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Chapter-4 Linear Programing Simplex

Optimization in construction

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Chapter-4 Linear Programing Simplex

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  1. 6 Linear Programming: A Geometric Approach • Graphing Systems of Linear Inequalities in Two Variables • Linear Programming Problems • Graphical Solutions of Linear Programming Problems • The Simplex Method: Standard Maximization Problems • The Simplex Method: Standard Minimization Problems

  2. 6.1 Graphing Systems of Linear Inequalities in Two Variables

  3. Graphing Linear Inequalities • We’ve seen that a linear equation in two variablesx and y has a solution set that may be exhibited graphically as points on a straight line in the xy-plane. • There is also a simple graphical representation for linear inequalities of two variables:

  4. Procedure for Graphing Linear Inequalities • Draw the graph of the equation obtained for the given inequality by replacing the inequality sign with an equal sign. • Use a dashed or dotted line if the problem involves a strict inequality, < or >. • Otherwise, use a solid line to indicate that the line itself constitutes part of the solution. • Pick a test point lying in one of the half-planes determined by the line sketched in step 1 and substitute the values of x and y into the given inequality. • Use the origin whenever possible. • If the inequality is satisfied, the graph of the inequality includes the half-plane containing the test point. • Otherwise, the solution includes the half-plane not containing the test point.

  5. Examples • Determine the solution set for the inequality2x + 3y 6. Solution • Replacing the inequality with an equality=, we obtain the equation 2x + 3y= 6, whose graph is: y 7 5 3 1 –1 2x + 3y= 6 x –5 –3 –1 1 3 5

  6. Examples • Determine the solution set for the inequality2x + 3y 6. Solution • Picking the origin as a test point, we find 2(0) + 3(0)  6, or 0  6, which is false. • Thus, the solution set is: y 7 5 3 1 –1 2x + 3y 6 2x + 3y= 6 (0, 0) x –5 –3 –1 1 3 5

  7. Graphing Systems of Linear Inequalities • The solution set of a system of linear inequalities in two variables x and y is the set of all points(x, y) that satisfy each inequality of the system. • The graphical solution of such a system may be obtained by graphing the solution set for each inequality independently and then determining the region in common with each solution set.

  8. Examples • Graph x – 3y> 0. Solution • Replacing the inequality> with an equality=, we obtain the equation x – 3y= 0, whose graph is: y 3 1 –1 –3 x – 3y= 0 x –5 –3 1 3 5

  9. Examples • Graph x – 3y> 0. Solution • We use a dashed line to indicate the line itself will not be part of the solution, since we are dealing with a strict inequality>. y 3 1 –1 –3 x – 3y= 0 x –5 –3 1 3 5

  10. Examples • Graph x – 3y> 0. Solution • Since the origin lies on the line, we cannot use the origin as a testing point: y 3 1 –1 –3 x – 3y= 0 (0, 0) x –5 –3 1 3 5

  11. Examples • Graph x – 3y> 0. Solution • Picking instead (3, 0) as a test point, we find (3) – 2(0) > 0, or 3 > 0, which is true. • Thus, the solution set is: y 3 1 –1 –3 x – 3y= 0 (3, 0) x –5 –3 1 3 5 x – 3y> 0

  12. Graphing Systems of Linear Inequalities • The solution set of a system of linear inequalities in two variables x and y is the set of all points(x, y) that satisfy each inequality of the system. • The graphical solution of such a system may be obtained by graphing the solution set for each inequality independently and then determining the region in common with each solution set.

  13. Example • Determine the solution set for the system Solution • The intersection of the solution regionsof the two inequalities represents the solution to the system: y 4x + 3y= 12 4 3 2 1 4x + 3y 12 x –1 1 2 3

  14. Example • Determine the solution set for the system Solution • The intersection of the solution regionsof the two inequalities represents the solution to the system: y 4 3 2 1 x – y  0 x – y = 0 x –1 1 2 3

  15. Example • Determine the solution set for the system Solution • The intersection of the solution regionsof the two inequalities represents the solution to the system: y 4x + 3y= 12 4 3 2 1 x – y = 0 x –1 1 2 3

  16. Bounded and Unbounded Sets • The solution set of a system of linear inequalities is bounded if it can be enclosed by a circle. • Otherwise, it is unbounded.

