📊 CPM, PERT & Construction Management 🔥 Day 1
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CPM & PERT — Construction Management

Ready to test your CPM & PERT knowledge? Practice chapter-wise mock tests with worked step-by-step solutions for GATE, ESE and SSC JE.
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CPM & PERT — Complete Study Notes
8 chapters · GATE, ESE (IES), SSC JE, State PSC
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CPM (Critical Path Method) and PERT (Program Evaluation and Review Technique) are the two foundational project-scheduling techniques used to plan, monitor, and control construction and engineering projects — from building the network diagram itself, through forward/backward pass and float calculations, time-cost trade-off (crashing), PERT's probabilistic treatment of uncertain durations, to Gantt charts and resource scheduling. Every formula, worked step-by-step example, diagram, and exam-pattern table is included.

Learning Objectives

After studying this chapter you will be able to:

  • Construct AOA and AON project networks, correctly identifying where dummy activities are required
  • Perform the forward pass (earliest times) and backward pass (latest times) through a network
  • Compute Total Float, Free Float, and Independent Float, and identify the critical path
  • Apply CPM time-cost trade-off (crashing) using cost slope to find the optimal project duration
  • Apply PERT's three-time-estimate model to compute expected duration, variance, and completion probability
  • Interpret Gantt charts and recognise their limitations relative to network techniques
  • Distinguish resource levelling from resource-constrained allocation and apply basic heuristic scheduling rules

Exam Relevance

ESE★★★★☆
GATE★★★★★
SSC JE★★★☆☆
State PSC★★★☆☆
💡 CPM-PERT numericals are a near-guaranteed GATE question every year — forward/backward pass, float calculations, PERT expected time and variance, and Z-score probability problems are the highest-yield topics for the effort invested.

Concept Map

Project Networks → Forward & Backward Pass
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Float & Critical Path → Time-Cost Trade-off (Crashing)
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PERT Probabilistic Analysis
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Gantt Charts → Resource Scheduling

Where This Fits

Prerequisite: Highway Engineering (large-scale road and pavement construction projects rely on the network scheduling techniques developed here to plan and monitor progress). CPM-PERT completes the civil engineering study material series on Mock Test Mitra — from here, head back to Study Material to revisit any subject or start practising topic-wise mock tests.

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1. Project Networks
Terminology · AOA vs AON · Network Construction Rules
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1.1 Network Terminology

TermDefinition
ActivityA task or work package that consumes time and/or resources
Event (node)A point in time representing the start or end of one or more activities; consumes no time
PredecessorAn activity that must be completed before the current activity can start
SuccessorAn activity that cannot start until the current activity is finished
Dummy activityA fictitious activity with zero duration used in AOA networks to show logical dependency without actual work
NetworkA directed graph of all activities and their precedence relationships

1.2 AOA vs AON Networks

FeatureAOA (Activity on Arrow)AON (Activity on Node / Precedence Diagram)
RepresentationActivities on arrows; nodes are eventsActivities on nodes; arrows show dependencies
Dummy activitiesRequired to avoid ambiguityNot needed
Start and endSingle start node, single end nodeSingle start node, single end node
LagsNot easily handledSupports SS, FF, SF, FS with lags
Software useOlder; PERT originated hereModern PM software (MS Project, Primavera)

AON networks have largely superseded AOA in modern practice specifically because they eliminate the need for dummy activities altogether and natively support the four dependency types (Finish-to-Start, Start-to-Start, Finish-to-Finish, Start-to-Finish) with lags — but AOA remains the standard teaching format for GATE/ESE since the classical PERT technique was originally developed on arrow diagrams.

1.3 Rules for AOA Network Construction

A valid AOA network follows strict conventions: each activity gets exactly one arrow; whenever two activities would otherwise share the same start and end node (making them indistinguishable), a dummy activity (drawn as a dashed arrow with zero duration and zero resource consumption) is inserted to break the ambiguity; the whole network must reduce to a single start event and single end event; no circular dependencies (loops) are permitted; and every event is uniquely numbered.

💡 Most GATE questions use AOA (arrow diagram). Recognise when a dummy activity is required — two activities with the same start and end nodes always need a dummy. Dummy = dashed arrow, zero duration.

AOA Network Rules

One arrow per activity; dummy (dashed, zero duration) needed when two activities share both start and end nodes

Network Validity Checks

Single start event, single end event; no loops; unique event numbering

Example 1 — Recognising the Need for a Dummy Activity

Given: Activities A and B both start at event 1 and both must finish before activity C can start (i.e., C depends on both A and B), and A and B have no other dependents. If drawn naively, A and B would both go from node 1 to node 2. What is needed?

Solution: Two activities cannot share the same start and end node pair (this makes them indistinguishable when referenced by their node numbers, e.g., both would be called "activity 1-2").

Answer: A dummy activity must be inserted — e.g., draw A as 1→2, B as 1→3, then a dummy 3→2 (zero duration) so C can start from node 2 depending on both A (directly) and B (via the dummy).

Example 2 — Identifying Predecessors from a Dependency Table

Given: Activity list: A (no predecessor), B (no predecessor), C (depends on A), D (depends on A and B). Describe the network logic.

Solution: A and B both start the project independently (from the start event). C depends only on A finishing. D depends on both A and B finishing — requiring either a merge node after both A and B, or a dummy if A and B are drawn to different end nodes.

Answer: The network has a burst from the start event into A and B, with C branching from A's end node, and D requiring a merge (possibly via dummy) of both A's and B's end nodes.

Example 3 — AOA vs AON Choice

Given: A project needs to model a Start-to-Start (SS) relationship with a 2-day lag between two activities. Which network type handles this natively?

Solution: AOA (Activity on Arrow) networks only support Finish-to-Start relationships directly through the node-and-arrow structure; SS/FF/SF relationships with lags require workarounds or aren't supported.

Answer: AON (Activity on Node / Precedence Diagram Method) natively supports SS, FF, SF, and FS relationships with lags — the correct choice here.

AOA network showing a dummy activity resolving two activities that would otherwise share start and end nodes 1 3 2 4 A (1→3) B (1→2) Dummy (3→2) C (2→4)

Fig. 1.1 — AOA network with a dummy activity: C depends on both A and B; since A ends at node 3 and B ends at node 2, a dummy (dashed, zero duration) connects node 3 to node 2 so C can correctly depend on both.

