Minimum cardinality non-anticipativity constraint sets for multistage stochastic programming

Abstract

We consider multistage stochastic programs, in which decisions can adapt over time, (i.e., at each stage), in response to observation of one or more random variables (uncertain parameters). The case that the time at which each observation occurs is decision-dependent, known as stochastic programming with endogeneous observation of uncertainty, presents particular challenges in handling non-anticipativity. Although such stochastic programs can be tackled by using binary variables to model the time at which each endogenous uncertain parameter is observed, the consequent conditional non-anticipativity constraints form a very large class, with cardinality in the order of the square of the number of scenarios. However, depending on the properties of the set of scenarios considered, only very few of these constraints may be required for validity of the model. Here we characterize minimal sufficient sets of non-anticipativity constraints, and prove that their matroid structure enables sets of minimum cardinality to be found efficiently, under general conditions on the structure of the scenario set.

Appendix

This appendix provides the proofs of Propositions 1 and 2.

Proof

(Proof of Proposition 1) Let $\{r,s\}\in \mathcal{{T}}$, let $\sigma \in \{r,s\}$ be chosen so that (8) holds, and define $\bar{\sigma }$ so that $\{\sigma ,\bar{\sigma }\} = \{r,s\}$. Thus (8) ensures that, for any given $i\in \mathcal{{I}}$,

$$\begin{aligned} \sum _{j\in \mathcal{{D}}(r,s)} \beta ^{\sigma }_{j,t} = 0\ \ \Rightarrow \ \ \beta ^s_{i,t+1} = \beta ^r_{i,t+1}, \ \forall t=1,\dots ,T-1 \end{aligned}$$

(9)

is satisfied. To show (6) holds, we must show

$$\begin{aligned} \sum _{j\in \mathcal{{D}}(r,s)} \beta ^{\bar{\sigma }}_{j,t} = 0\ \ \Rightarrow \ \ \beta ^s_{i,t+1} = \beta ^r_{i,t+1}, \ \forall t=1,\dots ,T-1 \end{aligned}$$

(10)

is also satisfied. We proceed by induction on t. For the case $t=1$, (7) ensures $\beta ^\sigma _{i,1} = 0$ so by (9) we have $\beta ^s_{i,2} = \beta ^r_{i,2}$, and (10) must hold for $t=1$. Make the inductive assumption that (10) holds for some $t \in \{1,\dots ,T-2\}$. Now suppose that $\sum _{j\in \mathcal{{D}}(r,s)} \beta ^{\bar{\sigma }}_{j,t+1} = 0$. Then by (2), $\sum _{j\in \mathcal{{D}}(r,s)} \beta ^{\bar{\sigma }}_{j,t} = 0$, so by the inductive assumption, it must be that $\beta ^s_{i,t+1} = \beta ^r_{i,t+1}$. Hence $\sum _{j\in \mathcal{{D}}(r,s)} \beta ^{\sigma }_{j,t+1} = 0$ and so by (9) it must be that $\beta ^s_{i,t+2} = \beta ^r_{i,t+2}$ as required; (10) holds for $t+1$. $\square$

Proof

(Proof of Proposition 2) We prove that ${\mathcal {C}}$ is a generator by showing that for every $e=\{r,s\}\in {\mathcal {T}}{\setminus }{\mathcal {C}}$ there is a dsc-path for e in ${\mathcal {C}}$. We proceed by induction on $d(\theta ^r,\theta ^s)$. For $d(\theta ^r,\theta ^s)=1$ the statement is vacuously true because there is no edge $e=\{r,s\}\in {\mathcal {T}}{\setminus }{\mathcal {C}}$ with $d(\theta ^r,\theta ^s)=1$. For $d(\theta ^r,\theta ^s)\geqslant 2$, pick $i\in {\mathcal {I}}$ with $\theta ^r_i\ne \theta ^s_i$, let j and $j'$ be the indices with $\theta ^r_i=h^i_{j}$ and $\theta ^s_i=h^i_{j'}$, and define $s'$ by $\theta ^{s'}_k=\theta ^{s'}_k$ for $k\ne i$ and

$$\begin{aligned} \theta ^{s'}_{i}={\left\{ \begin{array}{ll} h^i_{j'+1} &{} {\text {if }}j>j',\\ h^i_{j'-1} &{} {\text {if }}j<j'. \end{array}\right. } \end{aligned}$$

Then $d(\theta ^r,\,\theta ^{s'})=d(\theta ^r,\,\theta ^{s})-1$ and $d(\theta ^{s'},\theta ^{s})=1$, i.e., $\{s',s\}\in {\mathcal {C}}$. By induction, there is a dsc-path P for $e'=\{r,s'\}$ in ${\mathcal {C}}$ with ${\mathcal {D}}(e'')\subseteq {\mathcal {D}}(e')\subseteq {\mathcal {D}}(e)$, and together with the edge $\{s',s\}$ we obtain the required dsc-path for e.

Minimality of $\mathcal{{C}}$ is not difficult to establish. Consider $e=\{r,s\}\in \mathcal{{C}}$: without loss of generality suppose $\theta ^r = (h^1_k,\alpha _2,\dots ,\alpha _{\mu })$ (where $\mu =|\mathcal{{I}}|$) and $\theta ^s = (h^1_{k+1},\alpha _2,\dots ,\alpha _{\mu })$ for some $(\alpha _2,\dots ,\alpha _{\mu })\in \times \varTheta _{\mathcal{{I}}{\setminus }\{1\}}$ and some $k\in \{1,\dots ,n_1-1\}$. Then $\mathcal{{D}}(e) = \{1\}$ and any dsc-path for e in $\mathcal{{C}}$ must have the form $r=v_0, v_1,\dots ,v_m=s$ with $e_j = \{v_{j-1},v_{j}\}\in \mathcal{{C}}$ and $\mathcal{{D}}(e_j) = \{1\}$ for $j=1,\dots ,m$. Thus it must be that $\theta ^{v_j} = (h^1_{i_j},\alpha _2,\dots ,\alpha _{\mu })$ for $h^1_{i_j}\in \varTheta _1$ and $|i_j - i_{j+1}| = 1$, for all j. So $i_0 = k$, $i_{m} = k+1$, $i_j\in \{1,\dots ,n_1\}$ for all $j=0,\dots m$ and $|i_j - i_{j-1}| = 1$, for all $j=1,\dots ,m$. Since the path can be assumed to be simple, $\{i_j\}_{j=0}^m$ are distinct. The unique solution to these requirements is to take $m=1$. Thus $\{r,s\}$ is necessary; if it is removed from $\mathcal{{C}}$, the set will no longer be a generator. $\square$