02.3 Chapter2 Solving Linear Programs.pdf

8/10/2019 02.3 Chapter2 Solving Linear Programs.pdf

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PORTOFEUP FACULDADE DE ENGENHARIA

UNIVERSIDADE DO PORTODEPARTAMENTO DE ENGENHARIA E GESTO INDUST RIAL

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Linear Programming ageometrical preview

(FEUP | DEGI) Operations Research 2012/2013 1 / 103
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Linear ProgrammingLearning Objectives

Ability to formulate a linear program:dene the decision variables (elements under control of the decision maker whose values determinethe solution of the model);dene the objective function as linear combinations of the decision variables (criterion the decisionmaker will use to evaluate alternative solutions to the problem);dene the constraints of the problem as linear combinations of the decision variables (restrictionsimposed upon the values of the decision variables by the caracteristics of the problem under study);

Ability to represent graphically the decision space of a linear program.Ability to nd, based on the graphical representation, the optimal solution of the linearprogram.Ability to determine, based on the graphical representation, the shadow price on a constraint:How much does the objective function increase if there is a unit increase in a constraintresource?Ability to determine, based on the graphical representation, over what range can a particularobjective-function coefficient vary without changing the optimal solution.Ability to determine, based on the graphical representation, over what range can theright-hand-side of a constraint change in order to maintain the shadow price.

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Production Planning at BA

The factory of Barbosa & Almeida (BA) in Avintes produces glass containers through mouldinjection.BA recently won two new clients and intends to assign this production to one of its furnaces. Theorders were of bottles of Sandeman Ruby Port (75cl) and one liter bottles of olive oil Oliveira daSerra. Both customers buy all the bottles that BA can produce.

Due to differences in the production (number of cavities and different cycle times), each batch of portwine bottles takes 50 hours to produce, whereas each batch of oil bottles requires only 30hours. There are a total of 2000 hours available at the oven.Moreover, there are restrictions in the storage capacity of the bottles prior to their dispatch. Thewarehouse has 300 m3 available and each batch of portwine bottles occupies 6 m3 of warehousespace, while each batch of bottles of olive oil holds 5 m3.Finally we must also consider the capacity available in the decoration sector (e.g. labels,packaging), which is 200 hours. The portwine bottles spend 3 hours per batch and the olive oilbottles 5 hours per batch.Barbosa & Almeida wants to maximize the prot from these two orders, knowing that the protper batch is 50 e and 60 e , respectively.

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Production Planning at BALinear Programming Model

Decision variables

x VP Number of batches of Vinho do Porto Ruby Sandeman bottles to produce;x A Number of batches of Azeite Oliveira da Serra bottles to produce.

Objective:max Z = 50 x VP + 60 x A

Subject to:50x VP + 30 x A 2000 (time in the oven)

6x VP + 5 x A 300 (space in the warehouse)3x VP + 5 x A 200 (capacity in the decoration sector)

x VP , x A 0

In this model x VP and x A may not be integer.

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Production Planning at BAGraphical resolution Active and redundant constraints

0

10

20

30

40

50

60

70

80

0 10 20 30 40 50 60 70 80

X A

X VP

Tempo no Forno Espao no armazm

Capacidade na decorao

Funo objectivo

6X VP + 5X A = 300(restrio redundante)

3X VP + 5X A = 200

(restrio ativa)

50X VP + 30X A = 2000(restrio ativa)

X VP =0

X A =0

Sentido de crescimento de Z

Z = 50X VP + 60X A

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Production Planning at BAGraphical resolution (Sensitivity analysis Coefficients of the objectivefunction)

0

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40

50

60

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80

0 10 20 30 40 50 60 70 80



Funo objectivo

Z = 50X VP + 60X A

50X VP + 30X A = 2000

3X VP + 5X A = 200

6X VP + 5X A = 300

X A

X VP

The optimal solution ramains unchanged whilethe slope of the objective function is maintainedbetween the slope of the restriction time in theoven and the slope of the constaint decorationcapacity.

Varying the slope of the objective function theoptimal solution jumps from vertex to vertex.

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Production Planning at BAGraphical resolution (Sensitivity analysis Coefficients of the objectivefunction)

In order to calculate the variation of the coefficient of x VP (C VP x VP + 60 x A) we can write the two active constraintsin the optimal solution with coefficient 60 in x A:

100x VP + 60 x A 4000 (time in the oven)36x VP + 60 x A 2400 (decoration cap.)

The coefficients of x VP in the two constraints will then bethe limits for the variation of C VP such that the optimalsolution does not change.

36 C VP 100

36 C VP (initial value 50) 100

In order to calculate the variation of the coefficient of x A(50x VP + C Ax A) we can write the two active constraints inthe optimal solution with coefficient 50 em x VP :

50x VP + 30x A 2000 (time in the oven)

50x VP + 2503 x A 100000

3 (decoration cap.)

The coefficients of x A in the two constraints will then be thelimits for the variation of C A such that the optimal solutiondoes not change.

30 C A 250

3

30 C A(initial value 60) 250

3

In both cases, the initial value of the coefficients is within the calculated range of variation.If the initial values were not contained in this interval, which ranges should be considered?

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Production Planning at BAGraphical resolution (Sensitivity analysis RHS of the constraints)

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50

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0 10 20 30 40 50 60 70 80



Funo objectivo

Z = 50X VP + 60X A

50X VP + 30X A = b 1

3X VP + 5X A = 200

6X VP + 5X A = 300

X A

X VP

Active constraint with another slope:the optimal value changes but the vertex does

not change.

A change in the RHS value of a constraint

translates the constraint into a paralel position.

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Production Planning at BAGraphical resolution (Shadow prices 1 of the constraints)

The shadow price of one constraint represents the variation in the objective function value if we increase theRHS of the constraint by one unit.Example:By how much would the prot increase if we could have an additional hour in the decoration? (increase theRHS of the contraint from 200 to 201)

50x VP + 30 x A = 20003x VP + 5 x A = 201

x VP = 24 1316x A = 25 516

Optimal solution: x VP = 241316 , x

A = 25 516 , Z

= 2759 38

With an additional hour in the decoration (an increase from 200 to 201) the value of the optimal solutionincreased from 2750 to 2759 38 . The constraint decoration capacityhas therefore a shadow price of 9

38 .

The shadow price remains the same while the optimal vertex does not change(see the sensitivity analysis for the RHS of the constraints).

The shadow price of a non-active constraint is zero.

1The shadow prices are also called dual variables, marginal values or pi values.(FEUP | DEGI) Operations Research 2012/2013 12 / 103
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Linear ProgrammingGraphical resolution (Unique optimal solution)

Objective:max Z = 4 x 1 + x 2

Subject to:x 1 x 2 2x 1+ 2 x 2 8

x 1, x 2 0 x 1

x2

x1

+ 2x2

= 8

x 1 - x 2 = 2

Z = 4 x1

+ x2

04

8 1216

20

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Linear ProgrammingGraphical resolution (Non-unique optimal solution)

Objective:

max Z = x 1 x 2 30Subject to:x 1 x 2 2x 1+ 2 x 2 8

x 1, x 2 0 x1

x 2

x1

+ 2x2

= 8

x 1 - x 2 = 2

Z = x1

- x2

- 30

- 28

- 30

- 32

- 34

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Linear ProgrammingGraphical resolution (Unlimited solution)

Objective:

max Z = 4 x 1 + x 2Subject to:

x 1 x 2 2x 1, x 2 0

x1

x2

x 1 - x 2 = 2

Z = 4 x1

+ x2

04

8 1216

20

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Linear ProgrammingGraphical resolution (Without admissible solution)

Objective:max Z

Subject to:x 1 x 2 5x 1+ 2 x 2 8

x 1, x 2 0

x1

x2

x1

+ 2x2

= 8

x1

- x2

= -5


h l
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Mathematical Programming

We consider only very special Mathematical Programming Models:All the variables take values in R or in Z .There is only one objective to minimize or to maximize.

