Operations Management

This is a production lot-size model with shortages permitted. The production cost (Rs 2 per item) is the same for any lot size, so it does not affect the decision.

Given: $D=25$ units/month, $C_3=$ Rs 30 per run, $C_1=$ Rs 0.50 per unit per month, $C_2=$ Rs 3 per unit per month.

$Q^*=\sqrt{\frac{2C_3D}{C_1}\cdot\frac{C_1+C_2}{C_2}}=\sqrt{\frac{2(30)(25)}{0.5}\cdot\frac{3.5}{3}}=\sqrt{3500}\approx 59 \text{ units}$

$t^*=\frac{Q^*}{D}=\frac{59.16}{25}\approx 2.37 \text{ months}$

Maximum shortage $=Q^*\dfrac{C_1}{C_1+C_2}=59.16\times\dfrac{0.5}{3.5}\approx 8.5$ units.

So a production run of about 59 units should be made about every 2.4 months.

Inventory management is the process of planning, ordering, storing, controlling and monitoring a firm's stock of raw materials, work-in-process and finished goods so that the right quantity is available at the right time at the lowest total cost. It balances the cost of ordering and holding stock against the risk of stockouts, using techniques such as EOQ, ABC analysis and reorder levels.

Row minima: A1 = 1, A2 = -3, so maximin = 1. Column maxima: B1 = 4, B2 = 2, so minimax = 2.

Since maximin (1) is not equal to minimax (2), there is no saddle point, so the game is not strictly determinable (it needs mixed strategies).

Value of the game:

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

Since $v \neq 0$, the game is not fair; it favours Player A.

Operations strategy is a long-term plan for using a firm's resources to support its business strategy. It is called a competitive weapon because it builds the capabilities that win customers from rivals.

• Low cost: efficient processes, good capacity use, waste reduction and economies of scale allow lower prices than competitors. Example: Toyota's lean production.
• Quality: a strategy of quality improvement and TQM gives reliable, superior products, which builds reputation and loyalty.
• Speed and delivery: short lead times and dependable delivery attract time-sensitive customers.
• Flexibility: the ability to change volumes and products quickly lets a firm meet customer demands and survive market changes.
• Innovation: new product and process development gives a unique offering that is hard to imitate.
• Aligned capabilities: decisions on location, technology, supply chain and workforce form a combination of strengths that competitors cannot copy easily, and so make a sustainable advantage.

Hence a well-formulated operations strategy converts operational capabilities into market advantages, and so it acts as a competitive weapon.

Emerging issues in product and service design:

• Customer focus and customisation: products must meet changing needs, and mass customisation allows tailor-made options at low cost.
• Shorter product life cycles: rapid change requires faster development and quick introduction to the market (time-to-market).
• Technology: CAD/CAM, 3D printing, digital prototyping and the use of AI and IoT in design.
• Sustainability and environment: design for recycling and reuse, lower energy use, biodegradable materials and green regulation.
• Concurrent engineering and cross-functional teams: design, production and marketing work together to reduce changes and time.
• Design for manufacturing and assembly (DFMA), modularity and standardisation: fewer parts and common components reduce cost and complexity.
• Quality and reliability: quality function deployment (QFD), robust design and value analysis.
• Service design issues: customer involvement, intangibility, e-services and self-service, and consistency in service delivery.
• Global issues: designing for global markets, culture, safety standards and outsourcing of design.
• Ethics and safety: product liability, data privacy and legal compliance.

Row minima: 0, 2, 0, 0, so maximin = 2. Column maxima: 4, 4, 4, 8, so minimax = 4. These differ, so there is no saddle point. Reduce by dominance (Player B is the minimiser; Player A the maximiser).

Step 1. The average of columns B3 $(4,2,4,0)$ and B4 $(0,4,0,8)$ is $(2,3,2,4)$. Column B2 $(2,4,4,4)$ is greater than or equal to this average in every row, so B2 is dominated (B, the minimiser, would never play it). Delete B2.

Step 2. Row A3 $(0,4,0)$ is dominated by row A1 $(3,4,0)$. Delete A3.

Step 3. The average of B3 $(4,2,0)$ and B4 $(0,4,8)$ is $(2,3,4)$. Column B1 $(3,3,4)$ is greater than or equal to it in each row, so B1 is dominated. Delete B1.

Step 4. Row A2 $(2,4)$ is the average of A1 and A4, $(2,4)$, so A2 is dominated. Delete A2.

Reduced 2x2 game

No saddle point (maximin 0, minimax 4). Use the formulas.

$p_1=\frac{8-0}{4+8-0-0}=\frac{2}{3},\quad p_4=\frac{1}{3}$

$q_3=\frac{8-0}{12}=\frac{2}{3},\quad q_4=\frac{1}{3}$

$v=\frac{4(8)-0(0)}{12}=\frac{8}{3}\approx 2.67$

Optimal strategies

• Player A: $(A_1,A_2,A_3,A_4)=(2/3,\,0,\,0,\,1/3)$.
• Player B: $(B_1,B_2,B_3,B_4)=(0,\,0,\,2/3,\,1/3)$.
• Value of the game $=8/3$.

Check: against A's strategy, B1 gives $10/3$, B2 gives $8/3$, B3 gives $8/3$ and B4 gives $8/3$, so none gives B less than $8/3$.

Row minima: A1 = 1, A2 = 2, so maximin = 2. Column maxima: B1 = 4, B2 = 3, so minimax = 3.

Since maximin (2) is not equal to minimax (3), there is no saddle point, so the game is not strictly determinable.

Value of the game:

$v=\frac{(1)(2)-(3)(4)}{1+2-3-4}=\frac{-10}{-4}=2.5$

Since $v\neq 0$, the game is not fair.

Player A is the maximiser and B the minimiser. There is no saddle point: the maximin is 2 and the minimax is 4.

Step 1: Dominance.

1. Row A3 (4, 2, 4, 0) is at least as good as row A1 (3, 2, 4, 0) in every column, so A1 is dominated and is deleted.
2. With A1 removed, column B1 (4, 4, 0) is at least as large as column B3 (2, 4, 0) in every row. B wants smaller losses, so B1 is dominated and is deleted.
3. Column B2 (4, 2, 4) is at least as large as the average of columns B3 and B4, which is (3, 2, 4). So B2 is dominated and is deleted.
4. Row A2 (2, 4) equals the average of rows A3 (4, 0) and A4 (0, 8), which is (2, 4). So A2 is dominated and is deleted.

