Variable Eq z x 1 x 2 x 3 x 4 x 5 x 6 RHS z 0 1 1630M1130 113M 43 M 21M 36 x1 1

Variable eq z x 1 x 2 x 3 x 4 x 5 x 6 rhs z 0 1

This preview shows page 82 - 94 out of 121 pages.

Variable Eq. z x 1 x 2 x 3 x 4 x 5 x 6 RHS z (0) 1 0 (16/30)M+11/30 (11/3)M-4/3 0 M 0 2.1M 3.6 x 1 (1) 0 1 1/3 10/3 0 0 0 9 x 4 (2) 0 0 1/3 5/3 1 0 0 1.5 x 6 (3) 0 0 0.2 2 0 1 1 0.6
Image of page 82
artificial variable x 6 已變為 nonbasic variable, 所以這一行可不計算 Basic coeff of Variable Eq. z x 1 x 2 x 3 x 4 x 5 x 6 RHS z 1 0 0 (5/3)M+7/3 0 (5/3)M+11/6 (8/3)M 11/6 0.5M 4.7 x 1 (1) 0 1 0 20/3 0 5/3 -5/3 8 x 4 (2) 0 0 0 5/3 1 5/3 -5/3 0.5 x 2 (3) 0 0 1 -10 0 -5 5 3 Optimal solution: ( x 1 , x 2 ) = ( 7.5, 4.5 ) Basic coeff. of Variable Eq. z x 1 x 2 x 3 x 4 x 5 RHS z (0) 1 0 0 0.5 M 1.1 0 5.25 x 1 (1) 0 1 0 5 1 0 7.5 x 5 (2) 0 0 0 1 0.6 1 0.3 x 2 (3) 0 0 1 5 3 0 4.5
Image of page 83
Image of page 84
No feasible solution exists in the following example. Minimize z = 0.4 x 1 + 0.5 x 2 subject to 0.3 x 1 + 0.1 x 2 1.8 0.5 x 1 + 0.5 x 2 = 6 0.6 x 1 + 0.4 x 2 6 , x 1 0, x 2 0 0.3 x 1 + 0.1 x 2 2.7 is replaced by 0.3 x 1 + 0.1 x 2 1.8 0.3 x 1 + 0.1 x 2 1.8
Image of page 85
Minimize z = 0.4 x 1 + 0.5 x 2 subject to 0.3 x 1 + 0.1 x 2 1.8 0.5 x 1 + 0.5 x 2 = 6 0.6 x 1 + 0.4 x 2 6 , x 1 0, x 2 0 Minimize z = 0.4 x 1 + 0.5 x 2 + M x 4 subject to 0.3 x 1 + 0.1 x 2 + x 3 = 1.8 0.5 x 1 + 0.5 x 2 + x 4 = 6 0.6 x + 0.4 x - x = 6 1 2 5 x 1 0, x 2 0, x 3 0, x 4 0, x 5 0. Minimize z = 0.4 x 1 + 0.5 x 2 + M x 4 + M x 6 subject to 0.3 x 1 + 0.1 x 2 + x 3 = 1.8 0.5 x 1 + 0.5 x 2 + x 4 = 6 0.6 x 1 + 0.4 x 2 x 5 + x 6 = 6 x 1 0, x 2 0, x 3 0, x 4 0, x 5 0, x 6 0. x 3 : slack variable x 4 : artificial variable x 5 : surplus variable
Image of page 86
Complete set of tableau (using Big-M Method): Basic coeff of Variable Eq. z x 1 x 2 x 3 x 4 x 5 x 6 RHS z (0) -1 -1.1M+0.4 -0.9M+0.5 0 0 M 0 -12M x 3 (1) 0 0.3 0.1 1 0 0 0 1.8 x 4 (2) 0 0.5 0.5 0 1 0 0 6 x 6 (3) 0 0.6 0.4 0 0 1 1 6 Basic coeff of Variable Eq. z x 1 x 2 x 3 x 4 x 5 x 6 RHS z (0) -1 0 -(16/30)M+11/3 0 (11/3)M-4/3 0 M 0 -5.4M-2.4 x 1 (1) 0 1 1/3 10/3 0 0 0 6 x 4 (2) 0 0 1/3 5/3 1 0 0 3 x 6 (3) 0 0 0.2 2 0 1 1 2.4 maximize - z
Image of page 87
