Linear programming problem with no objective function I have a binary integer programming problem for which I only need a solution that meets all the constraints.  I do not have an objective function that I am trying to minimize or maximize.
I've been using lp_solve to solve this problem and it works well — I simply define my objective function to be
$$\begin{array}{ll} \text{maximize} & {\bf 0}^T {\bf x}\end{array}$$
However, this seems kind of silly and I keep wondering if there is a better way.
Is there a name for linear programming problems with no objective function?  If I don't have an objective function is there some technique more efficient than linear programming (in particular, branch and bound) that I should be using?

Suppose you have a (bounded) linear programming problem:

$\mathrm{maximize }\,{\bf c}^T x$
$s. t. \mathrm{A}\bf{x} \ge b$

Then you can determine a bound on the optimal solution, e.g. $\mathrm{maximize }\,{\bf c}^T x > n$ by constructing the following LP:

$\mathrm{maximize }\,{\bf 0}^T {\bf x}$
$s. t. \mathrm{A}\bf{x} \ge b$, ${\bf c}^T {\bf x} \geq n$

That LP will be feasible iff the optimal solution has at least value $n$, so you can find the true maximum of the objective function through binary search in a number of steps proportional to the logarithm of its value. So getting any feasible point can't be much easier than finding the optimum, or you could use it to make an actual quick LP solver too.
For binary integer programming in particular, there's a reduction from satisfiability to determining if a binary integer program is feasible. So checking if a BIP is feasible is still NP-complete.
That said, it's possible to find feasible solutions to some LPs relatively quickly. E.g.

$\mathrm{maximize }\,{\bf 0}^T {\bf x}$
$s. t. {\bf b}^T {\bf x} \ge d$, ${\bf x} \geq 0$

has feasible points iff the optimum of

$\mathrm{maximize }\,{\bf b}^T {\bf x}$
$s. t. {\bf x} \geq 0$

is greater than $d$, and this LP is generally unbounded if at least one element in $\bf b$ is greater than zero, otherwise the optimum is 0.
