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| Different Types of Linear Programming Problems (e.g., Manufacturing, Diet, Transportation) | Characteristics and Examples of Different LPP Types | |
Types of Linear Programming Problems
Different Types of Linear Programming Problems (e.g., Manufacturing, Diet, Transportation)
Contextual Categories of Linear Programming Problems
While all Linear Programming Problems (LPPs) adhere to the same fundamental mathematical structure – optimizing a linear objective function subject to linear constraints and non-negativity restrictions – they arise from and model a wide variety of real-world situations. Categorizing LPPs based on the context of the problem or the type of decision being made is helpful for recognizing the applicability of LP and understanding the typical patterns of variables, objectives, and constraints for different scenarios.
Understanding these common types can make the process of formulating an LPP from a verbal description more intuitive, as you can often recognize the underlying structure of the problem. Some of the prominent types of Linear Programming Problems encountered in various fields include:
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Manufacturing / Production Problems: These problems focus on determining the optimal quantities of various products to manufacture or produce within a given timeframe. The primary goals are usually to maximize profit or production output, while respecting limitations on resources such as raw materials, labour hours, machine capacity, and market demand.
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Diet Problems: Also known as nutrition problems, this type deals with finding the least expensive combination of different food items or ingredients that will satisfy a set of minimum nutritional requirements (e.g., vitamins, minerals, protein, calories).
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Transportation Problems: These problems address the challenge of efficiently moving goods or resources from multiple points of origin (sources) to multiple destinations (demands). The objective is typically to minimize the total transportation cost, distance, or time, subject to constraints on the supply available at each source and the demand required at each destination.
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Allocation / Assignment Problems: This category involves optimally assigning limited resources (such as workers, machines, funds, or time slots) to various tasks, jobs, or activities. The goal is often to maximize overall efficiency, productivity, or profit, or to minimize total cost, time, or completion time. Transportation problems can be viewed as a specific type of allocation problem (allocating transport capacity).
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Blending Problems: These problems involve determining the optimal proportions or quantities of different raw materials or ingredients to mix together to produce a final product with desired characteristics or specifications. The objective is usually to minimize the cost of the raw materials used or maximize the profit from the final blend. Diet problems are a particular application of blending problems.
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Scheduling Problems: These problems aim to create optimal schedules for activities or resources over time. This could involve scheduling jobs on machines, shifts for employees, or flights for an airline. The objective is often to minimize the total time taken, minimize costs (e.g., overtime pay, idle time), or maximize resource utilization, subject to constraints like resource availability, task durations, and precedence relationships.
Recognizing which category a real-world problem falls into can provide a useful starting point for identifying the decision variables and structuring the objective function and constraints during the mathematical formulation phase.
Characteristics and Examples of Different LPP Types
Detailed Examination of Common LPP Types
To further understand the practical application of Linear Programming, let's examine the typical characteristics, objectives, decision variables, and constraints for some of the common types of LPPs, accompanied by illustrative examples formulated from an Indian perspective.
1. Manufacturing / Production Problems
These problems are prevalent in industries and manufacturing units. They focus on determining the optimal production levels for different products to maximize profit or output, given constraints on resources like machinery, labour, and raw materials.
- Typical Objective: Primarily Maximize Profit. Other objectives could include maximizing revenue or production quantity, or minimizing production cost for a target output.
- Common Decision Variables: Represent the number of units of each product to be manufactured within a specified period (e.g., daily, weekly, monthly). If there are 'n' products, variables might be $x_1, x_2, \dots, x_n$.
- Typical Constraints: Usually represent limitations on the availability of resources used in production. These are often of the '$\le$' type (less than or equal to the available resource).
- Available machine hours on different machines.
- Available labour hours (possibly categorized by skill level).
- Limited supply of raw materials.
- Production capacity limits for specific items or departments.
- Market demand limits (maximum units that can be sold).
Example 1. A small workshop manufactures two types of decorative items: Wall Hanging (WH) and Table Piece (TP). Each WH requires 2 hours of carving and 1 hour of polishing. Each TP requires 3 hours of carving and 1.5 hours of polishing. The workshop has 48 hours of carving time and 30 hours of polishing time available per week. The profit is $\textsf{₹}600$ per WH and $\textsf{₹}800$ per TP. Formulate an LPP to determine the number of units of each item to produce weekly to maximize total profit.
