Chapter 3 LP Simplex Method

Quantitative Techniques in Management

4th Edition

N. D. Vohra

© 2010

Chapter 3

Linear Programming II: Simplex Method

Contents

1. Formulation Conditions for Application of Simplex Method

2. Assumptions Simplex Method – Maximization Problem

a) Standardization of the Problemb) Obtaining Initial Solutionc) Testing the Optimalityd) Revised, Improved Solution

3. Justification and Significance of Elements in the Simplex Tableau

4. Simplex Method – Minimization Problem

a) Big–M Methodb) Two-phase Method

Contents (…continued)

5. Special Topicsa) Multiple Optimal Solutionsb) Infeasibilityc) Unbounded Solutiond) Degeneracy

Conditions for using Simplex Method

1. Non-negative bi valuesMultiply a constraint on both sides by -1 if it involves a negative bi value

2. Non-negative variablesReplace every unrestricted variable by difference of two non-negative variables

Conditions for using Simplex Method (…continued) Maximize Z = 8x1 + 5x2 +12x3 Subject to 2x1 + 5x2 + 4x3 ≤ 44 3x1 – 7x2 – 6x3 ≥ –10 7x1 + 2x2 +4x3 = 54 x1, x2 ≥ 0 x3: unrestricted in

signShould change as follows:(i) Let x3 = x4 – x5, where x4, x5 ≥ 0(ii) 3x1 – 7x2 – 6x3 ≥ –10 be replaced as – 3x1 + 7x2 + 6x3 ≤ 10Revised LPP:Maximize Z = 8x1 + 5x2 +12x4 – 12x5

Subject to 2x1 + 5x2 + 4x4 – 4x5 ≤ 44 – 3x1 + 7x2 + 6x4 – 6x5 ≤ 10 7x1 + 2x2 + 4x4 – 4x5 = 54 x1, x2, x4, x5 ≥ 0

Simplex Method for a Max Problem (all constraints of ≤ type)

1. Standardise the problemIntroduce slack variables

2. Obtain initial solution (with slack variables)

3. Test if solution is optimalTo be optimal, ∆ij ≤ 0 for all variables

4. If optimal, stop and exit; else go to step 5

5. Obtain a revised solution and go back to step 3

An Example

Example 3.2 dataMax Z = 5x1 + 10x2 + 8x3 ContributionSt 3x1 + 10x2 + 2x3 ≤ 60 Fabrication Hrs 4x1 + 4x2 + 4x3 ≤ 72 Finishing Hrs 2x1 + 4x2 + 4x3 ≤ 100 Packaging Hrs x1, x2, x3 ≥ 0

Conditions for application of simplex method are both satisfied hereWe need 3 slack variables to convert the inequalities in to equationsThe problem becomes:

Max Z = 5x1 + 10x2 + 8x3 + 0S1 + 0S2 + 0S3

St 3x1 + 10x2 + 2x3 + S1 = 60 4x1 + 4x2 + 4x3 + S2 = 72 2x1 + 4x2 + 4x3 + S3 = 100 x1, x2, x3, S1, S2, S3 ≥ 0

Simplex Solution

Basis x1 x2 x3 S1 S2 S3 bi bi/aij

S1 0 3 5* 2 1 0 0 60 12

S2 0 4 4 4 0 1 0 72 18

S3 0 2 4 5 0 0 1 100 25

Cj 5 10 8 0 0 0

Sol 0 0 0 60 72 100 Z=0

Δj 5 10 8 0 0 0

Simplex Tableau 1

Includes variables that yield identity matrix

Key element

Key rowKey column

Simplex Solution

Basis x1 x2 x3 S1 S2 S3 bi bi/aij

x2 10 3/5 1 2/5 1/5 0 0 12 30

S2 0 8/5 0 12/5 -4/5 1 0 24 10

S3 0 -2/5 0 17/5 -4/5 0 1 52 260/17

Cj 5 10 8 0 0 0

Sol 0 12 0 0 24 52 120

Δj -1 0 4 -2 0 0

Simplex Tableau 2

To derive values in the next tableau,Divide each element of key row by key element: 3/5, 5/5, 2/5, 1/5 etc.For other rows (for example row 2):

4 – 4 × 3/5 = 8/5

4 – 4 × 1 = 04 – 4 × 2/5 = 12/50 – 4 × 1/5 = -4/5 etc.

