10.3 Earned Value Management (EVM), Decision Trees, and Risk Preference

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2 Chapter 2: Mathematics for Industrial Engineers

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9 Chapter 9: Stochastic Models & Simulation

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10 Chapter 10: Engineering Management, Project Planning & Performance Measurement

10.1 Organizational Structure, Leadership, and Worker Motivation Theories 10.2 Project Management: WBS, Critical Path Method (CPM), PERT, and Crashing 10.3 Earned Value Management (EVM), Decision Trees, and Risk Preference 10.4 Performance Measurement: Productivity, KPIs, Wage Systems, and the Balanced Scorecard

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12 Chapter 12: Production Planning, Inventory Control & Sequencing

12.1 Demand Forecasting Methods and Forecast Error Tracking 12.2 Aggregate Planning, Master Production Scheduling (MPS), and MRP 12.3 Deterministic Inventory Models: EOQ, EPQ, and Safety Stock 12.4 Job Sequencing, Dispatching Rules, and Flow Shop Scheduling

13 Chapter 13: Lean Systems & Theory of Constraints

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14 Chapter 14: Facilities Planning & Material Handling

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15 Chapter 15: Supply Chain Design & Logistics

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16 Chapter 16: Human Factors & Cognitive Ergonomics

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17 Chapter 17: Physical Ergonomics & Workplace Safety

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18 Chapter 18: Work Design & Measurement

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19 Chapter 19: Quality Management, Six Sigma & Analytical Tools

19.1 Quality Philosophies, Total Quality Management, and Cost of Quality 19.2 Six Sigma DMAIC Methodology and Quality Function Deployment 19.3 The Seven Basic Quality Tools, Root Cause Analysis, and Taguchi Loss Function

20 Chapter 20: Statistical Process Control

20.1 Statistical Process Control Principles, Variation, and Run Rules 20.2 Variable Control Charts: X-bar & R, X-bar & S, and Individual-Moving Range 20.3 Attribute Control Charts: Defectives (p, np) and Defects (c, u)

21 Chapter 21: Process Capability & Acceptance Sampling

21.1 Process Capability Indices and Defect Levels 21.2 Acceptance Sampling Concepts and Operating Characteristic Curves 21.3 Sampling Risks, AQL, LTPD, and Standard Sampling Plans

22 Chapter 22: Systems Engineering & Life-Cycle Management

22.1 The Systems Engineering V-Model, Stakeholder Requirements, and Functional Architecture 22.2 Configuration Management, Interface Control, and Verification & Validation 22.3 Life-Cycle Costing, Maintainability, and System Availability

23 Chapter 23: Reliability Engineering & Risk Analysis

23.1 Reliability Mathematics, Hazard Rates, and the Bathtub Curve 23.2 System Reliability Configurations: Series, Parallel, and k-out-of-n Systems 23.3 Failure Modes and Effects Analysis and Fault Tree Analysis

10.3 Earned Value Management (EVM), Decision Trees, and Risk Preference

Key Takeaways Explain This Topic AI

Earned Value Management (EVM) measures performance through three primary baseline metrics: Planned Value (PV = % planned × BAC), Earned Value (EV = % actual × BAC), and Actual Cost (AC).

Cost Variance (CV = EV - AC) and Schedule Variance (SV = EV - PV) indicate monetary overruns/underruns and schedule leads/delays; CPI = EV/AC and SPI = EV/PV define operational efficiency indices where values > 1.0 indicate favorable performance.

Under decision-making under complete uncertainty, non-probabilistic criteria include Maximax (optimistic maximum of maximums), Maximin (conservative maximum of minimums), Hurwicz (weighted realism criterion), and Minimax Regret (minimizing maximum opportunity loss).

Decision trees evaluate sequential choices by rolling back backward from terminal payoffs: calculating Expected Monetary Value (EMV = Σ p_j × X_ij) at chance nodes and selecting the maximum EMV alternative at decision nodes.

Expected Value of Perfect Information (EVPI = EV|PI - EMV_max) defines the maximum economic value of resolving uncertainty before committing to a decision, while utility curvature reflects risk-averse (concave), risk-neutral (linear), or risk-seeking (convex) attitudes.

