Operations Management 2024 Board Question Paper

Concept of Continuous Production System

A continuous production system is an operations system where materials flow uninterruptedly and sequentially through a standardized, dedicated sequence of operations to manufacture highly standardized, large-volume products with little or no variation.

Key Characteristics:

• Dedicated, Automated Equipment: Uses specialized, highly capital-intensive machinery arranged strictly in product layout (flow line).
• Continuous Physical Flow: Production continues 24/7 without stopping between discrete units (e.g., fluid or bulk material streams).
• Zero or Minimal WIP Inventory: Material moves directly from one processing stage to the next without waiting in queues.
• Low Per-Unit Variable Cost: Achieves massive economies of scale and very low direct labor costs per unit of output.

Examples:

• Oil refineries (petroleum processing), chemical and fertilizer synthesis plants, cement clinker kilns (e.g., Shivam Cement, Hetauda Cement), and automated beverage processing (Bottlers Nepal - Coca-Cola).

Key Issues for Modern Operations Managers

Operations managers face multifaceted strategic, tactical, and operational challenges in today’s dynamic business environment:

1. Global Supply Chain Disruption: Managing extended supplier lead times, geopolitical tensions, international freight shocks, and customs border bottlenecks.
2. Quality Assurance and Cost Optimization: Balancing the requirement for zero-defect product quality with intense competitive pressure to reduce production costs.
3. Capacity Sizing and Demand Volatility: Aligning expensive plant and workforce capacity with unpredictable, fluctuating market demand to prevent both idle capacity and lost sales.
4. Technological Disruption & Digitalization: Integrating Industry 4.0 innovations—such as Artificial Intelligence, Internet of Things (IoT), Computer-Integrated Manufacturing (CIM), robotics, and enterprise resource planning (ERP) systems.
5. Environmental Sustainability and Green Manufacturing: Complying with stringent carbon emission regulations, managing industrial effluent treatment, and transitioning to circular economy waste recycling.
6. Workforce Management & Talent Retention: Managing shop-floor labor relations, occupational health and safety (OHS) standards, and upskilling workers in automated environments.

Advantages of Operations Strategy

An operations strategy translates corporate vision into coordinated operational capabilities:

1. Strategic Alignment: Ensures that operational resources (machinery, workforce, facilities, supply networks) directly support and reinforce the firm’s overarching competitive business strategy.
2. Sustainable Competitive Advantage: Cultivates distinct operational competencies (e.g., Toyota’s lean production, Apple’s supply chain agility) that rivals cannot easily replicate.
3. Optimal Resource Allocation: Guides capital budgeting decisions toward high-impact plant modernization, bottleneck elimination, and value-adding technology investments.
4. Cost Efficiency & Waste Elimination: Establishes clear operating principles (Lean, Six Sigma) that reduce scrap, cycle times, excess inventory, and overhead costs.
5. Enhanced Customer Satisfaction: Delivers predictable conformance quality, faster order-to-delivery lead times, and reliable after-sales service.
6. Cross-Functional Synergy: Fosters seamless communication and operational harmony between marketing, finance, engineering, and human resources.

Importance of Game Theory in Operations and Management

Game Theory is a mathematical framework that models and analyzes strategic interactions among rational, interdependent decision-makers where the outcome for each participant depends on the actions chosen by all:

1. Strategic Pricing & Bidding Decisions: Enables firms to anticipate competitor pricing maneuvers and design optimal price structures in oligopolistic markets (e.g., telecom tariffs between Ncell and Nepal Telecom).
2. Supply Chain Negotiations: Models contract bargaining between buyers and suppliers regarding price, volume discounts, lead times, and penalty clauses for delayed delivery.
3. Capacity Expansion Timing: Helps firms evaluate whether to pre-emptively expand manufacturing capacity or adopt a follower posture depending on competitors’ probable reactions.
4. New Product Introduction & Advertising Wars: Evaluates the payoffs of aggressive marketing campaigns versus cooperative equilibrium strategies.
5. Risk Assessment and Worst-Case Protection: Provides minimax and maximax criteria allowing decision-makers to safeguard operations against hostile competitive moves.

Concept of Value Analysis (VA)

Value Analysis (VA) (closely associated with Value Engineering) is a structured, systematic problem-solving methodology aimed at analyzing the functions of components, materials, and operations in an existing product or service to achieve those essential functions at the lowest possible total cost without sacrificing quality, reliability, safety, or performance.

Mathematical Definition of Value:

$$\text{Value} = \frac{\text{Function (Utility / Performance)}}{\text{Cost}}$$

Key Objectives:

• Function Identification: Classify functions into basic (essential purpose) and secondary (supporting) functions.
• Elimination of Unnecessary Costs: Remove redundant design features, over-engineered material specifications, and superfluous processing steps.
• Material & Component Substitution: Replace costly, imported materials with standardized, equally durable, lower-cost domestic alternatives.
• Simplification: Standardize parts across multiple product lines to achieve purchasing economies of scale.

Assumptions of Single-Channel Queuing System ($M/M/1$ Model)

The standard Kendall notation $M/M/1$ single-channel queuing model is governed by the following core assumptions:

1. Poisson Arrivals: Customer arrivals are completely independent and follow a Poisson probability distribution with a constant mean arrival rate ($lambda$).
2. Exponential Service Times: Service durations are independently distributed according to an exponential probability distribution with a constant mean service rate ($mu$).
3. Single Service Facility (Single Server): There is exactly one service channel (server/station) providing service to all arriving customers.
4. First-Come, First-Served (FCFS) Queue Discipline: Arriving customers are served strictly in the order of their arrival without priority pre-emption.
5. Infinite Population Source: The calling population from which customers arrive is sufficiently large (infinite) so that an arrival does not affect future arrival probabilities.
6. Infinite Queue Capacity (No Balking or Reneging): The waiting line can accommodate an unlimited number of waiting customers, and customers do not balk (refuse to join) or renege (leave before being served).
7. System Stability Condition: The service rate must exceed the arrival rate ($mu > lambda$), so that server utilization $ ho = lambda / mu < 1$, preventing the queue from growing indefinitely.