The reduced 2x2 game is:

Step 2: Solve the 2x2 game. Let A play A3 with probability $p$ and A4 with probability $1-p$. Equating the payoffs against B3 and B4:

$4p = 8(1-p) \Rightarrow p = \tfrac{2}{3}$

Let B play B3 with probability $q$ and B4 with probability $1-q$:

$4q = 8(1-q) \Rightarrow q = \tfrac{2}{3}$

Optimal strategies.

• Player A: A1 = 0, A2 = 0, A3 = 2/3, A4 = 1/3.
• Player B: B1 = 0, B2 = 0, B3 = 2/3, B4 = 1/3.

Value of the game:

$V = 4\left(\tfrac23\right) = \tfrac{8}{3} \approx 2.67$

This favours player A.

Operations strategy is the long-term plan for using the firm's resources (capacity, technology, facilities, people, supply chain) to support the business strategy. It becomes a competitive weapon when the operations function is deliberately built to give the firm an edge over rivals on the dimensions customers value, called competitive priorities:

• Cost: Efficient processes, high productivity and low waste allow lower prices or higher margins.
• Quality: Consistent, reliable products build brand image and loyalty.
• Delivery speed and dependability: Short lead times and on-time delivery win customers.
• Flexibility: The firm can quickly change volume, product mix or design.
• Innovation and service: New products and good after-sales service differentiate the firm.

Because operations accounts for most of a firm's costs and assets, and because these capabilities are hard for competitors to copy, a sound operations strategy turns day-to-day operations into a lasting advantage rather than a mere support function.

Example: Toyota's lean production system (JIT, Kaizen, zero defects) gave it low cost and very high quality at the same time, which made it a world leader. Similarly, Dell used build-to-order and direct selling to deliver customised computers quickly with very little inventory. In Nepal, a dairy that invests in cold chains and fast distribution competes on freshness and delivery, and a restaurant chain that standardises its process competes on consistency and speed. In each case the operational choice, not just marketing, is what beats the rival.

Difference between product design and service design

Emerging issues in product and service design

1. Sustainability and green design: Recycling, remanufacturing, energy efficiency and use of eco-friendly materials, with cradle-to-cradle thinking.
2. Technology and digitalisation: Use of CAD/CAM, 3D printing, AI, IoT and self-service or mobile apps in service delivery.
3. Mass customisation: Meeting individual customer needs at near mass-production cost through modular design.
4. Speed to market: Concurrent engineering and shorter design cycles.
5. Design for manufacturing and assembly (DFMA), and value analysis: Simplified, cheaper products.
6. Quality function deployment (QFD): Translating customer needs into design specifications.
7. Globalisation and standards: Designs must meet varying customer tastes, regulations and ISO standards.
8. Customer experience and personalisation: Service design stresses user experience and multi-channel delivery.
9. Ethical, legal and safety concerns: Product liability, data privacy and security of customers.

A continuous production system is one in which standardised products are made in large volumes without interruption, using special-purpose machines arranged in a fixed sequence (product layout), with materials flowing continuously through the process. Examples are oil refining, cement, sugar and chemical plants.

Key issues for an operations manager:

• Deciding what to produce and the design of products and services.
• Capacity planning and facility location and layout.
• Quality management and control.
• Scheduling, inventory and materials management.
• Controlling cost and productivity.
• Managing the workforce and technology, and meeting customer delivery needs.

Advantages of operations strategy:

• Gives a clear direction and links operations to corporate and business goals.
• Builds competitive priorities such as cost, quality, speed and flexibility.
• Improves use of resources and capacity.
• Gives consistency in decisions, with better coordination between functions.
• Helps the firm respond to market and technology changes and gain competitive advantage.

Game theory is important because it:

• Gives a mathematical method for decision-making under competition and conflict, where the outcome depends on the choices of rivals.
• Helps firms choose the best strategy in pricing, advertising, bidding and product launch.
• Helps predict competitors' reactions and reduces risk.
• Is applied in business, economics, military planning and bargaining.

Value analysis is a systematic, function-oriented study of an existing product or component to find its essential functions and then achieve them at the lowest total cost, without reducing quality or reliability. It removes unnecessary cost by looking for cheaper materials, simpler design or better processes.

Assumptions of a single-channel queuing system:

• There is one service facility (server), and one customer is served at a time.
• Arrivals follow a Poisson distribution with a mean rate $\lambda$ and are independent.
• Service times are exponentially distributed with a mean rate $\mu$.
• The queue discipline is first come, first served (FCFS).
• The population (source) of customers is infinite and there is no limit on queue length.
• The service rate is greater than the arrival rate ($\mu > \lambda$), so the system reaches a steady state.
• There is no balking, reneging or jockeying.

Stages of the historical evolution of TQM:

1. Inspection (detecting defects after production).
2. Quality control (statistical process control).
3. Quality assurance (preventing defects through systems).
4. Total Quality Management (organisation-wide, customer-focused continuous improvement).

Independent demand is demand for finished goods or end items that comes from customers and is not related to the demand for any other item. It must be forecast, for example demand for cars or TVs.

Dependent demand is demand for items (raw materials, components, sub-assemblies) that is derived from the demand for a higher-level item. It is calculated, not forecast, for example the tyres needed for planned car production (4 per car).

The transformation process is the core of operations management. It converts inputs into outputs (goods or services) by adding value.

Components

• Inputs: materials, labour, capital, energy, information, land and equipment.
• Transformation: the activities that add value, such as physical (manufacturing), locational (transport), exchange (retailing), storage (warehousing), physiological (hospital) and informational (telecom).
• Outputs: finished goods and services delivered to customers.
• Feedback and control: measures of output (quality, cost, time) are compared with standards and used to adjust inputs and processes.

Example (bakery): Inputs are flour, sugar, yeast, electricity, bakers and ovens. The transformation is mixing, baking and packing. The outputs are bread and cakes sold to customers. Feedback comes from quality checks and customer complaints, which lead to a change in the recipe or the baking time.

Example (hospital): Inputs are patients, doctors, medicines and equipment. The treatment process transforms sick patients into healthy ones (output).

Given: $D=10{,}000$, ordering cost $S=$ Rs 5, holding cost $=10\%$ of unit price. Price is Rs 100 for $Q<200$ and Rs 95 for $Q\ge 200$.

Step 1: EOQ at each price

At Rs 95: $h=0.10\times95=9.5$

$EOQ=\sqrt{\frac{2(10000)(5)}{9.5}}=\sqrt{10526}\approx 102.6$

This is below 200, so it is not feasible at the Rs 95 price.

At Rs 100: $h=10$

$EOQ=\sqrt{\frac{2(10000)(5)}{10}}=100$

This is below 200, so it is feasible at the Rs 100 price.

