Linear Programming: The Simplex Method

1© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458

Linear Programming: The Simplex Linear Programming: The Simplex MethodMethod


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.


IntroductionWhy study the Simplex Method?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 becauseo it allows for understanding how to interpret LP computer

printouts.


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: X1 4 The slack variable for this constraint is defined to be S1=4 - X1

which is the amount of slack in the left-hand side of the inequality. Thus

X1+S1=4


Example

• Objective Function Max. Z=3X1+5X2

• Subject to: X1 4 X1+S1=4 2X2 12 2X2+S2=12 3X1+2X2 18 3X1+2X2+S3=18 X1,X2 0


The Standard Form of the Model

Z-3X1-5X2=0 ..………(0)

X1+S1=4 ...……...(1)

2X2+S2=12 ..………(2)

3X1+2X2+X5=18 ………...(3)


The Simplex Method in Tabular Form

Tabular form

Eq. Basic variable

Coefficient of: Right sideZ X1 X2 S1 S2 S3

(0) Z 1 -3 -5 0 0 0 0

(1) S1 0 1 0 1 0 0 4

(2) S2 0 0 2 0 1 0 12

(3) S3 0 3 2 0 0 1 18


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: Step1 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.


Summary of the Simplex Method/ continued

Step 2 Determine the leaving basic variable by applying the minimum ratio

test. Minimum Ratio Test1. 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.


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.)


Example

• Solve this model using the simplex method:

• Objective Function Max. Z=3X1+5X2

• Subject to: X1 4 2X2 12 3X1+2X2 18 X1,X2 0


Solution

• First, the Standard Form of the model:

• Z-3X1-5X2=0 ..……….(0)

• X1+S1=4 ...……....(1)• 2X2+S2=12 ..…….…(2)• 3X1+2X2+X5=18 ………...(3)


The Simplex Method in Tabular Form

Tabular form

Eq. Basic variable


(0) Z 1 -3 -5 0 0 0 0

(1) S1 0 1 0 1 0 0 4

(2) S2 0 0 2 0 1 0 12

(3) S3 0 3 2 0 0 1 18


# (0) Iteration

Tabular form

Eq. Basic variable


(0) Z 1 -3 -5 0 0 0 0

(1) S1 0 1 0 1 0 0 4

(2) S2 0 0 2 0 1 0 12

(3) S3 0 3 2 0 0 1 18


# (1) Iteration

Tabular form

Eq. Basic variable


(0) Z 1 -3 0 0 2\5 0 30

(1) S1 0 1 0 1 0 0 4

(2) X2 0 0 1 0 1\2 0 6

(3) S3 0 3 0 0 -1 1 6


# (2) Iteration

Tabular form

Eq. Basic variable


(0) Z 1 0 0 0 2\3 1 36

(1) S1 0 0 0 1 1\3 -1\3 2

(2) X2 0 0 1 0 1\2 0 6

(3) X1 0 1 0 0 -1\3 1\3 2


•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 X1=2, X2=6.


Artificial Starting Solution (The M-Technique)

• In our presentation of the simplex method we have used the slack variable as starting basic solutions. However, If the original constraint is an equation or of the type (), we no longer have a ready starting basic feasible solution.

• The idea of using artificial variables is quite simple.• The added variable will play the same role as that of a

slack variable, in providing a starting basic variable.• However, since such artificial variables have no physical

meaning from the standpoint of the original problem (hence the name “artificial”), the procedure will be valid only if we force these variables to be zero when the optimum is reached.


• In other words, we use them only to start the solution and must subsequently force them to be zero in the final solution; otherwise, the resulting solution will be infeasible.

• Let us consider the next example:


Example• Objective Function Min. Z=4X1+X2

• ST 3X1+X2=3 4X1+3X2 6 X1+2X2 4 X1,X2 0

The standard form is obtained by augmenting a surplus (A1) and adding a slack (S1) to the left sides of constraints 2 and 3.Thus, we have:


Example\continued

• The standard form:Min. Z=4X1+X2

ST 3X1+X2 =3

4X1+3X2 –A1 =6

X1+2X2 +S1= 4

X1,X2, A, S1 0

The first and second equations do not have variables that play the role of a slack. Hence we augment the two artificial variables R1 and R2 in these two equations as follows:


Example\continued 3X1+ X2+R1=3

4X1+3X2-A1+R2=6

We can penalize R1 and R2 in the objective function by assigning them very large positive coefficients in the objective function. Let M>0 be a very large constant; then the LP with its artificial variables becomes:

Min. Z=4X1+X2+MR1+MR2

ST 3X1+ X2 +R1 =3

4X1+3X2-A1 +R2 =6

X1+2X2 +S1=4

X1, X2, S1, A1, R1, R2 0


Example\continued• How does the M-technique change if we are maximizing

instead of minimizing? Using the same logic of penalizing the artificial variable, we

must assign them the coefficient (-M) in the objective function (M>0), thus making it unattractive to maintain the artificial variable at a positive level in the optimum solution.

• Having constructed a starting feasible solution, we must “condition” the problem so that when we put it in tabular form, the right-side column will render the starting solution directly. This is done by using the constraint equations to substitute out R1 and R2 in the objective function. Thus

R1=3-3X1-X2

R2=6-4X1-3X2+A1


Example\continued• The objective function thus becomes Z=4X1+X2+M(3-3X1-X2)+M(6-4X1-3X2+A1)

=(4-7M)X1+(1-4M)X2+MA1+9M• The sequence of tableaus leading to the optimum solution

is shown in the next slides.• Observe that this is a minimization problem so that the

entering variable must have the most positive coefficient in the Z-equation. The optimum is reached when all the nonbasic variables have nonpositive z-coefficients. (remember that M is a very large positive constant).


