On sample average approximation for two-stage stochastic programs without relatively complete recourse

Abstract

We investigate sample average approximation (SAA) for two-stage stochastic programs without relatively complete recourse, i.e., for problems in which there are first-stage feasible solutions that are not guaranteed to have a feasible recourse action. As a feasibility measure of the SAA solution, we consider the “recourse likelihood”, which is the probability that the solution has a feasible recourse action. For $\epsilon \in (0,1)$, we demonstrate that the probability that a SAA solution has recourse likelihood below $1-\epsilon$ converges to zero exponentially fast with the sample size. Next, we analyze the rate of convergence of optimal solutions of the SAA to optimal solutions of the true problem for problems with a finite feasible region, such as bounded integer programming problems. For problems with non-finite feasible region, we propose modified “padded” SAA problems and demonstrate in two cases that such problems can yield, with high confidence, solutions that are certain to have a feasible recourse decision. Finally, we conduct a numerical study on a two-stage resource planning problem that illustrates the results, and also suggests there may be room for improvement in some of the theoretical analysis.

Appendix

1.1 Proof of Corollary 5

For any $\nu \in (0,1)$,

$$\begin{aligned} N&\ge \frac{1}{1-\nu }\bigg [\frac{1}{\epsilon }\log \frac{1}{\beta } +\frac{n_1}{\epsilon }\Bigl (\log \frac{1}{\nu \epsilon }+\log \Bigl (\frac{m_1}{n_1}+2\Bigr )+1\Bigr )\\&\quad +\frac{2n_1(n_2+1)}{\epsilon }\Bigl (\log \Bigl (\frac{m_2}{n_2+1}\Bigr )+1\Bigr )+n_1\bigg ]\\ \Rightarrow (1-\nu )N&\ge \frac{1}{\epsilon }\log \frac{1}{\beta } +\frac{n_1}{\epsilon }\Bigl (\log \frac{1}{\nu \epsilon } +\log \Bigl (\frac{2n_1+m_1}{n_1}\Bigr )+1\Bigr )\\&\quad +\frac{2n_1(n_2+1)}{\epsilon } \Bigl (\log \Bigl (\frac{m_2}{n_2+1}\Bigr )+1\Bigr )+n_1\\ \Rightarrow N&\ge \frac{1}{\epsilon }\log \frac{1}{\beta } +\frac{n_1}{\epsilon }\Bigg (\log \frac{1}{\nu \epsilon }+\log n_1-1 +\frac{\nu N\epsilon }{n_1}\Bigg )\\&\quad +\frac{n_1}{\epsilon } \Bigl (\log \Bigl (\frac{2n_1+m_1}{n_1^2}\Bigr )+2\Bigr )\\&\quad +\frac{2n_1(n_2+1)}{\epsilon }\Bigl (\log \Bigl (\frac{m_2}{n_2+1}\Bigr )+1\Bigr )+n_1\\ \Rightarrow N&\ge \frac{1}{\epsilon }\log \frac{1}{\beta } +\frac{n_1}{\epsilon }\log N+\frac{n_1}{\epsilon } \Bigl (\log \Bigl (\frac{2n_1+m_1}{n_1^2}\Bigr )+2\Bigr )\\&\quad +\frac{2n_1(n_2+1)}{\epsilon }\Bigl (\log \Bigl (\frac{m_2}{n_2+1}\Bigr )+1\Bigr )+n_1\\ \Rightarrow \log \beta&\ge n_1\log N+n_1\Bigl (\log \Bigl (\frac{2n_1 +m_1}{n_1^2}\Bigr )+2\Bigr )+2n_1(n_2+1)\Bigl (\log \Bigl (\frac{m_2}{n_2+1}\Bigr )+1\Bigr )\\&\quad +\epsilon n_1-\epsilon N\\ \Rightarrow \log \beta&\ge n_1\log N+n_1\log (2n_1+m_1)+2n_1(n_2+1)\log m_2\\&\quad +\epsilon n_1-\epsilon N-2\log (n_1!)-2n_1\log \Bigl ((n_2+1)!\Bigr )\\ \Rightarrow \beta&\ge \frac{N^{n_1}(2n_1+m_1)^{n_1}m_2^{2n_1(n_2+1)} e^{-\epsilon (N-n_1)}}{(n_1!)^2\Bigl ((n_2+1)!\Bigr )^{2n_1}}\\&\ge \left( {\begin{array}{c}N\\ n_1\end{array}}\right) \frac{1}{n_1!}(2n_1+m_1)^{n_1} \biggl (\frac{m_2^{n_2+1}}{(n_2+1)!}\biggr )^{2n_1}(1-\epsilon )^{N-n_1}. \end{aligned}$$

