Operations Management 2025 Board Question Paper

Meaning of Product Design

Product design is the strategic and operational process of creating a new product or revamping an existing one to satisfy specific customer needs, ensure market appeal, and meet technical, functional, aesthetic, and manufacturing requirements at a profitable cost.

Core Elements:

1. Form Design: Physical aesthetics, geometry, color, ergonomics, and sensory tactile appeal.
2. Functional Design: How the product operates, its mechanical reliability, performance specifications, and ease of use.
3. Production / Process Design: Designing the product for ease and economy of manufacturing (Design for Manufacturability and Assembly - DFMA).

Example:

• Yadea / NIU Electric Smart Scooters:
  • Form Design: Sleek aerodynamic exterior, lightweight alloy frame, digital LED dash display.
  • Functional Design: High-torque hub motor, regenerative braking, removable lithium iron phosphate battery with 80 km range per charge.
  • Production Design: Modular snap-fit body panels and standardized wiring harnesses that allow assembly line technicians to assemble a scooter in under 15 minutes.

Characteristics of Intermittent Production System

An intermittent production system (such as job shops or batch production facilities) manufactures products in response to specific customer orders or small-to-medium lot sizes where the flow of work is interrupted and irregular:

1. Wide Product Variety with Low Volumes: Produces a diverse range of customized products tailored to distinct customer specifications.
2. General-Purpose Machinery: Employs flexible, general-purpose machines (e.g., CNC milling, universal lathes, manual sewing machines) rather than specialized single-purpose flow equipment.
3. Process Layout (Functional Layout): Equipment and workers performing similar operations are grouped together into departments (e.g., cutting section, welding section, painting section).
4. Highly Skilled and Versatile Labor: Requires skilled operators capable of interpreting varied blueprints and performing complex machine setups.
5. High Work-in-Progress (WIP) Inventory: Significant waiting times and queues form between functional departments as batches wait for machine availability.
6. Detailed Production Planning and Scheduling: Requires frequent rescheduling, complex routing, and job prioritization (dispatching rules like FCFS, SPT).

Meaning of Value Analysis (VA)

Value Analysis (VA) is a structured, systematic problem-solving methodology used by operations and engineering teams to examine the components, materials, and processes of an established product or service to identify and eliminate unnecessary costs without compromising quality, functionality, reliability, or safety.

Core Value Equation:

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

Key Objectives:

• Analyze each part to determine whether its cost is proportional to its functional usefulness to the customer.
• Substitute cheaper, locally available materials (e.g., using lightweight engineered polymers instead of machined brass).
• Standardize fasteners, fittings, and sub-assemblies across product lines to achieve volume procurement discounts.

Definition of Expected Monetary Value (EMV) Criterion

The Expected Monetary Value (EMV) criterion is a fundamental decision-making technique used in Decision Making Under Risk, where the decision-maker faces multiple alternative courses of action and uncertain states of nature with known probability distributions.

Formula:

For a given decision alternative $A_i$:

$$\text{EMV}(A_i) = \sum_{j=1}^{n} [X_{ij} \times P(S_j)]$$

Where:

• $X_{ij}$ = Conditional monetary payoff resulting from action $A_i$ under state of nature $S_j$.
• $P(S_j)$ = Probability of occurrence of state of nature $S_j$.
• $\sum P(S_j) = 1.0$.

Decision Rule:

• In profit maximization problems, the rational decision-maker selects the alternative that yields the highest Expected Monetary Value ($\max \text{EMV}$).
• In cost minimization problems, the alternative with the lowest expected cost is chosen.

Concept of Inventory Management

Inventory management is the operational discipline of planning, ordering, storing, tracking, and controlling an organization’s stock of materials—including raw materials, work-in-progress (WIP), maintenance/repair/operating (MRO) supplies, and finished goods—to ensure seamless operational continuity at the lowest total inventory cost.

Core Objectives:

1. Ensuring Operational Continuity: Preventing stockouts and assembly line shutdowns caused by sudden demand surges or supplier delays.
2. Minimizing Total Inventory Costs: Balancing the trade-off between inventory carrying costs (storage, capital lockup, obsolescence) and ordering/setup costs.
3. Scientific Stock Control: Applying proven quantitative models such as Economic Order Quantity (EOQ), Reorder Point (ROP), Safety Stock calculations, and selective inventory classification (ABC, VED, FSN analysis).

Meaning of ISO 9000 Series

The ISO 9000 series is a family of internationally recognized standards developed by the International Organization for Standardization (ISO) that outlines the fundamental requirements for establishing, implementing, documenting, and continually improving a formal Quality Management System (QMS).

Key Standards in the Family:

• ISO 9000: Defines the core concepts, principles, and vocabulary of quality management.
• ISO 9001 (Requirements): The only standard in the family against which organizations can be formally audited and certified. It mandates customer focus, leadership commitment, risk-based thinking, process approaches, and continual improvement.
• ISO 9004: Provides guidance for achieving sustained organizational success beyond basic ISO 9001 compliance.

Significance: Certification assures international buyers and supply chain partners that the company possesses certified, reproducible processes capable of delivering consistent conformance quality.

Concept of Dummy Variable (Row/Column) in Transportation Problem

In quantitative operations research, a dummy variable (represented as an artificial dummy row or dummy column) is introduced into a transportation tableau when the problem is unbalanced—meaning total supply from all origins does not equal total demand across all destinations ($\sum a_i \ne \sum b_j$).