Last updated: September 2026

10.3 Earned Value Management (EVM), Decision Trees, and Risk Preference

Controlling engineering projects requires real-time monitoring of cost and schedule health alongside formal mathematical decision frameworks to evaluate capital allocation under uncertainty. Earned Value Management (EVM) provides standardized, objective metrics to detect cost overruns and schedule slippages before they become irreversible. Concurrently, decision analysis—utilizing decision payoff tables, decision trees, Expected Monetary Value (EMV), Expected Value of Perfect Information (EVPI), and utility theory—enables engineering leaders to select optimal technological alternatives under varying risk attitudes.

1. Earned Value Management (EVM) Core Parameters

Earned Value Management integrates project scope, calendar schedule, and cost baselines. EVM tracks performance by comparing what was planned, what was physically accomplished, and what was actually spent. EVM Three Core Dimensions

• 1. Planned Value (PV / BCWS) ──> What we PLANNED to accomplish by today (Budgeted Cost)
• 2. Earned Value (EV / BCWP) ──> What we ACTUALLY accomplished by today (in Budgeted Terms)
• 3. Actual Cost (AC / ACWP) ──> What we ACTUALLY spent to accomplish that work
• Foundational EVM Terminology

Budget at Completion ( B A C BAC B A C ) : The total approved baseline budget allocated for the complete scope of project work.

Planned Value ( P V PV P V , historically BCWS - Budgeted Cost of Work Scheduled) : The authorized budget assigned to scheduled work up to the measurement date: P V = ( % Planned Complete ) × B A C PV = (\% \text{ Planned Complete}) \times BAC P V = ( % Planned Complete ) × B A C

Earned Value ( E V EV E V , historically BCWP - Budgeted Cost of Work Performed) : The budgeted value of work physically completed up to the measurement date: E V = ( % Actual Physical Complete ) × B A C EV = (\% \text{ Actual Physical Complete}) \times BAC E V = ( % Actual Physical Complete ) × B A C

Actual Cost ( A C AC A C , historically ACWP - Actual Cost of Work Performed) : The total realized direct and indirect expenditure incurred in executing the work completed up to the measurement date.

Variance Formulations

Variances measure deviations from the performance baseline in dollars. In EVM, positive values are favorable , while negative values are unfavorable :

Cost Variance ( C V CV C V ) : C V = E V − A C CV = EV - AC C V = E V − A C

C V > 0 CV > 0 C V > 0 : Project is under budget (favorable cost efficiency).

C V = 0 CV = 0 C V = 0 : Project is exactly on budget.

C V < 0 CV < 0 C V < 0 : Project is over budget (unfavorable cost overrun).

Schedule Variance ( S V SV S V ) : S V = E V − P V SV = EV - PV S V = E V − P V

S V > 0 SV > 0 S V > 0 : Project is ahead of schedule (accomplished more budgeted work than scheduled).

S V = 0 SV = 0 S V = 0 : Project is exactly on schedule.

S V < 0 SV < 0 S V < 0 : Project is behind schedule (accomplished less budgeted work than scheduled).

Performance Indices

Performance indices express project efficiency as dimensionless ratios. Values > 1.0 > 1.0 > 1.0 are favorable , while values < 1.0 < 1.0 < 1.0 are unfavorable :

Cost Performance Index ( C P I CPI CP I ) : C P I = E V A C CPI = \frac{EV}{AC} CP I = A C E V ​ Interpretation : A C P I CPI CP I of 0.85 0.85 0.85 means that for every $1.00 spent, the project achieves only $0.85 of budgeted value.

Schedule Performance Index ( S P I SPI SP I ) : S P I = E V P V SPI = \frac{EV}{PV} SP I = P V E V ​ Interpretation : An S P I SPI SP I of 1.15 1.15 1.15 means that work is progressing at 115% of the scheduled rate.

• 2. EVM Forecasting & To-Complete Performance Index (TCPI)
• Forecasting Estimate at Completion ( E A C EAC E A C )

The Estimate at Completion ( E A C EAC E A C ) forecasts the total expected project cost upon final delivery, depending on assumptions regarding future performance:

Case 1: Past variances are typical (historical C P I CPI CP I will continue for remaining work) : E A C = B A C C P I EAC = \frac{BAC}{CPI} E A C = CP I B A C ​

Case 2: Past variances were atypical anomalies (future remaining work will proceed at the original budgeted rate) : E A C = A C + ( B A C − E V ) EAC = AC + (BAC - EV) E A C = A C + ( B A C − E V )

Case 3: Both cost and schedule performance influence remaining work : E A C = A C + B A C − E V C P I × S P I EAC = AC + \frac{BAC - EV}{CPI \times SPI} E A C = A C + CP I × SP I B A C − E V ​

Estimate to Complete ( E T C ETC ETC ) and Variance at Completion ( V A C VAC V A C )

Estimate to Complete ( E T C ETC ETC ) : The projected cost required to finish all remaining work: E T C = E A C − A C ETC = EAC - AC ETC = E A C − A C

Variance at Completion ( V A C VAC V A C ) : V A C = B A C − E A C VAC = BAC - EAC V A C = B A C − E A C

V A C > 0 VAC > 0 V A C > 0 : Favorable projected cost surplus.

