Sequence independent lifting for mixed knapsack problems with GUB constraints

Abstract

In this paper, we consider the semi-continuous knapsack problem with generalized upper bound constraints on binary variables. We prove that generalized flow cover inequalities are valid in this setting and, under mild assumptions, are facet-defining inequalities for the entire problem. We then focus on simultaneous lifting of pairs of variables. The associated lifting problem naturally induces multidimensional lifting functions, and we prove that a simple relaxation in a restricted domain is a superadditive function. Furthermore, we also prove that this approximation is, under extra requirements, the optimal lifting function. We then analyze the separation problem in two phases. First, finding a seed inequality, and second, select the inequality to be added. In the first step we evaluate both exact and heuristic methods. The second step is necessary because the proposed lifting procedure is simultaneous; from where our class of lifted inequalities might contain an exponential number of these. We choose a strategy of maximizing the resulting violation. Finally, we test this class of inequalities using instances arising from electrical planning problems. Our tests show that the proposed class of inequalities is strong in the sense that the addition of these inequalities closes, on average, 57.70 % of the root integrality gap and 97.70 % of the relative gap while adding less than three cuts on average.

Appendix: Extended proofs

1.1 Proof of Proposition 1

• 1. $\forall j \in M$, define $\varvec{e_j}$ as the index vector of dimension $n$ with a unique one in position $j$. Let $\varvec{0_n}$ be the zero vector of size $n$. Let $\overline{a}$ be the maximum contribution of any binary variable to the knapsack constraint, i.e. $\overline{a}=\max _{j\in M}a_j$ and $0<\epsilon <b-\overline{a}$. Then, $\forall k \in M$, construct $\varvec{p^k}=(\varvec{e_k},\varvec{0_n})$ and $\varvec{q^k}=(\varvec{e_k},\epsilon \varvec{e_k})$. Define $\varvec{s}=(\varvec{0_n},\varvec{0_n})$. The points $\varvec{p^k}, \varvec{q^k}$ for $k\in M$, and $\varvec{s}$ belong to $X_G$ because of Eq. (1) and $\bar{a}<b$. These points are affinely independent because $\forall k \in M, \varvec{p^k}-\varvec{s}$ ($n$ points) and $\varvec{q^k}-\varvec{s}$ ($n$ points) are linearly independent. Since we have described $2n+1$ affinely independent points in $X_G$; we have shown that $X_G$ is full-dimensional.$\square$

• 2. To prove that $y_k\ge 0$ is facet-defining, we use the same definitions as before, but eliminate element $\varvec{q^k}$ from the set of valid points. Since we have described $2n$ affinely independent points in $X_G$ satisfying these inequalities at equality; we have shown that these inequalities are facet-defining.$\square$

• 3. To prove that $y_k\le x_k$ is facet-defining; we use the same definitions as in Proposition 1, but eliminate element $\varvec{p^k}$ from the set of valid points and redefine $\varvec{q^j}$ as $(\varvec{e_j},\varvec{e_j})$ for all $j\in M$. Again, since we have described $2n$ affinely independent points in $X_G$ and these points satisfy our inequality at equality; we have shown that these inequalities are facet-defining.$\square$

• 4. Note that $\overline{a}_g$ is the maximum contribution to the knapsack constraint when we choose any binary variable and the minimum contribution in $M_g$ at the same time. This allows us to prove that for each $g\in G, \sum _{k\in M_g}x_k\le 1$ is facet-defining, by constructing the points $\varvec{p}^k$ and $\varvec{q}^k$, for each $k\in M$ as follows:

  $$\begin{aligned} \varvec{p^{k}} = \left\{ \begin{array}{ll} (\varvec{e_{k}}+\varvec{e_{k_{o}}},\varvec{0_n}) &{} \quad \forall k\notin M_g \\ (\varvec{e_{k}},\varvec{0_n}) &{} \quad \forall k \in M_g \end{array} \right. \end{aligned}$$

  and

  $$\begin{aligned} \varvec{q^{k}} = \left\{ \begin{array}{ll} (\varvec{e_{k}}+\varvec{e_{k_{o}}},\epsilon \varvec{e_{k}}) &{} \quad \forall k\notin M_g \\ (\varvec{e_{k}},\epsilon \varvec{e_{k}}) &{} \quad \forall k \in M_g, \end{array} \right. \end{aligned}$$

  where $k_o\in \hbox {argmin}_{k\in M_g}\{a_{k}\}$ and $0<\epsilon <b-\overline{a}_g$. Since we have described $2n$ affinely independent points in $X_G$ and these points satisfy the inequality at equality; we have shown that these inequalities are facet-defining.$\square$

