Using Numerical Libraries for .NET from F#

With the release of version 5.0 of our Numerical Libraries for .NET, we’ve also considerably improved our support for the F# language. In this article I want to walk you through setting up F# Interactive to use the Extreme Optimization Numerical Libraries for .NET, and show off some of the new features. This tutorial will focus on F# 3.0 in Visual Studio 2012. Most of the functionality also works with F# 2.0 in Visual Studio 2010.

Setting up F# Interactive

The first step is to bring the Extreme Optimization Numerical Libraries for .NET assemblies into the F# Interactive environment:

#I @"c:\Program Files\Extreme Optimization\Numerical Libraries for .NET\bin\Net40\";;
#r "Extreme.Numerics.Net40.dll";;
#r "Extreme.Numerics.FSharp30.Net40.dll";;

open Extreme.Mathematics;;

fsi.AddPrinter(fun (A : Extreme.Mathematics.ISummarizable) -> A.Summarize());;

We do this by first adding the path to the Extreme Optimization binaries to the DLL search path using the #I command. (Note: the path depends on the installation folder and may be different on your machine.) We the main assembly Extreme.Numerics.Net40, and the F# specific assembly Extreme.Numerics.FSharp30.Net40. Finally, we tell fsi to use a pretty printer for objects that implement the ISummarizable interface. This will print out vectors and matrices in a user-friendly form.

Working with vectors and matrices

So let’s get our hands dirty right away. When we referenced the Extreme.Numerics.FSharp30.Net40 dll, this automatically opened several modules that make it easier to do linear algebra in F#. We can call Vector and Matrix factory methods, or we can use some convenient F# functions:

> let A = rand 3 3;;

val A : LinearAlgebra.DenseMatrix =
  3x3 DenseMatrix

> let c = !![1.0..10.0];;
val c : LinearAlgebra.DenseVector =

Notice the !! operator, which creates a vector from a list or sequence. We can access elements of vectors and matrices using the familiar F# syntax. Slices work too, as do integer sequences and predicates:

> A.[0,1];;
val it : float = 0.100694183
> A.[1,1]  A.[1..2,0..1];;
val it : Matrix = 2x2 DenseMatrix
> c.[1..3];;
val it : Vector = [2.0000,3.0000,4.0000]
> c.[fun x -> sin x > 0.0];;
val it : Vector = [1.0000,2.0000,3.0000,7.0000,8.0000,9.0000]
> c.[[2;4;5]];;
val it : Vector = [3.0000,5.0000,6.0000]
> c.[[for i in 1..7 -> i % 3]];;
val it : Vector = [2.0000,3.0000,1.0000,2.0000,3.0000,1.0000,2.0000]

Let’s do some simple calculations:

> A*b;;
 val it : LinearAlgebra.DenseVector = [3.7378,5.1463,0.8651]
 > A.T*b;;
 val it : Vector = [2.9263,2.1614,2.2214]
 > 2.0 * b;;
 val it : LinearAlgebra.DenseVector = [2.0000,4.0000,10.0000]
 > 1.0 + b;;
 val it : Vector = [2.0000,3.0000,6.0000]
 > log (1.0 + b);;
 val it : Vector = [0.6931,1.0986,1.7918]
 > det A;;
 val it : float = -0.1707208593
 > inv A;;
 val it : Matrix =
 3x3 DenseMatrix
> let b = Vector.Create(3, fun i -> 1.0 + (float i)**2.0);;
val b : LinearAlgebra.DenseVector = [1.0000,2.0000,5.0000]
> let x = A.Solve(b);;
val x : LinearAlgebra.DenseVector = [15.3876,-7.8258,0.8488]

The last command solved the system of equations represented by the matrix A and solved it for the right-hand side b. We can also work with matrix decompositions. There are actually two ways. Let’s use the SVD as an example. First, we can get the factors of the decomposition.

> let (U,Σ,V) = A.svd();;

val Σ : LinearAlgebra.DiagonalMatrix =
  3x3 DiagonalMatrix
val V : Matrix =
  3x3 DenseMatrix
val U : Matrix =
  3x3 DenseMatrix

Or, if we just want the singular values, we can get those as well:

> let σ = A.svd();;

val σ : LinearAlgebra.DenseVector = [1.2980,0.5101,0.2578]

The second way to work with decompositions is by creating decomposition objects. Let’s work with a non-square matrix this time:

> let C = rand 15 4;;

val C : LinearAlgebra.DenseMatrix =
  15x4 DenseMatrix

> let d = (rand 15 1).GetColumn(0);;

val d : Vector =

Here we see that the output is truncated, and only the 5 first and last rows are shown. We can find the least squares solution to C*y = d.
We’ll also compute the residuals and their norm:

> let y = svdC.GetPseudoInverse() * d;;

val y : Vector = [0.3221,0.4458,0.3383,0.1298]

> let r = d - C * y;;

val r : Vector =

> norm r;;
val it : float = 0.8214250504

Working with functions

You’d expect a functional language to work with functions. So let’s see what we can do.

