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IE521 Advanced Optimization
Lecture 8
Dr. Zeliha Akc
¸a
December 2011
1 / 23
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View Full Document Hw3 and Midterm 1 Grades
I
Hw3 is posted.
I
Midterm 1 was graded.
I
Max grade = 86
I
Average = 44.3
2 / 23
Recall Previous Lectures
I
Example:
For the following LP,
x
1
and
x
2
are
basic variables
in
the
optimal
tableau. Construct the optimal tableau.
min

3
x
1

x
2
2
x
1

x
2
≤
2

x
1
+
x
2
≤
4
x
1
,
x
2
≥
0
3 / 23
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View Full Document Recall Lecture 7: Motivation of Duality
I
Allow the constraint to be violated
for a price
.
⇒
Under a speciﬁc value of
p
, the presence or absence of the
constraint does not affect the optimal cost
I
These prices can be found by solving a
new linear program
called
dual problem
.
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We have seen the procedure to construct a dual problem from a
Primal Problem:
min
c
>
x
s.t.
Ax
=
b
x
≥
0
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We relax constraints, form function
g
(
p
)
, and construct and LP to
maximize the value of
g
(
p
)
, the
dual problem:
max
p
>
b
s.t.
p
>
A
≤
c
>
4 / 23
Recall: How to Come Up with the Dual Problem
I
Consider the
Primal Problem:
min
c
>
x
s.t.
Ax
=
b
x
≥
0
I
Relaxing
the constraints for a price
p
:
min
c
>
x
+
p
>
(
b

Ax
)
s.t.
x
≥
0
I
Deﬁne a function of
p
,
g
(
p
)
which is the optimal cost to the new
problem.
g
(
p
) =
min
x
≥
0
±
c
>
x
+
p
>
(
b

Ax
)
²
I
g
(
p
)
is a lower bound for the optimal solution of primal problem.
I
To ﬁnd the
best
lower bound
maximize
g
(
p
)
.
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Dual problem:
Maximize
g
(
p
)
.
5 / 23
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View Full Document Recall: Constructing the Dual Problem
I
We have,
g
(
p
)
=
min
x
≥
0
±
c
>
x
+
p
>
(
b

Ax
)
²
=
p
>
b
+
min
x
≥
0
(
c
>

p
>
A
)
x
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Note that
min
x
≥
0
(
c
>

p
>
A
)
x
=
(
0
,
if
c
>

p
>
A
≥
0
>
,
∞
,
otherwise
,
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We only need to consider the ﬁrst case.
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Therefore, maximization of
g
(
p
)
becomes the following
dual
problem:
max
p
>
b
s.t.
p
>
A
≤
c
>
6 / 23
Example
min

x
1

x
2
x
1
+
8
3
x
2
+
x
3
=
4
x
1
+
x
2
+
x
4
=
2
x
1
,
x
2
,
x
3
,
x
4
≥
0
Construct function
g
(
p
)
. Maximize
g
(
p
)
and construct dual problem.
7 / 23
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View Full Document From the Primal to the Dual
We can dualize general LPs as follows
PRIMAL
minimize
maximize
DUAL
≥
b
i
≥
0
constraints
≤
b
i
≤
0
variables
=
b
i
free
≥
0
≤
c
j
variables
≤
0
≥
c
j
constraints
free
=
c
j
8 / 23
From the Primal to the Dual
min
c
>
x
s.t.
a
>
i
x
≥
b
i
i
∈
M
1
a
>
i
x
≤
b
i
i
∈
M
2
a
>
i
x
=
b
i
i
∈
M
3
x
j
≥
0
j
∈
N
1
x
j
≤
0
j
∈
N
2
x
j
free
j
∈
N
3
max
p
>
b
s.t.
p
i
≥
0
i
∈
M
1
p
i
≤
0
i
∈
M
2
p
i
free
i
∈
M
3
p
>
A
j
≤
c
j
j
∈
N
1
p
>
A
j
≥
c
j
i
∈
N
2
p
>
A
j
=
c
j
i
∈
N
3
9 / 23
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This note was uploaded on 01/09/2012 for the course IE 521 taught by Professor Zelihaakça during the Fall '11 term at Fatih Üniversitesi.
 Fall '11
 ZelihaAkça
 Optimization

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