Study guide
9.4 Float Calculations & Schedule Compression
AACE Certified Cost Technician CCT
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1 Chapter 1: AACE Certification, Ethics & Total Cost Management Framework
1.1 About the AACE CCT Exam & Certification Pathway 1.2 AACE Canons of Ethics & Professional Practice 1.3 Total Cost Management (TCM) Framework Architecture 1.4 Cost Engineering Terminology & Core Concepts 1.5 Strategic Asset Management: Asset Performance, Assessment & Configuration Control
2 Chapter 2: Elements of Cost, Labor, Materials & Indirects
2.1 Cost Classification & Cost Behavior 2.2 Labor Costs, Craft Rates & Labor Burden 2.3 Materials & Equipment Cost Fundamentals 2.4 Job-Site Indirects & Home Office Overhead Allocation 2.5 Cost Dimensions: Life Cycles, Perspective, Influence & the Legal Dimension
3 Chapter 3: Engineering Economics: Time Value of Money & Depreciation
3.1 Time Value of Money & Cash Flow Diagrams 3.2 Uniform Series & Equivalence Factors 3.3 Nominal vs. Effective Interest Rates & Inflation 3.4 Depreciation Methods & Asset Write-Down
4 Chapter 4: Capital Budgeting, NPV, IRR & Life Cycle Costing
4.1 Net Present Value (NPV) & Present Worth Analysis 4.2 Internal Rate of Return (IRR) & MARR 4.3 Payback Period & Benefit-Cost Ratio Analysis 4.4 After-Tax Cash Flow Analysis & Life Cycle Costing
5 Chapter 5: Statistics, Probability & Decision Models
5.1 Descriptive Statistics & Measures of Dispersion 5.2 Probability Concepts & Distributions 5.3 Decision Trees & Expected Monetary Value (EMV) 5.4 Linear Programming & Optimization Models 5.5 Physical Measurement, Unit Systems & Quantity Conversions
6 Chapter 6: Organization, Communications & Project Management Basics
6.1 Organizational Structures & Project Roles 6.2 Communications Management & Project Reporting 6.3 Cost Data Management & Historical Normalization
7 Chapter 7: Cost Estimating Principles & Classification Systems
7.1 Estimating Fundamentals & Basis of Estimate 7.2 AACE 18R-97 Cost Estimate Classification System 7.3 Conceptual & Parametric Estimating Methods 7.4 Detailed Estimating & Estimate Closeout
8 Chapter 8: Allowances, Escalation, Location Factors & Contingency
8.1 Allowances vs. Contingency vs. Management Reserve 8.2 Location Factors & Escalation Analysis 8.3 Contingency Determination & Risk Analysis
9 Chapter 9: Project Planning, WBS & CPM Scheduling
9.1 Work Breakdown Structure (WBS) & Scope Definition 9.2 Activity Definition, Precedence & Dependencies 9.3 Critical Path Method (CPM) Forward & Backward Pass 9.4 Float Calculations & Schedule Compression 9.5 Requirements Elicitation, Plan Implementation & Plan Validation
10 Chapter 10: PERT Estimating, Progress Measurement & Schedule Control
10.1 PERT Three-Point Estimating & Network Variance 10.2 Progress Measurement & Rules of Credit 10.3 Schedule Baseline Maintenance & Change Control
11 Chapter 11: Earned Value Management (EVM): Variances, Indices & Forecasting
11.1 EVM Foundations & Core Parameters 11.2 Cost & Schedule Variances 11.3 Performance Indices 11.4 Forecasting: EAC, ETC, VAC & TCPI
12 Chapter 12: Procurement, Contract Types & Administration
12.1 Contract Types & Risk Allocation Spectrum 12.2 Procurement Process, Solicitation & Bidding 12.3 Surety Bonds & Project Insurance 12.4 Contract Administration, Changes & Claims
13 Chapter 13: Project Risk Management: Identification, Analysis & Response
13.1 Risk Management Planning & Identification 13.2 Qualitative & Quantitative Risk Analysis 13.3 Risk Response Strategies & Risk Control
14 Chapter 14: Quality Management, Value Engineering & Resource Management
14.1 Quality Concepts & Cost of Quality (COQ) 14.2 Quality Control Tools & Continuous Improvement 14.3 Value Engineering & Value Methodology 14.4 Resource Management, Leveling & Learning Curve Theory 14.5 Environmental, Health & Safety (EHS) Cost Management
9.4 Float Calculations & Schedule Compression
Key Takeaways Explain This Topic AI
Total Float (TF = LS - ES = LF - EF) measures the total duration an activity can be delayed without delaying the project completion date, and is shared across all activities on that specific network path.