The objective and the constraints are linear functions of the decision variables. Linear Programming Models if all the variables take values in R . Integer Programming Models if all the variables take values in Z .


M h i l P i f d l
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Mathematical Programming: fundamentals

Objective:min f (X )

Subject to:

g i (X ) 0 i {1,..., m}

hi (X ) = 0 i {1,..., l }

Admissible set All the points S R n that satisfy the constraints.Admissible solution Each X S is an admissible solution.

Optimal solution X S

f (X ) f (X ) X S


M th ti l P i
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Mathematical ProgrammingNon-linear programming example

x1

x2

x 1 + x 2 = 6 x 2 = x 12

Conjuntoadmissvel

(3, 6)

Curvas de nvel de f

Objective:min f (X ) = ( x 1 3)2 + ( x 2 6)2

Subject to: x 21 + x 2 0x 1 + x 2 6x 1, x 2 0

Optimal solution: x 1 = 32 , x 2 = 92 , f =

92

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Simplex Methodbased on:Bradley, Hax, and Magnanti; AppliedMathematical Programming, Addison-Wesley,1977

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Simplex MethodLearning Objectives

Canonical FormAbility to transform any linear program to a canonical form:

with nonnegative variablesby replacing each decision variable unconstrained in sign by a difference between two nonnegative variables. This replacement applies to all equations including the objective function.with equality constraintsby changing inequalities to equalities by the introduction of slack and surplus variables. For inequalities, let the nonnegative surplus variable represent the amount by which the lefthand side exceeds the righthand side; for inequalities, let the nonnegative slack variable represent the amount by which the righthand side exceeds the lefthand side.with non-negative righthand-side coefficientsby multiplying equations with a negative righthand side coefficient by -1.with one basic variable isolated in each constraintby adding a (nonnegative) articial variable to any equation that does not have an isolated variable readily apparent, and construct the BIG M objective function.

Simplex MethodAbility to use the simplex method in tableau formto solve a linear program represented in the Canonical Form, i.e.:

Ability to determine the starting point to initiate the simplex method.

Ability to write the linear program in tableau form.Ability to use the improvement mechanism for moving from a point to another point with a bettervalue of the objective function.

Improvement CriterionIn a maximization (minimization) problem, choose the nonbasic variable that has the most positive (negative)coefficient in the objective function of a canonical form. If that variable has a positive coefficient in some constraint, then a new basic feasible solution may be obtained by pivoting Ratio and Pivoting CriterionWhen improving a given canonical form by introducing variable x s into the basis, pivot in a constraint that

gives the minimum ratio of righthand-side coefficient to the corre sp onding x s c oe ffi cient. C o mpu te t hese rat io s only for constraints that have a positive coefficient for x s .Abilit to detect termination crite ria to indicate when a solution has been obtained.(FEUP | DEGI) Operations Research 2012/2013 21 / 103

Simplex method: A systematic procedure for solving linear
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Simplex method: A systematic procedure for solving linearprograms

The method proceeds by moving from one feasible solution to another, at each step improving

the value of the objective function and terminates after a nite number of such transitions.Two important characteristics of the simplex method:

The method is robust it solves any linear program;it detects redundant constraints in the problem formulation;it identies instances when the objective value is unbounded overthe feasible region;it solves problems with one or more optimal solutions;the method is also self-initiating: it uses itself either to generatean appropriate feasible solution, as required, to start the method,or to show that the problem has no feasible solution.

The method provides much more than just optimal solutions. .it indicates how the optimal solution varies as a function of theproblem data (cost coefficients, constraint coefficients, andrighthand-side data).it gives information intimately related with a linear program calledthe dual of the given problem: the simplex method automaticallysolves this dual problem along with the given problem.

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Production Planning at BALinear Programming Model


Subject to:50x VP + 30 x A 2000 (time in the oven)

6x VP + 5 x A 300 (space in the warehouse)3x VP + 5 x A 200 (capacity in the decoration sector)

x VP , x A 0

Adding slack variables s1 , s2 and s3 :

Objective:max 50x VP + 60 x A

Subject to:50x VP +30 x A + s1 = 2000

6x VP +5 x A + s2 = 3003x VP +5 x A + s3 = 200x VP , x A, s1 , s2 , s3 0

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Subject to:50x VP +30 x A + s1 = 20006x VP +5 x A + s2 = 3003x VP +5 x A + s3 = 200x VP , x A, s1 , s2 , s3 0

Basic (admissible) solution:

Non-basic variables: x VP = 0x A = 0

Basic variables:s1 = 2000s2 = 300s3 = 200

Z = 0

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0 10 20 30 40 50 60 70 80

50X VP + 30X A = 2000

Z = 50X VP + 60X A

3X VP + 5X A = 200

6X VP + 5X A = 300

X A

X VP

Objective: maximize Z ; choose the variablewith the most positive coecient in theobjective function: x As 3 is the rst variable that turns zero whenx A grows.

x A grows from 0 until 40, a value such thats 3 = 0; s 1 and s 2 remain 0

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Objective:max Z = 50 x VP +60 x A

= 50 x VP +60(40 35 x VP 15 x d )

= 2400 +14 x VP 12s 3

Subject to:32x VP + s1 6s 3 = 800

3x VP + s2 s 3 = 10035 x VP + xA +

15 s 3 = 40

x VP , x A, s1 , s2 , s3 0


Non-basic variables: x VP = 0s 3 = 0

Basic variables:s1 = 800s2 = 100xA = 40

Z = 2400

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30

40

50

60

70

80

0 10 20 30 40 50 60 70 80

50X VP + 30X A = 2000

Z = 50X VP + 60X A

3X VP + 5X A = 200

6X VP + 5X A = 300

X A

X VP

Objective: maximize Z ; choose the variablewith maximum coecient in the objectivefunction: x VP s 1 is the rst variable that turns zero whenx VP grows.

x VP grows from 0 until 25, a value such thats 1 = 0; x A and s 2 remain 0

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Objective:max Z = 2400 +14 x VP 12s 3

= 2400 +14(25 132 s 1 + 316 s 3) 12s 3

= 2750 716 s 1 758 s 3

Subject to:xVP + 132 s 1 316 s 3 = 25

332 s 1 + s2 516 s 3 = 25

+ xA 3160 s 1 7180 s 3 = 25

xVP , xA, s 1 , s2 , s 3 0


Non-basic variables: s

1 = 0s 3 = 0

Basic variables:xVP = 25s2 = 25xA = 25

Z = 2750

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30

40

50

60

70

80

0 10 20 30 40 50 60 70 80

50X VP + 30X A = 2000

Z = 50X VP + 60X A

3X VP + 5X A = 200

6X VP + 5X A = 300

X A

X VP

This solution is optimal because an increasein s 1 or s 3 reduction of Z.


