Opreation Research

8/13/2019 Opreation Research

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OPERATIONS RESEARCH

Micha Kulej

Business Information Systems

The development of the potential and academic programmes of Wrocaw University ofTechnology

Project co-financed by European Union within European Social Fund

Micha Kulej (Wroc Univ. of Techn.) OPERATIONS RESEARCH BIS 1 / 41
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Artificial Starting Solution The Big M Method

The linear programming problem in which all constraints are () withnonnegative right-hand sides offers a all-slack starting basic feasible

solution. Problems with(=) and/or()constraints need to use artificial

variables to the beginning of simplex algorithm. There are two meth-ods: the M-method(also called the Big M-method) and theTwo-Phase

method.

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Example

max Z =2x1+x2 3x3x1+x2+x3 62x1+x2=14

x1, x2, x3 0After converting to standard form we have:

max Z =2x1+x2 3x3x1+x2+x3 s1=6

2x1+x2=14x1, x2, x3, s1 0

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Adding Artificial Variables

The above system of equations is not in basic form - thereis notabasic variable in each equation.

If the equation idoes not have a slack (or a variable that can play the

role of a slack), an artificial variable, ai, is added to form a startingsolution similar to the all-slack basic solution. However, because theartificial variables are not part of the original linear model, they are as-

signed a very highpenaltyin the objective function, thus forcing them(eventually) to equal zero in the optimal solution.

In the considering example we add two artificial variablesa1 anda2.

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Penalty Rule for Artificial Variables.

GivenM, a sufficiently large positive value (mathematically, M ),the objective coefficient of an artificial variable represents anappropriate penalty if:

Artificial variable objective coefficient = M, in maximization problemsM, in minimization problems.

So we have:

max Z =2x1+x2 3x3 Ma1 Ma2a

1 +x

1+x

2+x

3 s

1=6

a2 +2x1+x2=14x1, x2, x3, s1, a1, a2 0


A ifi i l S i S l i Th Bi M M h d
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The Big M Method Calculations

First and second simplex tableau are as follows:

cB BV a1 a2 x1 x2 x3 s1 Solution

M a1 1 0 1 1 1 1 6 6/1 = 6

M a2 0 1 2 1 0 0 14 14/2 = 7Z 0 0 3M 2 2M 1 M+ 3 M 20M

2 x1 1 0 1 1 1 1 6

M a2 2 1 0 1 2 2 2 2/2 = 1

Z 3M+ 2 0 0 M+ 1 2M

1

2M

2

2M+ 12

The optimal simplex tableau:

cB BV a1 a2 x1 x2 x3 s1 Solution

2 x1 0

1

2 1

1

2 0 0 70 s1 1 1

2 0 12 1 1 1

Z M M+1 0 0 3 0 14

Optimal solution is: x1=7, s1=1,x2 =x3 =0 withZ =14.


A tifi i l St ti S l ti Th Bi M M th d
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No Feasible Solution - an Example

The use of the penalty Mwill not force an artificial variable to zero levelin the final simplex iteration if LPP does not have a feasible solution

(i.e. the constraints are not consistent). In this case, the final simplextableau will include at least one artificial variable at positive level.

Solve the problem :max Z =2x1+2x26x1+4x2 24x1 5x1, x2 0

The standard form:max Z =2x1+2x2s1 +6x1+4x2=24

x1 s2=5x1, x2, s1, s2 0


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Using The Big M Method to Solve the Problem

It is enough to add only one artificial variable :

max Z =2x1+2x2 Ma2s1 +6x1+4x2=24a2 +x1 s2=5

x1, x2, s1, s2 0


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The Big M Method Calculations

cB BV s1 a2 x1 x2 s2 Solution

0 s1 1 0 6 4 0 24 24/6=4M a2 0 1 1 0 1 5 5/1= 5

Z 0 0 M 2 2 M 5M

cB BV s1 a2 x1 x2 s2 Solution

2 x116 0 1

23 0 4

M a2 16 1 0

23 1 1

Z 16 M+ 13 0 0

23 M

23 M M+8

The last simplex tableau is optimal - all Z- row coefficients are

nonnegative. The artificial variablea2is basic variable with positivevalue so the problemis not consistent- it has no feasible solutions.


Artificial Starting Solution The Two-Phase Method
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The Essence of The Two-Phase Method

This method solves the linear programming problem in two phases:Phase I attempts to find a starting feasible solution and, if one is found,Phase II is used to solve the original problem.

