Calculating Radius of Robust Feasibility of Uncertain Linear Conic Programs via Semi-definite Programs

Abstract

The radius of robust feasibility provides a numerical value for the largest possible uncertainty set that guarantees robust feasibility of an uncertain linear conic program. This determines when the robust feasible set is non-empty. Otherwise, the robust counterpart of an uncertain program is not well defined as a robust optimization problem. In this paper, we address a key fundamental question of robust optimization: How to compute the radius of robust feasibility of uncertain linear conic programs, including linear programs? We first provide computable lower and upper bounds for the radius of robust feasibility for general uncertain linear conic programs under the commonly used ball uncertainty set. We then provide important classes of linear conic programs where the bounds are calculated by finding the optimal values of related semi-definite linear programs, among them uncertain semi-definite programs, uncertain second-order cone programs and uncertain support vector machine problems. In the case of an uncertain linear program, the exact formula allows us to calculate the radius by finding the optimal value of an associated second-order cone program.

Appendix: Proofs of Basic Properties of Admissible Sets

In this appendix, we provide the proof of the basic properties of the admissible set of parameters (Proposition 2.1).

Proof of Proposition 2.1

Proof

[Proof of (a)] Direct verification gives us that

It now follows from the well-known existence theorem for linear systems [11, Corollary 3.1.1] that $F_{r}({\overline{A}},{\overline{b}})\ne \emptyset$ if and only if

$$\begin{aligned} \left( 0_{n},-1\right) \notin \overline{{\mathrm{cone}}\,\left\{ \displaystyle \sum \limits _{i=1}^{m}\lambda _{i}\left( {\overline{a}}_{i}+{\varDelta } a_{i},-{\overline{b}}_{i}-{\varDelta } b_{i}\right) :\lambda \in {{\mathcal {B}}},\Vert ({\varDelta } a_{i},{\varDelta } b_{i})\Vert \le r_{i},\forall i\right\} }, \end{aligned}$$

which, in turn, is equivalent to the statement

$$\begin{aligned} \begin{array}{ll} \left( 0_{n},1\right) &{} \notin \overline{{\mathrm{cone}}\,\left\{ \displaystyle \sum \limits _{i=1}^{m}\lambda _{i}\left( {\overline{a}}_{i}+{\varDelta } a_{i},{\overline{b}}_{i}+{\varDelta } b_{i}\right) :\lambda \in {{\mathcal {B}}},\Vert ({\varDelta } a_{i},{\varDelta } b_{i})\Vert \le r_{i},\forall i\right\} } \\ &{} =\overline{{\mathrm{cone}}\,\left\{ \displaystyle \sum \limits _{i=1}^{m} \lambda _{i}\left( ({\overline{a}}_{i},{\overline{b}}_{i})+r_{i}{{\mathbb {B}}} _{n+1}\right) :\lambda \in {{\mathcal {B}}}\right\} } . \end{array} \end{aligned}$$

Thus, the conclusion follows.

[Proof of (b)] Let ${{\mathcal {B}}}$ be the compact base of $K^{*}$. Then, $0_{m}\notin {{\mathcal {B}}}$, and so,

$$\begin{aligned} \mu :=\min \{\min _{1\le i\le m}\big \vert {\lambda }_{i}\big \vert :\lambda \in {{\mathcal {B}}}\}>0. \end{aligned}$$

Define $M=\max \{\Vert \displaystyle \sum \limits _{i=1}^{m}{\lambda }_{i}( {\overline{a}}_{i},{\overline{b}}_{i})\Vert :\lambda \in {{\mathcal {B}}}\}<+\infty .$ We shall prove by contradiction that $C({\overline{A}},{\overline{b}})\subseteq \frac{M}{\mu }{{\mathbb {B}}}_{m}.$ If, in the contrary, $C({\overline{A}}, {\overline{b}}) \nsubseteq \frac{M}{\mu }{{\mathbb {B}}}_{m}$, then we can take $r\in C({\overline{A}},{\overline{b}})$ such that $\left\| r\right\| \ge \frac{M+\epsilon }{\mu }$ for some $\epsilon >0.$ Now, fix any $\widetilde{ \lambda }\in {{\mathcal {B}}}$. Note that

