DNN Relaxation of Quadratic Assignment Problem DNN Optimization DNN-based Branch-and-bound DNN-based Branch-and-bound for the Quadratic Assignment Problem *Koichi Fujii, Naoki Ito, Yuji Shinano NTT DATA Mathematical Systems Inc., , FAST RETAILING CO., LTD., Zuse Institute Berlin 2019/03/29 1 / 27
DNN Relaxation of Quadratic Assignment Problem DNN Optimization DNN-based Branch-and-bound Introduction of NTT Data Mathematical Systems Inc. Will be introduced at next talk : Takahito Tanabe ”Implementation issues of Interior-Point Method for real-world NLP problems” 2 / 27
DNN Relaxation of Quadratic Assignment Problem DNN Optimization DNN-based Branch-and-bound Summary : DNN-based Branch-and-bound for the Quadratic Assignment Problem Motivation • Quadratic assignment problems still remain as one of the most difficult combinatorial problems • Recent conic relaxation technique DNN updates the lower bounds of quadratic assignment problem Goal Improve branch-and-bound method for quadratic assignment problems Our Results First implementation of DNN-based branch-and-bound 3 / 27
DNN Relaxation of Quadratic Assignment Problem DNN Optimization DNN-based Branch-and-bound Agenda DNN Relaxation of Quadratic Assignment Problem 1 DNN Optimization 2 DNN-based Branch-and-bound 3 4 / 27
DNN Relaxation of Quadratic Assignment Problem DNN Optimization DNN-based Branch-and-bound Quadratic Assignment Problem { } x T ( B ⊗ A ) x | x ∈ { 0 , 1 } n , ( I ⊗ e T ) x = ( e T ⊗ I ) x = e , x i x j = 0 min , where B ⊗ A denotes the kronecker product of the matrices A and B . Known as having week LP/QP relaxation. 5 / 27
DNN Relaxation of Quadratic Assignment Problem DNN Optimization DNN-based Branch-and-bound Quadratic Assignment Problem as Polynomial Optimization Problem Relax linear constraints by Lagrangian multiplier λ . { � } x T ( B ⊗ A ) x + λ ∥ B ⊗ A ∥ � x T D ˜ � x ∈ [0 , 1] n , x i x j = 0 , ˜ min ˜ x = [1; x ] , x � ∥ D ∥ where ( d T d ) − d T C := (3) D − C T d C T C I ⊗ e T := (4) C d := [ e ; e ] (5) 6 / 27
DNN Relaxation of Quadratic Assignment Problem DNN Optimization DNN-based Branch-and-bound Quadratic Assignment Problem and DNN relaxation Polynomial optimization problem (POP) with non-negative variables min x { f 0 ( x ) | f i ( x ) = 0 ( i = 1 , 2 , . . . , m ) , x ≥ 0 } • 0-1 binary quadratic optimization problem • Optimal power flow, sensor network localization, ... Doubly non-negative (DNN) relaxation SDP relaxation + non-negative constraints • better lower bounds than SDP • very large O ( n 2 ) non-negative constraints POP lower bound DNN BP method relaxation BBCPOP improved lower bounds for QAPLIB instances. 7 / 27
