SOCPs involve the second-order cone which is defined by the property that for each of its members the first element is at least as large as the Euclidean
norm of the remaining elements. This corresponds to the case where C is the second-order cone (also known as the quadratic cone, Lorenz cone, and the ice-cream cone):
Cq :={x = (x1, x2, . . . , xn)∈ IRn: x1≥ k(x2, . . . , xn)k}. (9.2)
A portion of the second-order cone in 3 dimensions for x1 ∈ [0, 1] is
depicted in Figure 9.1. As seen from the figure, the second-order cone in three dimensions resembles an ice-cream cone that stretches to infinity. We observe that by “slicing” the second-order cone, i.e., by intersecting it with a hyperplane at different angles we can obtain spherical and ellipsoidal sets. Any convex quadratic constraint can be expressed using the second-order cone (or its rotations) and one or more hyperplanes.
−1 −0.5 0 0.5 1 −1 −0.5 0 0.5 1 0 0.2 0.4 0.6 0.8 1 x3 {(x1, x2, x3): x1≥ ||(x2, x3)||} x2 x1
Figure 9.1: The second-order cone
Exercise 9.1 Another important cone that appears in conic optimization formulations is the rotated quadratic cone defined as follows:
Cqr:={(x1, x2, x3, . . . , xn) : 2x1x2 ≥ n
X
j=3
x2j, x1, x2 ≥ 0.}. (9.3)
Show that x = (x1, x2, x3, . . . , xn)∈ Cqrif and only if y = (y1, y2, y3, . . . , yn)∈
Cq where y1 = √12(x1+ x2), y2 = √12(x1 − x2), and yj = xj, j = 3, . . . , n.
The vector y given here is obtained by rotating the vector x by 45 degrees in the plane defined by the first two coordinate axes. In other words, each element of the cone Cqr can be mapped to a corresponding element of Cq
through a 45 degree rotation (Why?). This is why the cone Cr
q is called the
Exercise 9.2 Show that the problem
min x32
s.t. x≥ 0, x ∈ S is equivalent to the following problem:
min t
s.t. x≥ 0, x ∈ S x2 ≤ t · u u2 ≤ x.
Express the second problem as an SOCP using Cr q.
Exercise 9.3 Consider the following optimization problem: min c1x1+ c2x2+ d1x 3 2 1 + d2x 3 2 2 s.t. a11x1+ a12x2 = b1, x1, x2 ≥ 0,
where d1, d2 > 0. The nonlinear objective function of this problem is a
convex function. Write this problem as a conic optimization problem with a linear objective function and convex cone constraints.
HINT: Use the previous exercise.
A recent review of second-order cone programming models and meth- ods is provided in [26]. One of the most common uses of second-order cone programs in financial applications is in the modeling and treatment of parameter uncertainties in optimization problems. After generating an ap- propriate description of the uncertainties, robust optimization models seek to find solutions to such problems that will perform well under many sce- narios. As we will see in Chapter 19, ellipsoidal sets are among the most popular structures used for describing uncertainty in such problems and the close relationship between ellipsoidal sets and second-order cones make them particularly useful. We illustrate this approach in the following subsection.
9.2.1 Ellipsoidal Uncertainty for Linear Constraints
Consider the following single-constraint linear program: min cTx
s.t. aTx + b ≥ 0.
We consider the setting where the objective function is certain but the con- straint coefficients are uncertain. We assume that the constraint coefficients [a; b] belong to an ellipsoidal uncertainty set:
U = {[a; b] = [a0; b0] +
k
X
j=1
Our objective is to find a solution that minimizes the objective function among the vectors that are feasible for all [a; b] ∈ U. In other words, we want to solve
min cTx
s.t. aTx + b ≥ 0, ∀[a; b] ∈ U.