Simplex Method


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Simplex MethodSteps of the algorithm

Initial Basic Admissible Solution (BAS) :Obtain an initial Basic Admissible Solution (BAS).

BAS is optimal? 2

Maximization if all the marginal costs are negative then the solution is optimal;Minimization if all the marginal costs are positive then the solution is optimal.If the BAS is not optimal then ITERATE.

ITERATE :1 Choose the variable that enters the basis: 3

Maximization the non-basic variable with the most positive marginal cost.

Minimization the non-basic variable with the most negative marginal cost.2 Choose the variable that enters the basis:

divide the RHS of all the equations by the positive coefficients of the variable that willenter the basis;the variable that will leave the basis will be the basic variable of the equation with thelowest ratio;if all the coefficients of the variable that will enter the basis are negative or zero thenthe solution is non-limited.

3 Change the lines in the tableau in order to obtain coecient equal to one for thepivotelement and zero for all the other elements of the column, including theobjective function.

4 BAS is optimal?2There are alternative optimal solutions when, in the optimal solution, one non-basic variable has a zero marginal

cost.

3If there is more than one variable that fullls the criterion choose one arbitrarilly .(FEUP | DEGI) Operations Research 2012/2013 27 / 103
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Production Planning at BASimplex method


Subject to:50x VP + 30 x A 2000

6x VP + 5 x A 3003x VP + 5 x A 200x VP , x A 0


Subject to:

50x VP + 30 x A+ s 1 = 20006x VP + 5 x A+ s 2 = 3003x VP + 5 x A+ s 3 = 200

x VP , x A, s 1, s 2 , s 3 0

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Production Planning at BASimplex method

x VP x A s 1 s 2 s 3s 1 50 30 1 0 0 2000s 2 6 5 0 1 0 300

s 3 3 5 0 0 1 200 Z 50 60 0 0 0 0

x VP x A s 1 s 2 s 3 s 1 1605 0 1 0

305 800

s 2 3 0 0 1 1 100x A 35 1 0 0

15 40

Z 14 0 0 0 12 2400

x VP x A s 1 s 2 s 3x VP 1 0 5160 0

316 25

s 2 0 0 15160 1 716 25

x A 0 1 3160 0 5

16 25 Z 0 0 716 0

758 2750

x A enters the basis because:max(50 , 60) = 60s 3 leaves the basis because:min( 200030 ,

3005 ,

2005 ) =

2005

x VP enters the basis because:max(14) = 14s 1 leaves the basis because:min( 800150

5, 1003 ,

4035

) = 8001505

All the coefficients of the objective functionare 0.Optimal solution:(x VP , x A) = (25 , 25) and Z = 2750What does s 2 = 25 mean?


Finding an initial Basic Admissible Solution (BAS)
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g ( )

An important condition for the Simplex method is the availability of an initial BasicAdmissible Solution in the canonical form.Sometimes this initial BAS is not evident or it may evn not exist (and sometimes there is noBasic Admissible Solution!).Solutions:

Trial and error solve the system for different sets of variables, reduce it to the canonical form andtest if the resulting solution is admissible.Using articial variables.

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3. Thearticial problem will only be equivalent to the original one if all the articial variableshave value zero.

Objective all the articial variables must leave the basis.

1

Two-fase method2 Big Mmethod


Big M method
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Assign the articial variables a very high cost ( M) (minimization problem) in the objectivefunction.The simplex method will take care, by improving the objective function, to expel the arcicialvariables from the basis articial variables equal to zero.

min n j =1 c j x j + mi = k +1 My i Subject to:

n

j =1aij x j = b i i {1,..., k }

n

j =1a ij x j + y i = b i i {k +1 ,..., m}

x j 0 j {1,..., n}y i 0 i {k +1 ,..., m}


Big M method
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example

Objective:min Z = 3x 1 + x 2 + x 3Subject to:

x 1 2x 2 + x 3 + s 1 = 11 4x 1 + x 2 + 2 x 3 s 2 = 3 2x 1 + x 3 = 1

x 1 , x 2 , x 3 , s 1 , s 2 0

Objective:min Z = 3x 1 + x 2 + x 3 + My1 + My2Subject to:

x 1 2x 2 + x 3 + s 1 = 11 4x 1 + x 2 + 2 x 3 s 2 + y1 = 3 2x 1 + x 3 + y2 = 1

x 1 , x 2 , x 3 , s 1 , s 2 , y1 , y2 0

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Z = 3x 1 + x 2 + x 3 + M (3 + 4 x 1 x 2 2x 3 + s 2) + M (1 + 2 x 1 x 3)= 4 M + ( 3 + 6 M )x 1 + (1 M )x 2 + (1 3M )x 3 + Ms 2

x 1 x 2 x 3 s 1 s 2 y 1 y 2s 1 1 2 1 1 0 0 0 11y 1 4 1 2 0 1 1 0 3

y 2 2 0 1 0 0 0 1 1 Z 3 1 1 0 0 0 0 0

6M M 3M 0 M 0 0 4M

x 1 x 2 x 3 s 1 s 2 y 1 y 2s 1 3 2 0 1 0 0 1 10

y 1 0 1 0 0 1 1 2 1

x 3 2 0 1 0 0 0 1 1 Z 1 1 0 0 0 0 1 10 M 0 0 M 0 3M M

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x 1 x 2 x 3 s 1 s 2 y 1 y 2

s 1 3 0 0 1 2 2 5 12x 2 0 1 0 0 1 1 2 1x 3 2 0 1 0 0 0 1 1

Z 1 0 0 0 1 1 +1 20 0 0 0 0 M M 0

x 1 x 2 x 3 s 1 s 2x 1 1 0 0 13

23 4

x 2 0 1 0 0 1 1x 3 0 0 1 23

43 9

Z 0 0 0 1313 2

Optimal solution: ( x 1 , x 2 , x 3 , s 1 , s 2) = (4 , 1, 9, 0, 0) with Z = 2


Some remarks
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The articial variables are only used to serve as basic variables in a given equation. Once replacedin the base by the original variables, the articial variables can be eliminated from the simplextableau (eliminating the respective columns).

If, in the optimal simplex tableau, some articial variables have a value > 0, it means that theoriginal problem has no admissible solution, and therefore it is an impossible problem.

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Sensitivity Analysisbased on:Bradley, Hax, and Magnanti; AppliedMathematical Programming, Addison-Wesley,1977Chapter 3, sections 3.1, 3.2, 3.3, 3.4 and 3.5


Sensitivity Analysis
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We have already been introduced to sensitivity analysis via the geometry of a simple example. Wesaw that the values of the decision variables and those of the slack and surplus variables remainunchanged even though some coefficients in the objective function are varied. We also saw thatvarying the righthand-side value for a particular constraint alters the optimal value of theobjective function in a way that allows us to impute a per-unit value, or shadow price, to thatconstraint. These shadow prices and the shadow prices on the implicit nonnegativity constraints,called reduced costs, remain unchanged even though some of the righthand-side values are varied.Since there is always some uncertainty in the data, it is useful to know over what range and underwhat conditions the components of a particular solution remain unchanged. Further, thesensitivity of a solution to changes in the data gives us insight into possible technologicalimprovements in the process being modeled. For instance, it might be that the available resourcesare not balanced properly and the primary issue is not to resolve the most effective allocation of these resources, but to investigate what additional resources should be acquired to eliminatepossible bottlenecks. Sensitivity analysis provides an invaluable tool for addressing such issues.

