Chapter 2 Linear Programming for business (1).pptx
- 1. CHAPTER 2 LINEAR PROGRAMMING PROBLEMS FORMULATIONAND SOLUTION 4/30/2023 1
- 2. 2.1.DEFINITION: Linear programming is widely applied mathematical technique that helps manager in decision making & planning for optimal allocation of limited resources. Linear Programming is a mathematical process that has been developed to help management in decision-making involving allocation of scares resources In other words linear programming is quantitative science that deals with reconciliation of scarce business resource and unlimited operation wants of the firms. 4/30/2023 2
- 3. 2.2. COMPONENTS OFLPP 1.Decision variables - mathematical symbols representing levels of activity of a firm 2.Objective function - in terms of decision variables (e.g., maximize the profit or minimize the cost) Optimization in linear programming implies either maximization (max) Profit, revenue, market share , resource utilization , production rate or minimization (min) Cost, time, distance, or a certain objective function, labour turnover, overtime, risk 2.Constraints - restrictions or limitations placed on the firm by the operating environment stated in linear relationships of the decision variables (restriction on decision making), scarce resources (such as labor supply, RMs, production capacity, machine time, storage space etc 3.Parameters - numerical coefficients, values and constants used in the objective function and constraint (are fixed values that specify the impact that one unit of each decision variable will have on the objective) 4/30/2023 3
- 4. 2.3.ASSUMPTIONS OF LINEAR PROGRAMMING A. Linearity/Proportionality/ The Objective Function and the constraints must be linear in nature in order to deal with a Linear Programming Problems (LPP). Here the term linearity implies proportionality and additively. With linear programs, we assume that the contribution of individual variables in the objective function and constraints is proportional to their value. That is, if we double the value of a variable, we double the contribution of that variable to the objective function and each constraint in which the variable appears. B. Certainty It is assumed that the decision maker here is completely certain (i.e., deterministic conditions) regarding all aspects of the situation, i.e., availability of resources, profit contribution of the Products, courses of action and their consequences etc. C. Divisibility It is assumed that the decision variables are continuous. It means that companies manufacture products in fractional units. For example, a company manufactures 2.5 vehicles, 3.2 barrels of oil etc. 4/30/2023 4
- 5. D.Additivity Additivity means that the total value of the objective function and each constraint function is obtained by adding up the individual contributions from each variable. E. Non- Negativity Indicate all variables are restricted to non-negative values (i.e., their numerical value will be ≥ 0) .i.e. negative values of variables are unrealistic or meaningless. 4/30/2023 5
- 6. 2.4. FORMULATION OFLPPM: 4/30/2023 6 General form of LPM of Maximization: Max z = c1x1 + c2x2 + ... + cnxn---objectivefunction subject to: a11x1 + a12x2 + ... + a1nxn (≤) b1 a21x1 + a22x2 + ... + a2nxn (≤) b2 : an1x1 + an2x2 + ... + annxn (≤) bn x1,x2,x3… xn ≥ 0 non-negativity Where: aij = constraintcoefficients xj = decision variables bi = constraint levels cj = objective function coefficients
- 7. General form of LPM of Minimization: Min z = c1x1 + c2x2 + ...+ cnxn---objectivefunction subject to: 4/30/2023 7 a11x1 + a12x2 + ...+ a1nxn ( ≥) b1 a21x1 + a22x2 + ...+ a2nxn ( ≥) b2 : an1x1 + an2x2 + ...+ annxn ( ≥)bn ≥ 0 non-negativity x1,x2,x3… xn Where:aij = constraintcoefficients xj = decision variables bi = constraintlevels cj= objective functioncoefficients