Basic coeff of Variable Eq. z x 1 x 2 x 3 x 4 x 5 x 6 RHS z (0) 1 0 0 M+0.5 1.6M-1.1 M 0 -0.6M-5.7 x 1 (1) 0 1 0 5 1 0 0 3 x 2 (2) 0 0 1 5 3 0 0 9 這一行可不計算 There exists no feasible solution. ( In the optimal tableau , the artificial variable x 6 is not 0.) x 6 (3) 0 0 0 1 0.6 1 1 0.6
Image of page 88
Solve the Big-M Problem Optimal Optimal is is finite unbounded X a * = 0. Optimal solution is founded. X a * not 0. No feasible solution X a * = 0. Optimal solution is unbounded. X a * not 0. No feasible solution
Image of page 89
The Two-phase method: Minimize z = 0.4 x 1 + 0.5 x 2 subject to 0.3 x 1 + 0.1 x 2 2.7 0.5 x 1 + 0.5 x 2 = 6 0.6 x 1 + 0.4 x 2 6, x 1 0, x 2 0 Phase 1: Solve The Two-phase method for the Radiation Therapy Example: Minimize z = x 4 + x 6 subject to 0.3 x 1 + 0.1 x 2 + x 3 = 2.7 0.5 x 1 + 0.5 x 2 + x 4 = 6 0.6 x 1 + 0.4 x 2 x 5 + x 6 = 6 x 1 0, x 2 0, x 3 0, x 4 0, x 5 0, x 6 0. ( There exists a feasible solution if and only if the optimal value of the phase 1 problem is 0, i.e. x 4 = x 6 = 0. )
Image of page 90
( Minimize z or maximize -z ) z 1.1 x 1 0.9 x 2 + x 5 = 12 0.3 x 1 + 0.1 x 2 + x 3 = 2.7 0.5 x 1 + 0.5 x 2 + x 4 = 6 0.6 x 1 + 0.4 x 2 x 5 + x 6 = 6 x 1 0, x 2 0, x 3 0, x 4 0, x 5 0, x 6 0. Basic coeff. of Variable Eq. z x 1 x 2 x 3 x 4 x 5 x 6 RHS z (0) 1 1.1 0.9 0 0 1 0 12 x 3 (1) 0 0.3 0.1 1 0 0 0 2.7 x 4 (2) 0 0.5 0.5 0 1 0 0 6 x 6 (3) 0 0.6 0.4 0 0 1 1 6 maximize -z
Image of page 91
Basic coeff. of Variable Eq. z x 1 x 2 x 3 x 4 x 5 x 6 RHS z (0) 1 0 16/30 11/3 0 1 0 2.1 x 1 (1) 0 1 1/3 10/3 0 0 0 9 x 4 (2) 0 0 1/3 5/3 1 0 0 1.5 x 6 (3) 0 0 0.2 2 0 1 1 0.6 Basic coeff. of Variable Eq. z x 1 x 2 x 3 x 4 x 5 x 6 RHS z (0) 1 0 0 5/3 0 5/3 8/3 0.5 x 1 (1) 0 1 0 20/3 0 5/3 5/3 8 x 4 (2) 0 0 0 5/3 1 5/3 5/3 0.5 x 2 (3) 0 0 1 10 0 -5 5 3 這一行可不計算
Image of page 92
Basic coeff. of Variable Eq. z x 1 x 2 x 3 x 4 x 5 RHS z (0) 1 0 0 0 1 0 0 x 1 (1) 0 1 0 0 4 5 6 x 3 (2) 0 0 0 1 3/5 1 0.3 x 2 (3) 0 0 1 0 6 5 6 這一行可不計算 x 1 5 x 5 = 6 x 3 + x 5 = 0.3 x 2 + 5 x 5 = 6 x 1 0, x 2 0,
Image of page 93
Image of page 94

You've reached the end of your free preview.

Want to read all 121 pages?

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture