1
5.7 Initialization Revisited
 Motivation:
a solution for the transformed system is feasible
for the original system if and only if all the
artificial variables are equal to zero.
 Two methods are available for this purpose:
Big M
Two-Phase
 They look quite different but are essentially
equivalent.
Big M
2
 Associate
with each artificial variable a
very unattractive cost coefficient (cj).
 Since we try to optimize the objective
function, the optimal solution generated by
such a model will make the artificial
variables as small as possible.
 If the problem is feasible, the smallest
feasible value of any artificial variable is
zero.
3
 Thus,
if the problems is feasible, this
approach will yield an optimal solution with
all the artificial variables equal to zero.
4
Example
max Z'   3x1  5 x2  Mx 5  M x 6
x
x1 
x4
2 x2 
x5
4
 12
3x1  2 x2  x 3 
x 6  18
x1, x2 , x3 , x4 , x 5 , x 6  0
(5.41)
BV
5
Eq. #
x4
1
2
3
Z'
x5
x6
Z'
 Initial
x1
1
0
3
3
x2
0
2
2
5
x3
0
0
-1
0
x4
1
0
0
0
x5
x6
0
1
0
M
0
0
1
M
tableau is not in canonical form:
 Have to set the reduced costs of the
artificial variables to zero.
 We use (legal) row operations for this
purpose.
RHS
4
12
18
0
BV
6
Eq. #
x4
1
2
3
Z'
x5
x6
Z'
BV
Eq. #
x4
x1
1
0
3
3
x2
0
2
2
5
x4
1
0
0
0
x3
0
0
-1
0
x1
x2
1
1
x5
x6
2
Z
x5
x6
0
1
0
M
0
0
1
M
RHS
4
12
18
0
x3
x4
x5
x6
RH S
0
0
1
0
0
4
0
2
0
0
1
0
12
3
3
2
- 1
0
0
1
18
Z
- 3M + 3
- 4M +5
M
0
0
0
- 3 0M
BV
Eq. #
x4
x3
x4
x5
x6
RH S
0
0
1
0
0
4
0
2
0
0
1
0
12
3
3
2
- 1
0
0
1
18
Z
Z
- 3M + 3
- 4M +5
M
0
0
0
- 3 0M
BV
Eq. #
x4
x2
7
x1
x2
1
1
x5
x6
2
x1
x2
x3
x4
x5
x6
RH S
1
1
0
0
1
0
0
2
2
0
1
0
0
1 /2
0
6
x6
3
3
0
- 1
0
- 1
1
2
Z
Z
- 3M + 3
0
M
0
2 M - 5 /2
0
- 6M - 30
BV
Eq. #
x4
x2
x1
Z'
x1
x2
x3
x4
x5
x6
RH S
1
0
0
1 /3
1
1 /3
- 1 /3
4
2
0
1
0
0
1 /2
0
6
3
1
0
- 1 /3
0
- 1/3
1 /3
6
Z'
0
0
1
0
M - 3 /2
M - 1
- 36
8BV
Eq. #
x4
x2
x1
Z'
x1
x2
x3
x4
x5
x6
RH S
1
0
0
1 /3
1
1 /3
- 1 /3
4
2
0
1
0
0
1 /2
0
6
3
1
0
- 1 /3
0
- 1/3
1 /3
6
Z'
0
0
1
0
M - 3 /2
M - 1
- 36
 Note
that, as expected, all the artificial
variables are nonbasic (thus equal to zero).
 The optimal solution is
x=(6,6,0,4,0,0)
9
 Once
Remark
an artificial variable is out of the basis,
we never put it back into the basis.
 Thus, once an artificial variable is out of the
basis, we can “ignore” its column.
 If you want to handle M numerically (i.e set
it to a given value) make sure that it is not
too large, but also not too small
 The Big M method is “not nice”.
 What happens if opt=min ?
10
2-Phase Method
 Phase
1: Find a basic feasible solution to
the original problem (i.e. take the artificial
variables out of the basis).
 Phase 2: Find an optimal solution to the
original problem, ignoring the artificial
variables.
11
Phase 1
 Let
w := sum of the artificial variables
w* := minimum value of w subject to
the constraints.