- 8. Steps: 1.Identify the decision variables (X1, X 2….. ) 2.Determine the objective function First decide whether the problem is maximization or minimization Second identify the coefficients of each decision variable. If the problem is a maximization problem, the profit per unit for each variable must be determined and cost per unit must be determined for minimization. 3. Identify the constraints First, express each constraint in words. Second identify the coefficients of the decision variables in the constraints; and the RHS values of the constraints. Determine the limits for the constraints i.e. see whether the constraint is of the form (≤), (≥) 4. Write the model in the standard form Using the above information (step 1 to 3), build the model. 4/30/2023 8
- 9. Example 1: A firm that assembles computers and computer equipment is about to start production two new microcomputers. Each type of microcomputers will require assembly time, inspection time and storage space. The amount of each of these resources that can be devoted to the production of these microcomputers is limited. The manager of the firm would like to determine the quantity of each microcomputer to produce in order to maximize the profit generated by sales of these microcomputers. Additionalinformation In order to develop a suitable model of the problem, the manager has met with design and manufacturing personnel. As a result of these meetings the manager has obtained the following information. 4/30/2023 9
- 10. The manager has also acquired information on the availability of company resources. These weekly resources are The manager also met with the firm‟s marketing manager and learned that demand for the microcomputers was such that what ever combination of these two types of microcomputers is produced, all of the outputs can be sold. 4/30/2023 10
- 11. Required: Formulate the LPPM of the problem. Solution Step 1: Identify the decision variable The quantity/amount/ units of each microcomputer (microcomputer type 1 and microcomputer type 2) to be produced produced Let X1=represent quantity of microcomputer type 1 to be X2=represent quantity of microcomputer type 2 to be produced Step 2: Identify the objective function The problem is maximization problem, as indicated in the problem To write the model both the objective function and the constraints summarize the given information in tabular form accordingly 4/30/2023 11
- 12. Hence,the objective function is Max Z = 60X1 + 50X2 Step 3:identify each constraints and write the equation 4X1 + 10X2 ≤ 100 ….........Assembly time constraint 2X1 + X2 ≤ 22 ……………Inspection time constraint 3X1+ 3X2 ≤ 39 ……………Storage spaceconstraint Non negativity constraint X1 and X2>0 The mathematicalmodel of the microcomputer problem is: Max Z= 60X1 + 50X2 Subject to 4X1 + 10 X2 ≤100 2X1 + X2 ≤22 3X1+ 3 X2 ≤ 39 And X1 and X2 > 0 4/30/2023 12
- 13. 2.5.Method for solving LPM 4/30/2023 13 Basically there are two methods for solving LPM Graphical method and Simplex method 2.5.1.Graphical method of solving LPP used to find solutions for LPP when the decision variables are only limited to two Steps in solving LPM Using GraphicalApproach 1.Convert each inequality of the constraints in to equality to determine intercepts (Set X1 = 0, and find the value for X2, Set X2 = 0, and find the value for X1) 2.Plot each of the constraint on the graph using intercepts (consider non negativity assumption) 3.Determine the area of feasibility (feasible region: a region which satisfies all the constraints in common) by shading using the inequalities of the constraints For minimization case (≥), the feasible region is the area that lies above constraint line (to the right sides of the constraint lines) For maximization case (≤), the feasible region is the area that lies below constraint line (to the left sides of constraintlines) 4.Determine the corner points bounding the feasible region (feasible solution space) and identify coordinates of each corner points
- 14. 5.Find optimum solution (that optimize the objective function) by substituting coordinates of the corner points in the objective function 4/30/2023 14 6.After all corner points have been evaluated in a similar fashion, select the one with the highest value of the objective function (for a maximization problem) or lowest value (for a minimization problem) as the optimal solution.
