LP special cases and Duality.pptx
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This document provides an overview of operations research and linear programming techniques. It begins with an introduction to the graphical method for solving linear programming problems with two variables by plotting the feasible region defined by the constraints. It then defines key terms like feasible solutions and optimal solutions. The document provides examples of using the graphical method to find the optimal solution for both maximization and minimization problems. It also discusses special cases that can occur with linear programs, such as alternative optimal solutions, unbounded solutions, infeasible solutions, and degenerate solutions. Finally, it provides an introduction to the concept of duality in linear programming.
LP special cases and Duality.pptx
- 1. Operations Research Presented by, Snehal S. Athawale Ph.D 1st year (Agricultural Economics) School of Social Sciences College of Post Graduate Studies in Agricultural Sciences, Umiam, Meghalaya. AEC-605 Linear Programming
- 3. Graphical Method For LP problems that have only two variables, it is possible that the entire set of feasible solutions can be displayed graphically by plotting linear constraints on a graph paper in order to locate the best (optimal) solution. The technique used to identify the optimal solution is called the graphical solution method (approach or technique) for an LP problem with two variables.
- 4. IMPORTANT DEFINITIONS Solution : The set of values of decision variables xj ( j = 1, 2, . . ., n) that satisfy the constraints of an LP problem is said to constitute the solution to that LP problem. Feasible solution : The set of values of decision variables xj ( j = 1, 2, . . ., n) that satisfy all the constraints and non-negativity conditions of an LP problem simultaneously is said to constitute the feasible solution to that LP problem. Infeasible solution : The set of values of decision variables xj ( j = 1, 2, . . ., n) that do not satisfy all the constraints and non-negativity conditions of an LP problem simultaneously is said to constitute the infeasible solution to that LP problem.
- 5. IMPORTANT DEFINITIONS Optimum basic feasible solution : A basic feasible solution that optimizes (maximizes or minimizes) the objective function value of the given LP problem is called an optimum basic feasible solution. Unbounded solution A solution that can increase or decrease infinitely the value of the objective function of the LP problem is called an unbounded solution
- 6. Extreme Point Solution Method Refers to the corner of the feasible region (or space), i.e. this point lies at the intersection of two constraint equations. The steps of the method are summarized as follows: Step 1 : Develop an LP model Step 2 : Plot constraints on graph paper and decide the feasible region. Step 3 : Examine extreme points of the feasible solution space to find an optimal solution
- 7. Examples on Maximization LP Problem Example Maximize Z = 15x1 + 10x2 subject to the constraints (i) 4x1 + 6x2 ≤ 360, (ii) 3x1 + 0x2 ≤ 180, (iii) 0x1 + 5x2 ≤ 200 Non-negativity x1, x2 ≥ 0 Solution : Compute the coordinates on X1X2 plane. From the 1st constraint 4x1 + 6x2 = 360 We get, X2 = 60 , when X1=0 X1= 90 , when X2=0
- 8. Examples on Maximization LP Problem Example Maximize Z = 15x1 + 10x2 subject to the constraints (i) 4x1 + 6x2 ≤ 360, (ii) 3x1 + 0x2 ≤ 180, (iii) 0x1 + 5x2 ≤ 200 Non-negativity x1, x2 ≥ 0 Solution : Compute the coordinates on X1X2 plane. From the 2st constraint ) 3x1 + 0x2 =180, We get, X2 = 0, when X1=0 X1= 60 , when X2=0
- 9. Examples on Maximization LP Problem Example Maximize Z = 15x1 + 10x2 subject to the constraints (i) 4x1 + 6x2 ≤ 360, (ii) 3x1 + 0x2 ≤ 180, (iii) 0x1 + 5x2 ≤ 200 Non-negativity x1, x2 ≥ 0 Solution : Compute the coordinates on X1X2 plane. From the 3st constraint ) 0x1 + 5x2 = 200 We get, X2 = 40, when X1=0 X1= 0 , when X2=0
- 10. Examples on Maximization LP Problem Since all constraints have been graphed, the area which is bounded by all the constraints lines including all the boundary points is called the feasible region (or solution space). The feasible region is shown in fig by the shaded area OABCD Note: feasible region is the overlapping area of constraints that satisfies all of the constraints on resources.