DNN Relaxation of Quadratic Assignment Problem DNN Optimization DNN-based Branch-and-bound Quadratic Assignment Problem and DNN relaxation DNN optimization problem min Z {⟨ F 0 , Z ⟩ | ⟨ H 0 , Z ⟩ = 1 , Z ∈ K 1 ∩ K 2 } where • F 0 ∈ S n and H 0 ∈ S n + ( i = 1 , 2 , . . . , m ) • K 1 = S n + and K 2 ⊆ S n ≥ 0 are convex cones • S n : space of symmetric matrices • S n + : space of symmetric positive semidefinite matrices • S n ≥ 0 : space of symmetric nonnegative matrices 8 / 27
DNN Relaxation of Quadratic Assignment Problem DNN Optimization DNN-based Branch-and-bound Quadratic Assignment Problem and DNN relaxation � { } x t Q ˜ � x ∈ [0 , 1] n , x i x j = 0 (( i , j ) ∈ Γ) , ˜ min x x = [1; x ] , (6) ⇓ DNN relaxation {⟨ Q , Z ⟩ | Z 00 = 1 , Z ∈ K 1 ∩ K 2 } (7) � Z αβ ≥ 0 nonnegativity � � Z ∈ S n +1 := Z 0 α = Z α 0 ≥ Z αα (8) K 2 � � Z αβ = 0 if ( α, β ) ∈ Γ � 9 / 27
DNN Relaxation of Quadratic Assignment Problem DNN Optimization DNN-based Branch-and-bound DNN Optimization : BP method [Kim, Kojima, & Toh, ’16] min Z {⟨ F 0 , Z ⟩ | ⟨ H 0 , Z ⟩ = 1 , Z ∈ K 1 ∩ K 2 } ⇕ Strong duality ∈ K ∗ 1 + K ∗ max y 0 { y 0 | F 0 − y 0 H 0 2 } � �� � G ( y 0 ) Feasible Infeasible y 0 y ∗ 0 : Opt. BP method : Bisection method to judge the feasibility of a point y 0 10 / 27
DNN Relaxation of Quadratic Assignment Problem DNN Optimization DNN-based Branch-and-bound DNN Optimization : BP Method [Kim, Kojima, & Toh, ’16] How to judge if G ( y 0 ) ∈ K ∗ 1 + K ∗ 2 ? ⇒ solve regression model f ∗ = min Y 1 , Y 2 {∥ G − ( Y 1 + Y 2 ) ∥ 2 | Y 1 ∈ K ∗ 1 , Y 2 ∈ K ∗ 2 } Y 2 {∥ ( G − Y 1 ) − Y 2 ∥ 2 | Y 2 ∈ K ∗ 2 } | Y 1 ∈ K ∗ = min Y 1 { min 1 } 2 ( G − Y 1 )) ∥ 2 | Y 1 ∈ K ∗ = min Y 1 {∥ ( G − Y 1 ) − Π K ∗ 1 } Y 1 {∥ Π K 2 ( Y 1 − G ) ∥ 2 | Y 1 ∈ K ∗ = min 1 } ( where Y 2 = Π K ∗ 2 ( G − Y 1 )) • Obviously, f ∗ = 0 ⇔ G ∈ K ∗ 1 + K ∗ 2 . • Apply accelerated proximal gradient (APG) to check if f ∗ = 0. → [Assumption 1] Π K 2 , Π K 1 can be computed efficiently. 11 / 27
DNN Relaxation of Quadratic Assignment Problem DNN Optimization DNN-based Branch-and-bound DNN Optimization : APG method Constrained optimization: min α ∈ S f ( α ) Gradient projection method (e.g., [Gold- stein, ’64]) Step 1: α k +1 = Π S ( α k − 1 ) L k ∇ f ( α k ) APG method [Beck and Teboulle, ’09] Step 1: α k = Π S ( β k − 1 ) L k ∇ f ( β k ) 1+ √ 1+4 t 2 Step 2: t k +1 = k 2 Step 3: β k +1 = α k + t k − 1 t k +1 ( α k − α k − 1 ) � �� � 12 / 27 momentum
DNN Relaxation of Quadratic Assignment Problem DNN Optimization DNN-based Branch-and-bound DNN Optimization : APG method Gradient projection method f ( α k ) − f ( α ∗ ) ≤ O (1 / k ) � �� � � �� � current opt. Accelerated proximal gradient (APG) method f ( α k ) − f ( α ∗ ) ≤ O (1 / k 2 ) e.g., [Beck and Teboulle, ’09] [Nesterov, ’03] 13 / 27
DNN Relaxation of Quadratic Assignment Problem DNN Optimization DNN-based Branch-and-bound DNN Optimization : Computing a Valid Lower Bound • BP method may output an UB of the opt. val. (infeasible sol.), because APG can fail to judge feasibility due to numerical error. • Can we compute a valid lower bound y ℓ 0 of DNN? Feasible Infeasible y 0 y ∗ 0 : Opt. 14 / 27