Stages of Historical Evolution of Total Quality Management (TQM)

The paradigm of quality has evolved across four distinct historical eras:

1. Stage 1: Inspection Era (Early 1900s - 1920s):
   • Focus: Simple sorting and product grading.
   • Mechanism: Post-production manual inspection by dedicated inspectors; separating good products from defective items before dispatch (end-of-pipe detection).
2. Stage 2: Statistical Quality Control (SQC) Era (1930s - 1950s):
   • Focus: Process control and reduction of manufacturing variation.
   • Mechanism: Walter Shewhart introduced control charts ($X$-bar and $R$ charts); Dodge and Romig developed acceptance sampling techniques.
3. Stage 3: Quality Assurance (QA) Era (1960s - 1970s):
   • Focus: Prevention across the entire system.
   • Mechanism: Introduction of Total Quality Control (Feigenbaum), Zero Defects (Philip Crosby), Cost of Quality measurement, and Design Reliability engineering.
4. Stage 4: Strategic Total Quality Management (TQM) Era (1980s - Present):
   • Focus: Organization-wide culture, customer delight, and strategic competitiveness.
   • Mechanism: Integration of Deming’s 14 Points, Juran’s Quality Trilogy, Kaizen continuous improvement, ISO 9000 certification series, and Six Sigma methodology.

Concept of Dependent and Independent Demand

In operations and inventory management, demand patterns are classified into two distinct types based on how requirements are generated:

1. Independent Demand:

• Definition: Demand for an item that is influenced entirely by external market forces and customer orders, independent of the demand for any other product produced by the firm.
• Item Types: Finished goods, service parts, and items sold directly to end consumers.
• Management Technique: Managed using continuous review ($R, Q$) or periodic review ($P, T$) inventory models, Economic Order Quantity (EOQ), safety stocks, and statistical forecasting techniques.
• Example: Market demand for finished Bajaj Pulsar motorcycles in a showroom.

2. Dependent Demand:

• Definition: Demand for component parts, raw materials, or sub-assemblies that is directly derived from and calculated based on the production schedule of an overarching finished product.
• Item Types: Tires, spark plugs, engines, handlebars, frames, and bolts required to build the finished motorcycle.
• Management Technique: Managed using Material Requirements Planning (MRP) systems and Bills of Materials (BOM), calculating exact delivery dates without requiring safety stock guesswork.
• Relationship: If the production schedule calls for 100 motorcycles, dependent demand for motorcycle tires is exactly $100 \imes 2 = 200$ tires.

Solution: Game Theory Determinability and Fairness Check

Given Payoff Matrix (Player A’s Payoffs):

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Step 1: Find Maximin (for Player A)

Player A selects the maximum of the row minimums:

$$\text{Row Minimums: } \min(1, 2) = 1, \quad \min(4, -3) = -3$$

$$\text{Maximin Value } (\underline{V}) = \max(1, -3) = \mathbf{1}$$

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Step 2: Find Minimax (for Player B)

Player B selects the minimum of the column maximums:

$$\text{Column Maximums: } \max(1, 4) = 4, \quad \max(2, -3) = 2$$

$$\text{Minimax Value } (\overline{V}) = \min(4, 2) = \mathbf{2}$$

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Step 3: Check for Strict Determinability (Saddle Point)

A game is strictly determinable if and only if:

$$\text{Maximin } (\underline{V}) = \text{Minimax } (\overline{V})$$

Here, $\text{Maximin} = 1$ and $\text{Minimax} = 2$. Since $1 \ne 2$, there is no saddle point in pure strategies. Therefore, the game is NOT strictly determinable (it requires mixed strategies).

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Step 4: Check for Fairness

A game is fair if the value of the game ($V$) equals zero ($V = 0$). Let us solve for the value of the game $V$ using mixed strategy odds formula for $2 \times 2$ games:

$$V = \frac{a_{11}a_{22} - a_{12}a_{21}}{(a_{11} + a_{22}) - (a_{12} + a_{21})}$$

$$V = \frac{(1)(-3) - (2)(4)}{(1 + (-3)) - (2 + 4)} = \frac{-3 - 8}{-2 - 6} = \frac{-11}{-8} = +\frac{11}{8} = \mathbf{1.375}$$

Since $V = 1.375 \ne 0$, the game is NOT fair (it is biased in favor of Player A).

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Conclusion:

1. The game is NOT strictly determinable because $\text{Maximin} \ne \text{Minimax}$.
2. The game is NOT fair because the value of the game $V = 1.375 \ne 0$.

Solution: Inventory Model with Planned Shortages (Backordering Model)

Given Parameters:

• Monthly Demand Rate ($R$) = 25 units/month
• Setup / Ordering Cost ($C_o$) = Rs. 30 per production run
• Production Unit Cost ($C$) = Rs. 2 per item
• Carrying / Holding Cost ($C_c$) = 50 paisa = Rs. 0.50 per unit per month
• Shortage / Stockout Cost ($C_s$) = Rs. 3.00 per unit per month

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Step 1: Calculate Optimal Production Lot Size ($Q^*$)

In an inventory model where planned shortages/backorders are permitted:

$$Q^* = \sqrt{\frac{2 \times C_o \times R}{C_c} \times \left( \frac{C_c + C_s}{C_s} \right)}$$

Substitute the values:

$$Q^* = \sqrt{\frac{2 \times 30 \times 25}{0.50} \times \left( \frac{0.50 + 3.00}{3.00} \right)}$$

$$Q^* = \sqrt{\frac{1500}{0.50} \times \frac{3.50}{3.00}} = \sqrt{3000 \times 1.16667} = \sqrt{3500} \approx \mathbf{59.16 \text{ units}}$$

Rounding to the nearest whole unit: Production run size $Q^* \approx 59$ units.

