A Full Nesterov–Todd Step Infeasible Interior-Point Method for Second-Order Cone Optimization

Abstract

After a brief introduction to Jordan algebras, we present a primal–dual interior-point algorithm for second-order conic optimization that uses full Nesterov–Todd steps; no line searches are required. The number of iterations of the algorithm coincides with the currently best iteration bound for second-order conic optimization. We also generalize an infeasible interior-point method for linear optimization to second-order conic optimization. As usual for infeasible interior-point methods, the starting point depends on a positive number. The algorithm either finds a solution in a finite number of iterations or determines that the primal–dual problem pair has no optimal solution with vanishing duality gap.

Appendix A: Technical Lemmas

Lemma A.1

For i=1,…,n, let f _i :ℝ₊→ℝ denote a convex function. Then we have, for any nonzero vector $z\in{\mathbb{R}}_{+}^{n}$, the following inequality:

$$\sum_{i=1}^n f_i(z_i) \le\frac{1}{e^Tz}\sum_{j=1}^n z_j \biggl(f_j\bigl(e^Tz\bigr)+ \sum _{i\ne j} f_i(0)\biggr). $$

Proof

We define the function $F:{\mathbb{R}}_{+}^{n}\rightarrow {\mathbb{R}}$ by

$$F(z) = \sum_{i=1}^n f_i(z_i), \quad z\ge0. $$

Letting e _j denote the jth unit vector in ℝⁿ, we may write z as a convex combination of the vectors (e ^T z)e _j , as follows.

$$z=\sum_{j=1}^n \frac{z_j}{e^Tz} \bigl(e^Tz\bigr)e_j, $$

Indeed, $\sum_{j=1}^{n} \frac{z_{j}}{e^{T}z}=1$ and z _j /e ^T z≥0 for each j. Since F(z) is a sum of convex functions, F(z) is convex in z, and hence we have

$$F(z) \le\sum_{j=1}^n \frac{z_j}{e^Tz} F \bigl( \bigl(e^Tz\bigr)e_j\bigr) =\sum _{j=1}^n \frac{z_j}{e^Tz} \sum _{i=1}^n f_i\bigl(\bigl(e^Tz \bigr) (e_j)_i\bigr). $$

Since (e _j ) _i =1 if i=j and zero if i≠j, we obtain

$$F(z) \le\sum_{j=1}^n \frac{z_j}{e^Tz} \biggl(f_j\bigl(e^Tz\bigr)+ \sum _{i\ne j} f_i(0)\biggr). $$

Hence the inequality in the lemma follows. □

Corollary A.1

Let f:ℝ₊→ℝ be a convex function such that f(0)=0. Then we have, for any vector $z\in{\mathbb{R}}_{+}^{n}$, the following inequality:

$$\sum_{i=1}^n f(z_i) \le f \Biggl(\sum_{i=1}^n z_i \Biggr). $$

Proof

In Lemma A.1, take f _i =f for each i; then the result follows. □

In the lemma below, we use that the cone $\mathcal{K}$ of squares in a Euclidean Jordan algebra defines a partial ordering $\preceq_{\mathcal{K}}$ of ℝⁿ according to the definition

$$x\preceq_{\mathcal{K}} s \quad\Leftrightarrow\quad s-x\in\mathcal{K}. $$

Lemma A.2

Let x,s∈ℝⁿ and x ^T s=0, then one has

• (i) $-\frac{1}{4}\| x+s \|^{2}_{F}e \preceq_{\mathcal{K}} x\circ s \preceq_{\mathcal{K}} \frac{1}{4}\| x+s \|^{2}_{F}e$;

• (ii) $\| x\circ s \|_{F} \le\frac{1}{2\sqrt{2}}\| x+s \|_{F}^{2}$.

Proof

We write

$$ x\circ s = \frac{1}{4}\bigl((x+s)^2 - (x-s)^2\bigr). $$

(88)

Since (x+s)² is a square, it belongs to $\mathcal{K}$. This means that $x\circ s + \frac {1}{4}(x-s)^{2}\in \mathcal{K}$, or, equivalently,

$$x\circ s \succeq_{\mathcal{K}} -\frac{1}{4}(x-s)^2. $$

Since $x\preceq_{\mathcal{K}} \| x \|_{F}e$ for every x, also using Lemma 2.5(i), we may write

$$(x-s)^2 \preceq_{\mathcal{K}} \| (x-s)^2 \|_Fe \preceq_{\mathcal {K}} \| x-s \|^2_Fe, $$

whence it follows that

$$x\circ s \succeq_{\mathcal{K}} -\frac{1}{4}(x-s)^2 \succeq_{\mathcal {K}} -\frac{1}{4}\| x-s \|^2_Fe. $$

In the same way one derives from (88) that

$$x\circ s \preceq_{\mathcal{K}} \frac{1}{4}(x+s)^2 \preceq_{\mathcal {K}} \frac{1}{4}\| x+s \|^2_Fe. $$

Thus we have shown that one has, for all x,s∈ℝⁿ,

$$-\frac{1}{4}\| x-s \|^2_Fe \preceq_{\mathcal{K}} x\circ s \preceq_{\mathcal{K}} \frac{1}{4}\| x+s \|^2_Fe. $$

Since x and s are orthogonal, we have ${\operatorname{tr}}(x\circ s) = 2x^{T}s = 0$, whence ∥x+s∥ _F =∥x−s∥ _F . Hence part (i) of the lemma follows.

For the proof of (ii) we return to (88). Using $\| z \|_{F}^{2} = {\operatorname{tr}}(z^{2})$, we obtain

Since (x+s)² and (x−s)² belong to $\mathcal{K}$, the trace of their product is nonnegative. Thus we obtain

Using Lemma 2.5(i) and ∥x+s∥ _F =∥x−s∥ _F again, we get

This implies (ii). Hence the proof of the lemma is complete. □