Step 2: Compare total costs

At $Q=100$ (Rs 100):

$TC=10000(100)+\frac{10000}{100}(5)+\frac{100}{2}(10)=1{,}000{,}000+500+500=\text{Rs }1{,}001{,}000$

At $Q=200$ (Rs 95):

$TC=10000(95)+\frac{10000}{200}(5)+\frac{200}{2}(9.5)=950{,}000+250+950=\text{Rs }951{,}200$

The cost is lowest at 200 units.

Economic lot size = 200 units (total cost Rs 951,200).

(i) Maximax: maximum payoff of each action: P1 = 8, P2 = 18, P3 = 14. The largest is 18, so choose P2.

(ii) Hurwitz (coefficient of pessimism 0.60, so optimism 0.40):

$H = 0.60(\text{minimum}) + 0.40(\text{maximum})$

• P1: $0.6(-10)+0.4(8)=-6+3.2=-2.8$
• P2: $0.6(-4)+0.4(18)=-2.4+7.2=4.8$
• P3: $0.6(8)+0.4(14)=4.8+5.6=10.4$

The highest value is 10.4, so choose P3.

(iii) Laplace (equal probability of 1/4 each):

• P1: $(8+0-10+6)/4=1$
• P2: $(-4+12+18-2)/4=24/4=6$
• P3: $(14+9+9+8)/4=40/4=10$

The highest is 10, so choose P3.

The table gives ranges for the first nine samples, so the first nine samples are used (the extra tenth mean, 16, has no range).

$\bar{\bar X}=\frac{15+17+15+18+17+14+18+15+17}{9}=\frac{146}{9}=16.22$

$\bar R=\frac{7+7+4+9+8+7+12+4+11}{9}=\frac{69}{9}=7.67$

Mean ($\bar X$) chart

• Central line $=16.22$
• $UCL=\bar{\bar X}+A_2\bar R=16.22+0.58(7.67)=16.22+4.45=20.67$
• $LCL=\bar{\bar X}-A_2\bar R=16.22-4.45=11.78$

Range (R) chart

• Central line $=7.67$
• $UCL=D_4\bar R=2.115(7.67)=16.22$
• $LCL=D_3\bar R=0$

Comment: all sample means (14 to 18) lie between 11.78 and 20.67, and all ranges (4 to 12) lie between 0 and 16.22. No point is outside the limits, so the process is in statistical control.

This is an assignment problem (minimisation). Use the Hungarian method.

Cost matrix (rows W, X, Y, Z; columns jobs 1 to 4):

Row reduction (subtract row minima 3, 1, 2, 1):

Column reduction (column minima 0, 0, 0, 1; subtract 1 from column 4):

Test: Y and Z have zeros only in column 3, so a full assignment is not possible. Three lines (row W, row X, column 3) cover all zeros, which is fewer than 4. The smallest uncovered element is 1. Subtract it from the uncovered elements and add it at the line intersections (W3 and X3):

Assignment on zeros: Z must take 3, then Y takes 1, X takes 2, and W takes 4.

Optimal assignment

Minimum total cost = 4 + 1 + 3 + 1 = 9.

Quality control is the operational process of measuring the quality of products or services against set standards, finding deviations and taking corrective action so that the output meets customer requirements. It includes inspection, testing and statistical monitoring.

Seven tools of quality (QC tools)

1. Check sheet: a simple form to record the frequency of defects or events as data are collected, which makes patterns visible.
2. Histogram: a bar chart that shows the frequency distribution of measurements, and shows the shape, centre and spread of the process.
3. Pareto chart: a bar chart in descending order with a cumulative line, based on the 80/20 rule. It identifies the "vital few" causes that give most of the defects.
4. Cause-and-effect (fishbone or Ishikawa) diagram: it lists possible causes of a problem under categories such as man, machine, material, method, measurement and environment.
5. Scatter diagram: it plots two variables to show whether there is a relationship (positive, negative or none) between them.
6. Control chart: it plots sample data over time against a central line and upper and lower control limits, and so shows whether a process is in statistical control.
7. Flow chart (or stratification): it shows the steps of a process in sequence and helps to find bottlenecks and unnecessary steps. (Stratification, which separates data into groups, is also given as the seventh tool.)

Use: these tools help to collect data, identify and analyse problems, find root causes and monitor improvement, and so support continuous improvement and TQM.

Total supply = 14 + 27 + 14 = 55 = total demand, so the problem is balanced.

Initial solution by Vogel's Approximation Method (VAM)

1. Largest penalty is column S2 (3). The cheapest cell is F1-S2 (7), so allocate 14. F1 is exhausted.
2. Largest penalty is column S1 (5). The cheapest cell is F2-S1 (8), so allocate 15. F2 has 12 left.
3. Largest penalty is column S3 (3). The cheapest cell is F2-S3 (9), so allocate 11. F2 has 1 left.
4. The remaining demand is S2 = 5 and S4 = 10. Allocate F2-S2 = 1, F3-S2 = 4 and F3-S4 = 10.

The number of allocated cells is 6, which equals $m+n-1=3+4-1=6$, so the solution is non-degenerate.

Test for optimality (MODI)

Put $u_1=0$. Then from the allocated cells: $v_2=7$, $u_2=11-7=4$, $v_1=8-4=4$, $v_3=9-4=5$, $u_3=10-7=3$, $v_4=10-3=7$.

Opportunity costs $c_{ij}-(u_i+v_j)$ of the empty cells:

• F1-S1: $9-4=5$
• F1-S3: $10-5=5$
• F1-S4: $8-7=1$
• F2-S4: $11-11=0$
• F3-S1: $13-7=6$
• F3-S3: $12-8=4$

None is negative, so the solution is optimal.

Minimum cost

$14(7)+15(8)+1(11)+11(9)+4(10)+10(10)=98+120+11+99+40+100$

Minimum transportation cost = Rs 468.

This is a single-period (newsboy) problem.

• Profit on each unit sold in season: $20-15=$ Rs 5.
• Loss on each unit unsold: cost 15, minus bargain price 9, plus holding cost 2 = Rs 8.

Conditional profit $=5\times(\text{units sold})-8\times(\text{units unsold})$.

Conditional profit table (Rs)

Example (stock 13): $0.2(52)+0.8(65)=10.4+52=62.4$.

Example (stock 14): $0.2(44)+0.2(57)+0.6(70)=8.8+11.4+42=62.2$.

Optimum stock level = 13 units, with maximum expected value (EMV) = Rs 62.40.