- 11. Examples on Maximization LP Problem Objective function Z is to be maximized from Table maximum value of Z = 1,100 is achieved at the point extreme B (60, 20).
- 12. Examples on Minimization LP Problem Example Minimize Z = 3x1 + 2x2 subject to the constraints (i) 5x1 + x2 ≥ 10, (ii) x1 + x2 ≥ 6, (iii) x1 + 4x2 ≥12 Non-negativity x1, x2 ≥ 0 Solution : Compute the coordinates on X1X2 plane. From the 1st constraint 5x1 + x2 = 10 We get, X2 = 10 , when X1=0 X1= 2 , when X2=0
- 13. Examples on Minimization LP Problem Example Minimize Z = 3x1 + 2x2 subject to the constraints (i) 5x1 + x2 ≥ 10, (ii) x1 + x2 ≥ 6, (iii) x1 + 4x2 ≥12 Non-negativity x1, x2 ≥ 0 Solution : Compute the coordinates on X1X2 plane. From the 2st constraint x1 + x2 = 6, We get, X2 = 6, when X1=0 X1= 6 , when X2=0
- 14. Examples on Minimization LP Problem Example Minimize Z = 3x1 + 2x2 subject to the constraints (i) 5x1 + x2 ≥ 10, (ii) x1 + x2 ≥ 6, (iii) x1 + 4x2 ≥12 Non-negativity x1, x2 ≥ 0 Solution : Compute the coordinates on X1X2 plane. From the 3st constraint x1 + 4x2 =12 We get, X2 = 3, when X1=0 X1= 12 , when X2=0
- 15. Examples on Minimization LP Problem The coordinates of the extreme points of the feasible region (bounded from below) The value of objective function at each of these extreme points is shown in Table
- 16. Examples on Minimization LP Problem The minimum (optimal) value of the objective function Z = 13 occurs at the extreme point C (1, 5). Hence, the optimal solution to the given LP problem is: x1 = 1, x2 = 5, and Min Z = 13.
- 17. SPECIAL CASES IN LINEAR PROGRAMMING 1) Alternative (or Multiple) Optimal Solution 2) Unbounded Solution 3) Infeasible Solution 4) Degenerate Solution
- 18. Alternative (or Multiple) Optimal Solution In certain cases, a given LP problem may have more than one solution yielding the same optimal objective function value. Each of such optimal solutions is termed as alternative optimal solution. Conditions (i) The slope of the objective function should be the same as that of the constraint forming the boundary of the feasible solutions region. (ii) The constraint should form a boundary on the feasible region in the direction of optimal movement of the objective function. In other words, the constraint should be an active constraint.
- 19. Example on Alternative optimum solution Example Maximize Z = 20x1 + 10x2 subject to the constraints (i) 10x1 + 5x2 ≤ 50, (ii) 6x1 + 10x2 ≤ 60, (iii) 4x1 + 12x2 ≤ 48 Non-negativity x1, x2 ≥ 0 Solution : Compute the coordinates on X1X2 plane. From the 1st constraint 10x1 + 5x2 = 50, We get, X2 = 10, when X1=0 X1= 5 , when X2=0
- 20. Example on Alternative optimum solution Example Maximize Z = 20x1 + 10x2 subject to the constraints (i) 10x1 + 5x2 ≤ 50, (ii) 6x1 + 10x2 ≤ 60, (iii) 4x1 + 12x2 ≤ 48 Non-negativity x1, x2 ≥ 0 Solution : Compute the coordinates on X1X2 plane. From the 2st constraint 6x1 + 10x2 = 60, We get, X2 = 6, when X1=0 X1= 10 , when X2=0
- 21. Example on Alternative optimum solution Example Maximize Z = 20x1 + 10x2 subject to the constraints (i) 10x1 + 5x2 ≤ 50, (ii) 6x1 + 10x2 ≤ 60, (iii) 4x1 + 12x2 ≤ 48 Non-negativity x1, x2 ≥ 0 Solution : Compute the coordinates on X1X2 plane. From the 3st constraint 4x1 + 12x2 = 48, We get, X2 = 4, when X1=0 X1= 12 , when X2=0
- 22. Examples on Alternative optimum solution The solution space is denoted by O(0,0), B(5,0), C(3.6,2.8) and D(0,4). The objective function value at the corner point B&C are same and maximum among all values. This indicates existence of multiple combinations of value of the decision variable for the same maximum objective function. Extrem e Point Coordinates (X1, X2) Objective Function Value Z = 20x1 + 10x2 O (0,0) 20(0)+ 10(0) = 0 B (5,0) 20(5)+ 10(0) = 100 C (3.6,2.8) 20(3.6)+ 10(2.8) = 100 D (0,4) 20(0) + 10(4) =40 Objective function is parallel to first constraint
- 23. Unbounded Solution • Infinite solution is referred as an unbounded solution. • It happens when value of certain decision variables and the value of the objective function (maximization case) are permitted to increase infinitely, without violating the feasibility condition.
