Unformatted text preview: Chapter 7
Chapter 7 Duality Theory
Duality Theory ).1 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. X The notion of duality within linear
The programming asserts that every linear
program has associated with it a related
linear program called its dual. The original
The
problem in relation to its dual is termed the
primal.
X it is the relationship between the primal
and its dual, both on a mathematical and
economic level, that is truly the essence of
duality theory.
).2 7.1 Examples
7.1 Examples
X ).3 There is a small company in Melbourne which has
There
recently become engaged in the production of office
furniture. The company manufactures tables, desks
and chairs. The production of a table requires 8 kgs
of wood and 5 kgs of metal and is sold for $80; 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.
wood Problem P1
Problem P1
max Z = 80 x1 + 60 x 2 + 50 x 3
x
8 x1 + 6 x 2 + 4 x3 ≤ 100 5x1 + 4 x2 + 4 x 3 ≤ 60
x1 , x 2 , x3 ≥ 0
).4 X Now consider that
Now ).5 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. X The challenge for this large company then is
The ).6 to develop a linear program which will
determine the appropriate amount of money
that 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.
company. Problem D1
Problem D1
min w = 100 y1 + 60 y2
y ).7 8 y1 + 5 y2 ≥ 80
6 y1 + 4 y2 ≥ 60
4 y1 + 4 y2 ≥ 50
y1, y2 ≥ 0 A Diet Problem
A Diet Problem
X ).8 An individual has a choice of two types of food to eat,
An
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 Problem P2
Problem P2
min Z = 10 x1 + 2 x2
x 80 x1 + 40 x 2 ≥ 400
20 x1 + 50 x 2 ≥ 200 ).9 30 x1 + 20 x 2 ≥ 100
x1 , x 2 ≥ 0 X Now consider
Now ).10 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 D2
Problem D2
max w = 400 y1 + 200 y2 + 100 y3
y 80 y1 + 20 y2 + 30 y3 ≤ 10
40 y1 + 50 y2 + 20 y3 ≤ 2
y1, y2 , y3 ≥ 0
).11 Comments
Comments
X Each of the two examples describes
Each ).12 some
kind of competition between two decision
makers.
makers.
X We shall investigate the notion of
We
“competition” more formally in 618261
under the heading “Game Theory”.
under
X We shall investigate the economic
We
interpretation of the primal/dual relationship
later in this chapter.
later 7.2 FINDING THE DUAL
7.2 FINDING THE DUAL
7.2
7.2
OF A STANDARD LINEAR
OF A STANDARD LINEAR
PROGRAM
PROGRAM
PROGRAM
PROGRAM
X In this section we formalise the intuitive
In feelings we have with regard to the the
relationship between the primal and dual
versions of the two illustrative examples we
examined in Section 7.1
examined
X The important thing to observe is that the
The
relationship  for the standard form  is
given as a definition.
).13 Standard form of the Primal
Standard form of the Primal
Standard
Standard
Problem
Problem
Problem
Problem
n
max Z = ∑ c j x j
x j =1 a11 x1 + a12 x 2 + . . . + a1n xn ≤ b1
a21 x 1 + a22 x 2 + . .. + a2 n x n ≤ b2 ).14 . ..
.. .
...
.. .
. ..
.. .
...
.. .
am1 x1 + a m2 x 2 + . .. + amn x n ≤ bm
x1 , x 2 , .. ., x n ≥ 0 Standard form of the Dual
Standard form of the Dual
Standard
Standard
Problem
Problem
Problem
Problem
m min w = ∑ bi yi
y i =1 a11 y1 + a21 y 2 + .. . + a m1 y m ≥ c1
a12 y1 + a 22 y 2 + . . . + a m2 ym ≥ c 2 ).15 . ..
.. .
...
.. .
. ..
.. .
...
.. .
