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1
15.053
Thursday, April 11
z
Some more applications of integer
programming
z
Cutting plane techniques for getting improved
bounds
Handouts:
Lecture Notes
2
Example:
Fire company location.
z
Consider locating fire companies in different
districts.
z
Objective:
place fire companies so that each
district either has a fire company in it, or one
that is adjacent, and so as to minimize cost.
3
Example for the
Fire Station Problem
123
4
56
7
89
11
10
12
14
15
13
16
Let x
j
= 1 if a fire
station is placed in
district j.
x
j
= 0
otherwise.
Let c
j
= cost of putting a
fire station is district j.
With partners,
formulate the fire
station problem.
4
Set covering problem
z
set S= {1, …, m} of items to be covered
–
districts that need to have a fire station or be next
to a district with one
z
n subsets of S
–
For each possible location j for a fire station, the
subset is district j plus the list of districts that are
adjacent to district j.
–
a
ij
= 1 if district i is adjacent to district j or if i = j.
jj
j
cx
∑
Minimize
1
for each i
ij
j
j
ax
≥
∑
is binary for each j
j
x
subject to
5
Covering constraints are common
z
Fleet routing for airlines:
–
Assigning airplanes to flight legs.
–
Each flight must be included
z
Assigning crews to airplanes
–
Each plane must be assigned a crew
z
Warehouse location
–
Each retailer needs to be served by some
warehouse
6
Independent Set (set packing)
4
7
11
10
12
14
15
13
16
What is the maximum
number of districts no
two of which have a
common border?
With your partner,
formulate the
independent set
problem.
List all
constraints containing
x
10
.
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Set Packing constraints arise often in
manufacturing and logistics
z
Cutting shapes out of sheet metal
z
Manufacturing many items at once which
share resources (how much work can be
done in parallel?)
8
Review from Last Lecture
Investment
1
2
3
4
5
6
Cash
Required
(1000s)
$5
$7
$4
$3
$4
$6
NPV
added
(1000s)
$16
$22
$12
$8
$11
$19
Investment budget = $14,000
maximize
16x
1
+ 22x
2
+ 12x
3
+ 8x
4
+11x
5
+ 19x
6
subject to
5x
1
+ 7x
2
+ 4x
3
+ 3x
4
+4x
5
+ 6x
6
≤
14
x
j
binary for j = 1 to 6
9
Branch and Bound
1
2
3
x
1
= 0
x
1
= 1
44
44
44
4
x
2
= 0
42
5
x
2
= 1
6
x
2
= 0
7
x
2
= 1
44
44
44
The
incumbent
solution has
value 43
x
3
= 0
x
3
= 1
x
3
= 0
x
3
= 1
12
x
3
= 0
13
x
3
= 1
43
43
43
43
44
Inf
8
9
10
11
Inf
Inf
14
15
16
17
44
44
18
19
Inf
38
10
On Bounds from Linear Programming
We found an incumbent with a value of 43.
The best LP solution value is
between 44 and 45.
Is there a way that we could have directly established
an upper bound between 43 and 44?
Perhaps we could
form a “better linear program.”
The closer the LP is to the Integer Program, the better.
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 Spring '05
 Prof.JamesOrlin

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