Project Planning

6% of exam

Quantity take-off, cost estimating, CPM schedules and activity sequencing, and economic/lifecycle comparison of alternatives (present worth).

4 concepts

Exam areas

A. Quantity take-off methods

Quantity Take-Off and Earthwork Volumes

Take quantities off plans, compute earthwork volumes by average-end-area and prismoidal methods, and balance cut and fill with a mass diagram while keeping bcy, ccy, and lcy straight.

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B. Cost estimating

Cost Estimating and Cost Indexes

Classify direct/indirect and fixed/variable costs, build unit-cost and parametric estimates, escalate historical prices with a cost index, and build up a bid with overhead, contingency, and profit.

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C. Project schedules

CPM Scheduling: Critical Path and Float

Build an activity-on-node network, run the forward and backward passes for ES/EF/LS/LF, compute total and free float, find the critical path, and track cost/schedule performance with earned value.

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E. Economic and sustainability analysis

Present Worth and Comparing Alternatives

Use time-value-of-money factors to compare alternatives by present worth, equivalent uniform annual cost, rate of return, benefit-cost ratio, payback, life-cycle cost, and the triple bottom line.

Every infrastructure decision is a money-over-time decision: a pump that costs more today but less to run, a pipe material that lasts twice as long, a green-infrastructure retrofit that pays back in avoided fees. Engineering economics is the language for comparing those alternatives on equal footing, and the PE Civil exam tests it directly under economic and sustainability analysis. The handbook collects the factors and definitions in

\S1.7

Engineering Economics; the whole subject reduces to moving cash flows to a common point in time with the right factor and then comparing apples to apples. This concept is the storefront walkthrough — slower and more complete — because the method here underwrites every other one.

Time value of money and the factors

A dollar today is worth more than a dollar next year because it can earn interest, so cash flows at different times cannot be added until they are moved to a common date at the interest rate

i

. The handbook's factor notation does exactly this:

(F/P,i,n)

grows a present amount

P

to a future amount

F

;

(P/F,i,n)

discounts it back;

(P/A,i,n)

and

(A/P,i,n)

convert between a uniform annual series

A

and a present sum;

(F/A,i,n)

and

(A/F,i,n)

link a series to a future sum; and the gradient factors

(P/G,i,n)

(A/G,i,n)

handle a uniformly increasing series. Master these six-plus factors and every problem is just bookkeeping.

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

Present worth — the master comparison

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

n

and compare totals. When lives differ, either evaluate over the least common multiple of lives (repeating each project) or, more simply, switch to an annual-cost basis.

\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 level annual amount over the project life using the capital-recovery factor

(A/P,i,n)

, with salvage credited through

(A/F,i,n)

. Its great virtue is comparing alternatives with different lives directly — an annual cost is already on a per-year basis, so no least-common-multiple gymnastics are needed. Pick the alternative with the lowest EUAC.

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

Rate of return and benefit-cost ratio

The rate of return (ROR) is the interest rate that makes an alternative's present worth zero — the break-even earning rate; accept the project if its ROR exceeds the minimum attractive rate of return (MARR). The benefit-cost ratio compares the present worth (or annual worth) of benefits to costs, and a project is economically justified when

B/C \ge 1

. For mutually exclusive alternatives, both methods must be applied incrementally: rank by cost and ask whether each added increment of investment earns at least the MARR, or returns

\Delta B/\Delta C \ge 1

. Comparing raw ROR or raw B/C between alternatives without the incremental step is the classic blunder.

\text{find } i^*\!: \ \text{PW}(i^*) = 0; \qquad \dfrac{B}{C} = \dfrac{\text{PW(benefits)}}{\text{PW(costs)}} \ge 1

Life-cycle cost and payback

Life-cycle cost (LCC) is the present worth (or annual worth) of all costs over an asset's life — acquisition, energy, operation, maintenance, replacement, and disposal, less salvage. It is the right lens for water and wastewater assets, where energy and O&M often dwarf first cost; the lowest-bid pump can be the most expensive over 20 years. Simple payback, by contrast, is the time for cumulative savings to repay the first cost, ignoring the time value of money; it is a quick screen, not a decision rule. Discounted payback restores the time value but still ignores everything after the payback date, so use LCC or PW for the actual choice.

