This example shows how to solve an optimization problem that has a linear or quadratic objective and quadratic inequality constraints. The example generates and uses the gradient and Hessian of the objective and constraint functions.

Suppose that your problem has the form

$\underset{x}{\mathrm{min}}\left(\frac{1}{2}{x}^{T}Qx+{f}^{T}x+c\right)$

subject to

$\frac{1}{2}{x}^{T}{H}_{i}x+{k}_{i}^{T}x+{d}_{i}\le 0,$

where 1 ≤ i ≤ m. Assume that at least one Hi is nonzero; otherwise, you can use quadprog or linprog to solve this problem. With nonzero Hi, the constraints are nonlinear, which means fmincon is the appropriate solver according to the Optimization Decision Table.

The example assumes that the quadratic matrices are symmetric without loss of generality. You can replace a nonsymmetric H (or Q) matrix with an equivalent symmetrized version (H + HT)/2.

If x has N components, then Q and Hi are N-by-N matrices, f and ki are N-by-1 vectors, and c and di are scalars.

### Objective Function

Formulate the problem using fmincon syntax. Assume that x and f are column vectors. (x is a column vector when the initial vector x0 is a column vector.)

y = 1/2*x'*Q*x + f'*x + c;
if nargout > 1
end

### Constraint Function

For consistency and easy indexing, place every quadratic constraint matrix in one cell array. Similarly, place the linear and constant terms in cell arrays.

jj = length(H); % jj is the number of inequality constraints
y = zeros(1,jj);
for i = 1:jj
y(i) = 1/2*x'*H{i}*x + k{i}'*x + d{i};
end
yeq = [];

if nargout > 2
for i = 1:jj
end
end

### Numeric Example

Suppose that you have the following problem.

Q = [3,2,1;
2,4,0;
1,0,5];
f = [-24;-48;-130];
c = -2;

rng default % For reproducibility
% Two sets of random quadratic constraints:
H{1} = gallery('randcorr',3); % Random positive definite matrix
H{2} = gallery('randcorr',3);
k{1} = randn(3,1);
k{2} = randn(3,1);
d{1} = randn;
d{2} = randn;

### Hessian

Create a Hessian function. The Hessian of the Lagrangian is given by the equation

${\nabla }_{xx}^{2}L\left(x,\lambda \right)={\nabla }^{2}f\left(x\right)+\sum {\lambda }_{i}{\nabla }^{2}{c}_{i}\left(x\right)+\sum {\lambda }_{i}{\nabla }^{2}ce{q}_{i}\left(x\right).$

fmincon calculates an approximate set of Lagrange multipliers λi, and packages them in a structure. To include the Hessian, use the following function.

hess = Q;
jj = length(H); % jj is the number of inequality constraints
for i = 1:jj
hess = hess + lambda.ineqnonlin(i)*H{i};
end

### Solution

Use the fmincon interior-point algorithm to solve the problem most efficiently. This algorithm accepts a Hessian function that you supply. Set these options.

options = optimoptions(@fmincon,'Algorithm','interior-point',...

Call fmincon to solve the problem.

x0 = [0;0;0]; % Column vector
[x,fval,eflag,output,lambda] = fmincon(fun,x0,...
[],[],[],[],[],[],nonlconstr,options);

Examine the Lagrange multipliers.

lambda.ineqnonlin
ans =

12.8412
39.2337

Both nonlinear inequality multipliers are nonzero, so both quadratic constraints are active at the solution.

### Efficiency When Providing a Hessian

The interior-point algorithm with gradients and a Hessian is efficient. View the number of function evaluations.

output
output =

iterations: 9
funcCount: 10
constrviolation: 0
stepsize: 5.3547e-04
algorithm: 'interior-point'
firstorderopt: 1.5851e-05
cgiterations: 0
message: 'Local minimum found that satisfies the constraints.

Optimization compl...'

fmincon takes only 10 function evaluations to solve the problem.

Compare this result to the solution without the Hessian.

options.HessianFcn = [];
[x2,fval2,eflag2,output2,lambda2] = fmincon(fun,[0;0;0],...
[],[],[],[],[],[],nonlconstr,options);
output2
output2 =

iterations: 17
funcCount: 22
constrviolation: 0
stepsize: 2.8475e-04
algorithm: 'interior-point'
firstorderopt: 1.7680e-05
cgiterations: 0
message: 'Local minimum found that satisfies the constraints.

Optimization compl...'

In this case, fmincon takes about twice as many iterations and function evaluations. The solutions are the same to within tolerances.

### Extension to Quadratic Equality Constraints

If you also have quadratic equality constraints, you can use essentially the same technique. The problem is the same, with the additional constraints

$\frac{1}{2}{x}^{T}{J}_{i}x+{p}_{i}^{T}x+{q}_{i}=0.$

Reformulate your constraints to use the Ji, pi, and qi variables. The lambda.eqnonlin(i) structure has the Lagrange multipliers for equality constraints.