Free Float (FF = min(ES of successors) - EF) measures the duration an activity can be delayed without delaying the early start of any immediate successor, strictly adhering to the mathematical rule FF <= TF.
Interfering Float equals Total Float minus Free Float, representing the portion of path slack that delays successor early start dates when consumed, without extending total project duration.
Negative float occurs whenever an imposed target finish date or contractual milestone constraint is set earlier than the mathematically calculated early finish date derived from the forward pass.
Schedule compression techniques include Crashing (trading direct cost for reduced duration on critical activities using cost slope = delta Cost / delta Time) and Fast-Tracking (overlapping sequential critical activities, incurring increased execution risk and rework without direct labor cost premiums).
Last updated: September 2026
9.4 Float Calculations & Schedule Compression
Quick Summary: Float (or Slack ) quantifies schedule flexibility. Total Float (TF) represents the time an activity can slip without delaying project completion ( T F = L S − E S = L F − E F TF = LS - ES = LF - EF TF = L S − ES = L F − EF ), and is a shared asset belonging to the entire path. Free Float (FF) is the time an activity can slip without delaying the early start of any immediate successor ( F F = min ( E S succ ) − E F FF = \min(ES_{\text{succ}}) - EF FF = min ( E S succ ) − EF ), strictly satisfying F F ≤ T F FF \le TF FF ≤ TF . When projects encounter delays or imposed deadlines, schedulers employ two primary Schedule Compression techniques: Crashing (shortening critical path activities by expending direct funds, governed by the Cost Slope formula) and Fast-Tracking (executing sequential critical tasks in parallel, trading execution risk and potential rework for time without adding direct labor premiums).
1. Float (Slack) Theory & Significance in Cost Engineering
In Total Cost Management, time and money are intrinsically linked. Float is not merely an abstract scheduling parameter; it represents a tangible project buffer that protects against delays, enables resource leveling, and dictates commercial claims liability. Understanding float mechanics allows cost engineers to determine which activities can absorb jobsite delays without financial damage and which require immediate capital intervention.
AACE International Recommended Practice 10S-90 recognizes several distinct classifications of float, each measuring schedule flexibility relative to different project milestones. +-----------------------------------------------------------------------------------+
- | THE TAXONOMY OF FLOAT |
- +-------------------+------------------------------------+--------------------------+
- | FLOAT TYPE | MATHEMATICAL DEFINITION | IMPACT WHEN CONSUMED |
- +-------------------+------------------------------------+--------------------------+
- | Total Float (TF) | TF = LS - ES = LF - EF | Delays final project |
- | | | completion if exceeded |
- +-------------------+------------------------------------+--------------------------+
- | Free Float (FF) | FF = min(ES_succ) - EF | Delays successor early |
- | | | starts; zero impact on TF|
- +-------------------+------------------------------------+--------------------------+
- | Interfering Float | IntF = TF - FF | Delays successor starts |
- | | | without delaying project |
- +-------------------+------------------------------------+--------------------------+
- | Independent Float | IndF = max(0, min(ES_succ) - | Zero impact on any other |
- | | max(LF_pred) - Duration) | activity in the network |
- +-------------------+------------------------------------+--------------------------+
- 2. Total Float (TF): Mathematical Formulation & Path Ownership
Total Float (TF) is the total amount of time an activity can be delayed from its Early Start date without delaying the planned project completion date (or violating a mandatory late finish constraint).
The Formulas TF = L S − E S = L F − E F \text{TF} = LS - ES = LF - EF TF = L S − ES = L F − EF
Both formulas ( L S − E S LS - ES L S − ES and L F − E F LF - EF L F − EF ) are mathematically identical because E F = E S + D EF = ES + D EF = ES + D and L S = L F − D LS = LF - D L S = L F − D : L F − E F = ( L S + D ) − ( E S + D ) = L S − E S LF - EF = (LS + D) - (ES + D) = LS - ES L F − EF = ( L S + D ) − ( ES + D ) = L S − ES
The Shared Asset Principle
A foundational concept on the AACE CCT exam is that Total Float belongs to the path, not to an individual activity .
