Complexity of branch-and-bound and cutting planes in mixed-integer optimization - II

We first give a proof of (i). Since $v$ is a feasible integer point, $0\leq\sum_{i\in I}v_{i}\leq\sum_{i=1}^{n}v_{i}\leq\left\lfloor\frac{n}{2}\right\rfloor$. Thus, there exists $v^{\prime}=(v^{\prime}_{1},v^{\prime}_{2},\ldots,v^{\prime}_{n})$, where $v_{i}^{\prime}=v_{i}$ for $i\in I$ and $\sum_{i=1}^{n}v_{i}^{\prime}=\left\lfloor\frac{n}{2}\right\rfloor$. So $v^{\prime}\in V\neq\emptyset$.

For each $v^{*}\in V$, we wish to show that $v^{*}\in P$. This will show that $v^{*}\in P_{I}$ and $V\subseteq P_{I}$. Consider any inequality describing $P$; if it is not the original defining inequality $\sum_{i=1}^{n}x_{i}\leq\frac{n}{2}$ or a 0/1 bound on a variable, then this inequality was introduced on the path from the root to $N$. A false split disjunction cannot remove $v^{*}$ since $v^{*}$ is integral. Consider an inequality coming from a true split disjunction. Let $\sum_{i\in S}a_{i}x_{i}\leq\delta^{*}$ for some $S\subseteq I$ be such an inequality. Since $v\in P_{I}$ and $v^{*}_{i}=v_{i}$ for $i\in I$, we observe that $\sum_{i\in S}a_{i}v_{i}=\sum_{i\in S}a_{i}v^{*}_{i}\leq\delta^{*}$.

We will prove (ii) by contradiction, so we assume the objective LP value of $N$ is strictly less than $\frac{n}{2}$. Let $P_{0}$ denote the relaxation corresponding to the root node. Assume $\ell\in\{1,2,\ldots,n\}\backslash I$.

Since $\lvert I\rvert\leq\left\lfloor\frac{n}{2}\right\rfloor-s$, there exists $v^{1}=(v_{1}^{1},v_{2}^{1},\ldots,v_{n}^{1})\in V$, where $v_{\ell}^{1}=0$. Define $v^{2}=(v_{1}^{2},v_{2}^{2},\ldots,v_{n}^{2})$, where $v_{\ell}^{2}=\frac{1}{2}$, and $v_{i}^{2}=v_{i}^{1}$ for $i\in\{1,2,\ldots,n\}\backslash\{\ell\}$. It is clear that $v^{2}\in P_{0}$, and $v^{2}\notin P$ since the LP value is assumed to be strictly less than $\frac{n}{2}$. Since $\ell\notin I$, there must be a halfspace $\hat{H}$ coming from a false split disjunction of $N$ that excludes $v^{2}$. The inequality describing this halfspace $\hat{H}$ must involve variable $x_{\ell}$, otherwise $v^{1}$ also violates $\hat{H}$, which leads to a contradiction since $\hat{H}$ comes from a false split disjunction and therefore cannot cut off any integer point. Hence assume the inequality describing $\hat{H}$ is $a_{\ell}x_{\ell}+\sum_{i\in S}a_{i}x_{i}\leq\delta$ for some $S\subseteq\{1,2,\ldots,n\}\backslash\{\ell\}$, and $\lvert S\rvert\leq s-1$ (since the sparsity of the disjunctions is restricted to be at most $s$). Since $\sum_{i\in I}v^{1}_{i}\leq\left\lfloor\frac{n}{2}\right\rfloor-s$, we have $\sum_{i\notin I\cup\{\ell\}}v^{1}_{i}\geq s$, and there exists $r\in\{1,2,\ldots,n\}\backslash(S\cup I\cup\{\ell\})$ such that $v_{r}^{1}=1$. Let $v^{3}=(v_{1}^{3},v_{2}^{3},\ldots,v_{n}^{3})$, where $v_{\ell}^{3}=1$, $v_{r}^{3}=0$, and $v_{i}^{3}=v_{i}^{1}$ for $i\neq\ell,r$. By definition of $V$, $v^{3}\in V$. Since $v^{1},v^{3}$ are integral, and $\hat{H}$ comes from a false split disjunction, $\hat{H}$ must be valid for $v^{1}$ and $v^{3}$. Thus, we have

Summing up (2.1) and (2.2) and dividing by 2, we get

which implies that $\hat{H}$ is valid for $v^{2}$. This is a contradiction. ∎

For a node $N$ of the branch-and-bound tree containing at least one integer point, if it is derived by exactly $m$ true split disjunctions, then we say it is a node of generation $m$. By Lemma 2.2, if $m\leq\frac{1}{s}\big{\lfloor}\frac{n}{2}\big{\rfloor}-1$, then a node $N$ of generation $m$ has LP objective value $\frac{n}{2}$, and in the subtree rooted at $N$ there must exist at least two descendants from generation $m+1$, since the leaf nodes must have LP values less than or equal to $\lfloor\frac{n}{2}\rfloor$. Therefore, there are at least $2^{m}$ nodes of generation $m$ when $m\leq\frac{1}{s}\big{\lfloor}\frac{n}{2}\big{\rfloor}-1$. This finishes the proof. ∎