Q1. A dummy activity in an AOA network has:
A dummy activity is purely logical — it shows a dependency relationship without representing actual work, so it has zero duration and consumes no resources.
Q2. A dummy activity is required in an AOA network when:
Without a dummy, two activities sharing identical start/end nodes would be indistinguishable when referenced by node numbers — the dummy breaks this ambiguity.
Q3. Compared to AOA, an AON (precedence diagram) network:
AON places activities on nodes and dependencies on the connecting arrows, eliminating the need for dummy activities and natively supporting all four relationship types with lags — this is why modern PM software (MS Project, Primavera) uses AON.
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2. Forward & Backward Pass
Event Times · Activity Times · Earliest & Latest Schedules
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2.1 Event Times

SymbolNameDefinition
\(E_i\) / \(TE_i\) / ESTEarliest Event TimeEarliest time an event can occur (finish of all preceding activities)
\(L_i\) / \(TL_i\) / LSTLatest Event TimeLatest time an event can occur without delaying the project

2.2 Activity Times

SymbolNameFormula
ESTEarliest Start Time\(EST(i\text{–}j)=E_i\)
EFTEarliest Finish Time\(EFT(i\text{–}j)=EST+D=E_i+D\)
LSTLatest Start Time\(LST(i\text{–}j)=LFT-D=L_j-D\)
LFTLatest Finish Time\(LFT(i\text{–}j)=L_j\)

2.3 Forward Pass (Earliest Times)

The forward pass sweeps through the network from the start event (time zero) toward the end, computing the earliest time each event can occur. Whenever a node has multiple incoming activities (a merge point), its earliest event time must be the maximum of all the incoming paths' earliest finish times — because an event genuinely cannot occur until every activity leading into it has completed, so the slowest incoming path governs. The earliest time of the final node directly gives the overall project duration.

2.4 Backward Pass (Latest Times)

The backward pass sweeps in the opposite direction, from the end event (set equal to the project deadline) back toward the start, computing the latest time each event can occur without delaying the project. Where a node has multiple outgoing activities (a burst point), its latest event time must be the minimum of all the outgoing paths' latest start times — since the event must be ready early enough to satisfy the most demanding (earliest-deadline) successor. A completed backward pass should always recover \(L_1=E_1=0\) at the start node — this is the standard consistency check that the calculation was done correctly.

💡 Forward pass uses MAX; backward pass uses MIN — this distinction causes many errors. Always verify \(L_1=0\) as a check. Tabulate all activity EST, EFT, LST, LFT in a table for clarity.

Forward Pass

\( E_1=0 \)

\( E_j = \max_{(i\to j)}[E_i+D(i\text{-}j)] \)

Project duration \(=E_n\)

Backward Pass

\( L_n = E_n \)

\( L_i = \min_{(i\to j)}[L_j-D(i\text{-}j)] \)

Check: \( L_1=E_1=0 \)

Activity Times

\( EST=E_i \); \( EFT=E_i+D \)

\( LFT=L_j \); \( LST=L_j-D \)

Example 1 — Forward Pass at a Merge Node

Given: Node 4 is reached by activity A (from node 2, \(E_2=5\), duration 4) and activity B (from node 3, \(E_3=7\), duration 3). Find \(E_4\).

Solution:

Via A: \(5+4=9\); Via B: \(7+3=10\)

\( E_4 = \max(9,10) = 10 \)

Answer: \(E_4=10\) — governed by the slower path (via B).

Example 2 — Backward Pass at a Burst Node

Given: Node 2 feeds activity C (to node 4, \(L_4=15\), duration 5) and activity D (to node 5, \(L_5=12\), duration 4). Find \(L_2\).

Solution:

Via C: \(15-5=10\); Via D: \(12-4=8\)

\( L_2 = \min(10,8) = 8 \)

Answer: \(L_2=8\) — governed by the more demanding successor (via D).

Example 3 — Complete Small Network

Given: Activities: A(1-2, 3 days), B(1-3, 5 days), C(2-4, 4 days), D(3-4, 2 days). Find project duration and \(L_1\) (as a check).

Solution:

Forward: \(E_1=0\); \(E_2=0+3=3\); \(E_3=0+5=5\); \(E_4=\max(3+4,\ 5+2)=\max(7,7)=7\)

Backward: \(L_4=7\); \(L_3=7-2=5\); \(L_2=7-4=3\); \(L_1=\min(3-3,\ 5-5)=\min(0,0)=0\)

Answer: Project duration \(=7\) days; \(L_1=0=E_1\) ✓ (check passes) — note both paths through node 4 are equally critical (two critical paths).

Forward and backward pass through a small network showing earliest and latest event times 1 E=0,L=0 2 E=3,L=3 3 E=5,L=5 4 E=7,L=7 A (D=3) B (D=5) C (D=4) D (D=2) Forward pass (MAX at merges) computes E; backward pass (MIN at bursts) computes L

Fig. 2.1 — Forward and backward pass: E (earliest) and L (latest) event times computed for a small network with two paths of equal length (7 days each), giving two critical paths.

Q1. At a merge node (multiple activities arriving), the forward pass takes the:
An event cannot occur until ALL activities leading into it are complete, so the earliest time is governed by the slowest (maximum) incoming path.
Q2. At a burst node (multiple activities leaving), the backward pass takes the:
The event's latest time must satisfy the most demanding successor (the one with the earliest required start), so the backward pass takes the minimum across all outgoing paths.
Q3. After completing a backward pass, the standard consistency check is that:
If the backward pass was computed correctly, tracing all the way back to the start node must recover L₁=0, matching E₁=0 — any mismatch signals an arithmetic error somewhere in the pass.
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3. Float & Critical Path
Total/Free/Independent Float · Float Relationships · Identifying the Critical Path
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3.1 Types of Float

Float TypeFormulaMeaning
Total Float (TF)\(TF=L_j-E_i-D(i\text{–}j)=LST-EST=LFT-EFT\)Maximum delay without delaying project completion
Free Float (FF)\(FF=E_j-E_i-D(i\text{–}j)=E_j-EFT\)Delay without delaying successor's earliest start
Independent Float (IF)\(IF=E_j-L_i-D(i\text{–}j)\ge0\) (negative \(=\) zero)Float available regardless of when predecessors finish or successors start
Interfering FloatInterfering Float \(=TF-FF\)Portion of TF that, if used, delays a successor

3.2 Relationships Between Float Types

The three float measures always satisfy \(IF\le FF\le TF\), reflecting progressively looser assumptions about the surrounding activities: Independent Float assumes the worst case for both neighbours (predecessors finish as late as possible, successors must start as early as possible), Free Float assumes predecessors finish at their earliest, and Total Float makes no allowance for neighbours at all — it's the delay possible if the activity had the entire path's slack to itself. Free Float can always be consumed without affecting any other activity's schedule, whereas Total Float is shared along a path — using it on one activity reduces the float remaining for others sharing that same slack.