Simplex Solution

Basis x1 x2 x3 S1 S2 S3 bi

x2 10 1/3 1 0 1/3 -1/6 0 8

x3 8 2/3 0 1 -1/3 5/12 0 10

S3 0 -8/3 0 0 1/3 -17/12 1 18

Cj 5 10 8 0 0 0

Sol 0 8 10 0 0 18 160

Δj -11/3 0 4 -2/3 -5/3 0

Simplex Tableau 3: Optimal Solution

Solution optimal as all Δj ≤ 0Optimal Product mix: x1 = 0, x2 = 8 and x3 = 10Maximum contribution = Rs 160Capacity utilization: Fabrication and Finishing – Full; Packaging – 18 hours unutilizedNo production of x1 since every unit of it produced would result in a loss of Rs 11/3Production of a unit of x1 will result in losing 1/3 unit of x2 and 2/3 unit of x3 together with a release of 8/3 hours of Packaging time

Solution to LPPs when some/all Constraints are not ≤ type: Big M

1. Standardise the problemIntroduce surplus and artificial variables for all “≥” constraintsIntroduce artificial variables for all “=“ constraintsIn the objective function assign co-efficients to each of the artificial variables [For max problems: assign –M and for min problems: assign M]

2. Obtain initial solution (with slack/artificial variables)

3. Test if solution is optimal (to be optimal, ∆ij ≥ 0 for all variables)

4. If optimal, stop and exit; else go to step 5

5. If not, obtain a revised solution and go back to step 3

Artificial Variables

Needed to be used to obtain initial solutionIntroduced where a constraint is of “ ≥ ” or “ = ” typeMay also be required in maximisation problemsExpected to leave the basis one-by-one in successive simplex iterationsSolution infeasible as long as one or more artificial variables present in the basisNot expected to be present in the basis of the final solutionIn case artificial variable present in basis of final solution, there is infeasibility

Solution to LPPs when some/all Constraints are not ≤ type: 2-Phase

Phase I1. Standardise the problem

For each artificial variable, assign co-efficients in the objective functionFor max problem: assign –1 For min problem: assign 1For every other variable, assign co-efficient of zero

2. Solve by Simplex Method. If the objective function value for optimal solution is zero, move to Phase II

Solution to LPPs when some/all Constraints are not ≤ type: 2-Phase

Phase II

1. Begin with Simplex Tableau of Phase I final solution, eliminate all entries for artificial variables and replace the zero co-efficients of decision variables by original co-efficient values

2. Solve the modified problem by Simplex Method

How to Interpret a Solution Obtained by Simplex Method

No artificial variable present in the basis

It is a feasible solution

It is a feasible solution and the ∆ij values of all non-basic variables are ≤ 0 for a max problem (or ≥ 0 for a min problem)

The solution is optimal

It is an optimal solution and the ∆ij values of all non-basic variables are < 0 for a max problem (or > 0 for a min problem)

The solution is optimal and unique

How to Interpret a Solution Obtained by Simplex Method

It is an optimal solution and the ∆ij value of one or more non-basic variables is equal to zero

The solution is not unique: the problem has multiple optimal solutions

It is final in terms of the ∆ij values and has an artificial variable in the basis

It indicates infeasibility

It is non-optimal and the elements of the key column are all zero/negative

The problem has unbounded solution

Degeneracy

A solution in which a basic variable has solution value equal to zero is degenerate solutionThe basic variable which has solution value of zero is called degenerate variableWhenever there is a tie in the replacement ratios (bi/aij) to be selected, the next solution will be a degenerate solutionA degenerate solution may or may not be optimalRevision of a non-optimal solution leads to no improvement (in terms of the Z-value) if the degenerate variable is the outgoing variable

Degeneracy: An Example

Basis x1 x2 S1 S2 S3 bi bi/aij

S1 0 6 3 1 0 0 18 6

S2 0 3 1 0 1 0 8 8

S3 0 4 5 0 0 1 30 6

Cj 28 30 0 0 0

Sol 0 0 18 8 30

Δj 28 30 0 0 0

Simplex Tableau 1

Max Z = 28x1 + 30x2 Stt 6x1 + 3x2 ≤ 18 3x1 + x2 ≤ 8 4x1 + 5x2 ≤ 30 x1, x2 ≥ 0With slack variables S1,S2 and S3, solution follows:

Next solution would be degenerate

Tie

Key column

Degeneracy: An Example (…continued)

Basis x1 x2 S1 S2 S3 bi bi/aij

S1 0 18/5 0 1 0 -3/5 0 0

S2 0 11/5 0 0 1 -1/5 2 10/11

X2 30 4/5 1 0 0 1/5 6 15/2

Cj 28 30 0 0 0

Sol 0 6 0 2 0 180

Δj 4 0 0 0 -6

Simplex Tableau 2: Non-optimal Solution

Basis x1 x2 S1 S2 S3 bi

x1 28 1 0 5/18 0 -1/6 0

S2 0 0 0 -11/18 1 1/6 2

x2 30 0 1 -2/9 0 1/3 6

Cj 28 30 0 0 0

Sol 0 6 0 2 180

Δj 0 0 -10/9 0 -16/3

Simplex Tableau 3: Optimal SolutionDegenerate outgoing variable

Degeneracy: An Example (…continued)

A tie in the minimum replacement ratios in Tableau 1 results in the solution in next tableau to be degenerate (even if the outgoing variable was chosen to be S1 instead of S3)Since all values in the key column are positive, the least ratio to be considered is 0 and the outgoing variable (S1) is degenerateAs a result, there is no change in the objective function value: it remains at 180The solution in Tableau 3 is also degenerateWhile the solution in Tableau 2 is non-optimal, the solution in Tableau 3 is optimal

Multiple Choice Questions

Mark the wrong statement: 1. An LPP with n variables and m

constraints gives nCm basic solutions.