\text{Payback} = \dfrac{P_0}{A_{\text{savings}}}, \qquad \text{LCC} = P_0 + \text{PW}(\text{O\&M} + \text{energy} + \text{repl.}) - \text{PW}(S)

Sustainability and the triple bottom line

Economic and sustainability analysis on the modern PE asks you to weigh more than dollars. The triple-bottom-line frames a decision across three accounts — economic (life-cycle cost, ROR), environmental (energy, carbon, water use, habitat), and social (safety, equity, resilience, community disruption). Some externalities can be monetized and folded into a B/C ratio (avoided flood damage, health costs); others are weighed qualitatively or with a weighted-criteria matrix. A green-infrastructure or low-impact-development alternative often shows a higher first cost but a stronger life-cycle and triple-bottom-line position once avoided fees, energy, and environmental benefits are counted — which is exactly why present-worth thinking, not first cost, must drive the comparison.

\text{Value}_{\text{TBL}} = f\bigl(\text{economic},\ \text{environmental},\ \text{social}\bigr)

Exam strategy

First decide the basis: equal lives → present worth; unequal lives → EUAC (or least common multiple of lives). Read whether the interest factors are given in a table or must be computed from the closed forms, and double-check

n

and

i

match the compounding period. Credit salvage with

(A/F)

in EUAC or

(P/F)

in PW — never forget it. For choosing among alternatives by ROR or B/C, always go incremental: order by increasing cost and test each step against the MARR or against

\Delta B/\Delta C \ge 1

. Treat payback as a screen only, use life-cycle cost for the real decision, and when the problem mentions sustainability, organize the non-dollar factors into the three triple-bottom-line accounts rather than hand-waving.

Key equations

Single-payment compound amount / present worth

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

$(F/P,i,n)$ and $(P/F,i,n)$ . Move a lump sum forward or back $n$ periods at rate $i$ .

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 $A$ for $n$ periods. Inverse is the capital-recovery factor $(A/P)$ .

Capital recovery factor (A/P)

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

Converts a present cost $P$ into an equivalent uniform annual amount; the workhorse of EUAC.

Sinking-fund factor (A/F)

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

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

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 different lives by lowest EUAC. $S$ = salvage.

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; choose least-cost (or greatest net-benefit) PW over equal lives.

Benefit-cost ratio

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

Project justified when $\ge 1$ ; for alternatives use incremental $\Delta B/\Delta C$ .

Rate of return condition

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

$i^*$ = interest rate that 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 maintenance of a structure.

Inflation-adjusted rate

d = i + f + i\,f

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

Simple payback period

\text{Payback} = \dfrac{P_0}{A_{\text{savings}}}

Years to recover first cost from annual savings; ignores time value — a screen, not a decision rule.

Worked examples

Comparing alternatives with different lives (EUAC)

Problem. Two pump alternatives serve the same duty at MARR $= 6\%$ . Alternative A: first cost $\$120{,}000$ , annual O&M $\$8{,}000$ , salvage $\$15{,}000$ , life 10 yr. Alternative B: first cost $\$180{,}000$ , annual O&M $\$5{,}000$ , salvage $\$25{,}000$ , life 15 yr. Which is more economical?

Solution. Lives differ (10 vs 15 yr), so compare on an annual basis with EUAC. Factors at 6%:

(A/P,6\%,10) = \dfrac{0.06(1.06)^{10}}{(1.06)^{10}-1} = 0.13587

;

(A/F,6\%,10) = 0.07587

(A/P,6\%,15) = 0.10296

;

(A/F,6\%,15) = 0.04296

. EUAC

_A = 120{,}000(0.13587) + 8{,}000 - 15{,}000(0.07587) = 16{,}304 + 8{,}000 - 1{,}138 = 23{,}166 \approx \$23{,}200/\text{yr}

. EUAC

_B = 180{,}000(0.10296) + 5{,}000 - 25{,}000(0.04296) = 18{,}533 + 5{,}000 - 1{,}074 = 22{,}459 \approx \$22{,}500/\text{yr}

. Alternative B has the lower EUAC by about

\$700/\text{yr}

— choose B. Sanity check: B's higher first cost is annualized over a longer life and offset by lower O&M, so the per-year totals come out close, with B slightly ahead. Both near

\$23{,}000/\text{yr}

, a believable spread.

