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

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

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

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

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

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

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

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

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. )

( 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

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

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,