1.2 Proof of Theorem 1

The validity of inequality (3) was shown by Van Roy and Wolsey [21]. We prove that (3) is facet-defining for $X_o:=X\cap \{x_i=0,i\notin C\}$ by constructing a set of $2s$ affinely independent points in $X_o$ satisfying it at equality, where $s=|C|$. For this, define $C_L=C{\setminus } C_U, C_k^+=\{j\in C_k:\xi _j>{\varGamma }\}$, and $\bar{C}_k^+=C_k{\setminus } C_k^+$ for $k=L,U$, and recall that ${\varGamma }=\xi (C)-b>0$ and our hypothesis is $m(C_U^+)>{\varGamma }$. Let $t_o\in \hbox {argmin}_{j \in C_U^+}\{\xi _j\}$ and assume that $\frac{1}{0}=\infty$. Note that $C_U^+\ne \emptyset$, and thus $t_o$ exists. Let $\varvec{e_j}$ be the index vector of dimension $s$ with a unique one in position $j$, for all $j\in C$. Let $\varvec{0_s}$ be the zero vector of size $s, \varvec{1_s}$ be the one vector of size $s, \varvec{1_{u^+}}:=\sum _{j\in C_U^+}\varvec{e_j}, \varvec{1_{\bar{u}^+}}:=\sum _{j\in \bar{C}_U^+}\varvec{e_j}$, and $\varvec{1_u}=\varvec{1_{u^+}}+\varvec{1_{\bar{u}^+}}$. Then, construct $\varvec{p^j}$ as

$$\begin{aligned} \varvec{p^j}=\left\{ \begin{array}{ll} (\varvec{1_s}-\varvec{e_j},\varvec{1_u}) &{}\quad {\mathrm {if\ }} j\in C_L^+\\ (\varvec{1_s}-\varvec{e_j},\varvec{1_u}-y_{oj}\varvec{1_{u^+}}) &{}\quad {\mathrm {if\ }} j\in \bar{C}_L^+\\ (\varvec{1_s}-\varvec{e_j},\varvec{1_u}-\varvec{e_j}) &{}\quad {\mathrm {if\ }} j\in C_U^+\\ (\varvec{1_s}-\varvec{e_j},\varvec{1_u}-\varvec{e_j}-y_{oj}\varvec{1_{u^+}}) &{}\quad {\mathrm {if\ }} j\in \bar{C}_U^+\end{array}\right. \end{aligned}$$

where $y_{oj}=\frac{{\varGamma }-\xi _j}{m(C_U^+)}$. Now, construct $\varvec{q^j}$ as

$$\begin{aligned} \varvec{q^j}=\left\{ \begin{array}{ll} p^{t_o}+(\varvec{0_s},\delta \varvec{e_j}) &{}\quad {\mathrm {if\ }} j\in C_L \\ (\varvec{1_s},\varvec{1_u}-y_{+\epsilon }\varvec{1_{u^+}}+\varepsilon _j\varvec{e_j}) &{}\quad {\mathrm {if\ }} j\in C_U^+\\ (\varvec{1_s},\varvec{1_u}-y_{-\epsilon }\varvec{1_{u^+}}-\varepsilon _j\varvec{e_j}) &{}\quad {\mathrm {if\ }} j\in \bar{C}_U^+\end{array}\right. \end{aligned}$$

where

$$\begin{aligned}&0<\delta <\min \left\{ 1,\min _{j\in C_L}\left\{ \frac{\xi _{t_o}-{\varGamma }}{m_j}\right\} \right\} ,\\&0<\epsilon <\min _{j\in C_U}\{m_j\}\min \{{\varGamma },m(C_U^+)-{\varGamma }\}/m(C_U^+), \end{aligned}$$