Phase I Put the problem in equation form, and add the necessary

artificial variables to the constraints to get starting feasible

basic solution. Next, find a basic solution of the resultingequations that, regardless of whether the problem ismaximization or minimization,alwaysminimizes the sum

of the artificial variables. If minimum value of the sum ispositive, the problem has no feasible solution, which endsthe process. Otherwise, proceed to Phase II.

Phase II Use the feasible solution from Phase I as starting feasiblebasic solution for original problem.


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Artificial Starting Solution The Two Phase Method

Example

We use this method to solve the following problem:

max Z =4x1+x23x1+x2=34x1+3x2 6x1+2x2 4

x1, x2 0

Usings2 as a surplus in the second constraint ands3 as a slack in thethird constraint, the standard form of the problem is given as:

max Z =4x1+x23x1+x2=34x1+3x2 s2=6x1+2x2+s3=4x1, x2, s2, s3 0


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Artificial Starting Solution The Two Phase Method

The Model with Arificial Variables - Phase I

The third equation has its basic variable (s3) but the first and second do

not. Thus we add the artificial variables a1 anda2 and we minimize thesum of the artificial variables a1+a2. The resulting model is given as:

min Za =a1+a23x1+x2 +a1 = 34x1+3x2 s2 +a2 = 6x1+2x2 +s3 = 4x1, x2, s2, s3, a1, a2 0

Taking a1, a2, s3 as a basic variables and using formula (2) forcomputing coefficients ofZa - row we obtain the first simplex tableau:

cB BV x1 x2 s2 a1 a2 s3 Solution

1 a1 3 1 0 1 0 0 31 a2 4 3 1 0 1 0 60 s3 1 2 0 0 0 1 4

Za

7 4 1 0 0 0 9Micha Kulej (Wroc Univ. of Techn.) OPERATIONS RESEARCH BIS 12 / 41

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g

The Optimal Simplex Tableau of Phase I

Now we use simplex algorithm for minimization problem and receive(after two iterations) the following optimum tableau:

cB BV x1 x2 s2 a1 a2 s3 Solution

0 x1 1 0

1

5

3

5

1

5 0

3

50 x2 0 1

35

45

35 0

65

0 s3 0 0 1 1 1 1 1Za 0 0 0 1 1 0 0

Because minimumZa

=0. Phase I produces the basic feasiblesolutionx1=

35 , x2=

65 , s3=1. We can eliminate columns fora1 and

a2 from tableau and move to Phase II.


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g

Phase II

After deleting the artificial columns, we write the original problem as

max Z =4x1+x2x1 +

15 s2 =

35

x2 35 s2 =

65

s2 +s3 = 1x1, x2, s2, s3 0

Now we can begin the simplex algorithm with the following simplextableau:

cB BV x1 x2 s2 s3 Solution

4 x1 1 0 1

5 0 3

5

1 x2 0 1 35 0 650 s2 0 0 1 1 1

Z 0 0 15 0 18

5

Because we are maximizing, this is the optimal tableau. The optimal

solution isx1 =

3

5 ,x

2 =

6

5andZ

=

18

5.


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Removal of Artificial Variables after Phase I

The removal of the artificial variables and their columns at the end of Phase I

can take place only when they are all nonbasic. If one or more artificialvariables are basic (at zero level) at the end of Phase I, then the following

additional steps must be undertaken to remove them prior to start of Phase II.

Step 1. Select a zero artificial variable to leave the basic solution

(choosing the leaving variable) and designate its row as the pivotrow. The entering variable can be any nonbasic variable with a

nonzero(positive or negative) coefficient in the pivot row. Perform theassociated simplex iteration.

Step 2. Remove the column of the (just-leaving) artificial variable fromthe tableau. If all the zero artificial variables have been removed, go toPhase II. Otherwise, go back to Step 1.

RemarkAll commercial packages use the two-phase method to solve

linear programming problems.Micha Kulej (Wroc Univ. of Techn.) OPERATIONS RESEARCH BIS 15 / 41

Sensitivity Analysis Changes in Objective Coefficients
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Changes in Objective Coefficients

The sensitivity analysisis concerned with how changes in an parame-

ters of the linear programming model affect the optimal solution.

