vehicle-operating savings — while costs are agency capital plus maintenance, often net of residual value. The rate of return is the companion idea: the interest rate

i^*

that drives present worth to zero; accept when

i^*

exceeds the minimum attractive rate of return (MARR). For choosing among mutually exclusive alternatives, neither raw

B/C

nor raw ROR may be compared directly — that requires the incremental test covered in the alternatives concept.

\dfrac{B}{C} = \dfrac{\text{PW (or AW) of benefits}}{\text{PW (or AW) of costs}} \ge 1

Perpetual services and capitalized cost

Some transportation assets serve effectively forever — a bridge corridor that will be maintained and periodically rehabilitated for as long as the route exists. The present worth of a perpetual uniform annual cost

A

is simply

P = A/i

, the capitalized cost. A recurring lump (a deck replacement every

k

years) is first turned into its equivalent annual amount with

(A/F,i,k)

and then capitalized. Capitalized cost is the right basis when comparing alternatives with essentially infinite or very long horizons against shorter-lived ones.

P = \dfrac{A}{i}

Exam strategy

First fix the reference date and the basis: equal lives invite present worth; unequal lives invite EUAC. Read whether the factors are tabulated or must be computed from the closed forms, and confirm

i

and

n

share the same period — convert any non-annual compounding with

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

before touching a factor. Never drop salvage: credit it with

(A/F)

in EUAC or

(P/F)

in PW. For justification, a single project lives or dies by

B/C \ge 1

i^* \ge \text{MARR}

; for a perpetual obligation reach for

P = A/i

, not a finite series. Write each factor with its

(i,n)

stated so you never substitute the wrong one.

Key equations

Single payment (F/P, P/F)

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

Grow ( $F/P$ ) or discount ( $P/F$ ) a lump sum $n$ periods at rate $i$ . $P$ , $F$ in dollars, $i$ per period, $n$ periods.

Uniform series present worth (P/A)

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

Present worth of a level annual amount $A$ over $n$ periods. Inverse is capital recovery $(A/P)$ .

Capital recovery factor (A/P)

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

Converts a present cost $P$ into a level annual amount; the core of every EUAC.

Sinking-fund factor (A/F)

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

Converts a future amount (e.g., salvage) into a level annual amount; credited in EUAC.

Uniform gradient present worth (P/G)

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

Present worth of a series that increases by a constant $G$ each period (first cash flow at end of period 2). $G$ in dollars/period.

Net present worth

\text{PW} = -P_0 + \sum_{t=1}^{n}\dfrac{A_t}{(1+i)^t} + \dfrac{S}{(1+i)^n}

Discount all cash flows to time zero; accept if $\text{PW} \ge 0$ , choose the largest among net-benefit alternatives over equal lives.

Equivalent uniform annual cost

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

Annualized total cost; compare alternatives of any lives by lowest EUAC. $S$ = salvage at end of life $n$ .

Benefit-cost ratio

\dfrac{B}{C} = \dfrac{\text{PW or AW of benefits}}{\text{PW or AW of costs}} \ge 1

A single project is justified when $\ge 1$ ; for mutually exclusive alternatives use incremental $\Delta B/\Delta C$ .

Rate of return condition

\text{PW}(i^*) = 0, \quad \text{accept if } i^* \ge \text{MARR}

$i^*$ zeroes present worth; compare to the minimum attractive rate of return.

Capitalized cost (perpetual)

P = \dfrac{A}{i}

Present worth of a perpetual annual cost $A$ (infinite life), e.g., permanent corridor maintenance.

Effective interest rate

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

Convert nominal annual rate $r$ with $m$ compounding periods/yr to an effective annual rate before using annual factors (handbook $\S1.7.2$ ).

Inflation-adjusted rate

d = i + f + i\,f

Combined rate when cash flows are in actual (then-current) dollars and inflation $f$ applies (handbook $\S1.7.4$ ).

Worked examples

Net present worth of a signal-system upgrade

Problem. A city evaluates an adaptive traffic-signal upgrade on a corridor. First cost is $\$250{,}000$ . It saves $\$45{,}000$ /yr in delay and fuel against $\$23{,}000$ /yr in added O&M, so the net benefit is $\$22{,}000$ in year 1 and grows by $\$4{,}000$ /yr as traffic increases (a uniform gradient). Salvage of the controllers is $\$25{,}000$ at the end of the 10-yr life. At MARR $= 6\%$ , is the upgrade justified?

Solution. Factors at 6%, 10 yr:

(P/A,6\%,10) = \dfrac{1.06^{10}-1}{0.06\,(1.06)^{10}} = 7.3601

;

(P/G,6\%,10) = 29.602

;

(P/F,6\%,10) = 0.55839

. Discount each piece to time zero: Uniform net benefit:

22{,}000(7.3601) = \$161{,}922

. Gradient:

4{,}000(29.602) = \$118{,}409

. Salvage:

25{,}000(0.55839) = \$13{,}960

\text{PW} = -250{,}000 + 161{,}922 + 118{,}409 + 13{,}960 = +\$44{,}291 \approx +\$44{,}300

. Since

\text{PW} > 0

, the upgrade earns more than 6% — it is justified. Sanity check: ignoring time value, undiscounted net benefits over 10 yr total roughly

22{,}000(10) + 4{,}000(45) = 400{,}000

, far above the

\$250{,}000

cost, so a positive (but much smaller) discounted NPW is the expected sign.