3.3 Critical Path

The critical path is the longest path through the network from start to finish, and every activity on it has zero total float (and consequently zero free float and zero independent float too) — meaning any delay on a critical activity delays the entire project. The critical path can be identified either by marking every activity with \(TF=0\) and tracing the connected chain from start to end, or by exhaustively listing every path through the network and comparing total durations. A network is not guaranteed to have only one critical path — multiple critical paths can coexist, all sharing the same (longest) duration, which is a critical consideration when crashing (since shortening only one critical path leaves the others still governing project duration).

💡 \(TF=L_j-E_i-D\) is the most frequently tested calculation. Critical path has \(TF=0\). Free Float \(=E_j-E_i-D\) — used when you want to know if an activity can be delayed without impacting any successor. \(IF\le FF\le TF\) always.

Total Float

\( TF = L_j - E_i - D = LST-EST \)

Free Float

\( FF = E_j - E_i - D \)

Independent Float

\( IF = E_j - L_i - D \) (\(\ge0\), else 0)

Float Relationship

\( IF \le FF \le TF \)

Critical activity: \( TF=0 \)

Example 1 — Total Float Calculation

Given: Activity (2-4) has \(E_2=3\), \(L_4=12\), duration \(D=6\). Find TF.

Solution:

\( TF = 12-3-6 = 3\ \text{days} \)

Answer: \(TF=3\) days — this activity can be delayed up to 3 days without delaying the project.

Example 2 — Free Float and Independent Float

Given: Activity (3-5) has \(E_3=5\), \(L_3=7\), \(E_5=14\), duration \(D=6\). Find FF and IF.

Solution:

\( FF = E_5-E_3-D = 14-5-6 = 3\ \text{days} \)

\( IF = E_5-L_3-D = 14-7-6 = 1\ \text{day} \)

Answer: \(FF=3\) days, \(IF=1\) day — confirming \(IF\le FF\).

Example 3 — Identifying the Critical Path

Given: A network has three paths from start to finish with durations 18, 22, and 22 days. Identify the critical path(s) and project duration.

Solution:

The critical path is the longest path — here two paths tie at 22 days, both longer than the 18-day path.

Answer: Project duration \(=22\) days, with TWO critical paths (both 22-day paths); the 18-day path has \(TF=22-18=4\) days of float on its non-shared activities.

Timeline diagram showing total float, free float, and independent float on a single activity Time → Activity duration D EST EFT Free Float Total Float Indep. Float LFT

Fig. 3.1 — Float types on an activity timeline: Free Float is the smallest (delay without affecting the immediate successor), Total Float extends to the latest finish time, and Independent Float is bounded by the most conservative assumptions about neighbouring activities.

Q1. An activity with Total Float = 0 is:
TF=0 means the activity has zero slack — any delay pushes back the entire project, which is exactly the definition of being on the critical path.
Q2. The relationship between the three float types is always:
Independent Float makes the most conservative assumptions (smallest value), Free Float is intermediate, and Total Float (no assumption about neighbours) is the largest — so IF ≤ FF ≤ TF always.
Q3. Can a network have more than one critical path?
Whenever two or more paths through the network tie for the maximum (longest) duration, all of them are critical paths — an important consideration for crashing, since shortening just one leaves the others still governing the project duration.
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4. CPM Time-Cost Trade-off (Crashing)
Normal vs Crash · Cost Slope · Crashing Procedure · Optimal Duration
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4.1 Normal vs Crash

ParameterNormalCrash
DurationNormal time \(t_n\) (maximum economical)Crash time \(t_c\) (minimum possible)
CostNormal cost \(C_n\) (minimum for normal time)Crash cost \(C_c\) (maximum for crash time)
Relationship\(t_n>t_c;\ C_nCrashing reduces duration but increases direct cost

4.2 Cost Slope

The cost slope expresses how much direct cost rises for every unit of time saved by crashing a particular activity — the additional resources (overtime, extra crews, expedited materials) needed to compress that activity's duration. Since the goal is to reduce project duration as cheaply as possible, the activity with the lowest cost slope among the critical activities should always be crashed first.

4.3 Crashing Procedure

Crashing proceeds iteratively: identify all currently critical activities (\(TF=0\)), select the one with the lowest cost slope, crash it by one time unit, then recheck the critical path — because crashing one activity can reduce its path's duration enough that a previously non-critical path becomes newly critical (or ties for critical) alongside it. When multiple critical paths exist simultaneously, activities on every critical path must be crashed together in that step (potentially by crashing a single shared activity, if one exists on all paths, or several activities in parallel otherwise) — crashing only one path leaves project duration unchanged since the other critical path(s) still govern.

4.4 Indirect Costs and the Optimal Duration

Direct costs (labour, materials, equipment for the activities themselves) rise as duration is compressed, while indirect costs (site overheads, supervision, equipment rental) fall as duration shortens, since less time on site means less overhead accumulated. The total project cost is the sum of both, and it traces out a U-shaped curve as duration varies — the optimal duration is the point of minimum total cost, found where the marginal cost of crashing one more day just equals the marginal indirect-cost saving from finishing one day sooner. Beyond this point, further crashing costs more than it saves.

💡 Cost slope \(=\dfrac{C_c-C_n}{t_n-t_c}\) — always crash the critical activity with minimum cost slope first. When multiple critical paths develop, the crashing cost equals the sum of cost slopes on all critical paths for that step.

Cost Slope

\( \text{Cost slope} = \dfrac{C_c-C_n}{t_n-t_c} \)

Units: cost per unit time (₹/day)

Total Project Cost

Total cost \(=\) Direct cost \(+\) Indirect cost

Optimal duration: minimum total project cost

Example 1 — Cost Slope Calculation

Given: An activity has \(t_n=10\) days (\(C_n=₹40{,}000\)) and \(t_c=6\) days (\(C_c=₹60{,}000\)). Find the cost slope.

Solution:

\( \text{Cost slope} = \dfrac{60{,}000-40{,}000}{10-6} = \dfrac{20{,}000}{4} = ₹\dfrac{5000}{\text{day}} \)

Answer: \(₹5000\) per day of crashing.

Example 2 — Choosing Which Activity to Crash First

Given: Two critical activities have cost slopes ₹3000/day and ₹4500/day respectively, both with available crash days remaining. Which is crashed first?

Solution: The activity with the lower cost slope achieves the same one-day reduction in project duration at lower additional cost.

Answer: Crash the ₹3000/day activity first — it is cheaper to compress.

