notes 5ie 3121 knapsack model intuitive idea: what is the most valuable collection of items that can...
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Notes 5 IE 312 1
Knapsack Model Intuitive idea: what is the most
valuable collection of items that can be fit into a backpack?
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Notes 5 IE 312 2
Race Car Features
1 2 3 4 5 6Cost (thousand) 10.2$ 6.0$ 23.0$ 11.1$ 9.8$ 31.6$ Speed increase (mph) 8 3 15 7 10 12
Proposed Feature (j )
Budget of $35,000
Which features should be added?
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Notes 5 IE 312 3
Decision variables
ILP
Formulation
otherwise0
added is feature if1 jx j
1or 0
356.318.91.110.230.62.10s.t.121071538max
654321
654321
jx
xxxxxxxxxxxx
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Notes 5 IE 312 4
LINGO FormulationMODEL:
SETS:
FEATURES /F1,F2,F3,F4,F5,F6/: INCLUDE,SPEED_INC,COST;
ENDSETS
DATA:
SPEED_INC = 8 3 15 7 10 12;
COST = 10.2 6.0 23.0 11.1 9.8 31.6;
BUDGET = 35;
ENDDATA
MAX = @SUM( FEATURES: SPEED_INC * INCLUDE);
@SUM( FEATURES: COST * INCLUDE) <= BUDGET;
@FOR( FEATURES: @BIN( INCLUDE));
END
Specifyindex sets
All theconstants
Objective
Constraints
Variables indexed bythis set
Decision variablesare binary
Note ; to end command: to begin an environment
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Notes 5 IE 312 5
Solve using Branch & Bound
5 0x 5 1x
Solution?
Candidate Problem
Relaxed Problem
1or 0
356.311.110.230.62.10s.t.1271538max
64321
64321
jx
xxxxxxxxxx
1 0
356.318.91.110.230.62.10s.t.121071538max
654321
654321
jx
xxxxxxxxxxxx
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Notes 5 IE 312 6
What is the Relative Worth?
1
2
3
4
5
6
8: 0.7810.23: 0.5615: 0.65237: 0.6311.110: 1.029.812: 0.3831.6
x
x
x
x
x
x
Want to add thisfeature first
Want to add thisfeature second
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Notes 5 IE 312 7
Solve Relaxed Problem
5 0x 5 1x
Solution:
Relaxed Problem
Objective <= 24.8
8 1 15 1 7 0.257 1 0
356.318.91.110.230.62.10s.t.121071538max
654321
654321
jx
xxxxxxxxxxxx
money ofout ,78.1
$1.8 $23.0 - $24.8 have still ,1
$24.8 $10.2 - $35 have still ,1
4
3
1
x
x
x
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Notes 5 IE 312 8
Now the other node…
5 0x 5 1x
Relaxed Problem
Solution:
1
3
1 Still have $25.2 $10.2 $15
15 0.65223
x
x
Objective <= 27.8
1 0
8.9356.311.110.230.62.10s.t.121071538max
64321
64321
jx
xxxxxxxxxx
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Notes 5 IE 312 9
Next Step?
5 0x 5 1x
Objective <= 24.8 Objective <= 27.8
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Notes 5 IE 312 10
Rule of Thumb: Better Value
5 0x 5 1x
Obj <= 24.8
1 0x 1 1x Solution:
315 0.65223x
Obj. <= 27.8
3
4
1 Still have $25.2 $23 $2.2
2.2 0.19811.1
x
x
Obj. <=26.4
Solution:
Relaxed Problem
1 0
2.256.311.110.230.6s.t.12715310max
6432
6432
jx
xxxxxxxx
Relaxed Problem
1 0
2.156.311.110.230.6s.t.12715318max
6432
6432
jx
xxxxxxxx
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Notes 5 IE 312 11
Next Level
5 0x 5 1x
Obj <= 24.8
1 0x 1 1x
Obj. <=26.4
3 0x 3 1x
InfeasibleObj. = 25
Now What?
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Notes 5 IE 312 12
Next Steps …
5 0x 5 1x
Obj <= 24.8
1 0x 1 1x
Obj. <= 26.43 0x 3 1x
InfeasibleObj. = 25
Still need to continue branching here.
Finally we will have
accounted for every solution!
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Notes 5 IE 312 13
Capital Budgeting 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)
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Notes 5 IE 312 14
NASA Capital Budgeting2000- 2005- 2010- 2015- 2020- Not Depends
j Mission 2004 2009 2014 2019 2024 Value With On1 Communication satellite 6 2002 Orbital microwave 2 3 33 Io lander 3 5 204 Uranus orbiter 2020 10 50 5 35 Uranus orbiter 2010 5 8 70 4 36 Mercury probe 1 8 4 20 37 Saturn probe 1 8 5 38 Infrared imaging 5 10 119 Ground-based SETI 4 5 200 14
10 Large orbital structures 8 4 15011 Color imaging 2 7 18 8 212 Medical technology 5 7 813 Polar orbital platform 1 4 1 1 30014 Geosynchronous SETI 4 5 3 3 185 9
Budget 10 12 14 14 14
Budget Requirements
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Notes 5 IE 312 15
Formulation Decision variables
Budget constraints
otherwise0
chosen is mission if1 jx j
1 2 3 7 9 126 2 3 1 4 5 10 (year 1)x x x x x x
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Notes 5 IE 312 16
Formulation Mutually exclusive choices
Dependencies
1
1
1
149
118
54
xx
xx
xx
34
211
xx
xx
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Notes 5 IE 312 17
Assignment Problems Assignment problems deal with
optimal pairing or matching of objects in two distinct sets
Decision variable
Let A be the set of allowed assignments and cij be the cost of assigning i to j.