$$\begin{aligned} \displaystyle \sum \limits _{i=1}^{m}\left| {\widetilde{\lambda }} _{i}\right| r_{i}\ge \mu \displaystyle \sum \limits _{i=1}^{m}r_{i}\ge \mu \sqrt{\displaystyle \sum \limits _{i=1}^{m}r_{i}^{2}}\ge M+\epsilon \ge \Vert \displaystyle \sum \limits _{i=1}^{m}{\widetilde{\lambda }}_{i}({\overline{a}} _{i},{\overline{b}}_{i})\Vert +\epsilon . \end{aligned}$$

It follows that

$$\begin{aligned} \epsilon {{\mathbb {B}}}_{n+1} \subseteq \displaystyle \sum \limits _{i=1}^{m} {\widetilde{\lambda }}_{i}({\overline{a}}_{i},{\overline{b}}_{i})+\displaystyle \sum \limits _{i=1}^{m}\left| {\widetilde{\lambda }}_{i}\right| r_{i} {{\mathbb {B}}}_{n+1}= & {} \displaystyle \sum \limits _{i=1}^{m}{\widetilde{\lambda }} _{i}(\left( {\overline{a}}_{i},{\overline{b}}_{i})+r_{i}{{\mathbb {B}}}_{n+1}\right) \\\subseteq & {} \displaystyle \bigcup \limits _{\lambda \in {{\mathcal {B}}}}\left\{ \displaystyle \sum \limits _{i=1}^{m}\lambda _{i}(\left( {\overline{a}}_{i}, {\overline{b}}_{i})+r_{i}{{\mathbb {B}}}_{n+1}\right) \right\} , \end{aligned}$$

and so

$$\begin{aligned} \left( 0_{n},1\right) \in {\mathrm{cone}}\,\left\{ \displaystyle \sum \limits _{i=1}^{m}\lambda _{i}\left( ({\overline{a}}_{i},{\overline{b}} _{i})+r_{i}{{\mathbb {B}}}_{n+1}\right) :\lambda \in {{\mathcal {B}}}\right\} . \end{aligned}$$

Then, by Proposition 2.1 part (1) , $r\notin C(\overline{ A},{\overline{b}})$ (contradiction). Hence, $C({\overline{A}},{\overline{b}})$ is bounded. Thus, the conclusion follows by the fact that $C({\overline{A}}, {\overline{b}})$ is radiant.

[Proof of (c)] Let us show the “$\left[ \Longrightarrow \right]$” direction first. Assume that $\sigma _{r}^{{{\mathcal {B}}}}$ satisfies the Slater condition. As $r\in {{\mathbb {R}}}^m_{++}$, there exists $\xi >0$ such that $r+\xi {{\mathbb {B}}}_{m}\subseteq {{\mathbb {R}}}_{++}^{m}$. Since ${{\mathcal {B}}}$ is a compact base (and so, $0_m \notin {{\mathcal {B}}}$), $\eta :=\max \limits _{\lambda \in {{\mathcal {B}}}}\max \limits _{1\le i\le m}\left| \lambda _{i}\right|$ is a positive real number. Consider the mapping

$$\begin{aligned} \begin{array}{ccccc} &{} &{} (m) &{} &{} \\ {\varPhi } : &{} {{\mathbb {R}}}^{m}\times &{} \overbrace{{{\mathbb {R}}}^{n+1}\times ...\times {{\mathbb {R}}}^{n+1}} &{} \longrightarrow &{} {{\mathbb {R}}}^{n+1} \\ &{} \lambda &{} \left( z^{1},...,z^{m}\right) &{} &{} \displaystyle \sum \limits _{i=1}^{m}\lambda _{i}z^{i} \end{array} \end{aligned}$$