Answer:
Decision Variables:
- Let $x$ = Number of Wall Hangings (WH) produced per week.
- Let $y$ = Number of Table Pieces (TP) produced per week.
Objective Function:
The goal is to maximize total profit.
Maximize $Z = 600x + 800y$
Constraints:
The limitations are the available carving and polishing times.
- Carving Time Constraint:
- Polishing Time Constraint:
Time for $x$ units of WH: $2x$ hours.
Time for $y$ units of TP: $3y$ hours.
Total carving time used: $2x + 3y$. Available: 48 hours.
$2x + 3y \leq 48$
[Carving Time Limit] ... (1)
Time for $x$ units of WH: $1x$ hour.
Time for $y$ units of TP: $1.5y$ hours.
Total polishing time used: $x + 1.5y$. Available: 30 hours.
$x + 1.5y \leq 30$
[Polishing Time Limit] ... (2)
Non-negativity Restrictions:
The number of items produced cannot be negative.
$x \ge 0$
... (3)
$y \ge 0$
... (4)
Complete Formulation:
Find $x, y$ to
Maximize $Z = 600x + 800y$
Subject to:
$2x + 3y \leq 48$
... (1)
$x + 1.5y \leq 30$
... (2)
$x \ge 0$
... (3)
$y \ge 0$
... (4)
2. Diet Problems
These problems involve formulating a diet or mixture of ingredients to meet specific nutritional requirements at the lowest possible cost. They are a subtype of blending problems.
- Typical Objective: Almost always Minimize Cost of the diet or mixture.
- Common Decision Variables: Represent the amount (e.g., kilograms, grams, litres) of each food item or ingredient to be included in the mixture. If there are 'k' foods, variables might be $f_1, f_2, \dots, f_k$.
- Typical Constraints: Ensure the mixture meets minimum (and sometimes maximum) requirements for various nutrients or properties. These are usually of the '$\ge$' type for minimum requirements and '$\le$' for maximum limits.
- Minimum required units of various vitamins (e.g., Vitamin A, C, E).
- Minimum required units of minerals (e.g., Calcium, Iron, Zinc).
- Minimum/Maximum required levels of macronutrients (e.g., protein, fat, carbohydrates).
- Constraints on total weight or volume of the mixture.
Example 1. A cattle feed manufacturer wants to produce a mix using two ingredients, Ingredient P and Ingredient Q. Each kg of Ingredient P costs $\textsf{₹}15$ and contains 8 units of fiber and 12 units of protein. Each kg of Ingredient Q costs $\textsf{₹}20$ and contains 6 units of fiber and 10 units of protein. A minimum quantity of the mixed feed must contain at least 96 units of fiber and 144 units of protein. Formulate this as an LPP to minimize the cost of producing a mixture that meets these nutritional requirements.
Answer:
Decision Variables:
- Let $x$ = Amount (in kg) of Ingredient P to use in the mixture.
- Let $y$ = Amount (in kg) of Ingredient Q to use in the mixture.
Objective Function:
The goal is to minimize the total cost of the ingredients.
Cost of $x$ kg of P: $\textsf{₹}15x$. Cost of $y$ kg of Q: $\textsf{₹}20y$.
Minimize $Z = 15x + 20y$
Constraints:
The mixture must meet the minimum requirements for fiber and protein.
- Fiber Constraint:
- Protein Constraint:
Fiber from $x$ kg of P: $8x$ units. Fiber from $y$ kg of Q: $6y$ units.
Total Fiber: $8x + 6y$. Minimum required: 96 units.
$8x + 6y \ge 96$
[Fiber Minimum] ... (1)
(Simplified by dividing by 2: $4x + 3y \ge 48$)
Protein from $x$ kg of P: $12x$ units. Protein from $y$ kg of Q: $10y$ units.
Total Protein: $12x + 10y$. Minimum required: 144 units.
$12x + 10y \ge 144$
[Protein Minimum] ... (2)
(Simplified by dividing by 2: $6x + 5y \ge 72$)
Non-negativity Restrictions:
Amounts of ingredients cannot be negative.
