Decision theory and PERT/CPM
Decision criteria depend on what is known: certainty, risk (EMV) or uncertainty. In a project network, the critical path is the longest path and fixes the project duration.
Decision environments
| Environment | What is known | Tools |
|---|---|---|
| Certainty | The outcome of each act | Pick the best payoff; linear programming |
| Risk | The probabilities of the states | EMV, EOL, EVPI, decision trees |
| Uncertainty | Only the possible states, with no probabilities | Maximax, maximin, Hurwicz, Laplace, minimax regret |
| Conflict | An intelligent opponent | Game theory |
Criteria under uncertainty
| Criterion | Attitude | Rule |
|---|---|---|
| Maximax | Optimist | Take each act's best payoff; choose the highest |
| Maximin (Wald) | Pessimist | Take each act's worst payoff; choose the highest |
| Minimax regret (Savage) | Avoids regret | Regret = best payoff in that state − the act's payoff; choose the act with the smallest maximum regret |
| Hurwicz | Realist | α × best + (1 − α) × worst, where α is the coefficient of optimism; choose the highest |
| Laplace | Equal likelihood | Average each act's payoffs (principle of insufficient reason); choose the highest |
Worked example 1: one payoff table, six criteria
A Tiruppur garment exporter chooses between A1 (expand), A2 (add a shift) and A3 (no change). Payoffs in ₹ lakh for high, medium and low demand: A1 = 80, 40, −20; A2 = 60, 45, 5; A3 = 30, 30, 30. Use α = 0.6 for Hurwicz and probabilities 0.3, 0.5 and 0.2 for EMV.
- Maximax: best payoffs 80, 60, 30, so choose A1 (80).
- Maximin: worst payoffs −20, 5, 30, so choose A3 (30).
- Hurwicz: A1 = 0.6 × 80 + 0.4 × (−20) = 48 − 8 = 40; A2 = 36 + 2 = 38; A3 = 30. Choose A1.
- Laplace: A1 = 100 ÷ 3 = 33.33; A2 = 110 ÷ 3 = 36.67; A3 = 30. Choose A2.
- Regret: column bests are 80, 45, 30. Regrets: A1 = 0, 5, 50; A2 = 20, 0, 25; A3 = 50, 15, 0. Maximum regrets 50, 25, 50, so choose A2.
- EMV: A1 = 24 + 20 − 4 = 40; A2 = 18 + 22.5 + 1 = 41.5; A3 = 30. Choose A2 (₹41.5 lakh).
- EVPI: with perfect information, 0.3 × 80 + 0.5 × 45 + 0.2 × 30 = 52.5. EVPI = 52.5 − 41.5 = ₹11 lakh, equal to A2's EOL (0.3 × 20 + 0.2 × 25 = 11).
Maximax and Hurwicz pick A1; maximin picks A3; Laplace, minimax regret and EMV pick A2. EVPI = ₹11 lakh.
Network basics
- An activity uses time and resources; an event (node) is a point in time when activities start or end.
- Activity-on-arrow (AOA) networks may need a dummy activity: zero time and zero resources, drawn dotted, to show logic only.
- Fulkerson's rule numbers the events so that every arrow runs from a lower to a higher number.
- Forward pass: ES = the largest EF of the predecessors; EF = ES + t. Backward pass: LF = the smallest LS of the successors; LS = LF − t.
- Total float = LS − ES = LF − EF. Free float = earliest start of the successors − EF. Independent float ≤ free float ≤ total float.
- Critical activities have zero total float. The critical path is the longest path, and there can be more than one.
PERT vs CPM
Worked example 2: critical path and floats
Activities (days) and their predecessors: A 3 (none), B 4 (none), C 2 (A), D 6 (A), E 3 (B, C), F 4 (D, E). Find the critical path, the project time and the floats.
- Paths: A-D-F = 3 + 6 + 4 = 13; A-C-E-F = 3 + 2 + 3 + 4 = 12; B-E-F = 4 + 3 + 4 = 11.
- Forward pass (ES, EF): A 0, 3; B 0, 4; C 3, 5; D 3, 9; E starts at max(4, 5) = 5 and ends at 8; F starts at max(9, 8) = 9 and ends at 13.
- Backward pass (LS, LF): F 9, 13; E 6, 9; D 3, 9; C 4, 6; B 2, 6. For A, LF = min(4, 3) = 3, so LS = 0.
- Total float (LS − ES): A 0, B 2, C 1, D 0, E 1, F 0.
- Free float: B = ES of E − EF of B = 5 − 4 = 1; C = 5 − 5 = 0; E = 9 − 8 = 1.
The critical path is A-D-F, 13 days. B can slip 2 days without delaying the project, but only 1 day without delaying E.
Worked example 3: PERT time and probability
The critical path has three activities with (a, m, b) in days: P (2, 5, 14), Q (4, 7, 16), R (3, 6, 9). Find the expected project time and the chance of finishing within 26 days.
- tₑ: P = (2 + 20 + 14) ÷ 6 = 6; Q = (4 + 28 + 16) ÷ 6 = 8; R = (3 + 24 + 9) ÷ 6 = 6. Project Tₑ = 20 days.
- Variances [(b − a) ÷ 6]²: P = (12 ÷ 6)² = 4; Q = (12 ÷ 6)² = 4; R = (6 ÷ 6)² = 1. Project variance = 9, so σ = 3 days.
- z = (26 − 20) ÷ 3 = 2. The area up to z = 2 is 0.9772.
- For 23 days: z = (23 − 20) ÷ 3 = 1, giving 0.8413.
Expected time 20 days and σ = 3 days. P(finish within 26 days) ≈ 97.72%; P(within 23 days) ≈ 84.13%.
NET gives a small payoff table and asks which act each criterion picks (maximax, maximin, Hurwicz, Laplace, regret, EMV), or a short network and asks for the critical path, floats, PERT tₑ and variance, and the probability of finishing on time.
Typical MCQ: “Activity times are a = 4, m = 7 and b = 16. Find tₑ and its variance.” tₑ = (4 + 28 + 16) ÷ 6 = 8; variance = (12 ÷ 6)² = 4. Trap: using the simple average (9), or giving σ = 2 as the variance.
Typical MCQ: “Which criterion uses a coefficient of optimism?” Hurwicz. Trap: Laplace uses equal probabilities and Savage uses regret.
Typical MCQ: “Three critical activities have SDs of 1, 2 and 2 days. What is the project SD?” Variance = 1 + 4 + 4 = 9, so σ = 3 days. Trap: adding the SDs to get 5.
Optimist: maximax. Pessimist: maximin. Realist: Hurwicz. Fair-minded: Laplace. Regretful: Savage. In PERT, add variances, never SDs.