Construction Engineering

A. Project administration

Project Administration and Delivery Methods

Choose among design-bid-build, design-build, and CM-at-risk, match the contract type to the risk, and run the paperwork: submittals, RFIs, change orders, progress payments, and OSHA safety.

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B. Construction operations and methods

Construction Operations, Methods, and Drawing Interpretation

Convert soil among bank, loose, and compacted states, compute equipment production and truck-fleet matches, and read plan-and-profile drawings and temporary-works requirements.

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C. Project controls

CPM Scheduling and Project Controls

Build an activity-on-node network, run the forward and backward passes for float and the critical path, then crash and track it with earned value (CPI/SPI).

Project controls is where the FE Civil rewards bookkeeping discipline, not cleverness. The Critical Path Method (CPM) turns a list of activities and their dependencies into a network you can pass over twice — once forward for the earliest you can do things, once backward for the latest you can do them without slipping the finish — and the difference between those two answers, the float, tells you exactly which activities the whole project waits on. The NCEES FE Reference Handbook (Civil Engineering chapter, Construction section) gives you the activity-on-node annotation and the bare nomenclature for

ES

,

EF

,

LS

,

LF

, and float; it expects you to supply the procedure. Points are lost here in two predictable ways: an off-by-one from mixing the 'day-0' and 'day-1' counting conventions, and confusing total float with free float.

The activity-on-node network

In activity-on-node (AON) — also called precedence diagramming — each box is an activity carrying a duration

D

, and the arrows are pure logic showing which activity must finish before the next can start (the default finish-to-start relationship). The handbook also lists start-to-start, finish-to-finish, and start-to-finish links, but the FE almost always uses finish-to-start. You read the network left to right; a path is any chain of activities from the single start to the single finish, and a path's length is the sum of its durations. The whole method exists to find the one longest path, because that length is the shortest time the project can possibly take.

Forward pass: early start and early finish

Sweep left to right. Set the project start time; the FE uses two conventions, and you must read which. In the 'day-0' (or 'finish-of-day') convention the first activity has

ES = 0

and

EF = ES + D

. In the 'day-1' (calendar-day) convention the first activity starts on day 1 and

EF = ES + D - 1

. Whichever you adopt, an activity's early start is the latest early finish among all its predecessors — you cannot start until every prerequisite is done. The early finish of the last activity is the project duration.

EF = ES + D, \qquad ES = \max_{\text{predecessors}} (EF)

Backward pass: late finish and late start

Now sweep right to left. Seed the last activity's late finish equal to the project duration you just found (do not invent a deadline unless the problem gives one), then

LS = LF - D

. Working backward, an activity's late finish is the smallest late start among all its successors — you must finish early enough not to delay any follower. When the forward and backward seeds match at the ends, every interior number falls out mechanically.

LS = LF - D, \qquad LF = \min_{\text{successors}} (LS)

Total float versus free float

Total float is how long an activity can slip without delaying the project finish:

TF = LS - ES = LF - EF

. Free float is how long it can slip without delaying any successor's early start:

FF = \min(ES_{\text{succ}}) - EF

. Free float is always less than or equal to total float, and it is the float you can spend without stealing from anyone downstream. The critical path is the chain of activities with zero total float — equivalently, the longest path through the network. Several activities can be critical; on rare networks more than one full path is critical at once.

TF = LS - ES = LF - EF, \qquad FF = \min(ES_{\text{succ}}) - EF

Crashing: buying time on the critical path

To shorten a project you compress activities, paying a premium. Each activity has a cost slope — the dollars per day to crash it — found from its normal and crash points. You can only shorten the project by crashing an activity that lies on the current critical path, and you always crash the cheapest critical activity first. As you compress, a once-slack path can become critical too; from that point you must crash both paths simultaneously (often two activities per day) until the target or the crash limits stop you.

\text{cost slope} = \dfrac{C_{\text{crash}} - C_{\text{normal}}}{D_{\text{normal}} - D_{\text{crash}}}

Earned-value management: PV, EV, AC

Earned value collapses cost and schedule into three dollar quantities at a status date. Planned value

PV

(the handbook's BCWS) is the budgeted cost of the work you scheduled to have done; actual cost

AC

(ACWP) is what you have actually spent; earned value

EV

(BCWP) is the budgeted cost of the work actually completed — the budget value you have 'earned.' Cost variance is

CV = EV - AC

and schedule variance is

SV = EV - PV

; negative means over budget or behind schedule. The dimensionless indices are the exam favorites: a

CPI

or

SPI

below 1.0 is bad news, and dividing the budget at completion by

CPI

gives a quick forecast of the final cost.