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Production Planning at BA with Favaios Bottles

The factory of Barbosa & Almeida (BA) in Avintes produces glass containers through mouldinjection.BA recently won two new clients and intends to assign this production to one of its furnaces. Theorders were of bottles of Sandeman Ruby Port (75cl), one liter bottles of olive oil Oliveira da Serraand one liter bottles of Favaios Moscatel. The customers buy all the bottles that BA can produce.Due to differences in the production (number of cavities and different cycle times), each batch of

portwine bottles takes 50 hours to produce, whereas each batch of oil bottles requires only 30hours and a batch of favaios takes 50 hours to produce. There are a total of 2000 hours availableat the oven.Moreover, there are restrictions in the storage capacity of the bottles prior to their dispatch. Thewarehouse has 300 m3 available and each batch of portwine bottles occupies 6 m3 of warehousespace, each batch of bottles of olive oil holds 5 m3 and each batch of favaios holds 6 m3 .Finally we must also consider the capacity available in the decoration sector (e.g. labels,packaging), which is 200 hours. The portwine bottles spend 3 hours per batch, the olive oilbottles 5 hours per batch and the favaios bottles also 5 hours per batch.Barbosa & Almeida wants to maximize the prot from these two orders, knowing that the protper batch is 50 e 60e and 20e , respectively.

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Production Planning at BA with Favaios Bottles

Decision variablesx VP Number of batches of Vinho do Porto Ruby Sandeman bottles to produce;

x A Number of batches of Azeite Oliveira da Serra bottles to produce;x F Number of batches of Favaios bottles to produce.

Objective:max Z = 50 x VP + 60 x A + 20 x F

Subject to:50x VP +30 x A +50 x F 2000 (time in the oven)

6x VP +5 x A +6 x F 300 (space in the warehouse)3x VP +5 x A +5 x F 200 (capacity in the decoration sector)x VP , x A, x F 0

In this model x VP , x A and x F may not be integer.

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Some remarks about the sensitivity analysis


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We wish to analyze the effect on the optimal solution of changing various elements of theproblem data without re-solving the linear program or having to remember any of theintermediate tableaux generated in solving the problem by the simplex method.

The type of results that can be derived in this way are conservative, in the sense that theyprovide sensitivity analysis for changes in the problem data small enough so that the samedecision variables remain basic, but not for larger changes in the data.

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Shadow PricesReduced Costs

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Production Planning at BAProducing Favaios Bottles

x VP x A x F s 1 s 2 s 3x VP 1 0 58

132 0

316 25

s 2 0 0 78 332 1

716 25

x A 0 1 58 3160 0

516 25

Z 0 0 48 34 716 0 9

38 2750

The complete optimal solution for the problem is

(x VP , x A, x F , s 1 , s 2, s 3) = (25 , 25, 0, 0, 25, 0) and Z = 2750

The binding constraints in this optimal solution are therefore:the rst one (time in the oven, s 1 = 0);

the third one (capacity in the decoration sector, s 3 = 0);and x F = 0.

50x VP +30 x A +50 x F = 20003x VP +5 x A +5 x F = 200

x F = 0

(x VP , x A, x F ) = (25 , 25, 0)Z = 2750

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Shadow PriceDefinitionThe shadow price associated with a particular constraint is thechange in the optimal value of the objective function per unitincrease in the righthand-side value of that constraint, all otherproblem data remaining unchanged.

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Production Planning at BAProducing Favaios Bottles Shadow Prices

What will be the increase of the prot per unit increase in the oven time?

50x VP +30 x A +50 x F = 20013x VP +5 x A +5 x F = 200

x F = 0

50x VP +30 x A = 20013x VP +5 x A = 200x F = 0

50x VP +30 x A = 200118x VP +30 x A = 1200

x F = 0

x VP = 25 132x A = 24 157160

x F = 0

(x VP , x A, x F ) = (25 132 , 24157160 , 0)

Z = 50 25 132 + 60 24157160 + 20 0 = 2750

716

Prot increase: 2750 716 2750 = 716

x VP x A x F s 1 s 2 s 3

x VP 1 0 5

81

32 0 316 25

s 2 0 0 78

332 1

716 25

x A 0 1 58 3160 0 516 25

Z 0 0 48 34 716 0 9

38 2750

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Production Planning at BAProducing Favaios Bottles Shadow Prices

What will be the increase of the prot per unit increase in the decoration capacity?50x VP +30 x A +50 x F = 2000

3x VP +5 x A +5 x F = 201x F = 0

50x VP +30 x A = 2000

3x VP +5 x A = 201x F = 0

50x VP +30 x A = 200018x VP +30 x A = 1206

x F = 0

x VP = 24 1316

x A = 25 83240

x F = 0

(x VP , x A, x F ) = (24 1316 , 25 516 , 0)

Z = 50 24 1316 + 60 516 + 20 0 = 2759

38

Prot increase: 2759 38 2750 = 9 38


x VP 1 0 5

81

32 0 316 25

s 2 0 0 78

332 1

716 25

x A 0 1 58 3160 0

516 25

Z 0 0 48 34 716 0 9

38 2750

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Reduced CostDefinitionThe reduced cost associated with the nonnegativity constraint foreach variable is the shadow price of that constraint (i.e., thecorresponding change in the objective function per unit increase inthe lower bound of the variable).

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Production Planning at BAProducing Favaios Bottles Reduced costs

What will be the reduction of the prot if we must produce at least one batch of favaios bottles?

50x VP +30 x A +50 x F = 20003x VP +5 x A +5 x F = 200

x F = 1

50x VP +30 x A +50 = 20003x VP +5 x A +5 = 200x F = 1

50x VP +30 x A = 19503x VP +5 x A = 195

x F = 1

x VP = 195 / 8x A = 195 / 8

x F = 1

(x VP , x A, x F ) = ( 1958 , 195

8 , 1)

Z = 50 1958 + 60 195

8 + 20 1 = 2701 , 25Prot reduction:2750 2701 , 25 = 48 , 75 = 48 34


x VP 1 0 5

81

32 0 316 25

s 2 0 0 78

332 1

716 25

x A 0 1 5

8 3160 0

516 25

Z 0 0 48 34 716 0 9

38 2750

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Production Planning at BAProducing Favaios Bottles Reduced costs

What will be the reduction of the prot if we must produce at least one batch of favaios bottles?

x VP x A x F s 1 s 2 s 3s 1 50 30 50 1 0 0 2000s 2 6 5 6 0 1 0 300s 3 3 5 5 0 0 1 200

Z 50 60 20 0 0 0 0

x VP x A x F s 1 s 2 s 3x VP 1 0 58

132 0

316 25

s 2 0 0 78 332 1 716 25x A 0 1 58

3160 0

516 25

Z 0 0 48 34 716 0 9

38 2750

We can compute the reduced cost of x F in a different way by computing theoportunity cost for producing 1 batch of favaios bottles:50 716 + 6 0+5 9

38 = 21

78 + 46

78 =

68 34As the revenue of 1 batch of favaiosbottles is 20, the prot will change by20 68 34 = 48

34


General DiscussionFundamental relationship between shadow prices, reduced

t d th bl d t
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costs, and the problem data.

