60 likes | 298 Views
Linear Programming – Simplex Method: Computational Problems. Breaking Ties in Selection of Non-Basic Variable – if tie for non-basic variable with largest relative profit ( ), arbitrarily select incoming variable. Ties in Minimum Ratio Rule (Degeneracy) – if more than one
E N D
Linear Programming – Simplex Method: Computational Problems Breaking Ties in Selection of Non-Basic Variable – if tie for non-basic variable with largest relative profit ( ), arbitrarily select incoming variable. Ties in Minimum Ratio Rule (Degeneracy) – if more than one basic variable have same minimum ratio, select either variable to leave the basis. This will result in a basic variable taking on a value of 0. When this occurs, the solution is referred to as a degenerate basic feasible solution. When this occurs, you may transition through more than one simplex tableau with the same objective (Z) value.
Linear Programming – Simplex Method: Computational Problems Unbounded Solutions – if when performing the minimum ratio rule, none of the ratios are positive, then the solution is unbounded (e.g Max Z = or Min = - ). See pages 44-46 for good examples.
Simplex Method – Finding an Initial Basic Feasible Solution Min Z = -3x1 + x2 + x3 s.t. x1 – 2x2 + x3 <= 11 -4x1 + x2 +2x3 >= 3 2x1 - x3 = -1 x1, x2, x3 >= 0 Standard Form: (-) Max Z = 3x1 - x2 - x3 s.t. x1 – 2x2 + x3 + x4 = 11 -4x1 + x2 +2x3 -x5 = 3 -2x1 + x3 = 1 x1, x2, x3, x4, x5 >= 0
Simplex Method – Finding an Initial Basic Feasible Solution (-) Max Z = 3x1 - x2 - x3 s.t. x1 – 2x2 + x3 + x4 = 11 -4x1 + x2 +2x3 -x5 = 3 -2x1 + x3 = 1 x1, x2, x3, x4, x5 >= 0 Only x4 is basic. Introduce artificial variables. s.t. x1 – 2x2 + x3 + x4 = 11 -4x1 + x2 +2x3 -x5 + x6 = 3 -2x1 + x3 + x7 = 1 x1, x2, x3, x4, x5, x6, x7 >= 0
Simplex Method – Solve Using Big-M Method Let M be an arbitrarily large number, then: (-) Max Z = 3x1 - x2 - x3 + 0x4 + 0x5 – Mx6 – Mx7 s.t. x1 – 2x2 + x3 + x4 = 11 -4x1 + x2 +2x3 -x5 + x6 = 3 -2x1 + x3 + x7 = 1 x1, x2, x3, x4, x5, x6, x7 >= 0 Note: If the simplex algorithm terminates with one of the artificial variables as a basic variable, then the original problem has no feasible solution.