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Unformatted text preview: MthSc810 – Mathematical Programming Fall 2011, Homework #9. Due Tuesday , November 8, 2011, 6PM EDT. From the syllabus: homework will be penalized 50% for each day they are late. After two days, they will not be accepted. No exception. Problem 1. Given the problem min − x 1 − x 2 x 1 + 5 x 2 ≤ 10 + α x 1 + x 2 ≤ 6 + β x 1 , x 2 ≥ 1. Compute the set M = { ( α,β ) ∈ R 2 : the basis B = { 1 , 2 } is optimal } . 2. For each ( α,β ) ∈ M , compute the change in objective function. 3. Is M convex? Is it a polyhedron? Is it a cone? Solutions. 1. Let’s put the problem in standard form: min − x 1 − x 2 x 1 + 5 x 2 + x 3 = 10 + α x 1 + x 2 + x 4 = 6 + β x 1 , x 2 , x 3 , x 4 ≥ The given basis B is optimal if it is primal feasible and dual feasible, i.e. (a) x B = B − 1 b ≥ 0. Now, B = parenleftbigg 1 5 1 1 parenrightbigg , hence B − 1 = parenleftBigg − 1 4 5 4 1 4 − 1 4 parenrightBigg and B − 1 (10 + α, 6 + β ) ⊤ = parenleftBigg − 5 2 − α 4 + 15 2 + 5 4 β 5 2 + α 4 − 3 2 − 1 4 β parenrightBigg = parenleftBigg 5 − α 4 + 5 4 β 1 + α 4 − 1 4 β parenrightBigg . Therefore, necessary and sufficient conditions for primal feasibility are as follows: braceleftBigg − 1 4 α + 5 4 β ≥ − 5 1 4 α − 1 4 β ≥ − 1 ≡ braceleftBigg α − 5 β ≤ 20 α − β ≥ − 4 . (b) This does not depend on α or β , as the vector of reduced cost does not depend on b . Hence B is dual feasible either for any α,β or for none. We need to check ¯ c ⊤ = c ⊤ N − c ⊤ B B − 1 N ≥ , or, since N = I 2 , c ⊤ N − c ⊤ B B − 1 ≥ , that is, (0 , 0) − ( − 1 , − 1) parenleftBigg − 1 4 5 4 1 4 − 1 4 parenrightBigg = (0 , 1) ≥ , and the basis is optimal (when feasible) for any α,β . To recap, M = { ( α,β ) : α − 5 β ≤ 20 ,α − β ≥ − 4 } . 2. The objective function value is c ⊤ B x B = c ⊤ B B − 1 b = ( − 1 , − 1) parenleftBigg 5 − α 4 + 5 4 β 1 + α 4 − 1 4 β parenrightBigg = − 6 − β, and it depends on β only, given that β relaxes (or restricts) the constraint whose orthogonal is parallel to the objective function’s gradient. 3. Since M = { ( α,β ) : α − 5 β ≤ 20 ,α − β ≥ − 4 } , it is not a cone as righthand sides are nonzero. However, it’s a polyhedron and thus convex. Problem 2. Find extreme points, extreme rays, and recession cone of each of the following polyhedra: – P 1 = { x ∈ R 2 : x 1 ≥ ,x 2 ≥ 1 ,x 2 − αx 1 ≥ } for any α ∈ R ; – P 2 = { x ∈ R n : ∑ n i =1  x i  ≤ 1 } ; – P 3 = { x ∈ R 2 : x 1 ≥ ,x 2 − k 2 ≥ (( k +1) 2 − k 2 )( x 1 − k ) , k ∈ { , 1 , 2 ... ,K }} . Solutions. 1. Since P 1 ⊂ R 2 , extreme points are points with two linearly independent active constraints. There are three constraints in total, pairwise linearly in dependent, so we need to check for intersections of the three pairs of con straints: – x 1 = 0 ,x 2 = 1: x 1 = (0 , 1) is feasible for the remaining constraint...
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This note was uploaded on 03/14/2012 for the course MTHSC 810 taught by Professor Staff during the Fall '08 term at Clemson.
 Fall '08
 Staff
 Math

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