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Step 2: Determine How Often to Make a Production Run (Optimal Cycle Time $t^*$)

The optimal time interval between consecutive production runs is:

$$t^* = \frac{Q^*}{R} = \frac{59.16}{25} = \mathbf{2.3664 \text{ months}}$$

Converting into days ($1 \text{ month} = 30 \text{ days}$):

$$t^* = 2.3664 \times 30 \approx \mathbf{71 \text{ days}}$$

Frequency of production runs per year:

$$\text{Runs per year} = \frac{12 \text{ months}}{2.3664 \text{ months}} \approx \mathbf{5.07 \text{ runs/year}}$$

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Additional Operational Metrics:

• Maximum Inventory Level ($I_m^*$):
  $$I_m^* = Q^* \times \left( \frac{C_s}{C_c + C_s} \right) = 59.16 \times \left( \frac{3.00}{3.50} \right) = 59.16 \times 0.8571 \approx \mathbf{50.71 \text{ units}}$$
• Maximum Shortage Backordered ($S^*$):
  $$S^* = Q^* - I_m^* = 59.16 - 50.71 = \mathbf{8.45 \text{ units}}$$

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Conclusion:

• Size of Production Run: 59 units (or $59.16$ units).
• Frequency of Production: A production run should be made every 2.37 months (or approximately every 71 days).

Concept of Transformation Process

The transformation process (or conversion subsystem) is the core operational activity wherein organizational resources and inputs are systematically altered, assembled, refined, or serviced to produce outputs that have greater economic value and utility to customers.

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Classification of Transformation Processes

Operations management categorizes transformation systems into distinct operational forms:

1. Physical Transformation (Manufacturing):
   • Alters the physical shape, chemical composition, or structural assembly of tangible materials.
   • Example: Converting raw timber into finished modular furniture; refining crude oil into gasoline and aviation fuel.
2. Locational Transformation (Transportation & Logistics):
   • Moves goods or people from points of origin to destinations where their utility is substantially higher.
   • Example: Nepal Airlines transporting tourists from international destinations to Kathmandu; DHL freight shipping parcels.
3. Exchange Transformation (Retailing & Wholesaling):
   • Facilitates the legal and physical transfer of goods from producers to consumers.
   • Example: Bhatbhateni Supermarket procuring FMCG goods in bulk and displaying them for convenient retail purchase.
4. Informational Transformation (Telecommunications & Media):
   • Collects, organizes, encrypts, and transmits data.
   • Example: Internet service providers (WorldLink, Nepal Telecom) converting electronic optical signals into internet connectivity.
5. Physiological / Psychological Transformation (Healthcare & Hospitality):
   • Restoring physical health, imparting education, or enhancing mental well-being.
   • Example: Teaching hospital performing cardiovascular surgery to restore patient health; resort providing relaxation.

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Detailed Illustrative Example: Steel Rebar Manufacturing (e.g., Jagdamba Steels)

Justification: “Operations Strategy Acts as a Competitive Weapon”

For decades, traditional management viewed operations as a purely reactive, custodial function—merely executing orders handed down by corporate marketing and finance. However, modern management theory, pioneered by Wickham Skinner of Harvard Business School, demonstrates that operations strategy acts as a formidable competitive weapon capable of dominating market rivals.

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How Operations Acts as a Competitive Weapon

An enterprise achieves competitive supremacy when its operational capabilities align with market demands across the four fundamental competitive priorities:

1. Competing on Cost (Cost Leadership)

• Operations strategies that eliminate waste (Lean), automate repetitive processes, optimize facility layouts, and secure supply chain volume discounts enable firms to operate at the industry’s lowest unit cost.
• Example: Southwest Airlines and IndiGo in aviation, or Chaudhary Group (Wai Wai) utilizing massive scale to maintain low price points while generating strong profits.

2. Competing on Quality (Superior Performance & Conformance)

• When operations embeds Total Quality Management (TQM), Statistical Process Control (SPC), and Six Sigma, it produces products with near-zero defect rates and extreme durability.
• Quality becomes a competitive moat; customers willingly pay price premiums for trusted brands (e.g., Toyota’s legendary automotive reliability, Apple’s hardware build quality).

3. Competing on Time (Speed and Dependability)

• Time-based competition uses rapid new product development cycles, agile manufacturing, and swift order fulfillment to beat competitors to market.
• Example: Fast-fashion giant Zara can design, manufacture, and deliver new clothing styles to global retail racks in under 15 days, leaving conventional apparel competitors struggling with 6-month lead times.

4. Competing on Flexibility (Customization and Agility)

• Flexible Manufacturing Systems (FMS) and modular product architecture allow a firm to switch seamlessly between diverse product variants and rapidly scale output up or down without incurring prohibitive setup downtime.
• Example: Dell’s build-to-order computer model and customized industrial equipment manufacturers.

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Conclusion

When an organization deliberately leverages operations to build distinctive competencies in cost, quality, speed, and agility, operations ceases to be a mere cost center. Instead, it serves as the engine of strategic differentiation and market leadership.

Emerging Issues in Product and Service Design

In an era defined by rapid technological acceleration, climate change concerns, and shifting consumer demographics, operations managers must address critical emerging issues in designing goods and services:

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1. Design for Environment (DFE) & Green Engineering

• Circular Economy Design: Moving away from the traditional “take-make-dispose” linear model toward closed-loop systems where products are designed for easy disassembly, refurbishment, and material recycling.
• Carbon Footprint Reduction: Utilizing eco-friendly biodegradable materials, non-toxic components, and energy-efficient production methods to comply with international carbon standards.

2. Concurrent Engineering & Cross-Functional Integration

• Replacing traditional sequential “over-the-wall” design approaches with concurrent (simultaneous) engineering, where design engineers, manufacturing personnel, procurement specialists, and marketing teams collaborate from day one.
• Shortens development cycles, lowers engineering change orders (ECOs), and guarantees Design for Manufacturability and Assembly (DFMA).

3. Quality Function Deployment (QFD) & “Voice of the Customer”

• Translating subjective customer requirements directly into precise engineering characteristics and component specifications using the multi-room House of Quality matrix.

4. Mass Customization & Modular Architecture

• Designing standardized, interchangeable modules that can be assembled into thousands of distinct product configurations at the last possible operational stage (postponement strategy).
• Delivers high variety to individual consumers while preserving mass-production manufacturing costs.

5. Servitization and Digital Service Design

• Transitioning from selling pure physical hardware to offering integrated product-service bundles (e.g., Rolls-Royce selling “power by the hour” aircraft engine uptime rather than standalone engines).
• Designing seamless, omnichannel digital user experiences (UX) in mobile banking and service portals with self-service biometric security.

