LESSON 2.4 — PERT and CPM Network Analysis

A. Standard Map

Topic	Governing Source	Exam Focus
CPM origin	Morgan Walker + James Kelley, DuPont, 1957	Year, organisation, deterministic model
PERT origin	US Navy Polaris Programme, 1958	Year, organisation, probabilistic model
AOA vs AON network	Standard project management	Activity representation; dummy activity logic
Fulkerson’s Rule	Standard CPM convention	Correct event numbering procedure
Forward / backward pass	IS / standard CPM	EST, EFT, LST, LFT definitions and calculation
Critical path	Standard CPM	Three simultaneous conditions; longest path
Float types	Standard CPM	Total float, free float, independent float
PERT time estimates	US Navy origin / standard PERT	te formula; variance formula
PERT probability	Standard normal distribution; Z-score	Project completion probability calculation
Gantt chart	H.L. Gantt, ~1910	Distinction from network; bar chart; no dependency logic

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B. Mechanism in Words

1. A project is broken into discrete activities with defined durations and logical dependencies.
2. Activities are arranged into a network diagram — the sequence of arrows represents the order in which work must happen.
3. The forward pass traces the network from start to finish, computing the earliest time each activity can start and finish.
4. The backward pass traces from finish to start, computing the latest time each activity can start and finish without delaying the project.
5. Activities where earliest and latest times coincide (zero float) form the critical path — delay any one of them and the project is delayed by the same amount.
6. Non-critical activities have float: time they can be delayed without affecting project completion.
7. PERT extends this logic to uncertain durations: instead of one time estimate, three estimates are used per activity; statistical analysis gives the project’s expected completion time and its probability distribution.

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C. Core Concept Explanations

C1. CPM vs PERT — Full Comparison (read this BEFORE doing any network math)

Property	CPM (Critical Path Method)	PERT (Programme Evaluation and Review Technique)
Year developed	1957	1958
Developed by	Morgan Walker + James Kelley at DuPont	US Navy for the Polaris ballistic missile programme
Project type	Repetitive industrial/construction — durations well known from historical data	R&D, defence — durations uncertain; no prior data
Time model	Deterministic — one single time estimate per activity	Probabilistic — three time estimates per activity (to, tm, tp)
Network orientation	Activity-oriented — arrows represent activities; nodes are junction events	Event-oriented — nodes represent events (milestones); arrows represent activities
Cost analysis	Yes — CPM includes cost-time trade-off (crashing)	No — PERT does not inherently include cost analysis
Float concept	Full float analysis (total, free, independent)	Less emphasis on float; focus is on probability of meeting a deadline
Output	Critical path, project duration, float for each activity	Expected project duration, standard deviation, probability of completion by a target date

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C2. AOA vs AON Networks

Property	Activity-on-Arrow (AOA)	Activity-on-Node (AON)
Activity representation	Arrow between two nodes	Node (box)
Event representation	Node (circle) = start/finish point	Arrow = dependency only
Dummy activities required?	Yes — sometimes needed to show logic without adding an unreal activity	No — logic is shown directly by arrows
CPM/PERT usage	Traditional PERT/CPM	Modern project management software (MS Project, Primavera)
Dummy activity	Dashed arrow; zero duration; zero resources; only a logical link	Not used

When is a dummy activity required in AOA?
A dummy is needed when two activities share the same start and end events (creating ambiguity) or when activity B depends on activity A but activity C depends on both A and D — and B does not depend on D.

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C3. Forward Pass — Earliest Times

The forward pass proceeds from the start event toward the end event, computing how early each activity can start and finish.

Symbol	Name	Calculation Rule
EST	Earliest Start Time	For the first activity: EST = 0. For all others: EST = maximum EFT of all immediate predecessor activities.
EFT	Earliest Finish Time	EFT = EST + Duration

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C4. Backward Pass — Latest Times

The backward pass proceeds from the end event toward the start event, computing how late each activity can start and finish without delaying the project.

