# Quadratic Programming QuickStart Sample (IronPython)

Illustrates how to solve optimization problems a quadratic objective function and linear constraints using classes in the Extreme.Mathematics.Optimization namespace in IronPython.

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```Python import numerics from System import Array from Extreme.Mathematics import * # The quadratic programming classes reside in their own namespace. from Extreme.Mathematics.Optimization import * # Illustrates solving quadratic programming problems # using the classes in the Extreme.Mathematics.Optimization # namespace of the Extreme Optimization Numerical Libraries for .NET. # This QuickStart Sample illustrates the quadratic programming # functionality by solving a portfolio optimization problem. # Portfolio optimization is a common application of QP. # For a collection of assets, the goal is to minimize # the risk (variance of the return) while achieving # a minimal return for a set maximum amount invested. # The variables are the amounts invested in each asset. # The quadratic term is the covariance matrix of the assets. # THere is no linear term in this case. # There are three ways to create a Quadratic Program. # The first is in terms of matrices. The coefficients # of the constraints and the quadratic terms are supplied # as matrices. The cost vector, right-hand side and # constraints on the variables are supplied as vectors. # The linear term in the objective function: c = Vector.CreateConstant(4, 0.0) # The quaratic term in the objective function: R = Matrix.CreateSymmetric(4, Array[float]([ \ 0.08,-0.05,-0.05,-0.05, \ -0.05, 0.16,-0.02,-0.02, \ -0.05,-0.02, 0.35, 0.06, \ -0.05,-0.02, 0.06, 0.35 ]), MatrixTriangle.Upper) # The coefficients of the constraints: A = Matrix([ [1, 1, 1, 1], [-0.05, 0.2, -0.15, -0.30 ]]) # The right-hand sides of the constraints: b = Vector([10000, -1000]) # We're now ready to call the constructor. # The last parameter specifies the number of equality # constraints. qp1 = QuadraticProgram(c, R, A, b, 0) # Now we can call the Solve method to run the Revised # Simplex algorithm: x = qp1.Solve() print "Solution: {0:.1f}".format(x) # The optimal value is returned by the Extremum property: print "Optimal value: {0:.1f}".format(qp1.OptimalValue) # The second way to create a Quadratic Program is by constructing # it by hand. We start with an 'empty' quadratic program. qp2 = QuadraticProgram() # Next, we add two variables: we specify the name, the cost, # and optionally the lower and upper bound. qp2.AddVariable("X1", 0.0) qp2.AddVariable("X2", 0.0) qp2.AddVariable("X3", 0.0) qp2.AddVariable("X4", 0.0) # Next, we add constraints. Constraints also have a name. # We also specify the coefficients of the variables, # the lower bound and the upper bound. qp2.AddLinearConstraint("C1", Vector.Create(1.0, 1.0, 1.0, 1.0), ConstraintType.LessThanOrEqual, 10000) qp2.AddLinearConstraint("C2", Vector.Create(0.05, -0.2, 0.15, 0.3), ConstraintType.GreaterThanOrEqual, 1000) # If a constraint is a simple equality or inequality constraint, # you can supply a QuadraticProgramConstraintType value and the # right-hand side of the constraint. # Quadratic terms must be set individually. # Each combination appears at most once. qp2.SetQuadraticCoefficient("X1", "X1", 0.08) qp2.SetQuadraticCoefficient("X1", "X2", -0.05 * 2) qp2.SetQuadraticCoefficient("X1", "X3", -0.05 * 2) qp2.SetQuadraticCoefficient("X1", "X4", -0.05 * 2) qp2.SetQuadraticCoefficient("X2", "X2", 0.16) qp2.SetQuadraticCoefficient("X2", "X3", -0.02 * 2) qp2.SetQuadraticCoefficient("X2", "X4", -0.02 * 2) qp2.SetQuadraticCoefficient("X3", "X3", 0.35) qp2.SetQuadraticCoefficient("X3", "X4", 0.06 * 2) qp2.SetQuadraticCoefficient("X4", "X4", 0.35) # We can now solve the quadratic program: x = qp2.Solve() print "Solution: {0:.1f}".format(x) print "Optimal value: {0:.1f}".format(qp2.OptimalValue) # Finally, we can create a quadratic program from an MPS file. # The MPS format is a standard format. qp3 = MpsReader.ReadQuadraticProgram(r"..\data\portfolio.qps") # We can go straight to solving the quadratic program: x = qp3.Solve() print "Solution: {0:.1f}".format(x) print "Optimal value: {0:.1f}".format(qp3.OptimalValue) ```