Semi-definite programming relaxation of quadratic assignment problems based on nonredundant matrix splitting

Abstract

Quadratic assignment problems (QAPs) are known to be among the most challenging discrete optimization problems. Recently, a new class of semi-definite relaxation models for QAPs based on matrix splitting has been proposed (Mittelmann and Peng, SIAM J Optim 20:3408–3426, 2010; Peng et al., Math Program Comput 2:59–77, 2010). In this paper, we consider the issue of how to choose an appropriate matrix splitting scheme so that the resulting relaxation model is easy to solve and able to provide a strong bound. For this, we first introduce a new notion of the so-called redundant and non-redundant matrix splitting and show that the relaxation based on a non-redundant matrix splitting can provide a stronger bound than a redundant one. Then we propose to follow the minimal trace principle to find a non-redundant matrix splitting via solving an auxiliary semi-definite programming problem. We show that applying the minimal trace principle directly leads to the so-called orthogonal matrix splitting introduced in (Peng et al., Math Program Comput 2:59–77, 2010). To find other non-redundant matrix splitting schemes whose resulting relaxation models are relatively easy to solve, we elaborate on two splitting schemes based on the so-called one-matrix and the sum-matrix. We analyze the solutions from the auxiliary problems for these two cases and characterize when they can provide a non-redundant matrix splitting. The lower bounds from these two splitting schemes are compared theoretically. Promising numerical results on some large QAP instances are reported, which further validate our theoretical conclusions.

Appendix: Proofs of Theorems 2 and 3

Proof of Theorem 2

Denote the optimal solution to the MTMS-PSD problem by $(B^*_1, B^*_2)$. We first show $(B^*_1, B^*_2)=(B^+, B^-)$. Let $P$ be the projection matrix defined by

$$\begin{aligned} P=\sum _{i:\lambda _i\ge 0} q_iq_i^T. \end{aligned}$$

It follows immediately that

$$\begin{aligned} \mathrm{Tr}(B^*_1) \ge \mathrm{Tr}(B^*_1 P) =\mathrm{Tr}(B^*_1P^2)=\mathrm{Tr}(PB^*_1P) \ge \mathrm{Tr}(P(B^*_1-B^*_2)P) = \mathrm{Tr}(B^+), \end{aligned}$$

where the first inequality follows from the relation

$$\begin{aligned} \mathrm{Tr}(B^*_1(I-P))\ge 0. \end{aligned}$$

Here $I$ denotes the identity matrix in $\mathfrak {R}^{n\times n}$. Similarly, one has

$$\begin{aligned} \mathrm{Tr}(B^*_2) \ge \mathrm{Tr}(B^*_2 (I-P)) =\mathrm{Tr}((I-P) B^*_2 (I-P)) \ge \mathrm{Tr}(B^-). \end{aligned}$$

Therefore, we have

$$\begin{aligned} \mathrm{Tr}(B_1^*)+\mathrm{Tr}(B_2^*)\ge \mathrm{Tr}(B^+)+\mathrm{Tr}(B^-), \end{aligned}$$

and the equality holds if and only if

$$\begin{aligned} \mathrm{Tr}(B^*_1(I-P))= 0, \quad \mathrm{Tr}(B_2^*P)=0. \end{aligned}$$

(40)

Since all the matrices $B_1^*, B_2^*, P$ and $I-P$ are positive semi-definite. Relation holds if and only if

$$\begin{aligned} B_1^* = B^*_1P=P B_1^*, \quad B_2^*P=P B_2^*=0. \end{aligned}$$

(41)

Since $B_1^*-B_2^* = B = B^+ - B^-$, we thus have

$$\begin{aligned} B_1^*=PBP=B^+, \quad B_2^*=-(I-P)B(I-P)=B^-. \end{aligned}$$

It remains to show that the matrix splitting $(B^+, B^-)$ is non-redundant. Suppose to the contrary that $(B^+, B^-)$ is a redundant splitting of $B$, i.e., there exists $R\ne 0 \succeq 0$ such that

$$\begin{aligned} B_1=B^+-R \succeq 0, \quad B_2=B^{-}-R\succeq 0, \quad B_1-B_2=B. \end{aligned}$$

Then we have

$$\begin{aligned} \mathrm{Tr}(B^+B^-)= \mathrm{Tr}((B_1+R)(B_2+R))\ge \mathrm{Tr}(B_1B_2)+\mathrm{Tr}(R^2)>0, \end{aligned}$$

which contradicts to the relation $\mathrm{Tr}(B^+B^-)=0$. This finishes the proof of the theorem.$\square$

We next cite a well-known result for matrix product from [31] that will be used in the proof of Theorem 3.

Lemma 3

Let $A,B\in {\textsc {S}}^n$ with the eigenvalues $\lambda _i(A)$ and $\lambda _i(B)$, $i=1,\ldots ,n$ listed in nonincreasing order. Then

$$\begin{aligned} \mathrm{Tr}(AB)\le \sum _{i=1}^n\lambda _i(A)\lambda _i(B), \end{aligned}$$

where the equality holds if and only if there is an orthogonal matrix $P$ whose columns form a common set of eigenvectors for A and B and are ordered with respect to $\{\lambda _i(A)\}_{i=1}^n$ and $\{\lambda _i(B)\}_{i=1}^n$, such that $P^{-1}AP$ and $P^{-1}BP$ are diagonal.