For a fixed x the “robust” version of the constraint is satisfied if and only if 0≤ min [a;b]∈Ua Tx + b≡ min u:kuk≤1α + u Tβ, (9.4)
where α = (a0)Tx + b0 and β = (β1, . . . , βk) with βj = (aj)Tx + bj.
The second minimization problem in (9.4) is easy. Since α is constant, all we need to do is to minimize uTβ subject to the constraintkuk ≤ 1. Recall that for the angle θ between vectors u and β the following trigonometric equality holds:
cos θ = u
Tβ
kukkβk,
or uTβ =kukkβk cos θ. Since kβk is constant, this expression is minimized when kuk = 1 and cos θ = −1. This means that u points in the opposite direction from β, namely −β. Normalizing to satisfy the bound constraint we obtain u∗=−kβkβ as shown in Figure 9.2. Substituting this value we find
min [a;b]∈Ua Tx + b = α− kβk = (a0)Tx + b0− v u u u t k X j=1 ((aj)Tx + bj)2, (9.5)
and we obtain the robust version of the inequality aTx + b≥ 0 as
(a0)Tx + b0− v u u u t k X j=1 ((aj)Tx + bj)2 ≥ 0. (9.6)
It is now easy to observe that (9.6) can be written equivalently using the second-order cone:
zj = (aj)Tx + bj, j = 0, . . . k
(z0, z1, . . . , zk)∈ Cq
The approach outlined above generalizes to multiple constraints as long as the uncertainties are constraint-wise, that is, if the uncertainty sets of parameters in different constraints are unrelated. Thus, robust optimization models for uncertain linear constraints with ellipsoidal uncertainty leads to SOCPs. The strategy outlined above is well-known and is used in, for example, [7].
β u:||u|| ≤ 1 u*= −β / ||β|| −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2
Figure 9.2: Minimization of a linear function over a circle
9.2.2 Conversion of quadratic constraints into second-order cone constraints
The second-order cone membership constraint (x0, x1, . . . , xk) ∈ ˆCq can be
written equivalently as the combination of a linear and a quadratic con- straint:
x0≥ 0, x20− x21− . . . − x2k ≥ 0.
Conversely, any convex quadratic constraint of an optimization problem can be rewritten using second-order cone membership constraints. When we have access to a reliable solver for second-order cone optimization, it may be desirable to convert convex quadratic constraints to second-order cone constraints. Fortunately, a simple recipe is available for these conversions.
Consider the following quadratic constraint:
xTQx + 2pTx + γ ≤ 0. (9.7) This is a convex constraint if the function on the left-hand-side is convex which is true if and only if Q is a positive semidefinite matrix. Let us assume Q is positive definite for simplicity. In that case, there exists an invertible matrix, say R, satisfying Q = RRT. For example, the Cholesky factor of Q
satisfies this property. Then, (9.7) can be written as
(RTx)T(RTx) + 2pTx + γ≤ 0. (9.8) Define y = (y1, . . . , yk)T = RTx + R−1p. Then, we have
Thus, (9.8) is equivalent to
∃y s.t. y = RTx + R−1p, yTy≤ pTQ−1p− γ.
From this equivalence, we observe that the constraint (9.7) can be satisfied only if pTQ−1p− γ ≥ 0–we will assume that this is the case.
Now, it is straightforward to note that (9.7) is equivalent to the following set of linear equations coupled with a second-order cone constraint:
y1 .. . yk = R Tx + R−1p, y0 = q pTQ−1p− γ, (y0, y1, . . . , yk)∈ Cq.
Exercise 9.4 Rewrite the following convex quadratic constraint in “conic form”, i.e., as the intersection of linear equality constraints and a second- order cone constraint:
10x21+ 2x1x2+ 5x22+ 4x1+ 6x2+ 1≤ 0.
Exercise 9.5 Discuss how the approach outlined in this section must be modified to address the case when Q is positive semidefinite but not positive definite. In this case there still exists a matrix R satisfying Q = RRT. But R is no longer invertible and we can no longer define the vector y as above.