If three sequential non-critical activities on a path each display 5 days of Total Float, there are not 15 total days of float available. There are only 5 cumulative days shared among the three activities.
If the first activity slips by 5 days, it consumes all the Total Float on that path. The remaining two activities immediately drop to zero float and become critical!
Commercial Significance: In construction contract disputes, who "owns" the float (the owner or the contractor) is a frequent source of litigation. AACE Recommended Practice 29R-03 establishes that, unless the contract explicitly dictates otherwise, float is an expiring project resource available to whichever party reaches it first to absorb delays.
3. Free Float (FF): Mathematical Formulation & Successor Independence
Free Float (FF) is the amount of time an activity can be delayed without delaying the Early Start date of any immediate successor activity.
The Formula FF i = min j ∈ Succ ( i ) { E S j − Lag i j } − E F i \text{FF}_i = \min_{j \in \text{Succ}(i)} \{ ES_j - \text{Lag}_{ij} \} - EF_i FF i = j ∈ Succ ( i ) min { E S j − Lag ij } − E F i
For standard Finish-to-Start relationships with zero lag: FF i = min ( E S immediate successors ) − E F i \text{FF}_i = \min(ES_{\text{immediate successors}}) - EF_i FF i = min ( E S immediate successors ) − E F i
The Universal Law: FF <= TF
Because delaying an immediate successor's early start might still fall within that successor's late dates, Free Float can never exceed Total Float : 0 ≤ FF ≤ TF \text{0} \le \text{FF} \le \text{TF} 0 ≤ FF ≤ TF
If an activity has Total Float = 0, its Free Float is mathematically guaranteed to be Free Float = 0.
Free Float represents an activity's "private float" . Consuming Free Float does not impact any downstream activity's early schedule. Once Free Float is exhausted, any further delay consumes Interfering Float and pushes downstream early start dates. FREE FLOAT VS. TOTAL FLOAT DYNAMICS:
- Activity A (D = 4): ES = 0, EF = 4, LS = 3, LF = 7 --> TF = 3
- Successor B: ES = 6, EF = 10, LS = 7, LF = 11
Calculations for Activity A:
- Total Float (TF) = LF - EF = 7 - 4 = 3 Days
- Free Float (FF) = ES(B) - EF(A) = 6 - 4 = 2 Days
- Interfering Float = TF - FF = 3 - 2 = 1 Day
Operational Interpretation:
- Activity A can delay by 2 days without affecting Activity B's early start (Free Float).
- If Activity A delays by 3 days, Activity B's early start is pushed from Day 6 to Day 7,
but the project completion date is NOT delayed (Interfering Float).
- If Activity A delays by 4 days, the overall project is delayed by 1 day.
- 4. Advanced Float Concepts: Interfering & Independent Float
- Interfering Float
Definition: The portion of Total Float that, if consumed, delays the Early Start of subsequent activities without delaying the ultimate project completion date.
Formula: Interfering Float = Total Float − Free Float \text{Interfering Float} = \text{Total Float} - \text{Free Float} Interfering Float = Total Float − Free Float
Independent Float
Definition: The amount of time an activity can be delayed without delaying any successor activity, even if all predecessor activities finish at their latest possible dates (LF).
Formula: Independent Float i = max ( 0 , min j ∈ Succ ( i ) { E S j } − max k ∈ Pred ( i ) { L F k } − D i ) \text{Independent Float}_i = \max \left( 0, \min_{j \in \text{Succ}(i)} \{ ES_j \} - \max_{k \in \text{Pred}(i)} \{ LF_k \} - D_i \right) Independent Float i = max ( 0 , min j ∈ Succ ( i ) { E S j } − max k ∈ Pred ( i ) { L F k } − D i )
It is the most conservative measure of slack and is rarely present in tightly scheduled networks.