- 15. Example: Consider the following Maximization problem and solve by using graphical method Max Z = 4X1 + 3X2 Subject to 4X1+ 3X2 ≤ 24 X1 ≤ 4.5 X2 ≤ 6 X1 ≥ 0 ,X2 ≥ 0 4/30/2023 15
- 17. MINIMIZATION LPP: General form of LPM of Minimization: Min z = c1x1 + c2x2 + ...+ cnxn---objective function subject to: a11x1 + a12x2 + ...+ a1nxn ( ≥) b1 a21x1 + a22x2 + ...+ a2nxn ( ≥) b2 : an1x1 + an2x2 + ...+ annxn ( ≥) bn x1,x2,x3… xn ≥ 0 non-negativity Where: aij= constraint coefficients xj= decisionvariables bi = constraintlevels cj= objective functioncoefficients 4/30/2023 17
- 18. Example1: Suppose that a machine shop has two different types of machines; machine 1 and machine 2, which can be used to make a single product. These machine vary in the amount of product produced per hr., in the amount of labor used and in the cost of operation. Assume that at least a certain amount of product must be produced and that we would like to utilize at least the regular labor force to meet the requirement. How much should we utilize each machine in order to utilize total costs and still meets the requirement? Required: a) Model the problem of this company b) Solve the model graphically 4/30/2023 18
- 19. a). Model the problem of thiscompany Zmin=25X1+30X2 Subject to: 20X1+15X2≥100 2X1+3X2≥15 X1, X2≥0 4/30/2023 19 Minimum required hours Machine 1 (X1) Machine (X2) Product produced/hr Labor/hr Operation Cost 20 2 $25 15 3 $30 100 15
- 20. b) Solve the model graphically 4/30/2023 20 Steps in solving LPM Using GraphicalApproach 1.Convert each inequality of the constraints in to equality to determine intercepts 2.Plot each of the constraint on the graph using intercepts (consider non negativity assumption) 3.Determine feasible region: a region which satisfies all the constraints in common) by shading using the inequalities of the constraints For minimization case (≥), the feasible region is the area that lies above constraint line (to the right sides of the constraintlines) NB. Important terminologies A feasible solution is a solution for which all the constraints are satisfied. An infeasible solution is a solution for which at least one constraint is violated. An optimal solution is a feasible solution that has the most favorable value of the objective function. The most favorable value is the largest value if the objective function is to be maximized, whereas it is the smallest value if the objective function is to be minimized.
- 21. 4.Determine the corner points bounding the feasible region (feasible solution space) and identify coordinates of each cornerpoints 5.Find optimum solution (that optimize the objective function) by substituting coordinates of the corner points in the objectivefunction 6.After all corner points have been evaluated in a similar fashion, select the one with the lowest value (for a minimization problem) as the optimal solution. 4/30/2023 21
- 22. 4/30/2023 22
- 23. Example 2. Minimization Case (Fertilizer case): 4/30/2023 23 The agricultural research institute recommended a farmer to spread out at least 5000 kg of phosphate fertilizer and not less than 7000 kg of nitrogen fertilizer to raise the productivity of his crops on thefarm. There are two mixtures A and B, weighs 100 kg each, from which these fertilizers can be obtained. The cost of each Mixture A and B is Birr 40 and 25respectively. Mixture A contains 40 kg of phosphate and 60 kg of nitrogen while the Mixture B contains 60 kg of phosphate and 40 kg ofnitrogen. This problem can be represented as a linear programming problem to find out how many bags of each type a farmer should buy to get the desired amount of fertilizers at the minimum cost.
- 24. Step 1. Suppose x1 and x2 are the number of bags of mixture A and mixture B. Step 2. The cost of both the mixture is 40X1 + 25X2 and thus, the objective function will be: Minimize Z = 40X1+25X2 BagA content 4/30/2023 24 Bag B content Minimum Requirement Phosphate 40kg 60kg 5000kg Nitrogen 60kg 40kg 7000kg Cost per bag $40 $25 Step 3. Constraints 40X1 + 60X2 ≥ 5000…………………. phosphate constraint 60X1 + 40X2 ≥ 7000………………… nitrogen constraint Where, Bag A contains 60 kg of nitrogen and Bag B contains 40 kg of nitrogen, and the minimum requirement of nitrogen is 7000 kg.
- 25. Step 4.Develop linear programming problem (model) 4/30/2023 25 Minimize Z = 40X1+25X2 (cost) Subject to: 40x1 60x1 + 60x2 + 40x2 ≥ 5000 (Phosphate ≥ 7000 (Nitrogen Constraint) Constraint) x1, x2 ≥ 0 (Non-negativity Restriction)
- 26. Practice, Practice,Practice! 4/30/2023 26 Thank you for Your Attention & Participation!