Linear Programming The Simplex Method 2003 by Prentice

Introduction § Graphical methods are fine for 2 variables. § But most LP problems are too complex for simple graphical procedures. § The Simplex Method: o examines corner points, like in graphing; o systematically examines corner points, using algebra, until an optimal solution is found; o does its searching iteratively. © 2003 by Prentice Hall, Inc. 2

Introduction Why study the Simplex Method? § The Simplex Method: o Provides the optimal solution to the Xi variables and the maximum profit (or minimum cost). o Provides important economic information. § Understanding how the Simplex Method works is important because o it allows for understanding how to interpret LP computer printouts. © 2003 by Prentice Hall, Inc. 3

Setting UP the Simplex Method • The algebraic procedure is based on solving systems of equations. • The first step in setting up the simplex method is to convert the functional inequality constraints to equivalent equality constraints. This conversion is accomplished by introducing slack variables. • To illustrate, consider this constraint: X 1 4 The slack variable for this constraint is defined to be S 1=4 - X 1 which is the amount of slack in the left-hand side of the inequality. Thus X 1+S 1=4 © 2003 by Prentice Hall, Inc. 4

Example • Objective Function Max. Z=3 X 1+5 X 2 • Subject to: X

The Standard Form of the Model Z-3 X 1 -5 X 2=0 . .

The Simplex Method in Tabular Form Tabular form Eq. Basic variable Coefficient of: Right side Z X 1 X 2 S 1 S 2 S 3 (0) Z 1 -3 -5 0 0 (1) S 1 0 1 0 0 4 (2) S 2 0 0 2 0 12 (3) S 3 0 3 2 0 0 1 18 © 2003 by Prentice Hall, Inc. 7

Summary of the Simplex Method • Initialization: Introduce slack variables, then constructing the initial simplex tableau. • Optimality Test: The current basic feasible solution is optimal if and only if every coefficient in row (0) is nonegative ( 0). If it is, stop; otherwise, go to an iteration to obtain the next basic feasible solution. • Iteration: Step 1 Determine the entering basic variable by selecting the variable with negative coefficient having the largest absolute value (i, e. , the “most negative” coefficient) in Eq. (0). Put a box around the column below this coefficient, and call this the pivot column. © 2003 by Prentice Hall, Inc. 8

Summary of the Simplex Method/ continued Step 2 Determine the leaving basic variable by applying the minimum ratio test. Minimum Ratio Test 1. Pick out each coefficient in the pivot column that is strictly positive (>0). 2. Divide each of these coefficients into the right side entry for the same row. 3. Identify the row that has the smallest of these ratios. 4. The basic variable for that row is the leaving basic variable, so replace that variable by the entering basic variable in the basic variable column of the next simplex tableau. Put a box around this row and call it the pivot row. Also call the number that is in both boxes the pivot number. © 2003 by Prentice Hall, Inc. 9

Summary of the Simplex Method/ continued Step 3 Solve for the new basic feasible solution by using elementary row operations (by applying Gauss-jordan method), The method effects a change in basis by using two types of computations: 1. Type 1 (pivot equation): new pivot Eq. =old pivot Eq. ÷ pivot number 2. Type 2 (all other eqautions, including Z): new Eq. = old Eq. – (its entering column coefficient) X (new pivot Eq. ) © 2003 by Prentice Hall, Inc. 10

Example • Solve this model using the simplex method: • Objective Function Max. Z=3

Solution • First, the Standard Form of the model: • Z-3 X 1 -5 X 2=0 . . ………. (0) • X 1+S 1=4 • 2 X 2+S 2=12 • 3 X 1+2 X 2+X 5=18 . . . ……. . (1). . ……. …(2) ………. . . (3) © 2003 by Prentice Hall, Inc. 12

The Simplex Method in Tabular Form Tabular form Eq. Basic variable Coefficient of: Right side Z X 1 X 2 S 1 S 2 S 3 (0) Z 1 -3 -5 0 0 (1) S 1 0 1 0 0 4 (2) S 2 0 0 2 0 12 (3) S 3 0 3 2 0 0 1 18 © 2003 by Prentice Hall, Inc. 13

# (0) Iteration Tabular form Eq. Basic variable Coefficient of: Z X 1 X 2 S 1 S 2 S 3 Right side (0) Z 1 -3 -5 0 0 (1) S 1 0 1 0 0 4 (2) S 2 0 0 2 0 12 (3) S 3 0 3 2 0 0 1 18 © 2003 by Prentice Hall, Inc. 14

# (1) Iteration Tabular form Eq. Basic variable Coefficient of: Z X 1 X 2 S 1 S 2 S 3 Right side (0) Z 1 -3 0 0 52 0 30 (1) S 1 0 1 0 0 4 (2) X 2 0 0 12 0 6 (3) S 3 0 0 -1 1 6 © 2003 by Prentice Hall, Inc. 15

# (2) Iteration Tabular form Eq. Basic variable Coefficient of: Z X 1 X 2 S 1 S 2 S 3 1 Right side (0) Z 1 0 0 0 32 (1) S 1 0 0 0 1 13 -13 2 (2) X 2 0 0 12 6 (3) X 1 0 0 -13 0 36 2 © 2003 by Prentice Hall, Inc. 16

• Going to the optimality test, we find that this solution is optimal because none of the coefficients in row (0) is negative, so the algorithm is finished. • Consequently, the optimal solution for this problem is X 1=2, X 2=6. © 2003 by Prentice Hall, Inc. 17

