Matrix minor reformulation and SOCP-based spatial branch-and-cut method for the AC optimal power flow problem

Abstract

Alternating current optimal power flow (AC OPF) is one of the most fundamental optimization problems in electrical power systems. It can be formulated as a semidefinite program (SDP) with rank constraints. Solving AC OPF, that is, obtaining near optimal primal solutions as well as high quality dual bounds for this non-convex program, presents a major computational challenge to today’s power industry for the real-time operation of large-scale power grids. In this paper, we propose a new technique for reformulation of the rank constraints using both principal and non-principal 2-by-2 minors of the involved Hermitian matrix variable and characterize all such minors into three types. We show the equivalence of these minor constraints to the physical constraints of voltage angle differences summing to zero over three- and four-cycles in the power network. We study second-order conic programming (SOCP) relaxations of this minor reformulation and propose strong cutting planes, convex envelopes, and bound tightening techniques to strengthen the resulting SOCP relaxations. We then propose an SOCP-based spatial branch-and-cut method to obtain the global optimum of AC OPF. Extensive computational experiments show that the proposed algorithm significantly outperforms the state-of-the-art SDP-based OPF solver and on a simple personal computer is able to obtain on average a $0.71\%$ optimality gap in no more than 720 s for the most challenging power system instances in the literature.

Appendices

A KKT Points for Arctangent Envelopes

Let us rewrite the optimization problem in (29) as

$$\begin{aligned} \Delta \gamma = \max \{ f(c,s) : (c, s) \in F \}, \end{aligned}$$

(46)

where

$$\begin{aligned} f(c,s):= \arctan \left( \frac{s}{c} \right) - (\gamma + \alpha c + \beta s ) \end{aligned}$$

and

$$\begin{aligned} F := \{(c,s) : (c, s) \in [{\underline{c}}, {\overline{c}}] \times [{\underline{s}}, {\overline{s}}], \ c \tan {\underline{\theta }} \le s \le \tan {\overline{\theta }} \}. \end{aligned}$$

Without loss of generality, let us assume that the relations $\arctan ({\underline{s}}/ {\underline{c}}) \le {\underline{\theta }}$ and ${\overline{\theta }} \le \arctan ({\overline{s}}/ {\underline{c}})$ hold between the variable bounds (otherwise, at least one of the bounds can be improved). Let us denote the optimal solution to problem (46) as $(c^*, s^*)$. First, we claim that $(c^*, s^*)$ is not in the interior of F. This is due to the fact that the Hessian of f at a point (c, s), which is given as

$$\begin{aligned} \frac{1}{(c^2+s^2)^2} \begin{bmatrix} 2cs&\quad s^2-c^2\\ s^2-c^2&\quad -2cs \end{bmatrix}, \end{aligned}$$

is an indefinite matrix. Therefore, an interior point of F will not satisfy the second-order necessary conditions of local optimality.

The above argument implies that $(c^*, s^*)$ belongs to the boundary of F, which is the union of the following six line segments:

• (i) $[({\underline{c}}, {\underline{c}} \tan {\underline{\theta }}), ({\underline{c}}, {\underline{c}} \tan {\overline{\theta }})]$

• (ii) $[({\underline{c}}, {\underline{c}} \tan {\overline{\theta }}), ({\overline{s}}/\tan {\overline{\theta }}, {\overline{s}})]$

• (iii) $[ ({\overline{s}}/\tan {\overline{\theta }}, {\overline{s}}), ({\overline{c}}, {\overline{s}})]$

• (iv) $[ ({\overline{c}}, {\overline{s}}), ({\overline{c}}, {\underline{s}})]$

• (v) $[ ({\overline{c}}, {\underline{s}}), ({\underline{s}}/\tan {\underline{\theta }}, {\underline{s}})]$

• (vi) $[ ({\underline{s}}/\tan {\underline{\theta }}, {\underline{s}}), ({\underline{c}}, {\underline{c}} \tan {\underline{\theta }})]$

Note that the line segments (ii) and (vi) cannot contain $(c^*, s^*)$ in their relative interior since the function f is linear along them. Hence, the problem reduces to four 1-dimensional optimization problems, which can be solved easily. The global optimal solution $(c^*, s^*)$ is the one that gives the largest objective value among the KKT points calculated by solving those four 1-dimensional optimization problems.