Let us consider the following problem (described in the basic form) of

manufacturing two productsW1andW2 from two row materials S1, S2:

max Z =3x1+2x2 [Maximizing the revenue]s1 +2x1+x2=100 [Limit on row materialS1]

s2 +x1+x2 =80 [Limit on row materialS2]s3 +x1=40 [Demand onW1]

x1, x2, s1, s2, s3 0

The optimal solution of the model is x1 =20, x2 =60 withZ =180.We want to determine the range of values of parameter (i.e. coefficientof objective function - unit price of product W1 in our example ) over

which the optimal solution remain unchanged.


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The Optimal Simplex Tableau of the Problem

cB BV s1 s2 s3 x1 x2 Solution3 x1 1 -1 0 1 0 20

2 x2 -1 2 0 0 1 600 s3 -1 1 1 0 0 20

Z 1 1 0 0 0 180Let us assumed that unit price of W1 equals 3 +. Then we have:

cB BV s1 s2 s3 x1 x2 Solution

3+ x1 1 1 0 1 0 20

2 x2 1 2 0 0 1 600 s3 1 1 1 0 0 20

Z 1+ +1 0 0 0 180


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Possibly Changes of Values ofc1

The optimal solution remains optimal as long as all the Z rowcoefficients are nonnegative, so we get:

1 + 0+1 0

Solving these system inequalities we obtain that [1, 1]. Thusoptimal solution will remain optimal for values of price of W1belonging

to the interval [2,4].


Sensitivity Analysis Changes in Right-Hand Side
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Changes in Right-Hand Side

Now we examine how the optimal solution to linear problem willchanges if the right-hand side of a constraint is changed. For example

we consider the constraint for row material S1(coefficientb1 =100).Changing inbiwill leaveZ- row of simplex tableau unchanged but will

change the the solution column of the simplex tableau. If at least onecoefficient becomes negative, the current solution is no longer feasible.

So we look for the value of for which the following system equalitiesis consistent:

2x1+x2 =100 +x

1+x

2=80

s3+x1=40x1, x2, s1, s2, s3 0


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Matrix Form of System Equations

We describe this system in matrix form:

2 1 01 1 0

1 0 1

x1x2s3

=

100 +80

40

Multiplying both sides of this equations by the inverse matrix of thecoefficients matrix we get:

x1x2s3

= 2 1 0

1 1 01 0 1

1

100 +8040

0

00


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The Inverse Matrix of Basis

The inverse matrix we could obtain from the optimal simplex tableau. Itis consisting of the columns in the optimal tableau that correspond tothe initial basic variablesBV = {s1, s2, s3}(taking in the same order):

2 1 01 1 0

1 0 1

1

=

1 1 01 2 0

1 1 1


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Hence we get:

1 1 01 2 01 1 1

100 +8040

000

For the current set of basic variable (basis) to remain optimal we

require that the following system of inequalities holds:

20+ 060 020 0

The solution remains optimal (basic variablex1, x2, s3) as long as [20, 20]or if between 80 and 120 row material S1 (b1 [80, 120],where the current amount isb1=100) are available.


Dual Problem Definition Of the Dual Problem
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Associated with any linear programming problem is another linear

problem, called thedual problem(Dual in short). Now we explain howto find the dual problem to the given LPP, we discuss the economicinterpretation of the dual problem and we discuss the relation that exist

between an LPP (called Primal) and its dual problem. We consider theLPP with normal (canonical) form :

max Z =c1x1+c2x2+ +cnxna11x1 + a12x2 + + a1nxn b1a21x1 + a22x2 + + a2nxn b2

... ...

...

am1x1 + am2x2 + + amnxn bm

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

The original problem is calledthe primal.


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Dual Problem of LPP if Primal is in Canonical Form

The dual problem is defined as follows:

min W =b1y1+b2y2+ +bmym

a11y1 + a21y2 + + am1ym c1a12y1 + a22y2 + + am2ym c2

... ...

...a1ny1 + a2ny2 + + amnym cn

yi 0(i=1, 2, . . . , m)


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The Construction of the Dual Problem from the Primal

Problem

max Z

min W (x1 0) (x2 0) (xn 0)

x1 x2 xn(y1 0) y1 a11 a12 a1n b1(y2 0) y2 a21 a22 a2n b2

... ...

... ...

... ...