(a) Major problems faced by the operations manager

• Product design effort slowed down after the early phase, and the design strategies tried did not last in the market.
• There was no quality measurement since the company's establishment. Quality is tested only at the end, before supply, and not built in.
• Trade unions obstructed employee reward and punishment, so discipline and motivation were weak.
• A new executive director was appointed to manage operations, which suggests weak continuity in management.
• Sales fell to only Rs 5,000,000, below the previous year, and the market position deteriorated.
• The company was finally bankrupt and closed by the government.

(b) How good product design helps maintain quality

• Quality is built in at the design stage, which is cheaper than finding defects later through inspection.
• Design specifications, tolerances and standard materials make the product consistent.
• Simple design with fewer parts (design for manufacturing) reduces errors in production.
• Tools such as QFD turn customer needs into design features, so the product satisfies customers.
• Product testing, prototypes and evaluation find faults before launch.
• Reliability and robust design reduce failure, warranty cost and complaints.

(c) Product design strategies to adopt change

• Imitation or adaptation strategy: copy or adapt successful designs of competitors.
• Innovation strategy: develop new products, supported by R&D, to lead the market.
• Standardisation and modular design: common components make changes cheap and fast.
• Customisation: designs adapted to customer needs and tastes (for example, saris of different patterns and fabrics).
• Concurrent engineering and cross-functional teams: marketing, production and design share information to shorten time to market.
• Value analysis and design for manufacture: reduce cost while keeping function.
• Continuous improvement (Kaizen) and regular product evaluation, with phasing out of old designs.

(d) How the operations manager can cope with the challenges

• Install a quality system (TQM, statistical process control, ISO standards) with inspection at every stage.
• Revive product design and R&D, using customer and market feedback.
• Negotiate with the trade unions, use a fair performance-based reward system and give training and participation to employees.
• Plan capacity, inventory and production schedule against demand forecasts.
• Control costs by value analysis and waste reduction, and review technology.
• Work with marketing to track the market and set clear targets for sales recovery.
• Set up regular performance monitoring and strong leadership for continuity.

Product design is the process of deciding the features, form, materials, dimensions and performance of a product so that it meets customer needs and can be produced economically. It turns an idea into specifications and drawings.

Example: designing a new smartphone, including its size, screen, battery, camera, material and price, according to customer needs and production capability.

Characteristics of an intermittent production system:

• Products are made in batches or lots, according to customer orders, and not continuously.
• A wide variety of products, with low to medium volume of each.
• General-purpose machines and equipment, arranged in a process layout.
• Flexible, skilled labour.
• Frequent set-up changes and changes in machine use.
• Higher work-in-process inventory and more complex scheduling.
• Higher unit cost than continuous production, but greater flexibility. Examples are job shops, bakeries and tailoring.

Value analysis is a systematic, organised study of the functions of an existing product or component, done in order to deliver the same required function at the lowest possible cost without lowering quality, reliability or performance. It removes unnecessary cost from a product that is already in production.

The Expected Monetary Value (EMV) criterion is a decision-making rule under risk. For each alternative, the payoffs for all states of nature are multiplied by their probabilities and summed:

$EMV=\sum p_i \times \text{payoff}_i$

The alternative with the highest EMV (or lowest, if the payoffs are costs) is chosen.

The ISO 9000 series is a family of international standards for quality management systems, issued by the International Organization for Standardization (ISO). It sets requirements for a documented system that ensures an organization consistently provides products and services meeting customer and regulatory requirements. ISO 9001 is the standard against which firms are certified. The series is based on principles such as customer focus, leadership, process approach and continual improvement.

A dummy variable (dummy source or destination) is added in a transportation problem when total supply is not equal to total demand (an unbalanced problem), to make it balanced.

• If supply is greater than demand, a dummy destination with the demand equal to the excess supply is added.
• If demand is greater than supply, a dummy source with the supply equal to the shortage is added.
• The transportation cost of the dummy row or column is taken as zero, since nothing is actually shipped.

A saddle point of a game is the position in the payoff matrix where the maximin of the row minima equals the minimax of the column maxima. This element is at once the smallest in its row and the largest in its column. It gives the value of the game, and the game has pure strategies for both players.

Operations Management (OM) is the management of the process of converting inputs (materials, labour, capital, information) into finished goods and services in the most efficient and effective way. It involves planning, organising, directing and controlling the production system so as to meet customer needs at the lowest cost.

Scope of operations management

• Product and service design and development: deciding what to produce and its features.
• Process selection and facility location and layout: choosing technology, site and arrangement of facilities.
• Capacity and aggregate planning: deciding how much to produce and when.
• Materials and inventory management: purchasing, storage and control of stock (EOQ, MRP, JIT).
• Production planning, scheduling and control.
• Quality management: inspection, quality control, TQM and ISO standards.
• Maintenance management and work study (methods and work measurement).
• Supply chain management and productivity improvement.
• Human resource and cost control within operations.

A firm's strategy works at three levels, and each level must fit with the others.

Corporate strategy is the top level. It decides what businesses the firm should be in, how resources are shared among them, and growth, diversification or retrenchment.

Business strategy is set for each business unit. It decides how to compete in a particular market, through cost leadership, differentiation or focus.

Operations (functional) strategy decides how the operations function will support the business strategy. It sets the competitive priorities (cost, quality, speed, flexibility) and the decisions on capacity, location, process technology, supply chain and workforce.

Linkage

• The strategy flows top-down: corporate goals shape business strategy, and business strategy sets the requirements that operations must fulfil.
• Operations strategy implements the higher-level strategies, and its capabilities in turn shape which business strategies are feasible (bottom-up feedback).
• They must be consistent: for example, a business that competes on low cost needs an operations strategy of high efficiency, standardisation and high volume.
• When all three are aligned, resources are used well, and the firm gains a sustainable competitive advantage.

Product design is the process of defining the features, specifications, materials, shape and performance of a product, to meet customer needs, be economical to produce and be of good quality. It links the customer's needs to the production system.

Principles of product design

• Customer orientation: the design must satisfy identified customer needs and preferences.
• Functionality and performance: the product should do its intended job reliably.
• Simplicity: fewer parts and simple design reduce cost and failures.
• Standardisation and modularity: common parts and modules allow variety with low cost, easy repair and interchangeability.
• Manufacturability (design for manufacturing): the design should suit available processes, materials and skills.
• Quality and reliability: built in from the design stage, with appropriate tolerances.
• Cost-effectiveness: the design should deliver value at an acceptable cost and price (value analysis).
• Safety, ergonomics and aesthetics: safe, easy to use and attractive.
• Environmental considerations: minimal waste, recyclable and energy-efficient.
• Maintainability and serviceability after sale.