Since ${\varPhi }$ is continuous (as it has quadratic components) and the set $D_{r}:=\mathcal {B\times }\prod \limits _{i=1}^{m}\left[ \left( {\overline{a}} _{i},{\overline{b}}_{i}\right) +r_{i}{{\mathbb {B}}}_{n+1}\right]$ is compact, $C_{r}:={\varPhi } \left( D_{r}\right)$ is a compact subset of ${{\mathbb {R}}}^{n+1}$ too. By the Slater condition, there exists $x^{0}\in {{\mathbb {R}}}^{n}$ such that $\left\langle \left( y,y_{n+1}\right) ,\left( x^{0},1\right) \right\rangle <0$ for all $\left( y,y_{n+1}\right) \in C_{r}.$ Let $\epsilon >0$ be such that

$$\begin{aligned} \begin{array}{ll} -\epsilon &{} =\max \left\{ \left\langle \left( y,y_{n+1}\right) ,\left( x^{0},1\right) \right\rangle :\left( y,y_{n+1}\right) \in C_{r}\right\} \\ &{} =\max \left\{ \left\langle {\varPhi } \left( d\right) ,\left( x^{0},1\right) \right\rangle :d\in D_{r}\right\} . \end{array} \end{aligned}$$

Given $t\in {{\mathbb {R}}}_{++}^{m},$ we consider an element of $D_{t}$ of the form

$$\begin{aligned} d^{t}=\left( \lambda ^{t},\left( {\overline{a}}_{1},{\overline{b}}_{1}\right) +t_{1}u^{1,t},...,\left( {\overline{a}}_{m},{\overline{b}}_{m}\right) +t_{m}u^{m,t}\right) , \end{aligned}$$

with $\lambda ^{t}\in {{\mathcal {B}}}$ and $u^{i,t}\in {{\mathbb {B}}}_{n+1},$ $i=1,...,m,$ arbitrarily chosen. We define the vector $d^{r}:=\left( \lambda ^{t},\left( {\overline{a}}_{1},{\overline{b}}_{1}\right) +r_{1}u^{1,t},...,\left( {\overline{a}}_{m},{\overline{b}}_{m}\right) +r_{m}u^{m,t}\right) \in D_{r}.$ Since

$$\begin{aligned}&\left\| {\varPhi } \left( d^{t}\right) -{\varPhi } \left( d^{r}\right) \right\| =\left\| \displaystyle \sum \limits _{i=1}^{m}\lambda _{i}\left( t_{i}-r_{i}\right) u^{i,t}\right\| \le m\eta \left\| t-r\right\| ,\\&\left| \left\langle {\varPhi } \left( d^{t}\right) -{\varPhi } \left( d^{r}\right) ,\left( x^{0},1\right) \right\rangle \right| \le m\eta \left\| \left( x^{0},1\right) \right\| \left\| t-r\right\| , \end{aligned}$$

and so $\left\langle {\varPhi } \left( d^{t}\right) ,\left( x^{0},1\right) \right\rangle \le m\eta \left\| \left( x^{0},1\right) \right\| \left\| t-r\right\| -\epsilon$. Hence, if

$$\begin{aligned}\left\| t-r\right\| \le \min \left\{ \frac{\epsilon }{2m\eta \left\| \left( x^{0},1\right) \right\| },\xi \right\} , \end{aligned}$$

then, one has

$$\begin{aligned} \max \left\{ \left\langle \left( y,y_{n+1}\right) ,\left( x^{0},1\right) \right\rangle :\left( y,y_{n+1}\right) \in C_{t}\right\} \le -\frac{ \epsilon }{2}<0, \end{aligned}$$

which shows that $x^{0}$ is a Slater point for $\sigma _{t}^{{{\mathcal {B}}}}.$ This implies that $t\in C({\overline{A}},{\overline{b}}).$ Thus, one has $r\in {\mathrm{int}}\,C({\overline{A}},{\overline{b}}).$

We now show the reverse direction “$\left[ \Longleftarrow \right] .$” Let $r\in {\mathrm{int}}\,C({\overline{A}}, {\overline{b}})$ and for all $\lambda \in {{\mathcal {B}}}$, $\displaystyle \sum \nolimits _{i=1}^{m}r_{i}\lambda _{i}>0.$ Then, there exists $\epsilon >0$ such that $r+\epsilon 1_{m}\in C({\overline{A}},{\overline{b}})$ and

$$\begin{aligned}\gamma :=\min \left\{ \displaystyle \sum \limits _{i=1}^{m}r_{i}\lambda _{i}:\lambda \in {{\mathcal {B}}}\right\} >0. \end{aligned}$$

