L19 LP part 5
• Review
• Two-Phase Simplex Algotithm
• Summary
• Test 3 results
1

Single Phase Simplex Method
• Finds global solutions, if they exist
• Identifies multiple solutions
• Identifies unbounded problems
• Identifies degenerate problems
And, as we shall see in two-phase Simplex
Method, it also…
• identifies infeasible problems
2

Multiple Solutions
Non-basic c i
’=0
Non-unique global solutions, ∆ f = 0
3

Unbounded problem

2 x
1

1 x x
2
2

1 x
3 x
3

0 x
1
Non-basic pivot column coefficients a ij
< 0
4

Degenerate basic solution
First Tableau row basic e x1 f g h x4 x5 c' x1
1
0
0
0 x2
2
3
4
-2 x3
1
2
-1
-4 x4
0
1
0
0 x5
0
0
1
0
0
0 b b/a_pivot
3 3/1
0/2 neg
2.25
f=-2.25
One or more basic variable(s) = 0 min n/a
Simplex method will move to a solution, slowly
In rare cases it can “cycle” forever.
5

What’s next?
Single-Phase Simplex Method handles
“ ≤” inequality constraints
But, LP problems have “=” and “ ≥ ” constraints!
Such a tableau would not be feasible!
6

Simplex Method requires an
Max F initial basic feasible solution
 
1
4 x
2 x x
2
1 x
1
 x x
1 2
2 x x
1 2 x

2
2
0

 
5
4
   
1
“=” and or “ ≥” constraints or bj<0 (neg resource limits) cause infeasibility problems!
i.e. need canonic form to start Simplex
7

Can we transform to canonic form?
Max F
 
1
4 x
2 x x
2
1

2 x x
1 2
2 x x
1 2 x x
1 2

0

 
5
4
   
1
Min f
  
1
4 x
2 x
1

2 x
2
 x
3

2 x
1
 x
2

4 x
1
 x
2
 x
4
 x x
1 2

0
1
5
8

Basic Feasible Solution?
Initial tableau requires 3 identity columns for 3 basic variables (m=3)!
Min f
  
4 x
2
2
1 x
1

2 x
2

1 x
3

0 x
4

5 x
1
1 x
1 x x

2 x

2 x
2
0

0 x
3 x

0 x
4 x

   
4
0 1 1
3 4
Missing third basic variable! Bad!
Need “+1”, Bad!
9

Need two phases
• Phase I - finds a feasible basic solution
• Phase II- finds an optimal solution, if it exists.
Two-phase Simplex Method using artificial variables!
10

What is so darn infeasible?
Recall, in LaGrange technique, how we insured that an equation is not “violated”, i.e. feasible…
Min f
We set h j
=0 and g j
+s j
=0
  
1
4 x
2 x
1

2 x
2
 x
3

2 x
1
 x
2

4 x
1
 x
2
 x
4
 x x
1 2

0
1
5 h
1 g
2
: 4 (2 x
1
 x
2
)

0
:1 ( x
1
 x
2
 x
4
)

0
11

Use Artificial Variables x5, x6 to
Obtaining feasibility!
x
5 x
6
4 (2 x
1
 x
2

0 x
3

0 x
4
)

0
1 ( x
1
 x
2

0 x
3

1 )
4

0
Use the simplex method to minimize an artificial cost function w (i.e. w=0).
Minimize w
 x
5
 x
6
Min w
Min w

(4 2 x x
1
 x x
2
) (1 x x x x
1 2 4
 
1

0
2

0
3

1 x
4
)
12

Tableau w/Artificial Variables x5,x6
Min f
  
1
4 x
2
1 x
1

2 x
2

1 x
3

0 x
4

0 x
5

0 x
6

5
2 x
1

1 x
2

0 x
3

0 x
4

1 x
5

0 x
6

4
1 x
1 x x
1 2

1
 x
0
2

0 x
3

1 x
4

0 x
5

1 x
6

1
Canonic form,
Phase I
Min w 5 3 x
1

0 x
2

0 x
3

1 x
4 i.e. feasible basic solution!
Simplex Tableau row a b basic x3 x5 c d e x6 cost art cost
1
-1
-3 x1
1
2
-1
-4
0 x2
2
1
0
0
0 x3
1
0
-1
0
1 x4
0
0
0
0
0 x5
0
1
1
0
0 x6
0
0
1
0
-5 b
5
4 b/a_pivot
13

Transforming Process
1. Convert Max to Min, i.e. Min f(x) = Min -F(x)
2. Convert negative bj to positive, mult by(-1)
3. Add slack variables
4. Add surplus variables
5. Add artificial var’s for “=” and or “ ≥” constraints
6. Create artificial cost function,

( x art 1
 x art 2 x artw
)
14

Table 8.17
x
1

5 / 3 x
2

2 / 3 x
3 f

2 x x x
4 5 6

0
 
13 / 3 after w=0, ignore art. var’s reduced cost
Feasible when w=0
15

Two-Phase Simplex method
Phase I
 Transform infeasible LP to feasible using artificial variables.
 Use Simplex Meth. To minimize artificial cost function (i.e. art. cost row).
If w ≠ 0, problem is infeasible!
Phase II
 Use Simplex meth to min. reduced cost function
(i.e.row)
Ignore art. Var’s when choosing pivot columns!
16

Transformation example Ex8.65
take out a sheet of paper…
Max F x

10 x
1

6 x
2
2 x
1

3 x
2

90
4 x
 x

1

2
15
2
80 x
2
5 x
1
 x
2
 x x
1 2

0
25
Min f x
 
10 x
1

6 x
2
2 x
1

3 x
2
 x
3

4 x
5
2 x x x
1

 x
5
2

2
 

15
25
4
 x i x x
1 2

0
90
80
17

Phase I canonical form
Min f x
 
10 x
1

6 x
2
2 x
1

3 x
2
 x
3

4 x
5
2 x x x
1

 x
5
2

2
 

15
25
4
 x i x x
1 2

0
90
80
Min f x
 
10 x
1

6 x
2
2
4 x x
1
1


3
2 x x
2
2

 x x
  
3
4


90
80 x x x
2 5 6
5
  
2 7
15
25 x i x
1 x x

0
18

Summary
• Need for initial basic feasible solution
(i.e. canonic form)
• Phase I – solve for min “artificial cost” use artifical var’s for “=” and or “ ≥” constraints
• Phase II –solve for min “cost”
• Simplex method determines:
Multiple solutions (think c’)
Unbounded problems (think pivot a ij
<0)
Degenerate Solutions (think b j
=0)
Infeasible problems (think w≠0)
19