Knapsack Model

Notes 6
Intuitive idea: what is the most valuable
collection of items that can be fit into a
backpack?
IE312
1
Race Car Features
Cost (thousand)
Speed increase (mph)
$
1
10.2 $
8
Proposed Feature (j )
2
3
4
6.0 $ 23.0 $ 11.1 $
3
15
7
5
9.8 $
10
6
31.6
12
Budget of $35,000
Which features should be added?
Notes 6
IE312
2
Formulation


Decision variables
ILP
1 if feature j is added
xj  
otherwise
0
max 8 x1  3x2  15 x3  7 x4  10 x5  12 x6
10.2 x1  6.0 x2  23.0 x3  11.1x4  9.8 x5  31.6 x6  35
s.t
x j  0 or 1
Notes 6
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3
LINGO Formulation
Variables indexed by
this set
MODEL:
SETS:
Specify
index sets
FEATURES /F1,F2,F3,F4,F5,F6/: INCLUDE,SPEED_INC,COST;
ENDSETS
DATA:
SPEED_INC = 8 3 15 7 10 12;
All the
constants
COST = 10.2 6.0 23.0 11.1 9.8 31.6;
Note
; to end command
: to begin an
environment
BUDGET = 35;
ENDDATA
Objective MAX = @SUM( FEATURES: SPEED_INC * INCLUDE);
Constraints
@SUM( FEATURES: COST * INCLUDE) <= BUDGET;
@FOR( FEATURES: @BIN( INCLUDE));
END
Notes 6
IE312
Decision variables
are binary
4
Solve using Branch & Bound
x5  0
x5  1
Candidate Problem
max 8 x1  3x2  15 x3  7 x4  12 x6
10.2 x1  6.0 x2  23.0 x3  11.1x4  31.6 x6  35
s.t
Solution?
x j  0 or 1
Relaxed Problem
max 8 x1  3x2  15 x3  7 x4  12 x6
10.2 x1  6.0 x2  23.0 x3  11.1x4  31.6 x6  35
s.t
x j  [0,1]
Notes 6
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5
What is the Relative Worth?
x1 : 8
 0.78
Want to add this
feature second
10.2
x2 : 3  0.5
6
x3 : 15  0.65
23
x4 : 7
 0.63
11.1
x5 : 10
 1.02
9.8
x6 : 12
 0.38
31.6
Notes 6
Want to add this
feature first
IE312
6
Solve Relaxed Problem
x5  0
x5  1
Relaxed Problem
max 8 x1  3x2  15 x3  7 x4  12 x6
10.2 x1  6.0 x2  23.0 x3  11.1x4  31.6 x6  35
s.t
x j  [0,1]
8 1 15 1  7  0.257
Solution:
x1  1  Still have $35  $10.2  $24.8
x3  1  Still have $24.8  $23.0  $1.8
x4  1.8  0.257
7
Notes 6
IE312
Objective  24.8
7
Now the other node…
x5  0
x5  1
Relaxed Problem
max 8 x1  3x2  15 x3  7 x4  12 x6  10
10.2 x1  6.0 x2  23.0 x3  11.1x4  31.6 x6  35  9.8
s.t
x j  [0,1]
Solution:
x1  1  Still have $25.2  $10.2  $15
x3  15
Notes 6
23
 0.652
Objective  27.8
IE312
8
Next Step?
x5  0
x5  1
Objective  24.8
Notes 6
Objective  27.8
IE312
9
Rule of Thumb: Better Value
x5  0
Relaxed Problem
x5  1
max 3x2  15 x3  7 x4  12 x6  18
6.0 x2  23.0 x3  11.1x4  31.6 x6  15
s.t
Obj  24.8
x1  0
x1  1
Obj.  26.4
x j  [0,1]
Obj.  27.8
Relaxed Problem
Solution:
x3  15
23
 0.652
max 3x2  15 x3  7 x4  12 x6  10
6.0 x2  23.0 x3  11.1x4  31.6 x6  25.2
s.t
x j  [0,1]
Solution: x3  1  Still have $25.2  $23  $2.2
x4  2.2
Notes 6
11.1
 0.198
IE312
10
Next Level
Candidate Problem
x5  0
max 3x2  7 x4  12 x6  18
6.0 x2  11.1x4  31.6 x6  15
s.t
x5  1
x j  0,1
Obj  24.8
x1  0
x1  1
Solution:
x4  1, x2  x6  0
Obj.  26.4
x3  0
x3  1
Obj. = 25
(This turns out
to be true.)
Infeasible
Now What?
Notes 6
IE312
11
Next Steps …
x5  0
Still need to continue
branching here.
x5  1
Obj  24.8
Obj  25
Obj.  26.4
x1  0
x1  1
x3  0
x3  1
Obj. = 25
Notes 6
Infeasible
IE312
Finally we will
have accounted
for every
solution!
12
Capital Budgeting