The third inequality can be justified by observing that $\frac{\nu N\epsilon }{n_1}\ge 1+\log \frac{\nu N\epsilon }{n_1}$. The fifth inequality can be justified by observing that $\log (k!)\ge k(\log k -1)$ for $k\ge 1$. The result follows by setting $\nu =1/2$.

1.2 Proof of Proposition 11

We only need to prove that $H(x,\xi )$ is Lipshitz continuous in $(W(\xi ),T(\xi ),h(\xi ))$ for all $x\in X$ under infinity (matrix) norm. Then the result follows from the assumption that $(W(\xi ),T(\xi ),h(\xi ))$ is Lipshitz continuous in $\xi$.

For fixed $x\in X$, assume $(y^*,\eta ^*)$ and $(y',\eta ')$ are optimal solutions of (11) for $M:=({\overline{W}},{\overline{T}},{\overline{h}}):=({\overline{W}}(\xi ),{\overline{T}}(\xi ),{\overline{h}}(\xi ))$ and $M':=({\overline{W}}',{\overline{T}}',{\overline{h}}'):=({\overline{W}}(\xi '),{\overline{T}}(\xi '),{\overline{h}}(\xi '))$, respectively. Consider the different conditions.

• 1. If the problem has only right-hand side randomness, then

  $$\begin{aligned} \begin{aligned} \eta ^*&=\min _y\Big \{\max _{i \in I} \{{\overline{h}}_i - {\overline{W}}_iy-{\overline{T}}_ix\Big \} : Dy \ge d - Cx \}\\&\le \max _{i \in I} \Big \{{\overline{h}}_i -{\overline{W}}_iy'-{\overline{T}}_ix\Big \}\\&\le \max _{i \in I} \Big \{{\overline{h}}'_i-{\overline{W}}_iy'-{\overline{T}}_ix\Big \}+\Vert M-M'\Vert _\infty \Vert \left( \begin{array}{c} 0\\ 0\\ 1 \end{array} \right) \Vert _\infty \\&\le \eta '+\Vert M-M'\Vert _\infty . \end{aligned} \end{aligned}$$

• 2. If the problem has fixed recourse and X is bounded, then

  $$\begin{aligned} \begin{aligned} \eta ^*&=\min _y\Big \{\max _{i \in I} \Big \{{\overline{h}}_i - {\overline{W}}_iy-{\overline{T}}_ix\Big \}:Dy\ge d-Cx \Big \}\\&\le \max _{i \in I} \{{\overline{h}}_i -{\overline{W}}_iy'-{\overline{T}}_ix\}\\&\le \max _{i \in I} \{{\overline{h}}'_i-{\overline{W}}_iy'-{\overline{T}}'_ix\}+\Vert M-M'\Vert _\infty \Vert \left( \begin{array}{c} 0\\ x\\ 1 \end{array} \right) \Vert _\infty \\&\le \eta '+R\Vert M-M'\Vert _\infty \end{aligned} \end{aligned}$$

  for some $R>0$ since X is bounded.

