Lecture 4 - solution Expression for all alternative...

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The Simplex Method Tie breaking And different solutions
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Tie Breaking and other situations 1. A tie for an entering variable ->break the tie arbitrarily. 2. A tie for leaving variable degenerate solution (a basic variable having zero value): break the tie arbitrarily. 1. No leaving variable the coefficients of the entering basic variable are all non-positive: an unbounded optimal solution. 1. In the optimal tableau, the obj. coefficient of a non-basic variable is zero: multiple optima.
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Example of ties for entering and leaving variables Consider the following LP: Max Z = 3 x 1 + 3 x 2 St: x 1 + x 2 6 2 x 2 12 3 x 1 + 2 x 2 18 x 1 0 x 2 0
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Solution to example 1 Z x 1 x 2 x 3 x 4 x 5 RHS Z 1 -3 -3 0 0 0 0 X 3 0 1 1 1 0 0 6 X 4 0 0 2 0 1 0 12 X 5 0 3 2 0 0 1 18 Z 1 -3 0 0 3/2 0 18 X 3 0 1 0 1 -1/2 0 0 X 2 0 0 1 0 1/2 0 6 X 5 0 3 0 0 -1 1 6 Z 1 0 0 3 0 0 18 X 1 0 1 0 1 -1/2 0 0 Tie for the entering variable! Pick x 2 ! Tie for the leaving variable! Degenerate solution! Alternative optimal solution!
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x 1 x 2 Graphical view of the alternative
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Unformatted text preview: solution Expression for all alternative solutions Z 1 3 18 x 1 1 0 -2 1 6 x 2 1 3-1 X 4 0 -6 1 2 12 X * = X 1 * + (1- ) X 2 * or X * = (0,6,0,0,6) + (1- ) (6,0,0,12,0) or 6(1- ) X * = 6 where 0 1 12 (1- ) 6 Unbounded Solution Max Z = 3 x 1 + 2 x 2 st: x 1 4 3 x 1- x 2 6 x 1 0 , x 2 0 Simplex solution The last tableau of the problem: Z x 1 x 2 x 3 x 4 RHS Z 1 6-1 18 x 2 1 3/2 -1/2 6 x 1 1 1 Non-positive coefficients for the entering variable. Unbounded solution! Exercise Solve the following problem by the simplex method: max Z = - 3 x 1 + 6 x 2 st: 5 x 1 + 7 x 2 35- x 1 + 2 x 2 2 x 1 , x 2...
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Lecture 4 - solution Expression for all alternative...

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