B Proof of Theorem 3.2

Our proof approach is based on identifying the extreme points of ${\mathcal {S}}_a$. Let us start with a proposition.

Proposition B.1

Let (x, y) be an extreme point of the set ${\mathcal {S}}_a$. Then, for a distinct pair of indices i and j, the following four statements hold:

• (i) either $x_i$ or $y_j$ is at one of its bounds,

• (ii) either $x_i$ or $y_i$ is at one of its bounds,

• (iii) either $x_j$ or $y_j$ is at one of its bounds,

• (iv) either $x_j$ or $y_i$ is at one of its bounds.

Proof

We only prove the first statement. The others can be proven using exactly the same reasoning.

Assume for a contradiction that ${\underline{x}}_i< x_i < {\overline{x}}_i$ and ${\underline{y}}_j< y_j < {\overline{y}}_j$. Consider the following cases:

• Case 1: $y_i \ne 0$ and $x_j \ne 0$

  • Case 1a: $\frac{a_i y_i}{a_jx_j} > 0$

    Let $\epsilon = \{x_i - {\underline{x}}_i, {\overline{x}}_i - x_i, \frac{a_jx_j}{a_i y_i} (y_i - {\underline{y}}_i), \frac{a_jx_j}{a_i y_i} ({\overline{y}}_i - y_i) \}$ and $\delta = \frac{a_i y_i}{a_jx_j} \epsilon$. Note that both $\epsilon$ and $\delta$ are positive. Now, construct $(x^+, y^-) = (x+\epsilon e_i, y-\delta e_j)$ and $(x^-, y^+) = (x-\epsilon e_i, y+\delta e_j)$ where $e_i$ is the i-th unit vector. Observe that both $(x^+, y^-)$ and $(x^-, y^+)$ belong to ${\mathcal {S}}_a$. Moreover, $(x,y) = \frac{1}{2} (x^+, y^-) + \frac{1}{2} (x^-, y^+)$. But, this is a contradiction to (x, y) being an extreme point of ${\mathcal {S}}_a$.

  • Case 1b: $\frac{a_i y_i}{a_jx_j} < 0$

    Let $\epsilon = \{x_i - {\underline{x}}_i, {\overline{x}}_i - x_i, \frac{a_jx_j}{a_i y_i} ({\underline{y}}_i - y_i), \frac{a_jx_j}{a_i y_i} ( y_i -{\overline{y}}_i) \}$ and $\delta = -\frac{a_i y_i}{a_jx_j} \epsilon$. Note that both $\epsilon$ and $\delta$ are positive. Now, construct $(x^+, y^+) = (x+\epsilon e_i, y+\delta e_j)$ and $(x^-, y^-) = (x-\epsilon e_i, y-\delta e_j)$. Observe that both $(x^+, y^+)$ and $(x^-, y^-)$ belong to ${\mathcal {S}}_a$. Moreover, $(x,y) = \frac{1}{2} (x^+, y^+) + \frac{1}{2} (x^-, y^-)$. But, this is a contradiction to (x, y) being an extreme point of ${\mathcal {S}}_a$.

  • Case 2: $y_i = 0$

    Let $\epsilon = \{x_i - {\underline{x}}_i, {\overline{x}}_i - x_i \}$. Note that $\epsilon$ is positive. Now, construct $(x^+, y) = (x+\epsilon e_i, y)$ and $(x^-, y) = (x-\epsilon e_i, y)$. Observe that both $(x^+, y)$ and $(x^-, y)$ belong to ${\mathcal {S}}_a$. Moreover, $(x,y) = \frac{1}{2} (x^+, y) + \frac{1}{2} (x^-, y)$. But, this is a contradiction to (x, y) being an extreme point of ${\mathcal {S}}_a$.

