Constraints - Wood 492 | Modeling for Decision Support

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WOOD 492
MODELLING FOR DECISION SUPPORT
Lecture 17
Integer Programming
Integer Programming (IP)
• For discrete inputs/outputs
– Number of workers required in a factory
• For binary variables (yes/no decisions)
– Building a facility
• For logical conditions (if {x} then {y})
– If sawing pattern 1 is selected, then sawing pattern 2 can not be
selected
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Example 9: Integer Formulation Example
• A facility location problem
– A manufacturing company has $10 million available capital and would like to
invest this in building new factories and warehouses in Los Angeles and San
Francisco in order to maximize Net Present Value (NPV).
– Decision Variables: which facilities to build? (either build or don’t build, binary)
– Constraints:
– Factories can be built in L.A. or S.F, or both
– Only one warehouse can be located and should be in a city where a new factory
is being built
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Yes/No Variable
NPV
Capital required
Factory in L.A.
X1
$9 m
$6 m
Factory in S.F
X2
$5 m
$3 m
Warehouse in L.A.
X3
$6 m
$5 m
Warehouse in S.F.
X4
$4 m
$2 m
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• This is a Binary Integer Program (BIP)
• Obj Z=9 X1 +5 X2 +6 X3 +4 X4 (Total NPV)
• Subject to:
6 X1+3 X2+ 5 X3+ 2 X4
X3+
-X1
X4
+ X3
- X2
+ X4
Xi
Xi






10 (Total Capital)
1 (One warehouse)
0 (warehouse and factory
0 in the same city)
1 (Upper bound)
0 (Lower Bound)
LP Matrix
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Using indicator variables
• Indicators are binary variables used for modelling
problems with:
–
–
–
–
Fixed costs
Either/OR constraints
If/Then constraints
…
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Fixed-Charge Problems
• When taking up an activity has a one-time fixed cost
– production line set-up cost
– Road establishment cost
• Example:
– A cabinet manufacturer can outsource the cabinet doors at
$20/door or make them in the factory with $500 line set-up cost
and then $10/door production cost.
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Fixed-Charge Problems
• Objective is to minimize costs
• Decision variables are
– X1 number of outsourced doors
– X2 number of manufactured doors
• Total cost for outsourced doors:
• 20 X1
• Total cost for the manufactured doors:
• 0
• 500 + 10 X2
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if X2=0
if X2>0
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Fixed-Charge Problems
• How can we force the $500 costs when X2 is greater
than 0?
• We define an indicator variable Y2 :
– if X2=0 then Y2=0
– otherwise Y2=1
– Now, rewrite the formula for production cost:
• Manufacturing cost: 500 Y2 + 10 X2
• Objective: Min Z= 20 X1 + 500 Y2 + 10 X2
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Fixed-Charge Problems
• Now, how to ensure that Y2 value changes according to X2 value?
• Big M method:
– Choose a big number M (relative to problem parameters, bigger
than any possible value of X2)
– Add the following to the existing constraints
• X2 <= M.Y2
• Y2 is binary
– How does this constraint work?
• If X2>0, then Y2 has to be 1, so that X2<=M
• If X2=0, then Y2 can be either 0 or 1, however since 500Y2 is a
term in the objective function, the solution algorithm always picks
Y2=0, because it results in the lower objective function value
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Either/OR Constraints
• When only one of two constraints must hold
– E.g. there are two suppliers for cabinet doors
– Each supplier has a certain number of doors available for
shipping
– Only one of the suppliers can be selected
• Variables: number of doors purchased (X)
• Supply constraints (depending on which supplier is elected):
– Either X <= 100
– OR X <= 150
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Either/OR Constraints
• Define an indicator variable Y:
– If first supplier is selected, Y=1 and first constraint is activated
– otherwise Y=0 and second constraint is activated
• Use the Big M Method
– Constraint 1: X <= 100 + M. (1-Y)
– Constraint 2: X <= 150 + M. Y
• How does this formulation work?
– If supplier 1 is selected, then Y=1, so the first constraint will be X<=100 and the
second constraint will be X<=150+M which in effect eliminates the second
constraint (since M is a really big number)
– If supplier 2 is selected, then Y=0 and the first constraint is eliminated
– The solution algorithm picks the Y value that results in the best value of the
objective function
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Lab 6 preview
• Same problem as in Lab 3
• Adding extra constraints:
– Building the roads to each cut block (if we harvest, we have to
build the road first)
– Selecting only one sawing pattern for all log sizes
– Selecting the optimal number of shifts to run the sawmill (1,2, or
3)
LP matrix
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Next Class
• Branch and bound
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