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Elementary Linear Algebra 6e - Larson, Edwards, Falvo - Chapter 9.5

# Elementary Linear Algebra 6e - Larson, Edwards, Falvo - Chapter 9.5

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9.5 THE SIMPLEX METHOD: MIXED CONSTRAINTS In Sections 9.3 and 9.4, you looked at linear programming problems that occurred in stan- dard form. The constraints for the maximization problems all involved inequalities, and the constraints for the minimization problems all involved inequalities. Linear programming problems for which the constraints involve both types of inequali- ties are called mixed-constraint problems. For instance, consider the following linear programming problem. Mixed-Constraint Problem: Find the maximum value of Objective function subject to the constraints where and Because this is a maximization problem, you would expect each of the inequalities in the set of constraints to involve Moreover, because the first inequality does involve you can add a slack variable to form the following equation. For the other two inequalities, a new type of variable, called a surplus variable, is intro- duced as follows. Notice that surplus variables are subtracted from (not added to) their inequalities. The vari- ables and are called surplus variables because they represent the amount that the left side of the inequality exceeds the right side. Surplus variables must be nonnegative. Now, to solve the linear programming problem, form an initial simplex tableau as follows. Basic x 1 x 2 x 3 s 1 s 2 s 3 b Variables 2 1 1 1 0 0 50 s 1 2 1 0 0 0 36 s 2 1 0 1 0 0 10 s 3 Departing 0 0 0 0 Entering You will soon discover that solving mixed-constraint problems can be difficult. One reason for this is that there is no convenient feasible solution to begin the simplex method. Note that the solution represented by the initial tableau above, x 1 , x 2 , x 3 , s 1 , s 2 , s 3 0, 0, 0, 50, 36, 10 , 2 1 1 1 1 s 3 s 2 x 1 x 3 s 3 10 2 x 1 x 2 s 2 36 2 x 1 x 2 x 3 s 1 50 , . x 3 0. x 2 0, x 1 0, x 1 x 3 10 2 x 1 x 2 36 2 x 1 x 2 x 3 50 z x 1 x 2 2 x 3 SECTION 9.5 THE SIMPLEX METHOD: MIXED CONSTRAINTS 557 Constraints

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is not a feasible solution because the values of the two surplus variables are negative. In fact, the values do not even satisfy the constraint equations. In order to elim- inate the surplus variables from the current solution, “trial and error” is used. That is, in an effort to find a feasible solution, arbitrarily choose new entering variables. For instance, in this tableau, it seems reasonable to select as the entering variable. After pivoting, the new simplex tableau becomes the following. Basic x 1 x 2 x 3 s 1 s 2 s 3 b Variables 1 1 0 1 0 1 40 s 1 2 1 0 0 0 36 s 2 Departing 1 0 1 0 0 10 x 3 1 0 0 0 20 Entering The current solution is still not feasible, so choose as the entering variable and pivot to obtain the following simplex tableau. Basic x 1 x 2 x 3 s 1 s 2 s 3 b Variables 0 0 1 1 1 4 s 1 Departing 2 1 0 0 0 36 x 2 1 0 1 0 0 10 x 3 3 0 0 0 56 Entering At this point, the following feasible solution is finally obtained.
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Elementary Linear Algebra 6e - Larson, Edwards, Falvo - Chapter 9.5

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