CELE Construction Management & Methods — Project Planning and Scheduling (CPM/PERT)Exam Answer Templates

Marks

1

Topic

Critical Path Definition

Difficulty

easy

Template Id

T1

Examiner Tip

Examiners check for two things: 'longest path' and either 'minimum duration' or 'zero float.' Include both in one or two lines to guarantee the mark.

Model Answer

The critical path is the longest path through the project network from start to finish. It determines the minimum possible project duration, and all activities on it have zero total float.

Question Type

very_short_answer

Answer Structure

• Line 1: State that the critical path is the LONGEST path through the network [½ mark]
• Line 2: State that it equals the minimum project duration and activities have zero total float [½ mark]

Scoring Breakdown

Common Mark Deductions

• Writing 'shortest path' instead of 'longest path' — immediate zero
• Defining it only as 'the most important path' without technical content
• Omitting the zero-float characteristic

Key Phrases To Include

• longest path
• minimum project duration
• zero total float
• start to finish

Marks

2

Topic

Float Definitions

Difficulty

easy

Template Id

T2

Examiner Tip

Board exam checkers look for the keyword 'successor' in the Free Float definition. Without it, you will likely lose that mark.

Model Answer

Total Float (TF) is the maximum time an activity can be delayed without delaying the overall project completion date. It is computed as TF = LS − ES = LF − EF. Free Float (FF) is the maximum time an activity can be delayed without delaying the Early Start of any of its immediate successors. FF ≤ TF for any activity.

Question Type

very_short_answer

Answer Structure

• Line 1–2: Define Total Float with formula TF = LS − ES [1 mark]
• Line 3–4: Define Free Float and state it does not delay any successor [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the same definition for both — zero marks for the confused entry
• Omitting the formula for Total Float when the question is numerical in nature
• Stating Free Float delays the project instead of the successor

Key Phrases To Include

• Total Float
• TF = LS − ES
• project completion date
• Free Float
• immediate successors
• Early Start of successor

Marks

2

Topic

PERT Formulas

Difficulty

easy

Template Id

T3

Examiner Tip

Write the formulas in standard notation first, then define variables below. This mirrors textbook format and is immediately recognizable to checkers.

Model Answer

In PERT, with optimistic duration a, most likely duration m, and pessimistic duration b: Expected duration: t_e = (a + 4m + b) / 6 Variance: σ² = [(b − a) / 6]² The weight of 4 is assigned to the most likely estimate m, reflecting its greater probability. The project path variance is the sum of individual activity variances along the critical path.

Question Type

very_short_answer

Answer Structure

• Line 1: Define a, m, b clearly [½ mark]
• Line 2: Write correct formula for t_e with weight 4 on m and divisor 6 [½ mark]
• Line 3: Write correct formula for σ² [½ mark]
• Line 4: Note that project variance = sum of critical-path variances [½ mark]

Scoring Breakdown

Common Mark Deductions

• Using weight 4 on b or a instead of m
• Writing divisor as 3 or 4 instead of 6
• Forgetting to square the term in the variance formula

Key Phrases To Include

• t_e = (a + 4m + b) / 6
• σ² = [(b − a) / 6]²
• optimistic
• most likely
• pessimistic
• weight 4

Marks

2

Topic

PERT Expected Duration and Variance

Difficulty

easy

Template Id

T4

Examiner Tip

Show every arithmetic step. If your final answer is wrong but your formula and substitution are correct, you may still earn 1 mark for method.

Model Answer

Given: a = 4 days, m = 6 days, b = 14 days Expected duration: t_e = (a + 4m + b) / 6 = (4 + 4(6) + 14) / 6 = (4 + 24 + 14) / 6 = 42 / 6 = 7 days Variance: σ² = [(b − a) / 6]² = [(14 − 4) / 6]² = [10 / 6]² = (1.667)² = 2.78 days² Answer: t_e = 7 days; σ² = 2.78 days²

Question Type

numerical

Answer Structure

• Line 1: State given values [0 mark, but sets up solution clearly]
• Line 2–5: Substitute into t_e formula and evaluate step-by-step [1 mark]
• Line 6–9: Substitute into σ² formula and evaluate step-by-step [1 mark]
• Line 10: State final answers clearly labeled

Scoring Breakdown

Common Mark Deductions

• Adding 4 + 6 + 14 and dividing by 3 (using simple average instead of PERT formula)
• Omitting the squaring step in the variance formula, giving σ² = 1.667 instead of 2.78
• Not showing the formula before substitution

Key Phrases To Include

• t_e = (a + 4m + b) / 6
• σ² = [(b − a) / 6]²
• t_e = 7 days
• σ² = 2.78 days²

Marks

3

Topic

Forward Pass, Backward Pass, and Float Computation

Difficulty

medium

Template Id

T5

Examiner Tip

Always verify TF using both formulas (LS − ES and LF − EF) and show that they agree. Checkers reward this self-verification habit.

