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Operations Research I-

Chapter V: Duality TheoryM. NACEUR AZAIEZ, Professor

Tunis Business School

Tunis University

https://sites.google.com/site/naceurazaiez/



Content Introduction & Motivation

Definition of the dual of LP

Primal-Dual Relationships

Results on duality theory

Dual of LP in non-canonical form

Economic interpretations



Introduction & Motivation The theory of duality is a very elegant

and important concept within the field

of operations research. This theory was first developed in

relation to linear programming.



Introduction & Motivation Next, it was found out that it has many

applications, and perhaps even a more

natural and intuitive interpretation, inseveral related areas such as

nonlinear programming, networks and

game theory



Introduction & Motivation The notion of duality within LP asserts that

every LP has associated with it a related LP

called its dual. The original problem in relation to its dual is

termed the primal.

It is the relationship between the primal andits dual, both on a mathematical and economic

level, that is truly the essence of duality theory.



Example 1 There is a small company in Melbourne

which has recently become engaged in the

production of office furniture. The company manufactures tables, desks and

chairs.

The production of a table requires 8 Kgs ofwood and 5 Kgs of metal and is sold for $80.



Example 1-continued A desk uses 6 Kgs of wood and 4 Kgs of

metal and is sold for $60; and a chair requires

4 Kgs of both metal and wood and is sold for$50.

We would like to determine the revenue

maximizing strategy for this company,given that their resources are limited to 100

Kgs of wood and 60 Kgs of metal.



Formulation-example 1max

x Z x x x 80 60 501 2 3

8 6 4 100

5 4 4 60

0

1 2 3

1 2 3

1 2 3

x x x

x x x

x x x

, ,



Duality-example 1 Now consider that there is a much bigger

company in Melbourne which has been the lone

producer of this type of furniture for manyyears.

They don't appreciate the competition from

this new company; so they have decided totender an offer to buy all of their competitor's

resources and therefore put them out of

business.



Formulating the dual LP

The challenge for this large company then is

to develop a linear program which will

determine the appropriate amount of moneythat should be offered for a unit of each type

of resource, such that the offer will be

acceptable to the smaller company while

minimizing the expenditures of the larger

company.



Problem D1

8 5 80

6 4 60

4 4 50

0

1 2

1 2

1 2

1 2

y y

y y

y y

y y

,

min y

w y y 100 601 2





Example 2-continued

Given that a kg of steak costs $10 and provides

80 units of protein, 20 units of carbohydrates

and 30 units of fat, and that a kg of potatoescosts $2 and provides 40 units of protein, 50

units of carbohydrates and 20 units of fat, he

would like to find the minimum cost diet

which satisfies his nutritional requirements.



Formulating example 2

80 40 400

20 50 200

30 20 100

0

1 2

1 2

1 2

1 2

x x

x x

x x

x x

,

min x

Z x x 10 21 2



The Dual of example 2

Now consider a chemical company which

hopes to attract this individual away from

his present diet by offering him syntheticnutrients in the form of pills.

This company would like to determine

prices per unit for their synthetic nutrientswhich will bring them the highest possible

revenue while still providing an acceptable

dietary alternative to the individual.



Dual Problem of example 2

max

y

w y y y 400 200 1001 2 3

80 20 30 101 2 3 y y y

40 50 20 21 2 3 y y y

y y y1 2 3 0, ,



Comments

Each of the two examples describes some kind

of competition between two decision makers.

The notion of “competition” could beinvestigated more formally in “Game Theory”.

We shall investigate the economic

interpretation of the primal/dual relationshiplater in this chapter.



Canonical form of the Primal

Problem

a x a x a x ba x a x a x b

a x a x a x b

x x x

n n

n n

m m mn n m

n

11 1 12 2 1 1

21 1 22 2 2 2

1 1 2 2

1 2 0

. . .. ..

. .. .. . . . . .. .

. .. .. . . . . .. .. ..

, , .. .,

max x

j j j

n Z c x

1



Canonical form of the Dual

Problem

a y a y a y ca y a y a y c

a y a y a y c

y y y

m m

m m

n n mn m n

m

11 1 21 2 1 1

12 1 22 2 2 2

1 1 2 2

1 2 0

.. .. . .

. .. .. . . . . .. .

. .. .. . . . . .. .

. ..

, , .. . ,

min y

ii

m

iw b y 1



Definition

z Z cx

s t

Ax b x

x*: max

. .