V A C < 0 VAC < 0 V A C < 0 : Unfavorable projected budget deficit.

To-Complete Performance Index ( T C P I TCPI TCP I )

T C P I TCPI TCP I calculates the cost efficiency required on all remaining work to achieve a targeted budget objective:

To achieve the original budget ( B A C BAC B A C ) : T C P I B A C = Remaining Work Remaining Funds = B A C − E V B A C − A C TCPI_{BAC} = \frac{\text{Remaining Work}}{\text{Remaining Funds}} = \frac{BAC - EV}{BAC - AC} TCP I B A C ​ = Remaining Funds Remaining Work ​ = B A C − A C B A C − E V ​

To achieve a revised management forecast ( E A C EAC E A C ) : T C P I E A C = B A C − E V E A C − A C TCPI_{EAC} = \frac{BAC - EV}{EAC - AC} TCP I E A C ​ = E A C − A C B A C − E V ​

Rule : If T C P I > 1.0 TCPI > 1.0 TCP I > 1.0 , the project team must perform with higher cost efficiency than originally planned; if T C P I > 1.10 TCPI > 1.10 TCP I > 1.10 , reaching the target is typically considered unachievable without reducing scope.

3. Decision Analysis Under Complete Uncertainty (Non-Probabilistic)

When decision alternatives A i A_i A i ​ must be selected across various states of nature S j S_j S j ​ without objective probability distributions ( p j p_j p j ​ unknown), engineering managers utilize non-probabilistic decision criteria applied to a payoff matrix ( X i j X_{ij} X ij ​ , where X i j X_{ij} X ij ​ represents profit or net return):

Alternative State S 1 S_1 S 1 ​ State S 2 S_2 S 2 ​ … \dots … State S n S_n S n ​ Criterion Evaluation

Alternative A 1 A_1 A 1 ​ X 11 X_{11} X 11 ​ X 12 X_{12} X 12 ​ … \dots … X 1 n X_{1n} X 1 n ​ Evaluated per criterion

Alternative A 2 A_2 A 2 ​ X 21 X_{21} X 21 ​ X 22 X_{22} X 22 ​ … \dots … X 2 n X_{2n} X 2 n ​ Evaluated per criterion

Classical Decision Criteria

Maximax Criterion (Optimistic / Aggressive) :

Assumes the best possible state of nature will occur for whichever alternative is selected. Identify the maximum payoff for each alternative, then choose the alternative that maximizes the maximums: Select A ∗ = arg ⁡ max ⁡ i ( max ⁡ j X i j ) \text{Select } A^* = \arg\max_i \left( \max_j X_{ij} \right) Select A ∗ = ar g i max ​ ( j max ​ X ij ​ )

Maximin Criterion (Pessimistic / Conservative / Wald's Criterion) :

Guarantees the best worst-case outcome. Identify the minimum payoff for each alternative, then choose the alternative that maximizes the minimums: Select A ∗ = arg ⁡ max ⁡ i ( min ⁡ j X i j ) \text{Select } A^* = \arg\max_i \left( \min_j X_{ij} \right) Select A ∗ = ar g i max ​ ( j min ​ X ij ​ )

Hurwicz Criterion (Criterion of Realism) :

Strikes a weighted balance between absolute optimism and pessimism using an index of optimism α ∈ [ 0 , 1 ] \alpha \in [0, 1] α ∈ [ 0 , 1 ] : Score ( A i ) = α × [ max ⁡ j ( X i j ) ] + ( 1 − α ) × [ min ⁡ j ( X i j ) ] \text{Score}(A_i) = \alpha \times [\max_j(X_{ij})] + (1 - \alpha) \times [\min_j(X_{ij})] Score ( A i ​ ) = α × [ j max ​ ( X ij ​ )] + ( 1 − α ) × [ j min ​ ( X ij ​ )] Select A ∗ = arg ⁡ max ⁡ i [ Score ( A i ) ] \text{Select } A^* = \arg\max_i [\text{Score}(A_i)] Select A ∗ = ar g i max ​ [ Score ( A i ​ )] (When α = 1 \alpha = 1 α = 1 , Hurwicz collapses to Maximax; when α = 0 \alpha = 0 α = 0 , it collapses to Maximin).

