Project Planning and Scheduling (CPM/PERT) - Exam Answer Templates
In the PRC Civil Engineer Licensure Examination, knowing the correct answer is only half the battle — writing it in the format examiners expect is what converts knowledge into marks. For Construction Management topics like CPM and PERT, examiners reward structured, formula-driven answers with clearly labeled computations and logical flow. A vague or disorganized answer on a 5-mark problem can cost you 2–3 marks even when your arithmetic is correct. These templates show you the exact format, key phrases, and scoring logic for every mark level so you can maximize your score on the board exam.
Answer templates
Define the critical path of a project network. [1 mark]
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.
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
Criteria
Correct identification as longest path AND connection to project duration or zero float
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
Differentiate Total Float from Free Float. [2 marks]
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.
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
Criteria
Correct definition of Total Float referencing project completion or formula TF = LS − ES
Criteria
Correct definition of Free Float referencing immediate successors (not the project)
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
State the PERT formula for expected activity duration and variance. [2 marks]
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.
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
Criteria
Correct t_e formula: (a + 4m + b)/6 with all three estimates named
Criteria
Correct σ² formula: [(b−a)/6]² and/or statement on summing variances
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
An activity has optimistic duration a = 4 days, most likely duration m = 6 days, and pessimistic duration b = 14 days. Compute the PERT expected duration and variance. [2 marks]
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
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
Criteria
Correct computation of t_e = 7 days using formula (a + 4m + b)/6
Criteria
Correct computation of σ² = 2.78 days² using formula [(b−a)/6]²
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
An activity has ES = 7 days, duration d = 3 days, LS = 10 days, and LF = 13 days. Compute: (a) Early Finish, (b) Late Start, and (c) Total Float. Is this activity on the critical path? [3 marks]
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.
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
Criteria
Correct EF = 10 days using formula EF = ES + d
Criteria
Correct TF = 3 days using TF = LS − ES or TF = LF − EF
Criteria
Correct conclusion: activity is NOT on the critical path because TF ≠ 0
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
A project network has the following paths and 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. Determine: (a) project duration, (b) critical path(s), and (c) total float of Path 1. [3 marks]
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.
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
Criteria
Project duration correctly identified as 14 days (the longest path)
Criteria
Both critical paths A→C→D and A→B→E correctly identified
Criteria
Total float of Path 1 = 2 days correctly computed
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
Explain the forward pass and backward pass procedures in CPM. [3 marks]
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.
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
Criteria
Correct forward pass procedure: EF = ES + d, starting from ES = 0, using largest predecessor EF at merges
Criteria
Correct backward pass procedure: LS = LF − d, starting from project duration LF, using smallest successor LS at bursts
Criteria
Correct merge/burst rules stated explicitly (largest for forward pass, smallest for backward pass)
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
The activities on the critical path of a project have PERT variances of 4, 1, 4, and 2.25 days². Compute the standard deviation of the project duration. [2 marks]
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.
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
Criteria
Correct summation: σ²_project = 4 + 1 + 4 + 2.25 = 11.25 days²
Criteria
Correct square root: σ_project = √11.25 = 3.35 days
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
Activity X has a = 5, m = 8, b = 17. Activity Y has a = 2, m = 5, b = 8. Both are on the critical path. Find: (a) t_e and σ² for each activity, (b) the expected project duration, and (c) the project standard deviation. [5 marks]
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
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
Criteria
Correct t_e(X) = 9 days with formula and substitution shown
Criteria
Correct σ²(X) = 4 days² with formula and substitution shown
Criteria
Correct t_e(Y) = 5 days with formula and substitution shown
Criteria
Correct σ²(Y) = 1 day² with formula and substitution shown
Criteria
Correct project duration = 14 days AND project σ = 2.24 days with working shown
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
The following activities define a project (Activity — Duration — Predecessors): A — 4 days — None; B — 6 days — None; C — 5 days — A; D — 3 days — A, B; E — 4 days — C, D. Perform a complete forward pass and backward pass, identify the critical path, and compute total float for all activities. [5 marks]
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.