Model Answer

Given: ES = 7, d = 3, LS = 10, LF = 13 (a) Early Finish: EF = ES + d = 7 + 3 = 10 days (b) Late Start (verification): LS = LF − d = 13 − 3 = 10 days ✓ (consistent with given LS = 10) (c) Total Float: TF = LS − ES = 10 − 7 = 3 days (Check: TF = LF − EF = 13 − 10 = 3 days ✓) Conclusion: TF = 3 days ≠ 0, therefore this activity is NOT on the critical path.

Question Type

numerical

Answer Structure

• Part (a): Apply EF = ES + d [1 mark]
• Part (b): Apply LS = LF − d or use given LS to verify [½ mark]
• Part (c): Apply TF = LS − ES with correct values [1 mark]
• Conclusion: State critical path status based on TF value [½ mark]

Scoring Breakdown

Common Mark Deductions

• Computing TF = LF − ES (spanning the entire window) — incorrect formula
• Stating the activity IS on the critical path despite TF = 3
• Omitting the conclusion about critical path status

Key Phrases To Include

• EF = ES + d
• LS = LF − d
• TF = LS − ES
• TF = 3 days
• NOT on the critical path
• TF ≠ 0

Marks

3

Topic

Critical Path Identification and Float

Difficulty

medium

Template Id

T6

Examiner Tip

A network CAN have multiple critical paths. Always check all paths against the maximum duration. Examiners test this specifically.

Model Answer

Path durations: Path 1: A→B→D = 3 + 5 + 4 = 12 days Path 2: A→C→D = 3 + 7 + 4 = 14 days Path 3: A→B→E = 3 + 5 + 6 = 14 days (a) Project Duration: Project duration = maximum path duration = max(12, 14, 14) = 14 days (b) Critical Path(s): Paths 2 and 3 both equal 14 days (the longest). Critical paths: A→C→D and A→B→E Note: Two critical paths exist; activities on either path have TF = 0. (c) Total Float of Path 1 (A→B→D): TF = Project duration − Path 1 duration = 14 − 12 = 2 days Path 1 can be delayed up to 2 days without extending the project.

Question Type

numerical

Answer Structure

• Step 1: List all path durations with arithmetic [½ mark]
• Part (a): Identify maximum as project duration [1 mark]
• Part (b): Identify all paths equal to maximum as critical paths [1 mark]
• Part (c): Compute float = project duration − path duration [½ mark]

Scoring Breakdown

Common Mark Deductions

• Selecting only one critical path when two tie — missing the second critical path costs a mark
• Computing float as 12 − 14 (negative) and misinterpreting the sign
• Calling Path 3 non-critical because 'only one critical path can exist' — incorrect assumption

Key Phrases To Include

• maximum path duration
• 14 days
• critical path: A→C→D and A→B→E
• TF = 0 on critical path
• float = 2 days

Marks

3

Topic

Forward and Backward Pass Procedures

Difficulty

medium

Template Id

T7

Examiner Tip

The merge/burst rules (largest vs. smallest) are what examiners specifically test in theory questions. State them explicitly with a brief justification.

Model Answer

Forward Pass (Computes Early Times): Begin at the project start with ES = 0 for the first activity. For each subsequent activity: EF = ES + d where d is the activity duration. When an activity has multiple predecessors, its ES equals the LARGEST EF among all predecessors. This ensures all preceding work is complete. The forward pass ends at the last activity; its EF equals the project duration. Backward Pass (Computes Late Times): Begin at the last activity with LF = project duration. For each activity moving backward: LS = LF − d When an activity has multiple successors, its LF equals the SMALLEST LS among all successors. This ensures no successor is delayed. The backward pass propagates back to the first activity. Float is then computed as: TF = LS − ES (or LF − EF). Critical activities have TF = 0.