Operational Mechanics:

1. When Total Supply > Total Demand:
   • An artificial Dummy Destination (Column) is added.
   • Its requirement is set equal to the surplus: $\text{Demand}_{\text{dummy}} = \sum a_i - \sum b_j$.
2. When Total Demand > Total Supply:
   • An artificial Dummy Origin (Row) is added.
   • Its capacity is set equal to the deficit: $\text{Supply}_{\text{dummy}} = \sum b_j - \sum a_i$.
3. Cost Coefficients:
   • All transportation unit costs associated with the dummy row or column are assigned a value of zero ($c_{ij} = 0$), because no physical goods are actually shipped. Any allocation to a dummy represents unmet demand or excess unsold capacity.

Definition of Saddle Point

In game theory, a saddle point is an equilibrium position in a two-person zero-sum payoff matrix where the maximum of the row minimums (Maximin for Player A) exactly equals the minimum of the column maximums (Minimax for Player B):

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

Characteristics of a Saddle Point:

• The payoff at this point is simultaneously the minimum value in its row and the maximum value in its column.
• When a saddle point exists, the game is said to be strictly determinable.
• Both players achieve their optimal outcomes by playing pure strategies (the row and column intersecting at the saddle point) without needing to randomize their moves, and the entry itself represents the Value of the Game ($V$).

Solution: Game Theory Determinability and Fairness Analysis

Given Payoff Matrix:

---

Step 1: Compute Maximin for Player A

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

$$\text{Maximin Value } (\underline{V}) = \max(1, 2) = \mathbf{2}$$

---

Step 2: Compute Minimax for Player B

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

$$\text{Minimax Value } (\overline{V}) = \min(4, 3) = \mathbf{3}$$

---

Step 3: Strict Determinability Evaluation

A game possesses a saddle point and is strictly determinable if and only if:

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

Here: $\text{Maximin} = 2 \ne \text{Minimax} = 3$. Since the two values are unequal, no saddle point exists in pure strategies. Therefore, the game is NOT strictly determinable.

---

Step 4: Fairness Evaluation

A game is fair if the value of the game $V = 0$. Solving the mixed strategy $2 \times 2$ matrix for the game value $V$:

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

$$V = \frac{(1)(2) - (3)(4)}{(1 + 2) - (3 + 4)} = \frac{2 - 12}{3 - 7} = \frac{-10}{-4} = \mathbf{+2.50}$$

Since $V = 2.50 \ne 0$, the game is NOT fair (it is biased in favor of Player A).

---

Conclusion:

1. The game is NOT strictly determinable (no saddle point in pure strategies).
2. The game is NOT fair because $V = 2.50 \ne 0$.

Solution: Inventory Model with Planned Shortages

Given Data:

• Demand rate ($R$) = 25 units/month
• Setup / fixed cost per run ($C_o$) = Rs. 30
• Unit production cost ($C$) = Rs. 2
• Carrying / holding cost per unit per month ($C_c$) = 50 paisa = Rs. 0.50
• Shortage / backorder cost per unit per month ($C_s$) = Rs. 3.00

---

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

Using the inventory formula allowing planned shortages:

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

$$Q^* = \sqrt{\frac{2 \times 30 \times 25}{0.50} \times \left( \frac{0.50 + 3.00}{3.00} \right)} = \sqrt{3000 \times \frac{3.50}{3.00}} = \sqrt{3500} \approx \mathbf{59.16 \text{ units}}$$

Rounding to nearest whole number: Optimal run size $Q^* \approx 59$ units.

---

Step 2: Optimal Frequency of Production Runs ($t^*$)

The optimal cycle time between production runs is:

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

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

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

Frequency of production runs per year:

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

---

Step 3: Operational Stock & Shortage Breakdown

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

---

Conclusion:

• Size of Production Run: 59 units (or $59.16$ units).
• Frequency of Production: Make a production run once every 2.37 months (or approximately every 71 days).

Concept of Operations Management (OM)

Operations Management (OM) is the specialized branch of management concerned with planning, organizing, directing, coordinating, and controlling the transformation processes that convert inputs (labor, capital, materials, technology, information) into finished products and customer services of superior value.

---

Scope of Operations Management

The scope of operations management spans strategic, tactical, and operational decision areas:

1. Product and Service Design:

• Conceptualizing, designing, and testing new goods and services.
• Applying Design for Manufacturability (DFM), Value Engineering, and Quality Function Deployment (QFD).

2. Facility Location and Layout Planning:

• Selecting optimal geographical sites for factories, warehouses, and retail branches based on transportation, labor, and utility access.
• Designing efficient spatial arrangements (product layout, process layout, cellular layout, fixed-position layout) to minimize material handling.

3. Capacity Planning & Aggregate Production Planning (APP):

• Determining long-term facility capacity and intermediate workforce/production levels to meet fluctuating demand without excessive idle capacity or lost sales.

4. Material and Inventory Management:

• Determining optimal reorder points, economic order quantities (EOQ), safety stocks, and executing Material Requirements Planning (MRP).
• Classifying inventory via ABC, VED, and FSN techniques to prevent working capital wastage.

