A = Chapter 6. Linear Programming: The Simplex Method. + 21x 3 x x 2. C = 16x 1. + x x x 1. + x 3. 16,x 2.

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1 Chapter 6 Linear rogramming: The Simple Method Section The Dual roblem: Minimization with roblem Constraints of the Form Learning Objectives for Section 6. Dual roblem: Minimization with roblem Constraints of the Form > The student will be able to formulate the dual problem. The student will be able to solve minimization problems. The student will be able to solve applications of the dual such as the transportation problem. The student will be able to summarize problem types and solution methods. Dual roblem: Minimization With roblem Constraints of the Form > Initial Matri Associated with each minimization problem with > constraints is a maimization problem called the dual problem. The dual problem will be illustrated through an eample. We wish to minimize the objective function C subject to certain constraints: C = ,, 0 We start with an initial matri A which corresponds to the problem constraints: A =

2 Transpose of Matri A To find the transpose of matri A, interchange the rows and columns so that the first row of A is now the first column of A transpose. The transpose is used in formulating the dual problem to follow. A T = Dual of the Minimization roblem The dual of the minimization problem is the following maimization problem: Maimize under the constraints: = y + y 16 + y 9 + y 1, y Formation of the Dual roblem Theorem 1: Fundamental rinciple of Duality Given a minimization problem with problem constraints, 1. Use the coefficients and constants in the problem constraints and the objective function to form a matri A with the coefficients of the objective function in the last row.. Interchange the rows and columns of matri A to form the matri A T, the transpose of A.. Use the rows of A T to form a maimization problem with problem constraints. A minimization problem has a solution if and only if its dual problem has a solution. If a solution eists, then the optimal value of the minimization problem is the same as the optimum value of the dual problem. 7 8

3 Forming the Dual roblem Form the Simple Tableau for the Dual roblem We transform the inequalities into equalities by adding the slack variables 1,, : + y + 1 = 16 + y + = 9 + y + = y + p = 0 The first pivot element is (in red) because it is located in the column with the smallest negative number at the bottom (-16), and when divided into the rightmost constants yields the smallest quotient (16/=8) y1 y Simple rocess Divide row 1 by the pivot element () and change the entering variable to y Simple rocess erform row operations to get zeros in the column containing the pivot element. Result is shown below. Identify the net pivot element (0.5) (in red) y y 1 1 y New pivot element ivot element located in this column because of negative indicator y y 1 1 y

4 Simple rocess Simple rocess Variable becomes new entering variable Divide row by 0.5 to obtain a 1 in the pivot position (circled.) y1 y 1 y Get zeros in the column containing the new pivot element. y 1 y We have now obtained the optimal solution since none of the bottom row indicators are negative Solution of the Linear rogramming roblem Solution of a Minimization roblem Solution: An optimal solution to a minimization problem can always be obtained from the bottom row of the final simple tableau for the dual problem. Minimum of is 16, which is also the maimum of the dual problem. It occurs at 1 = 4, = 8, = 0 y y 1 y y Given a minimization problem with non-negative coefficients in the objective function, 1. Write all problem constraints as inequalities. (This may introduce negative numbers on the right side of some problem constraints.). Form the dual problem.. Write the initial iti system of fthe dual problem, using the variables from the minimization problem as slack variables. 16

5 Solution of a Minimization roblem Application: 4. Use the simple method to solve the dual problem. 5. Read the solution of the minimization problem from the bottom row of the final simple tableau in step 4. Note: If the dual problem has no optimal solution, the minimization problem has no optimal solution. One of the first applications of linear programming was to the problem of minimizing the cost of transporting materials. roblems of this type are referred to as transportation problems. Eample: A computer manufacturing company has two assembly plants, plant A and plant B, and two distribution outlets, outlet I and outlet II. lant A can assemble at most 700 computers a month, and plant B can assemble at most 900 computers a month. Outlet I must have at least 500 computers a month, and outlet II must have at least 1,000 computers a month Transportation costs for shipping one computer from each plant to each outlet are as follows: $6 from plant A to outlet I; $5 from plant A to outlet II: $4 from plant B to outlet I; $8 from plant B to outlet II. Find a shipping schedule that will minimize the total cost of shipping the computers from the assembly plants to the distribution outlets. What is the minimum cost? Solution: To form a shipping schedule, we must decide how many computers to ship from either plant to either outlet. This will involve 4 decision variables: 1 = number of computers shipped from plant A to outlet I = number of computers shipped from plant A to outlet II = number of computers shipped from plant B to outlet I 4 = number of computers shipped from plant B to outlet II 19 0

6 Constraints are as follows: 1 + < 700 Available from A + 4 < 900 Available from B 1 + > 500 Required at I + 4 > 1,000 Required at II Total shipping charges are: C = Thus, we must solve the following linear programming problem: subject to 1 + Minimize C = < 700 Available from A + 4 < 900 Available from B + > 500 Required at I > 1,000 Required at II Before we can solve this problem, we must multiply the first two constraints by -1 so that all are of the > type. 1 The problem can now be stated as: Minimize C = subject to > > > > 1,000 1,,, 4 > A = , T A = ,

7 The dual problem is; Maimize = y +500y + 1,000y 4 subject to - + y < y 4 < 5 -y + y < 4 -y + y 4 < 8, y, y, y 4 > 0 Introduce slack variables 1,,, and 4 to form the initial system for the dual: - + y + 1 = y 4 + = 5 -y + y + = 4 -yy + y = y +500y + 1,000y 4 + = Solution Summary of roblem Types and Simple Solution Methods If we form the simple tableau for this initial system and solve, we find that the shipping schedule that minimizes the shipping charges is 0 from plant A to outlet I, 700 from plant A to outlet II, 500 from plant B to outlet I, and 00 from plant B to outlet II. The total shipping cost is $7,900. roblem Constraints Right-Side Coefficients of Method of of Type Constants Objective Solution Function Maimization < Nonnegative Any real number Minimization > Any real Nonnegative number Simple Method Form dual and solve by simple method 7 8

Gauss-Jordan Elimination for Solving Linear Equations Example: 1. Solve the following equations: (3)

The Simple Method Gauss-Jordan Elimination for Solving Linear Equations Eample: Gauss-Jordan Elimination Solve the following equations: + + + + = 4 = = () () () - In the first step of the procedure, we