Minimax Regret Criterion (Savage Criterion / Opportunity Loss) :

Minimizes the maximum regret (opportunity loss) of making the wrong choice.

Step 1: Find the maximum payoff in each state of nature column: X j ∗ = max ⁡ i ( X i j ) X_j^* = \max_i(X_{ij}) X j ∗ ​ = max i ​ ( X ij ​ ) .

Step 2: Compute the regret (opportunity loss) matrix: R i j = X j ∗ − X i j R_{ij} = X_j^* - X_{ij} R ij ​ = X j ∗ ​ − X ij ​ .

Step 3: For each alternative A i A_i A i ​ , find its maximum regret across all states: Max Regret i = max ⁡ j ( R i j ) \text{Max Regret}_i = \max_j(R_{ij}) Max Regret i ​ = max j ​ ( R ij ​ ) .

Step 4: Select the alternative with the minimum of these maximum regrets: Select A ∗ = arg ⁡ min ⁡ i ( max ⁡ j R i j ) \text{Select } A^* = \arg\min_i \left( \max_j R_{ij} \right) Select A ∗ = ar g i min ​ ( j max ​ R ij ​ )

Laplace Criterion (Principle of Insufficient Reason) :

Assumes all states of nature are equally likely ( p j = 1 / n p_j = 1/n p j ​ = 1/ n ). Compute the simple arithmetic average payoff for each alternative and select the maximum: Select A ∗ = arg ⁡ max ⁡ i ( 1 n ∑ j = 1 n X i j ) \text{Select } A^* = \arg\max_i \left( \frac{1}{n} \sum_{j=1}^n X_{ij} \right) Select A ∗ = ar g i max ​ ( n 1 ​ j = 1 ∑ n ​ X ij ​ )

4. Decision Trees, Backward Induction, and EVPI

When decision problems involve sequential choices or objective probabilistic distributions, decision trees provide a graphical and computational framework. Decision Tree Standard Symbology

┌───────┐

│ □ │ Decision Node (Square): Controlled choice; branches are alternatives.

└───────┘ Value = Maximum EMV among departing branches (Prune suboptimal branches).

○ Chance Node (Circle): Uncertain event; branches are states of nature (probs sum to 1.0).

Value = Expected Monetary Value (EMV = Σ p_j * X_j).

▷ Terminal Node (Triangle): Final payoff outcome.

Expected Monetary Value ( E M V EMV EM V ) and Rollback Procedure

Expected Monetary Value ( E M V EMV EM V ) : The probability-weighted average payoff for alternative A i A_i A i ​ across mutually exclusive states S j S_j S j ​ with probabilities p j p_j p j ​ (where ∑ p j = 1.0 \sum p_j = 1.0 ∑ p j ​ = 1.0 ): E M V ( A i ) = ∑ j = 1 m p j X i j EMV(A_i) = \sum_{j=1}^m p_j X_{ij} EM V ( A i ​ ) = j = 1 ∑ m ​ p j ​ X ij ​

Rollback (Backward Induction) Method :

Draw the tree left-to-right from the initial decision node to terminal outcomes.

Assign probabilities to chance branches and payoffs to terminal endpoints.

Solve the tree right-to-left (backward induction):

At each chance node (circle) , calculate the E M V EMV EM V by taking the expected value of all departing branches.

At each decision node (square) , compare the values of all departing alternative branches, select the branch with the highest E M V EMV EM V (for profits) or lowest expected cost, and prune the suboptimal branches.

Propagate the optimal values backward to the root decision node.

Expected Value of Perfect Information ( E V P I EVPI E V P I )

E V P I EVPI E V P I quantifies the maximum amount of money an engineering organization should pay to eliminate uncertainty (e.g., via market research, destructive testing, or pilot plants) before committing to a decision.