• 3. If $\{(x,y):x\in X,Dy\ge d-Cx \}$ is bounded, then similarly

  $$\begin{aligned} \begin{aligned} \eta ^*&=\min _y\Big \{\max _{i \in I} \{{\overline{h}}_i - {\overline{W}}_iy-{\overline{T}}_ix\Big \}:Dy\ge d-Cx \}\\&\le \max _{i \in I} \{{\overline{h}}_i -{\overline{W}}_iy'-{\overline{T}}_ix\}\\&\le \max _{i \in I} \{{\overline{h}}'_i-{\overline{W}}'_iy'-{\overline{T}}'_ix\}+\Vert M-M'\Vert _\infty \Vert \left( \begin{array}{c} y'\\ x\\ 1 \end{array} \right) \Vert _\infty \\&\le \eta '+R\Vert M-M'\Vert _\infty . \end{aligned} \end{aligned}$$

  for some $R>0$ since $\{(x,y):x\in X,Dy\ge d-Cx \}$ is bounded.

So for each case, there exists $R>0$ such that

$$\begin{aligned} H(x,\xi )-H(x,\xi ')=\eta ^*-\eta '\le R\Vert M-M'\Vert _\infty . \end{aligned}$$

Similarly, $H(x,\xi ')-H(x,\xi )\le R\Vert M-M'\Vert _\infty$. Therefore,

$$\begin{aligned} |H(x,\xi )-H(x,\xi ')|\le R\Vert M-M'\Vert _\infty , \end{aligned}$$

which implies that $H(x,\xi )$ is a Lipschitz continuous function in $(W(\xi ),T(\xi ),h(\xi ) )$ for all $x\in X$ under infinity norm. Therefore, Assumption 4 holds.

1.3 Proof of Proposition 12

First observe that the linear program (11) is always feasible. Thus,

$$\begin{aligned} \begin{aligned} H(x,\xi )&=\max _{\alpha } \quad \alpha ^T({\overline{h}}(\xi )-{\overline{T}}(\xi ){\hat{x}})+\beta ^T(d - C{\hat{x}}),\\&\qquad {\text {s.t.}}\quad {\overline{W}}(\xi )^T\alpha + D^T\beta =0,\\&\qquad \qquad e^T\alpha =1,~\alpha \ge 0,~\beta \ge 0, \end{aligned} \end{aligned}$$

where we adopt the convention that if the dual linear program is infeasible then the optimal value is defined to be $-\infty$. Thus,

$$\begin{aligned} \max \{ H({\hat{x}},\xi ^J): J\in [N]^d \}&= \max _{\alpha ,\beta , J}\quad \alpha ^T\big ({\overline{h}}(\xi ^J)-{\overline{T}}(\xi ^J){\hat{x}}\big )+\beta ^T(d-C{\hat{x}}),\\&\qquad {\text {s.t.}}\quad \alpha ^T{\overline{W}}(\xi ^J) + \beta ^T D=0,\\&\qquad \qquad e^T\alpha =1,~\alpha \ge 0,~\beta \ge 0, J\in [N]^d \end{aligned}$$

Next, introduce binary variables $\delta _{qj}$ for $q \in [d]$ and $j \in [N]$, where $\delta _{qj}=1$ implies that $J_q=j$. This leads to the mixed-integer nonlinear program:

$$\begin{aligned} \begin{aligned} \max _{\alpha ,\beta ,\xi ,\delta }&\quad \alpha ^T({\overline{h}}(\xi )-{\overline{T}}(\xi ){\hat{x}}) + \beta ^T(d - C{\hat{x}})\\ {\text {s.t.}}&\quad \xi _q=\sum _{j \in [N]}\xi ^j_q\delta _{qj},\quad q \in [d],\\&\quad \sum _{j \in [N]}\delta _{qj}=1,\quad q \in [d],\\&\quad \alpha ^T {\overline{W}}(\xi ) + \beta ^TD =0,\\&\quad e^T\alpha =1,\\&\quad \alpha \ge 0,~\beta \ge 0, ~\delta \in \{0,1\}^{d\times N}. \end{aligned} \end{aligned}$$

(36)

Using the assumptions that ${\overline{W}}(\xi )$, ${\overline{T}}(\xi )$ and ${\overline{h}}(\xi )$ are linear in $\xi$, problem (36) can be written as the following mixed-integer bilinear program:

$$\begin{aligned} \max _{\alpha ,\beta ,\xi ,\delta }&\quad \alpha ^T\Biggl ({\overline{H}}\xi -\sum _{k \in [n_1]} {\overline{T}}^k\xi {\hat{x}}_k \Biggr ) + \beta ^T(d - C{\hat{x}}) \end{aligned}$$

