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Infeasible LP problems

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• Infeasible problem
  • Infeasible problem:

    Infeasible problem = a problem that has no (valid) solution

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• Characteristics of infeasible LP problems
  • What makes a LP infeasible:

    A feasible LP problem has a feasible region that is not empty Example: max: z = x1 + x2 s.t.: x1 + 2x2 ≤ 8 3x1 + 2x2 ≤ 12 x1 + 3x2 ≥ 3 The feasibility region: A infeasible LP problem has a feasible region that is empty Example: Consider the following LP: max: z = x1 + x2 s.t.: x1 + 2x2 ≤ 8 3x1 + 2x2 ≤ 12 x1 + 3x2 ≥ 6 <--- 3 changed to 6 The feasibility region will shrink: If we push the line further up (by increasing the constant further), the feasible region will become empty For example: max: z = x1 + x2 s.t.: x1 + 2x2 ≤ 8 3x1 + 2x2 ≤ 12 x1 + 3x2 ≥ 13 <--- 6 changed to 13 The feasibility region will become empty:

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• How to determine if an LP is infeasible
  • Fact 1:

    If an LP is infeasible, then you cannot find a feasible solution (That's because the feasible region is empty --- see the above discussion)

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  • Fact 2:

    When the two-phase Simplex method stops and all the artificial variables have value = 0, we can remove the artificial variables and remaining variables will form a feasible solution for the original LP problem (We learned this in the previous webpage)

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  • Conclusion:

    If an LP is infeasible, then the two-phase Simplex method will stop with a solution where some artificial variable has a ≠ 0 value !!!!

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• Example: using 2-phase Simplex on an infeasible LP
  • The LP problem:

    max: z = x1 + x2 s.t.: x1 + 2x2 ≤ 8 3x1 + 2x2 ≤ 12 x1 + 3x2 ≥ 13

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  • Put in tableau form:

    Convert to "="-constraints by adding surplus variables: z = x1 + x2 ⇒ z - x1 - x2 = 0 x1 + 2x2 ≤ 8 ⇒ x1 + 2x2 + s1 = 8 3x1 + 2x2 ≤ 12 ⇒ 3x1 + 2x2 + s2 = 12 x1 + 3x2 ≥ 13 ⇒ x1 + 3x2 - s3 = 13 Write in matrix (table) form: -x1 - x2 + z = 0 x1 + 2x2 + s1 = 8 3x1 + 2x2 + s2 = 12 x1 + 3x2 - s3 = 13 Lined up: -x1 - x2 + z = 0 x1 + 2x2 + s1 = 8 3x1 + 2x2 + s2 = 12 x1 + 3x2 - s3 = 13 Matrix form: x1 x2 s1 s2 s3 z | RHS ===============================+==== -1 -1 0 0 0 1 | 0 1 2 1 0 0 0 | 8 3 2 0 1 0 0 | 12 1 3 0 0 -1 0 | 13 <--- "≥"-constraint needs artificial var. Add artificial variable(s): x1 x2 s1 s2 s3 a1 z | RHS ====================================+==== -1 -1 0 0 0 0 1 | 0 1 2 1 0 0 0 0 | 8 3 2 0 1 0 0 0 | 12 1 3 0 0 -1 1 0 | 13

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  • Construct the Simplex tableau of Phase 1 of the 2-phased Simplex Method:

    Use the same constraints: x1 x2 s1 s2 s3 a1 z | RHS ====================================+==== .. .. ... 1 2 1 0 0 0 0 | 8 3 2 0 1 0 0 0 | 12 1 3 0 0 -1 1 0 | 13 to maximize the following object function: max: z = -a1 Or: max: z + a1 = 0 (We want to find a feasible solution where a1 = 0) Simplex Tableau for this linbear program: x1 x2 s1 s2 s3 a1 z | RHS ====================================+==== 0 0 0 0 0 1 1 | 0 1 2 1 0 0 0 0 | 8 3 2 0 1 0 0 0 | 12 1 3 0 0 -1 1 0 | 13

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  • Phase 1:

    Fix the basic column: (subtract row 4 from row 1) x1 x2 s1 s2 s3 a1 z | RHS ====================================+==== 0 0 0 0 0 1 1 | 0 (row 1) <----- 1 2 1 0 0 0 0 | 8 (row 2) 3 2 0 1 0 0 0 | 12 (row 3) 1 3 0 0 -1 1 0 | 13 (row 4) <-- Result: x1 x2 s1 s2 s3 a1 z | RHS ====================================+==== -1 -3 0 0 1 0 1 | -13 (row 1) <----- 1 2 1 0 0 0 0 | 8 (row 2) 3 2 0 1 0 0 0 | 12 (row 3) 1 3 0 0 -1 1 0 | 13 (row 4) <-- We have a valid Initial Simplex Tableau (with 4 canonical unit vectors) Run the Simplex Algorithm: Iteration 1: Find the pivot element: Most neg. coef. | V x1 x2 s1 s2 s3 a1 z | RHS ====================================+==== -1 -3 0 0 1 0 1 | -13 1 2 1 0 0 0 0 | 8 (test ratio = 8/2 = 4) 3 2 0 1 0 0 0 | 12 (test ratio = 12/2 = 6) 1 3 0 0 -1 1 0 | 13 (test ratio = 13/3 = 4.3) Normalize the pivot row: x1 x2 s1 s2 s3 a1 z | RHS ====================================+==== -1 -3 0 0 1 0 1 | -13 0.5 1 0.5 0 0 0 0 | 4 3 2 0 1 0 0 0 | 12 1 3 0 0 -1 1 0 | 13 Sweep column x2: x1 x2 s1 s2 s3 a1 z | RHS ====================================+==== 0.5 0 1.5 0 1 0 1 | -1 0.5 1 0.5 0 0 0 0 | 4 2 0 -1 1 0 0 0 | 4 -0.5 0 -1.5 0 -1 1 0 | 1 Iteration 1: Find the pivot element: All object coef's are ≥ 0. DONE.

    Observe that:

    x1 x2 s1 s2 s3 a1 z | RHS ====================================+==== 0.5 0 1.5 0 1 0 1 | -1 0.5 1 0.5 0 0 0 0 | 4 2 0 -1 1 0 0 0 | 4 -0.5 0 -1.5 0 -1 1 0 | 1 ^ | Artificial variable a1 = 1 ≠ 0 !!!

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  • Infeasible solution

    Because Phase 1 ending with some artificial variable ≠ 0, the LP is infeasible