\text{EUAC}_B = 180{,}000(0.10296) + 5{,}000 - 25{,}000(0.04296) \approx \$22{,}500/\text{yr}

Benefit-cost ratio of a flood-control project

Problem. A flood-control levee costs $\$5.0$ M to build with $\$120{,}000$ /yr maintenance and provides $\$650{,}000$ /yr in avoided flood damage over a 30-yr life at $i = 4\%$ (no salvage). Find the benefit-cost ratio and state whether it is justified.

Solution. Annualize the capital cost:

(A/P,4\%,30) = \dfrac{0.04(1.04)^{30}}{(1.04)^{30}-1} = 0.05783

. Annual cost

= 5{,}000{,}000(0.05783) + 120{,}000 = 289{,}150 + 120{,}000 = \$409{,}150/\text{yr}

. Annual benefit

= \$650{,}000/\text{yr}

\dfrac{B}{C} = \dfrac{650{,}000}{409{,}150} = 1.59 \ge 1

— the project is economically justified. Sanity check: benefits exceed annualized costs by about 59%, comfortably above 1; equivalently the net annual benefit is

\$241{,}000/\text{yr}

, positive. Consistent with

B/C>1

\dfrac{B}{C} = \dfrac{650{,}000}{5{,}000{,}000(0.05783)+120{,}000} = \dfrac{650{,}000}{409{,}150} = 1.59

Payback and rate of return for a green retrofit

Problem. A low-impact-development stormwater retrofit costs $\$85{,}000$ and saves $\$14{,}500$ /yr in stormwater utility fees and pumping over a 10-yr life (negligible salvage). Find the simple payback and the rate of return, and judge it against a MARR of $5\%$ and its triple-bottom-line merits.

Solution. Simple payback

= \dfrac{P_0}{A} = \dfrac{85{,}000}{14{,}500} = 5.86\ \text{yr}

— under six years, ignoring interest. Rate of return: find

i^*

with PW

= 0

, i.e.

(P/A,i^*,10) = \dfrac{85{,}000}{14{,}500} = 5.862

. From the closed form,

(P/A,10\%,10) = 6.145

and

(P/A,12\%,10) = 5.650

; interpolating,

i^* \approx 11.1\%

. Since

i^* \approx 11.1\% > \text{MARR} = 5\%

, the retrofit is economically justified. On the triple bottom line it also scores on the environmental account (reduced runoff and pumping energy) and the social account (flood resilience, urban heat/amenity), reinforcing the economic case. Sanity check: a 5.86-yr simple payback on a 10-yr life implies a return comfortably above the discount rate, and

i^* \approx 11\%

is consistent with recovering the cost nearly twice over 10 years. Note payback alone would not have told us the return — only the screen.

(P/A,\,i^*,\,10) = \dfrac{85{,}000}{14{,}500} = 5.86 \ \Rightarrow\ i^* \approx 11.1\%

Common pitfalls

•Comparing present worths over unequal lives. PW is only valid over a common study period; for different lives use EUAC or the least-common-multiple of lives.
•Dropping salvage. Credit it with $(A/F,i,n)$ in EUAC or $(P/F,i,n)$ in PW; omitting salvage overstates cost and can flip the choice.
•Comparing raw ROR or raw B/C between alternatives. Mutually exclusive choices require incremental analysis ( $\Delta B/\Delta C \ge 1$ , or $\Delta$ ROR vs MARR) — the highest standalone B/C is not necessarily best.
•Confusing nominal and effective rates. With non-annual compounding, convert with $i_e=(1+r/m)^m-1$ before using annual factors; mixing periods corrupts every factor.
•Treating simple payback as a decision rule. It ignores the time value of money and everything past the payback date; use it only as a screen and decide with PW/LCC.
•Mixing real and inflated dollars. Discount actual (then-current) dollars with the inflation-adjusted rate $d=i+f+if$ , and constant dollars with the real rate $i$ — never one with the other.
•Forgetting capitalized cost for perpetual services. A permanent maintenance obligation is $P=A/i$ , not an $n$ -year series; using a finite $n$ understates its present worth.

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 — factor tables and incremental comparison methods
Envision / ISI Rating System (triple-bottom-line sustainability) — framework for non-monetary environmental and social criteria