  17. Example • The solution to the problem we just discussed is unbounded, since the solution setcannot be enclosed in a circle: y 4x + 3y= 12 4 3 2 1 x – y = 0 x –1 1 2 3

  18. Example • Determine the solution set for the system Solution • The intersection of the solution regionsof the four inequalities represents the solution to the system: y 7 5 3 1 x –1 1 3 5 9

  19. Example • Determine the solution set for the system Solution • Note that the solution to this problem is bounded, since it can be enclosed by a circle: y 7 5 3 1 x –1 1 3 5 9

  20. 6.2 Maximize Subject to Linear Programming Problems

  21. Linear Programming Problem • A linear programming problem consists of a linear objective function to be maximized or minimized subject to certain constraints in the form of linear equations or inequalities.

  22. Applied Example 1: A Production Problem • Ace Novelty wishes to produce two types of souvenirs: type-A will result in a profit of $1.00, and type-B in a profit of $1.20. • To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II. • A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II. • There are 3 hours available on machine I and 5 hours available on machine II. • How many souvenirs of each type should Ace make in order to maximize its profit?

  23. Applied Example 1: A Production Problem Solution • Let’s first tabulate the given information: • Let x be the number of type-A souvenirs and y the number of type-B souvenirs to be made.

  24. Applied Example 1: A Production Problem Solution • Let’s first tabulate the given information: • Then, the total profit (in dollars) is given by which is the objective function to be maximized.

  25. Applied Example 1: A Production Problem Solution • Let’s first tabulate the given information: • The total amount of time that machine I is used is and must not exceed 180 minutes. • Thus, we have the inequality

  26. Applied Example 1: A Production Problem Solution • Let’s first tabulate the given information: • The total amount of time that machine II is used is and must not exceed 300 minutes. • Thus, we have the inequality

  27. Applied Example 1: A Production Problem Solution • Let’s first tabulate the given information: • Finally, neither x nor y can be negative, so

  28. Applied Example 1: A Production Problem Solution • In short, we want to maximize the objective function subject to the system of inequalities • We will discuss the solution to this problem in section 6.4.

  29. Applied Example 2: A Nutrition Problem • A nutritionist advises an individual who is suffering from iron and vitamin B deficiency to take at least 2400 milligrams (mg) of iron, 2100 mg of vitamin B1, and 1500 mg of vitamin B2 over a period of time. • Two vitamin pills are suitable, brand-A and brand-B. • Each brand-A pill costs 6 cents and contains 40 mg of iron, 10 mg of vitamin B1, and 5 mg of vitamin B2. • Each brand-B pill costs 8 cents and contains 10 mg of iron and 15 mg each of vitamins B1 and B2. • What combination of pills should the individual purchase in order to meet the minimum iron and vitamin requirements at the lowest cost?

  30. Applied Example 2: A Nutrition Problem Solution • Let’s first tabulate the given information: • Let x be the number of brand-A pills and y the number of brand-B pills to be purchased.

  31. Applied Example 2: A Nutrition Problem Solution • Let’s first tabulate the given information: • The costC (in cents) is given by and is the objective function to be minimized.

  32. Applied Example 2: A Nutrition Problem Solution • Let’s first tabulate the given information: • The amount of iron contained in xbrand-A pills and ybrand-B pills is given by 40x + 10y mg, and this must be greater than or equal to 2400 mg. • This translates into the inequality

  33. Applied Example 2: A Nutrition Problem Solution • Let’s first tabulate the given information: • The amount of vitamin B1 contained in xbrand-A pills and ybrand-B pills is given by 10x + 15y mg, and this must be greater or equal to 2100 mg. • This translates into the inequality

  34. Applied Example 2: A Nutrition Problem Solution • Let’s first tabulate the given information: • The amount of vitamin B2 contained in xbrand-A pills and ybrand-B pills is given by 5x + 15y mg, and this must be greater or equal to 1500 mg. • This translates into the inequality

  35. Applied Example 2: A Nutrition Problem Solution • In short, we want to minimize the objective function subject to the system of inequalities • We will discuss the solution to this problem in section 6.4.