CPI = \dfrac{EV}{AC}, \quad SPI = \dfrac{EV}{PV}, \quad EAC = \dfrac{BAC}{CPI}

Exam strategy

First decide and write down your counting convention (

ES=0

vs

ES=1

) and keep it for the whole network — most CPM errors are off-by-one slips from switching mid-problem. Run the forward pass fully before starting the backward pass, then read float off both ends as a check (

LS-ES

must equal

LF-EF

). For 'how much can activity X slip' questions, ask whether they mean total or free float — free if it must not disturb a successor. For EVM, memorize that every variance is

EV

minus something and every index is

EV

over something; if the sign or ratio is unfavorable,

EV

is the smaller number. A 30-second sketch of the AON network beats trying to track dependencies in your head.

Key equations

Early finish (forward pass)

EF = ES + D

Earliest an activity can finish; $D$ = duration. In day-1 counting use $EF = ES + D - 1$ .

Early start

ES = \max_{\text{predecessors}}(EF)

An activity cannot start until its latest-finishing predecessor is done.

Late start (backward pass)

LS = LF - D

Latest an activity can start without delaying the project.

Late finish

LF = \min_{\text{successors}}(LS)

Must finish before the earliest-starting successor's late start; last activity seeded at project duration.

Total float

TF = LS - ES = LF - EF

Slack before the project finish slips. $TF = 0$ defines the critical path.

Free float

FF = \min(ES_{\text{succ}}) - EF

Slack before any successor's early start slips; $FF \le TF$ . The FE Reference Handbook 10.6 prints only total float (Float = LS − ES or LF − EF), not free float.

Crash cost slope

\text{slope} = \dfrac{C_{\text{crash}} - C_{\text{normal}}}{D_{\text{normal}} - D_{\text{crash}}}

Dollars per day to compress an activity; crash the cheapest critical activity first.

Cost variance

CV = EV - AC

Budgeted value earned minus actual cost; negative = over budget. $EV$ =BCWP, $AC$ =ACWP.

Schedule variance

SV = EV - PV

Earned minus planned value; negative = behind schedule. $PV$ =BCWS (all in dollars).

Cost performance index

CPI = \dfrac{EV}{AC}

Dimensionless cost efficiency; $<1$ means each dollar buys less than a dollar of budgeted work.

Schedule performance index

SPI = \dfrac{EV}{PV}

Dimensionless schedule efficiency; $<1$ means work is being earned slower than planned.

Estimate at completion

EAC = \dfrac{BAC}{CPI}

Forecast final cost assuming current $CPI$ holds; $BAC$ = budget at completion.

Worked examples

Forward/backward pass, float, and the critical path

Problem. A project has activities A(4 d), B(3 d), C(6 d), D(5 d), E(2 d). A is first. B and C both depend on A. D depends on B. E depends on C and D. Using $ES = 0$ at the start, find the project duration, the critical path, and the total and free float of activity C.

Solution. Forward pass (

EF = ES + D

,

ES = \max EF_{\text{pred}}

): A:

ES=0,\ EF=4

. B:

ES=4,\ EF=7

. C:

ES=4,\ EF=10

. D:

ES=7,\ EF=12

. E:

ES=\max(10,12)=12,\ EF=14

. Project duration

=EF_E = 14\ \text{days}

. Backward pass (seed

LF_E = 14

,

LS = LF - D

,

LF = \min LS_{\text{succ}}

): E:

LF=14,\ LS=12

. D:

LF=12,\ LS=7

. C:

LF=12,\ LS=6

. B:

LF=7,\ LS=4

. A:

LF=\min(4,6)=4,\ LS=0

. Total floats

LS-ES

: A=0, B=0, C=2, D=0, E=0. The zero-float chain A-B-D-E is the critical path (length

4+3+5+2 = 14

, confirming the duration). Activity C:

TF = LS-ES = 6-4 = 2\ \text{days}

. Free float

= \min(ES_{\text{succ}}) - EF_C = ES_E - EF_C = 12 - 10 = 2\ \text{days}

. Sanity check:

LF_C - EF_C = 12 - 10 = 2 = TF

, consistent. C's float equals 2 days both ways, and the critical path length matches the forward-pass duration.