a1n y1 + a 2 n y2 + . .. + a mn y m ≥ c n
y1 , y 2 , .. . , ym ≥ 0 7.2.1 Definition
7.2.1 Definition
Dual Problem Primal Problem z*: = max Z = cx w : m w = yb
* = in s. t . s.t. x Ax ≤ b
x ≥0
).16 x yA≥ c
y ≥0 b is not assumed to be nonnegative 7.2.2 Example
7.2.2 Example
Primal max Z = 5x1 + 3x 2 − 8 x3 + 0 x 4 + 12 x5
x 3 x1 − 8x 2 + 9 x 4 − 15x 5 ≤ 20 18x1 + 5x 2 − 8 x3 + 4 x4 + 12 x5 ≤ 30
).17 x1, x2 , x3 , x4 , x5 ≥ 0 Dual min w = 20 y1 + 30 y2
y 3y1 + 18 y2 ≥ 5 − 8 y1 + 5y 2 ≥ 3
− 8y2 ≥ −8
9y1 + 4 y2 ≥ 0
−15 y1 + 12 y2 ≥ 12
).18 y1, y 2 ≥ 0 Table 7.1: PrimalDual
Table 7.1: PrimalDual
Table
Table
relationship
rrelationship
relationship
elationship
Dual
(mi n w) y1 ≥ 0
y2 ≥ 0 .. .
ym ≥ 0
Z= ).19 x1 ≥ 0
a11
a21
. ..
am1
≥ c1 x2 ≥ 0
a12
a22 xn ≥ 0
a1n
a2n c2 a mn
≥
cn ...
am2
≥ w=
≤
≤
. ..
≤ b1
b2 . ..
bn 7.2.3 Example
7.2.3 Example
max Z = 4 x1 + 10 x2 − 9 x3
x 5 x1 − 18 x2 + 5 x3 ≤ 15
−8 x1 + 12 x2 − 8x3 ≤ 8 ).20 12 x1 − 4 x 2 + 8 x3 ≤ 10
2 x1
− 5 x3 ≤ 5
x1, x2 , x3 ≥ 0 Dual
(mi n w) y1 ≥ 0
y2 ≥ 0
y3 ≥ 0
y4 ≥ 0
Z= ).21 x1 ≥ 0
5
−8
12
2
≥
4 x2 ≥ 0
 18
12
4
0
≥
10 x3 ≥ 0
5
0
8
5
≥
9 ≤
≤
≤
≤ w=
15
8
10
5 Dual
Dual min w = 15 y1 + 8y2 + 10 y3 + 5y4
y 5 y1 − 8 y2 + 12 y 3 + 2 y 4 ≥ 4
−18 y1 + 12 y 2 − 4 y 3
≥ 10
5 y1 +
8y 3 − 5 y 4 ≥ −9
y1 , y2 , y 3 , y4 ≥ 0
).22 7.3 FINDING THE DUAL
7.3 FINDING THE DUAL
7.3
7.3
OF NONSTANDARD
OF NONSTANDARD
LINEAR PROGRAMS
LINEAR PROGRAMS
LINEAR
LINEAR X The approach here is similar to the one we
The ).23 used in Section 5.6 when we dealt with nonused
standard formulations in the context of the
standard
simplex method.
simplex
X There is one exception: we do not add
There
artificial variables. We handle “=“
constraints by writing them as “<=“
constraints.
constraints. X This is possible here because we do not require here that the RHS is nonnegative.
nonnegative. ).24 k ∑ di xi = e i=1
k ∑ di xi ≤ e i=1
k ∑ di xi ≥ e i=1 k ∑d x
i i ≤e i =1
k − ∑ di x i ≤ − e
i= 1 Standard form!
).25 7.3.1 Example
7.3.1 Example
max Z = x1 + x2 + x3
x 2 x2 − x 3 ≥ 4
x1 − 3x 2 + 4 x 3 = 5
x1 − 2 x 2
≤3
x1 , x 2 ≥ 0; x3 urs: = unrestricted sign
).26 Conversion
Conversion
X Multiply through the greaterthanorequal ).27 to inequality constraint by 1
X Use the approach described above to
Use
convert the equality constraint to a pair of
convert
inequality constraints.
X Replace the variable unrestricted in sign, ,
Replace
by the difference of two nonnegative
variables.
variables. max Z = x1 + x2 +
x ,
x3 − ,,
x3 ,
,,
− 2 x2 + x3 − x3 ≤ − 4
, − 4 x ,, ≤ 5
x1 − 3x 2 + 4 x3
3
,
− x1 + 3 x2 − 4 x3 + 4 x, , ≤ −5
3 x1 − 2 x 2
).28 , , x, ,
x1, x2 , x3 3 ≤3
≥0 Dual
Dual
min w = −4 y1 + 5 y2 − 5y3 + 3 y4
y y2 − y3 + y4 ≥ 1
−2 y1 − 3 y2 + 3 y3 − 2 y4 ≥ 1
y1 + 4 y2 − 4 y3
− y1 − 4 y2 + 4 y3 ).29 y1, y2 , y3, y4 ≥ 0 ≥1
≥ −1 Streamlining the conversion ...
