# How to force a bounded relationship to "become redundant" or "not needed"?

As an engineer who is currently working with some optimization problem I am currently running into a difficult reformulation problem. Here $$a$$ is a binary decision variable, $$\phi \in [0,1]$$ and $$d$$ are just two decision variables. In addition to that, $$B$$ and $$C$$ are just two positive numbers that describe the operational characteristic of the system and they are not decision variables.

And hence, my constraint system for situation 1 ($$a=1$$) is as follows:

$$\left\{ {\begin{array}{*{20}{c}} {a = 1 \Rightarrow \frac{B}{\phi } \le d \le C}&{\left( {c1} \right)}\\ {\phi \le a}&{\left( {c2} \right)}\\ {0 \le \phi \le 1}&{\left( {c3} \right)}\\ {d \ge 0}&{\left( {c4} \right)} \end{array}} \right.$$

In the future, I am not sure if constraint $$c_2$$ should be a hard constraint or soft constraint but there is still some room for compromise because the spec is not clear at this point. For now, let temporarily consider $$c_2$$ to be a hard constraint.

For this system it is quite easy to construct a big M formulation when $$a=1$$ for constraint $$c_1$$ as follows: $$\frac{B}{\phi } - M\left( {1 - a} \right) \le d \le C + M\left( {1 - a} \right)$$

However for the case of $$a=0$$ (situation 2), constraint $$c_2$$ ruining the mathematical formulation. Particularly, $$a = 0 \Rightarrow \phi = 0$$ which leads to an infeasible (division by zero) constraint $$\underbrace {\frac{B}{\phi }}_{ + \infty } - \underbrace {M\left( {1 - a} \right)}_{{\rm{finite}}} \le d \le \underbrace C_{{\rm{finite}}} + \underbrace {M\left( {1 - a} \right)}_{{\rm{finite}}}$$.

Therefore, my question is "Given that $$c_2$$ is a hard constraint, is there any way to construct a formulation such that when $$a=0$$ the bounded relationship $${\frac{B}{\phi } \le d \le C}$$ become redundant or not needed or maybe to turn $${\frac{B}{\phi } \le d \le C}$$ into a soft constraint when $$a=0$$ ?"

P/S: Note that in the future, $$c_2$$ may become a soft constraint after some compromise between the engineers.

• Your big-M formulation is not linear (because of $B/\phi$) Commented Jul 8 at 15:51
• Within the context of whatever engineering problem you are addressing, does $a=0$ mean that $d$ is free to take on any nonnegative value, no matter how large or small?
– prubin
Commented Jul 8 at 18:34
• Not linear is ok. But at least it is convex in some sense. Well within my context when $a=0$ then $d$ should take a small non-negative value. But If you are providing a case where $d$ is large then I am willing to listen to your advice ! Commented Jul 9 at 2:55
• @TuongNguyenMinh, for simplicity you can drop $d \leq C$ as a bound constraint. Now and if at the moment the non-linearity does not matter, why not try $((a=1) \implies (B \leq d \phi))$ that yields $(B \leq d \phi + M(1-a))$. By that if $a = 0$ then $\phi$ can be zero without complaint of any division by zero. Commented Jul 9 at 6:56
• Well sadly it is very difficult to drop the upper bound constraint Commented Jul 9 at 7:50