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 \begin{equation} h_i(x,\psi)\leq b_i,\quad \forall i=1,...,m \end{equation} for the decision vector $x$.

A joint chance constraint is then given by \begin{equation} P(\ h_i(x,\psi)\leq b_i , \quad \forall i=1,..,m\ )\geq \alpha \end{equation} 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 \begin{equation} P(\ h_i(x,\psi)\leq b_i \ )\geq \alpha, \quad \forall i=1,..,m \end{equation} 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?

  • $\begingroup$ 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. $\endgroup$
    – Ehsan
    Commented Nov 22, 2021 at 18:00
  • $\begingroup$ @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? $\endgroup$
    – Sune
    Commented Nov 22, 2021 at 21:44
  • $\begingroup$ @Sune Yeah, you're right. I didn't pay attention to that. $\endgroup$
    – Ehsan
    Commented Nov 23, 2021 at 5:22
  • $\begingroup$ @Ehsan thanks for adding the "with equal probability of occurrence". I see that it is not at all obvious from the original question. $\endgroup$
    – Djames
    Commented Nov 23, 2021 at 8:04
  • $\begingroup$ @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. $\endgroup$
    – k88074
    Commented Nov 24, 2021 at 12:37


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