18
Here is the advice in the IBM CPLEX documentation. So this pertains to CPLEX. I don't know to what extent it applies to other solvers.
First of all, indicator constraints may not be available in all situations:
Indicator Constraints in Optimization
The constraint must be linear; a quadratic constraint is not allowed to have an indicator constraint.
...
15
For Gurobi there seems to be a dual advantage of using general constraints (http://www.gurobi.com/documentation/8.1/refman/constraints.html#subsubsection:GeneralConstraints):
Benefit number one - models are easier to create and can be interpreted easily:
If a model contains general constraints, then Gurobi adds the respective MIP formulations for those ...
13
To the best of my knowledge the indicator constraints are just syntactic sugar for the user. Internally these indicator constraints are reformulated using computed big-M formulations or SOS constraints (special ordered set constraints).
It might be that you are better at computing the value of the big-M using additional knowledge that the solver does not ...
8
Question by me at the IBM CPLEX Forum: Are indicator constraints immune to trickle flow or other numerics-induced logic "errors"?
Are indicator constraints immune to trickle flow or other
numerics-induced logic "errors"?
As discussed at
IBM Technote: Why does a binary or integer variable take on a noninteger value in the solution?, depending
on ...
7
First question: Yes, your algebraic formulation is correct.
Second question: I would lean toward using the algebraic formulation, for two reasons. First, it is not solver-specific. Second, a reader not familiar with indicator constraints will find interpreting the algebraic formulation a bit easier, while a reader familiar with indicators is not likely to ...
7
Introduce binary variable $y_0$ and linear constraints:
\begin{align}
y_0 + y_1 + y_2 &= 1\\
1y_0 + 2y_1 + 3y_2 &= x
\end{align}
Equivalently, eliminating $y_0$:
\begin{align}
y_1 + y_2 &\le 1\\
1 + y_1 + 2y_2 &= x
\end{align}
7
You want to enforce
$$\left(\bigwedge_{i \in I} \lnot x_i\right) \implies \sum_{j \in J} x_j = n.$$
Introduce a new binary variable $y$ and enforce
$$\left(\bigwedge_{i \in I} \lnot x_i\right) \implies y$$
and
$$y \implies \sum_{j \in J} x_j = n.$$
For the first implication, conjunctive normal form yields
\begin{align}
\left(\bigwedge_{i \in I} \lnot x_i\...
6
Answers to the linked question mention both big-M constraints and semicontinuous variables. To speed up the big-M approach, you might consider introducing the constraints dynamically only as they are violated ("row generation" or "cut generation"). Explicitly:
Omit all big-M constraints and the associated binary variables.
Solve the ...
6
$a \geq M_1x$
$a \leq M_2 - (M_2+eps)x$
$b = K_1 + (K_2-K_1)x$
$M_1 < 0$ bigM for the lower bound on $a$
$M_2 > 0$ bigM for the upper bound on $a$
$x$ binary variable equal to $1$ if $a < 0$, $0$ otherwise (i.e. $a \geq 0$)
$eps$ tolerance
6
You can have a look at SCIP's implementation in cons_indicator. They say that:
An indicator constraint is given by a binary variable $z$ and an inequality $ax \le b$. It states that if $z=1$ then $ax \le b$ holds.
This constraint is handled by adding a slack variable $s: ax−s \le b$ with $s \le 0$. The constraint is enforced by fixing $s$ to $0$ if $z=1$.
...
6
Introduce a binary variable $y_{i,j}$ and linear constraints
\begin{align}
x_{i,j} &\le M y_{i,j} \tag1 \\
y_{i,j} + y_{j,i} &\le 1 \tag2
\end{align}
Constraint $(1)$ enforces $x_{i,j} > 0 \implies y_{i,j} = 1$.
Constraint $(2)$ enforces $y_{i,j} = 1 \implies y_{j,i} = 0$.
Constraint $(1)$ (with the roles of $i$ and $j$ interchanged) enforces $y_{...
5
Here's a big-M formulation that does not depend on the objective. (Minimization with positive objective coefficients for $u$ and $d$ could be exploited, but you don't have that here.) Let $\epsilon > 0$ be a small tolerance for positivity of $c$. Let $\underline{c}$ and $\overline{c}$ be bounds for $c$, with similar notation for $w$, $u$, and $d$. Let ...
5
If $M$ is a (small) upper bound on $x$, introduce a binary variable $z$ and big-M constraint $x \le M z$. The idea is that $x>0$ implies $z=1$. Now use $B z$ in the objective.
5
Let $M_i$ be an upper bound on $Q_i$, and impose linear big-M constraints
$0 \le Q_i \le M_i x_i$.
4
in OPL CPLEX you could directly use logical constraints and write
int m=7;
range r=1..m;
int n=2;
{int} I={i | i in r : i mod 2==1};
{int} J={i | i in r : i mod 2==0};
assert card(I inter J)==0;
dvar boolean x[r];
subject to
{
(sum(i in I) x[i]==0) => (sum(j in J) x[j]==n);
}
4
To get the second largest variable when all are nonnegative and at most two can be nonzero, just take the sum of all of them and subtract the largest.