0

w* : min x w yb s.t .

yA c

y 0

Primal Problem Dual Problem

b is not assumed to be non-negative



21

Max Z = 10X1 + 5X2

2X1 + 3X2 10

6X1 + X2 15

4X1

– 5X2

35

X1, X2 0

Example



22

Max ZX= X1 + X2

X1 X2

X1 X2

X1 X2

X1, X2 0

Min ZY = Y1 Y2 Y3

Y1 Y2 Y3

Y1 Y2 Y3

Y1

,Y2

,Y3

0

Y1

Y2

Y3

1st constraint 2nd constraint

10

15

35

10 5

2

6

4

3

+

– 5

+ +

+ +

+



Example

3 8 9 15 20

18 5 8 4 12 30

0

1 2 4 5

1 2 3 4 5

1 2 3 4 5

x x x x

x x x x x

x x x x x

, , , ,

max x Z x x x x x

5 3 8 0 121 2 3 4 5

Primal



min

y

w y y 20 301 2

3 18 5

8 5 3

8 8

9 4 0

15 12 12

0

1 2

1 2

2

1 2

1 2

1 2

y y

y y

y

y y

y y

y y

,

Dual



Primal-Dual relationship

x1 0 x2 0 xn 0 w =

y1 0 a11 a12 a n1 b1D u a l y

20 a

21 a

22 a

n2 b

2(m i n w ) . . . . . . . . . . . . . . .

ym 0 am1 am2 amn bn

Z = c1 c2 cn



Example

5 18 5 158 12 8 8

12 4 8 10

2 5 5

0

1 2 3

1 2 3

1 2 3

1 3

1 2 3

x x x x x x

x x x

x x

x x x

, ,

max x

Z x x x 4 10 91 2 3



x1 0 x2 0 x3 0 w=

1 0 5 - 18 5 15Dual y2 0 8 12 0 8

(min w) 3 0 12 - 4 8 10 y4 0 2 0 - 5 5

Z= 4 10 - 9



Dual

5 8 12 2 418 12 4 10

5 8 5 9

0

1 2 3 4

1 2 3

1 3 4

1 2 3 4

y y y y y y y

y y y

y y y y

, , ,

min y

w y y y y 15 8 10 51 2 3 4



FINDING THE DUAL OF NON-

CANONICAL LP





Handling variables unrestricted in

sign Replace the variable unrestricted in sign, , by

the difference of two nonnegative variables.



Example

2 4

3 4 5

2 3

0

2 3

1 2 3

1 2

1 2 3

x x

x x x

x x

x x x

, ; urs:= unrestricted sign

max x

Z x x x 1 2 3



Conversion

Multiply through the greater-than-or-equal-to

inequality constraint by -1

Use the approach described above to convertthe equality constraint to a pair of inequality

constraints.

Replace the variable unrestricted in sign, , bythe difference of two nonnegative variables.



Dual

y2

y3 y4

1

2 y13 y2 3 y3 2 y4 1

y1 4 y2 4 y3 1

y1 4 y2 4 y3 1

y1, y2, y3, y4 0

min y

w y y y y 4 5 5 31 2 3 4



Streamlining the conversion ...

An equality constraint in the primal

generates a dual variable that is

unrestricted in sign. An unrestricted in sign variable in the

primal generates an equality constraint in

the dual.



Example (Continued)

min y

w y y y y 4 5 5 31 2 3 4

y2

y3 y4

1

2 y13 y2 3 y3 2 y4 1

y1 4 y2 4 y3 1

y1 4 y2 4 y3 1

y1, y2, y3, y4 0



y y

y y y

y y

y y y

2 3

1 2 3

1 2

1 3 2

1

2 3 2 1

4 1

0

, ,

, , ,

, ,

, , ,, ;

urs

min, , , , y

w y y y 4 5 31 2 3

+

correction



Primal-Dual relationship

Primal Problem

opt=max

Constraint i :

<= form

= form

Variable j:

x j >= 0

x j urs

opt=min

Dual Problem

Variable i :

yi >= 0

yi urs

Constraint j:

>= form

= form



Example

3 8 66 5

8 100

1 2

1 2

1

2 1

x x x x

x x x

; urs

max x

Z x x 5 41 2



Equivalent non-standard form

3 8 6

6 5

8 10

0

1 2

1 2

1

2 1

x x

x x

x

x x; urs

max x

Z x x 5 41 2



Dual from the recipe

3 8 5

8 6 4

0

1 2 3

1 2

1 2 3

y y y

y y

y y y; , urs

min y

w y y y 6 5 101 2 3



What about opt=min ?

Can use the usual trick of multiplying the

objective function by -1 (remembering to

undo this when the dual is constructed.) It is instructive to use this method to construct

the dual of the dual of the standard form.

i.e, what is the dual of the dual of thestandard primal problem?