$y_{\pm \epsilon }=\frac{{\varGamma }\pm \epsilon }{m(C_U^+)}$, and $\varepsilon _j=\frac{\epsilon }{m_j}$ for $j\in C_U$. Given the above definitions, it is easy to prove that points $\varvec{p^j}$ and $\varvec{q^j}$ belong to $X_o$. Therefore, it is only necessary to evaluate the knapsack constraint of Eq. (1). Verifying the validity of the other constraints is straightforward. Let $LHS_{KN}$ be the left-hand side of the knapsack constraint of Eq. (1).

• Case 1 $\varvec{p^j}$ with $j\in C_L^+$

  $$\begin{aligned} LHS_{KN} = \xi (C)-a_j= b + {\varGamma }- a_j < b \end{aligned}$$

• Case 2 $\varvec{p^j}$ with $j\in \bar{C}_L^+$

  $$\begin{aligned} LHS_{KN} = \xi (C)-a_j-y_{oj}m(C_U^+) =b + {\varGamma }-a_j - \frac{{\varGamma }-a_j}{m(C_U^+)} m(C_U^+) = b \end{aligned}$$

• Case 3 $\varvec{p^j}$ with $j\in C_U^+$

  $$\begin{aligned} LHS_{KN} = \xi (C)-\xi _j= b + {\varGamma }- \xi _j < b \end{aligned}$$

• Case 4 $\varvec{p^j}$ with $j\in \bar{C}_U^+$

  $$\begin{aligned} LHS_{KN} = \xi (C)-\xi _j-y_{oj}m(C_U^+) = b + {\varGamma }-\xi _j - \frac{{\varGamma }-\xi _j}{m(C_U^+)}m(C_U^+) = b \end{aligned}$$

• Case 5 $\varvec{q^j}$ with $j\in C_L$

  $$\begin{aligned} LHS_{KN} \!=\! \xi (C)-\xi _{t_o}+\delta m_j = b+{\varGamma }- \xi _{t_o} + \delta m_j<b+{\varGamma }-\xi _{t_o}+\xi _{t_o}\!-\!{\varGamma }\!=\!b \end{aligned}$$

• Case 6 $\varvec{q^j}$ with $j\in C_U^+$

  $$\begin{aligned} LHS_{KN} = \xi (C)- y_{+\epsilon }m(C_U^+) + \varepsilon _j m_j = b +\varepsilon _j m_j - \epsilon = b \end{aligned}$$

• Case 7 $\varvec{q^j}$ with $j\in \bar{C}_U^+$

  $$\begin{aligned} LHS_{KN} = \xi (C)-y_{-\epsilon } m(C_U^+) - \varepsilon _j m_j = b -\varepsilon _j m_j + \epsilon = b \end{aligned}$$

By proceeding in the same manner but now evaluating $\varvec{p^j}$ and $\varvec{q^j}$ in Eq. (3), we can show that these points satisfy this inequality at equality. Remember that the left-hand side of (3) is $\sum _{j\in C}\min \{1,\frac{\xi _j}{{\varGamma }}\}\left( x_j-1\right) +\sum _{j\in C_U}\frac{m_j}{{\varGamma }}\left( y_j-x_j\right)$, and its right-hand side is $-$1.

