I am solving an LP by using GUROBI - C++ combo.

Assume I have a variable $p \in \mathbb{R}^n$. I am adding this as a vector:

GRBEnv* env;
GRBModel* model;
vector<GRBVar> p;

Now I have $a\in\mathbb{R}^n$ parameters, and I want to add a constraint $a^\top p\ \leq 0$, so I say:

GRBLinExpr sum_expo = 0;
for (int i = 0; i < p.size(); i++) {// add all the coeffs given
                        sum_expo += a[i] * p[i];
model->addConstr(sum_expo <= 0);

Obviously, this is too inefficient. Is there a way that I can:

  1. Define the whole $p$ vector as a multivariate GRBVar, rather than having a vector with all GRBVar values
  2. Add a constraint $a^\top p \leq 0$ in a single line.
  • $\begingroup$ Thanks for the answer. I somehow taught using addVars in one command and then somehow giving inner-product defined constraint would be more efficient. So is what I am doing above standard? $\endgroup$ Commented Feb 13, 2020 at 23:51

1 Answer 1


You can use addTerms() to construct a linear expression in one call. According to the documentation this is the most efficient way:

  • You should avoid using expr = expr + x in a loop. It will lead to runtimes that are quadratic in the number of terms in the expression.
  • Using expr += x (or expr -= x) is much more efficient than expr = expr + x. Building a large expression by looping over += statements is reasonably efficient, but it isn't the most efficient approach.
  • The most efficient way to build a large expression is to make a single call to addTerms.

Concerning adding multiple constraints in one call (documentation about addConstrs):

We recommend that you build your model one constraint at a time (using addConstr), since it introduces no significant overhead and we find that it produces simpler code. Feel free to use these methods if you disagree, though.

  • $\begingroup$ thanks for this answer. How can I call that? Something like:sum_expo::addTerms(a, p, p.size()); $\endgroup$ Commented Feb 14, 2020 at 14:10
  • 1
    $\begingroup$ ok fied by sum_expo.addTerms(&a[0], &p[0], X.size()); $\endgroup$ Commented Feb 14, 2020 at 14:15

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