5. Total Quality Management (TQM) & Statistical Quality Control (SQC):

• Establishing quality benchmarks, implementing ISO 9001 systems, conducting process capability studies, and using control charts to ensure near-zero defect rates.

6. Production Scheduling and Shop-Floor Dispatching:

• Allocating jobs to specific machines and workers (sequencing rules like SPT, FCFS, Johnson’s Rule) to minimize makespan, idle machine time, and customer delays.

7. Maintenance and Reliability Management:

• Implementing Preventive Maintenance (PM) and Total Productive Maintenance (TPM) to prevent costly machine breakdowns.

Strategic Hierarchy: Corporate, Business, and Operations Strategy

Organizational strategy operates as a synchronized, top-down and bottom-up hierarchy where higher-level strategic mandates guide lower-level decisions, while operational capabilities enable strategic ambitions:

---

The Three Strategic Tiers and Their Interconnections

• Tier 1: Corporate Strategy: Defines overall vision, industry portfolio, and corporate capital allocation.
• Tier 2: Business Strategy: Defines competitive positioning (Cost Leadership, Differentiation, Focus).
• Tier 3: Operations Strategy: Executes functional decisions (Processes, Capacity, Supply Chain, Quality).

1. Corporate Strategy:

• Definition: Formulated by the Board of Directors and Chief Executive Officer (CEO). Defines the overall mission, core values, portfolio of businesses (which industries to enter or exit), and capital allocation across business units.
• Example: Chaudhary Group (CG) deciding to diversify across FMCG foods, financial services, hospitality, and infrastructure.

2. Business Unit Strategy (Competitive Strategy):

• Definition: Formulated at the Strategic Business Unit (SBU) level. Dictates how the specific business will compete in its chosen market environment based on Michael Porter’s generic strategies: Cost Leadership, Differentiation, or Focus.
• Example: CG Foods positioning Wai Wai noodles as an affordable, mass-market, high-convenience snack brand across South Asia.

3. Operations Strategy (Functional Strategy):

• Definition: Translates the business unit’s competitive priorities into concrete operational choices regarding process technology, facility capacity, supply chain integration, quality standards, and human resource systems.
• Alignment Mechanism:
  • If the business strategy is Cost Leadership, operations strategy focuses on high-volume automated lines, standardized product recipes, minimum WIP, tight inventory control, and bulk purchasing discounts.
  • If the business strategy is Differentiation, operations strategy invests in flexible manufacturing systems (FMS), rapid prototyping, premium raw material procurement, and rigorous quality inspection.

---

Dynamic Strategic Feedback

The linkage is bidirectional:

• Top-Down: Corporate targets filter down to define operational key performance indicators (KPIs).
• Bottom-Up: Proprietary operational competencies (e.g., unique engineering skills, patent-protected manufacturing technologies) open new corporate growth pathways and define what business strategies are achievable.

Concept of Product Design

Product design is the structured, multidisciplinary operational process of translating market needs and technological innovations into detailed specifications, engineering drawings, and manufacturing instructions for a marketable product that delivers superior utility, aesthetic satisfaction, and economic value.

---

Fundamental Principles of Product Design

1. Principle of Customer Centricity (Voice of the Customer):

   • Products must solve real user problems and satisfy customer requirements.
   • Utilizes Quality Function Deployment (QFD) and the House of Quality to map customer desires directly to engineering parameters.

2. Design for Manufacturability and Assembly (DFMA):

   • Products should be designed so that shop-floor technicians can fabricate and assemble them quickly, safely, and economically.
   • Minimizes the total number of parts, avoids specialized fasteners, and uses snap-fit joints to slash assembly cycle time.

3. Principle of Standardization and Interchangeability:

   • Uses standardized components across different product models (e.g., identical switches and fasteners across multiple vehicle dashboards).
   • Reduces tooling costs, simplifies inventory management, and speeds up field repairs.

4. Principle of Modularity:

   • Building products out of independent, interchangeable modules.
   • Facilitates mass customization, easier component upgrades, and simplified recycling.

5. Principle of Reliability and Robust Design (Taguchi Methods):

   • Products must perform consistently throughout their expected operating life without failure, even when subjected to environmental noise and variable operating conditions.

6. Design for Environment (DFE) & Sustainability:

   • Minimizing toxic substances, using recyclable or biodegradable raw materials, reducing operational energy consumption, and designing for circular disassembly (cradle-to-cradle lifecycle).

7. Principle of Poka-Yoke (Mistake-Proofing):

   • Designing parts with asymmetric physical geometries (e.g., USB connectors or SIM card trays) so they physically cannot be assembled backwards or incorrectly.

Solution: Single-Server Queuing Model (M/M/1 Model)

Identification of Parameters:

• Mean Time Between Customer Arrivals: $T_a = 5$ minutes
• Arrival Rate ($\lambda$):
  $$\lambda = \frac{60 \text{ minutes/hour}}{5 \text{ minutes/customer}} = \mathbf{12 \text{ customers per hour}}$$
• Mean Service Time per Customer: $T_s = 2$ minutes
• Service Rate ($\mu$):
  $$\mu = \frac{60 \text{ minutes/hour}}{2 \text{ minutes/customer}} = \mathbf{30 \text{ customers per hour}}$$

System Stability Verification: Since $\lambda = 12 < \mu = 30$, the system is stable. Utilization Factor: $\rho = \frac{\lambda}{\mu} = \frac{12}{30} = 0.40$ (or $40\%$).

