L19 LP part 5

• Review• Two-Phase Simplex Algotithm• Summary• Test 3 results

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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 problems2

Multiple Solutions

3

Non-basic ci’=0

Non-unique global solutions, ∆f = 0

Unbounded problem

4

Non-basic pivot column coefficients aij < 0

1 2 3

2 3 1

2 1 1 00 2

x x xx x x

Degenerate basic solution

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First Tableaurow basic x1 x2 x3 x4 x5 b b/a_pivot

e x1 1 2 1 0 0 3 3/1f x4 0 3 2 1 0 0 0/2 ming x5 0 4 -1 0 1 0 neg n/ah c' 0 -2 -4 0 0 2.25

f=-2.25

One or more basic variable(s) = 0

Simplex method will move to a solution, slowlyIn rare cases it can “cycle” forever.

What’s next?

Single-Phase Simplex Method handles “≤” inequality constraints

But, LP problems have “=” and “≥ ” constraints! Such a tableau would not be feasible!

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1 2

1 2

1 2

1 2

1 2

( ) 4. .2 5

2 41

, 0

Max F x xs tx xx xx xx x

x

Simplex Method requires an initial basic feasible solution

i.e. need canonic form to start Simplex

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“=” and or “≥” constraintsor bj<0 (neg resource limits) cause infeasibility problems!

Can we transform to canonic form?

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1 2

1 2

1 2

1 2

1 2

( ) 4. .2 5

2 41

, 0

Max F x xs tx xx xx xx x

x 1 2

1 2 3

1 2

1 2 4

1 2

( ) 4. .2 5

2 41

, 0

Min f x xs tx x xx xx x xx x

x

Basic Feasible Solution?

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1 2

1 2 3 4

1 2 3 4

1 2 3 4

1 2

( ) 4. .1 2 1 0 52 0 0 41 0 1 1, 0

Min f x xs tx x x xx x x xx x x x

x x

xInitial tableau requires 3 identity columns for 3 basic variables (m=3)!

Missing third basic variable! Bad!Need “+1”, Bad!

Need two phases

• Phase I - finds a feasible basic solution• Phase II- finds an optimal solution, if it exists.

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Two-phase Simplex Method using artificial variables!

What is so darn infeasible?

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Recall, in LaGrange technique, how we insured that an equation is not “violated”, i.e. feasible…

We set hj=0 and gj+sj=0

1 1 2

2 1 2 4

: 4 (2 ) 0:1 ( ) 0

h x xg x x x

1 2

1 2 3

1 2

1 2 4

1 2

( ) 4. .2 5

2 41

, 0

Min f x xs tx x xx xx x xx x

x

Use Artificial Variables x5, x6 toObtaining feasibility!

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5 1 2 3 4

6 1 2 3 4

4 (2 0 0 ) 01 ( 0 1 ) 0

x x x x xx x x x x

Use the simplex method to minimize an artificial cost function w (i.e. w=0).

5 6

1 2 1 2 4

1 2 3 4

(4 2 ) (1 )5 3 0 0 1

Minimize w x x

Min w x x x x xMin w x x x x

Tableau w/Artificial Variables x5,x6

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1 2

1 2 3 4 5 6

1 2 3 4 5 6

1 2 3 4 5 6

1 2

( ) 4. .1 2 1 0 0 0 52 1 0 0 1 0 41 1 0 1 0 1 1, 0

Min f x xs tx x x x x xx x x x x xx x x x x xx x

x

1 2 3 45 3 0 0 1Phase IMin w x x x x

Canonic form,i.e. feasible basic solution!

Simplex Tableaurow basic x1 x2 x3 x4 x5 x6 b b/a_pivot

a x3 1 2 1 0 0 0 5b x5 2 1 0 0 1 0 4c x6 1 -1 0 -1 0 1 1d cost -1 -4 0 0 0 0 0e art cost -3 0 0 1 0 0 -5

Transforming Process1. Convert Max to Min, i.e. Min f(x) = Min -F(x)2. Convert negative bj to positive, mult by(-1)3. Add slack variables4. Add surplus variables5. Add artificial var’s for “=” and or “≥”

constraints6. Create artificial cost function,

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1 2( )art art artwx x x

Table 8.17

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Feasible when w=0

after w=0, ignore art. var’s reduced cost

1

2

3

4 5 6

5 / 32 / 32

, , 013 / 3

xxxx x xf

Two-Phase Simplex methodPhase ITransform 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 IIUse Simplex meth to min. reduced cost function (i.e.row)

Ignore art. Var’s when choosing pivot columns!

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Transformation example Ex8.65take out a sheet of paper…

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1 2

1 2

1 2

2

1 2

1 2

( ) 10 6. .2 3 904 2 80

155 25, 0

Max F x xs tx xx xxx xx x

x 1 2

1 2 3

1 2 4

2 5

1 2

( ) 10 6. .2 3 904 2 80

155 25

0i

Min f x xs tx x xx x xx xx xx

x

Phase I canonical form

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1 2

1 2 3

1 2 4

2 5

1 2

( ) 10 6. .2 3 904 2 80

155 25

0i

Min f x xs tx x xx x xx xx xx

x 1 2

1 2 3

1 2 4

2 5 6

1 2 7

( ) 10 6. .2 3 904 2 80

155 25

0i

Min f x xs tx x xx x xx x xx x xx

x

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 aij<0)Degenerate Solutions (think bj=0)Infeasible problems (think w≠0)

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