Slide 1 Math 1520, Lecture 10
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1 Slide 1 Math 1520, Lecture 10 In this lecture, we study the simplex method which is a powerful method for solving maximization/minimization problems developed in the late 1940 s. It has been implemented on computers and can be used on problems with a large number of variables and constraints. The simplex method is used in some of the management courses at RPI. The simplex method is based on the Gauss-Jordan method we studied in chapter 5. The simplex method is more involved and harder than the methods we have encountered thus far in this course.
2 Slide 2 A description of the simplex method Suppose we want to maximize a linear objective function. When there are many vertexes to the feasible set, it is very time consuming to find all the vertexes and test each one to find the maximum as we did in the method of corners. The simplex method shortens this process by iteratively traversing the feasible set in directions so as to make the objective function bigger. It organizies all this in a matrix called the simplex tableau.
3 Slide 3 Standard Maximization Problem The simplex method we describe first applies to standard maximization problems: The standard maximization problem is one in which 1. The goal is to maximize a linear objective function. 2. All the variables involved are non-negative (e.g. x 0 ) 3. All the other linear constraints may be written so that the expression is less than or equal to a non-negative constant. The following problem is not in standard form: Maximize P = 2x + y Subject to the constraints x 0, y 0, x y 4, x + 3y 6 But we can rewrite it to one that is in standard form as: Maximize P = 2x + y Subject to the constraints x 0, y 0, x + y 4, x + 3y 6
4 Slide 4 Slack Variables Consider this problem: Maximize P = x + 2y Subject to the constraints x 0, y 0, x + y 4, x + 3y 6 To solve it usinig the simplex method we first introduce slack variables called u and v to rewrite the last two inequality constraints as equalities to get x + y + u = 4, x + 3y + v = 6, u 0,, v 0 We then convert the objective function P = x + 2y into the form x 2y + P = 0 and put these into a system of linear equations x + y + u = 4 x + 3y + v = 6 x 2y + P = 0 where we want to maximize P subject to x 0, y 0, u 0, v 0.
5 Slide 5 iclicker Question Introduce slack variables for the standard maximization problem:. Maximize P = 3x + 2y Subject to the constraints x 0, y 0, 2x + 3y 12, 2x + y 8 A. B. C. D. E. None of the above 2x + 3y + u = 12 2x + y + v = 8 3x + 2y P = 0 2x + 3y + u = 8 2x + y + v = 12 3x + 2y P = 0 2x + 3y + u = 12 2x + y + v = 8 3x 2y + P = 0 2x + 3y + u = 8 2x + y + v = 12 3x 2y + P = 0
6 Answer to Question Introduce slack variables for the standard maximization problem:. A. Maximize P = 3x + 2y Subject to the constraints x 0, y 0, 2x + 3y 12, 2x + y 8 B. C. D. is the correct answer. E. None of the above 2x + 3y + u = 12 2x + y + v = 8 3x + 2y P = 0 2x + 3y + u = 8 2x + y + v = 12 3x + 2y P = 0 2x + 3y + u = 12 2x + y + v = 8 3x 2y + P = 0 2x + 3y + u = 8 2x + y + v = 12 3x 2y + P = 0
7 Slide 6 Simplex Tableau Once we introduce slack varibles, we take the resulting equations which were x + y + u = 4 x + 3y + v = 6 x 2y + P = 0 and from this set of equations, we write what is called the simplex tableau as In this tableau, the variables associated with unit columns are called basic variables and those with non-unit columns are called non-basic variables. The goal will be to go through a sequence of pivoting steps to make all the entries to the left of the vertical bar non-negative..
8 Slide 7 iclicker Question Write the simplex tableau for the following problem. Maximize P = 3x + 2y Subject to the constraints x 0, y 0, 2x + 3y 12, 2x + y 8. A. B. C. None of the above D. E
9 Answer to Question Write the simplex tableau for the following problem.. A. Maximize P = 3x + 2y Subject to the constraints B. x 0, y 0, 2x + 3y 12, 2x + y 8 is the correct answer. C. None of the above D. E
10 Slide 8 Pivoting For the tableau We proceed by making some of the columns to the left of the vertical bar into unit columns using Gauss-Jordan elimination. But we have to choose the pivots more carefully than we did with Gauss-Jordan elimination. To choose the pivot column, pick the column with the most negative entry to the left of the vertical bar in the last row. To choose the pivot row, divide each positive entry in the pivot column into the corroseonding entry in the column of constants. The pivot row is the row corresponding to the smallest ratio. We have a solution when all the entries to the left of the vertical bar in the last row are non-negative. This is called a simplex tableau in final form. In the above matrix the pivot entry is in the second column and second row. We pivot on this entry to make the second column into a unit column. 2/ / / / / /3 1 4
11 Slide 9 Pivoting (continued) We now have the matrix: 2/ / / / / /3 1 4 The next pivot entry is the first column and the first row. We pivot on this entry to get the first column to be a unit column /2 1/ /2 1/ /2 1/2 1 5 We can then read off the solution from matching the unit columns with the last column to get x = 3, y = 1 and P = 5 and the other variables are set to 0 so u = 0, v = 0.
12 Slide 10 Practice Try the same thing for the simplex tableau you found in the last iclicker question: We proceed by making some of the columns to the left of the vertical bar into unit columns using Gauss-Jordan elimination. But we have to choose the pivots more carefully than we did with Gauss-Jordan elimination. To choose the pivot column, pick the column with the most negative entry to the left of the vertical bar in the last row. To choose the pivot row, divide each positive entry in the privot column into the corroseonding entry in the column of constants. The pivot row is the row corresponding to the smallest ratio. We have a solution when all the entries to the left of the vertical bar in the last row are non-negative. This is called a simplex tableau in final form. We can read off the solution by matching the unit columns with the last column. All other variables associated with non-unit columns are set to zero in the solution.
13 Slide 11 iclicker Question Which of the following tableaus is in final form? A. B. C. D. E x y z u v P Constant
14 Answer to Question Which of the following tableaus is in final form? A. B. C. D. E. is the correct answer x y z u v P Constant
15 Slide 12 iclicker Question What is the solution for the following tableau? x y z u v P Constant A. x = 0, y = 4, z = 0, u = 6, v = 0 for a maximum of 15 B. x = 4, y = 0, z = 6, u = 0, v = 0 for a maximum of 15 C. x = 0, y = 4, z = 6, u = 0, v = 0 for a maximum of 15 D. x = 6, y = 4, z = 0, u = 0, v = 0 for a maximum of 15 E. None of the above
16 Answer to Question What is the solution for the following tableau? x y z u v P Constant A. x = 0, y = 4, z = 0, u = 6, v = 0 for a maximum of 15 is the correct answer. B. x = 4, y = 0, z = 6, u = 0, v = 0 for a maximum of 15 C. x = 0, y = 4, z = 6, u = 0, v = 0 for a maximum of 15 D. x = 6, y = 4, z = 0, u = 0, v = 0 for a maximum of 15 E. None of the above
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