This is an M/M/1 queue.

Arrival rate: $\lambda = \dfrac{1}{5} = 0.2$ customers per minute.

Service rate: $\mu = \dfrac{1}{2} = 0.5$ customers per minute.

Utilisation: $\rho = \dfrac{\lambda}{\mu} = 0.4$, which is below 1, so the queue is stable.

(a) Average number of customers in the queue:

$L_q = \frac{\lambda^2}{\mu(\mu-\lambda)} = \frac{0.2^2}{0.5(0.5-0.2)} = \frac{0.04}{0.15} \approx 0.27 \text{ customers}$

(b) Average time a customer spends waiting in the queue:

$W_q = \frac{L_q}{\lambda} = \frac{0.2667}{0.2} \approx 1.33 \text{ minutes}$

The total time in the system, including service, is $W_s = \frac{1}{\mu-\lambda} = 3.33$ minutes.

The strategies are the columns S1, S2 and S3, and the payoffs are gains. Over states A, B, C, D the payoffs are:

• S1: 4, 0, -5, 3
• S2: -2, 6, 9, 1
• S3: 7, 3, 2, 4

(i) Minimax Regret. The best payoff in each state is A = 7, B = 6, C = 9, D = 4. The regrets (best payoff minus actual payoff) are:

The smallest maximum regret is 7, so choose S3.

(ii) Laplace. Each state is equally likely, with probability 1/4.

• S1: $(4+0-5+3)/4 = 0.5$
• S2: $(-2+6+9+1)/4 = 3.5$
• S3: $(7+3+2+4)/4 = 4.0$

The highest expected payoff is 4.0, so choose S3.

(iii) Hurwitz with $\alpha = 0.4$. The index is $H = 0.4(\text{max}) + 0.6(\text{min})$.

• S1: $0.4(4) + 0.6(-5) = -1.4$
• S2: $0.4(9) + 0.6(-2) = 2.4$
• S3: $0.4(7) + 0.6(2) = 4.0$

The highest index is 4.0, so choose S3.

Step 1: Averages.

$\bar{\bar{X}} = \frac{442}{10} = 44.2, \qquad \bar{R} = \frac{58}{10} = 5.8$

Step 2: Control limits for the mean chart ($A_2 = 0.58$).

$UCL = \bar{\bar X} + A_2\bar R = 44.2 + 0.58(5.8) = 44.2 + 3.364 = 47.56$

$CL = 44.2$

$LCL = \bar{\bar X} - A_2\bar R = 44.2 - 3.364 = 40.84$

Step 3: Plot the ten sample means against these limits, with the sample number on the x-axis and the mean on the y-axis.

Comment. Samples 2 and 7 lie above the UCL of 47.56. Samples 3 and 6 lie below the LCL of 40.84. Four of the ten points are outside the limits, so the process average is not in statistical control. Assignable causes such as machine settings, operator or material changes should be investigated and removed, and the limits revised afterwards.

For reference, the range chart limits are $UCL_R = D_4\bar R = 2.115(5.8) = 12.27$ and $LCL_R = D_3\bar R = 0$. All ranges lie within them, so the variability is stable. The problem is with the process mean.

Operations strategy is the long-term plan for using the organisation's resources to produce goods and services in a way that supports its business strategy and competitive position. Operations is one of the three core functions, alongside marketing and finance. It is not only a support function. It is a source of competitive advantage.

Strategic roles of operations

1. Supports the corporate and business strategy. Operations decisions on capacity, location, process and technology are aligned with the firm's mission, so that what the firm promises to customers can actually be delivered.
2. Builds competitive priorities. Operations develops the capabilities that win orders: low cost, high quality, fast and reliable delivery, flexibility in volume and product mix, and innovation.
3. Makes long-term structural decisions. Capacity, facility location and layout, choice of technology and degree of vertical integration commit large resources for many years and are hard to reverse.
4. Manages infrastructure decisions. Workforce policies, quality systems, planning and control, and supplier relationships determine day-to-day performance.
5. Creates cost advantage. Better productivity, less waste, lean practices and good inventory control lower unit cost. This allows lower prices or higher margins.
6. Delivers quality and customer value. Through TQM, Six Sigma and ISO standards, operations produces products and services that meet or exceed customer expectations and builds loyalty.
7. Gives flexibility and responsiveness. Flexible processes and supply chains let the firm adapt to demand shifts, customisation and new product launches.
8. Drives innovation. Operations works with R&D and marketing, for example through concurrent engineering and QFD, to bring new products to market faster.
9. Integrates with other functions. It supplies marketing with products that can be delivered, supplies finance with cost and asset-use data, and coordinates with suppliers and customers across the supply chain.
10. Contributes to sustainability and risk management. Efficient resource use, waste reduction, ethical sourcing and resilient supply chains protect the firm's long-term position.

Conclusion. Operations turns strategy into results. A firm whose operations are aligned with its strategy gains durable advantages in cost, quality, speed and flexibility that competitors find hard to copy.

Total supply = 22 + 15 + 8 = 45 and total demand = 7 + 12 + 17 + 9 = 45, so the problem is balanced.

Step 1: Initial solution by Vogel's Approximation Method.

The initial allocation is:

There are 6 occupied cells, which equals $m+n-1 = 6$, so the solution is non-degenerate.

Step 2: Optimality test (MODI). Put $u_1 = 0$ and use $u_i + v_j = c_{ij}$ on the occupied cells:

$v_2 = 3,\ v_3 = 5,\ v_4 = 4,\ u_2 = -3,\ u_3 = 2,\ v_1 = 3$

The opportunity costs $c_{ij} - u_i - v_j$ of the empty cells are:

All are positive, so the solution is optimal and unique.

Step 3: Minimum cost.

$12(3) + 2(5) + 8(4) + 15(2) + 7(5) + 1(6) = 36 + 10 + 32 + 30 + 35 + 6 = 149$

The minimum transportation cost is 149 (in the units of the cost matrix).

Profit per book sold = 15 - 10 = Rs 5. Loss per unsold book = Rs 10 (the cost, since leftovers are worthless).

(a) Expected profit (EMV).

For a stock of $Q$ books, the expected profit is $5E[\min(D,Q)] - 10\,(Q - E[\min(D,Q)]) = 15E[\min(D,Q)] - 10Q$.

The maximum expected profit is Rs 52, so the vendor should buy 11 books.

(b) Expected opportunity loss (EOL).

Shortage loss is Rs 5 per book short (lost profit). Excess loss is Rs 10 per unsold book.