Example 3 — Checking Whether Further Crashing Is Worthwhile

Given: Crashing the project by one more day would cost ₹8000 (sum of cost slopes on all currently critical paths), while the indirect cost saving for finishing one day earlier is ₹6000. Should crashing continue?

Solution: The marginal crashing cost (₹8000) exceeds the marginal indirect-cost saving (₹6000), so this additional day of crashing increases total project cost by ₹2000 net.

Answer: No — crashing should stop here; the current duration (before this step) is at or near the optimal (minimum total cost) point.

Time-cost trade-off curves showing direct cost, indirect cost, total cost, and the optimal duration Project duration → Cost Direct cost (rises as duration shortens) Indirect cost (falls) Total cost Optimal duration (min total cost)

Fig. 4.1 — Time-cost trade-off: direct cost rises and indirect cost falls as project duration shortens; the total cost curve reaches its minimum at the optimal duration, where further crashing costs more than it saves.

Q1. When crashing, which critical activity should be crashed first?
Crashing the lowest-cost-slope critical activity first achieves the required duration reduction at the smallest possible increase in direct cost.
Q2. As project duration decreases due to crashing, indirect costs (site overhead, supervision) typically:
Indirect costs accumulate per unit time the project is active (overheads, supervision), so a shorter project duration means less indirect cost overall — the opposite trend to direct cost.
Q3. When multiple critical paths exist simultaneously, crashing to reduce project duration by one day requires:
If only one of several tied critical paths is shortened, the others (still at the original duration) continue to govern the project's overall duration — so every critical path must be addressed together in that crashing step.
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5. PERT Probabilistic Analysis
PERT vs CPM · Three Time Estimates · Expected Duration & Variance · Completion Probability
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5.1 PERT vs CPM

FeatureCPMPERT
Activity durationsDeterministic (single estimate)Probabilistic (three time estimates)
ApplicationConstruction, manufacturing (well-defined tasks)R&D, new product development (uncertain tasks)
FocusTime–cost trade-offProbability of completing by a deadline
OriginDuPont Corporation, 1956US Navy Polaris missile project, 1958

CPM assumes each activity's duration is known with confidence, since construction tasks are typically well-understood and repeatable; PERT was developed for exactly the opposite situation — genuinely novel R&D work (originally, missile development) where a single duration estimate would be unreliable, so PERT instead models each activity's duration as a random variable.

5.2 Three Time Estimates

EstimateSymbolDefinition
Optimistic time\(t_o\) (or \(a\))Minimum time if everything goes perfectly; probability \(\approx1\%\)
Most likely time\(t_m\) (or \(m\))Time that would occur most often if activity repeated many times; mode of distribution
Pessimistic time\(t_p\) (or \(b\))Maximum time if everything goes wrong; probability \(\approx1\%\)

5.3 Expected Time and Variance

PERT assumes each activity's duration follows a Beta distribution (bounded between \(t_o\) and \(t_p\), and unimodal at \(t_m\)) — chosen because it can flexibly represent skewed uncertainty (e.g., more room for things to go wrong than right) unlike a symmetric normal distribution. The expected duration weights the most-likely estimate four times as heavily as either extreme, since the mode is considered the single best guess, while the optimistic and pessimistic bounds each get equal but lesser weight. The variance is derived directly from the spread between the two extreme estimates — a wider gap between optimistic and pessimistic times signals greater uncertainty (larger variance) in that activity's actual duration.

5.4 Project Completion Probability

Because the Central Limit Theorem justifies treating the sum of many independent random activity durations as approximately normally distributed, the expected project duration is simply the sum of the individual expected activity times along the critical path, and — critically — the project's variance is the sum of the individual variances (never the sum of standard deviations, which is a very common mistake) along the same critical path. The probability of finishing by a target date is then found by converting the gap between the target and the expected duration into a standard normal Z-score, and reading the corresponding probability from the standard normal cumulative distribution.

💡 \(t_e=\dfrac{a+4m+b}{6}\) and \(\sigma^2=[\dfrac{b-a}{6}]^2\) are essential formulae — asked almost every year. \(Z=\dfrac{T_S-T_E}{\sigma_p}\) and using the Z-table to find probability is the standard numerical type. Always sum variances (not \(\sigma\)) along the critical path.

Expected (Mean) Activity Duration

\( t_e = \dfrac{t_o+4t_m+t_p}{6} \)

Variance & Standard Deviation

\( \sigma^2 = \left(\dfrac{t_p-t_o}{6}\right)^2 \)

\( \sigma = \dfrac{t_p-t_o}{6} \)

Expected Project Duration & Variance

\( T_E = \Sigma t_e \) (along critical path)

\( \sigma_p^2 = \Sigma\sigma^2 \) (sum of variances)

Completion Probability

\( Z = \dfrac{T_S-T_E}{\sigma_p} \)

\( P(T\le T_S) = \Phi(Z) \)

Example 1 — Expected Time and Variance for One Activity

Given: \(t_o=4\) days, \(t_m=6\) days, \(t_p=14\) days. Find \(t_e\) and \(\sigma^2\).

Solution:

\( t_e = \dfrac{4+4(6)+14}{6} = \dfrac{4+24+14}{6} = \dfrac{42}{6} = 7\ \text{days} \)

\( \sigma^2 = \left(\dfrac{14-4}{6}\right)^2 = \left(\dfrac{10}{6}\right)^2 \approx 2.78 \)

Answer: \(t_e=7\) days, \(\sigma^2\approx2.78\ \text{days}^2\).

Example 2 — Expected Project Duration and Variance

Given: The critical path has three activities with \(t_e\) values 7, 10, 5 days and variances 2.78, 1.0, 0.44. Find \(T_E\) and \(\sigma_p^2\).

Solution:

\( T_E = 7+10+5 = 22\ \text{days} \)

\( \sigma_p^2 = 2.78+1.0+0.44 = 4.22 \)

Answer: \(T_E=22\) days, \(\sigma_p^2=4.22\ \text{days}^2\), so \(\sigma_p=\sqrt{4.22}\approx2.05\) days.

Example 3 — Probability of Meeting a Target Date

Given: \(T_E=22\) days, \(\sigma_p=2.05\) days, target date \(T_S=25\) days. Find the probability of completing by the target.

Solution:

\( Z = \dfrac{25-22}{2.05} = \dfrac{3}{2.05} \approx 1.46 \)

From the standard normal table, \(\Phi(1.46)\approx0.928\)

Answer: Probability of completing by day 25 \(\approx92.8\%\).