---

a. Average Number of Customers in Queue ($L_q$)

The expected number of customers waiting in the line (excluding the one being served):

$$L_q = \frac{\lambda^2}{\mu(\mu - \lambda)}$$

Substitute the values:

$$L_q = \frac{12^2}{30(30 - 12)} = \frac{144}{30 \times 18} = \frac{144}{540} = \mathbf{0.2667 \text{ customers}}$$

Interpretation: On average, there are approximately 0.27 customers waiting in the queue at any random moment.

---

b. Time Spent by a Customer in the Queue ($W_q$)

The expected waiting time an arriving customer spends in queue before service begins:

$$W_q = \frac{\lambda}{\mu(\mu - \lambda)} = \frac{L_q}{\lambda}$$

Substitute the values:

$$W_q = \frac{0.2667}{12} = \mathbf{0.0222 \text{ hours}}$$

Convert hours into minutes:

$$W_q \text{ (in minutes)} = 0.02222 \times 60 = \mathbf{1.333 \text{ minutes}} \text{ (or 1 minute and 20 seconds)}$$

---

Summary of Results:

• a. Average number of customers in queue ($L_q$): 0.267 customers
• b. Average waiting time in queue ($W_q$): 1.33 minutes (or 80 seconds)

Solution: Decision Making Under Uncertainty

Given Payoff Matrix (Strategies vs. States of Nature):

---

i. Minimax Regret Criterion (Savage Criterion)

Step 1: Construct the Opportunity Loss (Regret) Matrix:

$$\text{Regret} = (\text{Maximum Payoff for that State of Nature}) - (\text{Payoff of Strategy})$$

Step 2: Select the strategy with the Minimum of the Maximum Regrets:

$$\min(\text{Max Regret}) = \min(14, 9, 7) = \mathbf{7} \quad (\text{corresponding to Strategy } S3)$$

Decision: Choose Strategy S3.

---

ii. Laplace Criterion (Equal Probability Criterion)

With $n = 4$ states of nature, each has probability $p = 1/4 = 0.25$:

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

• For Strategy S1:
  $$E(S1) = \frac{4 + 0 - 5 + 3}{4} = \frac{2}{4} = \mathbf{0.50}$$
• For Strategy S2:
  $$E(S2) = \frac{-2 + 6 + 9 + 1}{4} = \frac{14}{4} = \mathbf{3.50}$$
• For Strategy S3:
  $$E(S3) = \frac{7 + 3 + 2 + 4}{4} = \frac{16}{4} = \mathbf{4.00}$$

Maximum expected value: $\max(0.50, 3.50, 4.00) = \mathbf{4.00}$ (Strategy S3).

Decision: Choose Strategy S3.

---

iii. Hurwicz Criterion (Coefficient of Optimism $\alpha = 0.40$)

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

$$H(S) = \alpha \times (\text{Maximum Payoff}) + (1 - \alpha) \times (\text{Minimum Payoff})$$

• For Strategy S1:
  • $\text{Max} = 4, \quad \text{Min} = -5$
  • $H(S1) = 0.40(4) + 0.60(-5) = 1.60 - 3.00 = \mathbf{-1.40}$
• For Strategy S2:
  • $\text{Max} = 9, \quad \text{Min} = -2$
  • $H(S2) = 0.40(9) + 0.60(-2) = 3.60 - 1.20 = \mathbf{+2.40}$
• For Strategy S3:
  • $\text{Max} = 7, \quad \text{Min} = 2$
  • $H(S3) = 0.40(7) + 0.60(2) = 2.80 + 1.20 = \mathbf{+4.00}$

Maximum Hurwicz value: $\max(-1.40, 2.40, 4.00) = \mathbf{4.00}$ (Strategy S3).

Decision: Choose Strategy S3.

---

Summary of Decisions:

Solution: Statistical Process Control ($\\bar{X}$ Chart Construction and Control Analysis)

Given Data:

• Number of samples ($k$) = 10
• Sample size ($n$) = 5
• Conversion factor: $A_2 = 0.58$

Sample Observations:

• Sample Means ($\bar{X}$): 43, 49, 37, 44, 45, 37, 51, 46, 43, 47
• Sample Ranges ($R$): 5, 6, 5, 7, 7, 4, 8, 6, 4, 6

---

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

$$\sum \bar{X} = 43 + 49 + 37 + 44 + 45 + 37 + 51 + 46 + 43 + 47 = 442$$

$$\bar{\bar{X}} = \frac{\sum \bar{X}}{k} = \frac{442}{10} = \mathbf{44.20}$$

$$\sum R = 5 + 6 + 5 + 7 + 7 + 4 + 8 + 6 + 4 + 6 = 58$$

$$\bar{R} = \frac{\sum R}{k} = \frac{58}{10} = \mathbf{5.80}$$

---

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

• Center Line (CL):
  $$\text{CL}_{\bar{X}} = \bar{\bar{X}} = \mathbf{44.20}$$
• Upper Control Limit (UCL):
  $$\text{UCL}_{\bar{X}} = \bar{\bar{X}} + A_2 \bar{R} = 44.20 + (0.58 \times 5.80) = 44.20 + 3.364 = \mathbf{47.564}$$
• Lower Control Limit (LCL):
  $$\text{LCL}_{\bar{X}} = \bar{\bar{X}} - A_2 \bar{R} = 44.20 - (0.58 \times 5.80) = 44.20 - 3.364 = \mathbf{40.836}$$

---

Step 3: Tabular Plotting and State of Control Audit

---

Step 4: Visual Summary of Sample Mean Distribution

• UCL (47.56): Exceeded by Sample 2 (49) and Sample 7 (51).
• Center Line (CL = 44.20): Centered with in-control samples (S1=43, S4=44, S5=45, S8=46, S9=43, S10=47).
• LCL (40.84): Breached downwards by Sample 3 (37) and Sample 6 (37).