Shadow x 1 . . . x n s 1 . . . s m price

s 1 a11 . . . a1n 1 . . . 0 b 1 y 1...

......

......

......

s m am1 . . . amn 0 . . . 1 b m y m Z c 1 . . . c n 0 . . . 0 0

At the nal tableau:

Z c 1 . . . c n c n+1 . . . c n+ m z 0= . . . = y 1 . . . y m

Recall that, at each iteration of the simplex method, the objective function is transformed by subtracting fromit a multiple of the row in which the pivot was performed. Consequently, the nal form of the objective functioncould be obtained by subtracting multiples of the original constraints from the original objective function.Consider rst the nal objective coefficients associated with the original basic variables: s 1 , . . . , s m . Let 1 ,. . . , m be the multiples of each row that are subtracted from the original objective function to obtain its nalform. Since s i appears only in the i th constraint and has a +1 coefficient we should havec n+ i = 0 1 i = i = y i . Thus the shadow prices are the multiples i .Since these multiples can be used to obtain every objective coefficient in the nal form, the reduced cost c j of variable x j is given by c j = c j mi =1 aij y i ( j = 1 . . . n).Since c j = 0 for the m basic variables: 0 = c j mi =1 aij y i (for j basic). This is a system of m equations in munknowns that uniquely determines the values of the shadow prices y i .The current value of the objective function is z 0 = mi =1 b i y i , z 0 =

mi =1 b i y i

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Variation in the objective coefficientsHow much the objective-function coefficients can vary withoutchanging the values of the decision variables in the optimal solution?

We will make the changes one at a time, holding all othercoefficients and righthand-side values constant.

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Production Planning at BAProducing Favaios BottlesVariation in the objective coefficient of x

F

Suppose that we increase the objective function coefficient of x F in the original problemformulation by c F :In applying the simplex method, multiples of the rows were subtracted from the objective functionto yield the nal system of equations. Therefore, the objective function in the nal tableau willremain unchanged except for the addition of c F x F .

x VP x A x F s 1 s 2 s 3s 1 50 30 50 1 0 0 2000s 2 6 5 6 0 1 0 300s 3 3 5 5 0 0 1 200

Z 50 60 20 + c F 0 0 0 0

x VP x A x F s 1 s 2 s 3x VP 1 0 58

132 0

316 25

s 2 0 0 78 332 1

716 25

x A 0 1 58 3160 0

516 25

Z 0 0 48 34 + c F 716 0 9

38 2750

x F will become a candidate toenter the basis, only when itsobjective-function coefficient ispositive.The optimal solution remains un-

changed as long as: 48 34 + c F 0

c F 48 34

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Production Planning at BAProducing Favaios BottlesVariation in the objective coefficient of x

VP

x VP x A x F s 1 s 2 s 3s 1 50 30 50 1 0 0 2000s 2 6 5 6 0 1 0 300s 3 3 5 5 0 0 1 200

Z 50 60 20 0 0 0 0+ c

VP

x VP x A x F s 1 s 2 s 3x VP 1 0 58

132 0

316 25

s 2 0 0 78 332 1

716 25

x A 0 1 58 3160 0

516 25

Z c VP 0 48 34 716 0 9

38 2750

x VP x A x F s 1 s 2 s 3x VP 1 0 581

32 0 316 25

s 2 0 0 78 332 1

716 25

x A 0 1 58 3160 0

516 25

Z 0 0 48 34 716 0 9

38 2750

0 0 58 c VP 132 c VP 0 +

316 c VP 25 c VP

The current solution re-mains unchanged while:

48 34 58 c VP 0

( c VP 78) 716 132 c VP 0

( c VP 14) 9 38 +

316 c VP 0

( c VP 50)

Taking the most limiting in-equalities, the bounds on c VP are:

14 c VP 50 14+50 c new VP 50+50

36 c new VP 100


General DiscussionRanges of the cost coefficients in the optimal solution.
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s 1 a11 . . . a1n 1 . . . 0 b 1 y 1...

.

.....

.

.....

.

.....


At the nal tableau: Z c 1 . . . c n c n+1 . . . c n+ m z 0

If x j is a non-basic variable at the final tableau with objective function coefficient c j that waschanged to c new j = c j + c j , then the current solution remains unchanged so long as the new reducedcost c new j remains nonnegative, that is, c

new j = c j + c j

mi =1 aij y i = c j + c j 0.

The range on the variation of the objective-function coefficient of a nonbasic variable is then given by: < c j < c j ; ( < c j + c j = c new j < c j c j ).

If x r is a basic variable at the final tableau with objective function coefficient c r that was changed toc new r = c r + c r , then c

new r = c r + c r

mi =1 aij y i = c r + c r .

Since x r is a basic variable, c r = 0. So, to recover a canonical form with c new r = 0, we subtract c r times the rth constraint in the nal tableau from the nal form of the objective function, obtaining newreduced costs for all nonbasic variables: c new j = c j a rj c r , where arj is the coefficient of variable x j inthe rth constraint in the nal tableau.For all basic variables c new j = 0 and the current basis remains optimal if c

new j 0.

The range on the variation of the objective-function coefficient is:

Max j c j arj

| a rj > 0 c r Min j c j arj

| a rj < 0

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Variations on the righthand-side values


P d i Pl i BA
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Production Planning at BAProducing Favaios BottlesVariations on the righthand-side values

Varying the righthand-side value of a constraint that is non-binding in the optimal solution:the warehouse constraint.

x VP

x A

x F

s 1

s 2

s 3s 1 50 30 50 1 0 0 2000

s 2 6 5 6 0 1 0 300 + b 2s 3 3 5 5 0 0 1 200

Z 50 60 20 0 0 0 0


x VP 1 0 5

81

32 0 316 25s 2 0 0 78

332 1

716 25 + b 2

x A 0 1 58 3160 0

516 25

Z 0 0 48 34 716 0 9

38 2750

We add an amount b 2 to the right-hand side of the warehouse constraint.s 2 was basic in the initial tableau and isbasic in the optimal tableau.

6 25+5 25+6 0+ s 2 = 300+ b 2s 2 = 25 + b 2

In order to keep the solution feasible:

25 + b 2 0 b 2 25

b new 2 = 300 + b 2 275


P d ti Pl i t BA
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Production Planning at BAProducing Favaios BottlesVariations on the righthand-side valuesVarying the righthand-side value of a constraint that is binding in the optimal solution:the decoration constraint.

We add an amount b 3 to the righthand side

of the decoration constraint.This is equivalent to decreasing the value of theslack variable s 3 by b 3 , that is substituting s 3by s 3 b 3 in the original problem formulation.s 3 , which is zero in the nal solution, is changedto s 3 = b 3

x VP = 25 316 b 3 0 that is b 3 25 16

3

s 2 = 25 716 b 3 0 that is b 3

25 16

7x A = 25 + 516 b 3 0 that is b 3 5 1625 16

7 b 3 80200 + 25 167 b

new 3 200 80

x VP x A x F s 1 s 2 s 3s 1 50 30 50 1 0 0 2000s 2 6 5 6 0 1 0 300s 3 3 5 5 0 0 1 200 + b 3

Z 50 60 20 0 0 0 0

x VP x A x F s 1 s 2 s 3x VP 1 0

58

132 0

316 25

316 b 3

s 2 0 0 78

332 1

716 25

716 b 3

x A 0 1 5

8 3160 0

516 25 +

516 b 3

Z 0 0 48 34 716 0 9

38 2750 9

38 b 3

For b new 3 < 200 80 ( b 3 < 80) the basic variable x Abecomes negative.