5. Float Values: Zero, Positive, and Negative Float +-----------------------------------------------------------------------------------+
- | FLOAT VALUE STATES & MEANINGS |
- +-------------------+------------------------------------+--------------------------+
- | FLOAT VALUE | NETWORK STATUS | MANAGEMENT IMPLICATION |
- +-------------------+------------------------------------+--------------------------+
- | Positive (TF > 0) | Sub-critical / Flexible | Buffer exists; resources |
- | | Path completes ahead of deadline | can be diverted |
- +-------------------+------------------------------------+--------------------------+
- | Zero (TF = 0) | Critical Path | On schedule; any delay |
- | | EF equals required completion date | delays project completion|
- +-------------------+------------------------------------+--------------------------+
- | Negative (TF < 0) | Hyper-critical / Behind Schedule | Project will miss target;|
- | | Imposed finish earlier than EF | compression mandatory |
- +-------------------+------------------------------------+--------------------------+
- The Mechanism of Negative Float
In an unconstrained network, the minimum project duration calculated during the forward pass is assigned as the late finish of the final activity ( L F = E F LF = EF L F = EF ), resulting in T F = 0 TF = 0 TF = 0 on the critical path. Negative float cannot exist in an unconstrained network.
Negative Float occurs when an external constraint is imposed:
Contractual Target Completion Constraint: The client imposes a contract completion date ( L F imposed LF_{\text{imposed}} L F imposed ) that is earlier than the forward pass calculated early finish ( E F calculated EF_{\text{calculated}} E F calculated ): TF = L F imposed − E F calculated < 0 \text{TF} = LF_{\text{imposed}} - EF_{\text{calculated}} < 0 TF = L F imposed − E F calculated < 0
In-Progress Delay to Critical Tasks: An ongoing project experiences a 10-day field delay on a critical path activity while the contract baseline finish date remains frozen.
Meaning: A Total Float of -10 days means the project is forecasted to finish 10 days late unless immediate corrective schedule compression is implemented.
6. Schedule Compression: Crashing Methodology & The Cost Slope
When a project must be shortened—due to negative float, contract acceleration requirements, or milestone liquidated damages—the project team must compress the schedule. The first formal compression technique is Crashing .
Defining Crashing
Crashing is a technique used to shorten the schedule duration for the least incremental cost by adding resources to activities. Crashing trades direct capital (overtime premiums, extra shifts, additional labor crews, larger equipment, expediting delivery fees) for reduced calendar time.
The Cost-Time Relationship
Each compressible activity has two operational points:
Normal Point: Normal Duration ( D N D_N D N ) executed at Normal Direct Cost ( C N C_N C N ).
Crash Point: Crash Duration ( D C D_C D C ), representing the absolute physical minimum duration, executed at Crash Direct Cost ( C C C_C C C ). Maximum Crash Time Available = D N − D C \text{Maximum Crash Time Available} = D_N - D_C Maximum Crash Time Available = D N − D C Incremental Crash Cost = C C − C N \text{Incremental Crash Cost} = C_C - C_N Incremental Crash Cost = C C − C N DIRECT COST-TIME TRADE-OFF CURVE:
- Direct
- Cost ($)
- |
- C_c +----------------------------* Crash Point (D_c, C_c)
- | /
- | / <-- Cost Slope = Delta Cost / Delta Time
- | /
- C_n +------------------------* Normal Point (D_n, C_n)
- | |
- +------------------------+------------------ Duration (Days)
- D_c D_n
- The Cost Slope Formula
The Cost Slope (or Crash Cost per Unit Time ) quantifies the direct cost required to compress an activity by one time unit (e.g., one day): Cost Slope = Crash Cost − Normal Cost Normal Duration − Crash Duration = Δ Cost Δ Time = C C − C N D N − D C \text{Cost Slope} = \frac{\text{Crash Cost} - \text{Normal Cost}}{\text{Normal Duration} - \text{Crash Duration}} = \frac{\Delta \text{Cost}}{\Delta \text{Time}} = \frac{C_C - C_N}{D_N - D_C} Cost Slope = Normal Duration − Crash Duration Crash Cost − Normal Cost = Δ Time Δ Cost = D N − D C C C − C N
Systematic Crashing Rules of Engagement (AACE Protocol)
To minimize project expenditures, cost engineers must adhere to four strict algorithmic rules:
Crash CRITICAL Activities ONLY: Crashing an activity with positive float spends money without shortening the project completion date. Only activities on the Critical Path ( T F = 0 TF = 0 TF = 0 or minimum) will compress total project duration.
Crash the Critical Activity with the LOWEST Cost Slope First: Always select the critical activity that yields the cheapest daily acceleration cost.
