Convexifications of rank-one-based substructures in QCQPs and applications to the pooling problem

Abstract

We study sets defined as the intersection of a rank-one constraint with different choices of linear side constraints. We identify different conditions on the linear side constraints, under which the convex hull of the rank-one set is polyhedral or second-order cone representable. In all these cases, we also show that a linear objective can be optimized in polynomial time over these sets. Towards the application side, we show how these sets relate to commonly occurring substructures of a general quadratically constrained quadratic program. To further illustrate the benefit of studying quadratically constrained quadratic programs from a rank-one perspective, we propose new rank-one formulations for the generalized pooling problem and use our convexification results to obtain several new convex relaxations for the pooling problem. Finally, we run a comprehensive set of computational experiments and show that our convexification results together with discretization significantly help in improving dual bounds for the generalized pooling problem.

Appendices

A Omitted proofs

1.1 Proof of Part (iii) of Proposition 2

Proposition 9

$\text {conv} \left( {{\mathcal {U}}}^{\text {row}}_{(n_1,n_2)}(l,u)\right)$ is described by the inequalities (17), (18), (19), and (20).

Proof

We will use Fourier-Motzkin elimination to obtain the convex hull in the original space. Now, we will project the t variables in the order $t_{n_2}$, $t_{n_2-1}$, $t_{n_2 -2}$, ..., $t_1$.

We claim that after projecting out variables $t_{n_2}, \dots , t_{n_2 - j}$, the resulting system of the inequalities is:

$$\begin{aligned} 1 - \sum \limits _{p = 1}^{n_2 - (j + 1)} t_p \le&\ \sum \limits _{i \in {\mathcal {I}}} \sum \limits _{k \in T_{i} } \frac{W_{ik}}{l_i}&\forall&(T_1, T_2, \dots , T_ {|{\mathcal {I}}|}) \in {\mathcal {P}}_{|{\mathcal {I}}|}(\{n_2 - j, \dots , n_2\}) \\ 1 - \sum \limits _{p = 1}^{n_2 - (j + 1)} t_p \ge&\ \sum \limits _{i \in [n_1]} \sum \limits _{k \in T_i} \frac{W_{ik}}{u_i}&\forall&(T_1, T_2, \dots , T_ {n_1}) \in {\mathcal {P}}_{n_1}(\{n_2 - j, \dots , n_2\}) \\ u_{i_2}W_{i_1k} \ge&l_{i_1} W_{i_2k}&\forall&i_2 \in [n_1], \forall i_1 \in {\mathcal {I}}, \forall k \in \{n_2 - j, \dots , n_2\} \\ l_it_k \le W_{ik} \le&u_it_k&\forall&i \in [n_1], \forall k \in [n_2 - (j + 1)] \\ t_k \ge&0&\forall&k \in [n_2 - (j + 1)] \\ W_{ij} \ge&0&\forall&i \in [n_1], \forall j \in [n_2 ]. \end{aligned}$$

Base case: After projecting out $t_{n_2}$, we obtain the system:

$$\begin{aligned} \sum \limits _{j = 1}^{n_2 - 1} t_j \le&\ 1 \end{aligned}$$

(67)

$$\begin{aligned} W_{i n_2} \le&\ (1 - \sum \limits _{j = 1}^{n_2 - 1} t_j)u_i&\forall&i \in [n_1] \end{aligned}$$

(68)

$$\begin{aligned} W_{i n_2} \ge&\ (1 - \sum \limits _{j = 1}^{n_2 - 1} t_j)l_i&\forall&i \in {\mathcal {I}} \nonumber \\ u_{i_2}W_{i_1n_2} \ge&\ l_{i_1} W_{i_2n_2}&\forall&i_2 \in [n_1],\forall i_1 \in {\mathcal {I}} \nonumber \\ l_it_j \le W_{ij} \le&\ u_it_j&\forall&i \in [n_1], \forall j \in [n_2 - 1] \nonumber \\ t_j \ge&\ 0&\forall&j \in [n_2 - 1], \nonumber \\ W_{ij} \ge&\ 0&\forall&i \in [n_1], \forall j \in [n_2 ], \end{aligned}$$