Symbol	Name	Calculation Rule
LFT	Latest Finish Time	For the last activity: LFT = EFT (project completion time). For others: LFT = minimum LST of all immediate successor activities.
LST	Latest Start Time	LST = LFT − Duration

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C5. Critical Path — Definition and Three Conditions

An activity is critical if and only if all three conditions are simultaneously true:

Condition	Expression	Meaning
1	EST = LST	No slack at the start — cannot delay the start
2	EFT = LFT	No slack at the finish — cannot delay the finish
3	EFT − EST = LFT − LST = Duration	The two conditions above are consistent

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C6. Float Types

Float Type	Formula	Interpretation
Total Float (TF)	TF = LST − EST = LFT − EFT	Maximum delay to an activity without delaying project completion. May consume float shared with other activities.
Free Float (FF)	FF = min(EST of all successors) − EFT	Delay to an activity without delaying any immediate successor’s earliest start. Belongs exclusively to that activity.
Independent Float (IF)	IF = max(0, min(EST successors) − max(LFT predecessors) − Duration)	Delay possible under the most adverse predecessor/successor conditions. Never negative (set to zero if formula gives negative).

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C7. PERT Time Estimates and Formulas

For each activity in a PERT network, three time estimates are made:

Symbol	Name	Meaning
t_o	Optimistic time	Shortest possible duration if everything goes perfectly; probability ~1%
t_m	Most likely time	Duration under normal conditions; the mode of the beta distribution
t_p	Pessimistic time	Longest duration if everything goes wrong; probability ~1%

Expected time (te):

$$t_e = frac{t_o + 4t_m + t_p}{6}$$

The weight of 4 on t_m reflects its central importance in the beta distribution model used by PERT. The formula is a weighted average, not a simple arithmetic mean.

Variance (σ²) of an individual activity:

$$sigma^2 = left(frac{t_p – t_o}{6}right)^2$$

Project expected duration (T_E): Sum of t_e values along the critical path.

Project variance (V_E): Sum of σ² values of activities on the critical path only.

$$V_E = sum_{text{critical path}} sigma^2_i quad;quad sigma_{text{project}} = sqrt{V_E}$$

Probability of completing by a target date T_s:

$$Z = frac{T_s – T_E}{sqrt{V_E}}$$

Look up Z in the standard normal distribution table (Φ(Z)) to obtain the probability.

Key Z-values (standard normal table — required for exam):

Z	Probability Φ(Z)
0.84	80.0%
1.04	85.1%
1.28	90.0%
1.645	95.0%
2.05	98.0%
2.33	99.0%

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C8. Gantt Chart vs Network Diagram

Property	Gantt Chart	PERT/CPM Network
Invented by	Henry Gantt (~1910)	Walker/Kelley (1957) and US Navy (1958)
Representation	Bar chart: time on horizontal axis; activities as horizontal bars	Arrows and nodes representing logic and duration
Dependency logic	Not shown — dependencies between activities are not visible	Core feature — the entire network encodes precedence relationships
Critical path	Cannot be identified from a Gantt chart	Directly identified through forward/backward pass
Float visualisation	Can show float as extended bars	Computed precisely from float formulas
Best use	Communicating schedule to non-technical stakeholders; progress tracking	Planning, analysis, identifying critical activities, crashing

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D. Worked Problems

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PROBLEM 2.4-1 (CPM — Warm-Up, Linear Chain)

Activities in sequence: A(3) → B(5) → C(4) → D(2)

Activity	Duration	Predecessor	EST	EFT	LFT	LST	TF
A	3	—	0	3	3	0	0
B	5	A	3	8	8	3	0
C	4	B	8	12	12	8	0
D	2	C	12	14	14	12	0

Critical path: A–B–C–D. Project duration = 14 days. All floats = 0.

All activities are critical in a purely linear network — there are no parallel paths, so no float can exist.