Beta distribution for a PERT activity duration showing optimistic, most likely, and pessimistic time estimates Duration → t_o (optimistic) t_m (most likely, mode) t_p (pessimistic) Beta distribution: bounded [t_o, t_p], unimodal at t_m

Fig. 5.1 — PERT's Beta distribution for activity duration: bounded between the optimistic and pessimistic estimates, peaked (mode) at the most-likely estimate, with the expected time weighting the mode four times as heavily as either extreme.

Q1. In PERT's expected time formula \(t_e=\dfrac{t_o+4t_m+t_p}{6}\), the most likely time is weighted:
The most likely time (mode) is considered the single best estimate and is weighted 4 out of the total 6 units in the formula, while optimistic and pessimistic each get 1 unit.
Q2. When computing the project's overall variance along the critical path, you must:
Variances of independent random variables add; standard deviations do not. Summing standard deviations directly is a very common and serious error.
Q3. If the Z-score for a target completion date works out to exactly 0, the probability of completing by that date is:
Z=0 means the target date exactly equals the expected project duration; for a normal distribution, the probability of being at or below the mean is exactly 50%.
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6. Gantt Charts
Description · Features · Advantages & Limitations vs Network Methods
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6.1 Description

A Gantt chart (bar chart) is a horizontal bar graph representing the project schedule, developed by Henry Gantt around 1910–1915 — making it far older than CPM or PERT, yet it remains the most widely used tool for communicating schedules to non-technical stakeholders because of its intuitive visual format.

6.2 Features

FeatureDescription
Horizontal axisTime scale (days, weeks, months)
Vertical axisList of activities / work packages
BarsSpan from EST to EFT (or LST to LFT); length = duration
FloatOften shown as thin lines or hatching extending beyond the solid bar to LFT
MilestoneDiamond symbol ◆ marking a key event with zero duration
Progress trackingBars partially filled to show % complete

6.3 Advantages and Limitations

A Gantt chart's core strength is also the source of its core weakness: its simple, visual bar-per-activity format is easy to read at a glance and communicates well with clients and field staff, but that same simplicity means it does not clearly show the logical dependencies between activities, nor does it explicitly identify the critical path — two things a network diagram shows immediately. For large, complex projects (roughly beyond 50 activities), a Gantt chart also becomes visually unwieldy and difficult to update when the underlying logic changes. In practice, the two tools are complementary rather than competing: the CPM/PERT network is used to plan the logic and identify the critical path, while the Gantt chart derived from it is used to communicate and track the schedule day-to-day.

💡 Gantt chart limitations vs CPM advantages is a common theoretical question. Key point: Gantt charts do NOT show logical dependencies between activities, while CPM/PERT networks do.

Gantt Bar Placement

Bar spans EST to EFT (or LST to LFT); length = activity duration

Float Display

Float shown as a thin line/hatch extending the solid bar out to LFT

Example 1 — Constructing a Gantt Bar

Given: An activity has \(EST=5\) days, \(EFT=12\) days, \(LFT=15\) days. Sketch the Gantt bar description.

Solution: The solid bar spans from day 5 to day 12 (duration 7 days); a hatched/thin extension continues from day 12 to day 15, representing the 3 days of Total Float.

Answer: Solid bar: day 5–12; float extension: day 12–15 (3 days float).

Example 2 — Recognising a Gantt Chart Limitation

Given: Two activities' bars on a Gantt chart appear to overlap in time but the chart doesn't indicate whether one depends on the other. What tool would clarify this?

Solution: A Gantt chart shows only when activities occur, not their logical relationships — dependency information must come from elsewhere.

Answer: A CPM/PERT network diagram (AOA or AON) would explicitly show whether a predecessor-successor relationship exists between the two activities.

Example 3 — Milestone Representation

Given: A project has a key deliverable ("Foundation Approved") that marks a point in time with no associated work duration. How is this shown on a Gantt chart?

Solution: Since it has zero duration, it cannot be drawn as a bar in the usual sense.

Answer: It is shown as a milestone — a diamond (◆) symbol placed at the appropriate date on the chart.

Gantt chart example showing activity bars, float extensions, and a milestone diamond Activity A Activity B Activity C Milestone ◆ Foundation approved Time → Solid = duration Hatched = float

Fig. 6.1 — Gantt chart: horizontal bars represent each activity's scheduled duration, with a lighter extension showing available float, and a diamond marking a zero-duration milestone.

Q1. The main limitation of a Gantt chart compared to a CPM/PERT network is that it:
A Gantt chart shows WHEN activities occur but not WHY (i.e., their predecessor/successor logic) — a network diagram is needed to see dependencies explicitly.
Q2. On a Gantt chart, a milestone (zero-duration key event) is typically shown as a:
Since a milestone has zero duration, it cannot be drawn as a bar — the standard convention is a diamond symbol placed at the milestone date.
Q3. Gantt charts were developed by Henry Gantt around:
Gantt charts predate CPM (1956, DuPont) and PERT (1958, US Navy) by several decades, having been developed by Henry Gantt around 1910–1915.
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7. Resource Scheduling
Resource Levelling vs Allocation · Levelling Procedure · Heuristic Priority Rules
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7.1 Resource Levelling vs Resource Allocation

TechniqueObjectiveConstraint
Resource levellingMinimise fluctuations in resource usage; smooth the demand profileProject duration fixed; resources unlimited but peak usage minimised
Resource allocation (limited)Complete project with available resources; may extend durationResource limit fixed; project duration may extend

These two techniques represent opposite trade-offs: levelling holds the schedule (project duration) fixed and adjusts resource deployment within that constraint, whereas limited resource allocation holds the resource ceiling fixed and allows the schedule to stretch if the available workforce or equipment can't handle the theoretical peak demand.

7.2 Resource Levelling Procedure

Levelling begins from the earliest-start schedule, plotting a resource histogram (resource units required versus time) to visualise demand peaks and troughs. The key principle is that only non-critical activities may be shifted — shifting them later (within their available float) smooths the resource profile without delaying the project, whereas shifting a critical activity has no available float to absorb the shift and directly extends the project duration. Levelling quality is often measured by minimising the sum of squared resource usage across periods, since squaring penalises sharp peaks much more heavily than a simple sum would, favouring a genuinely smoother profile over one with occasional extreme spikes.

7.3 Heuristic Rules for Scheduling

When resources are limited and multiple activities compete for the same resource simultaneously, priority must be assigned by some rule — common heuristics include scheduling the activity with the least total float first (protecting the most schedule-critical work), the shortest-duration activity first (freeing up the resource quickly for others), or the earliest latest-finish-time first. Resource-constrained project scheduling is known to be an NP-hard combinatorial problem in general, meaning no single algorithm guarantees the mathematically optimal schedule efficiently for large projects — these heuristic priority rules instead give practical, near-optimal solutions used throughout the industry.