(37)

$$\begin{aligned} {\text {s.t.}}&\quad \xi _q=\sum _{j \in [N]}\xi ^j_q\delta _{qj},\quad q \in [d], \end{aligned}$$

(38)

$$\begin{aligned}&\quad \sum _{j \in [N]}\delta _{qj}=1,\quad q \in [d], \end{aligned}$$

(39)

$$\begin{aligned}&\quad \alpha ^T{\overline{W}}^k\xi +\beta ^T D^k =0,\quad k \in [n_2], \end{aligned}$$

(40)

$$\begin{aligned}&\quad e^T\alpha =1, \end{aligned}$$

(41)

$$\begin{aligned}&\quad \alpha \ge 0,~\beta \ge 0,~\delta \in \{0,1\}^{d\times N}. \end{aligned}$$

(42)

We next use (38) to substitute out the variables $\xi _q$ in this formulation. Observe that one this is done, the only nonlinear terms are of the form $\alpha _p\delta _{qj}$ for $p \in I, q \in [d], j \in [N]$. Thus, introduce new variables $z_{pqj}$ to represent this product for each $p \in I$, $q \in [d]$, and $j \in [N]$. Using constraint (41) we derive the linear constraints:

$$\begin{aligned} \sum _{p \in I} z_{pqj} \biggl (= \sum _{p \in I} \alpha _p\delta _{qj} \biggr ) = \delta _{qj}, \quad q \in [d],~j \in [N]. \end{aligned}$$

(43)

Using constraints (39) we derive the linear constraints:

$$\begin{aligned} \sum _{j \in [N]} z_{pqj} \biggl (= \sum _{j \in [N]} \alpha _p\delta _{qj} \biggr ) = \alpha _p, \quad p \in I, \ q \in [d]. \end{aligned}$$

(44)

Observe that constraints (43),(44) together with (42) and (39) are sufficient to imply $z_{pqj} = \alpha _p \delta _{qj}$. Indeed for any fixed q, suppose $j^*_q$ is the index such that $\delta _{qj^*_q}=1$. Then (43) implies that $z_{pqj}=0=\alpha _p\delta _{qj}$ for any $j \ne j^*_q$ and all p, and (44) implies that $\sum _{j \in [N]} z_{pqj} = z_{pqj^*_q} = \alpha _p = \alpha _p \delta _{qj^*_q}$ also for all p. Also note that constraints (41), (43) and (44) imply

$$\begin{aligned} \sum _{j\in [N]}\delta _{qj}=\sum _{j\in [N]}\sum _{p\in I}z_{pqj}=\sum _{p\in I}\Big (\sum _{j\in [N]}z_{pqj}\Big )=e^T\alpha =1 \end{aligned}$$

for $q\in [d]$. Therefore, constraints (39) are redundant when constraints (41), (43) and (44) are present. Thus the mixed-integer bilinear program (37)–(42) can be reformulated as the MILP given in Proposition 12 by using (38) to substitute out the variables $\xi$, using $z_{pqj}$ to replace the bilinear terms $\alpha _p\delta _{qj}$, and adding the constraints (43)–(44).

1.4 Proof of Proposition 13

Note that $H({\hat{x}},\cdot )$ is convex under the assumptions and the maximum of a convex function can only be attained at extreme points. Therefore, we can reformulate (27) as

$$\begin{aligned} \max \Big \{H({\hat{x}},\xi ):\xi \in \prod _{q\in [d]} \big \{\xi ^{\min }_{q},\xi ^{\max }_{q}\big \} \Big \}. \end{aligned}$$

(45)