  36. 6.3 Graphical Solutions of Linear Programming Problems

  37. Feasible Solution Set and Optimal Solution • The constraints in a linear programming problem form a system of linear inequalities, which have a solution setS. • Each point in S is a candidate for the solution of the linear programming problem and is referred to as a feasible solution. • The set S itself is referred to as a feasible set. • Among all the points in the set S, the point(s) that optimizes the objective function of the linear programming problem is called an optimal solution.

  38. Theorem 1 Linear Programming • If a linear programming problem has a solution, then it must occur at a vertex, or corner point, of the feasible setS associated with the problem. • If the objective functionP is optimized at two adjacent vertices of S, then it is optimized at every point on the line segment joining these vertices, in which case there are infinitely many solutions to the problem.

  39. Theorem 2 Existence of a Solution • Suppose we are given a linear programming problem with a feasible setS and an objective functionP = ax + by. • If S is bounded, then P has both a maximum and a minimum value on S. • If S is unbounded and both a and b are nonnegative, then P has a minimum value on S provided that the constraints defining S include the inequalities x 0 and y 0. • If S is the empty set, then the linear programming problem has no solution: that is, P has neither a maximum nor a minimum value.

  40. The Method of Corners • Graph the feasible set. • Find the coordinates of all corner points (vertices) of the feasible set. • Evaluate the objective function at each corner point. • Find the vertex that renders the objective function a maximum or a minimum. • If there is only one such vertex, it constitutes a unique solution to the problem. • If there are two such adjacent vertices, there are infinitely many optimal solutions given by the points on the line segment determined by these vertices.

  41. Applied Example 1: A Production Problem • Recall Applied Example 1 from the last section (3.2), which required us to find the optimal quantities to produce of type-A and type-B souvenirs in order to maximize profits. • We restated the problem as a linear programming problem in which we wanted to maximize the objective function subject to the system of inequalities • We can now solve the problem graphically.

  42. Applied Example 1: A Production Problem • We first graph the feasible setS for the problem. • Graph the solution for the inequality considering only positive values for x and y: y 200 100 (0, 180) (90, 0) x 100 200 300

  43. Applied Example 1: A Production Problem • We first graph the feasible setS for the problem. • Graph the solution for the inequality considering only positive values for x and y: y 200 100 (0, 100) (300, 0) x 100 200 300

  44. Applied Example 1: A Production Problem • We first graph the feasible setS for the problem. • Graph the intersection of the solutions to the inequalities, yielding the feasible setS. (Note that the feasible setS is bounded) y 200 100 S x 100 200 300

  45. Applied Example 1: A Production Problem • Next, find the vertices of the feasible setS. • The vertices are A(0, 0), B(90, 0), C(48, 84), and D(0, 100). y 200 100 D(0, 100) C(48, 84) S B(90, 0) A(0, 0) x 100 200 300

  46. Applied Example 1: A Production Problem • Now, find the values of P at the vertices and tabulate them: y 200 100 D(0, 100) C(48, 84) S B(90, 0) A(0, 0) x 100 200 300

  47. Applied Example 1: A Production Problem • Finally, identify the vertex with the highest value for P: • We can see that P is maximized at the vertex C(48, 84) and has a value of 148.8. y 200 100 D(0, 100) C(48, 84) S B(90, 0) A(0, 0) x 100 200 300

  48. Applied Example 1: A Production Problem • Finally, identify the vertex with the highest value for P: • We can see that P is maximized at the vertex C(48, 84) and has a value of 148.8. • Recalling what the symbols x, y, and P represent, we conclude that ACE Novelty would maximize its profit at $148.80 by producing 48 type-A souvenirs and 84type-B souvenirs.

  49. Applied Example 2: A Nutrition Problem • Recall Applied Example 2 from the last section (3.2), which asked us to determine the optimal combinationof pills to be purchased in order to meet the minimum iron and vitaminrequirements at the lowest cost. • We restated the problem as a linear programming problem in which we wanted to minimize the objective function subject to the system of inequalities • We can now solve the problem graphically.

  50. Applied Example 2: A Nutrition Problem • We first graph the feasible setS for the problem. • Graph the solution for the inequality considering only positive values for x and y: y (0, 240) 200 100 (60, 0) x 100 200 300

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