Expected Value with Perfect Information ( E V ∣ P I EV|PI E V ∣ P I ) :

The expected payoff if the decision maker knew with 100% certainty which state of nature would occur prior to making the choice: E V ∣ P I = ∑ j = 1 m p j × ( max ⁡ i X i j ) EV|PI = \sum_{j=1}^m p_j \times \left( \max_i X_{ij} \right) E V ∣ P I = j = 1 ∑ m ​ p j ​ × ( i max ​ X ij ​ )

Expected Value without Information ( E M V ∗ EMV^* EM V ∗ ) :

The maximum expected monetary value achievable under basic uncertainty: E M V ∗ = max ⁡ i { E M V ( A i ) } EMV^* = \max_i \{ EMV(A_i) \} EM V ∗ = i max ​ { EM V ( A i ​ )}

Formula for E V P I EVPI E V P I : E V P I = E V ∣ P I − E M V ∗ EVPI = EV|PI - EMV^* E V P I = E V ∣ P I − EM V ∗

Core Rule : E V P I EVPI E V P I can never be negative ( E V P I ≥ 0 EVPI \ge 0 E V P I ≥ 0 ). Furthermore, E V P I EVPI E V P I exactly equals the Expected Opportunity Loss ( E O L EOL EO L ) of the optimal decision under uncertainty.

5. Risk Attitudes and Utility Theory

Expected Monetary Value assumes decision makers are risk-neutral. However, real engineering managers and corporate executives exhibit subjective risk preferences , modeled by utility functions U ( X ) U(X) U ( X ) . Utility Function Curvature Shapes

• Utility U(X) ┌ Risk-Averse (Concave, U'' < 0)
• │ . - ' '
• │ . ' / Risk-Neutral (Linear, U'' = 0)
• │ . ' /
• │ / . '
• │ . ' Risk-Seeking (Convex, U'' > 0)
• └───────────────────► Wealth / Payoff ($)
• The Three Risk Attitudes

Risk-Averse (Concave Utility Function, U ′ ′ ( X ) < 0 U''(X) < 0 U ′′ ( X ) < 0 ) :

Exhibits diminishing marginal utility of wealth.

Certainty Equivalent ( C E CE CE ) : The guaranteed cash amount that the decision maker regards as exactly equivalent to an uncertain gamble.

For a risk-averse individual: C E < E M V CE < EMV CE < EM V .

Risk Premium ( R P RP RP ) : The amount of expected payoff a decision maker will forfeit to guarantee an outcome: R P = E M V − C E > 0 RP = EMV - CE > 0 RP = EM V − CE > 0

Risk-Neutral (Linear Utility Function, U ′ ′ ( X ) = 0 U''(X) = 0 U ′′ ( X ) = 0 ) :

Constant marginal utility of wealth. Decisions are made strictly to maximize E M V EMV EM V .

C E = E M V CE = EMV CE = EM V and R P = 0 RP = 0 RP = 0 .

Risk-Seeking / Risk-Prone (Convex Utility Function, U ′ ′ ( X ) > 0 U''(X) > 0 U ′′ ( X ) > 0 ) :

Increasing marginal utility of wealth. Prefers a risky gamble over a certain payoff equal to the expected value.

C E > E M V CE > EMV CE > EM V and R P = E M V − C E < 0 RP = EMV - CE < 0 RP = EM V − CE < 0 .

• 6. Step-by-Step Worked Engineering Calculations
• Worked Example 10.3.1: Comprehensive EVM Performance & Forecasting Analysis

Problem : A manufacturing facility initiates a capital robotics installation with a total authorized budget B A C = $ 600 , 000 BAC = \text{\textdollar}600,000 B A C = $ 600 , 000 and a planned duration of 12 months. At the end of Month 6, project tracking reveals:

• Planned percentage complete: 50%
• Actual physical progress complete: 40%
• Actual cumulative cost incurred: A C = $ 300 , 000 AC = \text{\textdollar}300,000 A C = $ 300 , 000

Calculate:

Planned Value ( P V PV P V ) and Earned Value ( E V EV E V ).

Cost Variance ( C V CV C V ) and Schedule Variance ( S V SV S V ).

Cost Performance Index ( C P I CPI CP I ) and Schedule Performance Index ( S P I SPI SP I ).

Estimate at Completion ( E A C EAC E A C ) assuming past cost efficiency is typical of future work.

To-Complete Performance Index to achieve the original budget ( T C P I B A C TCPI_{BAC} TCP I B A C ​ ).