  • Case 3: $x_j = 0$

    Let $\delta = \{y_i - {\underline{y}}_i, {\overline{y}}_i - y_i \}$. Note that $\delta$ is positive. Now, construct $(x, y^+) = (x, y+\delta e_j)$ and $(x, y^-) = (x, y-\epsilon e_j)$. Observe that both $(x, y^+)$ and $(x, y^-)$ belong to ${\mathcal {S}}_a$. Moreover, $(x,y) = \frac{1}{2} (x, y^+) + \frac{1}{2} (x, y^-)$. But, this is a contradiction to (x, y) being an extreme point of ${\mathcal {S}}_a$.

$\square$

Proposition B.1 implies the following corollary.

Corollary B.1

Let (x, y) be an extreme point of the set ${\mathcal {S}}_a$. Then, either $x_i$ and $y_i$ or $x_j$ and $y_j$ are at their bounds for a distinct pair of indices i and j.

Proof

Let $x_i = {\hat{x}}_i$ be a shorthand for “either $x_i = {\underline{x}}_i$ or $x_i = {\overline{x}}_i$”. Then, Proposition B.1 implies that

$$\begin{aligned} \begin{aligned}&(x_i = {\hat{x}}_i \vee y_j = {\hat{y}}_j) \wedge (x_i = {\hat{x}}_i \vee x_j = {\hat{x}}_j) \wedge (y_i = {\hat{y}}_i \vee y_j = {\hat{y}}_j) \wedge (y_i = {\hat{y}}_i \\&\quad \vee x_j = {\hat{x}}_j) \\&\quad = (x_i = {\hat{x}}_i \wedge y_i = {\hat{y}}_i) \vee (x_j = {\hat{x}}_j \wedge y_j = {\hat{y}}_j), \end{aligned} \end{aligned}$$

(47)

which is the desired conclusion. $\square$

An immediate consequence of Corollary B.1 is the following characterization of extreme points of ${\mathcal {S}}_a$:

Corollary B.2

All the extreme points of ${\mathcal {S}}_a$ are in one of the following sets:

• $D_0 = \{(x,y) \in {\mathcal {S}}_a: (x_i,y_i) = ({\hat{x}}_i, {\hat{y}}_i)\ \forall i \}$

• $D_k = \{(x,y) \in {\mathcal {S}}_a: (x_i,y_i) = ({\hat{x}}_i, {\hat{y}}_i) \ i \ne k, x_ky_k = -\frac{1}{a_k} \sum _{i \ne k} a_i {\hat{x}}_i {\hat{y}}_i, \ x_k\in [{\underline{x}}_k, {\overline{x}}_k], \ y_k\in [{\underline{y}}_k , {\overline{y}}_k] \} \quad k=1,\dots ,N$

Note that $D_0$ is a collection of at most $4^N$ singletons whereas $D_k$ is a collection of $4^{N-1}$ sets for each k. The projection of such a set onto $(x_k,y_k)$ is of the following form

$$\begin{aligned} T_\alpha = \{(x,y) \in {\mathbb {R}}^2 : xy = \alpha , \ \ x\in [{\underline{x}}, {\overline{x}}], \ y\in [{\underline{y}} , {\overline{y}}] \} \end{aligned}$$

(48)

for some constant $\alpha$.

Proposition B.2

Set conv$(T_\alpha )$ is second-order cone representable for any value of $\alpha$.

There are several cases based on parameter values. In the most complicated case, we need $xy \ge \alpha$ (which is conic representable) and McCormick envelopes.

Now, we are ready to prove the main result.

Proof of Theorem 3.2

Since the convex hull of all the disjunctions are second-order cone representable (could be polyhedral or singleton depending on parameter values), conv$({\mathcal {S}}_a)$ is also second-order cone representable. $\square$