The modules discussed in this section are not automatically opened when you load the F# compatibility assembly. The available modules are:

  • Special: special functions, such as Gamma functions, Bessel functions, etc.
  • Functional: numerical integration and differentiation, finding zeros, maxima and minima of functions.
  • Statistics: some functions that are useful when doing statistical analysis.
  • Random: random number streams.

These are only the modules with F# specific functions. The full functionality of the libraries is of course available from F# as well.

Here we’ll focus mainly on the second category. Special functions can be very handy, however, so let’s start with those:

> open Extreme.Numerics.FSharp.Special;;
> gamma 0.5;;
val it : float = 1.772453851
> BesselJ 0.0 1.0;;
val it : float = 0.7651976866
> let J0 = BesselJ 0.0;;

val J0 : (float -> float)

> J0 1.0;;
val it : float = 0.7651976866

Notice how we used partial application of the Bessel function to get a function that returns the Bessel function of order 0. We can do something similar in many other places, for example with binomial coefficients:

> let binom9 = binomial 9;;

val binom9 : (int -> float)

> [for i in 0..9 -> binom9 i];;
val it : float list =
  [1.0; 9.0; 36.0; 84.0; 126.0; 126.0; 84.0; 36.0; 9.0; 1.0]

Now to the functional stuff. Two of the most basic operations on functions are differentiation and integration. Differentiation is done using the d function. Integration is done using integrate1. Partial application is supported, so you can create a function that computes the numerical derivative at any value you give it.

> open Extreme.Numerics.FSharp.Functional;;
> let J1 = BesselJ 1.0;;

val J1 : (float -> float)

> d J0 0.5;;
val it : float = -0.2422684577
> J1 0.5;; // d J0 = -J1
val it : float = 0.2422684577
> let dJ0 = d J0;;

val dJ0 : (float -> float)

> dJ0 0.5;;
val it : float = -0.2422684577
> dJ0 1.5;;
val it : float = -0.5579365079

Integration works in a similar way. The bounds of the integration interval are supplied as a tuple. Infinite intervals are supported as well.

> integrate1 J1 (0.0,5.0);;
val it : float = 1.177596771
> J0 5.0 - J0 0.0;; // Should give - the previous result:
val it : float = -1.177596771
> let iJ1 = integrate1 J1;;

val iJ1 : (float * float -> float)

> iJ1 (0.0,5.0);;
val it : float = 1.177596771
> integrate1 (fun x -> exp -x) (0.0,infinity);;
val it : float = 1.0

Finding zeros of functions is easy. Let’s quickly find the first 5 zeros of the Bessel function J0:

> [for x0 in [2.0;5.0;8.0;11.0;14.0] -> findzero (fun x -> J0 x) x0];;
val it : float list =
  [2.404825558; 5.52007811; 8.653727913; 11.79153444; 14.93091771]

Finding the minimum or maximum of a function can be difficult because most software requires that you supply the gradient of the function. Although computing derivatives is high school calculus, it’s still very error prone. Automatic differentiation comes to the rescue here. We’ll use the SymbolicMath class, which contains all kinds of functions that take advantage of symbolic computations to obtain a solution.

One of the most famous optimization problems is the Rosenbrock function. We can find the minimum of this function in just one line of code:

> SymbolicMath.FindMinimum((fun x y -> (1.0 - x)**2.0 + 100.0*(y-x**2.0)**2.0), !![-1.0;-1.0]);;
val it : SolutionReport =
    {ConvergenceTest = null;
     Error = 2.778930274e-09;
     EvaluationsNeeded = 74;
     IterationsNeeded = 26;
     Result = [1.0000,1.0000];
     Status = Converged;}

It’s time now for you to play. Download the trial version of the Extreme Optimization Numerical Libraries for .NET here and have fun!

Using Quadratic Programming for Portfolio Optimization

This week’s update of the Extreme Optimization Numerical Libraries for .NET includes the ability to solve quadratic programming (QP) problems. These are optimization problems where the objective function is a quadratic function and the solution is subject to linear constraints.

Our QP solver uses an active set method, and can be used to solve programs with thousands of variables. Quadratic programming has many applications in finance, economics, agriculture, and other areas. It also serves as the core of some methods for nonlinear programming such as Sequential Quadratic Programming (SQP).

We created some QuickStart samples that demonstrates how to use the new functionality:

In this post, I’d like to illustrate an application in finance: portfolio optimization.

Say we have a set of assets. We know the historical average return and variances of each asset. We also know the correlations between the assets. With this information, we can look for the combination of assets that best meets our objectives. In this case, our goal will be to minimize the risk while achieving a minimal return. Now let’s see how we can model this.