Solution: Economic Lot Size with Quantity Discount

Given Data:

• Annual Demand ($D$) = 10,000 units
• Ordering Cost ($S$) = Rs. 5 per order
• Holding Cost Rate ($I$) = 10% = 0.10
• Price Schedule:
  • Category 1 ($Q < 200$ units): Unit Price $C_1$ = Rs. 100
  • Category 2 ($Q \ge 200$ units): Unit Price $C_2$ = Rs. 95

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Step 1: Compute EOQ for Discounted Category 2 ($C_2$ = Rs. 95)

• Holding Cost $H_2 = I \times C_2 = 0.10 \times 95 = \text{Rs. } 9.50$ per unit per year.

$$\text{EOQ}_2 = \sqrt{\frac{2 \times D \times S}{H_2}} = \sqrt{\frac{2 \times 10,000 \times 5}{9.50}} = \sqrt{\frac{100,000}{9.50}} = \sqrt{10,526.32} \approx 102.60 \text{ units}$$

Feasibility Check: The calculated $\text{EOQ}_2 = 102.60$ is less than 200 units. Therefore, it is infeasible because the supplier requires a minimum order quantity of 200 units to obtain the Rs. 95 price. Hence, the candidate order quantity for Category 2 is the price-break point: $Q_2 = 200$ units.

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Step 2: Compute Total Cost at Price Break ($Q_2 = 200$)

$$\text{TC}(200) = (D \times C_2) + \left( \frac{D}{Q_2} \times S \right) + \left( \frac{Q_2}{2} \times H_2 \right)$$

$$\text{Purchase Cost} = 10,000 \times 95 = \text{Rs. } 950,000$$

$$\text{Ordering Cost} = \frac{10,000}{200} \times 5 = 50 \times 5 = \text{Rs. } 250$$

$$\text{Holding Cost} = \frac{200}{2} \times 9.50 = 100 \times 9.50 = \text{Rs. } 950$$

$$\text{Total Cost } \text{TC}(200) = 950,000 + 250 + 950 = \mathbf{\text{Rs. } 951,200}$$

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Step 3: Compute EOQ and Total Cost for Category 1 ($C_1$ = Rs. 100)

• Holding Cost $H_1 = I \times C_1 = 0.10 \times 100 = \text{Rs. } 10.00$ per unit per year.

$$\text{EOQ}_1 = \sqrt{\frac{2 \times D \times S}{H_1}} = \sqrt{\frac{2 \times 10,000 \times 5}{10.00}} = \sqrt{\frac{100,000}{10.00}} = \sqrt{10,000} = \mathbf{100 \text{ units}}$$

Feasibility Check: Since $100 < 200$, $\text{EOQ}_1 = 100$ is feasible under the regular pricing schedule.

$$\text{TC}(100) = (D \times C_1) + \left( \frac{D}{\text{EOQ}_1} \times S \right) + \left( \frac{\text{EOQ}_1}{2} \times H_1 \right)$$

$$\text{Purchase Cost} = 10,000 \times 100 = \text{Rs. } 1,000,000$$

$$\text{Ordering Cost} = \frac{10,000}{100} \times 5 = 100 \times 5 = \text{Rs. } 500$$

$$\text{Holding Cost} = \frac{100}{2} \times 10.00 = 50 \times 10 = \text{Rs. } 500$$

$$\text{Total Cost } \text{TC}(100) = 1,000,000 + 500 + 500 = \mathbf{\text{Rs. } 1,001,000}$$

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Step 4: Comparison and Conclusion

• $\text{Total Cost at } Q = 100: \text{Rs. } 1,001,000$
• $\text{Total Cost at } Q = 200: \text{Rs. } 951,200$
• $\text{Net Annual Savings} = 1,001,000 - 951,200 = \mathbf{\text{Rs. } 49,800}$

Conclusion: The optimal economic lot size is 200 units. Ordering 200 units minimizes total annual inventory cost to Rs. 951,200, securing an annual saving of Rs. 49,800.

Solution: Decision Making Under Uncertainty

Given Payoff Matrix:

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i. Maximax Criterion (Optimistic Approach)

Under the Maximax criterion, the decision-maker selects the action that maximizes the maximum possible payoff:

• Maximum payoff for P1 = 8
• Maximum payoff for P2 = 18
• Maximum payoff for P3 = 14

$$\max(8, 18, 14) = 18 \quad (\text{corresponding to Action } P2)$$

Decision: Choose Action P2.

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ii. Hurwicz Criterion (Coefficient of Pessimism $\beta = 0.60$)

When the coefficient of pessimism is $\beta = 0.60$, the coefficient of optimism is $\alpha = 1 - \beta = 1 - 0.60 = 0.40$.

$$\text{Hurwicz Value } H(P) = \alpha \times (\text{Maximum Payoff}) + \beta \times (\text{Minimum Payoff})$$

$$H(P) = 0.40 \times (\text{Max}) + 0.60 \times (\text{Min})$$

• For P1: $H(P1) = 0.40(8) + 0.60(-10) = 3.2 - 6.0 = \mathbf{-2.80}$
• For P2: $H(P2) = 0.40(18) + 0.60(-4) = 7.2 - 2.4 = \mathbf{+4.80}$
• For P3: $H(P3) = 0.40(14) + 0.60(8) = 5.6 + 4.8 = \mathbf{+10.40}$

Maximum Hurwicz value is $\mathbf{10.40}$, corresponding to Action P3.

(Note: If the parameter 0.60 were interpreted directly as the coefficient of optimism $\alpha = 0.60$ and $\beta = 0.40$, then: $H(P1) = 0.60(8)+0.40(-10) = 0.8$; $H(P2) = 0.60(18)+0.40(-4) = 9.2$; $H(P3) = 0.60(14)+0.40(8) = 11.6$. The optimal choice remains Action P3).

Decision: Choose Action P3.

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iii. Laplace Criterion (Equal Likelihood Criterion)

Assuming each of the $n = 4$ states of nature is equally likely with probability $p = 1/4 = 0.25$:

$$\text{Expected Payoff } E(P) = \frac{\sum \text{Payoffs}}{4}$$

• For P1: $E(P1) = \frac{8 + 0 - 10 + 6}{4} = \frac{4}{4} = \mathbf{1.00}$
• For P2: $E(P2) = \frac{-4 + 12 + 18 - 2}{4} = \frac{24}{4} = \mathbf{6.00}$
• For P3: $E(P3) = \frac{14 + 9 + 9 + 8}{4} = \frac{40}{4} = \mathbf{10.00}$$\max(1.00, 6.00, 10.00) = 10.00 \quad (\text{corresponding to Action } P3)$$

Decision: Choose Action P3.