Let $P:=\{i\in\{1,\ldots,k\}:w_{i}>0\}$ and $N:=\{i\in\{1,\ldots,k\}:w_{i}<0\}$. By making the variable change $x_{i}=1-y_{i}$ for $i\in N$ and $x_{i}=y_{i}$ for $i\in P$, it is seen that the number of 0/1 solutions to $\sum_{i=1}^{k}w_{i}x_{i}=W$ is the same as the number of 0/1 solutions to $\sum_{i\in P}w_{i}y_{i}+\sum_{i\in N}(-w_{i})y_{i}=W-\sum_{i\in N}w_{i}$. Writing this a bit more cleanly, we want to upper bound the number of 0/1 solutions to $\sum_{i=1}^{k}w^{\prime}_{i}y_{i}=W^{\prime}$, where $w_{i}^{\prime}>0$ for all $i\in\{1,\ldots,k\}$ and $W^{\prime}\in{\mathbb{Z}}$. The collection of subsets $I\subseteq\{1,\ldots,k\}$ that are solutions to $\sum_{i=1}^{k}w^{\prime}_{i}y_{i}=W^{\prime}$ is an antichain in the lattice of subsets with set inclusion as the partial order because all the $w^{\prime}_{i}$ values are strictly positive. By Sperner’s Theorem [52], the size of this collection is at most ${k\choose\lfloor k/2\rfloor}$. ∎

We consider the instance from Theorem 1.6. For any split disjunction $D:=\{x:\langle a,x\rangle\leq b\}\cup\{x:\langle a,x\rangle\geq b+1\}$, we define $V(D)$ to be the set of all the optimal LP vertices (of the original polytope) that lie strictly in the corresponding split set $\{x:b\leq\langle a,x\rangle\leq b+1\}$. Let the support of $a$ be given by $T\subseteq\{1,\ldots,n\}$ with $t:=|T|\leq s\leq\lfloor n/2\rfloor$. Since $a\in{\mathbb{Z}}^{n}$ and $b\in{\mathbb{Z}}$, $V(D)$ is precisely the subset of the optimal LP vertices $\hat{x}$ such that $\langle a,\hat{x}\rangle=b+\frac{1}{2}$. Fix some $\ell\in T$ and consider those optimal LP vertices $\hat{x}\in V(D)$ where $\hat{x}_{\ell}=\frac{1}{2}$. This means that $\sum_{j\in T\setminus\{\ell\}}a_{j}\hat{x}_{j}=b+\frac{1}{2}-\frac{a_{\ell}}{2}$. Let $r_{i}$ be the number of 0/1 solutions to $\sum_{j\in T\setminus\{\ell\}}a_{j}\hat{x}_{j}=b+\frac{1}{2}-\frac{a_{\ell}}{2}$ with exactly $i$ coordinates set to 1. Then the number of vertices from $V(D)$ with the $\ell$-th coordinate equal to $\frac{1}{2}$ is

since ${n-t\choose\lfloor n/2\rfloor-i}\leq{n-t\choose\lfloor n/2\rfloor-\lfloor t/2\rfloor}$ for all $i\in\{0,\ldots,t-1\}$. Using Lemma 2.3, $\sum_{i=0}^{t-1}r_{i}\leq{t-1\choose\lfloor t/2\rfloor}$ and we obtain the upper bound ${t-1\choose\lfloor t/2\rfloor}{n-t\choose\lfloor n/2\rfloor-\lfloor t/2\rfloor}$ on the number of vertices from $V(D)$ with the $\ell$-th coordinate equal to $\frac{1}{2}$. Therefore, $|V(D)|\leq t{t-1\choose\lfloor t/2\rfloor}{n-t\choose\lfloor n/2\rfloor-\lfloor t/2\rfloor}=:p(t).$ Since $n$ is odd, we have

A direct calculation then shows that

Let $h$ be the largest even number not exceeding $s$. Since $p(1)={n-1\choose\lfloor n/2\rfloor}$, we obtain, for every $t\in\{1,\dots,s\}$,

where $m!!$ denotes the product of all integers from $1$ up to $m$ of the same parity as $m$. Using the fact that, for every even positive integer $\ell$,

(see, e.g., [54, 9]), we have (for $h\geq 1$, i.e., $s\geq 2$)

Thus, this is an upper bound on $|V(D)|$. Since the total number of optimal LP vertices of the instance is ${n{n-1\choose\lfloor n/2\rfloor}}$, we obtain the following lower bound of on the size of a branch-and-bound proof: $\frac{{n{n-1\choose\lfloor n/2\rfloor}}}{|V(D)|}=\Omega\left(\sqrt{\frac{n(n-s)}{s}}\right).$ ∎

