L20 LP part 6 Homework Review Postoptimality Analysis Summary 1

L20 LP part 6

• Homework• Review• Postoptimality Analysis• Summary

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H19

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H19 cont’d

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Two-phase Simplex Method

• Phase I - finds a feasible basic solution• Phase II- finds an optimal feasible basic

solution, if it exists.

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Use artificial variables!

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

Problem 8.58 (from lecture)

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1 2

1 2 3 4 5 6

1 2 3 4 5 6

1 2 3 4 5 6

( ) 4. .

1 2 1 0 0 0 52 1 0 0 1 0 41 1 0 1 0 1 1

0i

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

x

5 6

5 1 2

6 1 2 4

1 2 1 2 4

1 2 3 4

4 (2 )1 ( )

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

Minimize w x xx x xx x x x

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

1 2

1 2

1 2

1 2

1 2

( ) 4. .

2 52 4

1, 0

Max F x xs tx xx xx xx x

x

H19 cont’d

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Phase ISimplex Tableaurow basic x1 x2 x3 x4 x5 x6 b b/a_pivot

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

Simplex Tableauoperations row basic x1 x2 x3 x4 x5 x6 b b/a_pivot

a-(1)h f x3 0 3 1 1 0 -1 4 4/3b-(2)h g x5 0 3 0 2 1 -2 2 2/3

h=c h x1 1 -1 0 -1 0 1 1 negd-(-1)h i cost 0 -5 0 -1 0 1 1e-(-3)h j art cost 0 -3 0 -2 0 3 -2

H19 cont’d

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Simplex Tableauoperations row basic x1 x2 x3 x4 x5 x6 b b/a_pivot

f-(3)l k x3 0 0 1 -1 -1 1 2g/3 l x2 0 1 0 0.66667 0.33333 -0.6667 0.66667

h-(-1)l m x1 1 0 0 -0.3333 0.33333 0.33333 1.66667i-(-5)l n cost 0 0 0 2.33333 1.66667 -2.3333 4.33333 f=-4.333j-(-3)l o art cost 0 0 0 0 1 1 0 w=0

1 2

1 2

1 2

1 2

1 2

( ) 4. .

2 52 4

1, 0

Max F x xs tx xx xx xx x

x

Simplex Method Identifies:

• global solutions, if they exist• multiple solutions• unbounded problems• degenerate problems• infeasible problems

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Postoptimality Analysis

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What happens to our results if: The inputs change? The conditions of the business objective change ?The system/factory parameters change

What-if analyses

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Problem formulation: …. Make assumptions about Resources/capacities, Price coefficientsCost coefficients

Also what would happen if conditions should change such as :Environmental aspectsConditions of useMarket conditionsProduction capabilities

Determine Strategies to handle:In case things changeTo account for competitive reactionsConsider worst case possibilities

Sensitivity Analyses

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how sensitive are the:a. optimal value (i.e. f(x) and b. optimal solution (i.e. x)

… to the parameters (i.e. assumptions) in our model?

Model parameters

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1 1 2 2

11 1 1 1

21 1 2 2

1 1

( ). .

0, 10, 1

n n

n n

n n

m mn n m

i

j

Min f c x c x c xs ta x a x ba x a x b

a x a x b

b i to mx j to n

x

( ). .Min fs t

Tx c x

Ax bb 0x 0

Consider your abc’s, i.e. A, b and c

Recall during Formulation

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Constraints usually arise from :Laws of nature

(e.g. F=ma)

Laws of economics(e.g. profit=revenues-costs)

Laws of man(e.g. max work week = 40 hrs)

How accurate are our assumptions?

Any approximations in our formulation?

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1 1 2 2

11 1 1 1

21 1 2 2

1 1

( ). .

0, 10, 1

n n

n n

n n

m mn n m

i

j

Min f c x c x c xs ta x a x ba x a x b

a x a x b

b i to mx j to n

x

( ). .Min fs t

Tx c x

Ax bb 0x 0

Abc’s of sensitivity analyses

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( ). .Min fs t

Tx c x

Ax bb 0x 0

Let’s look at “b” first, i.e. “resource limits”

Recall Relaxing constraints(i.e. adding more resources)

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* ( )

* ( )

ii i

ij j

f fυ

b bf f

ue e

x*

x*

The instantaneous rate of change in the objective function with respect to relaxing a constraint IS the LaGrange multiplier!

