Class_5_IntProgram_II

# Class_5_IntProgram_II - Integer Programming Models: part II...

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EMSE 154-254: Applied Optimization Modeling Integer Programming Models: part II

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EMSE 154 - 254: Applied Optimization Modeling Topics Modeling Logical constraints If-then logic Either-or logic Modeling piecewise linear functions Branch and Bound Method for IPs 2
EMSE 154 - 254: Applied Optimization Modeling Either-or logical constraints Suppose we have two constraints, namely gx ( ) 0 fx ( ) 0 We want at least one of them to be true. That is, either or . ( ) 0 () 0 To achieve this, we introduce a binary variable y , and require g x ( ) M (1 y ) My 3 M represents a suitable constant that does not block possible values of f and g.

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EMSE 154 - 254: Applied Optimization Modeling Example: Dorian Auto Dorian Auto wants to maximize profit by selling compact, midsize, and large cars, with the following characteristics: ar Type Car Type Resource Compact Midsize Large Steel Required 1.5 tons 3 tons 5 tons Labor required 30 hours 25 hours 40 hours Currently 6,000 tons of steel, and 60,000 labor hours q Profit Yielded (\$) 2,000 3,000 4,000 are available. If a car model is to be produced, then at least 1,000 4 units should be produced.
EMSE 154 - 254: Applied Optimization Modeling Dorian Auto: Solution We define the usual production variables x i =number of cars of type i to be produced ( i =compact,midsize,large) Note that the constraint on the minimum production of each model requires that x i 0 or x i 1000 By defining g(x i )=x i , and f(x i )=1000-x i , the problem is one of an either- or nature. To capture this logic, we introduce a binary variable y i , to take the value of 1 if cars of type i are to be produced. Then, for each product of type i , we add the constraints: ii i xM y 5 1000 (1 ) 1, 2, 3 i x My i ≤− =

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EMSE 154 - 254: Applied Optimization Modeling Dorian Auto: Solution ii i xM y 1000 (1 ) 1, 2, 3 i x My i −≤ = We need to determine appropriate values for each M i so that the quantities x i , and 1000-x i , are not artificially constrained. An upper bound for the expressions 1000-x i , is 1000 (attainable if x i =0), so each M i must be at least 000. o bound the possible values of each we consider the production 1000. To bound the possible values of each x i , we consider the production capacity constraints: 123 1.5 3 5 6000 (Steel constraint) xx x ++≤ From the steel constraint we conclude that and from the labor constraint we have Thus, we can choose M = 2000. Similarly, we 30 25 40 60000 (Labor constraint) x 1 4000 x 2000. x 6 1 can set M 2 = 2000 and M 3 = 1200. Of course, higher values of M also work, but tighter formulations are easier to solve by the algorithms. (good optimization software packages also do a preprocess procedure to tighten the formulation). 1
EMSE 154 - 254: Applied Optimization Modeling Dorian Auto Completed 12 3 max 2 3 4 zxx x = ++ 11 2 . . 2000 1000 2000(1 ) 2000 s tx y xy −≤ 22 33 1000 ) 1200 123 1000 ) 1.5 3 5 6000 (S x y xx x ≤− ++≤ 23 teel constraint) 30 25 40 60000 (Labor constraint) x ( ) ,, 0 ;,, i n t e g e r 0 1 xxx yy y o r = 7 Optimal solution z=2000,x 2 =2000, y 2 =1, all other variables are zero.

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EMSE 154 - 254: Applied Optimization Modeling Dorian Auto Completed:
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## This note was uploaded on 01/02/2010 for the course EMSE 254 taught by Professor Hernanabeledo during the Fall '08 term at GWU.

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Class_5_IntProgram_II - Integer Programming Models: part II...

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