6.4 Construction Project Management, CPM/PERT, and Resource Scheduling

6.4 Construction Project Management, CPM/PERT, and Resource Scheduling

Project planning, scheduling, and control represent pivotal competencies assessed in the CELE Applied Mathematics and Construction cluster. Construction projects are complex, capital-intensive endeavors subject to strict contractual deadlines, financial liquidated damages, and resource constraints. Civil engineers utilize Critical Path Method (CPM) and Program Evaluation and Review Technique (PERT) network models to establish baseline project schedules, optimize equipment and labor allocations, manage risk, and execute schedule compression (crashing) at minimum additional cost.

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Project Planning & Work Breakdown Structure (WBS)

A successful project begins with structured decomposition. The Work Breakdown Structure (WBS) is a deliverable-oriented hierarchical decomposition of the total project scope into manageable, measurable work packages:

• Level 1: Total Project (e.g., Multi-Storey Reinforced Concrete Hospital Facility)
• Level 2: Major Subprojects / Deliverables (e.g., Substructure, Superstructure, MEPFS, Architectural Finishes)
• Level 3: Work Packages (e.g., Pile Cap Foundation, Ground Floor Suspended Slab, Column Pours)
• Level 4: Field Activities (e.g., Formwork Erection, Rebar Tying, Concrete Pouring, Curing)

Each terminal activity in the WBS must have an assigned duration, quantifiable resource demand (crew, equipment, materials), predecessor relationships, and a responsible engineering supervisor.

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Network Scheduling Logic: AOA vs AON

Two fundamental network diagramming conventions model activity dependencies:

1. Activity-on-Arrow (AOA) / Arrow Diagramming Method (ADM):
   • Arrows represent activities with defined durations.
   • Nodes (circles) represent discrete points in time (events or milestones).
   • Dummy Activities: Dashed arrows with zero duration and zero resource demand introduced strictly to maintain logical dependencies or prevent two activities from sharing identical starting and ending nodes.
2. Activity-on-Node (AON) / Precedence Diagramming Method (PDM):
   • Nodes (boxes) represent activities.
   • Arrows represent precedence relationships.
   • Precedence types: Finish-to-Start (FS) (standard), Start-to-Start (SS), Finish-to-Finish (FF), and Start-to-Finish (SF), optionally modified by positive leads or negative lags.

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The Critical Path Method (CPM): Forward & Backward Pass

CPM is a deterministic scheduling algorithm that computes project duration and activity float values through two consecutive mathematical sweeps:

1. The Forward Pass (Early Dates)

The forward pass moves chronologically from project initiation to completion, determining the earliest possible time each activity can start and finish:

• Early Start ($ES$): The earliest time an activity can commence, governed by the completion of all immediate predecessors: $ES_j = \max \{ EF_i \} \quad \text{for all predecessors } i$ (For initial project activities with no predecessors, $ES = 0$.)
• Early Finish ($EF$): The earliest time an activity can finish: $EF_j = ES_j + D_j$ Where $D_j$ is the activity duration.

2. The Backward Pass (Late Dates)

The backward pass moves counter-chronologically from project completion back to the start, establishing the latest allowable time each activity can finish and start without delaying the overall project completion target ($T_{\text{target}} = \max EF$):

• Late Finish ($LF$): The latest time an activity can finish without delaying any immediate successor: $LF_i = \min \{ LS_j \} \quad \text{for all successors } j$ (For terminal activities, $LF = EF_{\text{project}}$.)
• Late Start ($LS$): The latest time an activity can start without delaying project completion: $LS_i = LF_i - D_i$

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Mathematical Float / Slack Analysis

Float represents the scheduling flexibility of an activity. CELE board problems test four distinct mathematical classifications of float:

1. Total Float ($TF$)

Total Float is the total time an activity can be delayed from its Early Start without delaying the overall project completion deadline:

$TF = LS - ES = LF - EF$

2. Free Float ($FF$)

Free Float is the time an activity can be delayed without delaying the Early Start of any immediately succeeding activity:

$FF_i = \min_{j \in \text{succ}} \{ ES_j \} - EF_i$

3. Interfering Float ($IntF$)

Interfering Float is the difference between Total Float and Free Float. It represents the portion of Total Float whose consumption will delay the early start of subsequent activities without delaying overall project completion:

$IntF = TF - FF$

4. Independent Float ($IF$)

Independent Float is the slack available when all predecessors finish at their latest dates ($LF$) and all successors start at their earliest dates ($ES$):

$IF_i = \max \left( 0, \min_{j \in \text{succ}} \{ ES_j \} - \max_{k \in \text{pred}} \{ LF_k \} - D_i \right)$

Identification of the Critical Path

The Critical Path is the continuous sequence of connected activities through the network with zero Total Float ($TF = 0$). It is the longest path in terms of cumulative duration and dictates the absolute minimum time required to complete the project.

