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lec6 - 15.053 z Simplex Method Continued 1 Today's Lecture...

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1 15.053 February 27, 2007 z Simplex Method Continued
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Today’s Lecture z z z z z Review of the simplex algorithm. Formalizing the approach Alternative Optimal Solutions Obtaining an initial bfs Is the simplex algorithm finite? Quote of the day: Everyone designs who devises courses of action aimed at changing existing situations into preferred ones. -- Herbert Simon 2
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3 The simplex algorithm (for max problems) Start with a feasible corner point solution Is it optimal? quit with optimal solution Is the optimum unbounded from above? quit with proof of unboundedness find an improved corner point solution No Yes No Yes
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4 = = = 1 9 x 1 = 3 0 0 x 2 x 3 1 z 0 x 5 1 x 4 2 LP Canonical Form The basic feasible solution ( bfs ) is z = 3, x 1 = 0, x 2 = 1, x 3 = 9, x 4 = 0, and x 5 = 4. The basic variables are z, x 2 , x 3 and x 5 . The non-basic variables are x 1 and x 4 . 0 1 0 0 1 0 0 4 0 1 0 0 0 2 1 0 3 -1 2
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5 The Simplex Pivot Rule and Optimality Conditions Pivot in a variable whose cost-row coefficient is negative. = = = 1 9 x 1 = 3 0 0 x 2 x 3 1 z 0 x 5 1 x 4 2 If all z-row coefficients are non-negative then the current basic feasible solution is optimal. optimal 0 1 0 0 1 0 0 4 0 1 0 0 0 2 1 0 3 -1 2
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6 Optimality Conditions = = = 1 9 x 1 = 3 0 0 x 2 x 3 1 z 0 x 5 x 4 - c 1 - c 4 Opt. conditions : The bfs is optimal if 0 for all j. - c j The bar indicates that it is possibly the coefficient after some pivots Maximize z = c 1 + c 4 0 1 0 0 1 0 0 4 0 1 0 0 0 2 1 0 3 -1 2 Writing the objective function coefficients with a negative looks awkward. But there is a good reason for it. We usually think of this objective function as maximize z = 1x 1 + 2x 4 . In this case, we let c 1 = 1 and c 4 = 2. But in the tableau, we have the coefficients –c 1 and –c 4 . Remember that a bar over the letter indicates that it may be after a pivot or more. c 1 and c 4 without a bar denote the original data.
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7 Determining the exiting variable 1. Choose an entering variable with negative z-row coeff. 2. Set the value of the entering variable to Δ , and make Δ , as large as possible while maintaining feasibility. = = = x 1 = 3 0 1 0 0 1 0 0 x 2 x 3 0 1 z 0 0 x 5 0 1 0 0 0 1 2 1 0 x 4 -2 3 -1 2 x 3 = 9 – 3 Δ x 2 = 1 + Δ x 5 = 4 - 2 Δ Δ ≤ 9/3 Δ ≤ 4/2 1 9 4 By increasing the size of Δ , we are moving along an edge of the feasible region. In this case, we can let Δ be as large as 4/2 before a variable becomes negative. Setting Δ = 2 results in another corner point solution.
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8 The min ratio rule 1. To determine the exiting variable, take the min ratio of the RHS coefficient divided by the coefficient of the entering variable, among all those whose coefficient is strictly positive. = = = x 1 = 3 0 1 0 0 1 0 0 x 2 x 3 0 1 z 0 0 x 5 0 1 0 0 0 1 2 1 0 x 4 -2 3 -1 2 x 3 = 9 – 3 Δ x 2 = 1 + Δ x 5 = 4 - 2 Δ Δ ≤ 9/3 Δ ≤ 4/2 1 9 4 In order to carry out the simplex algorithm, we don’t really need the algorithm to work with Δ . We can shortcut the procedure by directly identifying the next component of the tableau on which to pivot.
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