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Unformatted text preview: C&O 350 Linear Optimization – Fall 2009 – Solutions to Assignment 3 Question # Max. marks Part marks 1 12 3 2 2 3 2 2 9 + (5) 6 (5) 3 3 5 4 6 3 3 5 6 2 2 2 6 (Bonus) (8) Total 38 + (13) 1. Geometry of Basic Feasible Solutions 12 marks = 3+2+2+3+2 Solution: (a) The polyhedron P is depicted in Figure ?? . Figure 1: The polyhedron P . (b) The extreme points of P are the points: α = (0 , 0) T β = (20 , 0) T γ = (80 / 3 , 40 / 3) T δ = (0 , 40) T . They are marked on Figure ?? . (c) Notice that any pair of columns of A are linearly independent. Therefore, every possible set of two columns forms a basis. So the set of bases is {{ 1 , 2 } , { 1 , 3 } , { 1 , 4 } , { 2 , 3 } , { 2 , 4 } , { 3 , 4 }} . 1 (d) The basic solutions determined by each basis are: { 1 , 2 } : (80 / 3 , 40 / 3 , , 0) T { 1 , 3 } : (20 , , 20 , 0) T { 1 , 4 } : (40 , , , 40) T { 2 , 3 } : (0 , 40 , 80 , 0) T { 2 , 4 } : (0 , 40 , , 80) T { 3 , 4 } : (0 , , 40 , 40) T . The solutions corresponding to the bases { 1 , 4 } and { 2 , 3 } contain negative coordinates, and hence are not feasible. The rest are feasible. (e) The points corresponding to the basic solutions are: α = (0 , , 40 , 40) T β = (20 , , 20 , 0) T γ = (80 / 3 , 40 / 3 , , 0) T δ = (0 , 40 , , 80) T = (0 , 40 , 80 , 0) T ρ = (40 , , , 40) T See Figure ?? for reference. 2. Computing a basic feasible solution 9+(5) marks = 6+(5)+3 Solution: (a) First, suppose { A j : ˆ x j 6 = 0 } is a linearly independent set. Then by Theorem 5.1, ˆ x is a basic solution. Since it’s also feasible, we conclude that ˆ x is a basic feasible solution of ( S ), and we’re done. Now suppose that { A j : ˆ x j 6 = 0 } is linearly dependent. Let ˜ x denote ˆ x + t * d . We’ll show that (i) ˜ x also satisfies ( S ), and (ii) ˜ x has strictly fewer nonzero entries than ˆ...
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This note was uploaded on 01/19/2010 for the course MATH co 350 taught by Professor Cheriyan during the Spring '09 term at Waterloo.
 Spring '09
 cheriyan
 Geometry

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