Proof of Theorem 3

Since $(\alpha ,\beta )$ is an optimal solution of problem (23), there exists $U\in {\textsc {S}}^n$ such that

$$\begin{aligned}&n-\mathrm{Tr}(UE)=0,\end{aligned}$$

(42)

$$\begin{aligned}&n-\mathrm{Tr}(U)=0,\end{aligned}$$

(43)

$$\begin{aligned}&\mathrm{Tr}(U(\alpha E +\beta I - B)) =0,\end{aligned}$$

(44)

$$\begin{aligned}&U\succeq 0, \quad \alpha E +\beta I - B\succeq 0. \end{aligned}$$

(45)

From (42) to (44), we obtain directly

$$\begin{aligned} n(\alpha +\beta )=\mathrm{Tr}(UB). \end{aligned}$$

(46)

From (44) and (45), we follow that

$$\begin{aligned} U(\alpha E +\beta I - B)=(\alpha E +\beta I - B)U=0. \end{aligned}$$

(47)

This implies that $U$ and $\alpha E +\beta I - B$ can commute. By Theorem 1.3.12 in [20], $U$ and $\alpha E +\beta I - B$ are simultaneously diagonalizable. Since $U\in {\textsc {S}}^n$, $\alpha E +\beta I - B\in {\textsc {S}}^n$, there is an orthogonal matrix $P$ such that $P^{-1}UP$ and $P^{-1}(\alpha E +\beta I - B)P$ are diagonal. So, we have

$$\begin{aligned} \mathrm{Tr}(U(\alpha E +\beta I - B))=\sum _{i=1}^n\lambda _i(U)\lambda _i(\alpha E +\beta I - B) =0, \end{aligned}$$

which in turn by (45) implies that

$$\begin{aligned} \lambda _i(U)\lambda _i(\alpha E +\beta I - B)=0,~i=1,\ldots ,n. \end{aligned}$$

(48)

Due to the minimal trace principle, we have $m=\mathrm{Rank}(\alpha E+\beta I -B) < n$. Since $\alpha E +\beta I - B\succeq 0$, we assume that $\lambda _i(\alpha E +\beta I - B)>0$, $i=1,\ldots ,m$. The above equality (48) then yields $\lambda _i(U)=0$, $i=1,\ldots ,m$.

We now prove $UE\ne EU$. Suppose to the contrary that $UE= EU$. By Theorem 1.3.12 in [20], $U$ and $E$ are simultaneously diagonalizeable. Let

$$\begin{aligned} \lambda _1(U)=\cdots =\lambda _s(U)=0<\lambda _{m+1}(U)\le \ldots \le \lambda _n(U). \end{aligned}$$

(49)

Note that the eigenvalues of $E$ are $0,\ldots ,0,n$. Therefore, we have

$$\begin{aligned} \mathrm{Tr}(UE)=n\lambda _n(U), \end{aligned}$$

which by (42) implies $\lambda _n(U)=1$. Hence, we infer from (49) that

$$\begin{aligned} \mathrm{Tr}(U)=\sum _{i=1}^n\lambda _i(U)\le n-m<n, \end{aligned}$$

this contradicts (43).

Because $UE\ne EU$, from (47) we obtain $UB\ne BU$. Since $U\in {\textsc {S}}^n$ and $B\in {\textsc {S}}^n$, by Theorem 1.3.12 in [20], $U$ and $B$ are not simultaneously diagonalizable. Now using Lemma 3, we have

$$\begin{aligned} \mathrm{Tr}(UB)<\sum _{i=1}^n\lambda _i(U)\lambda _i(B), \end{aligned}$$

which, together with (46), yields

$$\begin{aligned} n(\alpha +\beta )<\sum _{i=1}^n\lambda _i(U)\lambda _i(B). \end{aligned}$$

(50)

Let $\lambda _{\max }(B)$ be the largest eigenvalue of $B$. Note that $\sum _{i=1}^n\lambda _i(U)=\mathrm{Tr}(U)=n$. Also, $\lambda _i(U)\ge 0$ for all $i$ since $U\succeq 0$. It then follows from (50) that

$$\begin{aligned} \alpha +\beta < \lambda _{\max }(B). \end{aligned}$$

(51)

On the other hand, from (45), we have

$$\begin{aligned} B-\alpha (E-I) -(\alpha +\beta )I \preceq 0. \end{aligned}$$

This means that

$$\begin{aligned} \alpha +\beta \ge \lambda _{\max }( B-\alpha (E-I)). \end{aligned}$$

(52)

If $\alpha =0$, then the combination of (51) and (52) leads to a contradiction.

If $\alpha <0$. Let $\rho (B)$ be the spectral radius of $B$. Since $B\in {\textsc {S}}^n$ is non-negative, by Theorem 8.3.1 in [20], then $\rho (B)$ is an eigenvalue of $B$ and there exists nontrivial $\hat{x} \ge 0\in {\mathfrak {R}}^n$ such that $B\hat{x}=\rho (B)\hat{x}$. Without loss of generality, we can further assume that $\Vert \hat{x}\Vert _2=1$. Thus we have $\hat{x}^TB\hat{x}=\rho (B)$. Since $\hat{x}\ge 0$, it holds $\hat{x}^T(E-I)\hat{x}\ge 0$. It follows from (52) that

$$\begin{aligned} \alpha +\beta&\ge \lambda _{\max }( B-\alpha (E-I))\\&= \max \left\{ x^T( B-\alpha (E-I))x:x^{T}x=1\right\} \\&\ge \hat{x}^T( B-\alpha (E-I))\hat{x} \\&\ge \hat{x}^T B\hat{x}=\rho (B)\\&\ge \lambda _{\max }(B), \end{aligned}$$

which contradicts to (51). Therefore, we can conclude $\alpha >0$. This finishes the proof of the theorem.$\square$