# Chance constrained optimization - interpretation

Suppose that we have a stochastic vector $$\psi$$ and $$S$$ realisations of $$\psi$$ given by $$\psi_1,\dots,\psi_S$$ with equal probability of occurrence. In addition, we have constraints of the form $$$$h_i(x,\psi)\leq b_i,\quad \forall i=1,...,m$$$$ for the decision vector $$x$$.

A joint chance constraint is then given by $$$$P(\ h_i(x,\psi)\leq b_i , \quad \forall i=1,..,m\ )\geq \alpha$$$$ stating that we can accept that some (or all) of these constraints are violation with a probability of $$1-\alpha$$. We could also write single chance constraints as follows $$$$P(\ h_i(x,\psi)\leq b_i \ )\geq \alpha, \quad \forall i=1,..,m$$$$ stating that we will accept violations of the individual constraints with a probability of $$1-\alpha$$. Using binary variables $$z^s$$ equalling 0 iff all constraints are satisfied in realisation $$s$$, we can formulate the joint chance constraint as the MIP \begin{align} h_i(x,\psi_s)\leq b_i+Mz^s,&&\forall i=1,...,m,s=1,...,S\\ \sum_{s=1}^Sz^s\leq \lfloor (1-\alpha)S\rfloor \end{align}

Using binary variables $$z^s_i$$ equalling 0 iff constraint $$i$$ is statisfied in realisation $$s$$ we can formulate the single chance constraint version as follows: \begin{align} &h_i(x,\psi_s)\leq b_i+Mz^s_i,&&\forall i=1,...,m,s=1,...,S\\ &\sum_{s=1}^Sz^s_i\leq \lfloor (1-\alpha)S\rfloor,&&\forall i=1,...,m \end{align}

My question is, what is the interpretation of the following MIP \begin{align} &h_i(x,\psi_s)\leq b_i+Mz^s_i,&&\forall i=1,...,m,s=1,...,S\\ &\sum_{i=1}^m\sum_{s=1}^Sz^s_i\leq \lfloor (1-\alpha)S\rfloor \end{align} Does it have some sensible interpretation?

• I just fixed your deterministic equivalent constraints as they were missing the multiplication of probability of each scenario by each constraint violation variable. If this was not a typo, then perhaps this was the issue confusing you about the constraint interpretation. Nov 22 at 18:00
• @Ehsan I am no expert in stochastic optimization, but isn't the probability present in the form of $1/S$ on the right hand side? So an implicit equal probability of all realisation?
– Sune
Nov 22 at 21:44
• @Sune Yeah, you're right. I didn't pay attention to that. Nov 23 at 5:22
• @Ehsan thanks for adding the "with equal probability of occurrence". I see that it is not at all obvious from the original question. Nov 23 at 8:04
• @Djames Is the last set of constraints something you came up with or you read it somewhere? I cannot see a sensible interpretation. Perhaps, if you multiply the right-hand-side by $m$ then you might read it as constraining the probability that any of the constraints in any of the scenarios is violated. Nov 24 at 12:37