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Unformatted text preview: CO350 Linear Programming Chapter 6: The Simplex Method 3rd June 2005 Chapter 6: The Simplex Method 1 Recap Suppose A is an mby n matrix with rank m . ( P ) max. c T x s.t. Ax = b x ≥ On Wednesday, we learned • if ( P ) has an optimal solution, then it has an optimal solution that is basic. • the motivation for simplex method: we search for an optimal solution one bfs at a time. • how to obtain the basic solution determined by a given basis, while preserving the data of the LP. Chapter 6: The Simplex Method 2 Bases and Tableaux Suppose A is mby n with rank m . ( P ) max. ( z =) c T x s.t. Ax = b x ≥ (Defn ) Tableau A tableau for ( P ) corresponding to a basis B is the system of equations of the form z X j ∈ N ¯ c j x j = ¯ v x i + X j ∈ N ¯ a ij x j = ¯ b i ( i ∈ B ) derived by applying elementary row operations to z c T x = 0 Ax = b Example ( B = { 1 , 2 , 3 } ) z + 8 / 3 x 4 1 / 3 x 5 = 23 x 3 5 / 3 x 4 + 4 / 3 x 5 = 4 x 1 + 1 / 3 x 4 + 1 / 3 x 5 = 4 x 2 + 1 / 3 x 4 2 / 3 x 5 = 1 ¯ c 4 = 8 / 3 ¯ c 5 = 1 / 3 ¯ v = 23 ¯ a 34 = 5 / 3 ¯ a 25 = 2 / 3 ¯ b 2 = 1 zrow→ x irow→ ( zrow) ( x 3row) ( x 1row) ( x 2row) Chapter 6: The Simplex Method 3 A tableau has a wealth of information (Pg 71) Tableau corresponding to B = { 1 , 2 , 3 } : z + 8 / 3 x 4 1 / 3 x 5 = 23 x 3 5 / 3 x 4 + 4 / 3 x 5 = 4 x 1 + 1 / 3 x 4 + 1 / 3 x 5 = 4 x 2 + 1 / 3 x 4 2 / 3 x 5 = 1 1. The basic solution determined by B = { 1 , 2 , 3 } is x * = [4 , 1 , 4 , , 0] T In general, the basic solution determined by B is x * i = ¯ b i for i ∈ B x * j = 0 for j ∈ N The solution is a bfs if and only if ¯ b i ≥ for all i ∈ B . In this case, we say that the tableau is feasible , and B is a feasible basis . E.g., B = { 1 , 2 , 3 } is feasible but B = { 2 , 3 , 5 } is not feasible. 2. The objective value of the basic solution determined by B = { 1 , 2 , 3 } is z = ¯ v = 23 In general, the objective value is given by ¯ v . Chapter 6: The Simplex Method 4 A tableau has a wealth of information (cont’d) Tableau corresponding to B = { 1 , 2 , 3 } : z + 8 / 3 x 4 1 / 3 x 5 = 23 x 3 5 / 3 x 4 + 4 / 3 x 5 = 4 x 1 + 1 / 3 x 4 + 1 / 3 x 5 = 4 x 2 + 1 / 3 x 4 2 / 3 x 5 = 1 3. When we increase x 5 from while keeping the other...
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This note was uploaded on 06/11/2011 for the course C 350 taught by Professor Wolkowicz during the Fall '97 term at Waterloo.
 Fall '97
 Wolkowicz

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