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
Criteria
Complete and correct forward pass for all 5 activities with merge rule applied at D and E
Criteria
Complete and correct backward pass for all 5 activities with burst rule applied at A
Criteria
Correct TF = 0 for all activities and both critical paths identified
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
What is activity crashing in CPM? State the principle used to select which activity to crash first. [2 marks]
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.
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
Criteria
Definition: crashing shortens duration by adding resources at increased cost
Criteria
Principle: select critical-path activity with minimum cost slope (cheapest per unit time)
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
List FOUR key differences between CPM and PERT. [2 marks]
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:
- Duration Estimates: CPM uses a single deterministic duration per activity; PERT uses three estimates (optimistic a, most likely m, pessimistic b) for probabilistic analysis.
- 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.
- Focus: CPM emphasizes time-cost trade-off (crashing); PERT emphasizes statistical probability of meeting a deadline.
- Output: CPM produces a definite project duration; PERT produces an expected duration with an associated standard deviation and probability distribution.
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
Criteria
Any two correct and distinct differences stated clearly
Criteria
Two additional correct and distinct differences stated clearly
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
An activity has ES = 4 days, duration d = 6 days. The project has LF = 14 days for this activity. Compute EF, LS, and TF. [3 marks]
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.
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
Criteria
EF = 10 days correctly computed
Criteria
LS = 8 days correctly computed
Criteria
TF = 4 days correctly computed with critical path status stated
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
The critical path of a PERT network has activities with the following data: Act. 1: a=2, m=4, b=6; Act. 2: a=1, m=3, b=5; Act. 3: a=3, m=5, b=13. Compute the probability that the project will be completed within 15 days. Use Z = (T_target − T_expected) / σ_project; Φ(1.0) = 0.8413. [5 marks]
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.
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
Criteria
All three t_e values correctly computed: 4, 3, 6 days
Criteria
All three σ² values correctly computed: 0.444, 0.444, 2.778 days²
Criteria
Project expected duration = 13 days correctly summed
Criteria
σ_project = 1.914 days correctly computed via variance sum and square root
Criteria
Z ≈ 1.0 computed and probability = 0.8413 (84.13%) stated with conclusion
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
A project network has activities A(2), B(4), C(3), D(5), E(2) with dependencies: C follows A and B; D follows B; E follows C and D. Compute: (a) all path durations, (b) the critical path and project duration, (c) the total float of activity A. [5 marks]
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
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
Criteria
All three paths correctly enumerated with correct durations (7, 9, 11 days)
Criteria
Critical path B→D→E and project duration 11 days correctly identified
Criteria
Correct forward pass for all activities with merge rule at C and E
Criteria
Correct backward pass for all activities with burst rule at B
Criteria
TF(A) = 4 days correctly computed
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
Mark wise strategy
State the answer directly with the key technical term or formula. No preamble needed. One clear sentence containing the examiner's target phrase is sufficient.
Do
- Use the exact technical term (e.g., 'longest path', 'zero total float')
- State the formula if the question is computational
- Write legibly and directly — one precise sentence
- Include units for any numerical answers
Don't
- Do not write a lengthy paragraph — time is wasted and marks are not gained
- Do not use vague language like 'the most important path'
- Do not omit key distinguishing terms (e.g., confusing 'total float' with 'free float')
For definition questions: give two distinguishable points or one definition with one formula. For numerical questions: write the formula, substitute, and solve. Each mark corresponds to one clearly identifiable element.
