L19 LP part 5
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Transcript of L19 LP part 5
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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
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Multiple Solutions
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Non-basic ci’=0
Non-unique global solutions, ∆f = 0
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Unbounded problem
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Non-basic pivot column coefficients aij < 0
1 2 3
2 3 1
2 1 1 00 2
x x xx x x
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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.
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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!
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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
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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!
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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!
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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
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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
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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
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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
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Table 8.17
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Feasible when w=0
after w=0, ignore art. var’s reduced cost
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2
3
4 5 6
5 / 32 / 32
, , 013 / 3
xxxx x xf
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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
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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
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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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