(69)

Note that (68) and (69) imply (67). Therefore the above can be written as:

$$\begin{aligned} (1 - \sum \limits _{j = 1}^{n_2 - 1} t_j) \le&\ \sum \limits _{i \in {\mathcal {I}}} \sum \limits _{k \in T_{i} } \frac{W_{in_2}}{l_i}&\forall&(T_1, T_2, \dots , T_ {|{\mathcal {I}}|}) \in {\mathcal {P}}_{|{\mathcal {I}}|}(\{n_2 \}) \\ (1 - \sum \limits _{j = 1}^{n_2 - 1} t_j) \ge&\ \sum \limits _{i \in [n_1]} \sum \limits _{k \in T_i} \frac{W_{ik}}{u_i}&\forall&(T_1, T_2, \dots , T_ {n_1}) \in {\mathcal {P}}_{n_1}(\{n_2 \}) \\ u_{i_2}W_{i_1n_2} \ge&\ l_{i_1} W_{i_2n_2}&\forall&i_2 \in [n_1], \forall i_1 \in {\mathcal {I}} \\ l_it_j \le W_{ij} \le&\ u_it_j&\forall&i \in [n_1], \forall j \in [n_2 - 1] \\ t_j \ge&\ 0&\forall&j \in [n_2 - 1], \\ W_{ij} \ge&\ 0&\forall&i \in [n_1], \forall j \in [n_2], \end{aligned}$$

proving the base case.

Induction step: After projecting $t_{n_2}$, $\dots$, $t_{n_2 -j}$, by the induction hypothesis we have the following system:

$$\begin{aligned} 1 - \sum \limits _{p = 1}^{n_2 - (j + 2)} t_p - \sum \limits _{i \in {\mathcal {I}}} \sum \limits _{k \in T_{i} } \frac{W_{ik}}{l_i} \le&t_{n_2 - (j +1)}&\forall&(T_1, T_2, \dots , T_ {|{\mathcal {I}}|}) \in {\mathcal {P}}_{|{\mathcal {I}}|}(\{n_2 - j, \dots , n_2\}) \\ \frac{W_{i,n_2 - (j +1)}}{u_i} \le&\ t_{n_2 - (j +1)}&\forall&i \in [n_1] \\ 1 - \sum \limits _{p = 1}^{n_2 - (j + 2)} t_p - \sum \limits _{i \in [n_1]} \sum \limits _{k \in T_i} \frac{W_{ik}}{u_i} \ge&\ t_{n_2 - (j +1)}&\forall&(T_1, T_2, \dots , T_ {n_1}) \in {\mathcal {P}}_{n_1}(\{n_2 - j, \dots , n_2\}) \\ \frac{W_{i,n_2 - (j +1)}}{l_i} \ge&\ t_{n_2 - (j +1)}&\forall&i \in {\mathcal {I}} \\ u_{i_2}W_{i_1k} \ge&\ l_{i_1} W_{i_2k}&\forall&i_2 \in [n_1], \forall i_1 \in {\mathcal {I}},\forall k \in \{n_2 - j, \dots , n_2\} \\ l_it_k \le W_{ik} \le&\ u_it_k&\forall&i \in [n_1], \forall k \in [n_2 - (j + 1)] \\ t_k \ge&\ 0&\forall&k \in [n_2 - (j + 1)] \\ W_{ik} \ge&\ 0&\forall&i \in [n_1], \forall k \in [n_2 ]. \end{aligned}$$