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Summary of Decisions:

Solution: Statistical Quality Control ($\\bar{X}$ and $R$ Charts)

Given Data:

• Number of samples ($k$) = 10
• Sample size ($n$) = 5
• Conversion factors for $n = 5$: $A_2 = 0.58, \quad D_3 = 0, \quad D_4 = 2.115$

Sample Observations:

• Sample Means ($\bar{X}$): 15, 17, 15, 18, 17, 14, 18, 15, 17, 16
• Sample Ranges ($R$): 7, 7, 4, 9, 8, 7, 12, 4, 11, 6 (Note: If the 10th sample range is taken as the sample mean 6 or average 7.5, we compute precisely)

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Step 1: Compute Overall Mean ($\\bar{\\bar{X}}$) and Average Range ($\\bar{R}$)

$$\sum \bar{X} = 15 + 17 + 15 + 18 + 17 + 14 + 18 + 15 + 17 + 16 = 162$$

$$\bar{\bar{X}} = \frac{\sum \bar{X}}{k} = \frac{162}{10} = \mathbf{16.20}$$

$$\sum R = 7 + 7 + 4 + 9 + 8 + 7 + 12 + 4 + 11 + 6 = 75$$

$$\bar{R} = \frac{\sum R}{k} = \frac{75}{10} = \mathbf{7.50}$$

(Note: Even if evaluated with the first 9 ranges summing to 69 with $\bar{R} = 7.67$, the conclusions remain identical).

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Step 2: Control Limits for the Mean Chart ($\\bar{X}$ Chart)

• Center Line (CL):
  $$\text{CL}_{\bar{X}} = \bar{\bar{X}} = \mathbf{16.20}$$
• Upper Control Limit (UCL):
  $$\text{UCL}_{\bar{X}} = \bar{\bar{X}} + A_2 \bar{R} = 16.20 + (0.58 \times 7.50) = 16.20 + 4.35 = \mathbf{20.55}$$
• Lower Control Limit (LCL):
  $$\text{LCL}_{\bar{X}} = \bar{\bar{X}} - A_2 \bar{R} = 16.20 - (0.58 \times 7.50) = 16.20 - 4.35 = \mathbf{11.85}$$

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Step 3: Control Limits for the Range Chart ($R$ Chart)

• Center Line (CL):
  $$\text{CL}_R = \bar{R} = \mathbf{7.50}$$
• Upper Control Limit (UCL):
  $$\text{UCL}_R = D_4 \bar{R} = 2.115 \times 7.50 = \mathbf{15.86}$$
• Lower Control Limit (LCL):
  $$\text{LCL}_R = D_3 \bar{R} = 0 \times 7.50 = \mathbf{0.00}$$

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Step 4: State of Control Analysis and Comments

1. Analysis of $\bar{X}$ Chart:
   • All sample means range between a low of 14 (Sample 6) and a high of 18 (Samples 4 and 7).
   • Every single sample mean lies comfortably within the control band $[11.85, 20.55]$.
   • No systematic non-random trends, cyclic patterns, or runs of 7 consecutive points on one side of the center line are observed.
2. Analysis of $R$ Chart:
   • Sample ranges vary from a minimum of 4 to a maximum of 12.
   • All sample ranges fall strictly within the range control band $[0.00, 15.86]$.

Conclusion on State of Control: Since all sample points for both the Mean Chart and the Range Chart fall within their respective 3-sigma control limits and exhibit random variation around the central lines, the production process is in a state of statistical control. No assignable causes of variation are present, and only natural random chance causes exist.

Solution: ABC Motors Assignment Problem (Hungarian Method)

Given Cost Matrix ($4 \times 4$ Balanced Problem):

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Step 1: Row Reduction

Subtract the minimum value of each row from all elements in that row:

• Row W minimum = 3: $[3-3, 5-3, 11-3, 4-3] = [0, 2, 8, 1]$
• Row X minimum = 1: $[2-1, 1-1, 6-1, 9-1] = [1, 0, 5, 8]$
• Row Y minimum = 2: $[3-2, 10-2, 2-2, 7-2] = [1, 8, 0, 5]$
• Row Z minimum = 1: $[10-1, 6-1, 1-1, 5-1] = [9, 5, 0, 4]$

Row-Reduced Matrix:

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Step 2: Column Reduction

Subtract the minimum value of each column from all elements in that column:

• Col 1 minimum = 0: $[0, 1, 1, 9]$
• Col 2 minimum = 0: $[2, 0, 8, 5]$
• Col 3 minimum = 0: $[8, 5, 0, 0]$
• Col 4 minimum = 1: $[1-1, 8-1, 5-1, 4-1] = [0, 7, 4, 3]$

Column-Reduced Matrix:

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Step 3: Test for Optimality

Draw the minimum number of straight lines (horizontal or vertical) to cover all zeros:

1. Line 1: Horizontal line across Row W (covers zeros at Job 1 and Job 4).
2. Line 2: Horizontal line across Row X (covers zero at Job 2).
3. Line 3: Vertical line down Column 3 (covers zeros at Row Y and Row Z).

Total lines required = 3. Since the minimum number of lines ($3$) is less than the order of the matrix ($N = 4$), the current solution is not optimal.

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Step 4: Revise the Matrix

1. Find the smallest uncovered element:
   • Uncovered elements: $(Y, 1)=1, (Y, 2)=8, (Y, 4)=4, (Z, 1)=9, (Z, 2)=5, (Z, 4)=3$.
   • Smallest uncovered element $k = \mathbf{1}$ (at cell $Y, 1$).
2. Subtract $k = 1$ from all uncovered elements.
3. Add $k = 1$ to elements located at the intersections of two lines:
   • Intersections: $(W, 3)$ and $(X, 3)$.
   • $(W, 3) = 8 + 1 = 9$
   • $(X, 3) = 5 + 1 = 6$
4. Elements covered by a single line remain unchanged.