2. A basic solution with all non-negative variables is called basic feasible solution.

3. For any LPP, there is a unique extreme point of the feasible region corresponding to every basic solution.

4. The basic feasible solution that maximises or minimises, as the case may be, is called optimal solution.


Mark the correct alternative.A feasible solution is the one which

1. satisfies all constraints of the problem.

2. is necessarily an optimal solution.

3. makes use of all available resources.

4. yields more than one way to achieve the objective.


Which of the following statements is not true about application of simplex method?1. The RHS of each constraint should

be greater-than-or-equal-to zero.

2. All decision variables of the problem should be non-negative.

3. All constraints of the given problem should be either or type.

4. All constraints should be converted into “=” type, with slack/surplus variables which, like decision variables, are also non-negative.


Choose the incorrect statement:

1. Simplex method is an iterative process wherein variables are substituted until optimal solution is reached.

2. Successive simplex tableaus always yield better solution in terms of the objective function.

3. For a minimisation problem, the optimal solution is reached when all j 0, with no artificial variable in the basis.

4. The key column indicates the incoming variable and the key row represents outgoing variable.


If j = cj – zj is equal to 24 – 7M for x1; 28 – 7M for x2; 45 – 4M for x3 and 28 – 2M in respect of x4 in a minimisation problem, then the entering variable would be:

1. x1

2. x2

3. x3

4. x4


Mark the wrong statement:

1. First Key element can never be negative.

2. Key element cannot be zero.

3. Key element has to be positive.

4. Key element can be negative, zero or positive.


The bi values and the elements in key column are (12, 0, 18, 6) and (-2, 8, 1, 3) respectively. Which of the rows will be selected as the key row?

1. First

2. Second

3. Third

4. Fourth


Which of these is not true?1. All basic variables in an LPP have j =

0.

2. A slack variable cannot be there in the basis of optimal solution.

3. If a problem has an optimal solution, the artificial variables, if any, introduced must be driven out of the basis one-by-one.

4. An outgoing variable in a simplex tableau cannot re-enter the basis in follow-up iterations.


Which of the following statements is not true about artificial variables?1. They are introduced when initial

solution to a problem cannot be obtained for want of identity sub-matrix in the simplex tableau.

2. Each of such variables has to be assigned a very large negative co-efficient in the objective function when it is of the maximisation type.

3. They help to obtain an initial feasible solution to the problem.

4. They bear no tangible relationship with the decision problem.


Mark the wrong statement:

1. If a non-basic variable in the optimal solution to a problem has j = 0, it indicates multiple optimal solutions.

2. If an artificial variable is found in the basis of the final solution to an LPP, it implies feasibility.

3. A constraint with sign does not require artificial variable to be introduced.

4. A slack variable is used to convert ‘’ type and a surplus variable is used to convert ‘ ‘ type of constraint into equality.


Mark the wrong statement: 1. If there are no non-negative

replacement ratios in a simplex tableau, then unboundness is indicated.

2. In a tableau, the key column elements are –5, 0, 0, 7 while the corresponding bi values are 20, 6, 8 and 0. This indicates unbounded solution.

3. A tie in the minimum replacement ratio implies that the next solution would be degenerate.

4. It is possible for the initial solution to an LP to be degenerate.


Which of these is not correct?1. The solution to a maximisation LPP

is unique if j values for all non-basic variables are less than zero.

2. If infeasibility is present, it is detected in phase I of the two-phase method.

3. In two-phase method, the artificial variables are each assigned a co-efficient of 1 in case of minimisation, and –1 in case of maximisation problems.

4. If the given problem has an optimal solution, the artificial variables are all removed one by one in phase I of the two-phase method.


If there is no non-negative replacement ratio in a solution which is sought to be improved, then the solution is indicated to be

1. infeasible

2. unbounded

3. degenerate

4. unique optimal


Which of the following is not correct?1. As and when a degenerate

variable is outgoing, no change will take place in the objective function value.

2. Degeneracy may be a temporary phenomenon.

3. It is possible for the repetition of same tableaus in the course of successive iterations when degeneracy is encountered.

4. A degenerate solution cannot be optimal.


Degeneracy in LPP (1) renders the solution infeasible, (2) leads to multiple optimal solutions, (3) increases computations without affecting the solution as long as the outgoing variable happens to be degenerate. Which of these statements is/are correct?

1. 1, 2 and 32. 1 and 33. 2 and 34. 3 only

Documents

Chapter 3 LP Simplex Method