What variable should enter the basis to take its place?In order for the new basis to be an optimal solution, the en-tering variable must be chosen so that the reduced costs arenot allowed to become positive.



58

132 0

316 25

316 b 3

s 2 0 0 1 58

332 1

716 25

716 b 3
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2 8 32 16 16 3x A 0 1

58

3160 0

516 25 +

516 b 3

Z 0 0 48 34 716 0 9

38 2750 9

38 b 3

The nal tableau for b new 3

= 120 ( b 3 = 80) will be:


58

132 0

316 40

s 2 0 0 78

332 1

716 60

x A 0 1 5

8 3160 0

516 0

Z 0 0 48 34 716 0 9

38 2480

For b new 3 < 120 the basic variable x A becomes negative.

What variable should enter the basis to take its place?In order for the new basis to be an optimal solution, the entering variable must be chosen so that the reducedcosts are not allowed to become positive.Regardless of which variable enters the basis, the entering variable will be isolated in row 3 of the nal tableauto replace x A, which leaves the basis.To isolate the entering variable, we must perform a pivot operation, and a multiple, say t, of row 3 in the naltableau will be subtracted from the objective-function row.


x VP 1 0 5

81

32 0 316 40

s 2 0 0 78

332 1

716 60

x A 0 1 5

8 3160 0

516 0

Z 0 0 48 34 716 0 9

38 2480

t t 58 + t 3160 t

516 t 0

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x VP x A x F s 1 s 2 s 3x VP 1 0 58

132 0

316 40

s 2

0 0 78

332

1 716

60x A 0 1 58

3160 0

516 0

Z 0 0 48 34 716 0 9

38 2480

t t 58 + t 3160 t

516 t 0

t 0 that is t 0 48 34 t

58 0 that is t

2885

716

+ t 3160

0 that is t 703 9 38 t

516 0 that is t 30

0 t 703Since the coefficient of s 1 is most constraining on t, s 1 will enter the basis. Note that the rangeon the righthand-side value and the variable transitions that would occur if that range wereexceeded by a small amount are easily computed. However, the pivot operation actually

introducing s 1 into the basis and eliminating x A need not be performed.


General DiscussionVariations in the Righthand-Side values.
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x 1 . . . x n s 1 . . . s ms 1 a11 . . . a1n 1 . . . 0 b 1

......

......

......

s m am1 . . . amn 0 . . . 1 b m Z c 1 . . . c n 0 . . . 0 0

At the nal tableau:x 1 . . . x n s 1 . . . s m

s 1 a11 . . . a1n 11 . . . 1m b 1...

......

......

...s m am1 . . . amn m1 . . . mm b m

Z c 1 . . . c n c n+1 . . . c n+ m z 0As this is a canonical form, aij and ij in the nal tableau will be structured so that one basic variable isisolated in each contraint.We can change the coefficient b k in the kth righthand-side by b k with all the other data held xed, simply bysubstituting s k b k for s k in the initial tableau. To see how this change affects the updated righthand-sidecoefficients, we make the same substitution in the nal tableau. The terms ik s k become ik (s k b k ) = ik s k ik b k . As ik b k is a constant, we move it to the righthand side to give modiedrighthand-side values b i + ik b k for (i = 1 , 2, . . . , m).


As long as all of these values are nonnegative the basis specied by the nal tableau remains
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As long as all of these values are nonnegative, the basis specied by the nal tableau remainsoptimal, since the reduced costs have not been changed. Consequently, the current basis isoptimal whenever b i + ik b k 0 for (i = 1 , 2, . . . , m) or equivalently,

Max i b

i ik | ik > 0 b k Mini b

i ik | ik < 0When b k reaches either its upper or lower bound any further increase (or decrease) in its valuemakes one of the updated righthand sides, say b r + rk b k negative. At this point the basicvariable in row r leaves the basis, and we must pivot in row r of the nal tableau to nd thevariable to be introduced in its place. Since pivoting subtracts a multiple t of the pivot row fromthe objective equation, the new objective equation has coefficients c j ta rj ( j = 1 , 2, . . . , n).For the basis to be optimal, each of these coefficients must be nonpositive. Since c

j = 0 for the

basic variable being dropped and since its coefficient in costraint r is arj = 1, we must have t 0.For any nonnegative t, the updated coefficient c j ta rj for any other variable remains nonpositiveif arj 0. Consequently we need only to consider arj < 0, and t is given by:t = Min j

c j arj

| a rj < 0The index u giving this minimum has c u ta ru = 0, and the corresponding variable x u canbecome the new basic variable in row r by pivoting on aru . Note that the pivot is made on anegative coefficient.

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Alternative optimal solutionsAs in the case of the objective function and righthand-side ranges,the nal tableau of the linear program tells us somethingconservative about the possibility of alternative optimal solutions.

If all reduced costs of the nonbasic variables are strictly negative(positive) in a maximization (minimization) problem, then there isno alternative optimal solution, because introducing any variableinto the basis at a positive level would reduce (increase) the value of the objective function.

If one or more of the reduced costs are zero, there may exist

alternative optimal solutions.

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x VP 1 0 581

32 0 316 40

s 2 0 0 78 332 1

716 60

x A 0 1 5

8 3160 0

516 0

Z 0 0 48 34 716 0 9

38 2480

Since the righthand-side value in row 3 is zero, it is possible to take x A (basic variable in row 3)out of the basis as long as there is a variable to introduce into the basis.The candidates to be introduced into the basis are x F , s 1 and s 3 , the one that corresponds to:

Min j c j arj | arj < 0 = Min 716

3160 s 1 will enter the basis.

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Duality in Linear Programmingbased on:Bradley, Hax, and Magnanti; AppliedMathematical Programming, Addison-Wesley,

1977Chapter 4, 4.1, 4.2, 4.3, 4.4, 4.5, 4.6 and 4.7


Duality in Linear Programming
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In the preceding chapter on sensitivity analysis, we saw that the shadow-price interpretation of the

optimal simplex multipliers is a very useful concept.First, these shadow prices give us directly the marginal worth of an additional unit of any of theresources.Second, when an activity is priced out using these shadow prices, the opportunity cost of allocating resources to that activity relative to other activities is determined.Duality in linear programming is essentially a unifying theory that develops the relationshipsbetween a given linear program and another related linear program stated in terms of variableswith this shadow-price interpretation.The importance of duality is twofold:First, fully understanding the shadow-price interpretation of the optimal simplex multipliers canprove very useful in understanding the implications of a particular linear-programming model.Second, it is often possible to solve the related linear program with the shadow prices as thevariables in place of, or in conjunction with, the original linear program, thereby taking advantageof some computational efficiencies.

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Duality in Linear ProgrammingLearning Objectives

Write the dual problem of a linear programming problem.

Understand the:

weak duality property,optimality property,unboundedness property,strong duality property.

Verify that the complementary slackness conditions hold for a given problem.

Apply the dual simplex method.

Apply the parametric primal-dual algorithm.

