homework-9-solutions

homework-9-solutions - MthSc810 – Mathematical...

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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 right-hand 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.

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homework-9-solutions - MthSc810 – Mathematical...

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