$x \ge 0$
... (3)
$y \ge 0$
... (4)
Complete Formulation:
Find $x, y$ to
Minimize $Z = 15x + 20y$
Subject to:
$8x + 6y \ge 96$ (or $4x + 3y \ge 48$)
... (1)
$12x + 10y \ge 144$ (or $6x + 5y \ge 72$)
... (2)
$x \ge 0$
... (3)
$y \ge 0$
... (4)
3. Transportation Problems
These problems are concerned with the efficient and cost-effective movement of goods from supply points (sources) to demand points (destinations).
- Typical Objective: Usually Minimize Total Transportation Cost. Other objectives could be minimizing total distance or time.
- Common Decision Variables: Represent the quantity of goods to be shipped from each specific source $i$ to each specific destination $j$. With 'm' sources and 'n' destinations, there are $m \times n$ variables, often denoted as $x_{ij}$ where $i$ is the source index and $j$ is the destination index.
- Typical Constraints: Represent the limitations on supply at each source and the requirements of demand at each destination.
- Supply Constraints: For each source, the total quantity shipped out must not exceed its supply capacity. These are typically '$\le$' or '=' constraints.
- Demand Constraints: For each destination, the total quantity received must meet its demand requirement. These are typically '$\ge$' or '=' constraints.
Example 1. A company has two warehouses, W1 and W2, with surplus stock of 500 quintals and 600 quintals respectively. They need to transport this stock to three markets, M1, M2, and M3, with requirements of 250 quintals, 300 quintals, and 350 quintals respectively. The transportation cost per quintal from each warehouse to each market is given in the table below. Formulate an LPP to minimize the total transportation cost.
Transportation Costs per Quintal ($\textsf{₹}$):
| From \ To | Market 1 (M1) | Market 2 (M2) | Market 3 (M3) |
|---|---|---|---|
| Warehouse 1 (W1) | 10 | 12 | 15 |
| Warehouse 2 (W2) | 8 | 10 | 11 |
Answer:
Decision Variables:
We need to determine the quantity shipped from each warehouse to each market.
- Let $x_{11}$ = Quintals shipped from W1 to M1.
- Let $x_{12}$ = Quintals shipped from W1 to M2.
- Let $x_{13}$ = Quintals shipped from W1 to M3.
- Let $x_{21}$ = Quintals shipped from W2 to M1.
- Let $x_{22}$ = Quintals shipped from W2 to M2.
- Let $x_{23}$ = Quintals shipped from W2 to M3.
Objective Function:
The goal is to minimize the total transportation cost.
Cost $Z = 10x_{11} + 12x_{12} + 15x_{13} + 8x_{21} + 10x_{22} + 11x_{23}$.
Minimize $Z = 10x_{11} + 12x_{12} + 15x_{13} + 8x_{21} + 10x_{22} + 11x_{23}$
Constraints:
Limitations are warehouse supplies and market demands. Check if total supply equals total demand: Total Supply = 500 + 600 = 1100. Total Demand = 250 + 300 + 350 = 900. Total Supply > Total Demand. We will use $\le$ for supply and $\ge$ for demand.
- Supply Constraints:
- Demand Constraints:
From W1: Total shipped from W1 cannot exceed its capacity.
$x_{11} + x_{12} + x_{13} \leq 500$
[W1 Supply Limit] ... (1)
From W2: Total shipped from W2 cannot exceed its capacity.
$x_{21} + x_{22} + x_{23} \leq 600$
[W2 Supply Limit] ... (2)
To M1: Total received at M1 must meet its demand.
$x_{11} + x_{21} \ge 250$
[M1 Demand Minimum] ... (3)
To M2: Total received at M2 must meet its demand.
$x_{12} + x_{22} \ge 300$
[M2 Demand Minimum] ... (4)
To M3: Total received at M3 must meet its demand.
$x_{13} + x_{23} \ge 350$
[M3 Demand Minimum] ... (5)
Non-negativity Restrictions:
Quantities shipped cannot be negative.
$x_{ij} \ge 0$ for $i=1,2$ and $j=1,2,3$
... (6-11)
Complete Formulation:
Find $x_{11}, x_{12}, x_{13}, x_{21}, x_{22}, x_{23}$ to
Minimize $Z = 10x_{11} + 12x_{12} + 15x_{13} + 8x_{21} + 10x_{22} + 11x_{23}$
Subject to:
$x_{11} + x_{12} + x_{13} \leq 500$
... (1)
$x_{21} + x_{22} + x_{23} \leq 600$
... (2)
$x_{11} + x_{21} \ge 250$
... (3)
$x_{12} + x_{22} \ge 300$
... (4)
$x_{13} + x_{23} \ge 350$
... (5)
$x_{ij} \ge 0$ for $i=1,2; j=1,2,3$
... (6-11)
Since total supply (1100 quintals) exceeds total demand (900 quintals), 200 quintals of stock will remain at the warehouses in the optimal solution.