(ym 0) ym am1 am2 amn bm c1 c2 cn


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Example

The Furniture Company STYLE manufactures tables and chairs. Atable sells for $160, and chair sells for $200. The demand for tablesand chairs is unlimited. The manufacture of each type of furniturerequire labor, lumber, and inventory space. The amount of eachresource needed to make tables and chairs and daily limit of available

resources is given in the following table:

Resources needed Amount ofResource Table Chair resource available(hours)

Labor(hours) 2 4 40Lumber((feet)3) 18 18 216

Inventory space((feet)2) 24 12 240

STYLE wants to maximize total revenue.


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Primal and Dual of the Example Problem

Primal problem

max z=160x1+200x2

2x1+4x2 40

18x1+18x2 216

24x1+12x2 240x1, x2 0.

Dual problem

min w=40y1+216y2+40y32y1+18y2+24y3 160

4y1+18y2+12y3 200

y1, y2, y3 0.


Dual Problem Economic Interpretation of Duality
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Interpretation of Dual for STYLE

Suppose there is an entrpreneur who wants to purchase all of Stylesresourses i.e. 40 hours of labor, 210(feet)3 of lumber and 240(feet)2

of inventory space. Then he must determine the price he is willing to

pay for a unit of each of STYLEs resources. Let y1, y2iy3 be the pricefor one hour of labor, one cubic feet of lumber and one square feet of

inventory space. We show that the resource prices should bedetermine by solving the dual problem. The total price the

entrepreneur must pay for the resources is 40y1+216y2+240y3andsince he wish to minimize the cost of purchasing the resources hewants to:

minimizeW =40y1+216y2+240y3.

But he must set resource prices high enough to induce STYLE to sellits resources.



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Interpretation of Dual for STYLE

For example, he must offer STYLE at lest $160 for a combination of

resources that includes 2 hours of labor, 18 cubic feet of lumber and24 square feet of inventory space, because STYLE could use these

resources to manufacture table that can be sold for $160. Since he isoffering 2y1+18y2+24y3 for the resources used to produce table, hemust choosey1, y2, y3to satisfy

2y1+18y2+24y3 160.

Similar reasoning for the chair gives:

4y1+18y2+12y3 200.

The sign restrictionsy1, y2, y3 0 must also hold. So we see that thesolution to the dual of STYLE problem does yield prices for labor,lumber and inventory space.



Fi di h D l LPP
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Finding the Dual to any LPP

We illustrate how to find the dual problem on the following example:

max z=2x1+x2 min w=2y1+3y2+y3

x1+x2 =2 y1 - Unrestricted2x1 x2 3 y2 0x1 x2 1 y3 0

x1 0 y1+2y2+y3 2x2 Unrestricted y1 y2 y3 =1.



R l f C t ti f D l
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Rules for Construction of Dual

The general conclusion from the preceding example is that the

variables and constraints in the primal and dual problems are definedby rules in the following table:

Maximization problem Minimization problem

Constraints Variables

0 0= Unrestricted

Variables Constraints

0

0 Unrestricted =

Table:Rules for construction the Dual Problem


Dual Problem Primal-Dual Relationships

Th K R l ti hi b t th P i l d D l
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The Key Relationships between the Primal and Dual

TheoremThe dual of the dual problem yields the original primal problem.

Theorem (Weak duality property)

If we choose any feasible solution to the primal and any feasiblesolution to the dual, the W value for the feasible dual solution will beat least as large as the Z value for the feasible primal solution. Letx= [x1, x2, . . . , xn]

T be any feasible solution to the primal and

y= [y1, y2, . . . , ym]be any feasible solution to the dual. Then

(Z value forx) (W value fory)

From this theorem results two properties:


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Property

Ifx= (x1,x2, . . . ,xn)andy= (y1,y2, . . . ,ym)are feasible solutions ofprimal problem and dual problem respectively such that

Z =c1x1+c2x2+. . .+cnxn=b1y1+b2y2+. . .+bmym=W , thenxmust be optimal solution for primal problem andymust be optimal

solution for dual problem.

Property

If the primal(dual) is unbounded, then the dual(primal) problem is

infeasible.



Duality Theorem
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Duality Theorem

TheoremThe following are only possible relations between the primal and dual

problems:

1 If one problem has feasible solutions and a bounded objective

function(and so has optimal solution), then so does the otherproblem.

2 If one problem has feasible solutions and an unbounded objective

function(and so no optimal solution), then the other problem has

no feasible solutions.

