•INFEASIBILITY•UNBOUNDEDNESS•DEGENERACY•ALTERNATE OPTIMAL SOLUTIONS

Special Cases in Linear Programming

INFEASIBILITY

There exists no solution that satisfies all of the problem’s constraints

An infeasible solution would be apparent by looking at the final tableau – Solution seems optimal but artificial variable(s) still exist in the solution mix

Generally indicates an error in formulating the problem or in entering data

Problems with no feasible solution do exist in practice, most often because management’s expectations are too high or because too many restrictions have been placed on the problem

Infeasibility is independent of the objective function (a change in the coefficients of the objective function will not help, the problem will remain infeasible)

Illustration: maxP = 2x1 + x2

Subject to: x1 + x2 ≤ 3x1 + x2 ≥ 6x1 , x2 ≥ 0

Standard Form: maxP = 2x1 + x2 +0S1 - MA1

Subject to: x1 + x2 + S1 = 3x1 + x2 - S2 + A1 = 6

x1 , X2 ≥ 0

INFEASIBILITY

INFEASIBILITY

Outgoing Row

S1 3 / 1 = 3

A1 6 / 1 = 6

Tableau 1Cj Solution Solution 2 1 0 0 -M

  Variables Values x1 x2 S1 S2 A1

0 S1 3 1 1 1 0 0

-M A1 6 1 1 0 -1 1

  Zj -6M -M -M 0 M -M

  Cj - Zj   2+M 1+M 0 -M 0

INFEASIBILITY

Tableau 2Cj Solution Solution 2 1 0 0 -M

  Variables Values x1 x2 S1 S2 A1

2 x1 3 1 1 1 0 0

-M A1 3 0 0 -1 -1 1

  Zj 6-3M 2 2 2+M M -M

  Cj - Zj   0 -1 -2-M -M 0

**Notice that Tableau2 is already optimal since there is no positive entry is present in the Cj – Zj row. However, the solution is x1=3 and A1=3. Since the existence of an artificial variable in the solution makes the solution meaningless, this is not a real solution. In general, if the solution is optimal but there are artificial variables in the solution, the solution is infeasible.

Exists when the objective function can be made infinitely large (small) without violating any constraints

Unboundedness can be discovered prior to reaching the final tableau

Values in the row value ratios are all negative or undefined (positive or undefined) in which case, no outgoing variable

Generally indicates an error in data or in formulating the constraints or its omission

Also termed as managerial utopia; however, this cannot occur in real-world problems

A change in the objective function can cause a previously unbounded problem to become bounded, even though no changes have been made in the constraints

UNBOUNDEDNESS

Illustration: maxP = 2x1 + x2

Subject to: x1 ≥ 3x2 ≥ 6

x1 , x2 ≥ 0

Standard Form: maxP = 2x1 +x2 +0S1+ 0S2 - MA1 - MA2

Subject to:

x1 - S1 + A1 = 3

x2 - S2 + A2 = 6

x1 , x2 ≥ 0

UNBOUNDEDNESS

UNBOUNDEDNESS

Tableau 1Cj Solution Solution 2 1 0 0 -M -M

  Variables Values x1 x2 S1 S2 A1 A2

-M A1 3 1 0 -1 0 1 0

-M A1 6 0 1 0 -1 0 1

  Zj -9M -M -M M M -M -M

  Cj - Zj   2+M 1+M -M -M 0 0

Outgoing Row

A1 3 / 1 = 3

A1 6 / 0 = ∞

UNBOUNDEDNESS

Tableau 2Cj Solution Solution 2 1 0 0 -M -M

  Variables Values x1 x2 S1 S2 A1 A2

2 x1 3 1 0 -1 0 0 0

-M A1 6 0 1 0 -1 0 1

  Zj 6-6M 2 M -2 M 0 -M

  Cj - Zj   0 1+M 2 -M 0 0

Outgoing Row

A1 3 / 0 = ∞

A1 6 / 1 = 6

UNBOUNDEDNESS

Tableau 3Cj Solution Solution 2 1 0 0 -M -M

  Variables Values x1 x2 S1 S2 A1 A2

2 x1 3 1 0 -1 0 0 0

1 x1 6 0 1 0 -1 0 0

  Zj 12 2 1 2 -1 0 0

  Cj - Zj   0 0 2 1 0 0

Outgoing Row

A1 3 / -1=-3

A1 6 / 0 = ∞Notice that one row value is -3 and the other is undefined. This indicates that the “most constrained” point doesn’t exist and that the solution is unbounded.