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PROBLEM 2.4-2 (CPM — Fork and Merge, Float Calculation)

Activity	Duration	Predecessors
A	4	—
B	6	A
C	3	A
D	5	B, C

Forward pass:

Activity	EST	EFT
A	0	4
B	4	10
C	4	7
D	max(10, 7) = 10	15

Project duration = 15 days.

Backward pass (start from LFT_D = 15):

Activity	LFT	LST
D	15	10
B	min(LST_D) = 10	4
C	min(LST_D) = 10	7
A	min(LST_B, LST_C) = min(4, 7) = 4	0

Float table:

Activity	EST	LST	TF = LST − EST	Critical?
A	0	0	0	✓ Yes
B	4	4	0	✓ Yes
C	4	7	3	No — 3 days float
D	10	10	0	✓ Yes

Critical path: A–B–D. Project duration = 15 days.
Activity C has 3 days of total float. It can be delayed up to 3 days without affecting project completion.

Path check: A–B–D = 4+6+5 = 15 ✓; A–C–D = 4+3+5 = 12 ✓

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PROBLEM 2.4-3 (CPM — Two Independent Start Activities)

Activity	Duration	Predecessors
A	3	—
B	4	—
C	5	A
D	2	B
E	7	C, D

Forward pass:

Activity	EST	EFT
A	0	3
B	0	4
C	3	8
D	4	6
E	max(8, 6) = 8	15

Project duration = 15 days.

Backward pass:

Activity	LFT	LST
E	15	8
C	min(LST_E) = 8	3
D	min(LST_E) = 8	6
A	min(LST_C) = 3	0
B	min(LST_D) = 6	2

Float table:

Activity	EST	LST	TF	Critical?
A	0	0	0	✓ Yes
B	0	2	2	No — 2 days float
C	3	3	0	✓ Yes
D	4	6	2	No — 2 days float
E	8	8	0	✓ Yes

Critical path: A–C–E. Duration = 15 days.
B and D both have 2 days float each.

Path check: A–C–E = 3+5+7 = 15 ✓; B–D–E = 4+2+7 = 13 ✓

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PROBLEM 2.4-4 (CPM — Multi-Predecessor Network, Free Float vs Total Float)

Activity	Duration	Predecessors
A	4	—
B	2	—
C	3	A
D	5	A
E	4	B, C
F	3	D, E

Forward pass:

Activity	EST	EFT
A	0	4
B	0	2
C	4	7
D	4	9
E	max(2, 7) = 7	11
F	max(9, 11) = 11	14

Project duration = 14 days.

Backward pass:

Activity	LFT	LST
F	14	11
E	min(LST_F) = 11	7
D	min(LST_F) = 11	6
C	min(LST_E) = 7	4
B	min(LST_E) = 7	5
A	min(LST_C, LST_D) = min(4, 6) = 4	0

Float table:

Activity	EST	LST	TF	EFT	min(EST successors)	FF = min(EST_succ) − EFT	Critical?
A	0	0	0	4	min(EST_C, EST_D) = 4	4−4 = 0	✓ Yes
B	0	5	5	2	EST_E = 7	7−2 = 5	No
C	4	4	0	7	EST_E = 7	7−7 = 0	✓ Yes
D	4	6	2	9	EST_F = 11	11−9 = 2	No
E	7	7	0	11	EST_F = 11	11−11 = 0	✓ Yes
F	11	11	0	14	—	—	✓ Yes

Critical path: A–C–E–F. Duration = 14 days.

Observations:
– Activity B: TF = 5, FF = 5. Both equal because E’s EST (7) is not delayed by B’s scheduling.
– Activity D: TF = 2, FF = 2. D can slip 2 days without affecting F.