Streamlining the conversion ...
X An ).30 equality constraint in the primal
generates a dual variable that is
unrestricted in sign.
X An unrestricted in sign variable in the
primal generates an equality constraint in
the dual.
X Read the discussion in the lecture notes
X Good material for a question in the final
exam! Example 7.3.1 (Continued)
Example 7.3.1 (Continued)
min w = −4 y1 + 5 y2 − 5y3 + 3 y4
y y2 − y3 + y4 ≥ 1
−2 y1 − 3 y2 + 3 y3 − 2 y4 ≥ 1
y1 + 4 y2 − 4 y3
− y1 − 4 y2 + 4 y3 ).31 y1, y2 , y3, y4 ≥ 0 ≥1
≥ −1 min w =
,
X+ y correction
).32 ,
,
−4 y1 + 5 y2 ,
+ 3y3 ,
,
y2 + y3 ≥ 1
, − 3 y, − 2 y, ≥ 1
−2 y1
2
3
,
,
+
y1 − 4 y2 +
=1
, , y , ≥ 0; y, urs
y1 3
2 Table 7.2: PrimalDual
Table 7.2: PrimalDual
Table
rrelationship
elationship
Primal Problem Dual Problem opt=max opt=min Constraint i :
<= form
= form Variable i :
yi >= 0
yi urs Variable j:
xj > = 0
).33
xj urs Constraint j:
>= form
= form 7.3.3 Example
7.3.3 Example
max Z = 5x1 + 4 x 2
x 3x1 − 8 x2 ≥ −6
x1 + 6 x 2 = −5
8 x1 = 10 x 2 ≥ 0; x1 urs
).34 equivalent nonstandard form
equivalent nonstandard form
max Z = 5x1 + 4 x 2
x −3 x1 + 8 x 2 ≤ 6
x1 + 6 x2 = − 5
8 x1
).35 = 10 x 2 ≥ 0; x1 urs Dual from the recipe
Dual from the recipe
min w = 6 y1 − 5 y2 + 10 y3
y −3 y1 + y2 + 8 y3 = 5
8y1 + 6 y2
y1 ≥ 0; y2 , y3 urs
).36 ≥4 What about opt=min ?
What about opt=min ?
X Can use the usual trick of multiplying the
Can ).37 objective function by 1 (remembering to
undo this when the dual is constructed.)
undo
X It is instructive to use this method to
It
construct the dual of the dual of the
standard form.
standard
X i.e, what is the dual of the dual of the
i.e,
standard primal problem?
standard What is the dual of
What is the dual of w : m w= y
* = in
b
x st.
. ).38 y ≥c
A
y ≥0 max − w = − yb max − w = − yb s.t . s.t . y yA ≥ c
y ≥0
).39 y − yA ≤ −c
y ≥0 min − Z = −cx max Z = cx s. t . s. t . x − Ax ≥ − b
x ≥0
).40 x Ax ≤ b
x ≥0 Important Observation
Important Observation
X FOR
FOR ANY PRIMAL LINEAR
PROGRAM, THE DUAL OF THE
DUAL IS THE PRIMAL
DUAL ).41 Table 7.3: PrimalDual
Table 7.3: PrimalDual
Table
Table
Relationship
Relationship
Relationship
Relationship
Primal or Dual
opt=max
Constraint i :
<= form
= form
Variable j:
xj > = 0
).42
xj urs Dual or Primal
opt=min
Variable i :
yi >= 0
yi urs
Constraint j:
>= form
= form Example 7.3.4
Example 7.3.4
min Z = 6 x1 + 4 x2
x 3x1 − 5 x2 ≥ 12
x1 + 2 x 2 = −8
5x1 − x2 ≤ 10
x1, x1 ≥ 0 ).43 equivalent form
equivalent form
min Z = 6 x1 + 4 x2
x 3x1 − 5x 2 ≥ 12
x1 + 2 x2 = −8
−5 x1 + x2 ≥ −10
x1 , x 2 ≥ 0
).44 Dual
Dual
max w = 12 y1 − 8 y2 − 10 y3
y 3y1 + y2 − 5 y3 ≤ 6
−5 y1 + 2 y2 + y3 ≤ 4
y1, y3 ≥ 0; y2 urs
).45 ...
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This note was uploaded on 03/01/2012 for the course STRATEGY Om121 taught by Professor Abhinavdhar during the Spring '12 term at Harvard.
 Spring '12
 AbhinavDhar

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