Crash Iteratively (One Day at a Time): As a critical path is shortened, non-critical paths maintain their original length. Eventually, a secondary path will become critical. Once multiple parallel critical paths exist, shortening the project requires crashing activities on ALL critical paths simultaneously!
Respect the Crash Limit ( D C D_C D C ): An activity cannot be crashed beyond its crash duration. Once D C D_C D C is reached, the activity is exhausted, and the scheduler must move to the next lowest cost slope activity.
7. Schedule Compression: Fast-Tracking & Risk Analysis
The second primary compression technique is Fast-Tracking .
Defining Fast-Tracking
Fast-Tracking is a schedule compression technique in which activities or phases normally performed in sequence are reconfigured to be performed in parallel .
Rather than waiting for architectural engineering to finish 100% before starting civil excavation, excavation begins when foundation drawings reach 60% completion (modeled via Start-to-Start links with lags). +-----------------------------------------------------------------------------------+
- | CRASHING VS. FAST-TRACKING |
- +-------------------+--------------------------------+------------------------------+
- | ATTRIBUTE | CRASHING | FAST-TRACKING |
- +-------------------+--------------------------------+------------------------------+
- | Core Mechanism | Adds resources to critical | Reconfigures serial tasks |
- | | activities to shorten duration | to run concurrently |
- +-------------------+--------------------------------+------------------------------+
- | Cost Impact | **Increases direct cost** | Zero direct cost increase |
- | | (overtime, extra craft labor) | at inception |
- +-------------------+--------------------------------+------------------------------+
- | Risk Impact | Low-to-moderate technical risk | **High risk of rework**, |
- | | (known tasks executed faster) | design churn, and collisions |
- +-------------------+--------------------------------+------------------------------+
- | Logic Changes | Durations shorten; network | Logic links modified |
- | | topology remains identical | (FS converted to SS/FF) |
- +-------------------+--------------------------------+------------------------------+
- | Quality & Safety | Fatigue from prolonged overtime| Coordination overhead, |
- | Considerations | can reduce productivity | work executed prematurely |
- +-------------------+--------------------------------+------------------------------+
- 8. Step-by-Step Worked Crashing Problem: Pipeline Replacement
Scenario: A municipal water utility must compress a pipeline replacement schedule. The baseline network consists of four sequential critical activities ( A → B → C → D A \to B \to C \to D A → B → C → D ). An external regulatory mandate requires the project to be completed in 20 days instead of the baseline 25 days (a 5-day compression requirement).
Given Project Data
Activity ID Description Normal Duration ( D N D_N D N ) Crash Duration ( D C D_C D C ) Normal Cost ( C N C_N C N ) Crash Cost ( C C C_C C C )
- A Sawcut Pavement & Clear Site 4 Days 3 Days $4,000 $5,500
- B Excavate Trench & Shore 8 Days 6 Days $16,000 $22,000
- C Lay 36-Inch Ductile Iron Pipe 9 Days 6 Days $27,000 $34,500
- D Backfill Trench & Restore Road 4 Days 2 Days $8,000 $12,000
- TOTALS 25 Days $55,000
Note: A parallel non-critical electrical conduit installation path (Activity E) runs concurrently with Activities B and C, possessing a fixed duration of 15 days and a Total Float of 25 − ( 4 + 15 + 4 ) = 2 25 - (4 + 15 + 4) = 2 25 − ( 4 + 15 + 4 ) = 2 days.
