optimization lecture3
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Lecture 3 The Dual Problem
ua s an pro em t at s er ve mat emat ca yfrom a given primal model
.
eL
P
the primal is a problem the dual will be a
problem and visa v
max min
er
,If LP
sa.this lecture we introduce the of the dual problem
based on the
definition
standard form
,
.
In
such by defining the dual problem f standardrom thethe results will be
formconsistentwith the info
, ,Asrmation contained in
the simplex tableau.
automatically accounts for all the forms given in the othertreatments
.
LP
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o f a p r im als tan da rd formT h e L P
1 1 2 2 n n z c x c x c x
1 1 2 2 n n1 1 1 1
a x a x a x
b
1 1 2 22 n n2 2 2
., ,,
1 1 2 2 n
j
m m m mn
x 0 j = 1 n
T
0,A x = b x
S u bject to
( ., ,
1 2 n=
x x x x
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that the variables include the and , , , .Note jn x artifslacks icials
t e purpose o construct ng t e ua , we arrange t ecoefficients of the primal schematically as shown in
.or
Table 1
diagrThe am shows that the dual is obtained symmetrically
from the primal according to the following rules:a dual variable
a dual
every primal c there is
there
onstraint
every primal varia isble
.
For
For
1.
2. constraint.
constraint coefficients of a primal variable form the left -side coefficients of the corres ondin dual constraint;
The3.
and the objective coefficient of the same variable becomes
t
he right side of the dual constraint.( ) , e.g., the tinted column under .jee xS
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Table 1
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Table 2
dual problemrules indicate that the willThesehave variables ( ) and constraints , , ... ,
1 2 mnm y y y
, , ... , .1 2 n
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xample 1- con.
Standard Primal
1 2 3 4
Subject to
1 2 3 4
x 2x x x 10
.
, , ,1 2 3 4
1 2 3 4 x 0 x 0 x
0 x 0
N slackthat is a in the first constraint;otice 4x
function and the second constrain .t
,
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.Example 1- con
Dual
1 2810 Min y y w
: 5
Subject to
x 2
:2 1 2 12y x 2 y
::
1 23
4
xx
y y ( )implies that1 2 1 0 0y y y0
, unrestricted1 2 y y
t at " s om nate y
dual constraint associated with .
" unrestricted .
bserve
The1 1
4
y
x
y
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, the should read asdual problemThus
1 210 8Min w y y
1 2 5
2y y
1 2
1 2 4
y 3y
, unrestricted1 20y y
the cha ng es in the du a l show n if its p r im alInd ica te
E x e rc ise 1
[ :
m in im iz a tio n m a x im iz a tis ins tea d o f io n
. C ha ng es are , fir s t th ree co ns tra in t s are
.
M a x im izA wen s
]o f th e typ e , a n d .1 0y
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Example 2
Primal Form
1 2
1.5
nSubject to
x x
x x
1 2 5 2x 3x
x x 0
Standard F rmo
1 2
S
x x x xubj
1ect
.50to
1 2 3 4 2x 3x x x
x x x
5
x 0
0
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Example 3
Primal 1 2
Sub ect to
1 2x 2x 5
-1 2
1 2 4x 7x 8
unrestricted1 x
2 x
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.xample 3 - con
wstandard rim
erethe becomesal
, , .etThen
1 1 1 1 1x x x x x
Standard primal
1 1 2z 5x 5x 6xMax
u ect to
x x 2x x x00 5
- 1 1 2 3 4x x 5x x x0 3
1 1 2 3 4 4x 4x 7x x x 80
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.Example 3 - con
ua 1 2 35y 3y 8y wMin
( )implies that
1 2 31 2 3- -
y y yy y
4 54 54 5
y
1 2 3 2 5 7 6 y y y
2 2
3 y 0
that the first and second constraints can be
replaced by the equation
dual
.
Observe
1 2 3y y y4
5
will always be the case when the primal variable is
unrestricted, meaning that an unrestricted
This
primal variable
uaw a ways ea to a equat on rather inequalihan tyt .
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Exercise 2
the changes in the dual just shownIndicate
if the objective is minimization and the first
" " .
the type" ", and y ], .2 3y0 0
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3. The dual problem in matrix form
now provide a general matrix definitionWe
standard
primal model:LP
T TPrimal
I III II
Su
ect
o
I II
Ax Ix b
,I IIx x
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( ) be the vector.. dual.., , ,Let 1 2 mY y y y
ru es n y e e uao ow ng :T
e a eMin w Y b
T
Subject to
I
I
I Y C
Y unrestricted vector
the primal problem is changed to the dualminimization
.
