10.2 Project Management: WBS, Critical Path Method (CPM), PERT, and Crashing

10.2 Project Management: WBS, Critical Path Method (CPM), PERT, and Crashing

Project management in industrial and systems engineering demands rigorous mathematical methodologies to translate high-level deliverable requirements into executable activity networks. Whether launching an automated manufacturing facility, deploying supply chain software, or conducting lean re-engineering, engineers rely on the Work Breakdown Structure (WBS), Critical Path Method (CPM), Program Evaluation and Review Technique (PERT), and time-cost crashing optimization.

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1. Work Breakdown Structure (WBS) & Network Representations

The Work Breakdown Structure (WBS)

A Work Breakdown Structure (WBS) is a hierarchical, deliverable-oriented decomposition of the total scope of work to be carried out by the project team.

• The 100% Rule: The WBS must encompass exactly 100% of the work defined by the project scope, accounting for all internal, external, and project management deliverables. The sum of the work at each subordinate level must roll up to exactly 100% of the parent element, with no overlapping scope and no orphaned activities.
• Work Packages: The lowest decomposition level in the WBS is termed a work package. A work package represents an identifiable, independent unit of work that can be reliably scheduled, assigned to an individual owner or control account manager, budgeted with distinct direct costs, and quantitatively monitored.

Activity Network Conventions: AON vs. AOA

Once work packages are decomposed into discrete sequential activities, precedence logic is represented via project networks:

Activity Network Conventions
Activity-on-Node (AON / PDM)             Activity-on-Arrow (AOA / ADM)
       ┌───────────┐                            (Event i) ──Activity A──> (Event j)
       │ Activity  │                                  \                  /
       │   Node    │                                   \   (Dummy)      /
       └───────────┘                                    \- - - - - - - >
  Nodes = Activities / Tasks               Arrows = Activities
  Arrows = Precedence Dependencies         Nodes = Milestone Events (Requires Dummies)

1. Activity-on-Node (AON / Precedence Diagramming Method - PDM): Nodes represent the activities, and directed arrows represent precedence relationships. AON is the universal standard in modern project scheduling. It natively supports four dependency types:
   • Finish-to-Start (FS): Activity B cannot start until Activity A finishes (most common).
   • Start-to-Start (SS): Activity B cannot start until Activity A starts.
   • Finish-to-Finish (FF): Activity B cannot finish until Activity A finishes.
   • Start-to-Finish (SF): Activity B cannot finish until Activity A starts.
2. Activity-on-Arrow (AOA / Arrow Diagramming Method - ADM): Arrows represent activities, and circular nodes represent milestone events (the start or completion of activities). AOA networks only model Finish-to-Start logic and frequently require dummy activities (represented as dashed arrows with zero duration and zero cost) to maintain unique node numbering and prevent ambiguous logical precedence.

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2. Critical Path Method (CPM): Forward Pass, Backward Pass, and Float

The Critical Path Method is a deterministic scheduling algorithm that calculates the earliest and latest dates each activity can proceed without delaying the overall project.

Standard AON Node Convention

On the NCEES FE exam, an activity node is structured with the following standard six-box layout:

┌──────────────┬───────────────────┬──────────────┐
│ Early Start  │     Duration      │ Early Finish │
│     (ES)     │        (t)        │     (EF)     │
├──────────────┼───────────────────┼──────────────┤
│  Late Start  │ Total Float / Slack│ Late Finish  │
│     (LS)     │       (TF)        │     (LF)     │
└──────────────┴───────────────────┴──────────────┘

The Forward Pass (Early Dates: $ES$ and $EF$)

The forward pass moves from project start to project completion to determine the earliest possible start and finish times for each activity:

1. Initialize project start: For initial activities with no predecessors, set $ES = 0$.
2. For any activity $i$ with duration $t_i$: $EF_i = ES_i + t_i$
3. For any subsequent activity $j$ that has multiple immediate predecessors: $ES_j = \max_{p \in \text{Pred}(j)} \{ EF_p \}$ Rule: An activity cannot begin until all of its immediate predecessor activities have finished.

The Backward Pass (Late Dates: $LS$ and $LF$)

The backward pass moves from project completion back to project start to determine the latest possible times activities can occur without delaying the target project completion date:

1. For the terminal activity (or project deadline $T_{\text{project}}$): $LF_{\text{terminal}} = EF_{\text{terminal}}$
2. For any activity $i$ with duration $t_i$: $LS_i = LF_i - t_i$
3. For any activity $i$ that has multiple immediate successors: $LF_i = \min_{s \in \text{Succ}(i)} \{ LS_s \}$ Rule: An activity must finish in time to accommodate the earliest late start among all its succeeding tasks.