What is the dual of

w* :min x w yb

s.t .

yA c

y 0



max y

w yb

s.t .

yA c

y 0

max y

w yb

s.t .

yA c

y 0



min

. .

x

Z cx

s t

Ax b

x

0

max

. .

x Z cx

s t Ax b

x

0



Important Observation

FOR ANY PRIMAL LINEAR

PROGRAM, THE DUAL OF THEDUAL IS THE PRIMAL



Primal-Dual Relationship

Primal or Dual

opt=max opt=min

Dual or Primal

Variable i :

yi >= 0

yi urs

Constraint j:

>= form

= form

Constraint i :

<= form

= form

Variable j:

x j >= 0

x j urs



Example

3 5 122 8

5 100

1 2

1 2

1 2

1 1

x x x x

x x x x

,

min x

Z x x 6 41 2



Equivalent form

min x

Z x x 6 41 2

3 5 122 8

5 100

1 2

1 2

1 2

1 2

x x x x

x x x x

,



Dual

3 5 6

5 2 4

0

1 2 3

1 2 3

1 3 2

y y y

y y y

y y y

, ; urs

max y

w y y y 12 8 101 2 3

Maximization Minimization



52

Number of constraints Number of variables

constraint Variable positive or zero

constraint Variable negative or zero

constraint = Unconstrained Variable

Number of variables Number of constraints

Variable positive or zero constraint

Variable negative or zero constraint

Unconstrained Variable constraint =

Coefficient of the jth variable

In the objective functionRHS of the jth constraint

RHS of the ith constraintCoefficient of the ith variable

In the objective function

Technological coefficient of the

jth variable in the ith constraint

Technological coefficient of the

ith variable in the jth constraint



Example

Primal Dual

Max ZX = 5X1 + 4X2 –

2X3

2X1 + 3X2 20

X1 – 4X3 5

X1 + X2 + X3 = 30

X1, X2 0

Min ZY = 20Y1 + 5Y2 + 30Y3

2Y1 + Y2 + Y3

3Y1 + Y3

- 4Y2 + Y3

Y1 0

5

4

= - 2

, Y2 0, Y3 IR



54

Primal Dual

Min ZX = 15X1 + 2X2

X1 – X2 10

7X1 + 2X2 50

- X1 + 3X2 2

X1 0, X2 IR

Max ZY = 10Y1+50Y2+2Y3+120Y4

Y1 + 7Y2 - Y3 + 9Y4

-Y1 + 2Y2 +3Y3 + Y4

Y1 0

9X1 + X2 = 120

15

= 2

, Y2 0, Y3 0, Y4 IR

Note that the dual of the dual is the primal



Results

Consider the primal and dual LP:

55



56

Theorem

T

. The dual of the dual is the primal.

The dual problem LP2 is equivalent to the following problem:

Max -

s. t. -

0.By definition, the dual of th

Theore

is pro

m

Proof

blem is

Min c

:

T

T

b y

A y c

y

T

x

s. t. ( A x b

x 0,

or equivalently,

Max

s. t.

0.

which is the primal problem.

T c x

Ax b

x



57

Various theorems related to duality

(without proof)

If x is primal feasible and y is dual feasible,then cTx ≤ bTy (weak duality theorem).

If x* is primal feasible and y* is dual feasible

and cTx* ≥ bTy* then x* and y* are optimal If one of a pair of primal and dual problems

has an optimal solution, then the other alsohas an optimal solution and the optimal values

of their objective functions are equal.





59

Interpretation of Duality

The dual variable y* measures the change in optimal

profit due to a unit change in resource i.

The fact that profit would increase by yi* if an

additional unit of resourcei were available imputes a

value or price to resource i.

This value or price is called a shadow price. Thus,

the shadow price is the amount of contribution of anadditional unit of a resource to total profit.

More will be said later in this context.





Discussion-continued

Thus, the weak duality theorem states that the

profit is less than or equal to the worth of

resources. That is, the resources are not exploited

according to the best allocation except at

optimality.




The yi's are the shadow prices of the resources.

A shadow price can be interpreted as the

additional unit profit that could be made byacquiring additional units of resource i.

In particular, when a resource is not totally

consumed at optimal exploitation the relatedconstraint is inactive and the corresponding

shadow price must be zero.




Note that in that case, the corresponding slack

variable is basic

If however the constraint is active (i.e., theresource is entirely consumed at optimal

exploitation), then the corresponding shadow

price is positive and therefore it will be

profitable to acquire additional units of that

resource.