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A Preview of Duality


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g

The factory of Barbosa & Almeida (BA) in Avintes produces glass containers through mouldinjection.BA recently won two new clients and intends to assign this production to one of its furnaces. Theorders were of bottles of Sandeman Ruby Port (75cl), one liter bottles of olive oil Oliveira da Serraand one liter bottles of Favaios Moscatel. The customers buy all the bottles that BA can produce.Due to differences in the production (number of cavities and different cycle times), each batch of portwine bottles takes 50 hours to produce, whereas each batch of oil bottles requires only 30hours and a batch of favaios takes 50 hours to produce. There are a total of 2000 hours availableat the oven.Moreover we must also consider the capacity available in the decoration sector (e.g. labels,packaging), which is 200 hours. The portwine bottles spend 3 hours per batch, the olive oilbottles 5 hours per batch and the favaios bottles also 5 hours per batch.

Barbosa & Almeida wants to maximize the prot from these two orders, knowing that the protper batch is 50 e 60e and 20e , respectively.


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Decision variablesx VP Number of batches of Vinho do Porto Ruby Sandeman bottles to produce;

x A Number of batches of Azeite Oliveira da Serra bottles to produce;x F Number of batches of Favaios bottles to produce.


Subject to:50x VP +30 x A +50 x F 2000 (time in the oven)

3x VP +5 x A +5 x F 200 (capacity in the decoration sector)x VP , x A, x F 0

In this model x VP , x A and x F may not be integer.


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Shadow prices

x VP x A x F s O s D s O 50 30 50 1 0 2000s D 3 5 5 0 1 200

Z 50 60 20 0 0 0

At the nal tableau:x VP x A x F s O s D

x VP 1 0 581

32 316 25

x A 0 1 5

8 3160

516 25

Z 0 0 48 34 716 9

38 2750

The shadow prices, y O for the oven ca-pacity and y D for the decoration ca-

pacity can be determined from the naltableau as the negative of the reducedcosts associated with the slack variabless O and s D .Thus these shadow prices are y O = 716and y D = 9 38


Economic properties of the shadow prices associated withthe resources

Shadow
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s 1 a11 . . . a1n 1 . . . 0 b 1 y 1...

......

......

......


At the nal tableau: Z c 1 . . . c n c n+1 . . . c n+ m z 0

Reduced costs in terms of shadow prices:c j = c j mi =1 aij y i ( j = 1 . . . n)

aij is the amount of resource i used per unit of activity j , and y i is the imputed value of thatresource.The term

mi =1 aij y i is the total value of the resources used per unit of activity j and is thus the

marginal resource cost per unit of activity j .If we think of the objective coefficients c j as being marginal revenues, the reduced costs

c j = c j mi =1 aij y i ( j = 1 . . . n)

are simply net marginal revenues (i.e., marginal revenue minus marginal cost).


Production Planning at BAR d d ( i l ) f h f
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Reduced cost (net marginal revenue) for each type ofbottle (decision variable)


Z 50 60 20 0 0 0At the nal tableau:

x VP x A x F s O s D

x VP 1 0 58 132 316 25

x A 0 1 58 3160

516 25

Z 0 0 48 34 716 9

38 2750

For the basic variables the reduced costs arezero.The marginal revenue is equal to themarginal cost for these activities.For all nonbasic variables the reduced costsare negative.The marginal revenue is less than themarginal cost for these activities, so theyshould not be pursued.

Shadow price for the oven capacity: y O = 716 .Shadow price for the decoration capacity: y D = 9 38 .Net marginal revenue for the production of portwine bottles:c VP = c VP mi =1 aiVP y i = 50 (50

716 + 3 9

38 ) = 0

Net marginal revenue for the production of azeite bottles:c A = c A mi =1 aiAy i = 60 (30

716 + 5 9

38 ) = 0

Net marginal revenue for the production of favaios bottles:c F = c F mi =1 aiF y i = 20 (50

716 + 5 9

38 ) = 48

34


Production Planning at BASh d i ( i ) i d i h h
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Shadow prices (opportunity costs) associated with theconsumption of the resources (constraints)


Z 50 60 20 0 0 0At the nal tableau:

x VP x A x F s O s D

x VP 1 0 581

32 316 25

x A 0 1 58 3160

516 25

Z 0 0 48 34

716

9 38

2750

If we value the total resources at the shadow prices, we nd their value is exactly equal to theoptimal value of the objective function:

2000 716 + 200 938 = 2750


Production Planning at BAC h d i b d t i d di tl ?
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Can shadow prices be determined directly?

For a maximization decision problem, the shadow prices must satisfy the requirement thatmarginal revenue be less than or equal to marginal cost for all activities.The shadow prices must be nonnegative since they are associated with less-than-or-equal-toconstraints in a maximization decision problem.Imagine that BA does not own its productive capacity but has to rent it. In this case the

shadow prices could be seen as rent rates.


Subject to:

50x VP + 30 x A + 50 x F 20003x VP + 5 x A + 5 x F 200

x VP , x A, x F 0

Objective:min V = 2000 y O + 200 y D

Subject to:

50y O + 3 y D 5030y O + 5 y D 6050y O + 5 y D 20

y O , y D 0


Production Planning at BAPrimal and Dual Decision Problems
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Primal and Dual Decision Problems

PRIMAL Decision Problem:Objective:

max Z = 50 x VP + 60 x A + 20 x F Subject to:

50x VP + 30 x A + 50 x F 20003x VP + 5 x A + 5 x F 200

x VP , x A, x F 0

Optimal solution:x VP = 25x A = 25x F = 0Z = 2750with shadow prices:y O = 716y D = 9 38

DUAL Decision Problem:Objective:

min V = 2000 y O + 200 y D Subject to:50y O + 3 y D 50

30y O + 5 y D 6050y O + 5 y D 20

y O , y D 0

Optimal solution:y O = 716y D = 9 38V = 2750with shadow prices:x VP = 25x A = 25x F = 0


Production Planning at BARequirements of the optimality conditions of the Simplex
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Requirements of the optimality conditions of the SimplexMethod

The reduced costs of the basicvariables are zero.If a decision variable of the primal is positive, then thecorresponding constraint in the dual must hold withequality.

If x VP 0 then c VP = 50 (50y O + 3 y D ) = 0

If x A 0 then c A = 60 (30y O + 5 y D ) = 0

The variables with negativereduced costs are nonbasicvariables (zero valued)If a constraint holds as a strict inequality, then thecorresponding decision variable must be zero.

If c F = 20 (50y O + 5 y D ) < 0 then x F = 0

If a shadow price is positive,then the correspondingconstraint must hold withequality.

If y O > 0 then 50 x VP + 30 x A + 50 x F = 2000If y D > 0 then 3x VP + 5 x A + 5 x F = 200


Complementary Slackness Conditions
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If a primal variable ispositive,then its corresponding(complementary) dualconstraint holds withequality.

If a dual constraint holdswith strict inequality,then its corresponding(complementary) primalvariable must be zero.