Path check: A–C–E–F = 4+3+4+3 = 14 ✓; A–D–F = 4+5+3 = 12 ✓; B–E–F = 2+4+3 = 9 ✓

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PROBLEM 2.4-5 (CPM — Complex Network, Identify Critical Path from Multiple Paths)

Activity	Duration	Predecessors
A	4	—
B	3	—
C	2	A
D	5	A, B
E	4	C
F	6	C, D
G	3	E, F

Forward pass:

Activity	EST	EFT
A	0	4
B	0	3
C	4	6
D	max(4, 3) = 4	9
E	6	10
F	max(6, 9) = 9	15
G	max(10, 15) = 15	18

Project duration = 18 days.

Backward pass:

Activity	LFT	LST
G	18	15
F	min(LST_G) = 15	9
E	min(LST_G) = 15	11
D	min(LST_F) = 9	4
C	min(LST_E, LST_F) = min(11, 9) = 9	7
B	min(LST_D) = 4	1
A	min(LST_C, LST_D) = min(7, 4) = 4	0

Float table:

Activity	EST	LST	TF	Critical?
A	0	0	0	✓ Yes
B	0	1	1	No
C	4	7	3	No
D	4	4	0	✓ Yes
E	6	11	5	No
F	9	9	0	✓ Yes
G	15	15	0	✓ Yes

Critical path: A–D–F–G. Duration = 18 days.

Notable non-critical activities:
– B: TF = 1 (project can afford a 1-day slip in B)
– C: TF = 3 (3-day float)
– E: TF = 5 (largest float in the network)

Path check: A–D–F–G = 4+5+6+3 = 18 ✓; A–C–F–G = 4+2+6+3 = 15; A–C–E–G = 4+2+4+3 = 13; B–D–F–G = 3+5+6+3 = 17

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C9. PERT Probability — Worked Problems

Z-table reminder (provide in exam answer when using):
Z = 1.28 → P = 90%; Z = 1.645 → P = 95%; Z = 0.84 → P = 80%

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PERT PROBLEM 1 — Expected Duration and Probability

Three activities X, Y, Z lie on the critical path with the following estimates:

Activity	t_o	t_m	t_p	t_e = (t_o + 4t_m + t_p)/6	σ² = ((t_p − t_o)/6)²
X	2	4	6	(2 + 16 + 6)/6 = 4.00	((6−2)/6)² = (4/6)² = 0.44
Y	3	5	9	(3 + 20 + 9)/6 = 5.33	((9−3)/6)² = (6/6)² = 1.00
Z	1	2	3	(1 + 8 + 3)/6 = 2.00	((3−1)/6)² = (2/6)² = 0.11

Expected project duration: T_E = 4.00 + 5.33 + 2.00 = 11.33 days

Project variance: V_E = 0.44 + 1.00 + 0.11 = 1.55

Standard deviation: σ_project = √1.55 = 1.25 days

Q: What is the probability of completing the project within 13 days?

Z = (T_s − T_E) / √V_E = (13 − 11.33) / 1.25 = 1.67 / 1.25 = Z = 1.34

From standard normal table: Φ(1.34) ≈ 0.9099 → probability ≈ 91%

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PERT PROBLEM 2 — Project Variance Dominated by One Activity

Four critical path activities P, Q, R, S:

Activity	t_o	t_m	t_p	t_e	σ²
P	4	6	8	(4 + 24 + 8)/6 = 6.00	((8−4)/6)² = 0.44
Q	2	4	12	(2 + 16 + 12)/6 = 5.00	((12−2)/6)² = (10/6)² = 2.78
R	3	5	7	(3 + 20 + 7)/6 = 5.00	((7−3)/6)² = 0.44
S	1	2	3	(1 + 8 + 3)/6 = 2.00	((3−1)/6)² = 0.11

T_E = 6 + 5 + 5 + 2 = 18.00 days

V_E = 0.44 + 2.78 + 0.44 + 0.11 = 3.77; σ = √3.77 = 1.94 days

Q: What is the probability of completing by day 20?