Float (Slack) Formulations

Float represents the scheduling flexibility of an activity.

Total Float ($TF$)

Total Float (or Total Slack) is the maximum time an activity can be delayed from its early start date without delaying the overall project completion date:

$TF_i = LS_i - ES_i = LF_i - EF_i$

Free Float ($FF$)

Free Float (or Free Slack) is the maximum time an activity can be delayed without delaying the Early Start ($ES$) of any immediate successor activity:

$FF_i = \min_{s \in \text{Succ}(i)} \{ ES_s \} - EF_i$

Fundamental Float Axiom: Free Float is always less than or equal to Total Float ($FF_i \le TF_i$). If an activity has zero Total Float, its Free Float must also be zero ($TF_i = 0 \implies FF_i = 0$).

Critical Path Identification

The critical path is the sequence of dependent activities from network start to finish that possesses the longest cumulative duration. Key characteristics:

• It establishes the absolute minimum calendar duration required to complete the project.
• Activities on the critical path have zero Total Float ($TF = 0$), assuming the project target completion time equals the early finish of the terminal activity.
• Any delay to a critical path activity results in a one-for-one day delay to final project completion.
• A project network may contain multiple concurrent critical paths.

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3. Program Evaluation and Review Technique (PERT)

In complex, first-of-a-kind engineering projects (such as advanced aerospace development or novel automated systems), activity durations cannot be known deterministically. PERT models duration uncertainty using a three-point estimation framework governed by a beta distribution.

The Three-Point Duration Estimates

For each activity, three estimates are assessed:

1. Optimistic duration ($a$): The minimum time required if execution proceeds exceptionally well with zero anomalies (typically 1-in-100 probability).
2. Most likely duration ($m$): The modal duration observed under ordinary operating conditions.
3. Pessimistic duration ($b$): The maximum time required if severe adverse conditions, material delays, and technical roadblocks occur (typically 1-in-100 probability).

Activity Expected Value and Variance Formulas

Based on the standard PERT beta distribution approximation:

• Expected Activity Duration ($t_e$): $t_e = \frac{a + 4m + b}{6}$
• Activity Standard Deviation ($\sigma$): $\sigma = \frac{b - a}{6}$
• Activity Variance ($\sigma^2$): $\sigma^2 = \left(\frac{b - a}{6}\right)^2 = \frac{(b - a)^2}{36}$

Project-Level Parameters (Summation Along the Critical Path)

By invoking the Central Limit Theorem (CLT), if a critical path consists of numerous independent sequential activities, the distribution of the total project duration ($T$) approaches a Normal Distribution:

$T \sim \mathcal{N}(T_e, \sigma_P^2)$

1. Expected Project Completion Time ($T_e$): $T_e = \sum_{i \in CP} t_{e,i}$
2. Total Project Variance ($\sigma_P^2$): $\sigma_P^2 = \sum_{i \in CP} \sigma_i^2$
3. Total Project Standard Deviation ($\sigma_P$): $\sigma_P = \sqrt{\sigma_P^2} = \sqrt{\sum_{i \in CP} \sigma_i^2}$

CRITICAL NCEES EXAM RULE: Variances are additive; standard deviations are NEVER additive! $\sigma_P \ne \sum_{i \in CP} \sigma_i \quad \Longleftrightarrow \quad \sigma_P = \sqrt{\sum_{i \in CP} \sigma_i^2}$

Probability of Project Completion by Target Date ($T_d$)

To determine the probability that the project will finish on or before a specified due date $T_d$, compute the standard normal test statistic ($Z$) and reference the unit normal distribution table $\Phi(Z)$:

$Z = \frac{T_d - T_e}{\sigma_P} = \frac{T_d - T_e}{\sqrt{\sum_{i \in CP} \sigma_i^2}}$

$P(T \le T_d) = \Phi(Z)$

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4. Project Crashing (Time-Cost Optimization)

Project crashing is the systematic procedure of compressing total project duration for the lowest additional direct cost. Crashing involves deploying additional resources—such as assigning overtime, adding second shifts, hiring subcontractors, or renting larger machinery.

Crashing Parameters

For each project activity:

• Normal Time ($NT$) and Normal Cost ($NC$): Standard operational execution.
• Crash Time ($CT$) and Crash Cost ($CC$): Maximum possible acceleration limit and its corresponding total cost.
• Maximum Time Reduction ($\Delta t_{\max}$): $\Delta t_{\max} = NT - CT$.