Question Type

short_answer

Answer Structure

• Paragraph 1: Define forward pass — formula EF = ES + d, merge rule (largest EF of predecessors) [1.5 marks]
• Paragraph 2: Define backward pass — formula LS = LF − d, burst rule (smallest LS of successors) [1.5 marks]
• Optional summary line: Connect to float computation for completeness

Scoring Breakdown

Common Mark Deductions

• Reversing the merge rules: using 'smallest' for forward pass and 'largest' for backward pass
• Not stating the starting conditions (ES=0 for forward, LF=project duration for backward)
• Confusing ES/EF (forward pass outputs) with LS/LF (backward pass outputs)

Key Phrases To Include

• EF = ES + d
• LS = LF − d
• largest EF of predecessors
• smallest LS of successors
• ES = 0 at start
• LF = project duration at finish
• TF = LS − ES

Marks

2

Topic

PERT Project Standard Deviation

Difficulty

medium

Template Id

T8

Examiner Tip

In PERT, you sum VARIANCES (not standard deviations) along the critical path, then take one square root at the end. This is a frequently tested distinction.

Model Answer

Given variances of critical-path activities: σ²₁ = 4, σ²₂ = 1, σ²₃ = 4, σ²₄ = 2.25 days² Project variance (sum of critical-path variances): σ²_project = σ²₁ + σ²₂ + σ²₃ + σ²₄ = 4 + 1 + 4 + 2.25 = 11.25 days² Project standard deviation: σ_project = √σ²_project = √11.25 = 3.35 days Answer: The standard deviation of the project duration is 3.35 days.

Question Type

numerical

Answer Structure

• Step 1: List all given variances [0 mark, sets up computation]
• Step 2: Sum all critical-path variances for project variance [1 mark]
• Step 3: Take square root for project standard deviation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Summing standard deviations instead of variances before taking the root
• Taking the square root of the sum then squaring it — circular error
• Including non-critical path activity variances in the sum

Key Phrases To Include

• sum of critical-path variances
• σ²_project = 11.25 days²
• σ_project = √11.25
• 3.35 days

Marks

5

Topic

PERT Full Computation — Multi-Activity

Difficulty

medium

Template Id

T9

Examiner Tip

Organize your solution with clear headings for each activity. Board exam checkers grade part by part; a clearly organized solution prevents confusion and ensures partial marks are awarded correctly.

Model Answer

Given: Activity X: a = 5, m = 8, b = 17 Activity Y: a = 2, m = 5, b = 8 Both are on the critical path (sequential). (a) PERT Expected Duration and Variance: Activity X: t_e(X) = (a + 4m + b) / 6 = (5 + 4(8) + 17) / 6 = (5 + 32 + 17) / 6 = 54 / 6 = 9 days σ²(X) = [(b − a) / 6]² = [(17 − 5) / 6]² = [12 / 6]² = (2)² = 4 days² Activity Y: t_e(Y) = (2 + 4(5) + 8) / 6 = (2 + 20 + 8) / 6 = 30 / 6 = 5 days σ²(Y) = [(8 − 2) / 6]² = [6 / 6]² = (1)² = 1 day² (b) Expected Project Duration: T_project = t_e(X) + t_e(Y) = 9 + 5 = 14 days (c) Project Standard Deviation: σ²_project = σ²(X) + σ²(Y) = 4 + 1 = 5 days² σ_project = √5 = 2.24 days Summary: Activity X: t_e = 9 days, σ² = 4 days² Activity Y: t_e = 5 days, σ² = 1 day² Expected project duration: 14 days Project standard deviation: 2.24 days

Question Type

numerical

Answer Structure

• Part (a) — Activity X: Apply t_e formula correctly [1 mark]; Apply σ² formula correctly [1 mark]
• Part (a) — Activity Y: Apply t_e formula correctly [1 mark]; Apply σ² formula correctly [0.5 mark — sharing with part c]
• Part (b): Sum t_e values for project duration [0.5 mark]
• Part (c): Sum variances then take square root [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using simple average (a + m + b)/3 instead of PERT formula — costs 2 marks immediately
• Adding σ instead of σ² for project variance — gives wrong final answer
• Showing final answers only without step-by-step working — risks losing all method marks

Key Phrases To Include

• t_e = (a + 4m + b) / 6
• σ² = [(b − a) / 6]²
• t_e(X) = 9 days
• σ²(X) = 4 days²
• t_e(Y) = 5 days
• σ²(Y) = 1 day²
• project duration = 14 days
• σ_project = √5 = 2.24 days

Marks

5

Topic

Complete CPM Network Analysis

Difficulty

hard

Template Id

T10

Examiner Tip

Present the forward pass and backward pass as two separate, clearly labeled sections. Then present a float table. This three-section structure is exactly what board exam checkers expect for full-mark CPM solutions.

