Lecture27aa - Lecture Use of Simplex Tableaus Unbounded...

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Lecture 10/30/02 Use of Simplex Tableaus Unbounded LP’s LP’s with Equality or Constraints
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Steps in Solving LP’s (Revised) 1. Put problem in standard form 2. Find an intial bfs. 3. Determine if optimal. 4. If not determine entering BV a. If max LP, choose NBV having coefficient <0 in Row 0. b. If min LP, choose NBV having coefficient >0 in Row 0. 5. Use ratio test to determine which BV becomes NBV. 6. Use ERO’s to find next bfs. 7. Go to step 3.
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LP’s with Unbounded Solution Problem #3 on P.158 Although x 1 can enter basis LP is unbounded. Why? Given following tableau while solving a max LP: z x 1 x 2 x 3 x 4 RHS 1 -3 -2 0 0 0 0 1 -1 1 0 8 0 2 0 0 1 5 Answer: Bring x 2 into basis. column are non-positive, minimum ratio rule can’t be applied. x 2 and z can increase without bound. Since all entries in pivot => z=3 x 1 + 2 x 2 = 2 x 2 => x 3 = 8+ x 2 - x 1 = 8+ x 2 = > x 4 =5 - 2 x 1 =5
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LP’s with Unbounded Solution An unbounded LP for a max (min) problem occurs when a NBV with a negative (positive) coefficient in Row 0 has non-positive coefficients in each constraint.
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Lecture27aa - Lecture Use of Simplex Tableaus Unbounded...

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