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Unformatted text preview: CO350 Linear Programming Chapter 6: The Simplex Method 10th June 2005 Chapter 6: The Simplex Method 1 Recap On Wednesday, we learned four common choice rules for entering variables: • Largest coefficient rule (Dantzig’s rule); • Smallest subscript rule; • Largest improvement rule; • Steepest edge rule. We also learned that the simplex method solves the dual problem implicitly. Chapter 6: The Simplex Method 2 The Simplex Method and Duality (cont’d) Suppose at the end of the simplex method, we have an optimal solution x * determined by a basis B and the cor responding tableau ( T ) z X j ∈ N ¯ c j x j = ¯ v x i + X j ∈ N ¯ a ij x j = ¯ b i ( i ∈ B ) Recall that 1. The zrow is [ z c T x = 0 ] + [linear combination of Ax = b ] I.e., there is some ˆ y = [ˆ y 1 , ˆ y 2 ,..., ˆ y m ] T such that z P j ∈ N ¯ c j x j = ¯ v is equivalent to z c T x + y T Ax = y T b 2. Comparing coefficients gives c i A T i ˆ y = 0 ( i ∈ B ) c j A T j ˆ y = ¯ c j ( j ∈ N ) which shows that ˆ y is optimal for the dual problem. Chapter 6: The Simplex Method 3 Finding dual optimal solution The abovementioned dual optimal solution ˆ y satisfies c i A T i y = 0 ( i ∈ B ) i.e. A T B y = c B [Note: c B to denotes the column matrix [ c i : i ∈ B ] .] Since B is a basis, A B is nonsingular, and so is A T B . Thus the system A T B y = c B has the unique solution ˆ y . Chapter 6: The Simplex Method 4 Example (Not in notes) Recall the example LP. max. z = 5 x 1 + 3 x 2 s.t. 2 x 1 + 3 x 2 + x 3 = 15 2 x 1 + x 2 + x 4 = 9 x 1 x 2 + x 5 = 3 x 1 , x 2 , x 3 , x 4 , x 5 ≥ We have applied the simplex method to get the optimal basis B = {...
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 Fall '97
 Wolkowicz
 Linear Programming, Optimization, LP, Dual Optimal Solution

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