otherwise0
toassigned is if1 jixij
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Notes 5 IE 312 18
Formulation
A(i,jx
Njx
Nix
xc
ij
ij
Ajii
ij
Ajij
ijij
Aji
in ) allfor 1,0
allfor ,1
allfor ,1s.t.
min
2
in ),(:
in ),(:
in ),(
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Notes 5 IE 312 19
Example We must determine how jobs should be
assigned to machines to minimize setup times, which are given below:
Job 1 Job 2 Job 3 Job 4
Machine 1
14 5 8 7
Machine 2
2 12 6 5
Machine 3
7 8 3 9
Machine 4
2 4 6 10
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Notes 5 IE 312 20
Hungarian Algorithm Step 1: (a) Find the minimum
element in each row of the cost matrix. Form a new matrix by subtracting this cost from each row. (b) Find the minimum cost in each column of the new matrix, and subtract this from each column. This is the reduced cost matrix.
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Notes 5 IE 312 21
Example: Step 1(a)Job 1 Job 2 Job 3 Job 4
Machine 1
14 5 8 7
Machine 2
2 12 6 5
Machine 3
7 8 3 9
Machine 4
2 4 6 10Job 1 Job 2 Job 3 Job 4
Machine 1
9 0 3 2
Machine 2
0 10 4 3
Machine 3
4 5 0 6
Machine 4
0 2 4 8
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Notes 5 IE 312 22
Example: Step 1(b)
Job 1 Job 2 Job 3 Job 4
Machine 1
9 0 3 0
Machine 2
0 10 4 1
Machine 3
4 5 0 4
Machine 4
0 2 4 6
Job 1 Job 2 Job 3 Job 4
Machine 1
9 0 3 2
Machine 2
0 10 4 3
Machine 3
4 5 0 6
Machine 4
0 2 4 8
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Notes 5 IE 312 23
Hungarian Algorithm Step 2: Draw the minimum
number of lines that are needed to cover all the zeros in the reduced cost matrix. If m lines are required, then an optimal solution is available among the covered zeros in the matrix. Otherwise, continue to Step 3.
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Notes 5 IE 312 24
Example: Step 2
Job 1 Job 2 Job 3 Job 4
Machine 1
9 0 3 0
Machine 2
0 10 4 1
Machine 3
4 5 0 4
Machine 4
0 2 4 6We need 3<4 lines, so continue to Step 3
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Notes 5 IE 312 25
Hungarian Algorithm Step 3: Find the smallest nonzero
element (say, k) in the reduced cost matrix that is uncovered by the lines. Subtract k from each uncovered element, and add k to each element that is covered by two lines. Return to Step 2.
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Notes 5 IE 312 26
Example: Step 3Job 1 Job 2 Job 3 Job 4
Machine 1
9 0 3 0
Machine 2
0 10 4 1
Machine 3
4 5 0 4
Machine 4
0 2 4 6Job 1 Job 2 Job 3 Job 4
Machine 1
10 0 3 0
Machine 2
0 9 3 0
Machine 3
5 5 0 4
Machine 4
0 1 3 5
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Notes 5 IE 312 27
Example: Step 2 (again)
Job 1 Job 2 Job 3 Job 4
Machine 1
10 0 3 0
Machine 2
0 9 3 0
Machine 3
5 5 0 4
Machine 4
0 1 3 5Need 4 lines, so we have the optimal assignment and we stop
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Notes 5 IE 312 28
Example: Final SolutionJob 1 Job 2 Job 3 Job 4
Machine 1
10 0 3 0
Machine 2
0 9 3 0
Machine 3
5 5 0 4
Machine 4
0 1 3 5Optimal assignment
1,1,1,1 24413312 xxxx
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Notes 5 IE 312 29
Travelling Salesman Problem (TSP)
Ames
Fort Dodge
BooneCarrollMarshalltown
West Des Moines
Waterloo
What is the shortest route,starting in Ames, that visitseach city exactly ones?
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Notes 5 IE 312 30
TSP Solution
Ames
Fort Dodge
BooneCarrollMarshalltown
West Des Moines
Waterloo
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Notes 5 IE 312 31
Not a TSP Solution
Ames
Fort Dodge
BooneCarrollMarshalltown
West Des Moines
Waterloo
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Notes 5 IE 312 32
Applications
Routing of vehicles (planes, trucks,
etc.)
Routing of postal workers
Drilling holes on printed circuit boards
Routing robots through a warehouse,
etc.