3 If one problem has no feasible solutions, then the other problem

has either no feasible solution or an unbounded objective function.



Interpretation of Optimal Dual Decision Variables
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Interpretation of Optimal Dual Decision Variables

Now we can give an interpretation of the dual problem formaximization problem. We know that for optimal solutions xof primalproblem andyof dual problem the following equality holds:

Z =c1x1+c2x2+. . .+cnxn=b1y1+b2y2+. . .+bmym=W

So eachbiyican be interpreted as the contribution to profit by havingbiunits of resourceiavailable for the primal problem. Thus, theoptimal dual variable yi(it is called shadow price) is interpreted as the

contribution to profit per unit of resource i(i=1, 2, . . . , m).


Dual Problem Reding the Optimal Dual Solution if Primal is a Maximum Problem

Reading Optimal Dual from Z row of Optimal Simplex
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Reading Optimal Dual from Z-row of Optimal Simplex

Tableau for STYLE

After solving primal problem by simplex method we could read theoptimal dual solution from the optimal simplex tableau. Let us consider

the LPP for firm STYLE. The optimal simplex tableau is as follows:

Table:Optimal simplex tableau

s1 s2 s3 x1 x2200 x2

12

118 0 0 1 8

160 x1 12

19 0 1 0 4

0 s3 6 -2 1 0 0 48

Z 20 20

3 0 0 0 2240The basic variables areZB= {x2, x1, s3} and the basis is

B=

4 2 0

18 18 0

12 24 1

. Inverse matrix for Bis: B1 =

12

118 0

1219 0

6 2 1



The Optimal Solution of Dual Problem
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The Optimal Solution of Dual Problem

(y1, y2, y3)we can compute using matrix B1 as follows:

(y

1,y

2,y

3) =c

BB1

= (200

,160

,0

)

12

118 0

1

2

1

9 06 2 1

= (20

,

20

3 ,0

).

where the vectorcBcontains the coefficients of objective function

corresponding the basic variables. The optimal dual solution are

coefficients of variabless1, s2, s3 ofZ row optimal simplex tableau.



Reding Dual Solution from Optimal Simplex Tableau
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Reding Dual Solution from Optimal Simplex Tableau

If the primal problem is any form , then the optimal dual solution maybe read fromZ row optimal simplex tableau by using the followingrules:

Optimal value of dual variable yi if constraintiis a () equal to

coefficient ofsi inZ row optimal simplex tableau.Optimal value of dual variable yi if constraintiis a () equal to-(coefficient ofei inZ row optimal simplex tableau), whereei issurplus variable.

Optimal value of dual variable yi if constraintiis an equality (=)equal to (coefficient ofai inZ row optimal simplex tableau)-M,whereai is artificial variable.



Example
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Example

max z=3x1+2x2+5x5x1+3x2+2x3 15

2x2 x3 5

2x1+x2 5x3 = 10

x1, x2, x3 0.To solve the problem we use the M method:

max Z =3x1+2x2+5x5 Ma2 Ma3

x1+3x2+2x3+s1 = 15

2x2 x3 e2+a2 = 5

2x1+x2 5x3+a3 = 10

x1, x2, x3, s1, a2, a3 0.



The Big-M Method the Last Simplex Tableau
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The Big M Method, the Last Simplex Tableau

x1 x2 x3 s1 e2 a2 a35 x3 0 0 1

423

523

523

223

1523

2 x2 0 1 0 2

23 923

923

123

6523

3 x1 1 0 0 9

231723

1723

723

12023

Z 0 0 0 51

23

58

23 M 58

23 M+ 9

23

565

23

The dual problem has the following form:

min W =15y1+5y2+10y3

y1+2y3 33y1+2y2+y3 2

2y1 y2 5y3 5

y1 0, y2 0, y3 Unrestricted.



Optimal Dual Solution
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Optimal Dual Solution

Reading optimal dual solution from optimal simplex tableau we get:

The first constraint is inequality soy1 = 5123 (coefficient ofZ

row for columns1).

The second constraint is inequality soy2 = 5823 (- coefficient of

Z row for columne2).

The third constraint is equality = so y3= 923 (coefficient ofZ

row for columna3 minus M).

Optimal value of objective function of dual problem is

W =15 5123 +5( 5823 ) +10 923 = 56523 and equals optimal value ofobjective function of the primal problem.

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