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Program Evaluation and Review Technique (PERT)

Unlike deterministic CPM, PERT accounts for uncertainty in activity durations (e.g., severe weather, subsurface anomalies, supply chain delays) by modeling each task via a Beta probability distribution defined by three subjective time estimates:

1. Optimistic Time ($a$): The minimum possible duration assuming everything proceeds exceptionally well (1 in 100 probability of finishing faster).
2. Most Likely Time ($m$): The modal duration representing the most frequent duration under normal working conditions.
3. Pessimistic Time ($b$): The maximum duration assuming adverse conditions encounter continuous difficulties (1 in 100 probability of finishing slower).

1. PERT Expected Mean Duration ($t_e$)

The expected mean duration is a weighted average that assigns four times greater statistical weight to the modal value $m$:

$t_e = \frac{a + 4m + b}{6}$

2. Activity Variance ($\sigma^2$) and Standard Deviation ($\sigma$)

Assuming the range $(b - a)$ spans approximately six standard deviations ($6\sigma$) of the unimodal Beta distribution:

$\sigma = \frac{b - a}{6}$ $\sigma^2 = \left( \frac{b - a}{6} \right)^2 = \frac{(b - a)^2}{36}$

3. Total Project Variance & Central Limit Theorem

By the Central Limit Theorem, the sum of independent random variables along the critical path converges to a Normal Distribution, regardless of the underlying activity distributions:

• Expected Project Completion Time ($T_e$): $T_e = \sum_{i \in \text{critical}} t_{e,i}$
• Total Project Variance ($\sigma_{\text{proj}}^2$): $\sigma_{\text{proj}}^2 = \sum_{i \in \text{critical}} \sigma_i^2$
• Project Standard Deviation ($\sigma_{\text{proj}}$): $\sigma_{\text{proj}} = \sqrt{\sum_{i \in \text{critical}} \sigma_i^2}$

4. Probability of Meeting a Target Completion Date ($T_d$)

The probability of completing the project on or before a specified contract deadline $T_d$ is evaluated using the standard normal distribution $Z$-score:

$Z = \frac{T_d - T_e}{\sigma_{\text{proj}}}$

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Project Crashing & Cost-Slope Optimization

Project Crashing is the method of shortening project duration by allocating additional labor, equipment, or premium overtime to critical activities at minimum incremental direct cost.

Cost Slope Formulation

Each activity possesses a normal operating state and a crashed operating state:

• Normal Duration ($N_D$) and Normal Cost ($N_C$)
• Crash Duration ($C_D$) and Crash Cost ($C_C$)

The cost slope represents the marginal cost incurred to accelerate an activity by one unit of time (e.g., PHP per day):

$\text{Cost Slope} = \frac{\text{Crash Cost} - \text{Normal Cost}}{\text{Normal Duration} - \text{Crash Duration}} = \frac{\Delta C}{\Delta D}$

Crashing Algorithm Protocol

1. Identify the critical path(s) using normal durations.
2. Crash only critical path activities. Shortening non-critical activities increases cost without accelerating the project schedule.
3. Among eligible critical activities, select the activity with the lowest cost slope.
4. Shorten that activity up to its maximum crash limit ($N_D - C_D$) or until a parallel path becomes critical.
5. When multiple paths become critical simultaneously, shorten activities in parallel across all critical paths or accelerate an activity shared by all critical paths.
6. Continue until the target duration is achieved or all critical activities reach their crash limits.

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Step-by-Step Worked Problem Examples

Worked Example 1: Deterministic CPM Forward/Backward Pass

Problem: A reinforced concrete bridge pier construction package involves six activities with the following dependencies and durations:

Perform complete CPM forward and backward passes. Identify all early/late dates, floats, the critical path, and total project duration.

