Linear Programming: The Simplex Method

Chapter 4 SupplementLinear Programming:The Simplex MethodPart 2 Deterministic Decision ModelsLearning ObjectivesExplain the ways in which the simplex method is superior to the graphical method for solving linear programming problems.Solve small maximization problems manually using the simplex method.Interpret simplex solutions.Convert = and constraints into standard form.Solve maximization problems that have mixed constraints and interpret those solutions.After completing this chapter, you should be able to:>2Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Learning Objectives (cont’d)Solve minimization problems and interpret those solutions.Discuss unbound solutions, degeneracy, and multiple optimal solutions in terms of the simplex method and recognize infeasibility in a simplex solution.After completing this chapter, you should be able to:3Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Overview of the Simplex MethodAdvantages and CharacteristicsMore realistic approach as it is not limited to problems with two decision variablesSystematically examines basic feasible solutions for an optimal solution.Based on the solutions of linear equations (equalities) using slack variables to achieve equality.RuleLinear programming models have fewer equations than variables; unless the number of equations equals the number of variables, a unique solution cannot be found.4Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Developing the Initial Simplex TableauNotation used in the simplex tableau:5Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Figure 4S–1 Comparison of Server Model and General Simplex Notation6Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–1 Comparison of Server Model and General Simplex Notation (cont’d)7Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–2 Completed Initial Tableau for the Server ProblemEach tableau represents a basic feasible solution to the problem. Unit VectorA simplex solution in a maximization problem is optimal if the C–Z row consists entirely of zeros and negative numbers (i.e., there are no positive values in the bottom row). When this has been achieved, there is no opportunity for improving the solution.8Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–3 Determining the Entering and Exiting VariablesSelect the leaving variable as the one that has the smallest nonnegative ratio of quantity divided by substitution rate.9Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Figure 4S–2 The Next Corner Point Is Determined by the Most Limiting Constraint10Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–4 Starting the Second TableauTable 4S–5 Initial Tableau11Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–6 The Pivot Row of the Second TableauTable 4S–7 Revised First Row and Pivot Row of the Second Tableau12Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–8 Partially Completed Second Tableau13Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–9 Completed Second TableauInterpreting the Second TableauAt this point, variables s1, x1, and s3 are in solution. Not only are they listed in the basis, they also have a 0 in row C – Z. The solution at this point is s1 = 56, x1 = 11, and s3 = 6. Note, too, that x2 and s2 are not in solution. Hence, they are each equal to zero. The profit at this point is $660, which is read in the Quantity column in row Z. Also, note that each variable in solution has a unit vector in its column.14Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–10 Determining the Exiting Variable15Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Figure 4S–3 Moving to the Next Corner Point16Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–11 Pivot Row Values for the Third TableauTable 4S–12 Partially Completed Third Tableau17Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–13 Completed Third TableauInterpreting the Third TableauIn this tableau, all of the values in the bottom row are either negative or zero, indicating that no additional potential for improvement exists. Hence, this tableau contains the optimal simplex solution, which iss1 = 24x1 = 9x2 = 418Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Summary of the Simplex Procedure for a Maximization ProblemInitial TableauWrite each constraint so that all variables are on the left side and a nonnegative constant is on the right. Then add a slack variable to the left side, thereby making it an equality.Develop the initial tableau.List the variables across the top of the table and write the objective function coefficient of each variable just above it.There should be one row in the body of the table for each constraint. List slack variables in the basis column, one per row.In the C column, enter the objective function coefficient of 0 for each slack variable.Compute values for row Z.Compute values for row C – Z.19Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Summary of the Simplex Procedure for a Maximization Problem (cont’d)Subsequent TableausIdentify the variable with the largest positive value in row C – Z. This variable will come into solution next.Using the constraint coefficients in the entering variable’s column, divide each one into the corresponding Quantity column value. The smallest nonnegative ratio that results indicates which variable will leave the solution mix.Compute replacement values for the leaving variable: Divide each element in the row by the row element that is in the entering variable column. These are the pivot row values for the next tableau. Enter them in the same row as the leaving variable and label the row with the name of the entering variable. Write the entering variable’s objective function coefficient next to it in column C. 20Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Summary of the Simplex Procedure for a Maximization Problem (cont’d)Subsequent Tableaus (cont’d)Compute values for each of the other constraint equations:Multiply each of the pivot row values by the number in the entering variable column of the row being transformed (e.g., for the first row, use the first number in the entering variable’s column; for the third row, use the third number in the entering variable’s column).Then subtract the resulting equation from the current equation for that row and enter the results in the same row of the next tableau.Compute values for row Z: For each column, multiply each row coefficient by the row value in column C and then add the results. Enter these in the tableau. 6. Compute values for row C – Z: For each column, subtract the value in row Z from the objective function coefficient listed in row C at the top of the tableau.21Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Summary of the Simplex Procedure for a Maximization Problem (cont’d)Subsequent Tableaus (cont’d)Examine the values in the bottom row. If all values are zero or negative, the optimal solution has been reached. The variables that comprise the solution are listed in the basis column and their optimal values can be read in the corresponding rows of the quantity column. The optimal value of the objective function will appear in row Z in the Quantity column.If the solution is not optimal, repeat steps 1–7 of this section until the optimal solution has been attained.22Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–14 Summary of Use of Slack, Surplus, and Artificial Variables23Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Example 4S-1 Solve this maximization problem using the simplex approach:24Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Figure 4S–4 Graph for Example 4S-125Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–15 Initial Tableau for Example 4S-126Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–16 The Second Tableau for Example 4S-127Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–17 The Third Tableau for Example 4S-128Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–18 The Final Tableau for Example 4S-129Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Figure 4S–5 Sequence of Tableaus for Example 4S-130Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Example 4S-2 Solve this minimization problem using the simplex method:31Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Figure 4S–6 Graph of the Problem in Example 4S-232Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–19 Initial Tableau for Example 4S-233Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–20 Second Tableau for Example 4S-234Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–21 Third Tableau for Example 4S-235Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Figure 4S–7 Sequence of Tableaus for Solution of Example 4S-236Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Some Special IssuesUnbounded SolutionsA solution is unbounded if the objective function can be improved without limit. An unbounded solution will exist if there are no positive values in the pivot column.DegeneracyA conditions that occurs when there is a tie for the lowest nonnegative ratio which, theoretically, makes it possible for subsequent solutions to cycle (i.e., to return to previous solutions).37Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Example 4S–3 38Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–22 Second Tableau39Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–23 Final Simplex Tableau40Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Some Special Issues (cont’d)Multiple Optimal SolutionsOccur when the same maximum value of the objective function might be possible with a number of different combinations of values of the decision variables because the objective function is parallel to a binding constraint.41Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–24 Final Tableau for Modified Server Problem with anAlternative Optimal Solution42Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–25 The Alternate Optimal Solution for the ModifiedServer Problem43Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Some Special Issues (cont’d)InfeasibilityA problem in which no combination of decision and slack/surplus variables will simultaneously satisfy all constraints. Can be the result of an error in formulating a problem or it can be because the existing set of constraints is too restrictive to permit a solution.Recognized by the presence of an artificial variable in a solution that appears optimal (i.e., a tableau in which the signs of the values in row C – Z indicate optimality), and it has a nonzero quantity.44Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Table 4S–26 Simplex Tableaus for Infeasibility Problem for Example 4S-445Copyright © 2007 The McGraw-Hill Companies. All rights reserved. Example 4S–4 46Copyright © 2007 The McGraw-Hill Companies. All rights reserved.