Step 1: Calculate the Cost Slope for Each Critical Activity Cost Slope A = $ 5 , 500 − $ 4 , 000 4 − 3 = $ 1 , 500 1 = $ 1 , 500 / day ( Max crash: 1 day ) \text{Cost Slope}_A = \frac{\$5,500 - \$4,000}{4 - 3} = \frac{\$1,500}{1} = \$1,500/\text{day} \quad (\text{Max crash: 1 day}) Cost Slope A = 4 − 3 $5 , 500 − $4 , 000 = 1 $1 , 500 = $1 , 500/ day ( Max crash: 1 day ) Cost Slope B = $ 22 , 000 − $ 16 , 000 8 − 6 = $ 6 , 000 2 = $ 3 , 000 / day ( Max crash: 2 days ) \text{Cost Slope}_B = \frac{\$22,000 - \$16,000}{8 - 6} = \frac{\$6,000}{2} = \$3,000/\text{day} \quad (\text{Max crash: 2 days}) Cost Slope B = 8 − 6 $22 , 000 − $16 , 000 = 2 $6 , 000 = $3 , 000/ day ( Max crash: 2 days ) Cost Slope C = $ 34 , 500 − $ 27 , 000 9 − 6 = $ 7 , 500 3 = $ 2 , 500 / day ( Max crash: 3 days ) \text{Cost Slope}_C = \frac{\$34,500 - \$27,000}{9 - 6} = \frac{\$7,500}{3} = \$2,500/\text{day} \quad (\text{Max crash: 3 days}) Cost Slope C = 9 − 6 $34 , 500 − $27 , 000 = 3 $7 , 500 = $2 , 500/ day ( Max crash: 3 days ) Cost Slope D = $ 12 , 000 − $ 8 , 000 4 − 2 = $ 4 , 000 2 = $ 2 , 000 / day ( Max crash: 2 days ) \text{Cost Slope}_D = \frac{\$12,000 - \$8,000}{4 - 2} = \frac{\$4,000}{2} = \$2,000/\text{day} \quad (\text{Max crash: 2 days}) Cost Slope D = 4 − 2 $12 , 000 − $8 , 000 = 2 $4 , 000 = $2 , 000/ day ( Max crash: 2 days )
Ranking by Cost Slope (Lowest to Highest):
- Activity A: $1,500/day (Limit: 1 day)
- Activity D: $2,000/day (Limit: 2 days)
- Activity C: $2,500/day (Limit: 3 days)
- Activity B: $3,000/day (Limit: 2 days)
- Step 2: Systematic Iterative Crashing Execution
Target: Compress from 25 days to 20 days (5 days total reduction).
Iteration 1: Crash Activity A
Lowest cost slope is Activity A ($1,500/day).
Crash Activity A by its maximum limit of 1 day .
New Project Duration: 25 − 1 = 24 days 25 - 1 = 24\text{ days} 25 − 1 = 24 days .
Incremental Cost: 1 × $ 1 , 500 = + $ 1 , 500 1 \times \text{\textdollar}1,500 = +\text{\textdollar}1,500 1 × $ 1 , 500 = + $ 1 , 500 .
Cumulative Direct Cost: $ 55 , 000 + $ 1 , 500 = $ 56 , 500 \text{\textdollar}55,000 + \text{\textdollar}1,500 = \text{\textdollar}56,500 $ 55 , 000 + $ 1 , 500 = $ 56 , 500 .
Parallel Path Check: Path E duration = 3 + 15 + 4 = 22 3 + 15 + 4 = 22 3 + 15 + 4 = 22 days. Main path = 24 24 24 days. Main path remains sole critical path (Path E float = 2 2 2 days).
Iteration 2: Crash Activity D
Next lowest cost slope is Activity D ($2,000/day; limit 2 days).
Crash Activity D by 1 day .
New Project Duration: 24 − 1 = 23 days 24 - 1 = 23\text{ days} 24 − 1 = 23 days .
Incremental Cost: 1 × $ 2 , 000 = + $ 2 , 000 1 \times \text{\textdollar}2,000 = +\text{\textdollar}2,000 1 × $ 2 , 000 = + $ 2 , 000 .
Cumulative Direct Cost: $ 56 , 500 + $ 2 , 000 = $ 58 , 500 \text{\textdollar}56,500 + \text{\textdollar}2,000 = \text{\textdollar}58,500 $ 56 , 500 + $ 2 , 000 = $ 58 , 500 .
Parallel Path Check: Path E duration = 3 + 15 + 3 = 21 3 + 15 + 3 = 21 3 + 15 + 3 = 21 days. Main path = 23 23 23 days. Main path remains critical (Path E float = 2 2 2 days).
Iteration 3: Crash Activity D Again
Crash Activity D by its remaining 1 day (Activity D is now fully crashed at D = 2 D = 2 D = 2 ).
New Project Duration: 23 − 1 = 22 days 23 - 1 = 22\text{ days} 23 − 1 = 22 days .
Incremental Cost: 1 × $ 2 , 000 = + $ 2 , 000 1 \times \text{\textdollar}2,000 = +\text{\textdollar}2,000 1 × $ 2 , 000 = + $ 2 , 000 .