4. Allocation / Assignment Problems
These problems involve distributing or assigning limited resources (like people, machines, budget, time) to various competing tasks or activities to achieve the best possible outcome. Assignment problems are a special case where resources are assigned one-to-one to tasks.
- Typical Objective: Maximize total effectiveness, efficiency, or profit, or Minimize total cost, time, or penalties.
- Common Decision Variables: Can be continuous (amount of resource allocated) or binary (0 or 1, indicating if a resource is assigned to a task). For simple allocation, variables might be $x_i$ = amount of resource allocated to activity $i$. For assignment problems, variables $x_{ij}$ are often 1 if resource $i$ is assigned to task $j$, and 0 otherwise (these are Integer LP variables, but the formulation follows an LP structure).
- Typical Constraints: Ensure resources are not over-allocated and tasks are sufficiently covered.
- Resource Limits: The total amount of a specific resource allocated to all activities cannot exceed its total availability. Typically '$\le$'.
- Task Requirements: Activities must receive at least a minimum amount of allocated resource or effort. Typically '$\ge$'.
- Assignment Rules (for Assignment Problems): Each resource is assigned to at most (or exactly) one task, and each task is assigned to at most (or exactly) one resource. These are often sum constraints on binary variables.
Example 1. A marketing firm has a maximum budget of $\textsf{₹}1,00,000$ for an advertising campaign using two channels: social media and print media. Each unit spent on social media is expected to reach 1,500 people, and each unit spent on print media is expected to reach 1,200 people. The firm decides to spend at least $\textsf{₹}30,000$ on social media and at most $\textsf{₹}40,000$ on print media. Formulate an LPP to maximize the total number of people reached.
Answer:
Decision Variables:
We need to decide how much budget to allocate to each advertising channel.
- Let $x$ = Amount (in $\textsf{₹}$) spent on social media.
- Let $y$ = Amount (in $\textsf{₹}$) spent on print media.
Objective Function:
The goal is to maximize the total number of people reached.
People reached per rupee on social media: 1500. Total reached by social media: $1500x$.
People reached per rupee on print media: 1200. Total reached by print media: $1200y$.
Maximize $Z = 1500x + 1200y$
Constraints:
Limitations are the total budget, minimum spend on social media, and maximum spend on print media.
- Total Budget Constraint:
- Minimum Social Media Spend Constraint:
- Maximum Print Media Spend Constraint:
Total spent on both channels: $x + y$. Maximum budget: $\textsf{₹}1,00,000$.
$x + y \leq 100000$
[Total Budget Limit] ... (1)
Amount spent on social media must be at least $\textsf{₹}30,000$.
$x \ge 30000$
[Minimum Social Media Spend] ... (2)
Amount spent on print media must be at most $\textsf{₹}40,000$.
$y \leq 40000$
[Maximum Print Media Spend] ... (3)
Non-negativity Restrictions:
Amount spent cannot be negative.
$x \ge 0$
... (4)
$y \ge 0$
... (5)
Note: Constraint (2) ($x \ge 30000$) makes $x \ge 0$ redundant, and constraint (3) ($y \leq 40000$) combined with the budget constraint $x+y \le 100000$ and $x \ge 30000$ implies $y \ge 0$ in many feasible scenarios, but explicitly stating non-negativity is standard practice.
Complete Formulation:
Find $x, y$ to
Maximize $Z = 1500x + 1200y$
Subject to:
$x + y \leq 100000$
... (1)
$x \ge 30000$
... (2)
$y \leq 40000$
... (3)
$x \ge 0$
... (4)
$y \ge 0$
... (5)
This LPP can be solved to find the optimal budget allocation for maximizing reach.
5. Blending Problems
Blending problems involve mixing different raw materials or ingredients to produce a final product that meets specific quality standards or compositions, while typically minimizing the cost of the ingredients. Diet problems are a common sub-category of blending problems.
- Typical Objective: Usually Minimize Cost of raw materials. Can also be Maximize Profit from the final blend or maximize some quality characteristic.