💡 Resource levelling vs resource allocation distinction is commonly asked as a definition question. The key principle for levelling: shift non-critical activities within available float — never shift critical activities or the project is delayed.

Resource Usage Rate (Levelling Objective)

Minimise \( \Sigma(r_i^2\times d_i) \)

\(r_i\)=resource units in period \(i\); \(d_i\)=duration of period \(i\)

Levelling Constraint

Shift amount \(\le TF\) of the (non-critical) activity being shifted

Example 1 — Identifying a Shiftable Activity

Given: Activity X has Total Float \(TF=4\) days and is scheduled at its earliest start, causing a resource peak. Can it be shifted to smooth the peak?

Solution: Since X is non-critical (TF > 0), it has room to move without delaying the project.

Answer: Yes — X can be shifted up to 4 days later (its Total Float) to help smooth the resource histogram, as long as the shift doesn't create a new conflict elsewhere.

Example 2 — Recognising a Non-Shiftable Activity

Given: Activity Y is on the critical path (\(TF=0\)) and also contributes to a resource peak. Can it be shifted?

Solution: A critical activity has zero float, so any shift directly delays the project's completion date.

Answer: No — Y cannot be shifted without extending project duration; the resource peak it contributes to must be resolved by other means (e.g., shifting a different, non-critical activity, or adding resources).

Example 3 — Applying a Priority Heuristic

Given: Two activities compete for one crew on the same day: Activity P (TF=2 days, duration 5 days) and Activity Q (TF=0 days, duration 3 days). Which gets priority under the "minimum total float first" rule?

Solution: The rule prioritises the activity with the least float — Q has TF=0 (critical), while P has TF=2 (some slack available).

Answer: Activity Q gets priority (it is critical and cannot be delayed); Activity P, having float, can wait if the crew is unavailable.

Resource histogram before and after levelling, showing a smoothed demand profile Before Levelling Sharp peak at day 3 (resource=6) After Levelling Smoothed (resource≈3-4) Only non-critical activities (with float) are shifted to smooth peaks; critical activities (TF=0) must stay fixed — shifting them delays the project.

Fig. 7.1 — Resource levelling: shifting non-critical activities within their available float smooths a sharp resource-demand peak into a more even profile, without changing the overall project duration.

Q1. Resource levelling holds ______ fixed while smoothing resource usage, whereas resource-constrained allocation holds ______ fixed.
Resource levelling fixes the project duration and smooths resource usage within it; resource-constrained allocation fixes the resource ceiling and allows the schedule to extend if needed.
Q2. Which activities can be shifted during resource levelling without delaying the project?
Non-critical activities have positive total float and can be shifted within that float without affecting the project's overall duration; critical activities have zero float and cannot be shifted without delaying completion.
Q3. Resource-constrained project scheduling is, in general, a problem that is:
Resource-constrained scheduling is a well-known NP-hard combinatorial problem, so practical project management relies on heuristic priority rules (least float first, shortest duration first, etc.) to obtain good, though not provably optimal, solutions.
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9. Estimating & Costing
Types of estimates · quantity methods · earthwork volumes · rate analysis · BBS
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9.1 Purpose & Types of Estimate

An estimate is the anticipated cost of a work computed before execution, from drawings and rates. Its accuracy grows as the design matures:

TypeBasis / Use
Preliminary / approximateRough cost for a decision, from plinth area or cubic content (per m² / per m³ rate)
Plinth-area estimatePlinth area × plinth-area rate for that class of building
Cubic-content estimatePlinth area × height × unit cubic rate (more reliable than plinth area)
Detailed estimateQuantities of every item measured from drawings, multiplied by analysed rates — the accurate, tender-grade estimate
Revised / supplementaryPrepared when the sanctioned cost is exceeded, or extra work is added

9.2 Methods of Taking Out Quantities

For a building, the quantity of each item (earthwork, foundation concrete, masonry) is obtained by two standard methods:

  • Long-wall & short-wall (separate / individual wall) method: walls running in one direction are "long walls", those perpendicular are "short walls". Long-wall length is measured out-to-out = centre-line length + one breadth; short-wall length is measured in-to-in = centre-line length − one breadth. Quantity = length × breadth × depth for each.
  • Centre-line method: the total centre-line length of all walls is multiplied once by the breadth and depth of the item. It is quicker but requires a deduction of one breadth for each junction (T- or cross-junction) of the centre lines to avoid double counting.

9.3 Earthwork Volumes

Earthwork for roads, canals and embankments is computed from cross-sectional areas taken at a regular interval \(d\) using the mid-section, trapezoidal (average end area), or prismoidal (Simpson's) rule — the prismoidal rule is the most accurate but needs an odd number of sections. A bar bending schedule (BBS) lists the shape, size, number and cut length of every reinforcement bar, giving the steel weight (via \(\text{weight} = \dfrac{d^2}{162}\ \text{kg/m}\) for a bar of diameter \(d\) mm).

9.4 Analysis of Rates

Rate analysis builds up the unit rate of an item by adding the cost of materials, labour and machinery for a unit quantity, plus water charges, contractor's profit (~10%) and overheads. It is the bridge between measured quantities and money, and is what makes a detailed estimate a priced tender.

💡 Long wall = CL + 1 breadth (out-to-out); short wall = CL − 1 breadth (in-to-in). Centre-line method deducts one breadth per junction. Steel weight \(= \dfrac{d^2}{162}\) kg/m (\(d\) in mm). Trapezoidal vs prismoidal (Simpson's) for earthwork. Heavily tested in SSC JE Paper II.

9.5 Tendering, Contracts & Quality Control

Once an estimate is priced, work is awarded and executed through a management framework:

  • Tendering process: invitation (open / limited / single / global), issue of tender documents, submission of the earnest-money deposit (EMD), opening and comparative statement, and award to the lowest responsive bidder (L1).
  • Contract types: item-rate (schedule/unit-price), lump-sum, percentage-rate, cost-plus, and labour contracts — each allocating cost and risk differently between owner and contractor.
  • Quality control (QC): the operational techniques — sampling, testing (cube tests, slump, field density) and statistical acceptance criteria — used to verify that materials and workmanship meet specification, distinct from quality assurance (QA), which is the planned system of procedures that gives confidence the QC will be met. IS 456 acceptance of concrete uses running-mean and individual-result criteria on characteristic strength.
  • Productivity & operation cost: output per unit of labour/plant time; monitored to control the direct cost component and to feed realistic rate analysis.