- 24. Example on Unbounded solution Example Maximize Z = 3x1 + 2x2 subject to the constraints (i) x1 - x2 ≥ 1, (ii) x1 + x2 ≥ 3, Non-negativity x1, x2 ≥ 0 Solution : Compute the coordinates on X1X2 plane. From the 1st constraint x1 - x2 = 1, We get, X2 = -1, when X1=0 X1= 1, when X2=0
- 25. Example on Unbounded solution Example Maximize Z = 3x1 + 2x2 subject to the constraints (i) x1 - x2 ≥ 1, (ii) x1 + x2 ≥ 3, Non-negativity x1, x2 ≥ 0 Solution : Compute the coordinates on X1X2 plane. From the 2st constraint x1 + x2 = 3, We get, X2 = 3, when X1=0 X1= 3, when X2=0
- 26. Example on Unbounded solution Since the given LP problem is of maximization, there exist a number of points in the shaded region for which the value of the objective function is more than 8. Extreme Point Coordinat es (X1, X2) Objective Function Value Z = 3x1 + 2x2 A (0,3) 3(0)+ 2(3) = 6 B (2,1) 3(2)+ 2(1) = 8
- 27. Example on Unbounded solution For example, the point (2, 2) lies in the region and the objective function value at this point is 10 which is more than 8. Thus, as value of variables x1 and x2 increases arbitrarily large, the value of Z also starts increasing. Hence, the LP problem has an unbounded solution.
- 28. Infeasible Solution • An infeasible solution to an LP problem arises when there is no solution that satisfies all the constraints simultaneously. • This happens when there is no unique (single) feasible region. • This situation arises when a LP model that has conflicting constraints. • Any point lying outside the feasible region violates one or more of the given constraints.
- 29. Example on Infeasible solution Example Maximize Z = 6x1 – 4X2 subject to the constraints (i) 2x1 + 4x2 ≤ 4, (ii) 4x1 + 8x2 ≥ 16, Non-negativity x1, x2 ≥ 0 Solution : Compute the coordinates on X1X2 plane. From the 1st constraint 2x1 + 4x2 = 4, , We get, X2 = 1, when X1=0 X1= 2, when X2=0
- 30. Example on Infeasible solution Example Maximize Z = 6x1 – 4X2 subject to the constraints (i) 2x1 + 4x2 ≤ 4, (ii) 4x1 + 8x2 ≥ 16, Non-negativity x1, x2 ≥ 0 Solution : Compute the coordinates on X1X2 plane. From the 2st constraint 4x1 + 8x2 = 16, , We get, X2 = 2, when X1=0 X1= 4, when X2=0
- 31. Example on Infeasible solution The constraints are plotted on graph as usual as shown in Fig. Since there is no unique feasible solution space, therefore a unique set of values of variables x1 and x2 that satisfy all the constraints cannot be determined. Hence, there is no feasible solution to this LP problem because of the conflicting constraints.