- Common Decision Variables: Represent the amount (e.g., kg, litres, tons) of each raw material or ingredient to be used in the blend.
- Typical Constraints: Define the required properties of the final blend and the availability of input materials.
- Compositional Requirements: The final blend must meet certain percentage or level requirements for its components (e.g., minimum percentage of gold in an alloy, exact octane rating in gasoline, maximum sulphur content in fuel). These often involve ratios, which need to be carefully converted into linear inequalities or equations.
- Availability Limits: The amount of each raw material used cannot exceed its available supply. Typically '$\le$'.
- Total Quantity: A constraint might fix the total amount of the final blend to be produced. Typically an '=' constraint.
Example 1. An oil refinery produces two types of gasoline, Super and Premium, by blending three components, Component A, Component B, and Component C. Component A costs $\textsf{₹}20$ per litre, B costs $\textsf{₹}30$ per litre, and C costs $\textsf{₹}40$ per litre. The required composition for Super gasoline is at least 25% A, at least 30% B, and at most 40% C. The required composition for Premium gasoline is at least 40% A, at most 20% B, and at least 30% C. The refinery wants to produce 5,000 litres of Super and 8,000 litres of Premium gasoline. Formulate an LPP to determine the amount of each component to use to minimize the total cost, assuming the total amount of each component used for Super and Premium comes from the same available stock (though availabilities aren't given here, focusing on blending requirements).
Note: This example involves blending multiple components into multiple products, making it slightly more complex than a simple blend. Let's simplify to just producing one type of blend using two components to fit the two-variable theme often seen in introductory LP.
Revised Example 1. A manufacturer produces a special alloy by blending two metals, Metal P and Metal Q. Metal P costs $\textsf{₹}400$ per kg and contains 90% copper and 10% zinc. Metal Q costs $\textsf{₹}500$ per kg and contains 70% copper and 30% zinc. The manufacturer wants to produce 100 kg of the alloy that contains at least 85% copper. Formulate an LPP to determine the amount of each metal to use to minimize the total cost of the alloy.
Answer:
Decision Variables:
- Let $x$ = Amount (in kg) of Metal P to use.
- Let $y$ = Amount (in kg) of Metal Q to use.
Objective Function:
The goal is to minimize the total cost of the metals used.
Cost of $x$ kg of P: $\textsf{₹}400x$. Cost of $y$ kg of Q: $\textsf{₹}500y$.
Minimize $Z = 400x + 500y$
Constraints:
The constraints are the total weight of the alloy and the minimum copper percentage requirement.
- Total Weight Constraint:
- Copper Content Constraint:
The total amount of the alloy must be 100 kg.
$x + y = 100$
[Total Alloy Weight] ... (1)
The alloy must contain at least 85% copper. This means the total weight of copper must be at least 85% of the total weight of the alloy (100 kg).
Copper from $x$ kg of Metal P: 90% of $x = 0.90x$ kg.
Copper from $y$ kg of Metal Q: 70% of $y = 0.70y$ kg.
Total copper in the alloy: $0.90x + 0.70y$ kg.
Required minimum copper: 85% of 100 kg = $0.85 \times 100 = 85$ kg.
Constraint:
$0.90x + 0.70y \ge 85$
[Copper Minimum] ... (2)
(Simplified by multiplying by 100: $90x + 70y \ge 8500$, or by 10: $9x + 7y \ge 850$)
Non-negativity Restrictions:
Amounts of metals used cannot be negative.
$x \ge 0$
... (3)
$y \ge 0$
... (4)
Note: Since $x+y = 100$ and $x \ge 0, y \ge 0$, these non-negativity constraints are implied by the total weight constraint, but are typically included for completeness.
Complete Formulation:
Find $x, y$ to
Minimize $Z = 400x + 500y$
Subject to:
$x + y = 100$
... (1)
$0.90x + 0.70y \ge 85$ (or $9x + 7y \ge 850$)
... (2)
$x \ge 0$
... (3)
$y \ge 0$
... (4)
This LPP can be solved to find the optimal mix of Metal P and Metal Q.
6. Scheduling Problems
Scheduling problems involve determining the optimal timing, sequencing, and assignment of tasks or activities to resources over a period of time. The goal is to optimize an objective such as minimizing completion time, cost, or maximizing throughput, subject to constraints on resource availability, task durations, and dependencies.
