homework-9-solutions

homework-9-solutions - MthSc810 Mathematical Programming...

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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 0 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 0 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 0 , or, since N = I 2 , c N c B B 1 0 , that is, (0 , 0) ( 1 , 1) parenleftBigg 1 4 5 4 1 4 1 4 parenrightBigg = (0 , 1) 0 , 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.
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Problem 2. Find extreme points, extreme rays, and recession cone of each of the following polyhedra: P 1 = { x R 2 : x 1 0 , x 2 1 , x 2 αx 1 0 } for any α R ; P 2 = { x R n : n i =1 | x i | ≤ 1 } ; P 3 = { x R 2 : x 1 0 , x 2 k 2 (( k +1) 2 k 2 )( x 1 k ) , k ∈ { 0 , 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 x 2 αx 1 0 for any α , hence it is an extreme point.
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