We consider perturbations of the coefficients of $\sigma _{r}^{{{\mathcal {B}}}}$ preserving the index set $\mathcal {B\times }\displaystyle \prod \nolimits _{i=1}^{m}\left( r_{i}{{\mathbb {B}}}_{n+1}\right)$ and we measure such perturbations by means of the Chebyshev metric $d_{\infty }$. The coefficients vector of $\sigma _{r}^{{{\mathcal {B}}}}$ can be expressed as

$$\begin{aligned} \displaystyle \sum \limits _{i=1}^{m}\lambda _{i}\left[ \left( {\overline{a}}_{i}, {\overline{b}}_{i}\right) +r_{i}u^{i}\right] =\displaystyle \sum \limits _{i=1}^{m}\lambda _{i}\left( {\overline{a}}_{i},{\overline{b}}_{i}\right) +\displaystyle \sum \limits _{i=1}^{m}\lambda _{i}r_{i}u^{i}, \end{aligned}$$

(17)

where $\lambda \in {{\mathcal {B}}}$ and $u^{i}\in {{\mathbb {B}}}_{n+1},$ $i=1,...,m.$ Consider an additive perturbation $\delta w,$ with $0<\delta <\gamma \epsilon$ and $w\in {{\mathbb {B}}}_{n+1}$ of the coefficient vector in (17). Let $v:=\frac{w}{\displaystyle \sum \nolimits _{1\le i\le m}r_{i}\lambda _{i}}.$ Since $\left\| u^{i}+\delta v\right\| \le 1+\delta \left\| v\right\| \le 1+\frac{\delta }{\gamma }$ for all i,  one has

$$\begin{aligned} \begin{array}{ll} \displaystyle \sum \limits _{i=1}^{m}\lambda _{i}\left[ \left( {\overline{a}}_{i}, {\overline{b}}_{i}\right) +r_{i}u^{i}\right] +\delta w &{} =\displaystyle \sum \limits _{i=1}^{m}\lambda _{i}\left( {\overline{a}}_{i},{\overline{b}} _{i}\right) +\displaystyle \sum \limits _{i=1}^{m}\lambda _{i}r_{i}\left( u^{i}+\delta v\right) \\ &{} \in \displaystyle \sum \limits _{i=1}^{m}\lambda _{i}\left( {\overline{a}}_{i}, {\overline{b}}_{i}\right) +\displaystyle \sum \limits _{i=1}^{m}\lambda _{i}r_{i}\left( 1+\frac{\delta }{\gamma }\right) {{\mathbb {B}}}_{n+1} \\ &{} \subseteq \displaystyle \sum \limits _{i=1}^{m}\lambda _{i}\left( {\overline{a}} _{i},{\overline{b}}_{i}\right) +\displaystyle \sum \limits _{i=1}^{m}\lambda _{i}r_{i}\left( 1+\epsilon \right) {{\mathbb {B}}}_{n+1}. \end{array} \end{aligned}$$

So, the solution set of any perturbed system obtained from $\sigma _{r}^{ {{\mathcal {B}}}}$ by summing up vectors of Euclidean norm less than $\gamma \epsilon$ to each the coefficient vector contains $F_{r}({\overline{A}}, {\overline{b}})\ne \emptyset .$ In particular, summing up vectors of Chebyshev norm less than $\frac{\gamma \epsilon }{\sqrt{m}}$ to each coefficient vector of $\sigma _{r}^{{{\mathcal {B}}}}$ we get a feasible perturbed system. Hence, by [11, Theorem 6.1], $\sigma _{r}^{\mathcal { B}}$ has a strong Slater solution, which shows that $\sigma _{r}^{{{\mathcal {B}}} }$ satisfies the Slater condition. $\square$