\text{PW} = -250{,}000 + 22{,}000(7.3601) + 4{,}000(29.602) + 25{,}000(0.55839) \approx +\$44{,}300

Benefit-cost ratio of a highway bypass

Problem. A rural bypass costs $\$12.0$ M to construct, with $\$200{,}000$ /yr maintenance, and yields $\$1.30$ M/yr in user benefits (travel-time and crash-cost savings) over a 25-yr analysis period at $i = 4\%$ , salvage negligible. Find the benefit-cost ratio and state whether the project is justified.

Solution. Put costs on an annual basis. Capital recovery:

(A/P,4\%,25) = \dfrac{0.04(1.04)^{25}}{(1.04)^{25}-1} = 0.06401

. Annual cost

= 12{,}000{,}000(0.06401) + 200{,}000 = 768{,}144 + 200{,}000 = \$968{,}144

/yr. Annual benefit

= \$1{,}300{,}000

/yr.

\dfrac{B}{C} = \dfrac{1{,}300{,}000}{968{,}144} = 1.34 \ge 1

— the bypass is economically justified. Sanity check: benefits exceed annualized costs by about 34%; the net annual benefit is

\$332{,}000

/yr, positive, consistent with

B/C > 1

\dfrac{B}{C} = \dfrac{1{,}300{,}000}{12{,}000{,}000(0.06401) + 200{,}000} = \dfrac{1{,}300{,}000}{968{,}144} = 1.34

Capitalized cost of a perpetual bridge obligation

Problem. A bridge will carry routine inspection and minor maintenance of $\$15{,}000$ /yr indefinitely, plus a major deck replacement of $\$800{,}000$ every 25 years in perpetuity. At $i = 5\%$ , find the capitalized cost (present worth of the perpetual obligation).

Solution. Convert the recurring lump to an equivalent annual amount:

(A/F,5\%,25) = \dfrac{0.05}{(1.05)^{25}-1} = 0.020952

. Annual equivalent of the deck work

= 800{,}000(0.020952) = \$16{,}762

/yr. Total perpetual annual cost

= 15{,}000 + 16{,}762 = \$31{,}762

/yr. Capitalize:

P = \dfrac{A}{i} = \dfrac{31{,}762}{0.05} = \$635{,}000

(3 sig figs). Sanity check: the

\$800{,}000

every 25 yr alone, capitalized, is

16{,}762/0.05 = \$335{,}000

— less than its face value because it is far in the future and recurs, which is the expected direction. Units:

(\$/\text{yr}) \div (1/\text{yr}) = \text{dollars}

P = \dfrac{15{,}000 + 800{,}000\,(A/F,5\%,25)}{0.05} = \dfrac{31{,}762}{0.05} \approx \$635{,}000

Common pitfalls

•Adding cash flows that occur at different dates. A first cost, an annual O&M, and a salvage are dollars at different times — discount each to a common date before summing.
•Inverting the capital-recovery and series-present-worth factors. $(A/P,i,n)$ spreads a present sum into a uniform series, while $(P/A,i,n)$ does the reverse; they are reciprocals, so grabbing the wrong one (or reading the wrong table column) scales the answer by $(A/P)^2$ . Confirm the direction from the units before you multiply.
•Confusing nominal and effective rates. With non-annual compounding, convert to $i_e = (1+r/m)^m - 1$ before using any annual factor; mixing periods corrupts every factor.
•Dropping salvage or residual value. Credit it with $(A/F,i,n)$ in EUAC or $(P/F,i,n)$ in PW; omitting it overstates cost and can flip the decision.
•Misreading the gradient base. $(P/G,i,n)$ values a series whose first increment appears in period 2; the period-1 amount is carried by the uniform $(P/A)$ term, not the gradient.
•Using a finite $n$ for a perpetual service. A permanent maintenance obligation is $P = A/i$ ; a finite series understates its present worth.
•Reporting $B/C$ from raw (undiscounted) totals. Benefits and costs must be present-worth or annual-worth equivalents at the same $i$ and $n$ before forming the ratio.

References

NCEES PE Civil Reference Handbook — §1.7 Engineering Economics
NCEES PE Civil Reference Handbook — §1.7.9 Benefit-Cost Analysis
Newnan, Lavelle & Eschenbach, Engineering Economic Analysis — interest-factor tables and worked transportation economics
AASHTO, User and Non-User Benefit Analysis for Highways (the 'Red Book') — framework for monetizing highway user benefits in B/C studies

Life-Cycle Cost & Comparing Alternatives

Compare alternatives with unequal lives by equivalent uniform annual cost, handle salvage and replacement, apply incremental benefit-cost, and test the choice against discount-rate sensitivity.

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