On Valid Inequalities for Mixed Integer p-Order Cone Programming

Abstract

We discuss two families of valid inequalities for linear mixed integer programming problems with cone constraints of arbitrary order, which arise in the context of stochastic optimization with downside risk measures. In particular, we extend the results of Atamtürk and Narayanan (Math. Program., 122:1–20, 2010, Math. Program., 126:351–363, 2011), who developed mixed integer rounding cuts and lifted cuts for mixed integer programming problems with second-order cone constraints. Numerical experiments conducted on randomly generated problems and portfolio optimization problems with historical data demonstrate the effectiveness of the proposed methods.

Appendix: A Direct Derivation of Conic Mixed Integer Rounding Cuts for Mixed Integer p-Order Cone Programming Problems

Following Atamtürk and Narayanan [7], let us first consider a simple case of the following set:

$$T= \bigl\{(y,w,t,x) \in{\mathbb{R}}^3_+\times{\mathbb{Z}} : [x+y-w-b]_+\le t \bigr\}. $$

Let us denote by $\operatorname{relax}(T)$ the continuous relaxation of T and by $\operatorname{conv}(T)$ its convex hull. It can be seen that the extreme rays of $\operatorname{relax}(T)$ are as follows: (1,0,0,1), (−1,0,0,0), (1,0,1,0), (−1,1,0,0), and its only extreme point is (b,0,0,0). Let us also denote f=b−⌊b⌋. Clearly, the case of f=0 is not interesting, hence it can be assumed that f>0, whereby $\operatorname{conv}(T)$ has four extreme points: (⌊b⌋,0,0,0), (⌊b⌋,f,0,0), (⌈b⌉,0,1−f,0), (⌈b⌉,0,0,1−f). With these observations in mind we can formulate the following proposition.

Proposition A.1

Inequality

$$ (1-f) \bigl(x-\lfloor b \rfloor\bigr)\le t+w $$

(18)

is valid for T and cuts off all points in $\operatorname {relax}(T)\setminus\operatorname{conv}(T)$.

Proof

First, let us show the validity of (18). The base inequality for T is

$$ [x+y-w-b]_+\le t. $$

(19)

Now, let x=⌊b⌋−α and α≥0. In this case, (19) turns into t≥[y−w−f−α]₊ and (18) becomes t≥−(1−f)α−w. Observing that [y−w−f−α]₊−(−(1−f)α−w)=max{y−f−αf,(1−f)α+w}≥0, one obtains that (19) implies (18) for x≤⌊b⌋.

On the other hand, if x=⌈b⌉+α with α≥0, then (19) becomes t≥[y−w+(1−f)+α]₊ and (18) turns into t≥(1−f)(1+α)−w. Similarly to above,

which means that (19) implies (18) for x≥⌈b⌉. Hence, (18) is valid for T.

To prove the remaining part of the proposition, consider the polyhedron $\acute{T}$ defined by the inequalities

(20)

(21)

(22)

(23)

(24)

Since $\acute{T}$ has four variables, the basic solutions of $\acute {T}$ are defined by four of these inequalities at equality. They are:

• Inequalities (20), (21), (22), (23): (x,y,w,t)=(b,0,0,0) is infeasible if f≠0.

• Inequalities (20), (21), (22), (24): (x,y,w,t)=(⌈b⌉,0,1−f,0).

• Inequalities (20), (21), (23), (24): (x,y,w,t)=(⌊b⌋,f,0,0).

• Inequalities (20), (22), (22), (24): (x,y,w,t)=(⌈b⌉,0,0,1−f).

• Inequalities (21), (23), (22), (24): (x,y,w,t)=(⌊b⌋,0,0,0).

Hence, $\operatorname{conv}(T)$ has exactly the same extreme points as $\acute{T}$, which completes the proof. □

In the general case, let

(25)

and consider the following function:

Proposition A.2

For α≠0 the following inequality

(26)

where $f_{|\alpha|} = \frac{b}{|\alpha|} - \lfloor\frac{b}{|\alpha |}\rfloor$, is valid for $\hat{T}$.

Proof

First consider the case α=1. We can rewrite the base inequality for (25) as

where f _j =a _j −⌊a _j ⌋. Observe that

Hence, we can apply simple conic MIR inequality (18) with variables ($\acute{x}, \acute{y}, \acute{w}, t$):

Rewriting it with the help of function ϕ _f (a), we obtain that

So, by Proposition A.1 inequality (26) is valid for α=1. In order to see that the result holds for all α≠0 we only need to scale the base inequality:

 □

Proposition A.3

Inequalities (26) with α=a _j , j=1,…,n, are sufficient to cut off all fractional extreme points of $\operatorname{relax} (\hat{T})$.

Proof

The set $\operatorname{relax}(\hat{T})$ is defined by n+3 variables and n+4 constraints. Therefore, if x _j >0 in an extreme point, then the remaining n+3 constraints must be active. Thus, the continuous relaxation has at most n fractional extreme points (x ^j,0,0,0) of the form $x_{j}^{j} = \frac {b}{a_{j}} >0$, and $x_{i}^{j} = 0$, for i≠j. Such points are infeasible if $\frac{b}{a_{j}} \notin{\mathbb{Z}}$. Now, let a _j >0. For such a fractional extreme point inequality (26) reduces to

which by Proposition A.1 cuts off fractional extreme point with $x_{j}^{j} = \frac{b}{a_{j}}$.

Now, let us consider a _j <0. In this case we observe that the inequality (26) reduces to

which, again, cuts off fractional extreme point with $x_{j}^{j} = \frac{b}{a_{j}}$. □