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Linear programming is used to optimize problems that can be expressed as linear inequalities, equations, and constraints. It aims to find the combination of variables and values that maximizes an objective function while satisfying the constraints. Common applications include operations research problems like resource scheduling and cost-benefit analysis. An example problem maximizes profit by determining the optimal production levels of 4 products given constraints on labor, capital, and materials. The problem is expressed as linear inequalities and can be solved using methods like the Simplex algorithm.
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- 1. VI – Linear Programming
- 2. When do I use Linear Programming? Optimization problems Operations research Express problem in terms of linear inequalities, linear equations and constraints Attempt to find which combination of variables and values maximizes the objective function whilst satisfying the inequalities and constraints
- 3. Common Application Areas Operations Research problems • Resource planning (e.g. break-even analysis): seek to maximize returns, given linearly expressed expenses, payoffs, etc. • Resource scheduling (expressing limits such as non- consecutive slots allotted as linear equations) • Cost/benefit analysis Graph algorithms • Shortest path, network flow, traveling salesman, set cover, etc.
- 4. Example Problem Economics Application: Profit Maximization • Suppose that a manufacturing plant produces four products (x1, x2, x3, x4) • Each product requires a certain amount of labour (L), capital (K) and materials(M) • The total labour for all products must not exceed a certain limit • You are given the following equations:
- 5. Example Problem cont’d • L = 15x1 + 20x2 + 25x3 + 12x4 <= 1000 • K = 12x1 + 10x2 + 30x3 + 16x4 <= 900 • M = 15x1 + 20x2 + 30x3 + 9x4 <= 810 • Meaning that to produce 1 unit of x1, you need 15 units of labour, 12 units of capital and 15 units of material. • Associated with each unit of labour, capital and material, there is a cost • The unit cost of labour is 10 • The unit cost of capital is 15 • The unit cost of material is 8
- 6. Example Problem cont’d So what are the total costs for producing 1 unit of ? • x1: 450 • x2: 510 • x3: 940 • x4: 432 Given that the prices of the products are: • x1: 800 • x2: 650 • x3: 1100 • x4: 700
- 7. Example Problem cont’d What are the profits per unit product? • Profit = Price – Cost • x1: 800 – 450 = 350 • x2: 650 – 510 = 140 • x3: 1100 – 940 = 160 • x4: 700 – 432 = 268
- 8. Example Problem cont’d The principal objective (P) is to maximise profits as follows: • P = 350x1 + 140x2 + 160x3 + 268x4 Given the following constraints • L: 15x1 + 20x2 + 25x3 + 12x4 <= 1000 • K: 12x1 + 10x2 + 30x3 + 16x4 <= 900 • M: 15x1 + 20x2 + 30x3 + 9x4 <= 810 • And also x1>=0, x2>=0, x3>=0 and x4>=0 LINEAR PROGRAM
- 9. Standard Approach to Linear Programming Expressed using linear inequalities • Versus Slack which uses linear equalities More convenient As in profit maximisation example where units are expressed as not exceeding certain limits Can be solved using graphs if two variables x1 and x2 are involved Several points in the graph can be tested to find the one that yields an optimal solution
- 10. Slack Approach to Linear Programming The standard inequalities are converted to equalities Our example equations become: • P = 350x1 + 140x2 + 160x3 + 268x4 • P = 350x1 + 140x2 + 160x3 + 268x4 • L: 15x1 + 20x2 + 25x3 + 12x4 <= 1000 • Slack: s1 = 1000 – 15x1 – 20x2 – 25x3 – 12x4 • Introduce slack variables (e.g. s1) • Decision variables (x1, x2, x3, x4) • Inequality becomes equality
- 11. Methods for Solving Linear Programs One popular method known as the Simplex algorithm Will not be covered in this course Many programs have been written that carry out the Simplex Method. Basic Idea of the Simplex Method: The optimal solution occurs at the corner points of the region (when you plot the graphs of the inequations) Simplex method: can be used to detect when a solution is not feasible. Focus: how to express a problem as a linear program.