Notes 6
Multidimensional knapsack problems
are often called capital budgeting
problems
Idea: select collection of projects,
investments, etc, so that the value is
maximized (subject to some resource
constraints)
IE312
13
NASA Capital Budgeting
Budget Requirements
2000- 2005- 2010- 2015- 2020Not Depends
j Mission
2004 2009 2014 2019 2024 Value With
On
1 Communication satellite
6
200
2 Orbital microwave
2
3
3
3 Io lander
3
5
20
4 Uranus orbiter 2020
10
50
5
3
5 Uranus orbiter 2010
5
8
70
4
3
6 Mercury probe
1
8
4
20
3
7 Saturn probe
1
8
5
3
8 Infrared imaging
5
10
11
9 Ground-based SETI
4
5
200
14
10 Large orbital structures
8
4
150
11 Color imaging
2
7
18
8
2
12 Medical technology
5
7
8
13 Polar orbital platform
1
4
1
1
300
14 Geosynchronous SETI
4
5
3
3
185
9
Budget
10
12
14
14
14
Notes 6
IE312
14
Formulation

Decision variables
1 if mission j is chosen
xj  
otherwise
0

Budget constraints
6 x1  2 x2  3x3  1x7  4 x9  5x12  10 (year 1)
Notes 6
IE312
15
Formulation


Mutually exclusive choices
x4  x5  1
x8  x11  1
x9  x14  1
Dependencies
x11  x2
x4  x3

Notes 6
IE312
16
Set Packing, Covering, and
Partitioning
Notes 6
IE312
17
Select Locations
Notes 6
IE312
18
Ways of Splitting the Set

Set covering constraints
x
jJ

1
Set packing constraints
x
jJ

j
j
1
Set partitioning constraints
x
jJ
Notes 6
j
1
IE312
19
Example: Choosing OR
Software
Software Package (j )
Algorithm 1
2
3
4
LP
Yes Yes Yes Yes
IP
Yes
Yes
NLP
Yes Yes
Cost
3
4
6
14
Formulate a set covering problem to acquire the minimum cost
software with LP, IP, and NLP capabilities.
Formulate set partitioning and set packing problems. What goals
do they meet?
Notes 6
IE312
20
Maximum Coverage




Perhaps the budget only allows $9000
What can we then do
 Maximum coverage
How do we now formulate the problem?
Need new variables
1 if ALG not provided
yALG  
otherwise
0
Notes 6
IE312
21
Travelling Salesman Problem
(TSP)
Fort Dodge
Waterloo
Carroll
Boone
Ames
Marshalltown
West Des Moines
Notes 6
IE312
What is the shortest route,
starting in Ames, that visits
each city exactly ones?
22
TSP Solution
Fort Dodge
Waterloo
Carroll
Boone
Ames
Marshalltown
West Des Moines
Notes 6
IE312
23
Not a TSP Solution
Fort Dodge
Waterloo
Carroll
Boone
Ames
Marshalltown
West Des Moines
Notes 6
IE312
24
Applications

Routing of vehicles (planes, trucks, etc.)

Routing of postal workers

Drilling holes on printed circuit boards

Routing robots through a warehouse,
etc.
Notes 6
IE312
25
Formulating TSP

A TSP is symmetric if you can go both
ways on every arc
1 if the route includes leg between i and j
xi , j  
otherwise
0
min
 c
i
j i
x
j i
Notes 6
i, j
x
i, j i, j
2
IE312
26
Example
10
1
1
1
1
5
2
3
10
10
4
1
1
1
6
Formulate a TSP
Notes 6
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27
Subtours




Notes 6
It is not sufficient to have two arcs
connected to each node
Why?
Must eliminate all subtours
Every subset of points must be exited
IE312
28
How do we eliminate
subtours?
10
1
1
1
1
5
Notes 6
2
3
10
10
IE312
4
1
1
1
6
29
Asymmetric TSP

Now we have decision variables
1 if the route goes from i to j
xi , j  
otherwise
0

Constraints
x
j i
i, j
x
j i
Notes 6
j ,i
1
(leave i )
1
(enter i )
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30
Asymmetric TSP (cont.)

Each tour must enter and leave every
subset of points
 x
iS jS

Notes 6
i, j
1
Along with all variables being 0 or 1,
this is a complete formulation
IE312
31
Example
10
1
1
1
1
5
2
3
10
4
10
1
1
1
6
Assume a two unit penalty for passing from a high to lower
numbered node.
This is now an asymmetric TSP. Why?
Notes 6
IE312
32
Subtour Elimination


Making sure there are no subtours
involves a very large number of
constraints
Can obtain simpler constraints if we go
with a nonlinear objective function
y k ,i
Notes 6
1 if the kth point visi ted is i

otherwise
0
IE312
33
Quadratic Assignment
Formulation
min
 d  y
 y 1
i, j
i
j
k ,i
yk 1, j
k
k ,i
i
y
k ,i
1
k
yk ,i  0,1
Notes 6
IE312
34
Example: reformulate
10
1
1
1
1
5
Notes 6
2
3
10
10
IE312
4
1
1
1
6
35