Cost Slope (Crash Cost per Unit Time)

Assuming a linear relationship between time reduction and direct cost increase:

$\text{Cost Slope} = \frac{\Delta \text{Cost}}{\Delta \text{Time}} = \frac{CC - NC}{NT - CT}$

$\text{Crash Cost per Period} = \frac{\text{Crash Cost} - \text{Normal Cost}}{\text{Normal Duration} - \text{Crash Duration}}$

Systematic Least-Cost Crashing Algorithm

1. Identify the Critical Path(s): Calculate early and late dates for the entire network to identify all critical paths.
2. Identify Candidate Activities: Only activities located on the critical path can reduce project duration. (Crashing a non-critical activity incurs cost without shortening project completion).
3. Select Minimum Cost Slope: Among the critical activities that can still be crashed, select the activity with the lowest cost slope.
4. Determine Crashing Increment: Shorten the selected activity by the minimum of:
   • The activity's remaining crash capacity ($NT - CT$).
   • The amount of time reduction that causes a parallel non-critical path to become critical.
   • The remaining duration needed to meet the project objective.
5. Handle Multiple Critical Paths: If multiple parallel critical paths exist, you must shorten all critical paths simultaneously by the same amount. This is achieved either by crashing a common shared critical activity or by crashing a combination of activities across paths whose sum of cost slopes is minimal.
6. Iterate: Recalculate network float and repeat until the target project duration is reached or all critical activities are crashed to their limits.

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5. Step-by-Step Worked Engineering Calculations

Worked Example 10.2.1: Full CPM Forward/Backward Pass with Total and Free Float

Problem: An industrial engineering workcell integration project has the activities, durations, and precedence constraints shown in the table below. Determine the critical path, project completion time, and the Total Float ($TF$) and Free Float ($FF$) for all activities.

Solution:

1. Forward Pass ($ES, EF = ES + t$):

   • Activity A: No pred $\implies ES_A = 0, EF_A = 0 + 5 = 5$.
   • Activity B: No pred $\implies ES_B = 0, EF_B = 0 + 6 = 6$.
   • Activity C: Pred A $\implies ES_C = EF_A = 5, EF_C = 5 + 4 = 9$.
   • Activity D: Preds A, B $\implies ES_D = \max(EF_A, EF_B) = \max(5, 6) = 6, EF_D = 6 + 7 = 13$.
   • Activity E: Preds C, D $\implies ES_E = \max(EF_C, EF_D) = \max(9, 13) = 13, EF_E = 13 + 5 = 18$.
   • Activity F: Pred D $\implies ES_F = EF_D = 13, EF_F = 13 + 3 = 16$.
   • Project Completion Time: $\max(EF_E, EF_F) = \max(18, 16) = 18\text{ days}$.

2. Backward Pass ($LF, LS = LF - t$):

   • Terminal activities are E and F $\implies LF_E = 18, LF_F = 18$.
   • Activity E: $LS_E = 18 - 5 = 13$.
   • Activity F: $LS_F = 18 - 3 = 15$.
   • Activity D: Successors are E and F $\implies LF_D = \min(LS_E, LS_F) = \min(13, 15) = 13, LS_D = 13 - 7 = 6$.
   • Activity C: Successor is E $\implies LF_C = LS_E = 13, LS_C = 13 - 4 = 9$.
   • Activity B: Successor is D $\implies LF_B = LS_D = 6, LS_B = 6 - 6 = 0$.
   • Activity A: Successors are C and D $\implies LF_A = \min(LS_C, LS_D) = \min(9, 6) = 6, LS_A = 6 - 5 = 1$.

3. Float Calculations:

   • Activity A: $TF = LS - ES = 1 - 0 = 1$. $FF = \min(ES_C, ES_D) - EF_A = \min(5, 6) - 5 = 5 - 5 = 0$.
   • Activity B: $TF = 0 - 0 = 0$. $FF = ES_D - EF_B = 6 - 6 = 0$. (Critical)
   • Activity C: $TF = 9 - 5 = 4$. $FF = ES_E - EF_C = 13 - 9 = 4$.
   • Activity D: $TF = 6 - 6 = 0$. $FF = \min(ES_E, ES_F) - EF_D = \min(13, 13) - 13 = 0$. (Critical)
   • Activity E: $TF = 13 - 13 = 0$. $FF = 18 - 18 = 0$. (Critical)
   • Activity F: $TF = 15 - 13 = 2$. $FF = 18 - 16 = 2$.

4. Engineering Conclusion: The critical path is B $\to$ D $\to$ E with a total project duration of 18 days.

Worked Example 10.2.2: PERT Project Variance and Deadline Probability

Problem: A supply chain automation project has an identified critical path comprising three sequential activities: X, Y, and Z. The 3-point time estimates in weeks are:

• Activity X: $a = 4, m = 7, b = 16$
• Activity Y: $a = 6, m = 11, b = 22$
• Activity Z: $a = 5, m = 8, b = 17$

Calculate:

1. The expected duration ($t_e$) and variance ($\sigma^2$) for each activity.
2. The total expected project duration ($T_e$) and project standard deviation ($\sigma_P$).
3. The probability that the project finishes within 34 weeks ($T_d = 34$).