Solution :

Compute P V PV P V and E V EV E V : P V = 0.50 × $ 600 , 000 = $ 300 , 000 PV = 0.50 \times \$600,000 = \$300,000 P V = 0.50 × $600 , 000 = $300 , 000 E V = 0.40 × $ 600 , 000 = $ 240 , 000 EV = 0.40 \times \$600,000 = \$240,000 E V = 0.40 × $600 , 000 = $240 , 000

Compute Variances: C V = E V − A C = $ 240 , 000 − $ 300 , 000 = − $ 60 , 000 (Over budget) CV = EV - AC = \$240,000 - \$300,000 = -\$60,000\text{ (Over budget)} C V = E V − A C = $240 , 000 − $300 , 000 = − $60 , 000 (Over budget) S V = E V − P V = $ 240 , 000 − $ 300 , 000 = − $ 60 , 000 (Behind schedule) SV = EV - PV = \$240,000 - \$300,000 = -\$60,000\text{ (Behind schedule)} S V = E V − P V = $240 , 000 − $300 , 000 = − $60 , 000 (Behind schedule)

Compute Indices: C P I = E V A C = 240 , 000 300 , 000 = 0.800 CPI = \frac{EV}{AC} = \frac{240,000}{300,000} = 0.800 CP I = A C E V ​ = 300 , 000 240 , 000 ​ = 0.800 S P I = E V P V = 240 , 000 300 , 000 = 0.800 SPI = \frac{EV}{PV} = \frac{240,000}{300,000} = 0.800 SP I = P V E V ​ = 300 , 000 240 , 000 ​ = 0.800

Forecast E A C EAC E A C under typical assumptions: E A C = B A C C P I = $ 600 , 000 0.800 = $ 750 , 000 EAC = \frac{BAC}{CPI} = \frac{\$600,000}{0.800} = \$750,000 E A C = CP I B A C ​ = 0.800 $600 , 000 ​ = $750 , 000 Projected Overrun ( V A C ) = B A C − E A C = 600 , 000 − 750 , 000 = − $ 150 , 000 \text{Projected Overrun } (VAC) = BAC - EAC = 600,000 - 750,000 = -\$150,000 Projected Overrun ( V A C ) = B A C − E A C = 600 , 000 − 750 , 000 = − $150 , 000

Compute T C P I B A C TCPI_{BAC} TCP I B A C ​ : T C P I B A C = B A C − E V B A C − A C = 600 , 000 − 240 , 000 600 , 000 − 300 , 000 = 360 , 000 300 , 000 = 1.200 TCPI_{BAC} = \frac{BAC - EV}{BAC - AC} = \frac{600,000 - 240,000}{600,000 - 300,000} = \frac{360,000}{300,000} = 1.200 TCP I B A C ​ = B A C − A C B A C − E V ​ = 600 , 000 − 300 , 000 600 , 000 − 240 , 000 ​ = 300 , 000 360 , 000 ​ = 1.200

Engineering Conclusion : The project is significantly behind schedule and over budget ( C P I = S P I = 0.80 CPI = SPI = 0.80 CP I = SP I = 0.80 ). To complete the project within the original B A C BAC B A C of $600,000, remaining work must achieve an extraordinary cost efficiency of $1.20 (20% above original plan), which is generally unrealistic.

Worked Example 10.3.2: Multi-Criteria Decision Table Evaluation

Problem : An industrial plant must select one of three facility layouts ( A 1 , A 2 , A 3 A_1, A_2, A_3 A 1 ​ , A 2 ​ , A 3 ​ ) facing three possible market demand scenarios ( S 1 , S 2 , S 3 S_1, S_2, S_3 S 1 ​ , S 2 ​ , S 3 ​ ). The profit payoff matrix (in thousands of dollars) is shown below:

Layout Alternative Weak Demand ( S 1 S_1 S 1 ​ ) Moderate Demand ( S 2 S_2 S 2 ​ ) Strong Demand ( S 3 S_3 S 3 ​ )

• A 1 A_1 A 1 ​ (Modular) $70 $80 $90
• A 2 A_2 A 2 ​ (Standard) $30 $100 $120
• A 3 A_3 A 3 ​ (Automated) -$20 $60 $170

Evaluate and determine the optimal decision alternative under:

• Maximax criterion
• Maximin criterion
• Hurwicz criterion with α = 0.6 \alpha = 0.6 α = 0.6
• Minimax Regret criterion

Solution :

Maximax :

Maximum payoffs: A 1 = 90 A_1 = 90 A 1 ​ = 90 , A 2 = 120 A_2 = 120 A 2 ​ = 120 , A 3 = 170 A_3 = 170 A 3 ​ = 170 .

