Quadratic programming

Let us assume that we have data generated from a noisy linear regression \(y_t = a_tx + e_t\). The goal is to estimate the parameter \(x\), given the measurements \(y_t\) and \(a_t\), and we will try 3 different approaches based on linear and quadratic programming.

Create some noisy data with severe outliers to work with.

x = [1 2 3 4 5 6]';
t = (0:0.02:2*pi)';
A = [sin(t) sin(2*t) sin(3*t) sin(4*t) sin(5*t) sin(6*t)];
e = (-4+8*rand(length(t),1));
e(100:115) = 30;
y = A*x+e;

Data for regression problem

Define the variable we are looking for

xhat = sdpvar(6,1);

By using xhat and the regressors in A, we can define the residuals (which also will be an sdpvar object, parametrized in xhat)

residuals = y-A*xhat;

To solve the 1-norm regression problem (minimize sum of absolute values of residuals), we can define a variable that will serve as a bound on the absolute values of y-A*xhat (we will solve this problem much more conveniently below by simply using the norm operator)

bound = sdpvar(length(residuals),1);

Express that the bound variables are larger than the absolute values of the residuals (note the convenient definition of a double-sided constraint).

Constraints = [-bound <= residuals <= bound];

Call YALMIP to minimize the sum of the bounds subject to the constraints in Constraints. YALMIP will automatically detect that this is a linear program, and call any LP solver available on your path.


The optimal value is, as always, extracted using the overloaded value command.

x_L1 = value(xhat);

The 2-norm problem (least-squares) is easily solved as a QP problem without any constraints.

x_L2 = value(xhat);

YALMIP automatically detects that the objective is a convex quadratic function, and solves the problem using any installed QP solver. If no QP solver is found, the problem is converted to an SOCP, and if no dedicated SOCP solver exist, the SOCP is converted to an SDP.

Finally, we minimize the \(\infty\)-norm. This corresponds to minimizing the largest (absolute value) residual. Introduce a scalar to bound the largest value in the vector residual (YALMIP uses MATLAB standard to compare scalars, vectors and matrices)

bound = sdpvar(1,1);
Constraints  = [-bound <= residuals <= bound];

and minimize the bound.

x_Linf = value(xhat);

We plot the solutions, and notice that the 1-norm estimate worked very well and essentially managed to capture the underlying harmonics despite the severe measurement error after 2 seconds, whereas the two other estimates perform badly on this data set.

plot(t,[y A*x_L1 A*x_L2 A*x_Linf]);

Solution to regression problem

Let us finally state that some of the manipulations here can be performed much easier by using the nonlinear operator framework in YALMIP.


Large-scale quadratic programs

The 2-norm solution is most easily stated in the described QP formulation, although it in some cases is more efficient in YALMIP to express the problem using a 2-norm, which will lead to a second order cone problem.


The reason is that a quadratic function with \(n\) variables can be composed of up to \(n(n+1)/2\) monomials, which YALMIP has to work with symbolically. If you absolutely need to solve a large-scale quadratic program with YALMIP using a QP solver, introduce an auxiliary variable and equality constraints. This will make the quadratic term sparse (not only YALMIP but many QP solvers will be significantly faster after this transformation)

aux = sdpvar(length(residuals),1);
optimize([aux == residuals],aux'*aux);

Of course, in this example, this makes no difference, as there only are 6 decision variables but in scenarios where your objective is \(x^TQx\) and \(Q=R^TR\) is large, a better model might be

R = chol(Q);
z = sdpvar(length(x),1);
optimize([z == R*x],z'*z);

Even better, if you know \(Q\) is low-rank or there is some other structure that allows you to compute a low-rank possibly sparse factor, you should exploit that

R = my_smart_factorization(Q);
z = sdpvar(size(R,2),1);
optimize([z == R*x],z'*z);

The archetypical example is sum(x)^2 which leads to a completely dense quadratic model of rank 1. Absolutely catastrophical for large problems (it will most likely be indefinite for vectors of length larger than 10 in floating point numerics) and a waste of memory. The trivial model is

z = sdpvar(1);
optimize([z == sum(x)],z^2);

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