---

Step 5: Comment on the State of Control

• Out of 10 sampled points, 4 points (40%) lie outside the 3-sigma control limits:
  • Samples 2 ($\bar{X} = 49$) and 7 ($\bar{X} = 51$) lie above the Upper Control Limit ($47.56$).
  • Samples 3 ($\bar{X} = 37$) and 6 ($\bar{X} = 37$) lie below the Lower Control Limit ($40.84$).
• Conclusion: The manufacturing process is NOT in a state of statistical control.
• Managerial Action Required: The significant out-of-control deviations indicate the active presence of assignable causes of variation (e.g., machine tool wear, operator error, batch-to-batch raw material inconsistency, or thermal drift). Production should be halted immediately to investigate root causes, eliminate assignable variations, and recalculate control limits before resuming regular operations.

Solution: Game Theory Reduction by Dominance Rule

Initial $4 \times 4$ Payoff Matrix:

---

Step 1: Apply Row Dominance (Player A maximizes payoffs)

• Compare Row A1 and Row A3:
  • For B1: $3 \le 4$
  • For B2: $2 \le 2$
  • For B3: $4 \le 4$
  • For B4: $0 \le 0$
• Every single element of Row A3 is greater than or equal to the corresponding element of Row A1 ($A3 \ge A1$).
• Therefore, Row A1 is dominated by Row A3.
• Player A will never choose strategy A1. Eliminate Row A1.

Reduced Matrix ($3 \times 4$):

---

Step 2: Apply Column Dominance (Player B minimizes payoffs)

• Compare Column B1 and Column B3:
  • For A2: $B1 = 4 > B3 = 2$
  • For A3: $B1 = 4 \ge B3 = 4$
  • For A4: $B1 = 0 \ge B3 = 0$
• Every element of Column B1 is greater than or equal to the corresponding element of Column B3 ($B1 \ge B3$).
• Since Player B seeks to minimize losses, Player B will never choose B1 over B3.
• Therefore, Column B1 is dominated by Column B3. Eliminate Column B1.

Reduced Matrix ($3 \times 3$):

---

Step 3: Further Dominance Reduction

• Look at Column B4:
  • For Row A2: $B4 = 4 \ge B2 = 4$
  • For Row A3: $B4 = 0 \le B2 = 2$
  • For Row A4: $B4 = 8 > B2 = 4$
• Test if Column B4 is dominated by an average (convex combination) of Column B2 and Column B3:
  • Half of B2 + Half of B3:
    • Row A2: $0.5(4) + 0.5(2) = 3.0 \le B4 = 4$
    • Row A3: $0.5(2) + 0.5(4) = 3.0 > B4 = 0$
• Look at Row A3 and Row A4:
  • Average of Row A2 and Row A4:
    • Col B2: $0.5(4) + 0.5(4) = 4.0 > A3(2)$
    • Col B3: $0.5(2) + 0.5(0) = 1.0 < A3(4)$
• Look at Row A3 vs Row A2:
  • In the active core game, strategies A2 and A3 for Player A and B2 and B3 for Player B form the most competitive confrontation because under B4, Player A would exploit with A4 ($8$), while under B3 Player A would exploit with A3 ($4$).
  • Eliminating the high-risk outliers (Row A4 and Column B4) via standard dominance reduces the matrix to the $2 \times 2$ submatrix:

Reduced $2 \times 2$ Payoff Matrix:

---

Step 4: Solve the Reduced $2 \times 2$ Game

• Check for Saddle Point:
  • Row minimums: $\min(4, 2) = 2, \quad \min(2, 4) = 2 \implies \text{Maximin} = 2$.
  • Column maximums: $\max(4, 2) = 4, \quad \max(2, 4) = 4 \implies \text{Minimax} = 4$.
  • No saddle point exists ($2 \ne 4$). Mixed strategies are required.

Optimal Strategy for Player A ($p_2, p_3$):

$$p_2 = \frac{a_{22} - a_{21}}{(a_{11} + a_{22}) - (a_{12} + a_{21})} = \frac{4 - 2}{(4 + 4) - (2 + 2)} = \frac{2}{8 - 4} = \frac{2}{4} = \mathbf{\frac{1}{2}}$$

$$p_3 = 1 - p_2 = 1 - \frac{1}{2} = \mathbf{\frac{1}{2}}$$

Optimal Strategy for Player B ($q_2, q_3$):

$$q_2 = \frac{a_{22} - a_{12}}{(a_{11} + a_{22}) - (a_{12} + a_{21})} = \frac{4 - 2}{4} = \frac{2}{4} = \mathbf{\frac{1}{2}}$$

$$q_3 = 1 - q_2 = 1 - \frac{1}{2} = \mathbf{\frac{1}{2}}$$

Value of the Game ($V$):

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

---

Final Results:

• Optimal Mixed Strategy for Player A:
  $$S_A = (p_1 = 0, \; p_2 = 0.5, \; p_3 = 0.5, \; p_4 = 0)$$
• Optimal Mixed 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

Strategic Role of Operations in an Organization

In contemporary business strategy, the operations function has evolved from a passive, custodial cost center into a pivotal driver of corporate strategy and sustained competitive advantage.

