CODEWITHSUNDEEP
c++optimizationlinear-programmingcplex
Rémi L.
Rémi L.

Reputation: 11

Enumerate some extreme points near optimum solution

I am looking for a simple way to obtain a lot of "good" solutions in a LP problem (not MIP) with CPLEX, and not only (one of the) optimal basic solution(s). By "good" solutions I mean that the corresponding objective values are not so far from the real optimal value. Such pool of solutions could help the decision-maker...

More precisely, given a certain polyedron Ax<=b with x>=0 and an objective function z=cx I want to maximize, after running the LP, I can obtain the optimal value z*. Then I want to enumerate all the extreme points of the polyhedron given by the set of constraints

Ax <= b
cx >= z* - epsilon
x  >= 0

when epsilon is a given tolerance.

I know that CPLEX offers way to generate solution pool (see here), but it will not function because this method is for MIP : it enumerates all the solutions of an IP (or one solution for every given set of fixed integer variables if the problem is a MIP).

An interesting efficient way is to visit the adjacent solutions of the optimal basic solution, i.e. all the adjacent extreme points : if I suppose the polyhedron is not degenerative, for each pair of basic variable x_B and non-basic variable x_N, I compute the basic solution obtained when x_B leaves the basis and x_N enters in the basis. Then I throw the solutions with cx < z*-epsilon, and for the others I repeat the procedure. [I know that I could improve this algorithm, but this is the general idea].

The routine CPPXpivot of the Callable Library could help to do this pivoting operation, but I did not find an equivalent in the C++ API (concert technology). Does someone know if such an equivalent exist, or could propose me an other way to answer my original problem ?

Thanks a lot :) !

Rémi L.

Upvotes: 1

Views: 503

Answers (1)

Erwin Kalvelagen
Erwin Kalvelagen

Reputation: 16724

There is one interesting way to make this suitable for use with the Cplex solution pool. Use binary variables to encode the current basis, e.g. basis[k] = 0 meaning nonbasic and basis[k] = 1 indicating variable (or row) k is basic. Of course we have sum(k, basis[k]) = m (number of rows). Finally we have x[k] <= basis[k] * upperbound[k] (i.e. if nonbasic then zero -- assuming positive variables). When we add this to the LP model we end up with a MIP and can enumerate (all or some, optimal or near optimal) bases using the Cplex solution pool. See here and here.

Upvotes: 0

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