Applications of Optimization with XpressMP by Christelle GuA©ret Christian Prins Marc Sevaux PDF

# Applications of Optimization with XpressMP by Christelle GuA©ret Christian Prins Marc Sevaux PDF

By Christelle GuA©ret Christian Prins Marc Sevaux

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Extra resources for Applications of Optimization with XpressMP

Example text

But as we have discussed before, it almost certainly is not worth the effort, and the LP solver should be clever enough to do the work for us automatically anyway if it thinks it will reduce the solution time. 3 Simple resource constraints We have already seen examples of this sort of constraint in the chess set problem where, to remind you, we had limited amounts of lathe-hours and boxwood, and we had to create a production plan that used neither too many lathe-hours nor too much boxwood. If we generalize what we mean by a resource, all linear programs can be formulated as maximizing some objective function subject to not using any more of any resource than we have available.

A local optimum is a point where all the nearest neighbors are worse than it, but where we have no guarantee that there is not a better point some way away. A global optimum is a point which we know to be the best. ) Theoretically, models that can be built with any of the entities we have listed above can equally well be modeled solely with binary variables. The reason why modern IP systems have some or all of the extra entities is that they often provide significant computational savings in computer time and storage when trying to solve the resulting model.

This is a particularly long winded way of demonstrating the equivalence of the product term and the three linear equations and in fact now we have got it it is actually quite easy to see why these three inequalities are correct. Since b3 is b1 multiplied by something that is less than or equal to 1, b3 will always be less than or equal to b1 and by a similar argument b3 will always be less than or equal to b2 . The only further case we have to consider is when both b1 and b2 are equal to 1 and then we have to force b3 to be equal to 1.