 Because the artificial variables must satisfy the
nonnegativity constraint, w*=0 if and only if all
the artificial variables are equal to zero.
 Thus, the goal in Phase 1 is to minimize w
(regardless of what is the value of opt in the
original problem)
12Case

1: w*>0
The problem is not feasible!!! (why!)
 Case 2: w*=0 and all the artificial variables are
non-basic
A basic feasible solution to the original
problem has been generated. Continue with
Phase 2.
 Case 3: w*=0, but at least one artificial
variable is in the basis.
Using pivot operations, take all the artificial
variables out of the basis.
13
5.7.3 Example
max Z '   3x1  5x 2
x
x1 
x4
4
2x2 
x5
 12
3x1  2 x2  x 3 
x 6  18
x1, x2 , x3 , x4 , x 5 , x 6  0
14
Phase 1
min w  x 5  x 6
x1 
x
x4
2x2 
x5
4
 12
3x1  2 x2  x 3 
x 6  18
x1, x2 , x3 , x4 , x 5 , x 6  0
15
BV
x4
x5
x6
w
Eq. #
1
2
3
w
 We
x1
1
0
3
0
x2
0
2
2
0
x3
0
0
-1
0
x4
1
0
0
0
x5
0
1
0
-1
x6
0
0
1
-1
have to restore the canonical form (by
legal row operations)
RHS
4
12
18
0
16 BV
Eq. #
x1
x2
x3
x4
x5
x6
RH S
x4
1
1
0
0
1
0
0
4
x5
2
0
2
0
0
1
0
12
x6
3
3
2
- 1
0
0
1
18
w
w
0
0
0
0
- 1
- 1
0
BV
Eq. #
x4
x1
x2
x3
x4
x5
x6
RH S
1
1
0
0
1
0
0
4
x5
2
0
2
0
0
1
0
12
x6
3
3
2
- 1
0
0
1
18
w
w
3
4
- 1
0
- 1
0
- 1
0
30
corrections!!!
17BV
Eq. #
x1
x2
x3
x4
x5
x6
RH S
x4
x2
x6
1
1
0
0
1
0
0
4
2
0
1
0
0
1 /2
0
6
3
3
0
- 1
0
- 1
1
6
w
w
3
0
- 1
0
- 2
0
6
x1
x2
x3
x4
x5
x6
RH S
BV
Eq. #
x4
x2
x1
1
0
0
1 /3
1
1 /3
0
2
2
0
1
0
0
1 /2
0
6
3
1
0
- 1 /3
0
- 1/3
1
2
w
w
0
0
0
0
- 1
0
0
 End
of Phase 1: All the artificial
variables are out of the basis.
Phase 2
18
BV
Eq. #
x4
x2
x1
w
x1
x2
x3
x4
x5
x6
RH S
1
0
0
1 /3
1
1 /3
0
2
2
0
1
0
0
1 /2
0
6
3
1
0
- 1 /3
0
- 1/3
1
2
w
0
0
0
0
- 1
0
0
 We
now have to restore the original
objective function:
z’ = 3x1  5x2
BV
Eq. #
x4
x2
x1
Z'
x1
x2
x3
x4
1
0
0
1/3
1
2
2
0
1
0
0
6
3
1
0
- 1 /3
0
2
Z'
3
0
0
1
0
- 36
5
0
x5
x6
RH S
0
BV
19
Eq. #
x4
x2
x1
Z'
x1
x2
x3
x4
1
0
0
1/3
1
2
2
0
1
0
0
6
3
1
0
- 1 /3
0
2
Z'
3
0
0
1
0
- 36
5
0
x5
x6
RH S
0
 This
is not in canonical form, so we use legal
row operations to restore the canonical form.
BV
x4
x2
x1
Z'
Eq. #
1
2
3
Z'
x1
0
0
1
0
x2
0
1
0
0
x3
1/3
0
- 1/3
1
x4
1
0
0
0
x5
x6
RHS
2
6
2
- 36
20
Remark
 Read
the material in the lecture notes
concerning the relationship between the Big
M method and the 2-Phase Method and
make sure you understand why there are
“equivalent” and why the 2 Phase Method is
better. (end of section 5.7)
21
5.8 Algorithm Complexity
 Worst
case is very bad
 In practice : surprisingly well!!!