Lecture Introduction to Management Science with Spreadsheets: Chapter 4S - Stevenson, Ozgur

Chapter 4 - Supplement Linear programming: The simplex method, after completing this chapter, you should be able to: Explain 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,...

Stevenson and Ozgur

First Edition

Introduction to Management Science

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Learning Objectives

1 Explain the ways in which the simplex method is

superior to the graphical method for solving linear programming problems

2 Solve small maximization problems manually using the simplex method

3 Interpret simplex solutions

4 Convert = and constraints into standard form

5 Solve maximization problems that have mixed

constraints and interpret those solutions

After completing this chapter, you should be able to:

>

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6 Solve minimization problems and interpret those

solutions

7 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:

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Overview of the Simplex Method

Overview of the Simplex Method

• Advantages and Characteristics

–More realistic approach as it is not limited to problems with two decision variables

–Systematically examines basic feasible solutions for

an optimal solution

–Based on the solutions of linear equations (equalities) using slack variables to achieve equality

• Rule

–Linear programming models have fewer equations

than variables; unless the number of equations equals

the number of variables, a unique solution cannot be found

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Developing the Initial Simplex Tableau

Developing the Initial Simplex Tableau

• Notation used in the simplex tableau:

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Figure 4S–1 Comparison of Server Model and General Simplex Notation Figure 4S–1 Comparison of Server Model and General Simplex Notation

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Table 4S–1 Comparison of Server Model and General Simplex Notation (cont’d) Table 4S–1 Comparison of Server Model and General Simplex Notation (cont’d)

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Table 4S–2 Completed Initial Tableau for the Server Problem

Table 4S–2 Completed Initial Tableau for the Server Problem

Each tableau represents a basic feasible solution to the problem

Unit Vector

A 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.

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Table 4S–3 Determining the Entering and Exiting Variables

Table 4S–3 Determining the Entering and Exiting Variables

Select the leaving variable as the one that has the smallest nonnegative ratio of quantity divided by substitution rate.

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Figure 4S–2 The Next Corner Point Is Determined by the Most Limiting

Constraint Figure 4S–2 The Next Corner Point Is Determined by the Most Limiting

Constraint

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Table 4S–4 Starting the Second Tableau

Table 4S–4 Starting the Second Tableau

Table 4S–5 Initial Tableau

Table 4S–5 Initial Tableau

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Table 4S–7 Revised First Row and Pivot Row of the Second Tableau Table 4S–7 Revised First Row and Pivot Row of the Second Tableau

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Table 4S–8 Partially Completed Second Tableau

Table 4S–8 Partially Completed Second Tableau

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Interpreting the Second Tableau

At 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.

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Table 4S–10 Determining the Exiting Variable

Table 4S–10 Determining the Exiting Variable

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Figure 4S–3 Moving to the Next Corner Point

Figure 4S–3 Moving to the Next Corner Point

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Table 4S–11 Pivot Row Values for the Third Tableau

Table 4S–11 Pivot Row Values for the Third Tableau

Table 4S–12 Partially Completed Third Tableau

Table 4S–12 Partially Completed Third Tableau

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Interpreting the Third Tableau

In 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 is

s1 = 24

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Summary of the Simplex Procedure for a

2 Develop the initial tableau.

1 List the variables across the top of the table and write the  objective function coefficient of each variable just above it.

2 There should be one row in the body of the table for each  constraint. List slack variables in the basis column, one per row.

3 In the C column, enter the objective function coefficient of 0 for  each slack variable.

4 Compute values for row Z.

5 Compute values for row C – Z.

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Summary of the Simplex Procedure for a

Maximization Problem (cont’d)

Summary of the Simplex Procedure for a

Maximization Problem (cont’d)

• Subsequent Tableaus

1 Identify the variable with the largest positive value in row C – Z This

variable will come into solution next.

2 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.

3 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

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Summary of the Simplex Procedure for a

Maximization Problem (cont’d)

Summary of the Simplex Procedure for a

Maximization Problem (cont’d)

• Subsequent Tableaus (cont’d)

4 Compute values for each of the other constraint equations:

1 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).

2 Then subtract the resulting equation from the current equation for that row  and enter the results in the same row of the next tableau.

4 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

5 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.

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Summary of the Simplex Procedure for a

Maximization Problem (cont’d)

Summary of the Simplex Procedure for a

Maximization Problem (cont’d)

• Subsequent Tableaus (cont’d)

6 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.

7 If the solution is not optimal, repeat steps 1–7 of this section until the optimal solution has been attained.

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Table 4S–14 Summary of Use of Slack, Surplus, and Artificial Variables Table 4S–14 Summary of Use of Slack, Surplus, and Artificial Variables

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Example 4S-1 Solve this maximization problem using the simplex

approach:

Example 4S-1 Solve this maximization problem using the simplex

approach:

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Table 4S–15 Initial Tableau for Example 4S-1

Table 4S–15 Initial Tableau for Example 4S-1

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Table 4S–16 The Second Tableau for Example 4S-1

Table 4S–16 The Second Tableau for Example 4S-1

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Table 4S–17 The Third Tableau for Example 4S-1

Table 4S–17 The Third Tableau for Example 4S-1

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Table 4S–18 The Final Tableau for Example 4S-1

Table 4S–18 The Final Tableau for Example 4S-1

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Figure 4S–5 Sequence of Tableaus for Example 4S-1

Figure 4S–5 Sequence of Tableaus for Example 4S-1

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Example 4S-2 Solve this minimization problem using the simplex method:

Example 4S-2 Solve this minimization problem using the simplex method:

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Figure 4S–6 Graph of the Problem in Example 4S-2

Figure 4S–6 Graph of the Problem in Example 4S-2

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Table 4S–19 Initial Tableau for Example 4S-2

Table 4S–19 Initial Tableau for Example 4S-2

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Table 4S–20 Second Tableau for Example 4S-2

Table 4S–20 Second Tableau for Example 4S-2

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Table 4S–21 Third Tableau for Example 4S-2

Table 4S–21 Third Tableau for Example 4S-2

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Figure 4S–7 Sequence of Tableaus for Solution of Example 4S-2

Figure 4S–7 Sequence of Tableaus for Solution of Example 4S-2

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Some Special Issues

Some Special Issues

• Unbounded Solutions

–A 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

• Degeneracy

–A 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)

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Example 4S–3

Example 4S–3

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Table 4S–23 Final Simplex Tableau

Table 4S–23 Final Simplex Tableau

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Some Special Issues (cont’d)

Some Special Issues (cont’d)

• Multiple Optimal Solutions

–Occur 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

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Table 4S–24 Final Tableau for Modified Server Problem with an

Alternative Optimal Solution Table 4S–24 Final Tableau for Modified Server Problem with an

Alternative Optimal Solution

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Table 4S–25 The Alternate Optimal Solution for the Modified

Server Problem Table 4S–25 The Alternate Optimal Solution for the Modified

Server Problem

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Some Special Issues (cont’d)

Some Special Issues (cont’d)

• Infeasibility

–A 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

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Table 4S–26 Simplex Tableaus for Infeasibility Problem for Example 4S-4 Table 4S–26 Simplex Tableaus for Infeasibility Problem for Example 4S-4

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Example 4S–4

Example 4S–4