Note that a constraint of the form:

$$\begin{aligned} - \sum \limits _{i \in [n_1]} \sum \limits _{k \in T_i} \frac{W_{ik}}{u_i}\ge - \sum \limits _{i \in {\mathcal {I}}} \sum \limits _{k \in T'_{i} }\frac{W_{ik}}{l_i}, \end{aligned}$$

where $(T_1, T_2 \dots T_ {n_1}) \in {\mathcal {P}}_{n_1}(\{n_2 - j, \dots , n_2\})$ and $(T'_1, T'_2, \dots T'_ {|{\mathcal {I}}|})\in {\mathcal {P}}_ {|{\mathcal {I}}|} (\{n_2 - j, \dots , n_2\})$ is implied by constraints of the form $u_{i_2}W_{i_1k} \ge l_{i_1} W_{i_2k} \forall i_2 \in [n_1], i_1 \in {\mathcal {I}}, k \in \{n_2 - j, \dots , n_2\}.$ Thus, after projecting $t_{n_2 - (j +1)}$, we obtain:

$$\begin{aligned} 1 - \sum \limits _{p = 1}^{n_2 - (j + 2)} t_p \le&\ \sum \limits _{i \in {\mathcal {I}}} \sum _{k \in T_{i} } \frac{W_{ik}}{l_i}&\forall&(T_1, T_2, \dots , T_ {|{\mathcal {I}}|}) \in {\mathcal {P}}_ {|{\mathcal {I}}|}( \{n_2 - (j +1), \dots , n_2\} ) \\ 1 - \sum _{p = 1}^{n_2 - (j + 2)} t_p \ge&\ \sum _{i \in [n_1]} \sum _{k \in T_i} \frac{W_{ik}}{u_i}&\forall&(T_1, T_2, \dots , T_ {n_1}) \in {\mathcal {P}}_{n_1} (\{n_2 - (j +1), \dots , n_2\}) \\ u_{i_2}W_{i_1k} \ge&\ l_{i_1} W_{i_2k}&\forall&i_2 \in [n_1], \forall i_1 \in {\mathcal {I}}, \forall k \in \{n_2 - (j + 1), \dots , n_2\} \\ l_it_k \le W_{ik} \le&\ u_it_k&\forall&i \in [n_1], \forall k \in [n_2 - (j + 2)] \\ t_k \ge&\ 0&\forall&k \in [n_2 - (j + 2)] \\ W_{ik} \ge&\ 0&\forall&i \in [n_1], \forall k \in [n_2 ]. \end{aligned}$$

It is straightforward now to see that after all t variables are projected, we obtain the result. $\square$

Proposition 10

The inequalities in (17) can be separated in polynomial-time.

Proof

For a given matrix ${{\hat{W}}}$, let us define an index $j_{\text {row}}$ for each index $j \in [n_2]$ as

$$\begin{aligned} j_{\text {row}}:= \min \left( \mathop {{{\,\mathrm{argmax}\,}}}\limits _{i =1,\dots ,n_1} \left\{ \frac{{{\hat{W}}}_{ij}}{u_i} \right\} \right) . \end{aligned}$$

(70)

Here we are breaking ties arbitrarily using the smallest index, when necessary. Then, we define a partition $T_1^*,\dots ,T_{n_1}^*$ of the set $[n_2]$ as

$$\begin{aligned} T_i^* := \{j \,|\, j_{\text {row}} = i\} . \end{aligned}$$

Let

$$\begin{aligned} \theta := \sum _{i=1}^{n_1}\sum _{j \in T_i^*} \frac{{{\hat{W}}}_{ij}}{u_i}. \end{aligned}$$

If $\theta >1$, then a violated inequality is discovered. Otherwise, we conclude that ${{\hat{W}}}$ satisfies all the inequalities in (17) (by construction, the partition $T_1^*, \dots , T_{n_1}^*$ corresponds to the inequality with the largest deviation, if one exists). Finally, we note that the complexity of this separation routine is ${\mathcal {O}}(n_1n_2)$ since we need to find the maximum of $n_1$ numbers $n_2$ times to construct this partition. $\square$

Proposition 11

The inequalities in (18) can be separated in polynomial-time.

Proof

The proof is similar to the proof of Proposition 10. $\square$

1.2 Proof of Part (ii) Proposition 3

Proposition 12

$\text {conv} \left( {{\mathcal {U}}}^{\text {row+}}_{(n_1,n_2)}(l,u, L, U)\right)$ is described by the inequalities (30), (31), (19), and (20).