Solution:

1. Evaluate individual activity parameters:

   • Activity X: $t_{e,X} = \frac{4 + 4(7) + 16}{6} = \frac{4 + 28 + 16}{6} = \frac{48}{6} = 8.0\text{ weeks}$ $\sigma_X = \frac{16 - 4}{6} = 2.0 \implies \sigma_X^2 = 2.0^2 = 4.00\text{ weeks}^2$
   • Activity Y: $t_{e,Y} = \frac{6 + 4(11) + 22}{6} = \frac{6 + 44 + 22}{6} = \frac{72}{6} = 12.0\text{ weeks}$ $\sigma_Y = \frac{22 - 6}{6} = \frac{16}{6} = 2.667 \implies \sigma_Y^2 = \left(\frac{16}{6}\right)^2 = 7.111\text{ weeks}^2$
   • Activity Z: $t_{e,Z} = \frac{5 + 4(8) + 17}{6} = \frac{5 + 32 + 17}{6} = \frac{54}{6} = 9.0\text{ weeks}$ $\sigma_Z = \frac{17 - 5}{6} = 2.0 \implies \sigma_Z^2 = 2.0^2 = 4.00\text{ weeks}^2$

2. Compute project-level parameters: $T_e = \sum t_{e,i} = 8.0 + 12.0 + 9.0 = 29.0\text{ weeks}$ $\sigma_P^2 = \sum \sigma_i^2 = 4.00 + 7.111 + 4.00 = 15.111\text{ weeks}^2$ $\sigma_P = \sqrt{15.111} = 3.887\text{ weeks}$

3. Determine probability of completion by $T_d = 34$ weeks: $Z = \frac{T_d - T_e}{\sigma_P} = \frac{34 - 29.0}{3.887} = \frac{5.0}{3.887} = +1.286 \approx 1.29$ From the standard normal cumulative probability table: $P(T \le 34) = \Phi(1.29) = 0.9015 = 90.15\%$

4. Engineering Conclusion: The project has a 90.15% probability of completing within 34 weeks.

Worked Example 10.2.3: Least-Cost Project Crashing Schedule

Problem: A plant upgrade network has two parallel paths: Path 1 is A $\to$ C (Durations: $A = 5$ wks, $C = 6$ wks; Path length = 11 wks), and Path 2 is B $\to$ D (Durations: $B = 4$ wks, $D = 5$ wks; Path length = 9 wks). Path 1 is currently the unique critical path (11 weeks). Activity data is given below:

Executive leadership requires the project to be completed in 9 weeks (a 2-week reduction). What is the optimal sequence of activities to crash, and what is the minimum additional cost incurred?

Solution:

1. Calculate cost slopes for all activities:

   • $\text{Cost Slope}_A = \frac{16,000 - 10,000}{5 - 3} = \frac{6,000}{2} = $3,000 / \text{wk}$
   • $\text{Cost Slope}_C = \frac{34,000 - 24,000}{6 - 4} = \frac{10,000}{2} = $5,000 / \text{wk}$
   • $\text{Cost Slope}_B = \frac{12,000 - 8,000}{4 - 2} = \frac{4,000}{2} = $2,000 / \text{wk}$
   • $\text{Cost Slope}_D = \frac{21,000 - 15,000}{5 - 3} = \frac{6,000}{2} = $3,000 / \text{wk}$

2. Evaluate crashing from 11 weeks down to 9 weeks:

   • The critical path is Path 1 (A $\to$ C, duration = 11 wks). Non-critical Path 2 (B $\to$ D) has duration = 9 wks.
   • To reduce project duration, we must crash activities on the critical path. The candidates on Path 1 are Activity A ($3,000/wk) and Activity C ($5,000/wk).
   • Select Activity A because it has the lowest cost slope ($3,000/wk).
   • Activity A can be crashed by up to 2 weeks. Crashing Activity A by 2 weeks reduces Path 1 duration from 11 weeks to 9 weeks.
   • At 9 weeks, Path 1 (9 wks) and Path 2 (9 wks) both become critical simultaneously.
   • Total additional crashing cost: $2\text{ weeks} \times $3,000 / \text{week} = $6,000$.

3. Engineering Conclusion: Crash Activity A by 2 weeks for a total additional cost of $6,000. Crashing Activity B or D is unnecessary because Path 2 was not critical.