New answers tagged constraint
2
You could define more than three tasks, the additional tasks being bundles of the original tasks. You would then make the interval variables optional, with the requirement that for each task exactly one of the intervals in which it would occur (either solo or as part of a bundle) must be present.
3
The constraint is defined only for triples $(i,j,k)$ satisfying the domain conditions, including the requirement that $k\in K(i,j)$. So I think it is fine to just point out that $K(i,j)$ can be empty and remind the reader that $K(i,j)=\emptyset$ implies there are no instances of this constraint for that combination of $i$ and $j$. In other words, I think ...
4
One approach is to define a sparse set $T$ of triples:
$$T=\{i\in [I], j \in [I], k\in [I]: A_i \cap A_j \cap A_k \not= \emptyset\}$$
or
$$T=\{(i,j,k)\in [I]^3: A_i \cap A_j \cap A_k \not= \emptyset\}.$$
Then write the constraint as
$$f_{i,j,k}(x) \le 0 \quad \text{for $(i,j,k)\in T$}$$
1
As far as i know you will have to
resort to a bi-level optimzation problem:
$\min_{A,B} x $ subject to ($\max_x$ subject to $x\leq A$ $x \leq B$)
solve a series of non linear problems where you approximate the $\max$ term by something non-linear (a soft max) and let that converge against $\max$
be fine with an $\frac{1}{m}$ error and turn it into an Mixed ...
6
If $y^s_{fg}=1$, then $y^s_{gf}=0=x^s_{fg}$. So the right-hand sides of (ii), (iii) and (iv) become $-M$ (think $-\infty$ here), which means any values of the left-hand side variables will satisfy them. So, in this case, constraints (ii)-(iv) have no influence on the solution. If, say, $y^s_{gf}=1$ and $y^s_{fg}=0=x^s_{fg}$, then (ii) becomes relevant and (i)...
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