What is the role of recourse variable in stochastic programming?

I often see two-stage stochastic programming problems with recourse, written as follows:

Stage 1 \begin{equation} \begin{array}{rrclcl} \displaystyle \min_{x} \,\,\,{c^T x} +E_{\zeta}[Q(x,\zeta)]\\ \textrm{s.t.} & A x & \leq & b \\ &\displaystyle \sum_{i=0}^{n} x_i & = & 1 \\ \end{array} \end{equation}

Stage 2 \begin{equation} \begin{array}{rrclcl} \displaystyle \min_{y} \,\,\,Q(x,\zeta)={q^T y} \\ \textrm{s.t.} & Tx+Wy=h \\ &\displaystyle y\ge0 \\ \end{array} \end{equation}

source: https://en.wikipedia.org/wiki/Stochastic_programming#Two-stage_problems

I’m noticing a few things about this formulation:

  1. The first stage actually depends on a distribution of results from the second stage.
  2. The "recourse" in this case seems synonymous with the word "scenario". Is that correct?

I cannot determine what we are referring to as a “recourse” in the context of this standard two-stage stochastic programming problem.

What is the role of "recourse" here and in what way is the solution to the second stage the "recourse action"? I'm having trouble imagining how such a problem would be solved without recourse. What would this problem look like if it were called "two-stage stochastic programming problem" (i.e. just drop the word recourse)?

  • $\begingroup$ Thanks for your edits. Any suggestion on how to answer this? $\endgroup$ – nundo May 9 at 7:23

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