\text{Critical path: A}\rightarrow\text{B}\rightarrow\text{D}\rightarrow\text{E},\quad T = 14\ \text{days}

Crashing to a target duration

Problem. For the same network (critical path A-B-D-E, 14 days; near-critical path A-C-E, 12 days), crash data are: A can crash 1 day at $300/day; B at $500/day; D at $400/day; E at $700/day. Find the least-cost way to deliver in 12 days.

Solution. Only critical-path activities (A, B, D, E) reduce the 14-day duration. Cheapest is A at $300. Day 1: crash A by 1 day (

\$300

). Durations: A=3. Path A-B-D-E

= 3+3+5+2 = 13

; path A-C-E

= 3+6+2 = 11

. A-B-D-E still critical at 13 days. A is now at its crash limit. Of the remaining critical activities, cheapest is D at $400 (B is $500, E is $700). Day 2: crash D by 1 day (

\$400

). Durations: D=4. Path A-B-D-E

= 3+3+4+2 = 12

; path A-C-E

= 11

. Target met at 12 days. Total crash premium

= \$300 + \$400 = \$700

. Sanity check: we never crashed the slack path A-C-E (it stayed below 12), and each day removed came off whatever path was longest, so 12 days is achieved for the minimum added cost of

\$700

.

\Delta C_{\min} = \$300 + \$400 = \$700

Earned-value status and forecast

Problem. A project has $BAC = \$500{,}000$ . At the status date the planned value is $PV = \$200{,}000$ , the earned value is $EV = \$180{,}000$ , and the actual cost is $AC = \$210{,}000$ . Compute CV, SV, CPI, SPI, and the estimate at completion. Is the job over or under budget, ahead or behind?

Solution. Cost variance:

CV = EV - AC = 180{,}000 - 210{,}000 = -\$30{,}000

(over budget). Schedule variance:

SV = EV - PV = 180{,}000 - 200{,}000 = -\$20{,}000

(behind schedule). Indices:

CPI = EV/AC = 180/210 = 0.857

;

SPI = EV/PV = 180/200 = 0.900

. Both below 1.0 confirm the variances. Forecast:

EAC = BAC/CPI = 500{,}000/0.857 = \$583{,}000

(3 s.f.), i.e., a projected overrun of about

\$83{,}000

versus the

\$500{,}000

budget. Sanity check:

EV

(

\$180

k) is the smallest of the three dollar figures, so both variances are negative and both indices are under 1 — the project is over budget and behind schedule, consistent.

CPI = \dfrac{180}{210} = 0.857,\qquad EAC = \dfrac{500{,}000}{0.857} \approx \$583{,}000

Common pitfalls

•Mixing the day-0 and day-1 counting conventions mid-network. Pick one ( $ES=0,\ EF=ES+D$ OR $ES=1,\ EF=ES+D-1$ ) and hold it; switching produces off-by-one durations and floats.
•On the forward pass taking the minimum predecessor $EF$ instead of the maximum. $ES$ is the LATEST predecessor finish — you wait for the slowest prerequisite.
•Seeding the backward pass with an arbitrary deadline. Set $LF$ of the last activity equal to the project duration from the forward pass unless the problem states a hard deadline.
•Confusing total float with free float. $TF$ protects the project finish; $FF$ protects successors. They differ whenever an activity feeds a path with its own slack.
•Crashing a non-critical activity. Compressing a slack activity wastes money and shortens nothing; only critical-path activities reduce duration, and a second path may go critical partway through.
•Sign errors in EVM. Every variance is $EV$ minus the other term ( $CV=EV-AC$ , $SV=EV-PV$ ); writing $AC-EV$ flips the sign and reverses the over/under-budget conclusion.
•Reading $SPI$ as a time. $SPI$ and $CPI$ are dimensionless dollar ratios; an $SPI$ of 0.9 does not mean '90% of the days,' it means earned value is accruing at 90% of the planned rate.

References

NCEES FE Reference Handbook — Civil Engineering (Construction: CPM precedence, ES/EF/LS/LF, float, earned-value analysis)
Project Management Institute, A Guide to the Project Management Body of Knowledge (PMBOK) — Earned-value terminology PV/EV/AC and the CPI/SPI indices.
Hinze, Construction Planning and Scheduling — AON network analysis, float, and least-cost crashing procedure.

D. Construction estimating

Construction Estimating and Quantity Take-Off

Turn drawings into measured quantities, price them with unit costs and crew productivity, then build the bid from direct cost, overhead, contingency, and profit.

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