- 32. Degeneracy A basic feasible solution is called degenerate if value of at least one basic variable is zero. Example Maximize Z = 100x1 + 50x2 subject to the constraints (i) 4x1 + 6x2 ≤ 24, (ii) x1 ≤ 4, (iii) x2 ≤ 4/3 Non-negativity x1, x2 ≥ 0 Solution : Compute the coordinates on X1X2 plane. From the 1st constraint 4x1 + 6x2 = 24 We get, X2 = 4, when X1=0 X1= 6, when X2=0
- 33. Example on Degeneracy Example Maximize Z = 100x1 + 50x2 subject to the constraints (i) 4x1 + 6x2 ≤ 24, (ii) x1 ≤ 4, (iii) x2 ≤ 4/3 Non-negativity x1, x2 ≥ 0 Solution : Compute the coordinates on X1X2 plane. From the 2st constraint x1 = 4 From the 3st constraint x2 = 4/3 by plotting these coordinates on graph,
- 34. Example on Degeneracy The Closed polygon ABCD is the feasible region. The intersection of two lines will define a corner point of feasible solution. But at corner point C, three lines intersect. This shows the presence of degeneracy in the problem. The coordinates the corner point C are (4. 4/3). Hence, Z(A) = 0 Z(B)= 400 Z(C)= 1400/3 Z(D)= 66.67 = 466.67
- 35. Duality In linear programming, duality implies that each linear programming problem can be analyzed in two different ways but would have equivalent solutions. Any LP problem (either maximization and minimization) can be stated in another equivalent form based on the same data. The new LP problem is called dual linear programming problem or in short dual.
- 36. Duality Every Linear Programming problem is associated with another linear programming problem called the Dual of the problem. The Dual problem of the LPP is obtained by, 1) Transposing the coefficient matrix 2) Interchanging the role of constant terms and the coefficient of the objective function 3) Reverting the inequalities 4) Minimization of objective function instead of maximizing it Primal : Maximize Z(P) = CX constraints Ax ≤ b non-negativity x ≥0 Dual Problem : Minimize Z(D) = b’w constraints A’w ≤ c’ non-negativity w ≥0 Where, w = (w1, w2, w3……wm)
- 37. Duality Rules for Constructing the Dual from Primal 1. A dual variable is defined corresponds to each constraint in the primal LP problem and vice versa. Thus, for a primal LP problem with m constraints and n variables, there exists a dual LP problem with m variables and n constraints and vice-versa. 2. The right-hand side constants b1, b2, . . ., bm of the primal LP problem becomes the coefficients of the dual variables y1, y2, . . ., ym in the dual objective function Z y. Also the coefficients c1, c2, . . ., cn of the primal variables x1, x2, . . ., xn in the objective function become the right-hand side constants in the dual LP problem. 3. For a maximization primal LP problem with all ≤ (less than or equal to) type constraints, there exists a minimization dual LP problem with all ≥ (greater than or equal to) type constraints and vice versa. Thus, the inequality sign is reversed in all the constraints except the non-negativity conditions.
- 38. Duality Rules for Constructing the Dual from Primal 4. The matrix of the coefficients of variables in the constraints of dual is the transpose of the matrix of coefficients of variables in the constraints of primal and vice versa. 5. If the objective function of a primal LP problem is to be maximized, the objective function of the dual is to be minimized and vice versa. 6. If the ith primal constraint is = (equality) type, then the ith dual variables is unrestricted in sign and vice versa.
- 39. Duality The primal-dual relationships may also be memorized by using the following table
- 40. Duality Example Maximize Z = 4x1 + 10x2 + 25x3 subject to the constraints (i) 2x1 + 4x2 + 8x3 ≤ 25, (ii) 4x1 + 9x2 + 8x3 ≤ 30, (iii) 6x1 + 8x2 + 2x3 ≤ 40 Non-negativity x1, x2, x3 ≥ 0 Solution :Let Yi be the dual variable associated with the ith constraints of the primal problem. Minimize Z’= 25Y1 + 30Y2 + 40Y3 subject to the constraints (i) 2Y1 + 4Y2 + 6Y3 ≥ 4, (ii) 4Y1 + 9Y2 + 8Y3 ≥ 10, (iii) 8Y1 + 8Y2 + 2Y3 ≥ 25 Non-negativity Y1, Y2, Y3 ≥ 0
- 41. A summary of the general relationships between primal and dual LP problems is given in Table Duality
- 42. t You can find me at – [email protected]