Select max ⁡ ( 90 , 120 , 170 ) = 170 ⟹ \max(90, 120, 170) = 170 \implies max ( 90 , 120 , 170 ) = 170 ⟹ Select A 3 A_3 A 3 ​ .

Maximin :

Minimum payoffs: A 1 = 70 A_1 = 70 A 1 ​ = 70 , A 2 = 30 A_2 = 30 A 2 ​ = 30 , A 3 = − 20 A_3 = -20 A 3 ​ = − 20 .

Select max ⁡ ( 70 , 30 , − 20 ) = 70 ⟹ \max(70, 30, -20) = 70 \implies max ( 70 , 30 , − 20 ) = 70 ⟹ Select A 1 A_1 A 1 ​ .

Hurwicz ( α = 0.6 \alpha = 0.6 α = 0.6 ) :

Score ( A 1 ) = 0.6 ( 90 ) + 0.4 ( 70 ) = 54 + 28 = 82 \text{Score}(A_1) = 0.6(90) + 0.4(70) = 54 + 28 = 82 Score ( A 1 ​ ) = 0.6 ( 90 ) + 0.4 ( 70 ) = 54 + 28 = 82

Score ( A 2 ) = 0.6 ( 120 ) + 0.4 ( 30 ) = 72 + 12 = 84 \text{Score}(A_2) = 0.6(120) + 0.4(30) = 72 + 12 = 84 Score ( A 2 ​ ) = 0.6 ( 120 ) + 0.4 ( 30 ) = 72 + 12 = 84

Score ( A 3 ) = 0.6 ( 170 ) + 0.4 ( − 20 ) = 102 − 8 = 94 \text{Score}(A_3) = 0.6(170) + 0.4(-20) = 102 - 8 = 94 Score ( A 3 ​ ) = 0.6 ( 170 ) + 0.4 ( − 20 ) = 102 − 8 = 94

Select max ⁡ ( 82 , 84 , 94 ) = 94 ⟹ \max(82, 84, 94) = 94 \implies max ( 82 , 84 , 94 ) = 94 ⟹ Select A 3 A_3 A 3 ​ .

Minimax Regret :

Column maximums ( X j ∗ X_j^* X j ∗ ​ ): S 1 = 70 S_1 = 70 S 1 ​ = 70 , S 2 = 100 S_2 = 100 S 2 ​ = 100 , S 3 = 170 S_3 = 170 S 3 ​ = 170 .

Regret table ( R i j = X j ∗ − X i j R_{ij} = X_j^* - X_{ij} R ij ​ = X j ∗ ​ − X ij ​ ):

A 1 A_1 A 1 ​ : S 1 = 70 − 70 = 0 S_1 = 70 - 70 = 0 S 1 ​ = 70 − 70 = 0 ; S 2 = 100 − 80 = 20 S_2 = 100 - 80 = 20 S 2 ​ = 100 − 80 = 20 ; S 3 = 170 − 90 = 80 S_3 = 170 - 90 = 80 S 3 ​ = 170 − 90 = 80 . Max Regret = 80 80 80 .

A 2 A_2 A 2 ​ : S 1 = 70 − 30 = 40 S_1 = 70 - 30 = 40 S 1 ​ = 70 − 30 = 40 ; S 2 = 100 − 100 = 0 S_2 = 100 - 100 = 0 S 2 ​ = 100 − 100 = 0 ; S 3 = 170 − 120 = 50 S_3 = 170 - 120 = 50 S 3 ​ = 170 − 120 = 50 . Max Regret = 50 50 50 .

A 3 A_3 A 3 ​ : S 1 = 70 − ( − 20 ) = 90 S_1 = 70 - (-20) = 90 S 1 ​ = 70 − ( − 20 ) = 90 ; S 2 = 100 − 60 = 40 S_2 = 100 - 60 = 40 S 2 ​ = 100 − 60 = 40 ; S 3 = 170 − 170 = 0 S_3 = 170 - 170 = 0 S 3 ​ = 170 − 170 = 0 . Max Regret = 90 90 90 .

Minimize maximum regret: min ⁡ ( 80 , 50 , 90 ) = 50 ⟹ \min(80, 50, 90) = 50 \implies min ( 80 , 50 , 90 ) = 50 ⟹ Select A 2 A_2 A 2 ​ .