Introduce binary variables $\delta _{q1}$ and $\delta _{q2}$ for $q\in [d]$, where $\delta _{q1}=1$ implies that $\xi _q=\xi ^{\min }_{q}$ and $\delta _{q2}=1$ implies that $\xi _q=\xi ^{\max }_{q}$. We can then rewrite (45) as a mixed-integer bilinear program:

$$\begin{aligned} \begin{aligned} \max _{\alpha ,\beta ,\xi ,\delta }&\quad \alpha ^T\biggl ({\overline{H}}\xi -\sum _{k \in [n_1]}{\overline{T}}^k\xi {\hat{x}}_k \biggr ) + \beta ^T(d - C{\hat{x}})\\ {\text {s.t.}}&\quad \xi _q=\xi ^{\min }_q\delta _{q1}+\xi ^{\max }_q\delta _{q2},\quad q \in [d],\\&\quad \delta _{q1}+\delta _{q2}=1,\quad q \in [d],\\&\quad \alpha ^T {\overline{W}}+ \beta ^TD =0,\\&\quad e^T\alpha =1,\\&\quad \alpha \ge 0,~\beta \ge 0, ~\delta \in \{0,1\}^{d\times 2}. \end{aligned} \end{aligned}$$

(46)

Similar to Proposition 12, we introduce variables $z_{pqj}$ to represent $\alpha _p\delta _{qj}$ for $p\in I,q\in [d],j\in \{1,2\}$. Then problem (45) can be written as the following mixed-integer linear program:

$$\begin{aligned} \max _{\alpha ,\beta ,\delta ,z}&\quad \sum _{p\in I}\sum _{q\in [d]} \Bigg [\Bigg ({\overline{H}}_{pq}-\sum _{k\in [n_1]}{\hat{x}}_k {\overline{T}}^k_{pq}\Bigg )(\xi _q^{\min }z_{pq1} +\xi _q^{\max }z_{pq2}) \Bigg ]+ \beta ^T(d - C{\hat{x}}) \end{aligned}$$

(47)

$$\begin{aligned} {\text {s.t.}}&\quad \sum _{p\in I}z_{pqj}=\delta _{qj},\quad q\in [d],~j\in \{1,2\},\end{aligned}$$

(48)

$$\begin{aligned}&\quad z_{pq1}+z_{pq2}=\alpha _p,\quad p\in I,~q\in [d],\end{aligned}$$

(49)

$$\begin{aligned}&\quad \alpha ^T {\overline{W}}+ \beta ^TD =0,\end{aligned}$$

(50)

$$\begin{aligned}&\quad e^T\alpha =1,\end{aligned}$$

(51)

$$\begin{aligned}&\quad \alpha \ge 0,~\beta \ge 0,~z\ge 0, ~\delta \in \{0,1\}^{d\times 2}. \end{aligned}$$

(52)

Finally, we apply the reformulation-linearization technique to strengthen the MILP formulation (47)–(52). We introduce new variables $w_{pqj}\ge 0$ to represent the product $\beta _p\delta _{qj}$ for $p\in [m_2]{\setminus } I,q\in [d],j\in \{1,2\}$. Note that $w_{pqj}=\beta _p\delta _{qj}$ and $\alpha ^T{\overline{W}}+\beta ^TD=0$ imply

$$\begin{aligned}&\sum _{p\in I}{\overline{W}}_{pk}z_{pqj}+\sum _{p\in [m_2]{\setminus } I}D_{pk} w_{pqj}\Bigg (=\delta _{qj}\Bigg (\sum _{p\in I}{\overline{W}}_{pk}\alpha _p+\sum _{p\in [m_2]{\setminus } I}D_{pk}\beta _p \Bigg )\Bigg )=0,\nonumber \\&\quad k\in [n_1],q\in [d],j\in \{1,2\}, \end{aligned}$$

(53)

and constraints $\delta _{q1}+\delta _{q2}=1$ for $q\in [d]$ imply

$$\begin{aligned} w_{pq1}+w_{pq2}\Big (=\beta _p(\delta _{q1}+\delta _{q2})\Big )=\beta _p,\quad p\in [m_2]{\setminus } I,~q\in [d]. \end{aligned}$$

(54)

Note that constraints (49), (53) and (54) imply (50). Therefore, adding new variables $w\ge 0$ together with constraints (53) and (54) into the original MILP formulation yields a strengthened formulation in the lifted space, as the MILP given in Proposition 13.