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This is an optimization problem. The first step is to work out more precisely what your objective function is. In other words, given a candidate assignment of pilots, you need a well-specified way t …

Define a variable $b$ that represents the imbalance between the lengths of the parking lanes. For instance, if $b=5$, that means that the lengths of every pair of parking lanes differs by at most 5. …

With linear programming: you can't. With integer linear programming, here is one solution: introduce zero-or-one variable $t$, and the inequalities $$x \le y - 1 + (1-t)n, \qquad x \ge y+1 - tn.$$ Int …

Yuval Filmus suggests a counterexample: $x_1 + 2x_2 = 2$, $0 \le x_1 \le 1$, $0 \le x_2 \le 1$. Here there is a feasible solution to the LP instance with $x_1=1$, namely, $x_1=1$, $x_2=1/2$. However t …

One approach: replace each inequality $a_1 x_1 + \dots + a_n x_n \le b$ by $a_1 x_1 + \dots + a_n x_n \le b + t$, replace each inequality $a_1 x_1 + \dots + a_n x_n \ge b$ by $a_1 x_1 + \dots + a_n x_ …

A heuristic, using linear programming One approach might be to pick a random objective function, and maximize it. Then repeat, with a different set of objective functions each time. In other words, …

I recommend you use an integer linear programming (ILP) solver to approach this. It will be relatively easy to code this up, and the resulting solution will probably out-perform any other simple sche …

Here's a solution that uses two temporary variables. Let $t,u$ be integer zero-or-one variables, with the intended meaning that $t=1$ if $x \ge 0$, $u=1$ if $x \le 0$, and $y=\neg(t \land u)$. These …

Introduce variables $m_1,m_2,m_3$ to represent the three maxes. Add the linear inequality $m_1 + m_2 + m_3 \ge q$. Then, add the following extra inequalities for $m_1$: $0 \le m_1 \le 1$ $m_1 \le …

Yes (assuming you truly did mean "if" rather than "if and only if"). Add any objective function you want (e.g., 1, or the sum of the variables, or anything), then feed it to a MILP solver and see whe …

This can be expressed with just the equation $X=Y$. Since $X,Y$ are zero-or-one variables, the only possible assignments that are consistent with your condition are $X=Y=0$ and $X=Y=1$. See Express …

If $X$ and $Y$ are zero-or-one (binary) integer variables, then this is encoded as $$X \ge Y.$$ Why does this work? If $Y=1$, then this enforces the constraint $X \ge Y$, as you wanted. If $Y=0$, …

You can solve this by introducing some temporary variables ($t,u,v$) and applying the big-M method. For each expression L <= b <= U, we'll introduce 0-or-1 variables $t,u,v$, with the intent that $u= …

If your measure of "better" is number of variables or number of constraints, I don't believe it is possible to do better. You obviously can't use fewer variables as all of them are already defined; yo …

It can be done with two auxiliary variables, y1,y2: x1 >= y1 x1 >= y2 x1 <= y1 + y2 x2 + x3 + x4 >= 3*(y2 - y1) x2 + x3 + x4 <= 2 + y1 + y2 x5 + x6 >= 2*(y1 - y2) x5 + x6 <= 1 + y1 + y2 x7 >= y1 + …