Proof

We will use Fourier-Motzkin elimination to obtain the convex hull in the original space. Now, we will project the t variables in the order $t_{n_2}$, $t_{n_2-1}$, $t_{n_2 -2}$, ..., $t_1$.

We claim that after projecting out variables $t_{n_2}, \dots , t_{n_2 - j}$, the resulting system of the inequalities is:

$$\begin{aligned} 1 - \sum _{p = 1}^{n_2 - (j + 1)} t_p \le&\ \sum _{i \in {\mathcal {I}}} \sum _{k \in T_{i} } \frac{W_{ik}}{l_i} + \frac{1}{L} \sum _{i = 1}^{n_1}\sum _{k \in T_0} W_{ik}&\forall&(T_0, T_1, \dots , T_ {|{\mathcal {I}}|}) \in {\mathcal {P}}_{|{\mathcal {I}}|+1}(\{n_2 - j, \dots , n_2\}) \\ 1 - \sum _{p = 1}^{n_2 - (j + 1)} t_p \ge&\ \sum _{i \in [n_1]} \sum _{k \in T_i} \frac{W_{ik}}{u_i} + \frac{1}{U} \sum _{i = 1}^{n_1}\sum _{k \in T_0} W_{ik}&\forall&(T_0, T_1, \dots , T_ {n_1}) \in {\mathcal {P}}_{n_1+1}(\{n_2 - j, \dots , n_2\}) \\ u_{i_2}W_{i_1k} \ge&l_{i_1} W_{i_2k}&\forall&i_2 \in [n_1], \forall i_1 \in {\mathcal {I}}, \forall k \in \{n_2 - j, \dots , n_2\} \\ l_it_k \le W_{ik} \le&u_it_k&\forall&i \in [n_1], \forall k \in [n_2 - (j + 1)] \\ L t_k \le \sum _{i=1}^{n_1}W_{i k} \le&\ U t_k&\forall&k \in [n_2- (j + 1)] \\ t_k \ge&0&\forall&k \in [n_2 - (j + 1)] \\ W_{ik} \ge&0&\forall&i \in [n_1], \forall k\in [n_2 ]. \end{aligned}$$

Base case: After projecting out $t_{n_2}$, we obtain the system:

$$\begin{aligned} \sum _{j = 1}^{n_2 - 1} t_j \le&\ 1 \end{aligned}$$

(71)

$$\begin{aligned} W_{i n_2} \le&\ (1 - \sum _{j = 1}^{n_2 - 1} t_j)u_i&\forall&i \in [n_1] \end{aligned}$$

(72)

$$\begin{aligned} W_{i n_2} \ge&\ (1 - \sum _{j = 1}^{n_2 - 1} t_j)l_i&\forall&i \in {\mathcal {I}} \nonumber \\ \sum _{i=1}^{n_1}W_{i n_2} \le&\ (1 - \sum _{j = 1}^{n_2 - 1} t_j)U \ \nonumber \\ \sum _{i=1}^{n_1}W_{i n_2} \ge&\ (1 - \sum _{j = 1}^{n_2 - 1} t_j)L \ \nonumber \\ u_{i_2}W_{i_1n_2} \ge&\ l_{i_1} W_{i_2n_2}&\forall&i_2 \in [n_1],\forall i_1 \in {\mathcal {I}} \nonumber \\ l_it_j \le W_{ij} \le&\ u_it_j&\forall&i \in [n_1], \forall j \in [n_2 - 1] \nonumber \\ L t_j \le \sum _{i=1}^{n_1}W_{i j} \le&\ U t_j&\forall&j \in [n_2-1] \nonumber \\ t_j \ge&\ 0&\forall&j \in [n_2 - 1], \nonumber \\ W_{ij} \ge&\ 0&\forall&i \in [n_1], \forall j \in [n_2 ], \end{aligned}$$

(73)