Long-wall / Short-wall Lengths

Long wall (out-to-out) \(= L_{CL} + 1\times b\)

Short wall (in-to-in) \(= L_{CL} - 1\times b\)

Trapezoidal (Average End Area)

\( V = d\left[\dfrac{A_1+A_n}{2} + (A_2+A_3+\dots+A_{n-1})\right] \)

Prismoidal (Simpson's) Rule

\( V = \dfrac{d}{3}\big[(A_1+A_n) + 4(A_2+A_4+\dots) + 2(A_3+A_5+\dots)\big] \)

Needs an odd number of sections

Mid-section Formula

\( V = A_m \times L \) (\(A_m\) = area at mid-section)

Reinforcement Steel Weight

\( w = \dfrac{d^2}{162}\ \text{kg/m} \) (\(d\) in mm)

Example 1 — Long-wall & short-wall lengths

Given: A room with centre-line lengths 5 m (long walls, two of them) and 4 m (short walls, two of them); wall breadth at foundation \(b=0.4\ \text{m}\). Find the lengths to use for the two pairs.

Solution: Long wall (out-to-out) \(= 5 + 0.4 = 5.4\ \text{m}\) each; short wall (in-to-in) \(= 4 - 0.4 = 3.6\ \text{m}\) each.

Answer: 2 long walls of 5.4 m and 2 short walls of 3.6 m.

Example 2 — Earthwork by the prismoidal rule

Given: Cross-sectional areas at 30 m intervals along an embankment: 20, 40, 60, 50, 30 m² (5 sections). Find the volume by Simpson's rule.

Solution: \( V = \dfrac{30}{3}\big[(20+30) + 4(40+50) + 2(60)\big] = 10\big[50 + 360 + 120\big] = 10\times530 = 5300\ \text{m}^3 \)

Answer: Volume ≈ 5300 m³.

Example 3 — Steel weight from a bar bending schedule

Given: 40 bars of 12 mm diameter, each 6 m long. Find the total steel weight.

Solution: Weight per metre \(= \dfrac{12^2}{162} = \dfrac{144}{162} = 0.889\ \text{kg/m}\). Total length \(= 40\times6 = 240\ \text{m}\). Weight \(= 240\times0.889 \approx 213\ \text{kg}\).

Answer: ≈ 213 kg of steel.

Q1. In the long-wall/short-wall method, the length of a long wall is measured as:
Long walls run continuously past the short walls, so their length is taken out-to-out = centre-line length + one breadth; short walls fit between them, in-to-in = centre-line length − one breadth.
Q2. The prismoidal (Simpson's) rule for earthwork requires:
Simpson's rule pairs intervals, so it needs an even number of equal intervals — i.e. an odd number of cross-sections; it is more accurate than the trapezoidal rule.
Q3. The unit weight of a reinforcing bar of diameter \(d\) mm is:
Derived from the density of steel, the standard shortcut is \(w = \dfrac{d^2}{162}\) kg per metre for a bar of diameter \(d\) in mm — a staple of bar-bending-schedule calculations.
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10. Valuation & Depreciation
Value terms · depreciation methods · sinking fund · years' purchase
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10.1 Cost, Price & Value

Cost is the money spent to produce a property; value is its present worth in the market. Common value terms:

TermMeaning
Market valuePrice obtainable in the open market at a given time
Book valueCost less total depreciation to date (shown in the books)
Scrap valueValue of the dismantled materials at the end of life (≈ 10% of cost), less demolition cost
Salvage valueValue at end of utility period without being dismantled (useful to another user)
Assessed valueValue fixed by an authority for levying tax
Distress / forced valueLower value realised under a forced or urgent sale

10.2 Depreciation & Obsolescence

Depreciation is the gradual loss in value of a property due to age, wear, and decay. Obsolescence is loss of value due to the asset going out of date (better designs, changed fashion, new regulations) even while physically sound. Common methods to compute depreciation:

  • Straight-line method: equal depreciation each year; annual depreciation \(D = \dfrac{C - S}{n}\).
  • Constant-percentage (declining-balance) method: a fixed percentage of the reducing book value is written off each year, so book value \(= C(1-r)^t\).
  • Sinking-fund method: depreciation set aside is invested to accumulate the replacement cost over the life.
  • Quantity-survey method: depreciation assessed by a detailed survey of the actual physical deterioration of each part.

10.3 Sinking Fund & Years' Purchase

A sinking fund is an amount set aside annually (and invested at compound interest) so that a required sum — the cost of replacement, less scrap value — is accumulated by the end of the asset's life. Years' purchase (YP) is the capital sum whose interest equals one unit of annual net income; for a perpetual income \(YP = \dfrac{1}{i}\). The capitalised value of a property = net annual income × YP.

💡 Straight-line: \(D=\dfrac{C-S}{n}\). Declining balance: book value \(=C(1-r)^t\). Sinking fund instalment \(I = \dfrac{S\,i}{(1+i)^n-1}\). Perpetual YP \(=\dfrac{1}{i}\); capitalised value = net income × YP. Scrap value ≈ 10% of cost.

Straight-line Depreciation

\( D = \dfrac{C-S}{n} \) per year

Book value after \(t\) years \(= C - tD\)

Declining-balance (Constant %)

Book value \(= C(1-r)^t\)

\(r\) = fixed annual depreciation rate

Sinking-fund Annual Instalment

\( I = \dfrac{S\,i}{(1+i)^n - 1} \)

\(S\) = amount to accumulate, \(i\) = rate, \(n\) = years

Years' Purchase & Capitalised Value

Perpetual \( YP = \dfrac{1}{i} \)

Capitalised value \(=\) net annual income \(\times\ YP\)

Example 1 — Straight-line depreciation

Given: A building costs ₹20,00,000, has a life of \(n = 50\) years and a scrap value of ₹2,00,000. Find the annual depreciation and the book value after 20 years.

Solution: \( D = \dfrac{20{,}00{,}000 - 2{,}00{,}000}{50} = \dfrac{18{,}00{,}000}{50} = 36{,}000 \) per year. Book value after 20 years \(= 20{,}00{,}000 - 20\times36{,}000 = 12{,}80{,}000\).

Answer: ₹36,000 per year; book value ₹12,80,000 after 20 years.

Example 2 — Capitalised value from years' purchase

Given: A property yields a net annual income of ₹1,20,000; the expected rate of interest is \(i = 6\%\). Find its capitalised value.