Example

Consider the following LP and its dual :

Max ZX = 5X1 + 2X2 + 7X3

X1 + X2 + X3 50

2X1 + 4X2 + 3X3 75

X1, X2, X3 0

Min ZY = 50Y1 + 75Y2

Y1 + 2Y2

Y1 + 4Y2

Y1 + 3Y2

Y1 0

5

2

≥ 7

, Y2 0



65

CJ 5 2 7 0 0

Basis X1 X2 X3 S1 S2 bI

0

5

S1

X1

0

1

-1

2

-1/2

3/2

1

0

-1/2

1/2

12,5

37,5

J 0 -8 -1/2 0 -5/2 187,5

The optimal tableau is given by

DualSolution

E1 E2 E3 Y1 Y2 ZY

0 8 1/2 0 5/2 187,5



66

Economic Interpretation of thedual problem• Assume that in the context of the previous LP, 3

products P1, P2 et P3 are to be manufactured.

• The production process requires the use of 2

resources R1 & R2 , which are available in limited

quantities of 50 et 75 units respectively.



67

Let Xi be the quantity of Pi to produce, i=1,3. As

the producer seeks to maximize his profit, his LP

can be written as:

Max ZX = 5X1 + 2X2 + 7X3

X1 + X2 + X3 50

2X1 + 4X2 + 3X3 75

X1, X2, X3 0





69

• However, when the producer sells his resources,he will renounce to produce P1, P2 & P3.

• Hence, he will lose the corresponding profit.

• Consequently, the producer will not accept the

offer unless he gets sufficient compensation for his

lost profit.

• The manufacturing of P1 requires one unit



• The manufacturing of P1 requires one unit

of R1 and 2 units of R2.

• Hence, the producer is willing to sell oneunit of R1 and two units of R2 for an

amount of at least 5 dinars (unit profit of

P1).

• As Yi is the amount of money proposed

against one unit of Ri, i =1,2,then the sum

received against the amounts of resources

necessary for manufacturing 1 unit of P1 is

Y1+2Y2.

70

• We must ensure that:



• We must ensure that:

Y1 + 2Y2 5

• Similarly, the sale of 1 unit of R1 and 4 unitsof R2 must provide at least a profit of 2

dinars.

• In addition, the sale of 1 unit of R1 and 3units of R2 must provide a profit of at least 7

dinars. This leads to the following two

constraints:Y1 + 4Y2 2

Y1 + 3Y2 771

Fi ll i th t Y &Y t f



72

• Finally, given that Y1 &Y2 are amounts of money,

they cannot be negative.

• It follows that the problem is formulated as:

Min ZY = 50Y1 + 75Y2

Y1 + 2Y2 5

Y1 + 4Y2 2

Y1 + 3Y

2 7

Y1, Y2 0

• Clearly, this LP is the dual of the original LP.

• We Call the marginal value of a resource the



We Call the marginal value of a resource the

amount of change (increase or decrease) in

the value of Z due to the use of an additional

unit of the resource.

• If the problem is a maximization of profit, the

marginal value of a resource is also called

marginal gain or shadow price (introduced

above)

• If the problem is a minimization of cost, the

marginal value is also called marginal cost

or reduced cost.

73



Reconsider the following example

Max ZX = 5X1 + 2X2 + 7X3

X1 + X2 + X3 50 (R1)

2X1 + 4X2 + 3X3 75 (R2)

X1, X2, X3 0

CJ 5 2 7 0 0

Basis X1 X

2 X

3 S

1 S

2 b

i

0

5

S1

X1

0

1

-1

2

-1/2

3/2

1

0

-1/2

1/2

12,5

37,5

CJ-ZJ 0 -8 -1/2 0 -5/2 187,5



• From the corresponding optimal Tableau,

the marginal value or shadow price of R1 is 0 and the shadow price of R2 is 2.5.

• That is, an additional unit of R1 has no

effect on the value of Z, whereas anadditional unit of R2 will increase Z by 2.5

dinars.

75

• The producer will not be ready to cede a



The producer will not be ready to cede a

unit of R1 (respectively R2) unless he

obtains against it at least 0 (respectively2.5 dinars) additional gain.

• Equivalently, in order to acquire one more

unit of R1 (respectively R

2), the producer

will be willing to pay additional cost of 0

(respectively 2.5 dinars).

76



The CJ-ZJ of a decision variable

• CJ-ZJ is called the reduced cost of thecorresponding decision variable

• It is interpreted as the deficit for the

decision variable to become basic

• In the last tableau, C2-Z2 =-8. This says

that the unit profit of P2 must increase at

least by 8 dinars for P2 to becomeprofitable.

77



The CJ-ZJ of a decision variable

• Equivalently, by forcing one unit of P2 to beproduced, the profit will be reduced by 8

dinars.

• Similarly, the unit profit of P3 must increaseat least by 0.5 dinars for P3 to become

profitable