The minimum EOL is Rs 7.50, so the vendor should buy 11 books.

(c) Comment. Both criteria recommend 11 books. This is expected, because EOL and expected profit are linked by

$\text{EMV} + \text{EOL} = \text{expected profit under perfect information} = \text{Rs } 59.50$

for every stock level (for example, 52 + 7.50 = 59.50). Maximising EMV is therefore the same as minimising EOL. The critical ratio $\frac{5}{5+10} = 0.33$ agrees: the smallest $Q$ with $P(D \le Q) \ge 0.33$ is 11, since $P(D \le 10) = 0.20$ and $P(D \le 11) = 0.40$. The value of perfect information is Rs 7.50.

This is a maximisation assignment problem. Convert it to a minimisation by subtracting every entry from the largest value (14) to get the opportunity-loss matrix.

Row reduction. The row minima are 0, 0, 2, 2, 4, 0. After subtracting them:

Column reduction. The column minima are 0, 1, 2, 2, 0, 0. After subtracting them:

Test. The zeros can be covered by 5 lines (columns A5 and A6, and rows D, E and F). That is fewer than 6, so no complete assignment exists yet.

The smallest uncovered element is 1. Subtract it from the uncovered elements and add it to the elements at the line intersections:

A complete assignment on zeros is now possible: F to A1, D to A2, E to A3, C to A4, B to A5, A to A6.

Optimal assignment.

$\text{Maximum total sales} = 14 + 14 + 9 + 11 + 8 + 14 = 70 \text{ hours per year}$

(a) Coordinating production and service functions

Ms. Shrestha treated BIROI as one transformation system in which inputs (materials, labour, information) are converted into goods and services.

• Mapping the transformation process. She traced the flow from inputs through conversion to outputs and feedback, which showed where the factory floor and customer care depended on each other.
• Classifying production. Because of the variety of customisable items she classified production as intermittent (job and batch). This justified flexible layouts, general-purpose equipment and skilled workers, and it showed why scheduling and service promises needed close coordination.
• Measuring productivity. She examined productivity by department (single-factor and multifactor), which exposed bottlenecks and idle capacity.
• Resolving the manufacturing-service conflict. Production wanted long runs and low cost. Service wanted speed, customisation and responsiveness. She set shared objectives, a customer-centred service approach and a hybrid production strategy (make-to-stock for standard components, make-to-order for customised ones). This reduced waste and aligned both with customer value.
• Concurrent engineering and QFD. Product and service were designed together, so customer requirements were built into both the appliance and its support.

(b) Changes in decision-making through operations strategy

Before the review, decisions were made department by department and reactively. After the operations plan, decisions were made against common business objectives of cost, quality, delivery and flexibility.

• Product design decisions used customer feedback loops and high-value features (QFD).
• Production decisions used a hybrid make-to-stock and make-to-order approach and the intermittent process.
• Inventory decisions followed economic order quantities adjusted for fluctuating demand.
• Quality decisions rested on data from control charts, Six Sigma and ISO 9000, not on inspection alone.
• Supply-chain and logistics decisions used decision theory and operations research (transportation and assignment).
• Finance and marketing decisions, such as pricing and distributor negotiation, were informed by game theory and operations data.

Decisions became cross-functional, data-driven and long-term, with the operations strategy as the common reference.

(c) Inventory control and quality systems

Inventory control. Economic order quantity balances ordering and holding costs. Safety stock and reorder levels absorb fluctuating demand. Classification such as ABC focuses control on high-value items. Together these cut carrying costs, avoid stock-outs and free working capital.

Quality systems. Six Sigma reduces defects and variation using the DMAIC cycle. ISO 9000 standardises and documents processes. Control charts monitor the process and signal assignable causes before defects occur. These give lower rework, scrap and warranty costs and higher customer satisfaction.

The two systems together improved effectiveness through lower cost, fewer defects, more reliable delivery and a better reputation.

(d) Game theory and decision theory in strategic planning and negotiation

Decision theory. Under uncertainty, BIROI can compare alternatives using expected monetary value (EMV), expected opportunity loss (EOL) and criteria such as maximin, maximax, Laplace, Hurwitz and minimax regret. It can also value perfect information. This supports choices on order quantity, capacity, supplier and shipping under uncertain demand, and it makes risk explicit.

Game theory. In the pricing dispute, BIROI and the distributor are rival players with conflicting interests. A payoff matrix lets BIROI anticipate the other party's options. If a saddle point exists, there is a pure strategy for each side and a stable value of the game. If not, a mixed strategy (found by dominance, algebraic or graphical methods) gives the best expected outcome. This helps BIROI choose its bargaining position, anticipate responses and avoid concessions that leave it worse off.

Used together, the tools make planning more rational, reduce the cost of wrong decisions and improve the firm's negotiating strength, which supports a sustained competitive advantage.

Operations management is the management of the process that converts inputs (materials, labour, machines, information) into goods and services. It involves planning, organising, directing and controlling these activities to deliver products of the required quality, in the required quantity, at the right time and at the lowest cost.

Operations strategy is the long-term plan for using the resources of the operations function to support the overall business strategy and give a competitive advantage. It sets the direction on the main priorities (cost, quality, delivery, flexibility), and on decisions such as capacity, location, process technology, facilities and supply chain.

Service design is the planning and creation of a service so that it meets customer needs and the firm's objectives. It covers the service concept, the service package (facilities, goods, explicit and implicit services), the delivery process and system, the physical layout, and the customer contact points. Good service design makes the service useful, consistent, efficient and satisfying to the customer.

The types of production system are:

1. Job production: one or a few items made to a customer's order, such as a bridge or tailor-made suit.
2. Batch production: a limited quantity made in lots, such as bakery items or medicines.
3. Mass (flow) production: large volumes of standard items on assembly lines, such as cars or appliances.
4. Continuous (process) production: non-stop production, such as oil refining, cement and chemicals.
5. Project production (sometimes added): a single large unique project, such as a dam or building.

The main objective of Six Sigma is to improve quality by reducing defects and variation in processes to a level of no more than 3.4 defects per million opportunities. This leads to near-perfect products and services, lower cost, higher customer satisfaction and better profit.

Just-in-time (JIT) delivers materials and produces goods only when needed. Its main benefits are:

• Lower inventory and storage costs, with less capital tied up.
• Less waste and scrap, and fewer defects.
• Higher quality and productivity.
• Shorter lead times and quicker response to customer demand.
• Better use of space and closer supplier relations.