4
$$
x \le z + M(1-\beta) \\
x \ge z - M(1-\beta) \\
x' \le z + M\beta \\
x' \ge z - M\beta \\
$$
If $\beta=1$, we have
$$
x \le z \\
x \ge z \\
x' \le z + M \\
x' \ge z - M \\
$$
which leads to $x=z$ and $x'$ unconstrained. And likewise if $\beta=0$.
4
I would like to remind that CPLEX can handle "if then" directly through logical constraints.
In OPL for example:
int nbKids=300;
float costBus40=500;
float costBus30=400;
dvar int+ nbBus40;
dvar int+ nbBus30;
minimize
costBus40*nbBus40 +nbBus30*costBus30;
subject to
{
40*nbBus40+nbBus30*30>=nbKids;
// with if nb buses 40 more than 3 then nb ...
4
Consider the following tiny example. You have two factories, one warehouse and two product. Factory 1 can produce both goods in sufficient quantity to meet demand but has a very large cost coefficient. Factory 2 only produces the second product, with adequate capacity and a small cost coefficient. The optimal solution to the original problem is to ship ...
3
The answer to the first part is yes, provided that you are using a solver that supports indicator constraints. As far as I know, there is no "standard" notation for it. Something like $$a_1 x_1 \le b \implies x_2 = 0$$would seem reasonable to me. The "else" part is tricky, since it deals with the case $a_1 x_1 > b$ and strict inequalities are a no-no. You ...
3
It sounds like you want to enforce the following logical proposition:
$$\bigvee_{i=6}^{10} (x_i>0) \implies \bigwedge_{j=1}^{5} (x_j=1)$$
You can model this by introducing a binary variable $y$ and linear constraints:
\begin{align}
x_i &\le y&&\text{for $i\in\{6,\dots,10\}$}\\
y&\le x_j &&\text{for $j\in\{1,\dots,5\}$}\\
\end{...
3
This type of constraint is called a complementarity constraint, and there are several ways of modeling it, two of them you already mentioned.
There is no silver-bullet formulation: some will work better than others depending on your instance, your solver, etc.
Some solvers support complementarity constraints directly, for instance Knitro or PATH.
If you ...
3
You can use conjunctive normal form to derive the desired constraints. The first one is:
$$a \ge b \implies a\ge c\\
(b \implies a) \implies (c \implies a)\\
\lnot(\lnot b \lor a) \lor (\lnot c \lor a)\\
(b \land \lnot a) \lor (\lnot c \lor a)\\
(b \lor \lnot c \lor a)
\land (\lnot a \lor \lnot c \lor a)\\
(b \lor \lnot c \lor a)\\
b+ 1- c +a \ge 1\\
a+b \...
3
To enforce $x = 1 \implies y = 1$ for binary variables $x$ and $y$, impose linear constraint $x \le y$. You can derive this constraint via conjunctive normal form:
$$
x \implies y \\
\lnot x \lor y \\
(1 - x) + y \ge 1 \\
x \le y
$$
3
I assume that $w_i$ is a continuous variable with $0 \le w_i \le 1$ and $w_i^\text{start}$ is a constant with $0 \le w_i^\text{start} \le 1$.
You want to enforce $$|w_i-w_i^\text{start}| > 0 \implies |w_i-w_i^\text{start}| \ge 0.02.$$
You can introduce binary variables $y_i^+$ and $y_i^-$ and linear big-M constraints:
\begin{align}
0.02 y_i^+ \le w_i - ...
3
Let $y$ be a binary variable and let $f$ be a linear function bounded above by some constant $M$. The standard approach to enforce $y=1 \implies f\le 0$ is to impose linear big-M constraint $$f\le M(1-y)\tag1.$$ All four of your implications are of this form. For the first one, take $y=\beta$ and $f=z-x$ in $(1)$, yielding $z-x\le M(1-\beta)$. For the ...
3
For sake of completeness: An alternative to using an additional binary variable and a big M would be to use an SOS constraint. An SOS constraint specifies a set of variables, at most one of which may be non-zero.
3
You can introduce a binary variable $x_k$ and linear constraints
\begin{align}
\sum_k x_k &\ge N\tag1\\
-t_k+T_k&\le M_k(1-x_k) &&\text{for $k\in K$}\tag2
\end{align}
Here, the “big-M” constant $M_k$ is a small upper bound on $-t_k+T_k$. Because $t_k\ge 0$, you can take $M_k=T_k$, and the constraint simplifies to $t_k\ge T_k x_k$.
Constraint $...
2
A non-linear formulation is given for $ x \geq 0$, $ x’ \geq 0$, $z \geq 0$ and $ \beta $ binary.
$\left\{ \begin{array}{l}
x \geq \beta \cdot z \\
x’ \geq (1- \beta) \cdot z \\
\end{array} \right. $
If $ \beta = 1 $ we get
$\left\{ \begin{array}{l}
x \geq z \\
x’ \geq 0 \\
\end{array} \right. $
If $ \beta = 0 $ we get
$\left\{ \begin{array}{l}
x \geq 0 \\...
2
Do you mean $i$ and $j$ instead of $v$ and $v’$?
If so, the constraints you want are $\alpha_{i,j}\le A_i$ and $\alpha_{i,j}\le A_j$.
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