Model Answer

Step 1 — Forward Pass (Compute ES and EF): Activity A: ES=0, EF = 0+4 = 4 Activity B: ES=0, EF = 0+6 = 6 Activity C: ES = EF(A) = 4, EF = 4+5 = 9 Activity D: ES = max[EF(A), EF(B)] = max[4, 6] = 6; EF = 6+3 = 9 Activity E: ES = max[EF(C), EF(D)] = max[9, 9] = 9; EF = 9+4 = 13 Project Duration = EF(E) = 13 days Step 2 — Backward Pass (Compute LS and LF): Activity E: LF = 13 (project end), LS = 13−4 = 9 Activity C: LF = LS(E) = 9, LS = 9−5 = 4 Activity D: LF = LS(E) = 9, LS = 9−3 = 6 Activity A: LF = min[LS(C), LS(D)] = min[4, 6] = 4; LS = 4−4 = 0 Activity B: LF = LS(D) = 6; LS = 6−6 = 0 Step 3 — Total Float (TF = LS − ES): Activity A: TF = 0 − 0 = 0 → CRITICAL Activity B: TF = 0 − 0 = 0 → CRITICAL Activity C: TF = 4 − 4 = 0 → CRITICAL Activity D: TF = 6 − 6 = 0 → CRITICAL Activity E: TF = 9 − 9 = 0 → CRITICAL Critical Path: A→C→E = 4+5+4 = 13 days AND B→D→E = 6+3+4 = 13 days Both paths are critical. Project duration = 13 days.

Question Type

numerical

Answer Structure

• Step 1 — Forward Pass: Compute ES and EF for all 5 activities with merge-node rule shown [2 marks]
• State project duration = 13 days [included in Step 1]
• Step 2 — Backward Pass: Compute LS and LF for all 5 activities with burst-node rule shown [2 marks]
• Step 3 — Float Table: TF for each activity and critical path identification [1 mark]

Scoring Breakdown

Common Mark Deductions

• At node D: using min instead of max for ES — produces wrong cascade of values
• At node A (backward): using max instead of min for LF — fundamental backward pass error
• Missing the second critical path (B→D→E) and only identifying A→C→E

Key Phrases To Include

• ES = max[EF of predecessors]
• LF = min[LS of successors]
• EF = ES + d
• LS = LF − d
• TF = LS − ES
• project duration = 13 days
• critical path: A→C→E and B→D→E

Marks

2

Topic

Crashing and Time-Cost Trade-off

Difficulty

medium

Template Id

T11

Examiner Tip

The phrase 'critical path activity with lowest cost slope' is the exact phrase examiners look for. Include it explicitly.

Model Answer

Activity crashing is the deliberate reduction of an activity's duration by adding resources (additional labor, equipment, or overtime), which typically increases direct costs. The goal is to shorten the overall project duration at minimum additional cost. Selection Principle: Crash the critical-path activity with the lowest crash cost per unit time (lowest cost slope) first. The cost slope is defined as: Cost slope = (Crash cost − Normal cost) / (Normal duration − Crash duration) Crashing non-critical activities does NOT shorten the project duration.

Question Type

short_answer

Answer Structure

• Sentence 1–2: Define crashing — reduce duration by adding resources, increases cost [1 mark]
• Sentence 3–4: State the least-cost principle — crash critical-path activity with lowest cost slope first [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating that any activity can be crashed — must specify critical-path activities only
• Omitting the cost slope concept — missing the selection criterion costs 1 mark
• Confusing crashing with normal schedule compression (re-sequencing activities)

Key Phrases To Include

• reduce duration
• additional resources
• increased direct cost
• critical-path activity
• lowest cost slope
• cost per unit time

Marks

2

Topic

CPM vs. PERT Comparison

Difficulty

easy

Template Id

T12

Examiner Tip

Number your points clearly (1, 2, 3, 4). Examiners count — four numbered points signals a complete answer and prevents them from stopping at two.