Degeneracy can be discovered during the computation for the outgoing variable

Results from a tie in the minimum positive (negative) replacement ratio for determining the outgoing variable – choice can be made arbitrarily

The presence of degeneracy sometimes result to cycling Cycling - A sequence of pivots that goes through the same tableaus

and repeats itself indefinitely

In practice, degeneracy frequently occurs although cycling is rare; computer programs generally have no difficulty reaching the optimum even when degeneracy occurs (LINDO, LPSBA)

DEGENERACY

Illustration:

DEGENERACY

maxP = 2x1 + 3x2

Subject to: 3x1 + 2x2 ≤ 12x2 ≤ 3

x1 + 2x2 ≤ 8x1 , x2 ≥ 0

Standard Form: maxP = 2x1+3x2+0S1+0S2+0S3

Subject to: 3x1+2x2+ S1 = 12x2 + S2 = 3

x1 +2x2+ S3 = 8x1 , x2 = 0

Tableau 1Cj Solution Solution 2 3 0 0 0

  Variables Values x1 x2 S1 S2 S3

0 S1 12 3 2 1 0 00 S2 3 0 1 0 1 00 S3 8 1 2 0 0 1  Zj 0 0 0 0 0 0  Cj - Zj   2 3 0 0 0

DEGENERACY

Outgoing Row

S112

/ 2= 6

S2 3 / 1= 3S3 8 / 2= 4

Tableau 2Cj Solution Solution 2 3 0 0 0  Variables Values x1 x2 S1 S2 S3

0 S1 6 3 0 1 -2 03 x2 3 0 1 0 1 00 S3 2 1 0 0 -2 0  Zj 9 0 3 0 3 0  Cj - Zj   2 0 0 -3 0

DEGENERACY

Outgoing Row

S1 6 / 3 = 2

S2 3 / 0 = ∞

S3 2 / 1 = 2

Tableau 3Cj Solution Solution 2 3 0 0 0  Variables Values x1 x2 S1 S2 S3

2 x1 2 1 0 1/3 - 2/3 03 x2 3 0 1 0 1 00 S3 0 0 0 - 1/3 -1 1/3 0  Zj 13 0 3 2/3 1 2/3 0  Cj - Zj   2 0 - 2/3 -1 2/3 0

DEGENERACY

Decision: x1 = 2x2 = 3Zj = 13

Condition in which more than one solution is available for a linear programming problem each of which maximize or minimize the objective function

Indicated by a situation under which a non-basic variable in the final simplex tableau showing optimal solution has a net zero contribution – at least one of the non-basic variable is zero in the Cj-Zj row

We can discover the alternate solution by using the column of that non-basic variable as the pivot column to make another tableau

In practice, this situation is generally good for the manager or decision maker for it means that several combinations of the decision variables are optimal and that the manager can select the most desirable optimal solution

ALTERNATE OPTIMAL SOLUTIONS

Illustration:maxP =

1000x1 +2000x

2

Subject to: 3x1 + 2x2 ≤182x1 + 4x2 ≤20

x1 ≤ 5x1 , x2 ≥ 0

Standard Form:maxP =

1000x1

+2000x

2+0S1 + 0S2 + 0S3

Subject to: 3x1 +2x2 + S1 = 182x1 +4x2 + S2 = 20

x1 + S3 = 5x1 , x2 ≥ 0

ALTERNATE OPTIMAL SOLUTIONS

Tableau 1Cj Solution Solution 1000 2000 0 0 0

Variables Values x1 x2 S1 S2 S3

0 S1 18 3 2 1 0 00 S2 20 2 4 0 1 00 S3 5 1 0 0 0 1  Zj 0 0 0 0 0 0  Cj - Zj   1000 2000 0 0 0

Outgoing Row

S118

/ 2= 9

S220

/ 4= 5

S3 5 / 0= ∞

ALTERNATE OPTIMAL SOLUTIONS

Tableau 2Cj Solution Solution 1000 2000 0 0 0  Variables Values x1 x2 S1 S2 S3

0 S1 8 2 0 1 - 1/2 02000 x2 5 1/2 1 0 1/4 0

0 S3 5 1 0 0 0 1  Zj 10000 1000 2000 0 500 0  Cj - Zj   0 0 0 -500 0

**At this point, we can see that we have already reached our final tableau, but notice that a non-basic variable (x1) not included in the solution mix has a value of zero at the Cj – Zj row

ALTERNATE OPTIMAL SOLUTIONS

Tableau 2Cj Solution Solution 1000 2000 0 0 0  Variables Values x1 x2 S1 S2 S3

0 S1 8 2 0 1 - 1/2 02000 x2 5 1/2 1 0 1/4 0

0 S3 5 1 0 0 0 1  Zj 10000 1000 2000 0 500 0  Cj - Zj   0 0 0 -500 0

ALTERNATE OPTIMAL SOLUTIONS

We continue to make another tableau using the column on the non-basic variable with zero value as the pivot element to determine the alternative optimal solution.

Outgoing Row

S1 8/ 2= 4

x2 5 / 1/2= 10S3 5 / 1= 5

Tableau 3Cj Solution Solution 1000 2000 0 0 0  Variables Values x1 x2 S1 S2 S3

1000 x1 4 1 0 1/2 - 1/4 02000 x2 3 0 1 -1/4 3/8 0

0 S3 1 0 0 -1/2 -1/4 1  Zj 10000 1000 2000 0 500 0  Cj - Zj   0 0 0 -500 0

ALTERNATE OPTIMAL SOLUTIONS

Decision:

Solution in T2 Alternative Optimal Solution

x1 = 0 x1 = 4x2 = 5 x2 = 3Zj = 10,000 Zj = 10,000