Example\continued

• The final standard form:

Z-(4-7M)X1-(1-4M)X2-MA1=9M -------(0)

ST

3X1+ X2 +R1 =3 --------(1) 4X1+3X2-A1 +R2 =6 --------(2) X1+2X2 +S1=4 --------(3) X1, X2, S1, A1, R1, R2 0


Example\continued(The tabular form)

Tabular form

Eq. Basic variable

Coefficient of: Right side

X1 X2 A1 R1 R2 S1

(0) Z -4+7M -1+4M -M 0 0 0 9M

(1) R1 3 1 0 1 0 0 3

(2) R2 4 3 -1 0 1 0 6

(3) S1 1 2 0 0 0 1 4


Example\continued(Iteration #1)

Tabular form

Eq. Basic variable


X1 X2 A1 R1 R2 S1

(0) Z 0 (1+5M)/3 -M (4-7M)/3 0 0 4+2M

(1) X1 1 13/ 0 1/3 0 0 1

(2) R2 0 5/3 -1 -4/3 1 0 2

(3) S1 0 5/3 0 -1/3 0 1 3


Example\continued(Iteration #2)

Tabular form

Eq. Basic variable


X1 X2 A1 R1 R2 S1

(0) Z 0 0 1/5 8/5-M -1/5-M 0 18/5

(1) X1 1 0 1/5 3/5 -1/5 0 3/5

(2) X2 0 1 -3/5 -4/5 3/5 0 6/5

(3) S1 0 0 1 1 -1 1 1


Example\continued(Iteration #3 optimum)

Tabular form

Eq. Basic variable


X1 X2 A1 R1 R2 S1

(0) Z 0 0 0 7/5-M -M -1/5 17/5

(1) X1 1 0 0 2/5 0 -1/5 2/5

(2) X2 0 1 0 -1/5 0 3/5 9/5

(3) A1 0 0 1 1 -1 1 1


Example\continued

• The optimum solution is X1=2/5, X2=9/5, Z=17/5.• Since it contains no artificial variables at positive level, the

solution is feasible with respect to the original problem before the artificials are added. (If the problem has no feasible solution, at least one artificial variable will be positive in the optimum solution).


Example #1• Consider the following linear program problem: Min. Z=3X1+8X2+X3

ST 6X1 +2X2+6X3 6 6X1 +4X2 =12 2X1 -2X2 2 X1,X2, X301. Construct the starting feasible solution (Set Up The Initial

Simplex Tableau).2. Determine the entering column and the leaving row.


Example #2• Consider the following linear program problem: Max. Z=3X1+5X2

ST X1 4 2X2 12 3X1+2X2=18 X1,X2 01. Construct the starting feasible solution (Set Up The Initial

Simplex Tableau).2. Determine the entering column and the leaving row.


Special Cases in Simplex Method Application

• We will consider special cases that can arise in the application of the simplex method, which include:

1. Degeneracy.2. Alternative optima (more than one optimum solution).3. Unbounded solutions.4. Nonexisting (or infeasible) solutions.


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.


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.


Degeneracy/example

Max. Z=3X1+9X2

ST X1+4X2 8

X1+2X2 4

X1,X2 0


Degeneracy/example

Iteration Basic X1 X2 S1 S2 R.S.

0)starting(

X2 entersS1 leaves

Z -3 -9 0 0 0

S1 1 4 1 0 8

S2 1 2 0 1 4

1X1 entersS2 leaves

Z 4/3- 0 4/9 0 18

X2 4/1 1 4/1 0 2

S2 2/1 0 2/1- 1 0

2)optimum(

Z 0 0 2/3 2/3 18

X2 0 1 2/1 2/1- 2

X1 1 0 1- 2 0


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.


Alternative Optima/example

Max. Z=2X1+4X2

ST X1+2X2 5

X1+ X2 4

X1,X2 0


Alternative Optima/example

Iteration Basic X1 X2 S1 S2 R.S.

0)starting(

X2 entersS1 leaves

Z -2 -4 0 0 0

S1 1 2 1 0 5

S2 1 1 0 1 4

1)optimum(

X1 entersS2 leaves

Z 0 0 2 0 10

X2 2/1 1 2/1 0 2/5

S2 2/1 0 2/1- 1 2/3

2 )alternate

optimum(

Z 0 0 2 0 10

X2 0 1 1 1- 1

X1 1 0 1- 2 3


(2, 6)

(4, 3)

Every point on this darker line segment is optimal, each with z = 18.

As in this case, any problem having multiple optimal solutions will have an infinite number of them, each with the same optimal value of 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.


Unbounded Solution/example

Max. Z=2X1+X2

ST X1-X2 10

2X1 40

X1,X2 0


Unbounded Solution/example

Basic X1 X2 S1 S2 R.S.

Z -2 -1 0 0 0

S1 1 -1 1 0 10

S2 2 0 0 1 40


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.


Infeasible Solution/example

Max. Z=3X1+2X2

ST 2X1+X2 2

3X1 +4X2 12

X1,X2 0


Infeasible Solution/example

Iteration Basic X1 X2 S1 A1 R R.S.

0)starting(

X2 entersS1 leaves

Z -3-3M -2-4M 0 M 0 0

S1 2 1 1 0 0 2

R 3 4 0 1- 1 12

1)optimum(

X1 entersS2 leaves

Z 1+5M 0 2+4M M 0 4-4M

X2 2 1 1 0 0 2

R 5- 0 4- 1- 1 4

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Linear Programming: The Simplex Method

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