Advantages/disadvantages of different representations of non-anticipativity constraints

Question

When reading various papers about two-stage (or multi-stage) stochastic programs with recourse, a common representation of the non-anticipativity constraints is: $$ \sum_{i}H_ix_i=h, $$ where $i$ indexes the scenarios, and the matrices $H_i$ and vector $h$ are typically not specified (see, e.g., this paper by Carøe and Schultz).

Question: What are some common choices of $H_i$ and $h$ used to enforce non-anticipativity? What advantages and disadvantages do these choices have when used algorithmically (in decomposition approaches, etc.)?

Answer 1

Edit: If one define $H_i$ as an operator, not necessarily a matrix, I can think of what happens in progressive hedging and go for something like: $H_i(x_i)=I(x_i−x^*)+1/2(x_i−x^*)^TI(x_i−x^*)$. If in the matrix format: $H_i=1/rIx_i$ and $h=x^*$.

Answer 2

Suppose there are three scenarios $i$ in the second stage and two time steps $t$. With $$ x_1 = \begin{pmatrix} x_{11} \\ x_{12} \\ \end{pmatrix},\quad x_2 = \begin{pmatrix} x_{21} \\ x_{22} \\ \end{pmatrix},\quad x_3 = \begin{pmatrix} x_{31} \\ x_{32} \\ \end{pmatrix}, $$ one obvious choice would be $$ H_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{pmatrix},\quad H_2 = \begin{pmatrix} -1 & 0 \\ 1 & 0 \\ 0 & -1 \\ 0 & 1 \\ \end{pmatrix},\quad H_3 = \begin{pmatrix} 0 & 0 \\ -1 & 0 \\ 0 & 0 \\ 0 & -1 \\ \end{pmatrix}. $$ Another possibility:

$$ H_1 = \begin{pmatrix} 1 & 0 \\ 1 & 0 \\ 0 & 1 \\ 0 & 1 \\ \end{pmatrix},\quad H_2 = \begin{pmatrix} -1 & 0 \\ 0 & 0 \\ 0 & -1 \\ 0 & 0 \\ \end{pmatrix},\quad H_3 = \begin{pmatrix} 0 & 0 \\ -1 & 0 \\ 0 & 0 \\ 0 & -1 \\ \end{pmatrix}. $$ Both with $h = 0$.

Visually, the block structure is more obvious in the latter formulation. I don't see any major advantages or disadvantages by using one or the other.