- Typical Objective: Minimize completion time (Makespan), Minimize total cost (including idle time, penalties), Maximize throughput, Maximize resource utilization.
- Common Decision Variables: Can include continuous variables (e.g., start times, duration of allocation) or binary variables (e.g., indicating if a task is assigned to a resource, or if one task precedes another). Complex scheduling problems often require integer or mixed-integer linear programming.
- Typical Constraints: Reflect the operational requirements and resource limitations of the scheduling environment.
- Resource Capacity Constraints: A resource (like a machine or worker) can only handle a certain amount of work at any given time.
- Task Duration Constraints: Define the time required to complete each task.
- Precedence Constraints: Some tasks must be finished before others can begin.
- Time Window Constraints: Tasks must start or be completed within specific time periods.
Example 1. A construction company needs to plan the allocation of labour-hours to two critical phases of a project: Phase 1 (Foundation) and Phase 2 (Structure). Phase 1 requires a total of 500 labour-hours, and Phase 2 requires 700 labour-hours. The project must be completed within 20 weeks. The company can allocate a maximum of 80 labour-hours per week across both phases combined. The cost of labour for Phase 1 is $\textsf{₹}300$ per labour-hour, and for Phase 2 is $\textsf{₹}350$ per labour-hour. Assume labour-hours can be allocated continuously per week. Formulate an LPP to determine the weekly allocation of labour-hours to each phase to minimize the total labour cost, while ensuring both phases are completed within the timeframe and total weekly labour does not exceed the limit.
Answer:
Decision Variables:
We need to decide how many labour-hours to allocate to each phase per week. Assuming a uniform allocation over the 20 weeks simplifies this for standard LP.
- Let $x$ = Average labour-hours allocated to Phase 1 per week.
- Let $y$ = Average labour-hours allocated to Phase 2 per week.
Objective Function:
The goal is to minimize the total labour cost over the 20 weeks.
Total labour-hours for Phase 1 over 20 weeks = $x \times 20$. Total labour-hours for Phase 2 over 20 weeks = $y \times 20$.
Total Cost = (Cost per hour for Phase 1 $\times$ Total hours on Phase 1) + (Cost per hour for Phase 2 $\times$ Total hours on Phase 2).
Total Cost $Z = 300 \times (20x) + 350 \times (20y) = 6000x + 7000y$.
Minimize $Z = 6000x + 7000y$
Constraints:
Limitations are total weekly labour capacity and the total labour-hours required for each phase within the 20 weeks.
- Total Weekly Labour Constraint:
- Phase 1 Total Labour-Hours Constraint:
- Phase 2 Total Labour-Hours Constraint:
Total labour-hours allocated per week: $x + y$. Maximum available per week: 80.
$x + y \leq 80$
[Weekly Labour Limit] ... (1)
Total labour-hours for Phase 1 over 20 weeks ($20x$) must meet the requirement of 500 hours.
$20x \ge 500$
[Phase 1 Total Hours] ... (2)
(Simplified by dividing by 20: $x \ge 25$)
Total labour-hours for Phase 2 over 20 weeks ($20y$) must meet the requirement of 700 hours.
$20y \ge 700$
[Phase 2 Total Hours] ... (3)
(Simplified by dividing by 20: $y \ge 35$)
Non-negativity Restrictions:
Labour hours allocated cannot be negative.
$x \ge 0$
... (4)
$y \ge 0$
... (5)
Note: Constraints (2) and (3) make (4) and (5) redundant here, but they are included in the formal structure.
Complete Formulation:
Find $x, y$ to
Minimize $Z = 6000x + 7000y$
Subject to:
$x + y \leq 80$
... (1)
$20x \ge 500$ (or $x \ge 25$)
... (2)
$20y \ge 700$ (or $y \ge 35$)
... (3)
$x \ge 0$
... (4)
$y \ge 0$
... (5)
This formulation represents a simplified scheduling problem using average allocation. Note that for a feasible solution, $x \ge 25$ and $y \ge 35$, so $x+y \ge 25+35=60$. Since $x+y \le 80$, there is a feasible region. The problem assumes a continuous allocation of labour-hours, which is common in LP approximations of scheduling problems.
By classifying real-world problems into these types, we can better understand their underlying structure and apply the principles of Linear Programming to find optimal solutions for efficient resource management and decision-making.