,If
II
problem sens maxime of optimization is changed izatto
and t
n
h
io
e first two sets of constraints are changed to withY
remaining unrestricted.
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Duality - Conclusion
DualPrimal
I
T T
I II IIMinMax
Sub ect to
w z C X C X Y b
I II
AX IX b IA CY
,I II
unrestricted vectorY
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the ofDual nonstaa ndardFinding LP
c x c x c x
Max problem
It s dual
Subject to
Min problem
w b b b
2 2 2 2
n n
1 1 2 2 n n
a x a x a x b
Subject to
m m1 1 2 2a x a x
1 1 2 2 m m
m
1 1 1 1
n mn
a x a
=
b
1 1 2 2 m m2 2 2 2a a cy y y
.
It s dual
1 1 2 2n n n nm m a ay y y
=
a c
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Example 4
Primal Dual
1 2 1 z 5x 2x wMax Min 12y 9
Subjec
t to
2 3y 20y
Subject to
1 2
1 2 2
13
4
x x
x x
12 y
9 y
1
3
1 2 3y y y 5
2
1
3 4 8
3 7
1 2 38x x
x 0
y27 0
, ,1 2 3y 0 y y
Original constraints Dual constraints
1 2
1 2
x x
x x4 3
1 2 3y y y 5
91 3 7
1 2x x 1 2 3
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Examp le 5
t e o ow ng pa r o pr ma an ua pro emsPrimal Prim
onsal
er
1 2z 0.4x 0.5x z 0Min Min
1 2.4x 0.5x
1 20.3 0.1x x 2.7
1 2 1y2.70.3 0.1x x- -
1 2
1 2
0.50.6
x xx x
0.50.4
66
-
- -
1 2
1 2
2
2
y6y6
0.50.5
0.50.5
x xx x
,1 2
x x 0 1 320.0.6x x y
x
4 6
0
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.Dual
Example 5 - con
-( )1 2 1 2 2 3
2.7y 6 y y 6ywz 0.4x 0.5x -MaxMi n
1 2 10.1 y2.7x x0.3
--
u ec o
-
-
( )1 2 2 3
y y y y 0.40.3 0- .5 0.6
- - -
1 2
1 2
2
2
0.5
0.
y60.5
0.5
x x
x x5 y6
-
-
, , ,
1 2 2 3
1 2 2 3
y .
0
y y y. . .
y y y y
,
1 2
2
3
1
0.6x x
x x 0
0
y6.4
Dual
1 2 32.7y 6yw - 6yMax
1 20.3 0.5 0- y y
-
3
0.1 0.5 0.4
y. 0.46
.50
,1 3 0yy
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Primal DualRelationship Between And Objective Values
satisfy the following relationship:
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carefully that these say nothing about whichtwo resu tslObserve
is the sense of optimization ( )
.
maximization and minimizationIt
t at matters n t s case
prove
.
To the validity of these results let ( ) and be the, ,I IIX YXfeasible primal and dual solutions corresponding to the primal
-dual definitions given in matrix form.
premultiplying the primal constra,Then ints by we get,Y
I IIX X w( Y b Y
we get
,T T T T T I I II II
IIX A C X X XY Y C
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that ; hence the direction of the inequality
remains unchan ed
,
.
Observe I IIX 0 X 0
adding the two constraints yields
T T T T T
,Then
the left - hand sides of th
I II I II
Since
I II z
e and identities above are equal,w zwe conc u e t at
z w
to show that at the optimum solutions observe
.
,Now z w that isz
means that seeks the highest value among all feasible ( )
and seeks the lowest value amon all feas lib e
,
.
This IIz X
w Y
X
( ) for all feasible solutions including the optima , the two
roblems will reach o timalit onl when .m
z w
wax
Si ce
min
n
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Example 6
e o ow ng pa r o pr ma an pro emua sons er
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feasible solutions given above are determined by inspectingThe
objective value in the maximization problem ( ) is less
.
The dual
result means that
.
This
the range ( ) is relatively narrow we can actually
to ,Since
14 15
think of the two feasible solutions above as being near optimalessence the given inequality can be used to test the goodness
. , " "In
two limit
of the feasible solutions
the ha en to be e ual the corres ondin solutionss
.
Ifare optimal.
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the of the followin robldu emal.Write3
1 2z x x 5 2 Max
Subject to
1 2
1 2x x 93 4
1 2x x 2087
,1 2
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