Value analysis is a systematic, function-oriented technique to study an existing product or part to find ways of reducing its cost without lowering its quality or function. It asks what the item does, what it costs, and whether the same function can be achieved at lower cost by changing the design, material or method. It is used to remove unnecessary cost and increase the value of a product to the customer.

Independent demand is demand for an item that is not derived from the demand for any other item. It comes directly from customers and is therefore forecast rather than calculated. Finished goods, spare parts and service items are typical examples, whereas raw materials and components have dependent demand.

Productivity is the relationship between the output produced and the inputs (labour, materials, capital, energy) used to produce it.

$\text{Productivity}=\frac{\text{Output}}{\text{Input}}$

Higher productivity means more output from the same resources, or the same output from fewer resources. It can be measured as partial, multifactor or total productivity, and it is the key source of lower costs and competitiveness.

Operations management has evolved through the following stages.

• Craft production (before 1750): Skilled artisans made goods one at a time to customer order. Output was small, costly and not uniform.
• Industrial Revolution (1770s onwards): Machines and steam power replaced manual work. Factories, division of labour (Adam Smith) and the idea of interchangeable parts (Eli Whitney) appeared.
• Scientific management (early 1900s): F. W. Taylor applied time and motion study, standard methods and incentives. Frank and Lillian Gilbreth refined motion study, and Henry Gantt developed scheduling charts.
• Mass production (1910s onwards): Henry Ford introduced the moving assembly line and standardised products, which gave high volume and low unit cost.
• Human relations movement (1930s-1960s): The Hawthorne studies, Maslow and McGregor stressed worker motivation and participation.
• Management science / quantitative era (1940s-1970s): Operations research methods such as linear programming, queuing, simulation and inventory models were developed. Computers supported MRP.
• Quality and lean revolution (1970s-1990s): Japanese practices (JIT, Kanban, TQM, Kaizen) and Deming, Juran and Crosby's quality ideas spread worldwide.
• Modern era (1990s onwards): Business process reengineering, ERP, supply chain management, e-commerce, globalisation, mass customisation, Six Sigma, sustainability and automation or AI now shape operations.

Total Quality Management (TQM) is a company-wide, customer-focused approach in which every employee works continuously to improve the quality of products, services and processes. Its main principles are as follows.

1. Customer focus: Quality is defined by the customer. The organisation must understand and meet or exceed customer needs and expectations.
2. Top management commitment and leadership: Management sets the quality vision, provides resources and leads by example.
3. Total employee involvement and empowerment: All employees take responsibility for quality. Quality circles and teamwork are encouraged.
4. Continuous improvement (Kaizen): Processes are improved step by step using the PDCA (Plan-Do-Check-Act) cycle.
5. Process approach: Quality is built into the process. Defects are prevented rather than inspected out afterwards.
6. Fact-based decision making: Decisions rest on data and statistical tools such as control charts, Pareto and cause-and-effect diagrams.
7. Supplier partnership: Long-term, trust-based relationships with suppliers ensure the quality of inputs.
8. Training and education: Employees receive continuous training in quality tools and skills.
9. Benchmarking and a quality culture: The firm compares itself with the best performers and makes quality part of its culture.

Given: daily use = 60 packs, working days = 275, holding cost $H$ = Rs 5 per pack per year, ordering cost $S$ = Rs 8 per order.

Annual demand:

$D = 60 \times 275 = 16{,}500 \text{ packs}$

(a) Optimal order size (EOQ)

$Q^* = \sqrt{\frac{2DS}{H}} = \sqrt{\frac{2 \times 16500 \times 8}{5}} = \sqrt{52800} \approx 229.8 \approx 230 \text{ packs}$

(b) Total annual cost

Annual carrying cost:

$\frac{Q}{2}H = \frac{229.78}{2}\times 5 \approx \text{Rs } 574.46$

Annual ordering cost:

$\frac{D}{Q}S = \frac{16500}{229.78}\times 8 \approx \text{Rs } 574.46$

$\text{Total annual cost} = 574.46 + 574.46 \approx \text{Rs } 1{,}148.91$

At the EOQ, carrying cost equals ordering cost. The total cost of about Rs 1,149 per year excludes the purchase cost of the items.

Mean of sample means:

$\bar{\bar{X}}=\frac{14+17+15+18+17+13+18+16+17+19}{10}=\frac{164}{10}=16.4$

Mean of ranges:

$\bar{R}=\frac{6+7+5+10+8+12+5+12+11+5}{10}=\frac{81}{10}=8.1$

Control limits with $A_2 = 0.58$:

$A_2\bar{R}=0.58\times 8.1=4.698$

$UCL=\bar{\bar{X}}+A_2\bar{R}=16.4+4.698=21.10$

$CL = 16.4$

$LCL=\bar{\bar{X}}-A_2\bar{R}=16.4-4.698=11.70$

Mean chart: Plot the sample numbers 1 to 10 on the horizontal axis and the sample means on the vertical axis. Draw three horizontal lines: UCL = 21.10, CL = 16.4 and LCL = 11.70. Then join the plotted points 14, 17, 15, 18, 17, 13, 18, 16, 17, 19.

Comment: All ten sample means (the lowest is 13 and the highest is 19) lie between the LCL of 11.70 and the UCL of 21.10. There is no point outside the limits and no obvious trend, so the process average is in statistical control, and only chance variation is present. (The ranges should also be checked on an R chart before the process is fully declared in control.)

Use the Hungarian method on the cost matrix.

Step 1: Row reduction (subtract the row minimum)

Step 2: Column reduction (column minima are 0, 1, 1)

Step 3: Test and assign. The minimum number of lines needed to cover all zeros is 3, which equals the order of the matrix, so an optimal assignment is possible. Starting from the unique zeros, C is assigned to Z. Then B is assigned to Y (its other zero, in X, is left for A), and A is assigned to X.

Minimum total cost = Rs 12.

Payoff table (Rs):

(i) Maximax (optimistic). Row maxima are A = 4000, B = 2500, C = 2500. The largest is 4000, so choose A.

(ii) Maximin (pessimistic). Row minima are A = 1750, B = 2250, C = 1000. The largest is 2250, so choose B.

(iii) Minimax regret. Column maxima are X = 4000, Y = 2500, Z = 2500.

The smallest maximum regret is 750, so choose A.

(iv) Laplace (equal probability of 1/3).

• A = (4000 + 1750 + 2500)/3 = 2750
• B = (2250 + 2250 + 2500)/3 = 2333.33
• C = (1000 + 2500 + 2000)/3 = 1833.33

Choose A.

