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# handout17 - Math 171A Linear Programming Lecture 17...

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Unformatted text preview: Math 171A: Linear Programming Lecture 17 Degenerate Constraints and the Simplex Method Philip E. Gill c 2011 http://ccom.ucsd.edu/~peg/math171a Monday, February 14th, 2011 Recap: the simplex method One step moves from a vertex to an adjacent vertex . Each step requires the solution of two sets of equations: A T k λ k = c and A k p k = e s where A k is the nonsingular working-set matrix. Given a nonsingular A , all subsequent A k are nonsingular. Now we focus on what can happen at a degenerate vertex . UCSD Center for Computational Mathematics Slide 2/41, Monday, February 14th, 2011 Example: minimize 2 x 1 + x 2 subject to the constraints: constraint #1: x 1 + x 2 ≥ 1 constraint #2: x 2 ≥ constraint #3: x 1 ≥ constraint #4:- x 1 ≥ - 2 constraint #5: x 1 + 2 x 2 ≥ - 2 Written in the form min c T x subject to Ax ≥ b , we get c = 2 1 ! , A = 1 1 1 1- 1 1 2 , b = 1- 2 1 UCSD Center for Computational Mathematics Slide 3/41, Monday, February 14th, 2011 x 1 x 2 x * #3 #1 #2 #5 #4 x Execute the steps of the simplex method, starting at x = 1 At x , we have A ( x ) = { 1 , 2 , 5 } , so that if A = 1 1 1 1- 1 1 2 , then A a = 1 1 0 1 1 2 If we choose W = { 2 , 5 } , then A = 0 1 1 2 ! , b = 1 ! UCSD Center for Computational Mathematics Slide 5/41, Monday, February 14th, 2011 STEP 1: Check for optimality: A T λ = c ⇒ 0 1 1 2 ! λ = 2 1 ! ⇒ λ =- 3 2 ! ← s = 1 W = { w 1 , w 2 } = { 2 ↑ ws =2 , 5 } [ λ ] 1 =- 3 is associated with constraint #2, i.e., x 2 ≥ 0. UCSD Center for Computational Mathematics Slide 6/41, Monday, February 14th, 2011 STEP 2: Compute the search direction: A p = e 1 ⇒ 0 1 1 2 ! p = 1 ! ⇒ p =- 2 1 ! UCSD Center for Computational Mathematics Slide 7/41, Monday, February 14th, 2011 STEP 3: Take a step: Find the constraint residuals and decreasing constraints: r = Ax- b = 1 1 1 1- 1 1 2 1 !- 1- 2 1 = 1 1 Ap = 1 1 1 1- 1 1 2 - 2 1 ! = - 1 1- 2 2 UCSD Center for Computational Mathematics Slide 8/41, Monday, February 14th, 2011 STEP 3: (continued) Take a step: r = 1 1 , Ap = - 1 1- 2 2 σ 1 = [ r ] 1- [ Ap ] 1 =- (- 1) = 0 σ 3 = [ r ] 3- [ Ap ] 3 = 1- (- 2) = 1 2 σ 2 = σ 4 = σ 5 = + ∞ UCSD Center for Computational Mathematics Slide 9/41, Monday, February 14th, 2011 α = min { σ 1 ,σ 2 ,σ 3 ,σ 4 ,σ 5 } = min { , + ∞ , 1 2 , + ∞ , + ∞} = σ 1 = 0 In this case:...
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handout17 - Math 171A Linear Programming Lecture 17...

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