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Column Generation
Teo ChungPiaw (NUS)
25
th
February 2003, Singapore
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View Full Document 1L
e
c
t
u
r
e
1.1
Outline
•
Cutting
Stock
Problem
•
Classical
Integer
Programming
Formulation
•
Set
Covering
Formulation
•
Column
Generation
Approach
•
Connection
with
Lagrangian
Relaxation
•
Computational
issues
2
Cutting
Stock
Problem
2.1
Introduction
2.1.1
Example
Slide 1
Slide 2
•
A
paper
company
has
a
supply
of
large
rolls
of
paper,
each
of
width
W
.
Slide 3
•
Customers
demand
n
i
rolls
of
width
w
i
(
i
=1
,...,m
).
(
w
i
≤
W
)
Example:
Quantity
Ordered
n
i
Order
Width
(inches)
w
i
97
45
610
36
395
31
211
14
Slide 4
•
The
demand
can
be
met
by
slicing
a
large
roll
in
a
certain
way,
called
a
pattern
.
•
For
example,
a
large
roll
of
width
100
can
be
cut
into
1
±
±
±
–
4
rolls
each
of
width
25,
or
–
2
rolls
each
of
width
35,
with
a
waste
of
30.
3
Solution
Approach
I
3.1
L. V. Kantorovich
3.1.1
Formulation
Slide 5
(1939
Russian,
1960
English)
“Mathematical
Methods
of
Planning
and
Organ
ising
Production”
Management Science
,
6
,
366422.
• K
:
Set
of
available
rolls.
•
y
k
:
1
if
roll
k
is
cut,
0
otherwise.
•
x
k
:
number
of
times
item
i
is
cut
on
roll
k
.
i
Objective:
To
minimize
the
number
of
rolls
used
to
meet
all
the
demand
k
min
y
k
∈K
Slide 6
Constraints
•
Total
number
of
times
item
i
is
cut
is
not
less
than
the
demand.
k
x
i
≥
n
i
k
∈K
•
The
width
of
a
roll
is
at
most
W
k
w
i
x
i
≤
Wy
k
i
2
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View Full Document ∑
∑
∑
Slide 7
k
min
k
∈K
y
k
s.t.
k
∈K
x
i
≥
n
i
for
a
ﬁxed
item
i,
k
≤
Wy
k
i
w
i
x
i
for
ﬁxed
roll
k,
x
k
≥
0
,
0
≤
y
k
≤
1
i
Integrality
constraints
on
all
variabes
3.1.2
Quality of Solution
Slide 8
Scenario I
:
N1
•
n
i
:
uniform,
between
1
and
100
(rand(100)+1);
•
w
i
:
uniform,
between
1
and
30
(rand(30)+1);
•
Width
of
Roll,
W
=
3000;
Rolls
Items
constr
variables
CPU
(s)
30
60
90
50
100
150
100
200
300
200
400
600
1830
2.8
5050
14.33
20100
179
80200
3048
Note 1
Code
The
OPL
code
for
the
problem:
3
∑
±
±
±
Slide 9
Scenario II
:
Change
width
of
the
roll
from
3000
to
150.
Rolls
Items
constr
variables
CPU
(s)
70
10
80
770
3.18
 never
140
20
160
2940
58.28
 never
210
30
240
6510
Out
of
memory
The
performance
has
deteriorated.
Why?
Slide 10
How
good
is
the
LP
relaxation?
w
i
n
i
Observation
:
Z
LP
=
i
W
.
N2
This
bound
is
trivial:
The
objective
is
to
k
min
y
k
∈K
the
optimal
solution
will
satisfy:
•
Choose
y
k
as
small
as
possible.
Therefore
k
w
i
x
i
=
Wy
k
i
for
all
k
Slide 11
k
•
Choose
x
i
as
small
as
possible.
Therefore
k
x
i
=
n
i
k
4
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View Full Document ±
∑
∑
±
±
∑
±
±
±
±
±
for
all
i
.
±
y
k
=
±
∑
i
w
i
x
k
i
W
=
±
±
w
i
W
x
k
i
k
∈K
k
∈K
i
k
∈K
=
±
w
i
W
±
x
k
i
=
±
w
i
n
i
W
i
k
∈K
i
Note 2
Proof
We
have
the
constraints
k
w
i
x
i
≤
Wy
k
.
i
In
the
LP,
since
the
objective
is
to
minimize
k
y
k
,
the
optimal
LP
solution
will
be
such
that
k
i
w
i
x
i
=
y
k
.
W
k
y
k
will
be
small
if
the
x
i
values
are
small.
At
the
same
time,
because
k
x
i
≥
n
i
,
k
∈K
k
to
make
x
i
small,
at
the
optimal
solution,
we
must
have
k
x
i
=
n
i
.
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This note was uploaded on 12/04/2011 for the course ESD 15.094 taught by Professor Jiesun during the Spring '04 term at MIT.
 Spring '04
 JieSun

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