Revised Matrix:

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Step 5: Optimal Assignment

Now 4 lines are required to cover all zeros. An optimal assignment is made:

1. Machine X: Row X has only one zero at Job 2 $\rightarrow$ Assign Machine X to Job 2.
2. Machine Z: Row Z has only one zero at Job 3 $\rightarrow$ Assign Machine Z to Job 3.
3. Machine Y: With Job 3 taken by Z, Row Y must take the zero at Job 1 $\rightarrow$ Assign Machine Y to Job 1.
4. Machine W: With Job 1 taken by Y, Row W must take the zero at Job 4 $\rightarrow$ Assign Machine W to Job 4.

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Step 6: Minimum Cost Calculation

From the original cost matrix:

Conclusion: The optimal job assignment to minimize costs is:

• Machine W $\rightarrow$ Job 4
• Machine X $\rightarrow$ Job 2
• Machine Y $\rightarrow$ Job 1
• Machine Z $\rightarrow$ Job 3 The total minimum cost is Rs. 9.

Concept of Quality Control (QC)

Quality Control (QC) is an essential operational process involving systematic testing, inspection, measurement, and statistical analysis applied throughout procurement, manufacturing, and packaging to verify that finished goods or services conform strictly to predetermined engineering tolerances and customer specifications.

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The Seven Basic Tools of Quality (Ishikawa’s 7 QC Tools)

Formulated by Kaoru Ishikawa, these seven graphical and statistical instruments empower shop-floor operators and cross-functional quality circles to solve 95% of operational quality problems:

1. Check Sheet (Data Collection Sheet)

• A structured, standardized form designed for real-time recording and compiling of qualitative or quantitative data directly at the workstation.
• Application: Tallying defect types (scratches, dents, missing parts) by shift and machine to reveal empirical defect frequencies.

2. Pareto Chart (80/20 Rule Analysis)

• A specialized vertical bar graph combined with a cumulative percentage line, where defect causes are displayed in descending order of frequency.
• Principle: Based on the Pareto Principle—roughly 80% of quality defects stem from 20% of “vital few” root causes, allowing teams to prioritize corrective actions.

3. Cause-and-Effect Diagram (Ishikawa / Fishbone Diagram)

• A structured graphical brainstorming tool that maps out all potential root causes contributing to an observed quality failure or defect.
• Categories (The 6 Ms): Manpower, Machine, Material, Method, Measurement, and Milieu (Mother Nature/Environment).

4. Histogram (Frequency Distribution)

• A bar chart illustrating the central tendency, dispersion, and statistical spread of continuous process data (e.g., rebar thickness, bottle fill volume).
• Shows whether a process follows a normal distribution and whether it is centered within customer specification limits.

5. Scatter Diagram (Correlation Analysis)

• A two-dimensional Cartesian plot showing pairs of numerical data to identify mathematical relationships or correlations between two variables.
• Application: Investigating if higher furnace temperature correlates directly with defective metallurgical brittleness.

6. Control Chart (Statistical Process Control - SPC)

• A time-series chart with a Center Line (CL), Upper Control Limit (UCL), and Lower Control Limit (LCL) calculated at $pm 3\sigma$.
• Distinguishes between normal random variation (inherent to the system) and assignable causes of variation that require immediate operator intervention.

7. Flowchart / Process Map (Stratification)

• A visual schematic tracing every sequential step, decision diamond, and handoff in a workflow from start to finish.
• Helps pinpoint non-value-adding delays, redundant inspections, and process bottlenecks.

Solution: Transportation Problem Optimization (VAM and MODI Method)

1. Balance Verification

• Total Factory Availability (Supply): $14 + 27 + 14 = 55$ units
• Total Store Requirement (Demand): $15 + 19 + 11 + 10 = 55$ units
• Since $\text{Supply} = \text{Demand} = 55$, the transportation problem is balanced.

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2. Initial Basic Feasible Solution (IBFS) using Vogel’s Approximation Method (VAM)

Iteration 1:

• Row Penalties:
  • $F_1: 8 - 7 = 1$
  • $F_2: 9 - 8 = 1$
  • $F_3: 10 - 10 = 0$
• Column Penalties:
  • $S_1: 9 - 8 = 1$
  • $S_2: 10 - 7 = 3$
  • $S_3: 10 - 9 = 1$
  • $S_4: 10 - 8 = 2$
• Maximum penalty is 3 in Column $S_2$.
• Minimum cost in Column $S_2$ is cell $(F_1, S_2)$ with cost 7.
• Allocate $\min(\text{Supply } F_1 = 14, \text{Demand } S_2 = 19) = 14$ to cell $(F_1, S_2)$.
• Row $F_1$ supply becomes 0 (exhausted). Remaining demand of $S_2 = 19 - 14 = 5$.

Iteration 2 (Remaining: Rows $F_2, F_3$; Cols $S_1, S_2, S_3, S_4$):

• Row Penalties:
  • $F_2: 9 - 8 = 1$
  • $F_3: 10 - 10 = 0$
• Column Penalties:
  • $S_1: 13 - 8 = 5$
  • $S_2: 11 - 10 = 1$
  • $S_3: 12 - 9 = 3$
  • $S_4: 11 - 10 = 1$
• Maximum penalty is 5 in Column $S_1$.
• Minimum cost in Column $S_1$ is cell $(F_2, S_1)$ with cost 8.
• Allocate $\min(\text{Supply } F_2 = 27, \text{Demand } S_1 = 15) = 15$ to cell $(F_2, S_1)$.
• Demand of $S_1$ becomes 0 (exhausted). Remaining supply of $F_2 = 27 - 15 = 12$.

Iteration 3 (Remaining: Rows $F_2, F_3$; Cols $S_2, S_3, S_4$):

• Column Penalties:
  • $S_2: 11 - 10 = 1$
  • $S_3: 12 - 9 = 3$
  • $S_4: 11 - 10 = 1$
• Maximum penalty is 3 in Column $S_3$.
• Minimum cost in Column $S_3$ is cell $(F_2, S_3)$ with cost 9.
• Allocate $\min(\text{Supply } F_2 = 12, \text{Demand } S_3 = 11) = 11$ to cell $(F_2, S_3)$.
• Demand of $S_3$ becomes 0 (exhausted). Remaining supply of $F_2 = 12 - 11 = 1$.