Special Cases in Simplex Method Application • 1. 2. 3. 4. We will consider special cases that can arise in the application of the simplex method, which include: Degeneracy. Alternative optima (more than one optimum solution). Unbounded solutions. Nonexisting (or infeasible) solutions. © 2003 by Prentice Hall, Inc. 18

Degeneracy • In the application of the feasibility condition, a tie for the minimum ratio may be broken for the purpose of determining the leaving variable. When this happens, however, one or more of the basic variables will necessarily equal zero in the next iteration. • In this case, we say that the new solution is degenerate. (In all LP examples we have solved so far, the basic variables always assumed strictly positive values). • The degeneracy has two implications: The first deals with the phenomenon of cycling or circling (If you look at iterations 1 and 2 in the next example you will find that the objective value has not improved (Z=18)), in general, the simplex procedure would repeat the same sequence of iterations, never improving the objective value and never terminating the computations. © 2003 by Prentice Hall, Inc. 19

Degeneracy • The second theoretical point arises in the examination of iterations 1 and 2. Both iterations, although differing in classifying the variables as basic and nonbasic, yield identical values of all variables and objective. • An argument thus arises as to the possibility of stopping the computations at iteration 1 (when degeneracy first appears), even though it is not optimum. This argument is not valid because, in general, a solution may be temporarily degenerate. © 2003 by Prentice Hall, Inc. 20

Degeneracy/example Max. Z=3 X 1+9 X 2 ST X 1+4 X 2 8 X

Degeneracy/example Iterat Basic io n 0 (starti ng ) X 2 en ter s S 1 lea ve s 1 X 1 en ter X 1 X 2 S 1 S 2 R. S. Z -3 -9 0 0 0 S 1 1 4 1 0 8 S 2 1 2 0 1 4 Z 4/3 - 0 4/9 0 18 X 2 4/1 1 4/1 0 2 © 2003 by Prentice Hall, Inc. 22

Alternative Optima • When the objective function is parallel to a binding constraint, the objective function will assume the same optimal value at more than one solution point. For this reason they are called alternative optima. • Algebraically, after the simplex method finds one optimal basic feasible (BF) solution, you can detect if there any others and, if so, find them as follows: Whenever a problem has more than one optimal BF solution, at least one of the nonbasic variables has a coefficient of zero in the final row (0), so increasing any such variable will not change the value of Z. Therefore, these other optimal BF solutions can be identified (if desired) by performing additional iterations of the simplex method, each time choosing a nonbasic variable with a zero coefficient as the entering basic variable. © 2003 by Prentice Hall, Inc. 23

Alternative Optima/example Max. Z=2 X 1+4 X 2 ST X 1+2 X 2 5

Alternative Optima/example Iteration Basic X 1 X 2 S 1 S 2 R. S. 0 (starting) X 2 enters S 1 leaves Z -2 -4 0 0 0 S 1 1 2 1 0 5 S 2 1 1 0 1 4 Z 0 0 2 0 10 X 2 2/1 1 2/1 0 2/5 S 2 2/1 0 2/1 - 1 2/3 Z 0 0 2 0 10 X 2 0 1 1 1 - 1 X 1 1 0 1 - 2 3 1 (optimum) X 1 enters S 2 leaves 2 ( alternate optimum) © 2003 by Prentice Hall, Inc. 25

(2, 6) Every point on this darker line segment is optimal, each with z = 18. (4, 3) As in this case, any problem having multiple optimal solutions will have an infinite number of them, each with the same optimal value of © 2003 by Prentice Hall, Inc. 26 the objective function.

Unbounded Solution • In some LP models, the values of the variables may be increased indefinitely without violating any of the constraints, meaning that the solution space is unbounded in at least one direction. • Unboundedness in a model can point to one thing only. The model is poor constructed. • The general rule for recognizing unboundedness is as follows: If at any iteration the constraint coefficients of a nonbasic variable are nonpositive, then the solution space is unbounded in that direction. © 2003 by Prentice Hall, Inc. 27

Unbounded Solution/example Max. Z=2 X 1+X 2 ST X 1 -X 2 10 2

Unbounded Solution/example Basic X 1 X 2 S 1 S 2 R. S. Z

Infeasible Solution • If the constraints cannot be satisfied, the model is said to have no feasible solution. This situation can never occur if all the constraints are of the type (assuming nonegative right-side constants), since the slack variable always provides a feasible solution. • However, when we employ the other types of constraints, we resort to the use of artificial variables which may do not provide a feasible solution to the original model. • Although provisions are made to force the artificial variables to zero at the optimum, this can occur only if the model has a feasible space. If it does not, at least one artificial variable will be positive in the optimum iteration. • See the next example, the artificial variable R is positive (=4) in the optimal solution. © 2003 by Prentice Hall, Inc. 30

Infeasible Solution/example Max. Z=3 X 1+2 X 2 ST 2 X 1+X 2 2

Infeasible Solution/example Iteration Basic X 1 X 2 S 1 A 1 R R. S. 0 (starting) X 2 enters S 1 leaves Z -3 -3 M -2 -4 M 0 0 S 1 2 1 1 0 0 2 R 3 4 0 1 - 1 12 Z 1+5 M 0 2+4 M M 0 4 -4 M X 2 2 1 1 0 0 2 R 5 - 0 4 - 1 - 1 4 1 (optimum) X 1 enters S 2 leaves © 2003 by Prentice Hall, Inc. 32