(v) Hurwicz, $\alpha = 0.7$. Value = $0.7 \times \text{max} + 0.3 \times \text{min}$.

• A = 0.7(4000) + 0.3(1750) = 2800 + 525 = 3325
• B = 0.7(2500) + 0.3(2250) = 1750 + 675 = 2425
• C = 0.7(2500) + 0.3(1000) = 1750 + 300 = 2050

Choose A.

Strategy A is preferred under every criterion except Maximin, where B is chosen.

Cost = Rs 10, selling price = Rs 15. Profit per copy sold = Rs 5, and loss per unsold copy = Rs 10.

(a) Optimum stock level

Cumulative probabilities: 20 → 0.19, 21 → 0.35, 22 → 0.60, 23 → 0.87, 24 → 1.00.

Critical ratio:

$\frac{MP}{MP+ML}=\frac{5}{5+10}=0.333$

The smallest stock level whose cumulative probability is at least 0.333 is 21 copies. This is confirmed by the expected profits below, where profit = $15\min(Q,D) - 10Q$.

Optimum stock level = 21 copies.

(b) Maximum EMV

$EMV_{21}=0.19(90)+0.16(105)+0.25(105)+0.27(105)+0.13(105)=\text{Rs } 102.15$

(c) EVPI

With perfect information the vendor would stock exactly the demand, earning Rs 5 on each copy. Expected demand:

$20(0.19)+21(0.16)+22(0.25)+23(0.27)+24(0.13)=21.99$

$EPPI = 5 \times 21.99 = \text{Rs } 109.95$

$EVPI = EPPI - \text{max EMV} = 109.95 - 102.15 = \text{Rs } 7.80$

The cost matrix (Rs) is as follows. Total supply (22 + 15 + 8) equals total demand (7 + 12 + 17 + 9), so the problem is balanced at 45.

Basic feasible solution. Allocating by the least-cost method with improvements (the row and column penalties of VAM point to the same cells): B-III = 15 (cost 2), A-II = 12 (cost 3), A-IV = 8 (cost 4), C-I = 7 (cost 5), then A-III = 2 and C-IV = 1 to complete the supplies and demands.

Allocation: A-II = 12, A-III = 2, A-IV = 8, B-III = 15, C-I = 7, C-IV = 1. This has 6 occupied cells ($m + n - 1 = 6$), so it is non-degenerate.

Optimality test by MODI. Let $u_A = 0$. From the occupied cells, $v_{II}=3$, $v_{III}=5$, $v_{IV}=4$, then $u_B = 2-5 = -3$, $u_C = 6-4 = 2$, and $v_I = 5-2 = 3$.

Opportunity cost for the empty cells ($d = c - u - v$):

All values are positive, so the solution is optimal (and unique).

Optimum solution

Minimum total cost = Rs 149.

Payoff matrix for A (the maximiser):

Maximin = -3 and minimax = 7. They are unequal, so there is no saddle point and mixed strategies are needed.

Dominance.

• A1 dominates A3 (12 ≥ 9, 11 ≥ 10, -4 ≥ -4), so delete A3.
• For B (the minimiser), B2 is better than B1 against both remaining rows (11 ≤ 12, -3 ≤ 6), so delete B1.

The reduced game is:

This has no saddle point (maximin = -3, minimax = 7). Let A play A1 with probability $p$ and A2 with $1-p$. Equating the expected payoffs against B2 and B3:

$11p - 3(1-p) = -4p + 7(1-p)$

$14p - 3 = 7 - 11p \Rightarrow 25p = 10 \Rightarrow p = \frac{2}{5}$

So A1 = 2/5, A2 = 3/5. Let B play B2 with probability $q$ and B3 with $1-q$:

$11q - 4(1-q) = -3q + 7(1-q)$

$15q - 4 = 7 - 10q \Rightarrow q = \frac{11}{25}$

So B2 = 11/25, B3 = 14/25.

Value of the game:

$V = 11\left(\tfrac{2}{5}\right) - 3\left(\tfrac{3}{5}\right) = \frac{22-9}{5} = 2.6$

Optimal strategies:

• Player A: (A1, A2, A3) = (2/5, 3/5, 0)
• Player B: (B1, B2, B3) = (0, 11/25, 14/25)
• Value of the game = 2.6 in favour of A.

(a) Increasing production capacity

• Expand the existing plant or add a second shift, and move to a larger or better-designed site in stages, so that the traditional handmade quality is not lost.
• Buy semi-automatic machines for the repetitive steps (mixing, moulding, baking and packing), while keeping artisan skills for finishing and for premium lines.
• Use subcontracting or contract baking for standard products at peak times, and use overtime or part-time staff in a planned rather than emergency way.
• Do capacity planning from demand forecasts so that capacity is added ahead of demand. Remove bottlenecks first using the theory of constraints.
• Train workers across several tasks and schedule maintenance so that machines do not idle.

(b) Changes to the operations strategy to raise profit and cut costs

• Adopt lean principles: remove waste (scrap, waiting, rework), use value-stream mapping and Kaizen, and shorten lead times.
• Standardise recipes and work methods, and use quality control and TQM to raise yield.
• Reorganise the layout into a product-focused, flow-based line for the best-selling biscuits, and keep a job-shop style for specialty items. This gives a mixed strategy.
• Improve productivity through better materials purchasing, preventive maintenance and energy use, and reduce expensive expedited shipping through better scheduling.
• Choose the competitive priorities clearly: for example, compete on quality and heritage with reliable delivery, rather than trying to be the cheapest.

(c) Improving inventory management

• Use ABC analysis to classify items and control the high-value items tightly.
• Apply EOQ and reorder-point models with safety stock for flour, sugar, butter and packaging, based on forecast demand and supplier lead time.
• Introduce computerised inventory records or an ERP or MRP system, with regular cycle counts, to give accurate stock data.
• Use FIFO because biscuit ingredients and finished goods are perishable, and build good supplier relationships (vendor-managed inventory or JIT delivery) to reduce both shortage and overstock.
• Improve demand forecasting and sales and operations planning, so that production matches expected sales.

(d) Increasing agility and adaptability

• Use flexible, multi-purpose equipment with quick changeover (SMED) and cross-trained workers.
• Create a modular product range with common ingredients, so that varieties can be switched quickly.
• Shorten the supply chain, develop several suppliers, and share information with retailers to detect demand changes early.
• Use make-to-order for special lines and make-to-stock for fast movers, and adopt demand-driven scheduling.
• Encourage a culture of empowerment and quick decisions, and review the market regularly through customer feedback and data.

Together, these changes would let XYZ grow while keeping the quality that is the heart of its brand.