DNN Relaxation of Quadratic Assignment Problem DNN Optimization DNN-based Branch-and-bound CDNN Optimization : Computing a Valid Lower Bound [Arima, Kim, Kojima & Toh, ’17] � { } � ⟨ H 0 , Z ⟩ = 1 , Z ∈ K 1 ∩ K 2 min ⟨ F 0 , Z ⟩ . Z ⇕ with I ∈ K 1 and large enough ρ ≥ 0 Z {⟨ F 0 , Z ⟩ | ⟨ H 0 , Z ⟩ = 1 , ⟨ I , Z ⟩ ≤ ρ, Z ∈ K 1 ∩ K 2 } min ⇕ Strong duality µ I ∈ K ∗ 1 + K ∗ max y 0 ,µ { y 0 + ρµ | F 0 − y 0 H 0 2 , µ ≥ 0 } � �� � G ( y 0 ) ⇕ y 0 ,µ, Y 2 { y 0 + ρµ | G ( y 0 ) − Y 2 − µ I ∈ K ∗ 1 (= S n + ) , Y 2 ∈ K ∗ max 2 , µ ≥ 0 } 15 / 27
DNN Relaxation of Quadratic Assignment Problem DNN Optimization DNN-based Branch-and-bound DNN Optimization : Summary min Z {⟨ F 0 , Z ⟩ | ⟨ H 0 , Z ⟩ = 1 , ( i = 1 , 2 , . . . , m ) , Z ∈ K 1 ∩ K 2 } . Dual of Lagrangian relaxation with parameter ρ ≥ 0 y 0 ,µ, Y 2 { y 0 + ρµ | G ( y 0 ) − Y 2 − µ I ∈ K ∗ 1 , Y 2 ∈ K ∗ max 2 , µ ≥ 0 } We searches • y 0 by bisection method • Y 1 ∈ K ∗ 1 and Y 2 ∈ K ∗ 2 by APG (to judge feasibility of y 0 ) • µ : minimal eigenvalue of G ( y 0 ) − Y 2 → always gives a valid LB [Assumption 1 ] Π K 2 , Π K ∗ 1 can be computed efficiently. [Assumption 2 ] We have a tight ρ ≥ 0. 16 / 27
DNN Relaxation of Quadratic Assignment Problem DNN Optimization DNN-based Branch-and-bound BBCPOP : Matlab implementation [Naoki Ito, Kim, Kojima, Takeda and Toh, 2018] 17 / 27
DNN Relaxation of Quadratic Assignment Problem DNN Optimization DNN-based Branch-and-bound DNN Optimization : the case of Quadratic Assignment Problem DNN formulation min {⟨ F 0 , Z ⟩ | ⟨ H 0 , Z ⟩ = 1 , Z ∈ K 1 ∩ K 2 } . (9) { } Z ∈ S n +1 := (10) K 1 + � Z αβ ≥ 0 nonnegativity � � Z ∈ S n +1 := Z 0 α = Z α 0 ≥ Z αα (11) K 2 � � if ( α, β ) ∈ Γ Z αβ = 0 � 18 / 27
DNN Relaxation of Quadratic Assignment Problem DNN Optimization DNN-based Branch-and-bound DNN Optimization : the case of Quadratic Assignment Problem [Assumption 1 ] Π K 2 , Π K ∗ 1 can be computed efficiently. • Π K ∗ is the projection on to symmetric cones 1 • Π K 2 is defined as: Π K 2 ( Z αβ ) := max (0 , Z α,β ) if ( α, β ) ∈ γ Π K 2 ( Z αα ) := avg ( Z αα , Z α 0 , Z 0 α ) if Z α 0 < Z αα (12) Π K 2 ( Z αβ ) := Z αβ otherwise 19 / 27
DNN Relaxation of Quadratic Assignment Problem DNN Optimization DNN-based Branch-and-bound DNN Optimization : the case of Quadratic Assignment Problem Example: 9 . 51 4 . 10 5 . 23 5 . 30 3 . 96 4 . 10 1 . 76 2 . 25 2 . 28 1 . 70 Z = 5 . 23 2 . 25 2 . 88 2 . 91 2 . 18 (13) 5 . 30 2 . 28 2 . 91 2 . 95 2 . 20 3 . 96 1 . 70 2 . 18 2 . 20 8 . 65 9 . 51 4 . 10 5 . 23 5 . 30 5 . 52 4 . 10 1 . 76 0 0 1 . 70 Π K 2 ( Z ) = 5 . 23 0 2 . 88 2 . 91 0 (14) 5 . 30 0 2 . 91 2 . 95 0 5 . 52 1 . 70 0 0 5 . 52 20 / 27
Recommend
More recommend