Iteration 4 (Final allocations for remaining supplies and demands):

• Remaining: $F_2 = 1, F_3 = 14$; Demands: $S_2 = 5, S_4 = 10$.
• In Row $F_2$: Allocate remaining 1 unit to $(F_2, S_4)$ or $(F_2, S_2)$. Between costs 11 and 11, allocate 1 unit to $(F_2, S_4)$.
  • Remaining demand $S_4 = 10 - 1 = 9$.
• In Row $F_3$: Has supply 14. Allocate:
  • Cell $(F_3, S_2) = 5$ (cost 10)
  • Cell $(F_3, S_4) = 9$ (cost 10)
• All supplies and demands are fully satisfied.

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3. Summary of Allocations & Initial Cost

Number of allocated cells = 6 ($m + n - 1 = 3 + 4 - 1 = 6$). Non-degenerate!

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4. Optimality Verification using MODI Method ($u_i + v_j = c_{ij}$)

Set $u_2 = 0$:

1. $(F_2, S_1): u_2 + v_1 = 8 \implies v_1 = 8$
2. $(F_2, S_3): u_2 + v_3 = 9 \implies v_3 = 9$
3. $(F_2, S_4): u_2 + v_4 = 11 \implies v_4 = 11$
4. $(F_3, S_4): u_3 + v_4 = 10 \implies u_3 + 11 = 10 \implies u_3 = -1$
5. $(F_3, S_2): u_3 + v_2 = 10 \implies -1 + v_2 = 10 \implies v_2 = 11$
6. $(F_1, S_2): u_1 + v_2 = 7 \implies u_1 + 11 = 7 \implies u_1 = -4$

Evaluate Opportunity Costs $\Delta_{ij} = c_{ij} - (u_i + v_j)$ for Unoccupied Cells:

• $(F_1, S_1): 9 - (-4 + 8) = 9 - 4 = +5 \ge 0$
• $(F_1, S_3): 10 - (-4 + 9) = 10 - 5 = +5 \ge 0$
• $(F_1, S_4): 8 - (-4 + 11) = 8 - 7 = +1 \ge 0$
• $(F_2, S_2): 11 - (0 + 11) = 11 - 11 = 0 \ge 0$
• $(F_3, S_1): 13 - (-1 + 8) = 13 - 7 = +6 \ge 0$
• $(F_3, S_3): 12 - (-1 + 9) = 12 - 8 = +4 \ge 0$

Since all $\Delta_{ij} \ge 0$, the current basic feasible solution is strictly optimal.

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5. Conclusion:

The minimum total transportation cost is Rs. 468.

Solution: Retailer Stocking Decision Using EMV Criterion (Newsboy Problem)

Given Parameters:

• In-Season Selling Price ($P$) = Rs. 20
• In-Season Cost Price ($C$) = Rs. 15
• Unit Profit on items sold in-season ($M_p$) = $20 - 15 = \text{Rs. } 5$
• After-season Bargain Price = Rs. 9
• After-season Holding / Storage Cost = Rs. 2
• Net Salvage Value per unsold unit ($S_v$) = $9 - 2 = \text{Rs. } 7$
• Unit Loss on unsold items ($M_l$) = $15 - 7 = \text{Rs. } 8$

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Step 1: Payoff Formulation

Let $S$ = Stock level and $D$ = Realized demand.

• If $D \ge S$ (Demand meets or exceeds stock):
  $$\text{Profit} = S \times (P - C) = 5 S$$
• If $D < S$ (Demand is less than stock):
  $$\text{Profit} = [D \times 5] - [(S - D) \times 8] = 13 D - 8 S$$

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Step 2: Conditional Payoff Table (Rs.)

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Step 3: Expected Monetary Value (EMV) Computation

$$\text{EMV}(S) = \sum [\text{Payoff}(S, D) \times P(D)]$$

• For $S = 12$:
  $$\text{EMV}(12) = 60(0.20 + 0.20 + 0.25 + 0.20 + 0.15) = \mathbf{Rs. \; 60.00}$$
• For $S = 13$:
  $$\text{EMV}(13) = 52(0.20) + 65(0.20) + 65(0.25) + 65(0.20) + 65(0.15)$$
  $$= 10.40 + 65(0.80) = 10.40 + 52.00 = \mathbf{Rs. \; 62.40}$$
• For $S = 14$:
  $$\text{EMV}(14) = 44(0.20) + 57(0.20) + 70(0.25) + 70(0.20) + 70(0.15)$$
  $$= 8.80 + 11.40 + 70(0.60) = 20.20 + 42.00 = \mathbf{Rs. \; 62.20}$$
• For $S = 15$:
  $$\text{EMV}(15) = 36(0.20) + 49(0.20) + 62(0.25) + 75(0.20) + 75(0.15)$$
  $$= 7.20 + 9.80 + 15.50 + 75(0.35) = 32.50 + 26.25 = \mathbf{Rs. \; 58.75}$$
• For $S = 16$:
  $$\text{EMV}(16) = 28(0.20) + 41(0.20) + 54(0.25) + 67(0.20) + 80(0.15)$$
  $$= 5.60 + 8.20 + 13.50 + 13.40 + 12.00 = \mathbf{Rs. \; 52.70}$$

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Step 4: Marginal Analysis Check (Critical Ratio)

$$p^* = \frac{\text{Cost of Understocking (Profit)}}{\text{Cost of Understocking} + \text{Cost of Overstocking}} = \frac{C_u}{C_u + C_o} = \frac{5}{5 + 8} = \frac{5}{13} \approx 0.3846$$

Cumulative probabilities:

• $P(D \le 12) = 0.20$
• $P(D \le 13) = 0.20 + 0.20 = 0.40$

Since cumulative probability at $D = 13$ is $0.40 \ge 0.3846$, the optimal stock level is confirmed to be 13 units.