Constraint Variation Sensitivity TheoremFrom LaGrange theory

Simplex LaGrange Multipliers

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the right side paramter of the th constraintthe LaGrange multiplier of the th constraint

* ( )( )

i

i

i i i i new oldi i

e iy i

f fy f y e y e e

e e

x*

Constraint Type≤ = ≥slack either surplus

c’ column “regular” artificial artificial

0iy iy 0iy

Find the multipliers in the final tableau (right side)

Prob 8.58

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Simplex Tableau slack surplus art'l art'loperations row basic x1 x2 x3 x4 x5 x6 b b/a_pivot

f-(3)l k x3 0 0 1 -1 -1 1 2g/3 l x2 0 1 0 0.666667 0.333333 -0.66667 0.666667

h-(-1)l m x1 1 0 0 -0.33333 0.333333 0.333333 1.666667i-(-5)l n cost 0 0 0 2.333333 1.666667 -2.33333 4.333333 f=-4.333j-(-3)l o art cost 0 0 0 0 1 1 0 w=0

1 2

1 2 3 4 5 6

1 2 3 4 5 6

1 2 3 4 5 6

( ) 4. .

1 2 1 0 0 0 52 1 0 0 1 0 41 1 0 1 0 1 1

0i

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

x1 2

1 2

1 2

1 2

1 2

( ) 4. .

2 52 4

1, 0

Max F x xs tx xx xx xx x

x

“=“ “≥”reg. cols art’l col’s

1 3 1 1

1 5 2 2

2 6 3 3

(" "), : 0, 0(" "), : 0, 5 / 3(" "), : 0, 7 / 3

( )i i i new old

g x y yh x y yg x y yf y e y e e

Excel “shadow prices”

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Variable CellsFinal Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease$C$13 x1 1.666666667 0 -1 1E+30 7$C$14 x2 0.666666667 0 -4 3.5 1E+30

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease$C$16 g1 3 0 5 1E+30 2$C$17 h1 4 -1.666666667 4 2 2$C$18 g2 1 2.333333333 1 1 2

If Minimizing w/Excel…Reverse sign of LaGrange Multipliers!

Variable CellsFinal Reduced Objective Allowable Allowable

Cell Name Value Cost Coefficient Increase Decrease$C$13 x1 1.666666667 0 1 7 1E+30$C$14 x2 0.666666667 0 4 1E+30 3.5

ConstraintsFinal Shadow Constraint Allowable Allowable

Cell Name Value Price R.H. Side Increase Decrease$C$16 g1 3 0 5 1E+30 2$C$17 h1 4 1.666666667 4 2 2$C$18 g2 1 -2.333333333 1 1 2

If Maximizing w/Excel… signs are the same as our tableaus

Let’s minimize f even further

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1 1 2 2 3 3

1 2 3

2

3

1

(0) (5 / 3) ( 7 / 3)1

1(0) (5 / 3)(1) ( 7 / 3)( 1)12 / 3 4

f y e y e y ef e e eeef ef

Increase/decrease ei to reduce f(x)

Excel solution

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Objective Cell (Min)Cell Name Original Value Final Value

$C$15 c -4.333333333 -8.333333333

Variable CellsCell Name Original Value Final Value Integer

$C$13 x1 1.666666667 1.666666667 Contin$C$14 x2 0.666666667 1.666666667 Contin

ConstraintsCell Name Cell Value Formula Status Slack

$C$16 g1 5 $C$16<=$E$16 Binding 0$C$17 h1 5 $C$17=$E$17 Binding 0$C$18 g2 0 $C$18>=$E$18 Binding 0

Objective Cell (Min)Cell Name Original Value Final Value

$C$15 c -4.333333333 -4.333333333

Variable CellsCell Name Original Value Final Value Integer

$C$13 x1 1.666666667 1.666666667 Contin$C$14 x2 0.666666667 0.666666667 Contin

ConstraintsCell Name Cell Value Formula Status Slack

$C$16 g1 3 $C$16<=$E$16 Not Binding 2$C$17 h1 4 $C$17=$E$17 Binding 0$C$18 g2 1 $C$18>=$E$18 Binding 0

Abc’s

• Changes in cost coefficients, c• Changes in coefficient matrix A

Often times it’s simpler to re-run LP Solver

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Ranging-limits on changes

• RHS parameters, b• Cost coefficients, c • A coefficient matrix

Yes, formulas are available… often times it’s much easier to just re-run your LP Solver!

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Summary• Simplex method determines:

Multiple solutions (think c’)Unbounded problems (think pivot aij<0)Degenerate Solutions (think bj=0)Infeasible problems (think w≠0)

• Sensitivity Analyses add important value to your LP solutions, can provide “strategies”

• Sensitivity is as simple as Abc’s• Constraint variation sensitivity theorem can

answer simple resource limits questions

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