Solution:

1. Forward Pass ($ES, EF$):
   • Activity A: $ES = 0, \; EF = 0 + 5 = 5$
   • Activity B: $ES = 5, \; EF = 5 + 8 = 13$
   • Activity C: $ES = 5, \; EF = 5 + 4 = 9$
   • Activity D: $ES = 13, \; EF = 13 + 6 = 19$
   • Activity E: $ES = 9, \; EF = 9 + 5 = 14$
   • Activity F: $ES = \max(EF_D, EF_E) = \max(19, 14) = 19, \; EF = 19 + 7 = 26$ Project Duration = 26 Days.
2. Backward Pass ($LF, LS$) with $LF_F = 26$:
   • Activity F: $LF = 26, \; LS = 26 - 7 = 19$
   • Activity D: $LF = LS_F = 19, \; LS = 19 - 6 = 13$
   • Activity E: $LF = LS_F = 19, \; LS = 19 - 5 = 14$
   • Activity B: $LF = LS_D = 13, \; LS = 13 - 8 = 5$
   • Activity C: $LF = LS_E = 14, \; LS = 14 - 4 = 10$
   • Activity A: $LF = \min(LS_B, LS_C) = \min(5, 10) = 5, \; LS = 5 - 5 = 0$
3. Float Computations:

• Critical Path: A — B — D — F (Total duration = 26 Days).
• Note that Activity C has $TF = 5\text{ days}$, but $FF = 0\text{ days}$ because delaying C immediately delays the early start of its successor E.

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Worked Example 2: PERT Probabilistic Duration & Project Crashing

Problem: A critical path comprises three sequential activities with the following duration and cost parameters:

1. Calculate expected project duration $T_e$, project variance $\sigma_{\text{proj}}^2$, and standard deviation $\sigma_{\text{proj}}$.
2. Determine the probability of completing the project within a contract deadline of $T_d = 30.0\text{ days}$.
3. Determine the minimum direct cost to crash the project by 2 days.

Solution:

1. PERT Expected Durations & Variances:
   • Activity 1: $t_{e,1} = \frac{5 + 4(8) + 17}{6} = \frac{54}{6} = 9.0\text{ days}$, $\sigma_1 = \frac{17 - 5}{6} = 2.0\text{ days}$, $\sigma_1^2 = 4.0\text{ days}^2$.
   • Activity 2: $t_{e,2} = \frac{8 + 4(11) + 20}{6} = \frac{72}{6} = 12.0\text{ days}$, $\sigma_2 = \frac{20 - 8}{6} = 2.0\text{ days}$, $\sigma_2^2 = 4.0\text{ days}^2$.
   • Activity 3: $t_{e,3} = \frac{4 + 4(7) + 10}{6} = \frac{42}{6} = 7.0\text{ days}$, $\sigma_3 = \frac{10 - 4}{6} = 1.0\text{ day}$, $\sigma_3^2 = 1.0\text{ days}^2$.
   • Expected Total Duration: $T_e = 9.0 + 12.0 + 7.0 = 28.0\text{ days}$.
   • Project Variance: $\sigma_{\text{proj}}^2 = 4.0 + 4.0 + 1.0 = 9.0\text{ days}^2$.
   • Project Standard Deviation: $\sigma_{\text{proj}} = \sqrt{9.0} = 3.0\text{ days}$.
2. Probability of Finishing within $T_d = 30.0\text{ days}$: $Z = \frac{T_d - T_e}{\sigma_{\text{proj}}} = \frac{30.0 - 28.0}{3.0} = \frac{2.0}{3.0} = +0.67$ From the standard normal cumulative table, $P(Z \le +0.67) \approx 0.7486 = 74.86\%$.
3. Project Crashing for 2-Day Reduction: Calculate cost slope for each critical activity:
   • Activity 1: $\text{Slope} = \frac{160,000 - 120,000}{9 - 6} = \frac{40,000}{3} = 13,333.33\text{ PHP/day}$ (can crash 3 days).
   • Activity 2: $\text{Slope} = \frac{240,000 - 180,000}{12 - 9} = \frac{60,000}{3} = 20,000.00\text{ PHP/day}$ (can crash 3 days).
   • Activity 3: $\text{Slope} = \frac{130,000 - 90,000}{7 - 5} = \frac{40,000}{2} = 20,000.00\text{ PHP/day}$ (can crash 2 days).
   • Activity 1 has the lowest cost slope ($13,333.33\text{ PHP/day}$). Crash Activity 1 by 2 days.
   • Incremental Crash Cost: $2 \times 13,333.33 = 26,666.67\text{ PHP}$.
   • Total Direct Project Cost: $(120,000 + 180,000 + 90,000) + 26,666.67 = 416,666.67\text{ PHP}$.

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