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bilinear_vb.vb
' Copyright 2024, Gurobi Optimization, LLC */ ' This example formulates and solves the following simple bilinear model: ' ' maximize x ' subject to x + y + z <= 10 ' x * y <= 2 (bilinear inequality) ' x * z + y * z == 1 (bilinear equality) ' x, y, z non-negative (x integral in second version) Imports Gurobi Class bilinear_vb Shared Sub Main() Try Dim env As New GRBEnv("bilinear.log") Dim model As New GRBModel(env) ' Create variables Dim x As GRBVar = model.AddVar(0, GRB.INFINITY, 0, GRB.CONTINUOUS, "x") Dim y As GRBVar = model.AddVar(0, GRB.INFINITY, 0, GRB.CONTINUOUS, "y") Dim z As GRBVar = model.AddVar(0, GRB.INFINITY, 0, GRB.CONTINUOUS, "z") ' Set objective Dim obj As GRBLinExpr = x model.SetObjective(obj, GRB.MAXIMIZE) ' Add linear constraint: x + y + z <= 10 model.AddConstr(x + y + z <= 10, "c0") ' Add bilinear inequality: x * y <= 2 model.AddQConstr(x * y <= 2, "bilinear0") ' Add bilinear equality: x * z + y * z == 1 model.AddQConstr(x * z + y * z = 1, "bilinear1") ' Optimize model model.Optimize() Console.WriteLine(x.VarName & " " & x.X) Console.WriteLine(y.VarName & " " & y.X) Console.WriteLine(z.VarName & " " & z.X) Console.WriteLine("Obj: " & model.ObjVal & " " & obj.Value) x.Set(GRB.CharAttr.VType, GRB.INTEGER) model.Optimize() Console.WriteLine(x.VarName & " " & x.X) Console.WriteLine(y.VarName & " " & y.X) Console.WriteLine(z.VarName & " " & z.X) Console.WriteLine("Obj: " & model.ObjVal & " " & obj.Value) ' Dispose of model and env model.Dispose() env.Dispose() Catch e As GRBException Console.WriteLine("Error code: " & e.ErrorCode & ". " & e.Message) End Try End Sub End Class