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Duality Theory - Introduction to Operations Research - Lecture Slides, Slides of Operational Research

These are the important key points of lecture slides of Introduction to Operations Research are:Duality Theory, Theory of Duality, Important Concept, Operations Research, Linear Programming, Mathematical and Economic Level, Small Company, Company in Melbourne, Large Company, Smaller Company

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Duality Theory
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, but it has many
applications, and perhaps even a more
natural and intuitive interpretation, in
several related areas such as nonlinear
programming, networks and game
theory.
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Duality Theory

The theory of duality is a very elegantand important concept within the field of operations research. This theorywas first developed in relation to linear programming, but it has manyapplications, and perhaps even a more natural and intuitive interpretation, inseveral related areas such as nonlinear programming, networks and gametheory.

programming asserts that every linearThe notion of duality within linear program has associated with it a related linearprogram called its dual. The original problem in relation to its dual is termed the primal. it is the relationship between the primal and its dual, both on a mathematical andeconomic level, that is truly the essence of duality theory.

Problem P

max x

Z = 80 x 1 (^) + 60 x (^) 2 + 50 x 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

Now consider that there is a much bigger company in Melbourne which has been the lone producer of this type of furniture for many years. They don't appreciate the competition from this new company; so they have decided to tender an offer to buy all of their competitor's resources and therefore put them out of business.

Problem D

1 2 1 2 1 2 1 2

y y y y y y y y

min y

w = 100 y 1 (^) + 60 y 2

A Diet Problem An individual has a choice of two types of food to eat, meat and potatoes, each offering varying degrees of nutritional benefit. He has been warned by his doctor that he must receive at least 400 units of protein, 200 units of carbohydrates and 100 units of fat from his daily diet. 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 potatoes costs $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

Now consider a chemical company which hopes to attract this individual away from his present diet by offering him synthetic nutrients in the form of pills. This company would like determine prices per unit for their synthetic nutrients which will bring them the highest possible revenue while still providing an acceptable dietary alternative to the individual.

Problem D

max y

w = 400 y 1 (^) + 200 y 2 (^) + 100 y 3

80 y 1 (^) + 20 y 2 (^) + 30 y 3 ≤ 10 40 y 1 (^) + 50 y 2 (^) + 20 y 3 ≤ 2 y 1 (^) , y 2 (^) , y 3 ≥ 0

7.2 FINDING THE DUAL OF A

STANDARD LINEAR

PROGRAM

In this section we formalise the intuitive feelings we have with regard to the the relationship between the primal and dual versions of the two illustrative examples we examined in Section 7. The important thing to observe is that the relationship - for the standard form - is given as a definition.

Standard form of the Primal

Problem

a x a x a x b a 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 Z = ∑ jn = 1 c xj j

7.2.1 Definition

z Z cx

s t Ax b x

*: max x

..

= =

≤ ≥ 0

w * := min x w = yb s. t. yAc y ≥ 0

Primal Problem Dual Problem

b is not assumed to be non-negative Docsity.com

7.2.2 Example

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 = 5 x 1 (^) + 3 x (^) 2 − 8 x (^) 3 + 0 x (^) 4 + 12 x 5

Primal

Table 7.1: Primal-Dual relationship

y 1 ≥ 0^ x^1 a^11 ≥^0 x^2 a^12 ≥^0 x^ na^ 1 ≥ n 0 ≤ w= b 1 (min w)^ Dual^ y^2 ...^ ≥ 0 a ...^21 a ...^22 a^^2 n^ ...≤^ ... b^2 y (^) m ≥ 0 a ≥ (^) m 1 am ≥ 2 a (^) mn ≥ ≤ bn Z= (^) c 1 c 2 cn

7.2.3 Example

5 18 5 15 8 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 = 4 x 1 (^) + 10 x 2 (^) − 9 x 3