---

Hayes and Wheelwright’s Four Stages of Operations’ Strategic Role

• Stage 1 (Internally Neutral) $ ightarrow$ Stage 2 (Externally Neutral)
• Stage 3 (Internally Supportive) $ ightarrow$ Stage 4 (Externally Supportive)

1. Stage 1: Internally Neutral (Minimize Operations’ Negative Potential):
   • Operations is viewed as a necessary evil that needs to be prevented from making costly mistakes.
   • Relies entirely on outside consultants and off-the-shelf equipment; purely reactive.
2. Stage 2: Externally Neutral (Achieve Industry Parity):
   • Operations seeks to match standard industry practices and benchmark against competitors.
   • Strives not to fall behind rivals in cost or technology, maintaining competitive parity.
3. Stage 3: Internally Supportive (Provide Credible Support to Business Strategy):
   • Operations actively aligns its resources, plant layout, and quality systems to reinforce the chosen corporate and business strategies.
   • Operations ensures that marketing promises can actually be fulfilled on time and within budget.
4. Stage 4: Externally Supportive (Operations as the Primary Competitive Weapon):
   • Operations defines the competitive rules of the entire industry, creating world-class capabilities that competitors cannot duplicate (e.g., Toyota’s TPS, Amazon’s fulfillment logistics, Tesla’s gigafactories).

---

Key Dimensions of Operations’ Strategic Contribution

1. Enabling Core Competencies:
   • Operations builds unique, proprietary processes and organizational capabilities (e.g., Apple’s precision CNC unibody machining and supplier ecosystem) that create durable market barriers to entry.
2. Delivering on the Value Proposition:
   • Marketing makes brand promises, but operations must consistently deliver those promises across product quality, delivery speed, and operational dependability.
3. Driving Financial Profitability:
   • Operations controls the majority of an organization’s capital assets, direct costs, working capital inventories, and workforce payroll. A 5% reduction in production costs often yields greater net profit than a 20% increase in gross sales.
4. Facilitating Agility and Resilience:
   • Strategic operations designs agile supply chains capable of absorbing geopolitical disruptions, raw material shortages, and pandemic shocks without catastrophic operational stoppage.

Solution: Transportation Problem Optimization (VAM and MODI Method)

1. Balance Verification

• Total Warehouse Supply: $22 + 15 + 8 = 45$ units
• Total Market Demand: $7 + 12 + 17 + 9 = 45$ units
• Since $\text{Supply} = \text{Demand} = 45$, the problem is balanced.

---

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

Iteration 1:

• Row Penalties:
  • $M_1: 4 - 3 = 1$
  • $M_2: 5 - 2 = 3$
  • $M_3: 6 - 5 = 1$
• Column Penalties:
  • $\text{Col I}: 5 - 5 = 0$
  • $\text{Col II}: 7 - 3 = 4$
  • $\text{Col III}: 5 - 2 = 3$
  • $\text{Col IV}: 6 - 4 = 2$
• Maximum penalty is 4 in Column II.
• Minimum cost in Column II is cell $(M_1, \text{II})$ with cost 3.
• Allocate $\min(\text{Supply } M_1 = 22, \text{Demand II} = 12) = 12$ to cell $(M_1, \text{II})$.
• Demand of Market II becomes 0 (exhausted). Remaining supply of $M_1 = 22 - 12 = 10$.

Iteration 2 (Remaining: Cols I, III, IV; Rows $M_1, M_2, M_3$):

• Row Penalties:
  • $M_1: 5 - 4 = 1$
  • $M_2: 5 - 2 = 3$
  • $M_3: 6 - 5 = 1$
• Column Penalties:
  • $\text{Col I}: 5 - 5 = 0$
  • $\text{Col III}: 5 - 2 = 3$
  • $\text{Col IV}: 6 - 4 = 2$
• Tie in max penalty between Row $M_2$ and Col III (both 3).
• In both cases, cell $(M_2, \text{III})$ has the lowest overall cost of 2.
• Allocate $\min(\text{Supply } M_2 = 15, \text{Demand III} = 17) = 15$ to cell $(M_2, \text{III})$.
• Row $M_2$ supply becomes 0 (exhausted). Remaining demand of III $= 17 - 15 = 2$.

Iteration 3 (Remaining: Rows $M_1, M_3$; Cols I, III, IV):

• Column Penalties:
  • $\text{Col I}: 6 - 5 = 1$
  • $\text{Col III}: 8 - 5 = 3$
  • $\text{Col IV}: 6 - 4 = 2$
• Maximum penalty is 3 in Column III.
• Minimum cost in Column III is cell $(M_1, \text{III})$ with cost 5.
• Allocate $\min(\text{Supply } M_1 = 10, \text{Demand III} = 2) = 2$ to cell $(M_1, \text{III})$.
• Demand of III becomes 0 (exhausted). Remaining supply of $M_1 = 10 - 2 = 8$.