The unknowns in our model will be the amount to be invested in each asset, which we’ll denote by xi. All values together form the vector x. We will assume that all xi ≥ 0.

We can use the variance of the value of the portfolio as a measure for the risk. A larger variance means that the probability of a large loss also increases, which is what we want to avoid. If we denote the variance-covariance matrix of the assets by R, then the variance of the asset value is xTRx. This is what we want to minimize.

Now let’s look at the constraints. We obviously have a budget, so the sum of all amounts must be less than or equal to our budget. Say our budget is $10,000, then we must have Σ xi ≤ 10,000. We also want a minimum expected return on our investment, say 10%. If ri is the return associated with asset i, then the total expected return for our investment is Σ ri xi, and we want this value to be at least 10% of our total budget, or $1,000.

So, in summary, our model looks like this:

Minimize xTRx
Subject to Σ xi ≤ 10000
Σ ri xi ≥ 1000 xi ≥ 0

The last thing we need to do is formulate this quadratic program in terms of the optimization classes in the Numerical Libraries for .NET. The QuickStart samples show three ways of doing this: you can define the model directly in terms of vectors and matrices; you can build the model from scratch, or you can read it from a model definition file in MPS format. Since the problem above is already pretty much expressed in terms of vectors and matrices, we’ll use that approach here.

Quadratic programming problems are represented by the QuadraticProgram class. One of the constructors lets you create a model in so-called standard form. A quadratic program in standard form looks like this:

Minimize c0 + cTx + ½xTHx
Subject to AEx = bE

where c0 is a scalar, H, AE, and AI are matrices, and all other values are vectors. The subscript E denotes quality constraints, while the subscript I denotes inequality constraints. In our case, we don’t have equality constraints.

As a concrete example, we will work with four assets with returns 5%, -20%, 15% and 30%, respectively. Our budget is $10,000, and the minimum return is 10%. The implementation is quite straightforward. Noteworthy is that, since in the standard form the right hand side is an upper bound, we have to change the sign of the minimum return constraint which had a lower bound right-hand side.

Here’s the C# code:

// The linear term in the objective function: 
Vector c = Vector.CreateConstant(4, 0.0); 
// The quaratic term in the objective function: 
Matrix R = Matrix.CreateSymmetric(4, new double[] 
        -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: 
Matrix A = Matrix.Create(2, 4, new double[] 
        1, 1, 1, 1, 
        -0.05, 0.2, -0.15, -0.30
    }, MatrixElementOrder.RowMajor);
// The right-hand sides of the constraints: 
Vector b = Vector.Create(10000, -1000);
// We're now ready to call the constructor. 
// The last parameter specifies the number of equality 
// constraints. 
QuadraticProgram qp1 = new QuadraticProgram(c, R, A, b, 0); 
// Now we can call the Solve method to run the algorithm: 
Vector x = qp1.Solve();

This is how it looks in Visual Basic:

' The linear term in the objective function: 
Dim c As Vector = Vector.CreateConstant(4, 0.0) 
' The quaratic term in the objective function: 
Dim R As Matrix = Matrix.CreateSymmetric(4, _ 
    New Double() _
        { _
            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: 
Dim A As Matrix = Matrix.Create(2, 4, New Double() _
    { _
        1, 1, 1, 1, _
        -0.05, 0.2, -0.15, -0.3 _
    }, MatrixElementOrder.RowMajor) 
' The right-hand sides of the constraints: 
Dim b As Vector = Vector.Create(10000, -1000) 
' We're now ready to call the constructor. 
' The last parameter specifies the number of
' equality constraints. 
Dim qp1 As New QuadraticProgram(c, R, A, b, 0) 
' Now we can call the Solve method to run the algorithm: 
Dim x As Vector = qp1.Solve()

And finally in F#:

// The linear term in the objective function:
let c = Vector.CreateConstant(4, 0.0); 
// The quaratic term in the objective function:
let R = Matrix.CreateSymmetric(4,
        -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: 
let A = Matrix.Create(2, 4, 
        1.0; 1.0; 1.0; 1.0;
        -0.05; 0.2; -0.15; -0.30
    |], MatrixElementOrder.RowMajor) 
// The right-hand sides of the constraints: 
let b = Vector.Create(10000.0, -1000.0) 
// We're now ready to call the constructor. 
// The last parameter specifies the number of 
// equality constraints. 
let qp1 = new QuadraticProgram(c, R, A, b, 0)
// Now we can call the Solve method to run the algorithm:
let x = qp1.Solve()

After we’ve run this code, the vector x will contain the optimal amounts for each asset. It gives us: $3,453, $0, $1,069, $2,223. Not surprisingly, the second asset, which has had a negative return on average, should not be part of our portfolio. However, it turns out that the asset with the lowest positive return should have the largest share.