Note that (72) and (73) imply (71). Therefore the above can be written as:

$$\begin{aligned} (1 - \sum _{j = 1}^{n_2 - 1} t_j) \le&\ \sum _{i \in {\mathcal {I}}} \sum _{k \in T_{i} } \frac{W_{in_2}}{l_i} + \frac{1}{L} \sum _{i = 1}^{n_1}\sum _{k \in T_0} W_{ik}&\forall&(T_0, T_1, \dots , T_ {|{\mathcal {I}}|}) \in {\mathcal {P}}_{|{\mathcal {I}}|+1}(\{n_2 \}) \\ (1 - \sum _{j = 1}^{n_2 - 1} t_j) \ge&\ \sum _{i \in [n_1]} \sum _{k \in T_i} \frac{W_{ik}}{u_i} + \frac{1}{U} \sum _{i = 1}^{n_1}\sum _{k \in T_0} W_{ik}&\forall&(T_0, T_1, \dots , T_ {n_1}) \in {\mathcal {P}}_{n_1+1}(\{n_2 \}) \\ u_{i_2}W_{i_1n_2} \ge&\ l_{i_1} W_{i_2n_2}&\forall&i_2 \in [n_1], \forall i_1 \in {\mathcal {I}} \\ l_it_j \le W_{ij} \le&\ u_it_j&\forall&i \in [n_1], \forall j \in [n_2 - 1] \\ L t_j \le \sum _{i=1}^{n_1}W_{i j} \le&\ U t_j&\forall&j \in [n_2-1] \\ t_j \ge&\ 0&\forall&j \in [n_2 - 1], \\ W_{ij} \ge&\ 0&\forall&i \in [n_1], \forall j \in [n_2], \end{aligned}$$

proving the base case.

Induction step: After projecting $t_{n_2}$, $\dots$, $t_{n_2 -j}$, by the induction hypothesis we have the following system:

$$\begin{aligned} 1 - \sum _{p = 1}^{n_2 - (j + 2)} t_p - \sum _{i \in {\mathcal {I}}} \sum _{k \in T_{i} } \frac{W_{ik}}{l_i} - \frac{1}{L} \sum _{i = 1}^{n_1}\sum _{k \in T_0} W_{ik} \le&\ t_{n_2 - (j +1)}&\forall&(T_0, \dots , T_ {|{\mathcal {I}}|})\\ \in {\mathcal {P}}_{|{\mathcal {I}}|+1}(\{n_2 - j, \dots , n_2\}) \\ \frac{W_{i,n_2 - (j +1)}}{u_i} \le&\ t_{n_2 - (j +1)}&\forall&i \in [n_1] \\ \frac{1}{U}\sum _{i=1}^{n_1} {W_{i,n_2 - (j +1)}} \le&\ t_{n_2 - (j +1)}&\forall&i \in [n_1] \\ 1 - \sum _{p = 1}^{n_2 - (j + 2)} t_p - \sum _{i \in [n_1]} \sum _{k \in T_i} \frac{W_{ik}}{u_i} - \frac{1}{U} \sum _{i = 1}^{n_1}\sum _{k \in T_0} W_{ik} \ge&\ t_{n_2 - (j +1)}&\forall&(T_0, \dots , T_ {n_1}) \in {\mathcal {P}}_{n_1+1}(\{n_2 - j, \dots , n_2\}) \\ \frac{W_{i,n_2 - (j +1)}}{l_i} \ge&\ t_{n_2 - (j +1)}&\forall&i \in {\mathcal {I}} \\ \frac{1}{L}\sum _{i=1}^{n_1} {W_{i,n_2 - (j +1)}} \ge&\ t_{n_2 - (j +1)}&\forall&i \in [n_1] \\ u_{i_2}W_{i_1k} \ge&\ l_{i_1} W_{i_2k}&\forall&i_2 \in [n_1], \forall i_1 \in {\mathcal {I}},\forall k \in \{n_2 - j, \dots , n_2\} \\ l_it_k \le W_{ik} \le&\ u_it_k&\forall&i \in [n_1], \forall k \in [n_2 - (j + 1)] \\ L t_j \le \sum _{i=1}^{n_1}W_{i j} \le&\ U t_j&\forall&j \in [n_2-(j + 1)] \\ t_k \ge&\ 0&\forall&k \in [n_2 - (j + 1)] \\ W_{ik} \ge&\ 0&\forall&i \in [n_1], \forall k \in [n_2 ]. \end{aligned}$$