• Case 1 $\varvec{p^j}$ with $j\in C_L^+$

• Case 3 $\varvec{p^j}$ with $j\in C_U^+$

• Case 5 $\varvec{q^j}$ with $j\in C_L$

  $$\begin{aligned} LHS = -\min \{1,\frac{\xi _j}{{\varGamma }}\} = -1 \end{aligned}$$

• Case 2 $\varvec{p^j}$ with $j\in \bar{C}_L^+$

• Case 4 $\varvec{p^j}$ with $j\in \bar{C}_U^+$

  $$\begin{aligned} \begin{array}{rcl} LHS= & {} -\min \left\{ 1,\frac{\xi _j}{{\varGamma }}\right\} -y_{oj}\frac{m(C_U^+)}{{\varGamma }} = -\frac{\xi _j}{{\varGamma }}-\left( 1-\frac{\xi _j}{{\varGamma }}\right) = -1 \end{array} \end{aligned}$$

• Case 6 $\varvec{q^j}$ with $j\in C_U^+$

  $$\begin{aligned} LHS=-y_{+\epsilon }\frac{m(C_U^+)}{{\varGamma }}+\varepsilon _j\frac{m_j}{{\varGamma }}=-\frac{{\varGamma }-\epsilon +\epsilon }{{\varGamma }}=-1 \end{aligned}$$

• Case 7 $\varvec{q^j}$ with $j\in \bar{C}_U^+$

  $$\begin{aligned} LHS=-y_{-\epsilon }\frac{m(C_U^+)}{{\varGamma }}-\varepsilon _j\frac{m_j}{{\varGamma }}=-\frac{{\varGamma }+\epsilon -\epsilon }{{\varGamma }}=-1 \end{aligned}$$

Moreover, these points are affinely independent because $\forall j \in C{\setminus }\{t_o\}, \varvec{p^j}-\varvec{p^{t_o}}$, and $\varvec{q^j}-\varvec{p^{t_o}}$ are linearly independent. To show this, observe the structure of these points:

$$\begin{aligned} \left[ \begin{array}{ll} \varvec{p^j}-\varvec{p^{t_o}}, &{} \; j\in C_L^+ \\ \varvec{p^j}-\varvec{p^{t_o}}, &{} \; j\in \bar{C}_L^+ \\ \varvec{p^j}-\varvec{p^{t_o}}, &{} \; j\in \bar{C}_U^+ \\ \varvec{p^j}-\varvec{p^{t_o}}, &{} \; j\in C_U^+{\setminus }\{t_o\} \\ \varvec{q^j}-\varvec{p^{t_o}}, &{} \; j\in C_L^+ \\ \varvec{q^j}-\varvec{p^{t_o}}, &{} \; j\in \bar{C}_L^+ \\ \varvec{q^j}-\varvec{p^{t_o}}, &{} \; j\in \bar{C}_U^+ \\ \varvec{q^j}-\varvec{p^{t_o}}, &{} \; j\in C_U^+{\setminus }\{t_o\} \end{array}\right] = \left[ \begin{array}{lcl} \varvec{e_{to}}-\varvec{e_j} &{} \qquad &{} \varvec{e_{to}} \\ \varvec{e_{to}}-\varvec{e_j} &{} \qquad &{} \varvec{e_{to}}-y_{oj}\varvec{1_{u^+}} \\ \varvec{e_{to}}-\varvec{e_j} &{} \qquad &{} \varvec{e_{to}}-\varvec{e_j}-y_{oj}\varvec{1_{u^+}} \\ \varvec{e_{to}}-\varvec{e_j} &{} \qquad &{} \varvec{e_{to}} \\ \varvec{0_s} &{} \qquad &{} \delta \varvec{e_j} \\ \varvec{0_s} &{} \qquad &{} \delta \varvec{e_j} \\ \varvec{e_{to}} &{} \qquad &{} \varvec{e_{to}}-\varepsilon _j\varvec{e_j}-y_{-\epsilon }\varvec{1_{u^+}} \\ \varvec{e_{to}} &{} \qquad &{} \varvec{e_{to}}+\varepsilon _j\varvec{e_j}-y_{+\epsilon }\varvec{1_{u^+}} \end{array}\right] \end{aligned}$$

where, if elements $x_{t_o}$ and $y_{t_o}$ are placed in the last position in the columns, it is possible to obtain a upper-triangular matrix (using the first $2s-1$ columns) whose diagonal elements are non-zero. Since $2s$ affinely independent points in $X_o$ have been described and these points satisfy inequality (3) at equality, it has been shown that these inequalities are facet-defining for $conv\{X_o\}$.$\square$