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Notes 5 IE 312 33
Formulating TSP A TSP is symmetric if you can go
both ways on every arc
otherwise0
and between leg includes route theif1,
jix ji
i ij
jiji xc ,,min
2, ij
jix
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Notes 5 IE 312 34
Example1
5
2
6
3 4
10
10
10
11
1
1
1 1
Formulate a TSP
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Notes 5 IE 312 35
Subtours It is not sufficient to have two arcs
connected to each node Why?
Must eliminate all subtours Every subset of points must be
exited
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Notes 5 IE 312 36
How do we eliminate subtours?
1
5
2
6
3 4
10
10
10
11
1
1
1 1
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Notes 5 IE 312 37
Asymmetric TSP Now we have decision variables
Constraints
otherwise0
to from goes route theif1,
jix ji
)(enter 1
) (leave 1
,
,
ix
ix
ijij
ijji
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Notes 5 IE 312 38
Asymmetric TSP (cont.) Each tour must enter and leave
every subset of points
Along with all variables being 0 or 1, this is a complete formulation
1, Si Sj
jix
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Notes 5 IE 312 39
Example1
5
2
6
3 4
10
10
10
11
1
1
1 1
Assume a two unit penalty for passing from a high to lowernumbered node.
This is now an asymmetric TSP. Why?
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Notes 5 IE 312 40
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
otherwise0
is tedpoint visith theif1,
iky ik
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Notes 5 IE 312 41
Quadratic Assignment Formulation
1,0
1
1
min
,
,
,
,1,,
ik
kik
iik
i j kjkikji
y
y
y
yyd
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Notes 5 IE 312 42
Example: reformulate1
5
2
6
3 4
10
10
10
11
1
1
1 1
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Notes 5 IE 312 43
Solving TSP We can use branch-and-bound to get
an exact solution to the TSP problem As always, the key to implementing
branch-and-bound is to relax the problem so that we can easily solve the relaxed problem, but we still get good bounds
How can we relax the TSP?
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Notes 5 IE 312 44
Relaxing the TSP
)(enter 1
) (leave 1
,
not
,
not
ix
ix
ij
ij
ji
ij
1,
in not in
ji
SiSi
x
1,0, jix
jiji
iji
xc ,,
not
min
Reduces to the assignment problem
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Notes 5 IE 312 45
Branching
0, jix 0, ijx
Which cities should be selected?
What is the most important variable?
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Notes 5 IE 312 46
Example: Solution to Assignment Problem
Ames
DSM
CR
IC
1,,,, CRICICCRAmesDSMDSMAmes xxxx
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Notes 5 IE 312 47
Branching
0, AmesDSMx 0, DSMAmesx
Always branch to split up the subtours!
Solving the assignment problem will hopefully yield a feasible solution soon.
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Notes 5 IE 312 48
Solution to New Assignment Problem (Left Branch)
Ames
DSM
CR
IC
This makes both subtours impossible, so we get
Thus, optimal solution is found by solving a total of three assignment problems!
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Notes 5 IE 312 49
Poor Branching Examples
0, AmesCRx 0, CRAmesx
0, AmesCRx 1, AmesCRx
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Notes 5 IE 312 50
Set Packing, Covering, and Partitioning
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Notes 5 IE 312 51
Select Locations
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Notes 5 IE 312 52
Ways of Splitting the Set Set covering constraints
Set packing constraints
Set partitioning constraints
Jj
jx 1
Jj
jx 1
Jj
jx 1
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Notes 5 IE 312 53
Example: Choosing OR Software
Algorithm 1 2 3 4LP Yes Yes Yes YesIP Yes YesNLP Yes YesCost 3 4 6 14
Software Package (j )
Formulate a set covering problem to acquire the minimum costsoftware with LP, IP, and NLP capabilities.
Formulate set partitioning and set packing problems. What goalsdo they meet?
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Notes 5 IE 312 54
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
otherwise0
providednot ALG if1ALGy
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Notes 5 IE 312 55
Useful Formulation Tricks! Integer variables can be used to
model numerous things that seem difficult Fixed-charge IP Either-or constraints If-then constraints
All of these involve introducing a (dummy) binary variable and a “very large number”
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Fixed-Charge Constraint We need to determine how many Gadgets to make
We have variable cost of $7/ Gadget made, but also a fixed cost of $500 if we make any Gadgets (but we could also choose not to make any, in which case we would have no cost
How do we represent the cost?
Here, M is a very large number, e.g. M = 1000
yMx
yxCost
y
5007
otherwise0
Gadgets make weif1
made Gadgets ofNumber x
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Notes 5 IE 312 57
What-If Constraints
1,0
)1(100
100or 0
y
yMx
Myx
xx
Need to model (continuous variable):
Trick: add a binary variable y, and a very large M
Here, M = 200 would do nicely
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Notes 5 IE 312 58
If-Then Constraints We want to represent
Similar to before:
Here, M=3 would be big enough!
0 then 1 4321 xxxx
1,0
)1(1
432
y
yMx
yMxxx