Worked Example 10.3.3: Decision Tree Rollback with EVPI Determination

Problem : Using the facility layout alternatives from Example 10.3.2, assume the market research group establishes probability estimates for the three demand scenarios: P ( S 1 ) = 0.25 P(S_1) = 0.25 P ( S 1 ​ ) = 0.25 , P ( S 2 ) = 0.50 P(S_2) = 0.50 P ( S 2 ​ ) = 0.50 , and P ( S 3 ) = 0.25 P(S_3) = 0.25 P ( S 3 ​ ) = 0.25 .

Calculate the Expected Monetary Value ( E M V EMV EM V ) for each alternative and identify the optimal decision without additional information.

Calculate the Expected Value with Perfect Information ( E V ∣ P I EV|PI E V ∣ P I ).

Calculate the Expected Value of Perfect Information ( E V P I EVPI E V P I ).

Solution :

Compute E M V EMV EM V for each alternative: E M V ( A 1 ) = 0.25 ( 70 ) + 0.50 ( 80 ) + 0.25 ( 90 ) = 17.5 + 40 + 22.5 = $ 80.0 thousand EMV(A_1) = 0.25(70) + 0.50(80) + 0.25(90) = 17.5 + 40 + 22.5 = \$80.0\text{ thousand} EM V ( A 1 ​ ) = 0.25 ( 70 ) + 0.50 ( 80 ) + 0.25 ( 90 ) = 17.5 + 40 + 22.5 = $80.0 thousand E M V ( A 2 ) = 0.25 ( 30 ) + 0.50 ( 100 ) + 0.25 ( 120 ) = 7.5 + 50 + 30 = $ 87.5 thousand EMV(A_2) = 0.25(30) + 0.50(100) + 0.25(120) = 7.5 + 50 + 30 = \$87.5\text{ thousand} EM V ( A 2 ​ ) = 0.25 ( 30 ) + 0.50 ( 100 ) + 0.25 ( 120 ) = 7.5 + 50 + 30 = $87.5 thousand E M V ( A 3 ) = 0.25 ( − 20 ) + 0.50 ( 60 ) + 0.25 ( 170 ) = − 5.0 + 30 + 42.5 = $ 67.5 thousand EMV(A_3) = 0.25(-20) + 0.50(60) + 0.25(170) = -5.0 + 30 + 42.5 = \$67.5\text{ thousand} EM V ( A 3 ​ ) = 0.25 ( − 20 ) + 0.50 ( 60 ) + 0.25 ( 170 ) = − 5.0 + 30 + 42.5 = $67.5 thousand

Optimal decision without information: Select A 2 A_2 A 2 ​ with E M V ∗ = $ 87.5 thousand EMV^* = \text{\textdollar}87.5\text{ thousand} EM V ∗ = $ 87.5 thousand .

Compute E V ∣ P I EV|PI E V ∣ P I (if state of nature is known in advance, pick the best layout):

If S 1 S_1 S 1 ​ occurs (prob 0.25), select A 1 A_1 A 1 ​ with payoff 70 70 70 .

If S 2 S_2 S 2 ​ occurs (prob 0.50), select A 2 A_2 A 2 ​ with payoff 100 100 100 .

If S 3 S_3 S 3 ​ occurs (prob 0.25), select A 3 A_3 A 3 ​ with payoff 170 170 170 . E V ∣ P I = 0.25 ( 70 ) + 0.50 ( 100 ) + 0.25 ( 170 ) = 17.5 + 50 + 42.5 = $ 110.0 thousand EV|PI = 0.25(70) + 0.50(100) + 0.25(170) = 17.5 + 50 + 42.5 = \$110.0\text{ thousand} E V ∣ P I = 0.25 ( 70 ) + 0.50 ( 100 ) + 0.25 ( 170 ) = 17.5 + 50 + 42.5 = $110.0 thousand

Compute E V P I EVPI E V P I : E V P I = E V ∣ P I − E M V ∗ = 110.0 − 87.5 = $ 22.5 thousand = $ 22 , 500 EVPI = EV|PI - EMV^* = 110.0 - 87.5 = \$22.5\text{ thousand} = \$22,500 E V P I = E V ∣ P I − EM V ∗ = 110.0 − 87.5 = $22.5 thousand = $22 , 500

Engineering Conclusion : The firm should pay no more than $22,500 for any market survey or forecasting tool designed to predict demand.