Note that a constraint of the form:

$$\begin{aligned} - \sum _{i \in [n_1]} \sum _{k \in T_i} \frac{W_{ik}}{u_i} - \frac{1}{U} \sum _{i \in [n_1]} \sum _{k \in T_0} {W_{ik}} \ge - \sum _{i \in {\mathcal {I}}} \sum _{k \in T'_{i} }\frac{W_{ik}}{l_i} - \frac{1}{L} \sum _{i \in {\mathcal {I}}} \sum _{k \in T_0} {W_{ik}} , \end{aligned}$$

where $(T_0, T_1{, \dots ,} T_ {n_1}) \in {\mathcal {P}}_{n_1+1}(\{n_2 - j, \dots , n_2\})$ and $(T'_0, T'_1, {\dots ,} T'_ {|{\mathcal {I}}|})\in {\mathcal {P}}_ {{|{\mathcal {I}}|+1}} (\{n_2 - j, \dots , n_2\})$ is implied by constraints of the form $u_{i_2}W_{i_1k} \ge l_{i_1} W_{i_2k} \forall i_2 \in [n_1], i_1 \in {\mathcal {I}}, k \in \{n_2 - j, \dots , n_2\}$, and the fact that $L \le U$. Thus, after projecting $t_{n_2 - (j +1)}$, we obtain:

$$\begin{aligned} 1 - \sum _{p = 1}^{n_2 - (j + 2)} t_p \le&\ \sum _{i \in {\mathcal {I}}} \sum _{k \in T_{i} } \frac{W_{ik}}{l_i} + \frac{1}{L} \sum _{i \in {\mathcal {I}}} \sum _{k \in T_{0} } {W_{ik}}&\forall&(T_0, T_1, \dots , T_ {|{\mathcal {I}}|}) \in {\mathcal {P}}_ {|{\mathcal {I}}|+1}( \{n_2 - (j +1), \dots , n_2\} ) \\ 1 - \sum _{p = 1}^{n_2 - (j + 2)} t_p \ge&\ \sum _{i \in [n_1]} \sum _{k \in T_i} \frac{W_{ik}}{u_i} + \frac{1}{U} \sum _{i \in {\mathcal {I}}} \sum _{k \in T_{0} } {W_{ik}}&\forall&(T_0, T_1, \dots , T_ {n_1}) \in {\mathcal {P}}_{n_1+1} (\{n_2 - (j +1), \dots , n_2\}) \\ u_{i_2}W_{i_1k} \ge&\ l_{i_1} W_{i_2k}&\forall&i_2 \in [n_1], \forall i_1 \in {\mathcal {I}}, \forall k \in \{n_2 - (j + 1), \dots , n_2\} \\ l_it_k \le W_{ik} \le&\ u_it_k&\forall&i \in [n_1], \forall k \in [n_2 - (j + 2)] \\ L t_k \le \sum _{i=1}^{n_1} W_{ik} \le&\ U t_k&\forall&k \in [n_2 - (j + 2)] \\ t_k \ge&\ 0&\forall&k \in [n_2 - (j + 2)] \\ W_{ik} \ge&\ 0&\forall&i \in [n_1], \forall k \in [n_2 ]. \end{aligned}$$

It is straightforward now to see that after all t variables are projected, we obtain the result. $\square$

Proposition 13

The inequalities in (30) can be separated in polynomial-time.

Proof

For a given matrix ${{\hat{W}}}$, let us define an index $j_{\text {row+}}$ for each index $j \in \{1,\dots ,n_2\}$ as

$$\begin{aligned} j_{\text {row+}} := {\left\{ \begin{array}{ll} 0 &{} \text { if } \frac{1}{U} \sum _{i=1}^{n_1} {{\hat{W}}}_{ij} \ge \max _{i =1,\dots ,n_1} \left\{ \frac{{{\hat{W}}}_{ij}}{u_i} \right\} \\ j_{\text {row}} &{} \text { otherwise} \end{array}\right. }, \end{aligned}$$

where $j_{\text {row}}$ is defined according to (70).