Model Answer

CPM vs. PERT — Key Differences: 1. Duration Estimates: CPM uses a single deterministic duration per activity; PERT uses three estimates (optimistic a, most likely m, pessimistic b) for probabilistic analysis. 2. Application: CPM is suited to well-defined repetitive projects (e.g., construction, manufacturing); PERT is suited to research and development or novel projects with high duration uncertainty. 3. Focus: CPM emphasizes time-cost trade-off (crashing); PERT emphasizes statistical probability of meeting a deadline. 4. Output: CPM produces a definite project duration; PERT produces an expected duration with an associated standard deviation and probability distribution.

Question Type

short_answer

Answer Structure

• Point 1: Duration estimates — deterministic vs. probabilistic [½ mark]
• Point 2: Application domain — repetitive vs. uncertain projects [½ mark]
• Point 3: Focus — time-cost trade-off vs. probability of completion [½ mark]
• Point 4: Output — definite duration vs. expected duration with variance [½ mark]

Scoring Breakdown

Common Mark Deductions

• Repeating the same difference in different words — only one mark awarded
• Vague statements like 'CPM is older than PERT' — no technical content, no mark
• Confusing which method uses which approach (e.g., writing that PERT uses single estimates)

Key Phrases To Include

• deterministic
• probabilistic
• three time estimates
• single duration estimate
• time-cost trade-off
• probability of completion

Marks

3

Topic

CPM Activity Time Computations

Difficulty

easy

Template Id

T13

Examiner Tip

This three-part problem maps perfectly to one mark per part. Always show the formula, substitution, and result for each part on its own line.

Model Answer

Given: ES = 4 days, d = 6 days, LF = 14 days Step 1 — Early Finish: EF = ES + d = 4 + 6 = 10 days Step 2 — Late Start: LS = LF − d = 14 − 6 = 8 days Step 3 — Total Float: TF = LS − ES = 8 − 4 = 4 days Verification: TF = LF − EF = 14 − 10 = 4 days ✓ Conclusion: EF = 10 days, LS = 8 days, TF = 4 days. Since TF = 4 ≠ 0, this activity is NOT on the critical path.

Question Type

numerical

Answer Structure

• Step 1: EF = ES + d = 10 days [1 mark]
• Step 2: LS = LF − d = 8 days [1 mark]
• Step 3: TF = LS − ES = 4 days with critical path conclusion [1 mark]

Scoring Breakdown

Common Mark Deductions

• Computing TF = LF − ES = 10 — using total window instead of float formula
• Not stating whether the activity is on the critical path
• Omitting verification step (not penalized but missed opportunity to show thoroughness)

Key Phrases To Include

• EF = ES + d
• LS = LF − d
• TF = LS − ES
• EF = 10 days
• LS = 8 days
• TF = 4 days
• not on the critical path

Marks

5

Topic

PERT Probability of Completion

Difficulty

hard

Template Id

T14

Examiner Tip

In 5-mark PERT probability problems, present exactly 5 distinct computational steps. Each step earns one mark. A numbered, structured solution eliminates checker guesswork.

Model Answer

Step 1 — Expected Durations: t_e(1) = (2 + 4×4 + 6)/6 = (2+16+6)/6 = 24/6 = 4 days t_e(2) = (1 + 4×3 + 5)/6 = (1+12+5)/6 = 18/6 = 3 days t_e(3) = (3 + 4×5 + 13)/6 = (3+20+13)/6 = 36/6 = 6 days Step 2 — Variances: σ²(1) = [(6−2)/6]² = [4/6]² = (0.667)² = 0.444 days² σ²(2) = [(5−1)/6]² = [4/6]² = (0.667)² = 0.444 days² σ²(3) = [(13−3)/6]² = [10/6]² = (1.667)² = 2.778 days² Step 3 — Expected Project Duration: T_E = t_e(1) + t_e(2) + t_e(3) = 4 + 3 + 6 = 13 days Step 4 — Project Standard Deviation: σ²_project = 0.444 + 0.444 + 2.778 = 3.666 days² σ_project = √3.666 = 1.914 days Step 5 — Z-Score and Probability: Z = (T_target − T_E) / σ_project = (15 − 13) / 1.914 = 2 / 1.914 = 1.045 ≈ 1.0 P(T ≤ 15) = Φ(1.0) = 0.8413 = 84.13% Conclusion: There is an 84.13% probability that the project will be completed within 15 days.