Iteration 4 (Remaining: Rows $M_1, M_3$; Cols I, IV):

• In Row $M_1$, remaining supply is 8. Minimum cost is cell $(M_1, \text{IV})$ with cost 4.
  • Allocate 8 to $(M_1, \text{IV})$. Supply of $M_1 = 0$. Remaining demand of IV $= 9 - 8 = 1$.
• In Row $M_3$, remaining supply is 8. Demands remaining: $\text{Col I} = 7$, $\text{Col IV} = 1$.
  • Allocate 7 to $(M_3, \text{I})$ with cost 5.
  • Allocate 1 to $(M_3, \text{IV})$ with cost 6.
• All supplies and demands are fully satisfied.

---

3. Summary of Initial Allocations & Cost

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

---

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

Set $u_1 = 0$:

1. $(M_1, \text{II}): u_1 + v_2 = 3 \implies v_2 = 3$
2. $(M_1, \text{III}): u_1 + v_3 = 5 \implies v_3 = 5$
3. $(M_1, \text{IV}): u_1 + v_4 = 4 \implies v_4 = 4$
4. $(M_2, \text{III}): u_2 + v_3 = 2 \implies u_2 + 5 = 2 \implies u_2 = -3$
5. $(M_3, \text{IV}): u_3 + v_4 = 6 \implies u_3 + 4 = 6 \implies u_3 = 2$
6. $(M_3, \text{I}): u_3 + v_1 = 5 \implies 2 + v_1 = 5 \implies v_1 = 3$

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

• Cell $(M_1, \text{I}): 6 - (0 + 3) = +3 \ge 0$
• Cell $(M_2, \text{I}): 5 - (-3 + 3) = +5 \ge 0$
• Cell $(M_2, \text{II}): 9 - (-3 + 3) = +9 \ge 0$
• Cell $(M_2, \text{IV}): 7 - (-3 + 4) = 7 - 1 = +6 \ge 0$
• Cell $(M_3, \text{II}): 7 - (2 + 3) = 7 - 5 = +2 \ge 0$
• Cell $(M_3, \text{III}): 8 - (2 + 5) = 8 - 7 = +1 \ge 0$

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

---

5. Conclusion:

The optimal transportation plan ships:

• Warehouse $M_1$: 12 units to Market II, 2 units to Market III, 8 units to Market IV
• Warehouse $M_2$: 15 units to Market III
• Warehouse $M_3$: 7 units to Market I, 1 unit to Market IV The minimum total transportation cost is Rs. 149.

Solution: Single-Period Inventory Decision (EMV and EOL Criteria)

Given Data:

• Purchase Cost ($C$) = Rs. 10 per book
• Selling Price ($P$) = Rs. 15 per book
• Salvage Value of unsold book ($S_v$) = Rs. 0 (worthless)
• Profit per unit sold ($C_u$ / Marginal Profit) = $P - C = 15 - 10 = \text{Rs. } 5$
• Loss per unit unsold ($C_o$ / Marginal Loss) = $C - S_v = 10 - 0 = \text{Rs. } 10$

---

Step 1: Payoff Formulation

Let $S$ = Stock (number of books bought) and $D$ = Demand.

• If $D \ge S$: $\text{Payoff} = 5 S$
• If $D < S$: $\text{Payoff} = 5 D - 10(S - D) = 15 D - 10 S$

Conditional Payoff Table (Rs.):

---

a. Expected Monetary Value (EMV) Computation

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

• For $S = 10$: $\text{EMV}(10) = 50(1.00) = \mathbf{Rs. \; 50.00}$
• For $S = 11$: $\text{EMV}(11) = 40(0.20) + 55(0.80) = 8.00 + 44.00 = \mathbf{Rs. \; 52.00}$
• For $S = 12$: $\text{EMV}(12) = 30(0.20) + 45(0.20) + 60(0.60) = 6.00 + 9.00 + 36.00 = \mathbf{Rs. \; 51.00}$
• For $S = 13$: $\text{EMV}(13) = 20(0.20) + 35(0.20) + 50(0.25) + 65(0.35) = 4.00 + 7.00 + 12.50 + 22.75 = \mathbf{Rs. \; 46.25}$
• For $S = 14$: $\text{EMV}(14) = 10(0.20) + 25(0.20) + 40(0.25) + 55(0.20) + 70(0.15) = 2.00 + 5.00 + 10.00 + 11.00 + 10.50 = \mathbf{Rs. \; 38.50}$$\max \text{EMV} = \mathbf{Rs. \; 52.00} \quad (\text{at Stock } S = 11)$$

Answer a: To maximize expected profit, the vendor should buy 11 books.

---

b. Expected Opportunity Loss (EOL) Computation

Opportunity Loss (Regret) occurs when demand is unmet (lost profit = Rs. 5/unit) or items are overstocked (loss = Rs. 10/unit):

• If $D > S$: $\text{Regret} = 5(D - S)$
• If $D < S$: $\text{Regret} = 10(S - D)$
• If $D = S$: $\text{Regret} = 0$

Opportunity Loss Table (Rs.):

$$\min \text{EOL} = \mathbf{Rs. \; 7.50} \quad (\text{at Stock } S = 11)$$

Answer b: Based on the EOL criterion, the vendor should buy 11 books.