Cumulative Direct Cost: $ 58 , 500 + $ 2 , 000 = $ 60 , 500 \text{\textdollar}58,500 + \text{\textdollar}2,000 = \text{\textdollar}60,500 $ 58 , 500 + $ 2 , 000 = $ 60 , 500 .
Parallel Path Check: Path E duration = 3 + 15 + 2 = 20 3 + 15 + 2 = 20 3 + 15 + 2 = 20 days. Main path = 22 22 22 days. Main path remains critical.
Iteration 4: Crash Activity C
Activity A and D are fully crashed. Next lowest slope is Activity C ($2,500/day; limit 3 days).
Crash Activity C by 1 day .
New Project Duration: 22 − 1 = 21 days 22 - 1 = 21\text{ days} 22 − 1 = 21 days .
Incremental Cost: 1 × $ 2 , 500 = + $ 2 , 500 1 \times \text{\textdollar}2,500 = +\text{\textdollar}2,500 1 × $ 2 , 500 = + $ 2 , 500 .
Cumulative Direct Cost: $ 60 , 500 + $ 2 , 500 = $ 63 , 000 \text{\textdollar}60,500 + \text{\textdollar}2,500 = \text{\textdollar}63,000 $ 60 , 500 + $ 2 , 500 = $ 63 , 000 .
Iteration 5: Crash Activity C Again (Final Target Day)
Crash Activity C by 1 additional day (Activity C now has 1 day crash capacity remaining).
New Project Duration: 21 − 1 = 20 days 21 - 1 = 20\text{ days} 21 − 1 = 20 days ( TARGET ACHIEVED! ).
Incremental Cost: 1 × $ 2 , 500 = + $ 2 , 500 1 \times \text{\textdollar}2,500 = +\text{\textdollar}2,500 1 × $ 2 , 500 = + $ 2 , 500 .
Final Cumulative Direct Cost: $ 63 , 000 + $ 2 , 500 = $ 65 , 500 \text{\textdollar}63,000 + \text{\textdollar}2,500 = \text{\textdollar}65,500 $ 63 , 000 + $ 2 , 500 = $ 65 , 500 .
Summary Table of Crashing Strategy
- Compression Step Activity Crashed Days Crashed Incremental Cost Resulting Duration Cumulative Project Cost
- Baseline None 0 $0 25 Days $55,000
- Step 1 Activity A 1 Day $1,500 24 Days $56,500
- Step 2 Activity D 1 Day $2,000 23 Days $58,500
- Step 3 Activity D 1 Day $2,000 22 Days $60,500
- Step 4 Activity C 1 Day $2,500 21 Days $63,000
- Step 5 Activity C 1 Day $2,500 20 Days $65,500
Economic Result: The project is compressed by 5 days for a minimum total direct cost expenditure of $10,500 ($65,500 total direct cost).
9. CCT Exam Watch: High-Yield Float Traps & Crashing Calculation Errors
Crashing Non-Critical Activities: A classic trick question asks candidate to identify which activity to crash from a table, providing a non-critical activity with an enticingly low cost slope ($500/day) and a critical activity with a higher cost slope ($1,500/day). Never crash the non-critical activity! Shortening an activity with float wastes money and does not compress the project duration.
The Cost Slope Formula Inversion: Ensure you place Cost in the numerator and Time in the denominator: Cost Slope = Δ Cost Δ Duration = Crash Cost − Normal Cost Normal Duration − Crash Duration \text{Cost Slope} = \frac{\Delta \text{Cost}}{\Delta \text{Duration}} = \frac{\text{Crash Cost} - \text{Normal Cost}}{\text{Normal Duration} - \text{Crash Duration}} Cost Slope = Δ Duration Δ Cost = Normal Duration − Crash Duration Crash Cost − Normal Cost Inverting the formula to Delta D / Delta C results in an immediate mathematical failure.
Free Float Exceeding Total Float: On multiple-choice questions, verify that Free Float <= Total Float. Any option showing Free Float greater than Total Float violates mathematical scheduling laws and can be eliminated instantly.
Emergence of Dual Critical Paths During Crashing: When crashing a project through multiple days, keep track of sub-critical paths. If Path 1 is compressed to equal Path 2, further compression requires crashing an activity on Path 1 AND an activity on Path 2. The effective cost slope becomes Cost Slope 1 + Cost Slope 2.