Dual Basic Solutions Consider LP in standard form with A 2 R m n,rank(a) =m, and dual LP: Observation 5.7. AbasisB yields min c T x max p T b s.t. A x = b s.t. p T A apple c T x 0 aprimalbasicsolutiongivenbyx B := B 1 b and adualbasicsolutionp T := c B T B 1. Moreover, a the values of the primal and the dual basic solutions are equal: b p is feasible if and only if c 0; c c B T x B = c B T B 1 b = p T b ; reduced cost c i = 0correspondstoactivedualconstraint; d p is degenerate if and only if c i = 0forsomenon-basicvariablex i. 155
Dual Simplex Method Let B be a basis whose corresponding dual basic solution p is feasible. 156
Dual Simplex Method Let B be a basis whose corresponding dual basic solution p is feasible. f also the primal basic solution x is feasible, then x, p are optimal. 156
Dual Simplex Method Let B be a basis whose corresponding dual basic solution p is feasible. f also the primal basic solution x is feasible, then x, p are optimal. Assume that x B(`) < 0andconsiderthe`th row of the simplex tableau (x B(`), v 1,...,v n ) (pivot row) 156
Dual Simplex Method Let B be a basis whose corresponding dual basic solution p is feasible. f also the primal basic solution x is feasible, then x, p are optimal. Assume that x B(`) < 0andconsiderthe`th row of the simplex tableau (x B(`), v 1,...,v n ) (pivot row) Let j 2{1,...,n} with v j < 0and c j v j = min c i i:v i <0 v i 156
Dual Simplex Method Let B be a basis whose corresponding dual basic solution p is feasible. f also the primal basic solution x is feasible, then x, p are optimal. Assume that x B(`) < 0andconsiderthe`th row of the simplex tableau (x B(`), v 1,...,v n ) (pivot row) Let j 2{1,...,n} with v j < 0and c j v j = min c i i:v i <0 v i Performing an iteration of the simplex method with pivot element v j yields new basis B 0 and corresponding dual basic solution p 0 with c B 0 T B 0 1 A apple c T and p 0T b p T b (with > if c j > 0). 156
Dual Simplex Method Let B be a basis whose corresponding dual basic solution p is feasible. f also the primal basic solution x is feasible, then x, p are optimal. Assume that x B(`) < 0andconsiderthe`th row of the simplex tableau (x B(`), v 1,...,v n ) (pivot row) Let j 2{1,...,n} with v j < 0and c j v j = min c i i:v i <0 v i Performing an iteration of the simplex method with pivot element v j yields new basis B 0 and corresponding dual basic solution p 0 with c B 0 T B 0 1 A apple c T and p 0T b p T b (with > if c j > 0). f v i 0foralli 2{1,...,n}, thentheduallpisunboundedandtheprimallpis infeasible. 156
Dual Simplex Example x 1 x 2 x 3 x 4 x 5 0 2 6 10 0 0 x 4 = 2 2 4 1 1 0 x 5 = 1 4 2 3 0 1 157
Dual Simplex Example x 1 x 2 x 3 x 4 x 5 0 2 6 10 0 0 x 4 = 2 2 4 1 1 0 x 5 = 1 4 2 3 0 1 Determine pivot row (x 5 < 0) 157
Dual Simplex Example x 1 x 2 x 3 x 4 x 5 0 2 6 10 0 0 x 4 = 2 2 4 1 1 0 x 5 = 1 4 2 3 0 1 Determine pivot row (x 5 < 0) Find pivot column. Column 2 and 3 have negative entries in pivot row. 157
Dual Simplex Example x 1 x 2 x 3 x 4 x 5 0 2 6 10 0 0 x 4 = 2 2 4 1 1 0 x 5 = 1 4 2 3 0 1 Determine pivot row (x 5 < 0) Find pivot column. Column 2 and 3 have negative entries in pivot row. Column 2 attains minimum. 157
Dual Simplex Example x 1 x 2 x 3 x 4 x 5 0 2 6 10 0 0 x 4 = 2 2 4 1 1 0 x 5 = 1 4 2 3 0 1 Determine pivot row (x 5 < 0) Find pivot column. Column 2 and 3 have negative entries in pivot row. Column 2 attains minimum. Perform basis change: x5 leaves and x 2 enters basis. Eliminate other entries in the pivot column. Divide pivot row by pivot element. 157
Dual Simplex Example x 1 x 2 x 3 x 4 x 5 3 14 0 1 0 3 x 4 = 2 2 4 1 1 0 x 5 = 1 4 2 3 0 1 Determine pivot row (x 5 < 0) Find pivot column. Column 2 and 3 have negative entries in pivot row. Column 2 attains minimum. Perform basis change: x5 leaves and x 2 enters basis. Eliminate other entries in the pivot column. Divide pivot row by pivot element. 157
Dual Simplex Example x 1 x 2 x 3 x 4 x 5 3 14 0 1 0 3 x 4 = 0 6 0 5 1 2 x 5 = 1 4 2 3 0 1 Determine pivot row (x 5 < 0) Find pivot column. Column 2 and 3 have negative entries in pivot row. Column 2 attains minimum. Perform basis change: x5 leaves and x 2 enters basis. Eliminate other entries in the pivot column. Divide pivot row by pivot element. 157
Dual Simplex Example x 1 x 2 x 3 x 4 x 5 3 14 0 1 0 3 x 4 = 0 6 0 5 1 2 x 2 = 1/2 2 1 3/2 0 1/2 Determine pivot row (x 5 < 0) Find pivot column. Column 2 and 3 have negative entries in pivot row. Column 2 attains minimum. Perform basis change: x5 leaves and x 2 enters basis. Eliminate other entries in the pivot column. Divide pivot row by pivot element. 157
Remarks on the Dual Simplex Method Dual simplex method terminates if lexicographic pivoting rule is used: Choose any row ` with xb(`) < 0tobethepivotrow. Among all columns j with vj < 0choosetheonewhichislexicographically minimal when divided by v j. Dual simplex method is useful if, e. g., dual basic solution is readily available. Example: Resolve LP after right-hand-side b has changed. 158