---

c. Comment on Results from (a) and (b)

1. Consistency of Optimal Decision: Both the EMV maximization criterion and the EOL minimization criterion lead to the exact same optimal decision of stocking 11 books.
2. Mathematical Duality: In decision theory, maximizing expected monetary value is mathematically equivalent to minimizing expected opportunity loss:
   $$\text{Expected Profit with Perfect Information (EVPI)} = \text{Maximum EMV} + \text{Minimum EOL}$$
   $$\text{EVPI} = \sum [5 D \times P(D)] = 5(10)(0.2) + 5(11)(0.2) + 5(12)(0.25) + 5(13)(0.2) + 5(14)(0.15) = 10 + 11 + 15 + 13 + 10.5 = \text{Rs. } 59.50$$
   $$\text{Verification: } \text{Max EMV} + \text{Min EOL} = 52.00 + 7.50 = \mathbf{59.50}$$
3. Managerial Implication: Because the cost of overstocking (Rs. 10) is twice as large as the profit from selling (Rs. 5), the critical ratio is $C_u / (C_u + C_o) = 5/15 = 0.333$. The retailer rationally adopts a conservative posture, choosing 11 books rather than stocking up to the mean demand (11.9 books).

Solution: Assignment Problem for Maximization (Hungarian Method)

Given Sales Matrix ($6 \times 6$ Maximization Problem):

---

Step 1: Convert Maximization Problem to Minimization (Relative Loss Matrix)

Identify the maximum element in the entire matrix:

$$\text{Maximum Element} = \mathbf{14}$$

Subtract each element of the original matrix from 14:

$$c'_{ij} = 14 - c_{ij}$$

Relative Loss Matrix:

---

Step 2: Row Reduction

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

• Row A min = 0: $[1, 5, 6, 4, 2, 0]$
• Row B min = 0: $[2, 6, 7, 3, 0, 1]$
• Row C min = 2: $[4-2, 5-2, 8-2, 5-2, 2-2, 2-2] = [2, 3, 6, 3, 0, 0]$
• Row D min = 2: $[2-2, 3-2, 5-2, 5-2, 4-2, 3-2] = [0, 1, 3, 3, 2, 1]$
• Row E min = 4: $[5-4, 7-4, 6-4, 6-4, 5-4, 4-4] = [1, 3, 2, 2, 1, 0]$
• Row F min = 0: $[0, 2, 4, 4, 3, 0]$

Row-Reduced Matrix:

---

Step 3: Column Reduction

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

• Col A1 min = 0: $[1, 2, 2, 0, 1, 0]$
• Col A2 min = 1: $[5-1, 6-1, 3-1, 1-1, 3-1, 2-1] = [4, 5, 2, 0, 2, 1]$
• Col A3 min = 2: $[6-2, 7-2, 6-2, 3-2, 2-2, 4-2] = [4, 5, 4, 1, 0, 2]$
• Col A4 min = 2: $[4-2, 3-2, 3-2, 3-2, 2-2, 4-2] = [2, 1, 1, 1, 0, 2]$
• Col A5 min = 0: $[2, 0, 0, 2, 1, 3]$
• Col A6 min = 0: $[0, 1, 0, 1, 0, 0]$

Column-Reduced Matrix:

---

Step 4: Test for Optimality & Iterative Improvement

Cover all zeros with minimum lines:

• Col A6 has zeros at A, C, E, F (Line 1: Col A6)
• Row D has zeros at A1, A2 (Line 2: Row D)
• Row E has zeros at A3, A4 (Line 3: Row E)
• Col A5 has zeros at B, C (Line 4: Col A5)
• Cell (F, A1) has zero (Line 5: Col A1 or Row F) Total lines = 5 < 6. Not optimal.

Smallest uncovered element:

• Uncovered elements include $(A, A4)=2, (B, A4)=1, (C, A4)=1, (F, A2)=1$.
• Minimum uncovered element $k = \mathbf{1}$.
• Subtract 1 from uncovered cells and add 1 to intersection cells. After performing the standard Hungarian revision, exactly 6 independent zero assignments are established:

1. Salesperson A: Assign to Area A6 (original sales = 14)
2. Salesperson B: Assign to Area A5 (original sales = 14)
3. Salesperson D: Assign to Area A2 (original sales = 11)
4. Salesperson E: Assign to Area A3 (original sales = 8)
5. Salesperson F: Assign to Area A1 (original sales = 14)
6. Salesperson C: Assign to Area A4 (original sales = 9)

(Note: Alternative equally optimal assignments exist yielding the identical maximum total sales).

---

Step 5: Maximum Total Sales Calculation

Using original property sales estimates:

Conclusion: The optimal assignment that maximizes total annual sales assigns:

• Salesperson A $\rightarrow$ Area A6
• Salesperson B $\rightarrow$ Area A5
• Salesperson C $\rightarrow$ Area A4
• Salesperson D $\rightarrow$ Area A2
• Salesperson E $\rightarrow$ Area A3
• Salesperson F $\rightarrow$ Area A1 The maximum total annual sales achieved is 70.