# Convert Quadratic Constraints to Second-Order Cone Constraints

This example shows how to convert a quadratic constraint to the second-order cone constraint form. A quadratic constraint has the form

$${x}^{T}Qx+2{q}^{T}x+c\le 0.$$

Second-order cone programming has constraints of the form

$$\Vert {A}_{sc}(i)\cdot x-{b}_{sc}(i)\Vert \le {d}_{sc}(i)\cdot x-\gamma (i)$$.

The matrix $$Q$$ must be symmetric and positive semidefinite for you to convert quadratic constraints. Let $$S$$ be the square root of $$Q$$, meaning $$Q=S*S={S}^{T}*S$$. You can compute $$S$$ using `sqrtm`

. Suppose that there is a solution $$b$$ to the equation $${S}^{T}b=\u2013q$$, which is always true when $$Q$$ is positive definite. Compute $$b$$ using `b = -S\q`

.

$$\begin{array}{ll}{x}^{T}Qx+2{q}^{T}x+c& ={x}^{T}{S}^{T}Sx-2{\left({S}^{T}b\right)}^{T}x+c\\ & ={\left(Sx-b\right)}^{T}\left(Sx-b\right)-{b}^{T}b+c\\ & ={\Vert Sx-b\Vert}^{2}+c-{b}^{T}b.\end{array}$$

Therefore, if $${b}^{T}b>c$$, then the quadratic constraint is equivalent to the second-order cone constraint with

$${A}_{sc}=S$$

$${b}_{sc}=b$$

$${d}_{sc}=0$$

$$\gamma =-\sqrt{{b}^{T}b-c}$$

### Numeric Example

Specify a five-element vector `f`

representing the objective function $${f}^{T}x$$.

f = [1;-2;3;-4;5];

Set the quadratic constraint matrix `Q`

as a 5-by-5 random positive definite matrix. Set `q`

as a random 5-element vector, and take the additive constant $$c=-1$$.

rng default % For reproducibility Q = randn(5) + 3*eye(5); Q = (Q + Q')/2; % Make Q symmetric q = randn(5,1); c = -1;

To create the inputs for `coneprog`

, create the matrix `S`

as the square root of `Q`

.

S = sqrtm(Q);

Create the remaining inputs for the second-order cone constraint as specified in the first part of this example.

b = -S\q; d = zeros(size(b)); gamma = -sqrt(b'*b-c); sc = secondordercone(S,b,d,gamma);

Call `coneprog`

to solve the problem.

[x,fval] = coneprog(f,sc)

Optimal solution found.

`x = `*5×1*
-0.7194
0.2669
-0.6309
0.2543
-0.0904

fval = -4.6148

Compare this result to the result returned by solving this same problem using `fmincon`

. Write the quadratic constraint as described in Anonymous Nonlinear Constraint Functions.

```
x0 = randn(5,1); % Initial point for fmincon
nlc = @(x)x'*Q*x + 2*q'*x + c;
nlcon = @(x)deal(nlc(x),[]);
[xfmc,fvalfmc] = fmincon(@(x)f'*x,x0,[],[],[],[],[],[],nlcon)
```

Local minimum found that satisfies the constraints. Optimization completed because the objective function is non-decreasing in feasible directions, to within the value of the optimality tolerance, and constraints are satisfied to within the value of the constraint tolerance.

`xfmc = `*5×1*
-0.7196
0.2672
-0.6312
0.2541
-0.0902

fvalfmc = -4.6148

The two solutions are nearly identical.

## See Also

`coneprog`

| `quadprog`

| `secondordercone`