We first show a branch-and-bound algorithm with size $O(n)$. Let the root node be $N_{0}$. The objective LP value of $N_{0}$ is $\frac{n}{2}$. Let $N_{1}^{0}$ and $N_{1}^{1}$ be the children of $N_{0}$ produced by branches $x_{1}\leq 0$ and $x_{1}\geq 1$ respectively. Then the LP values of $N_{1}^{0}$ and $N_{1}^{1}$ are $\left\lfloor\frac{n}{2}\right\rfloor$ and $\frac{n}{2}$. Therefore $N_{1}^{0}$ is a leaf node. Recursively, let $N_{j+1}^{0}$ and $N_{j+1}^{1}$ be children of $N_{j}^{1}$ produced by $x_{j+1}\leq 0$ and $x_{j+1}\geq 1$ for $1\leq j\leq\left\lfloor\frac{n}{2}\right\rfloor$. Note that this is well defined since the LP values of $N_{j}^{0}$ and $N_{j}^{1}$ are $\left\lfloor\frac{n}{2}\right\rfloor$ and $\frac{n}{2}$ for $1\leq j\leq\left\lfloor\frac{n}{2}\right\rfloor$. It is clear that node $N_{j+1}^{0}$ is a leaf for $1\leq j\leq\left\lfloor\frac{n}{2}\right\rfloor$. Node $N_{\left\lceil\frac{n}{2}\right\rceil}^{1}$ is an infeasible leaf since there are $\left\lceil\frac{n}{2}\right\rceil$ variables set to be $1$. Therefore, the whole branch-and-bound tree has $n+2$ nodes.

Next, we show that any cutting plane for the problem with sparsity $s\leq\left\lfloor\frac{n}{2}\right\rfloor$ is valid for $[0,1]^{n}$. We will use the fact that $H\cap\{0,1\}^{n}=\{(x_{1},x_{2},\ldots,x_{n})\in\{0,1\}^{n}:\sum_{i=1}^{n}x_{i}\leq\left\lfloor\frac{n}{2}\right\rfloor\}$.

Let $S\subseteq\{1,\ldots,n\}$ be the set of indices for the non-zero coefficients in an inequality defining the cutting plane, i.e., the inequality is given by $\sum_{i\in S}a_{i}x_{i}\leq\delta$. Since this is a cutting plane it must be valid for all points in $H\cap\{0,1\}^{n}$. Let $V_{S}=\{(x_{1},x_{2},\ldots,x_{n})\in\{0,1\}^{n}:x_{i}=0,i\not\in S\}$. Since $|S|\leq s\leq\left\lfloor\frac{n}{2}\right\rfloor$, we have $V_{S}\subseteq H\cap\{0,1\}^{n}$. Therefore $\sum_{i\in S}a_{i}x_{i}\leq\delta$ is valid for all of $V_{S}$. Since the inequality only involves $x_{i}$, $i\in S$, it must also be a valid inequality for all of $\{0,1\}^{n}$. ∎

Let the cutting plane proof be $H_{1},H_{2},\ldots,H_{L}$, and the sequence of the corresponding disjunctions deriving it be $D_{1},D_{2},\ldots,D_{L}\in\mathcal{D}$. Moreover, assume $H_{i}$ is $\langle\alpha_{i},x\rangle\leq\delta_{i}$ for $1\leq i\leq L$. Since we assume all cutting planes are rational, we may assume $\alpha_{i}\in{\mathbb{Z}}^{n+d}$ and $\delta_{i}\in{\mathbb{Z}}$. Let $H^{\prime}_{i}$ be $\langle\alpha_{i},x\rangle\geq\delta_{i}+1$. Since $H_{i}$ is valid for $C\cap D_{i}$, we must have that $(C\cap H^{\prime}_{i})\cap D_{i}=\emptyset$.

Let $N_{0}=C$ be the root node of the branch-and-bound tree. Recursively, we define $N_{i}$ and $N_{i}^{\prime}$ be the children of $N_{i-1}$ generated by applying the split disjunction $H_{i}\cup H^{\prime}_{i}$ for $1\leq i\leq L$. Applying the disjunction $D_{i}$ on $N_{i}^{\prime}$ only generates infeasible nodes as noted above. Meanwhile, $N_{i}$ shows the validity of $H_{i}$. Thus, we have replaced the cut $H_{i}$ with $k+1$ nodes of the branch-and-bound tree: $k$ of these are infeasible and one is feasible. Therefore, we get a branch-and-bound tree of size $(k+1)L$.

A well-known family of instances in $\mathbb{R}^{3}$, given by $\operatorname*{conv}\{(0,0,0),(2,0,0),(0,2,0),(\frac{1}{2},\frac{1}{2},h)\}$ for $h\in{\mathbb{N}}$, from [18] can be solved by branch-and-bound in $O(1)$ iterations with just variable disjunctions; however, there is a $\operatorname*{poly}(\log(h))$ lower bound on the split rank [15], and therefore, on the length of proofs based on split cuts. ∎