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

2

x

2  1

x

3

x

3  0 0 2

x

1 Non-basic pivot column coefficients

a ij

< 0 4

Degenerate basic solution First Tableau row e f g h basic x1 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 b 3 0 0 2.25

f=-2.25

b/a_pivot 3/1 0/2 neg 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

1 2

x

1

x

1   2 1 , 2

x x

2 2

x

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

1 2

x

1

x

1   2 1 , 2

x x

2 2   5 4

x

2  0   1

Min f

   1 4

x

2

x

1  2

x

1

x

1   2

x

2 

x x

1 , 2

x

2

x

2   0 

x

4

x

3 4   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

x

1 1

x

1   2 

x

2

x

2   , 2 

x

2 0  0 0

x

1

x

3 3 

x

3   0 0

x

4

x

4  1

x

4   4 1 5 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

1

x

1   2

x

2 

x x

1 , 2

x

2

x

2   0 

x

4

x

3 4   1 5

h g

1 2

x

1

x

1  

x

2

x

2  ) 

x

4 ) 0  0 11

Use Artificial Variables x5, x6 to Obtaining feasibility!

x

5

x

6

x

1

x

1  

x

2

x

2   0 0

x

3

x

3   0

x

4

x

4 )   0 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 x

1

x

1   0

x

2

x

2  0

x

3 

x

1 1

x

4

x

2

x

4 ) 12

Tableau w/Artificial Variables x5,x6

Min f

   1 4

x

2 1 2 1

x x x

1 1 1

x x

1    2 1

x

2 1 

x

2

x

0 2 2    1 0 0

x x x

3 3 3    0 0

x x

4 4 1

x

4  0

x

5   1

x

5 0

x

5    0 0

x x

6 6 1

x

6   4 5  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 x1 1 2 1 -1 -3 x2 2 1 -1 -4 0 x3 1 0 0 0 0 x4 0 0 -1 0 1 x5 0 1 0 0 0 x6 0 0 1 0 0 b 5 4 1 0 -5 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 x x

3 1 2    5 / 3 2 / 3 2

x x x

4 , 5 , 6

f

   13 / 3 0 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

 10

x

1  6

x

2 2 4

x

2 5 1

x

1

x

1

x

1 ,     2 3 2 15

x

 2

x

2

x

2 0    90 80 25

Min f

  10

x

1  6

x

2 2

x

1 5

x i x

1   4

x

2

x

1   0 3

x

2 2 

x

5

x

2

x

2     15

x

3

x

25 4   90 80 17

Phase I canonical form

Min f

  10

x

1  6

x

2 2

x

1 5

x i x

1   4

x

2

x

1   0 3

x

2 2 

x

5

x

2

x

2     15

x

3

x

25 4   90 80

Min f

  10

x

1  6

x

2 2 4

x

2

x

1

x

1 5

x i x

1       0 3 2

x

5

x

2

x

2

x

2 

x

3  

x

6

x

7

x

4     15 90 80 25 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