Thursday 15 April 2010

mathematical optimization - Integer Programming unequal constraint -


I am trying to model the following barrier in MIP:

  x_1 + X_2 + ... + x_n! = D   

The idea is to apply the variable J1, if x_1 + x_2 + ... + x_n = d and to add constraint

  z   

But how can I prepare the barrier

  (x_1 + x_2 + ... + x_n = D) ==> ; Z = 1   

In an integer program.

I think all x_i are integers l And you give such a constant value

  l   

and y A binary variable, these constraints express what you are looking for:

  x_1 + X_2 + ... + x_n & gt; = D + 1 + (LD-1) Y x_1 + x_2 + ... + x_n & lt; = D-1 + (Ud + 1) (1-y)   

If y = 0 then the first forced x_1 + x_2 + ... + X_n & gt; = D + 1 and second bound x_1 + x_2 + ... + x_n & lt; = U is satisfied with the definition of U .

If y = 1 then the second forced x_1 + x_2 + ... + x_n & lt; = D-1 should be held and the first bound is x_1 + x_2 + ... + x_n & gt; = L L is satisfied with the definition.

(Please check the typo.)


This integer programming has a notorious big M method and may be worse off. Bad situation can occur.


For further tips, look for Google's "Integer Programming Tricks", especially for this big M method move.

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