Z = (20 − 18) / 1.94 = 2.00 / 1.94 = Z = 1.03

Φ(1.03) ≈ 0.848 → probability ≈ 85%

Q: What is the probability of completing by day 22?

Z = (22 − 18) / 1.94 = 4.00 / 1.94 = Z = 2.06

Φ(2.06) ≈ 0.980 → probability ≈ 98%

Observation: Activity Q (pessimistic time = 12, optimistic = 2) dominates the project variance (σ² = 2.78 of 3.77 total = 74%). Reducing uncertainty in Q would most effectively reduce project risk.

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PERT PROBLEM 3 — Find Target Date for Required Probability

A project has three critical path activities:

Activity	t_o	t_m	t_p	t_e	σ²
α	2	4	6	(2+16+6)/6 = 4.00	(4/6)² = 0.44
β	1	3	5	(1+12+5)/6 = 3.00	(4/6)² = 0.44
γ	3	4	11	(3+16+11)/6 = 5.00	(8/6)² = 1.78

T_E = 4 + 3 + 5 = 12.00 days

V_E = 0.44 + 0.44 + 1.78 = 2.67; σ = √2.67 = 1.63 days

Q: What is the target completion date that gives a 90% probability of meeting it?

For P = 90%: Z = 1.28

T_s = T_E + Z × σ = 12 + 1.28 × 1.63 = 12 + 2.09 = 14.09 ≈ 14.1 days

The project should be scheduled for 14.1 days (round up to 15 days in practice) to have a 90% chance of completion on time.

Q: What target date gives 95% confidence?

Z = 1.645: T_s = 12 + 1.645 × 1.63 = 12 + 2.68 = 14.68 ≈ 15 days

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E. Common Confusions

Confusion	Correct Distinction
CPM before PERT	CPM (1957, DuPont) preceded PERT (1958, US Navy). They were not developed simultaneously.
Critical path = shortest path	Critical path = LONGEST path through the network. It equals the minimum project duration.
PERT variance divides by 2	σ² = ((t_p − t_o)/6)². The denominator is 6, not 2.
All activities on the critical path have the same duration	Critical path is the longest path; individual activity durations on it vary. What they share is zero float.
Project variance = variance of longest path activity	Project variance = SUM of variances of ALL critical path activities. Each activity’s σ² is added.
Free float = Total float	FF ≤ TF. They are equal only when the head event of the activity is on the critical path (head event slack = 0).
Gantt chart shows dependencies	Gantt charts are bar charts — they show duration and schedule but NOT precedence (logical) relationships between activities.
PERT is more modern / advanced	CPM and PERT are contemporaries (1957–1958). Neither is “newer” in a meaningful way. They serve different purposes.

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F. Exam Traps

Trap	Incorrect Assumption	Correct Answer
T19	PERT was developed before CPM	CPM (1957) came first. PERT (1958) followed by approximately one year.
T20	σ² = ((t_p − t_o) / 2)²	The correct formula divides by 6: σ² = ((t_p − t_o)/6)² — this is a direct GATE-level trap.
T21	The critical path is the path with the fewest activities	Critical path = LONGEST total duration. Activity count is irrelevant. A 2-activity path at 20 days is more critical than a 10-activity path at 15 days.
T22	Project variance includes non-critical activities	Only critical path activities contribute to project variance (V_E). Non-critical activities are excluded.
T23	Dummy activity has a short duration	A dummy activity has ZERO duration and ZERO resource use. It is a logical link only.
T24	A Gantt chart can identify the critical path	Gantt charts cannot identify the critical path — they have no dependency logic. Only a PERT/CPM network can.