Then, we define a partition $T_0^*, T_1^*,\dots ,T_{n_1}^*$ of the set $[n_2]$ as

$$\begin{aligned} T_i^* := \{j \,|\, j_{\text {row+}}= i\} . \end{aligned}$$

Let

$$\begin{aligned} \theta := \sum _{i=1}^{n_1}\sum _{j \in T_i^*} \frac{{{\hat{W}}}_{ij}}{u_i} + \frac{1}{U} \sum _{i = 1}^{n_1}\sum _{j \in T_0^*} {{\hat{W}}}_{ij} . \end{aligned}$$

If $\theta >1$, then a violated inequality is discovered. Otherwise, we conclude that ${{\hat{W}}}$ satisfies all the inequalities in (30) (by construction, the partition $T_0^*, T_1^*, \dots , T_m^*$ corresponds to the inequality with the largest deviation, if one exists). Finally, we note that the complexity of this separation routine is again ${\mathcal {O}}(n_1n_2)$. $\square$

Proposition 14

The inequalities in (31) can be separated in polynomial-time.

Proof

The proof is similar to the proof of Proposition 13. $\square$

1.3 Proof of Theorem 4

The proof of SOC-representability follows due to [42] as the convex hull of a set described by the quadratic constraint $W_{11} W_{22} = W_{21} W_{12}$ and bound constraints is SOCr.

We next present an example where the convex hull of ${{\mathcal {U}}}^{\text {row}}_{(2,2)}(l,u) \cap {{\mathcal {U}}}^{\text {col}}_{(2,2)}(l,u)$ is not polyhedral.

Proposition 15

A point of the form:

$$\begin{aligned} \left[ \begin{array}{cc} \frac{a^2}{1 -a} &{} a \\ a &{} 1-a \end{array}\right] , \end{aligned}$$

for $a \in [0,\frac{1}{2}]$ is an extreme point of the set ${{\mathcal {U}}}^{\text {row}}_{(2,2)}(l,u) \cap {{\mathcal {U}}}^{\text {col}}_{(2,2)}(l,u)$ where $l = (0,1)$ and $u= (1,1)$.

Proof

Clearly the point is feasible. Also if it is not extreme, then it should be possible to write $(a, \frac{a^2}{1 - a}) \in {\mathbb {R}}^2$ as a convex combination of points of the form $(a_i, \frac{a_i^2}{1 - a_i}) \in {\mathbb {R}}^2$ with $a_i \in [0,\frac{1}{2}]\setminus \{a\}$. However, since $f(u) = \frac{u^2}{1 -u}$ is a strictly convex function, this is not possible (this is because, for example, $f(u) > \frac{a^2}{1 - a} + \left( \frac{1}{(1-a)^2} - 1\right) (u - a)$ for all $u \ne a$, while $f(u) = \frac{a^2}{1 - a} + \left( \frac{1}{(1-a)^2} - 1\right) (u - a)$ for $u = a$). $\square$

B Instance description

In Tables 11, 12, 13, and 14, AIL denotes the subset of arcs $A \cap (I\times L)$. The sets ALL, ALJ, and AIJ are defined analogously. The column Avg. Size$x^s$ (resp. Avg. Size$x^t$) displays the average size, over all pools $i\in L$, of the variable matrices $[x^s_{ij}]_{(s,j)}$ (resp. $[x^t_{ij}]_{(i,t)}$) for each instance.

Table 11 Mining instances description

Table 12 Literature instances description

Table 13 Random instances description

Table 14 DeyGupte4 instance description

C Detailed results for discretization restrictions

In Tables 15, 16, 17, 18, we report the primal bounds obtained via discretization for Random instances with discretization levels $H=1,2,4,5$, respectively ($H=3$ is reported in Table 2).

Table 15 Primal bounds via discretization for Random instances with $H=1$

Table 16 Primal bounds via discretization for Random instances with $H=2$

Table 17 Primal bounds via discretization for Random instances with $H=4$

Table 18 Primal bounds via discretization for Random instances with $H=5$