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Finding the dual in generalDenition of the dual problem


PRIMAL is a maximization problem
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PRIMAL

Objective:max Z =

n

j =1c j x j

Subject to:n

j =1

aij x j b i (i = 1 , 2, . . . , m )

n

j =1aij x j b i (i = m , 2, . . . , m )

n

j =1aij x j = b i (i = m , 2, . . . , m)

x j 0 ( j = 1 , 2, . . . , n)

DUAL

Objective:min V =

m

i =1b i y i +

m

i = m +1

b i y i +m

i = m +1

b i y i

Subject to:

m

i =1aij y i +

m

i = m +1aij y i +

m

i = m +1aij y i c j ( j = 1 , 2, . . . , n)

y i 0 (i = 1 , . . . , m )

y i 0 (i = m + 1 , . . . , m )

y i R (i = m + 1 , . . . , m)


PRIMAL is a minimization problem
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PRIMAL

Objective:min Z =

n

j =1c j x j

Subject to:n

j =1

aij x j b i (i = 1 , 2, . . . , m )

n

j =1aij x j b i (i = m , 2, . . . , m )

n

j =1aij x j = b i (i = m , 2, . . . , m)

x j 0 ( j = 1 , 2, . . . , n)

DUAL

Objective:max V =

m

i =1b i y i +

m

i = m +1

b i y i +m

i = m +1

b i y i

Subject to:

m

i =1aij y i +

m

i = m +1aij y i +

m

i = m +1aij y i c j ( j = 1 , 2, . . . , n)

y i 0 (i = 1 , . . . , m )

y i 0 (i = m + 1 , . . . , m )

y i R (i = m + 1 , . . . , m)


Summary of the correspondences between PRIMAL andDUAL decision problems
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Primal (Maximize) Dual (Minimize)i th constraint i th variable 0i th constraint i th variable 0i th constraint = i th variable unrestricted

j th variable 0 j th constraint j th variable 0 j th constraint j th variable unrestricted j th constraint =

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The fundamental dualityproperties


Weak duality property

PRIMAL DUAL
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PRIMALObjective:

max Z = n j =1 c j x j Subject to:n

j =1a ij x j b i (i = 1 , 2, . . . , m)

x j 0 ( j = 1 , 2, . . . , n)

Objective:

min V = mi =1 b i y i Subject to:m

i =1aij y i c j ( j = 1 , 2, . . . , n)

y i 0 (i = 1 , . . . , m)

n j =1 c j x j

mi =1 b i y i

If x j , j = 1 , 2, . . . , n is a feasible solution to

the primal problemand y i , i = 1 , 2, . . . , m is a feasible solutionto the dual problem.

DUAL feasible V decreasing

PRIMAL feasible Z increasing


Optimality property

PRIMAL DUAL


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PRIMALObjective:

max Z = n j =1 c j x j Subject to:n

j =1a ij x j b i (i = 1 , 2, . . . , m)

x j 0 ( j = 1 , 2, . . . , n)

Objective:min V = mi =1 b i y i Subject to:

m

i =1aij y i c j ( j = 1 , 2, . . . , n)

y i 0 (i = 1 , . . . , m)

If x j , j = 1 , 2 , . . . , n is a feasible solution to the primal problemand y i , i = 1 , 2 , . . . , m is a feasible solution to the dual problemand n j =1 c j x j =

mi =1 b i

y i

then x j , j = 1 , 2, . . . , n is an optimal solution for the primal problemand y i , i = 1 , 2, . . . , m is an optimal solution for the dual problem


Unboundedness Property
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If the PRIMAL problem hasan unbounded solutionthen the DUAL problem isinfeasible

If the DUAL problem has anunbounded solutionthen the PRIMAL problem isinfeasible

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The complementary slackness


Complementary Slackness Property

I ti l l ti f li

For the PRIMAL linear programposed as a maximization problem with
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In an optimal solution of a linear program: posed as a maximization problem withless-than-or-equal-to constraints, thismeans:

If the value of the dual variable(shadow price) associated with aconstraint is nonzero, then thatconstraint must be satisfied with

equality.

If y i > 0, then n j =1 aij x j = b i

If a constraint is satisfied withstrict inequality, then itscorresponding dual variable must be

zero.

If n j =1 aij

x j < b i then

y i = 0

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Optimality conditions


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If x j , j = 1 , 2 , . . . , n is a feasible solution to thePRIMAL problemand y i , i = 1 , 2 , . . . , m , is a feasible solution to theDUAL problem,then they are optimal solutions to these problems if,and only if, the complementary-slackness conditionshold for both the primal and the dual problems.



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The dual Simplex Algorithm


The DUAL Simplex Algorithm
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Solving the primal problem, moving throughsolutions (simplex tableaux) that are dual feasiblebut primal unfeasible. a

a Different from solving the DUAL problem with the (PRIMAL) simplex method

PRIMAL feasible: b i 0DUAL feasible: c i 0

An optimal solution is a solution that is both primaland dual feasible.

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DUAL Simplex Algorithm

Initialization: problem in canonical form with all c j 0.Optimality test :
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Optimality test :if b i 0 for all i = 1 , 2, . . . , m, then stop optimal solution found ;else:

Iterate :

Choose the variable to leave the basis by 4:

b r = min i b i |b i < 0 .

If arj 0 for j = 1 , 2, . . . , n, then stop the PRIMAL problem is infeasible (DUAL is unbounded) ;else choose the variable to enter the basis by 5:

c s ars = min i

c j arj

| arj < 0 .

Replace the basic variable in row r with variable in column s and reestablishthe canonical form (i.e. pivot on the coefficient ars ).

Return to the optimality test.

4nd the most promissing candidate for an improvement in feasibility5guarantee that c s 0 in each iteration

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The DUAL Simplex Algorithm An example

x 1 x 2 x 3 x 41 1 1 0 1 x4 leaves the basis because:
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x 3 1 1 1 0 1

x 4 2 -3 0 1 2 Z 3 1 0 0 0

x 1 x 2 x 3 x 4

x 3 13 0 1 13

13

x 2 23 1 0 13 23

Z 73 0 0 13

23

x 1 x 2 x 3 x 4x 4 1 0 3 1 1x 2 1 1 1 0 1

Z 2 0 1 0 1

x 4 leaves the basis because:mini b i | b i < 0 = min { 1, 2} = 2x 2 enters the basis because:min j

c j arj

| a rj < 0 = min 3 2 , 1 3 =

13

x 3 leaves the basis because:mini b i | b i < 0 = min 13 =

13

x 4 enters the basis because:min j

c j arj

| a rj < 0 = min 73 13

, 13 13

= min {7, 1} = 1

All the b i are 0.Optimal solution:

(x 1 , x 2 , x 3 , x 4) = (0 , 1, 0, 1) and Z = 1

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Parametric PRIMAL-DUALAlgorithmPrimal-dual algorithms are useful whensimultaneous changes of both righthand-side andcost coefficients are imposed on a previousoptimal solution, and a new optimal solution isto be recovered.

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Parametric PRIMAL-DUAL Algorithm an example IObjective:

max Z = 2x 1 + 3 x 2Subject to: This example can easily be put in canonical
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j

x 1 + x 2 6 x 1 + 2 x 2

12

x 1 3x 2 1x 1 , x 2 0

x 1 x 2 x 3 x 4 x 5x 3 1 1 1 0 0 6x 4 1 2 0 1 0 12x 5 1 3 0 0 1 1

Z 2 3 0 0 0 0

x 1 x 2 x 3 x 4 x 5

x 3 1 1 1 0 0 6x 4 1 2 0 1 0 12 + x 5 1 3 0 0 1 1 +

Z 2 3 0 0 0 0

p y pform by addition of slack variables.

However, neither primal feasibility nor primaloptimality conditions will be satised.

We will arbitrarily consider the example as afunction of the parameter .If we choose large enough, i.e. 3,this system of e

02.3 Chapter2 Solving Linear Programs.pdf

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