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G. Answer-Writing Cues

For CPM vs PERT identification:

For PERT probability:

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H. PYQ Linkage Note

Topic	Exam Appearance	Pattern
CPM year + organization	GATE, UPSC-CPWD, state PSC	MCQ: “CPM was developed in _ by _”
PERT year + organization	Same exams	MCQ: “PERT was developed for ____”
te formula	GATE, ISRO, UPSC-CPWD	MCQ: recall formula; NAT: compute te given three time estimates
σ² formula (denominator = 6)	GATE multiple years	MCQ: identify correct formula; NAT: compute σ²
Critical path identification	GATE (NAT)	NAT: given durations, find project duration; identify critical activities
Float calculation	GATE	MCQ: which activity has the most float; NAT: compute TF for a named activity
PERT probability	GATE, UPSC-CPWD	NAT: compute Z; find probability or target date
Gantt vs network	UPSC-CPWD	MCQ: “Which technique shows activity dependencies?”

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I. Mini-Check — Lesson 2.4 (5 Questions)

Q1 (MCQ): CPM was developed in:
(A) 1958 by the US Navy (B) 1957 by DuPont (C) 1960 by NASA (D) 1958 by DuPont

A1: (B) 1957 by DuPont. CPM was developed first (1957, Morgan Walker + James Kelley, DuPont). PERT came second (1958, US Navy Polaris). Options A and D reverse the year or organisation.

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Q2 (MCQ): An activity in a PERT network has t_o = 2, t_m = 5, t_p = 14. What is its expected time t_e?
(A) 7 (B) 6 (C) 5.33 (D) 8

A2: (B) 6. t_e = (t_o + 4t_m + t_p)/6 = (2 + 20 + 14)/6 = 36/6 = 6 days.

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Q3 (NAT): A project’s critical path has three activities with variances σ² = 0.44, σ² = 1.00, and σ² = 0.25. Expected project duration T_E = 20 days. What is the probability of completing the project by day 22? (Use Z = 1.57 → Φ = 0.942. Answer as a percentage, one decimal place.)

A3:
– V_E = 0.44 + 1.00 + 0.25 = 1.69
– σ = √1.69 = 1.30 days
– Z = (22 − 20) / 1.30 = 2.00 / 1.30 = 1.54
– Φ(1.54) ≈ 0.938 → P ≈ 93.8%

(Note: Using the given hint Z=1.57 implies slightly different rounding: Z = 2/√1.69 = 2/1.30 = 1.538 ≈ 1.54; Φ(1.54) = 93.8%.)

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Q4 (NAT): A network has four activities. Find the project duration and identify the critical path.

Activity	Duration	Predecessors
A	5	—
B	4	—
C	6	A, B
D	3	A

(Project ends when both C and D are complete.)

A4:
Forward pass:
– EST_A = 0, EFT_A = 5; EST_B = 0, EFT_B = 4
– EST_C = max(5, 4) = 5, EFT_C = 11
– EST_D = 5, EFT_D = 8
– Project duration = max(EFT_C, EFT_D) = max(11, 8) = 11 days

Backward pass (LFT of both terminal activities = 11):
– LFT_C = 11, LST_C = 5; LFT_D = 11, LST_D = 8
– LFT_A = min(LST_C, LST_D) = min(5, 8) = 5, LST_A = 0
– LFT_B = min(LST_C) = 5, LST_B = 1

Floats: A=0 ✓; B=1; C=0 ✓; D=3

Project duration = 11 days. Critical path: A–C.

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Q5 (MCQ): Which statement correctly distinguishes Total Float from Free Float?
(A) Total Float ≤ Free Float always
(B) Free Float is the delay allowable without delaying the project; Total Float is the delay allowable without delaying any successor
(C) Free Float ≤ Total Float; Free Float is the delay allowable without delaying any immediate successor’s earliest start
(D) Total Float and Free Float are identical for all activities on the critical path and off it

A5: (C). Free Float ≤ Total Float (always). Free Float = delay to an activity without affecting any immediate successor’s earliest start. Total Float = delay without affecting project completion (may consume float shared with other activities). On the critical path, both = 0 — making (D) half-true but misleading for non-critical activities.

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