Solution: \( YP = \dfrac{1}{i} = \dfrac{1}{0.06} = 16.67 \). Capitalised value \(= 1{,}20{,}000 \times 16.67 = 20{,}00{,}000\).

Answer: Capitalised value ≈ ₹20,00,000.

Example 3 — Sinking-fund instalment

Given: A sum of ₹5,00,000 must be accumulated in \(n = 20\) years at \(i = 5\%\). Find the annual sinking-fund instalment.

Solution: \( I = \dfrac{S\,i}{(1+i)^n - 1} = \dfrac{5{,}00{,}000 \times 0.05}{(1.05)^{20} - 1} = \dfrac{25{,}000}{2.653 - 1} = \dfrac{25{,}000}{1.653} \approx 15{,}125 \).

Answer: ≈ ₹15,125 per year.

Q1. The scrap value of a building is usually taken as about:
Scrap value — the worth of dismantled materials at end of life — is conventionally taken as about 10% of the construction cost (net of demolition charges).
Q2. Loss of value because a building has gone out of date (superseded by better designs), though still physically sound, is called:
Obsolescence is loss of value due to becoming outdated (new technology, changed regulations or fashion), distinct from depreciation, which is physical wear with age.
Q3. For a property with a perpetual net annual income, the years' purchase (YP) equals:
For a perpetual income at rate \(i\), \(YP = \dfrac{1}{i}\); the capitalised value is then net annual income × YP.
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8. Quick Revision
Key Formulae · Exam-Angle Summary · Mnemonics · Practice
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A. Key Formulae at a Glance

Forward Pass

\( E_j = \max(E_i+D) \)

Backward Pass

\( L_i = \min(L_j-D) \)

Total Float

\( TF = L_j-E_i-D = LST-EST \)

Free Float

\( FF = E_j-E_i-D \)

Independent Float

\( IF = E_j-L_i-D\ (\ge0) \)

Float Relationship

\( IF\le FF\le TF \); critical: \(TF=0\)

Cost Slope

\( \dfrac{C_c-C_n}{t_n-t_c} \)

PERT Expected Time

\( t_e = \dfrac{t_o+4t_m+t_p}{6} \)

PERT Variance

\( \sigma^2 = \left(\dfrac{t_p-t_o}{6}\right)^2 \)

Completion Probability

\( Z=\dfrac{T_S-T_E}{\sigma_p} \); \(P=\Phi(Z)\)

Resource Levelling Objective

Minimise \(\Sigma(r_i^2\times d_i)\)

Crash Priority

Crash critical activity with min cost slope first

B. Exam-Angle Summary

TopicGATE FocusESE FocusSSC JE Focus
Project networksRecognising dummy activity need; AOA constructionComplete network diagrams; AOA vs AON theoryBasic terminology; dummy activity concept
Forward/backward passFull numerical: E and L for all nodesComplete tabulation with EST/EFT/LST/LFTBasic forward pass calculation
Float & critical pathTF/FF/IF numericals; multiple critical pathsComplete float table; critical path identification methodsTF calculation; critical path definition
CrashingCost slope numericals; optimal durationComplete crashing procedure with multiple pathsCost slope formula; crash vs normal concept
PERTt_e, variance, Z-score numericals (near-certain)Complete probabilistic analysis; Beta distribution theoryt_e formula; basic probability concept
Gantt chartsRarely numerical; limitations vs CPMTheory questions on advantages/limitationsBasic Gantt chart reading
Resource schedulingLevelling vs allocation distinctionResource histogram; heuristic rules applicationDefinitions of levelling and allocation

C. Mnemonics & Memory Aids

💡 "Forward is Fast (MAX), Backward is Blocking (MIN)": forward pass takes the maximum (slowest predecessor governs); backward pass takes the minimum (most demanding successor governs).
💡 Float order "I Fit From Total" (IF ≤ FF ≤ TF): Independent is smallest (most conservative), Free is intermediate, Total is largest (no neighbour assumption).
💡 "Crash Cheap first": always crash the critical activity with the lowest cost slope first.
💡 PERT weights "1-4-1": \(t_e=\dfrac{t_o+4t_m+t_p}{6}\) — optimistic and pessimistic get 1 part each, most likely gets 4 parts, divided by 6 total.
💡 "Variances Add, Deviations Don't": always sum σ² (variance) along the critical path, never sum σ (standard deviation) directly.
💡 "Gantt shows When, Network shows Why": Gantt charts display timing but not logical dependencies; CPM/PERT networks show both.

Practice Questions

Q1. Activity (2-5) has \(E_2=8\), \(L_5=20\), duration \(D=7\). Find the Total Float.

Show Answer
\( TF = 20-8-7 = 5\ \text{days} \)

Q2. An activity has \(t_n=8\) days (\(C_n=₹20{,}000\)), \(t_c=5\) days (\(C_c=₹35{,}000\)). Find the cost slope.

Show Answer
\( \text{Cost slope}=\dfrac{35{,}000-20{,}000}{8-5}=\dfrac{15{,}000}{3}=₹\dfrac{5000}{\text{day}} \)

Q3. An activity has \(t_o=3\), \(t_m=5\), \(t_p=13\) days. Find \(t_e\) and \(\sigma\).

Show Answer
\( t_e=\dfrac{3+20+13}{6}=\dfrac{36}{6}=6\ \text{days} \); \( \sigma=\dfrac{13-3}{6}\approx1.67\ \text{days} \)

Q4. \(T_E=30\) days, \(\sigma_p=3\) days, target \(T_S=27\) days. Find Z and interpret.

Show Answer
\( Z=\dfrac{27-30}{3}=-1.0 \) — negative Z means the target is before the expected duration, so probability of meeting it is below 50% (specifically \(\Phi(-1.0)\approx15.9\%\))

Q5. A network has two paths of 25 and 25 days (tied) and one path of 20 days. How many critical paths exist, and what is the project duration?

Show Answer
Two critical paths (both 25 days, the longest); project duration = 25 days; the 20-day path has 5 days of float on its non-shared activities

Mini Quiz — Full Chapter Recap

Q1. The forward pass computes earliest event times using which operation at a merge node?
An event cannot occur until all incoming activities are complete, so the forward pass takes the maximum of all incoming earliest finish times.
Q2. PERT's expected activity time formula weights the most likely estimate by a factor of:
t_e = (t_o + 4t_m + t_p)/6 — the most likely time gets a weight of 4 out of the total divisor of 6.
Q3. Resource levelling shifts which activities to smooth resource demand?
Only non-critical activities have float available to shift without delaying the project — critical activities (TF=0) must remain fixed.
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