Do
- Show the formula explicitly before substituting values
- Number your points if answering a 'differentiate' or 'list' question
- Include units at every numerical step
- Use technical vocabulary from the CPM/PERT framework
Don't
- Do not repeat the same point twice in different words
- Do not skip the formula and go straight to numbers — risks zero marks if answer is wrong
- Do not write more than 5–6 lines — unnecessary content does not earn extra marks
Structure your answer as three explicit steps or parts, one per mark. For theory questions, use a paragraph per concept. For numerical, use the format: Given → Formula → Substitution → Answer → Conclusion.
Do
- Label your steps (Step 1, Step 2, Step 3) or answer parts (a), (b), (c) separately
- Show all intermediate arithmetic — partial marks are available
- State a conclusion sentence at the end (e.g., 'Activity is NOT on the critical path')
- Verify your answer using an alternative formula where possible (e.g., TF = LF − EF as a check)
Don't
- Do not combine all steps into one dense block of text
- Do not omit the conclusion — the interpretation of the result often carries a dedicated mark
- Do not round intermediate values excessively — carry at least 3 decimal places until the final answer
Treat a 5-mark question as five 1-mark questions in sequence. Structure your solution with clearly labeled sections (Forward Pass, Backward Pass, Float Table, Conclusion or Step 1 through Step 5 for PERT). Every section earns one mark. Show complete working — no steps can be skipped.
Do
- Begin by listing all given data in a table or list format
- Use section headings: 'Forward Pass', 'Backward Pass', 'Float Computation', 'Critical Path'
- For PERT problems, compute t_e and σ² for EACH activity separately before combining
- State the final answer as a boxed or underlined conclusion sentence
- Verify numerical answers using an alternative formula or path-sum check
Don't
- Do not skip the forward or backward pass and guess — they carry 2 marks each
- Do not sum standard deviations instead of variances in PERT problems
- Do not identify only one critical path when multiple paths tie — read the network carefully
- Do not leave the answer as a bare number — state its meaning (e.g., '... = 2.24 days, which is the project standard deviation')
General answer writing tips
- Always state the relevant formula FIRST before substituting values — examiners award a mark for the correct formula even if arithmetic errors follow.
- Label every variable you use (e.g., 'Let ES = Early Start, EF = Early Finish') — unlabeled variables confuse checkers and lose marks.
- For CPM network problems, always show the forward pass AND backward pass separately and in sequence; do not skip directly to float.
- Use box-and-arrow or table format when presenting activity data — organized data presentation signals professional competence and is rewarded.
- For PERT problems, compute t_e and σ² for each critical-path activity separately before summing — partial marks are awarded per activity.
- State your conclusion in a complete sentence (e.g., 'The critical path is A→C→E with a project duration of 18 days') — do not leave numerical answers without interpretation.
- When asked about float, always specify whether it is Total Float or Free Float — confusing the two is the single most common CPM mark deduction.
- Double-check your critical path by verifying that ALL activities on it have TF = 0; if even one has TF ≠ 0, re-examine your backward pass.
Diagrams
CPM Computation Procedure — Decision Flowchart
Step-by-step decision flowchart showing the complete CPM procedure from network setup through forward pass, backward pass, float computation, and critical path identification. The merge rule (MAX for forward pass) and burst rule (MIN for backward pass) are explicitly shown at decision nodes.
PERT Problem-Solving Procedure
Flowchart for solving PERT problems from raw data inputs to final probability output. Highlights the critical decision of which activities to include in the project variance sum (critical path only) and the correct sequence: sum variances THEN take square root.
CPM and PERT — Complete Topic Mind Map
Comprehensive mind map covering all major topics in the CPM/PERT chapter: network setup, forward and backward pass procedures, float types, critical path characteristics, crashing principles, and PERT three-estimate formulas. Use this as a pre-exam topic checklist.
Typical PRC Board Exam Mark Distribution — CPM/PERT Topic Areas
Estimated distribution of marks across CPM/PERT subtopics in PRC Civil Engineering board exams based on historical question patterns. Forward/backward pass and PERT computation together account for over half of all marks, making them the highest-priority study areas.