# Tag Info

2

I recommend you to try very powerful utility LRS, http://cgm.cs.mcgill.ca/~avis/C/lrs.html. It is an open source application (a set of applications) to handle polyhedrals. B.t.w. one of LRS applications, called redund "removes redundant inequalities from an H-representation" (from description) - seems that what you need. There are parallel (MPI-...

3

First of all, you should determine the sign of the multipliers based on the objective function direction and how the complicating constraints are violated. Then you have to use a standard method like subgradient optimization to solve the lagrangian dualized problem to determine the optimal value of the multipliers. For more details: Marshall L. Fisher, An ...

2

you could try constraint programming / scheduling within CPLEX and use noOverlap to model the time matrix. In OPL that gives using CP; execute { cp.param.timelimit=10; } {string} nodes={"O","A","B","C","D","E","T"}; tuple edge { key string o; key string d; int time; } {edge} ...

2

I think your third constraint should be + 1, not - 1, on the right hand side. As stated, it says you enter destination nodes one time fewer than you exit them. You want to enter one time more. Fixing that will make the optimal solution feasible, but it will not make the model correct. There still remains the possibility of a solution that is not a contiguous ...

3

If the set $S$ of nodes to be visited is not too large, you can solve $|S|$ shortest path problems with additional constraints imposing a visit to some nodes. With your example, $|S|=|\{A,C \}|=2$ so it is not too bad. 1/ Find the shortest path from $O$ to $A$, while imposing a visit to node $C$. 2/ Then find the shortest path from $O$ to $C$, while ...

4

To be clear, you have a set $S$ of nodes of a graph $G=(V,A)$, with $S\subseteq V$, which must be visited. There is a special node $O$, which must be the starting point of a tour. A tour visiting the nodes in $S$ starting from $O$ (but not returning to $O$) at minimum length must be found? If that is the case, I think the easiest way is to compute an all-...

2

Can't you add a parameter to your model which defines the nature of your variables ? Something like : def my_model(model, continuous): ... if continuous: model.my_variable = Var(within=NonNegativeReals) else: model.my_variable = Var(within=PositiveIntegers) ... and then model.solve(). I am not so familiar with Pyomo but with PuLP,...

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The first issue is that "LP" format means different things to lpsolve and CPLEX. See the warning about that near the top of the lpsolve page describing their format. There are converters available, but I don't know if there are any packaged for R. So you may need to install one and then convert the CPLEX .lp file to lpsolve format outside of R (or ...

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