Question Type

numerical

Answer Structure

• Step 1: Compute t_e for all 3 activities [1 mark]
• Step 2: Compute σ² for all 3 activities [1 mark]
• Step 3: Sum t_e values for project expected duration = 13 days [1 mark]
• Step 4: Sum variances, take square root for σ_project = 1.914 days [1 mark]
• Step 5: Apply Z formula, use table, state probability = 84.13% [1 mark]

Scoring Breakdown

Common Mark Deductions

• Summing standard deviations instead of variances — gives wrong σ_project
• Forgetting to take the square root of the summed variance
• Using T_E − T_target in the Z formula (reversed sign) — gives negative Z and wrong probability
• Not including a concluding probability statement — loses the interpretation mark

Key Phrases To Include

• t_e = (a + 4m + b)/6
• σ² = [(b−a)/6]²
• T_E = 13 days
• σ_project = 1.914 days
• Z = (T_target − T_E) / σ_project
• Z ≈ 1.0
• Φ(1.0) = 0.8413
• 84.13% probability

Marks

5

Topic

Complete CPM with Float — Multi-Path Network

Difficulty

hard

Template Id

T15

Examiner Tip

Draw the network before computing anything — even a rough sketch. Examiners cannot read your mind; showing the network structure demonstrates understanding and earns method marks even if subsequent arithmetic is flawed.

Model Answer

Network structure: Path analysis from activity dependencies: - A and B are start activities (no predecessors) - C requires both A and B to be complete (merge node) - D requires B - E requires both C and D to be complete (merge node) Step 1 — Identify all paths from Start to Finish: Path 1: A → C → E = 2 + 3 + 2 = 7 days Path 2: B → C → E = 4 + 3 + 2 = 9 days Path 3: B → D → E = 4 + 5 + 2 = 11 days Step 2 — Project Duration and Critical Path: Longest path = Path 3 = 11 days Critical Path: B → D → E (TF = 0 for B, D, E) Project Duration = 11 days Step 3 — Forward Pass (to find ES values for float computation): B: ES=0, EF=4 A: ES=0, EF=2 C: ES = max[EF(A), EF(B)] = max[2, 4] = 4; EF = 4+3 = 7 D: ES = EF(B) = 4; EF = 4+5 = 9 E: ES = max[EF(C), EF(D)] = max[7, 9] = 9; EF = 9+2 = 11 Step 4 — Backward Pass: E: LF=11, LS = 11−2 = 9 D: LF = LS(E) = 9; LS = 9−5 = 4 C: LF = LS(E) = 9; LS = 9−3 = 6 B: LF = min[LS(C), LS(D)] = min[6, 4] = 4; LS = 4−4 = 0 A: LF = LS(C) = 6; LS = 6−2 = 4 (c) Total Float of Activity A: TF(A) = LS(A) − ES(A) = 4 − 0 = 4 days Summary: (a) Path durations: 7, 9, 11 days (b) Critical Path: B→D→E; Project Duration = 11 days (c) TF(A) = 4 days

Question Type

numerical

Answer Structure

• Step 1: Identify and enumerate all three paths with correct durations [1 mark]
• Step 2: Identify longest path as critical path; state project duration = 11 days [1 mark]
• Step 3: Complete forward pass with merge-node rules applied correctly [1 mark]
• Step 4: Complete backward pass with burst-node rules applied correctly [1 mark]
• Part (c): TF(A) = 4 days correctly computed [1 mark]

Scoring Breakdown

Common Mark Deductions

• Missing the path A→C→E or B→C→E — incomplete path enumeration
• Applying merge rule incorrectly at node C (A and B both feed C) — cascading errors
• Applying burst rule incorrectly at B (B feeds both C and D) — wrong LF for B

Key Phrases To Include

• longest path
• B→D→E
• project duration = 11 days
• ES = max[EF of predecessors]
• LF = min[LS of successors]
• TF(A) = 4 days