I try to model the following problem: given $z\in\{0,1\}^m$ and a linear system $Ax\le b(z), x\in\Bbb R^n, A\in\Bbb R^{d\times n}, b(z)\in\Bbb R^{d}$, where $b(z)$ means that some entries of $b$ are functions of $z$. Specifically, the entries in $b$ that depend on $z$ are in the form of $c + 1-z$, where $c$ indicates some constant.

The statement I want to model is: minimize $\|z\|_0$ such that the linear system $Ax\le b(z)$ is infeasible. It is already known that this linear system is feasible when $z = 0$ and there exists a $z$ making the system infeasible. Is there any way to model it into MILP?

  • $\begingroup$ Out of curiosity, could you give some background regarding the nature of this problem ? I have never encountered a problem that must remain infeasible. $\endgroup$
    – Kuifje
    Apr 29, 2021 at 7:21
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    $\begingroup$ I can give an application (paper in my answer). Controllers for dynamical systems can be formulated as repeated solutions to optimization problems (often called MPC). To study stability of these interconnected systems w.r.t disturbances and model error, a first step is to ensure that the optimization problems actually remain feasible as time goes (you don't want to see "Infeasible" on the dash-board during a take-over in your Tesla). Hence, we might be interested in the smallest possible disturbance which renders the problem infeasible at some state. The larger, the better. $\endgroup$ Apr 29, 2021 at 7:37
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    $\begingroup$ @Kuifje Thanks for your comments. The application background for this problem is a network secured sensor deployment problem, which aims to find the optimal secured sensor placement with minimal cost such that any attacks satisfying certain constraints will not cause a specified consequence. The constraints on the attacker plus the consequence to be avoided together are modelled by the linear system. We hope to find the optimal placement to make such linear system infeasible. $\endgroup$ Apr 29, 2021 at 17:13

2 Answers 2


By Farkas lemma, infeasibility of $Ax\leq b$ is equivalent to feasibility of $A^Ty = 0, y^Tb < 0, y\geq 0$, or more practically useful $A^Ty=0, y^Tb \leq -1, y\geq 0$.

Unfortunately, this will lead to a bilinear model when you parameterize $b(z)$. It is fairly similar to an application I worked on a decade ago Oops! I cannot do it again: Testing for recursive feasibility in MPC where we started with bilinear stuff coming from Farkas lemma, and it was possible to trade that bilinearity with a complementarity condition (Sec 4.2) which (at the time) was more efficiently solved.

EDIT: Missed the fact that $z$ is binary. In that case, you can linearize $y^Tb(z)$ using standard big-M methods, and you end up with a MILP.

Here is a small proof-of-concept implementation in the MATLAB Toolbox YALMIP (disclaimer, developed by me). I was too lazy to manually linearize so I used a built-in command to derive the big-M model

% Feasible model
m = 50;n = 5;
A = randn(m,n);
b0 = A*randn(n,1) + 0.99

% Perturb
z = binvar(m,1);
b = b0 - z;

% Solve
y = sdpvar(m,1);
[linear_b_y,cut]=binmodel(b'*y,[-100 <= y <= 100]);
Model = [A'*y == 0, y >= 0, linear_b_y == -1, cut];

% Check that it is infeasible
x = sdpvar(n,1);
  • $\begingroup$ Thanks so much for your answer and the reference. $\endgroup$ Apr 30, 2021 at 19:58

I think this can be approached using a constraint generation technique (variant of Benders decomposition), although I have no idea if it would efficient. By reordering the rows of $A$, we can assume that $$b(z)=\left[\begin{array}{c} \hat{b}\\ c+e-z \end{array}\right]$$where $\hat{b}\in\mathbb{R}^{d-m}$, $c\in \mathbb{R}^m$, $e=(1,\dots,1)^\prime \in \mathbb{R}^m$ and $z\in \lbrace 0,1 \rbrace^m$.

You start out minimizing $\parallel z\parallel_{0}$ subject only to $z\in \lbrace 0,1 \rbrace^m$. Each time the solver finds what it thinks is a new incumbent $\tilde{z}$ (satisfying all constraints added to that point), you solve the inequality $Ax\le b(\tilde{z})$. If the inequality is infeasible, accept the new incumbent and keep chugging along. If a feasible solution $\tilde{x}$ is found, let $$A\tilde{x}=\left[\begin{array}{c} \tilde{u}\\ \tilde{v} \end{array}\right],$$noting that $\tilde{u}\le \hat{b}$ (about which you can do nothing) and $\tilde{v}\le c + e - \tilde{z}$. Let $J=\lbrace j : \tilde{z}_j = 0 \land \tilde{v}_j > c_j\rbrace$. If $J=\emptyset$ you are screwed: there is no way to make $\tilde{x}$ infeasible. Otherwise, you can add the constraint $$\sum_{j\in J} z_j \ge 1$$to the IP model, making $\tilde{z}$ infeasible, and resume solving.


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