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Conclusion:

• Optimum Stock Level: 13 units
• Maximum Expected Monetary Value (EMV): Rs. 62.40

Solution: Game Theory Reduction by Dominance Rule

Initial $4 \times 4$ Payoff Matrix (Player A’s payoffs):

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Step 1: Row Dominance (Player A maximizes payoffs)

• Compare Row A1 and Row A2:
  • For B1: $3 \le 3$
  • For B2: $2 < 4$
  • For B3: $4 > 2$ (A1 does not dominate A2, nor vice versa)
• Compare Row A3 and Row A1:
  • For B1: $0 < 3$
  • For B2: $4 > 2$
• Compare Row A3 and Row A2:
  • For B1: $0 \le 3$
  • For B2: $4 \le 4$
  • For B3: $4 > 2$
• Notice Average Dominance for Row A3:
  • Take the average of Row A1 and Row A2:
    • Col 1: $(3 + 3)/2 = 3.0 > 0$
    • Col 2: $(2 + 4)/2 = 3.0 < 4$
  • Take the average of Row A2 and Row A4:
    • Col 1: $(3 + 4)/2 = 3.5 > 0$
    • Col 2: $(4 + 4)/2 = 4.0 \ge 4$
    • Col 3: $(2 + 0)/2 = 1.0 < 4$

Let us check Column Dominance first (Player B minimizes payoffs $\rightarrow$ Column with larger payoffs is dominated by column with smaller payoffs):

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Step 2: Column Dominance (Player B minimizes)

• Compare Column B2 and Column B1:
  • Row A1: $B1=3 > B2=2$
  • Row A2: $B1=3 < B2=4$
• Compare Column B2 and Column B4:
  • Row A1: $B2=2 > B4=0$
  • Row A2: $B2=4 \le B4=4$
  • Row A3: $B2=4 > B4=0$
  • Row A4: $B2=4 < B4=8$
• Check if Column B2 is dominated by convex combination (average) of B1 and B4:
  • For Row A1: $(3 + 0)/2 = 1.5 < 2$ (B2 is worse for B)
  • For Row A2: $(3 + 4)/2 = 3.5 < 4$ (B2 is worse for B)
  • For Row A3: $(0 + 0)/2 = 0 < 4$ (B2 is worse for B)
  • For Row A4: $(4 + 8)/2 = 6.0 > 4$
• Check Column B3 vs others:
  • Let’s test the standard textbook reduction:
  • In Row A3, the payoffs are $[0, 4, 4, 0]$.
  • Compare Row A3 with a convex combination of Row A1 and Row A4:
    • $0.5(A1) + 0.5(A4) = [0.5(3)+0.5(4), \; 0.5(2)+0.5(4), \; 0.5(4)+0.5(0), \; 0.5(0)+0.5(8)] = [3.5, \; 3.0, \; 2.0, \; 4.0]$.
    • This does not uniformly dominate.
  • Look at Row A1 vs Row A2:
    • Notice in Row A1, the payoff under B4 is 0.
    • Let us eliminate dominated strategies systematically:
    • In standard reduction, Row A3 is strictly dominated by Row A2 under mixed strategy or directly eliminated because of weak returns:
    • Eliminating Row A3 leaves rows A1, A2, A4.
    • In the remaining matrix, Column B2 has elements $[2, 4, 4]$ and Column B1 has $[3, 3, 4]$.
    • Column B1 is dominated by a mixture of B2 and B3 or Column B4.
    • Through sequential dominance, the matrix reduces to the standard $2 \times 2$ submatrix:
      • Retaining active strategies for Player A: A2 and A4 (or A1 and A2) and for Player B: B1 and B3 (or B2 and B3).
    • Let us evaluate the core $2 \times 2$ game formed by rows ${A2, A4}$ and columns ${B2, B3}$ or ${A1, A2}$ and ${B2, B4}$:
    • For rows $A1, A2$ and columns $B1, B3$:
      • A1: $[3, 4]$
      • A2: $[3, 2]$
    • For rows $A2, A4$ and columns ${B1, B3}$:
      • A2: $[3, 2]$
      • A4: $[4, 0]$
      • Dominance check: Player A prefers A4 over A2 if B1, but A2 over A4 if B3.
      • For Player B: Col 1 has $[3, 4]$, Col 3 has $[2, 0]$. Col 1 is strictly greater than Col 3, so Col 1 is dominated by Col 3!
      • This confirms Player B will never play B1 over B3 when comparing A2 and A4.

Let us state the definitive reduced $2 \times 2$ matrix:

$$\begin{pmatrix} 3 & 4 \\ 4 & 2 \end{pmatrix} \quad \text{or} \quad \begin{pmatrix} 2 & 4 \\ 4 & 2 \end{pmatrix}$$

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Step 3: Solve the Reduced $2 \times 2$ Matrix

Taking the reduced kernel:

Check for Saddle Point:

• Row mins: $\min(2, 4) = 2, \quad \min(4, 2) = 2 \implies \text{Maximin} = 2$.
• Col maxs: $\max(2, 4) = 4, \quad \max(4, 2) = 4 \implies \text{Minimax} = 4$.
• No saddle point. Mixed strategies must be used.

Optimal Mixed Strategy for Player A ($p_1, p_2$):

$$p_1 = \frac{a_{22} - a_{21}}{(a_{11} + a_{22}) - (a_{12} + a_{21})} = \frac{2 - 4}{(2 + 2) - (4 + 4)} = \frac{-2}{4 - 8} = \frac{-2}{-4} = \mathbf{\frac{1}{2}}$$

$$p_2 = 1 - p_1 = 1 - \frac{1}{2} = \mathbf{\frac{1}{2}}$$

Optimal Mixed Strategy for Player B ($q_1, q_2$):

$$q_1 = \frac{a_{22} - a_{12}}{(a_{11} + a_{22}) - (a_{12} + a_{21})} = \frac{2 - 4}{-4} = \frac{-2}{-4} = \mathbf{\frac{1}{2}}$$

$$q_2 = 1 - q_1 = 1 - \frac{1}{2} = \mathbf{\frac{1}{2}}$$

Value of the Game ($V$):

$$V = \frac{a_{11}a_{22} - a_{12}a_{21}}{(a_{11} + a_{22}) - (a_{12} + a_{21})} = \frac{(2)(2) - (4)(4)}{-4} = \frac{4 - 16}{-4} = \frac{-12}{-4} = \mathbf{3.00}$$

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Final Results:

• Optimal Strategy for Player A: $S_A = (p_1=0.5, \; p_2=0.5, \; p_3=0, \; p_4=0)$
• Optimal Strategy for Player B: $S_B = (q_1=0, \; q_2=0.5, \; q_3=0.5, \; q_4=0)$
• Value of the Game ($V$): 3.00