Disconnecting Networks via Node Deletions Exact Interdiction Models - PowerPoint PPT Presentation

Node Deletion and Node Disconnection Disconnecting Networks via Node Deletions Exact Interdiction Models and Algorithms Siqian Shen 1 J. Cole Smith 2 R. Goli 2 1 IOE, University of Michigan 2 ISE, University of Florida 2012 INFORMS Optimization Society Conference, Miami FL 1 / 27

Node Deletion and Node Disconnection Outline 1 Introduction 2 Exact MIP Interdiction Models Maximizing the Number of Components (MaxNum) Minimizing the Largest Component Size (MinMaxC) 3 MIP Bounds and Inequalities Just Solve the MIP... Valid Inequalities from Partitions CPU Time Comparison 4 Summary and Future Research 2 / 27

Node Deletion and Node Disconnection Introduction MaxNum and MinMaxC on General Graphs? ( B = 1) Counterexamples: MinMaxC MaxNum 3 / 27

Node Deletion and Node Disconnection Introduction Motivation and Contributions The MaxNum and MinMaxC on general graphs: NP -hard. The MaxNum and MinMaxC on specially structured graphs: Polynomial-time Dynamic Programming Algorithms (Shen and Smith (2011)) This study will: 4 / 27

Node Deletion and Node Disconnection Introduction Motivation and Contributions The MaxNum and MinMaxC on general graphs: NP -hard. The MaxNum and MinMaxC on specially structured graphs: Polynomial-time Dynamic Programming Algorithms (Shen and Smith (2011)) This study will: 1 Formulate two-stage interdiction MIPs having LP subproblems 2 Take the subproblem dual s, and integrate the two stages 3 Linearize the monolithic MIP, and solve it to optimality 4 / 27

Node Deletion and Node Disconnection Introduction Motivation and Contributions The MaxNum and MinMaxC on general graphs: NP -hard. The MaxNum and MinMaxC on specially structured graphs: Polynomial-time Dynamic Programming Algorithms (Shen and Smith (2011)) This study will: 1 Formulate two-stage interdiction MIPs having LP subproblems 2 Take the subproblem dual s, and integrate the two stages 3 Linearize the monolithic MIP, and solve it to optimality 4 Reformulate the MIP based on subgraph partitions of G , and generate valid inequalities by using intermediate polynomial-time optimal DP solutions from each partition. 4 / 27

Node Deletion and Node Disconnection Exact MIP Interdiction Models Master Problem (MaxNum) ( n ) η ( x , y ) − 1 X ( 1 − x i ) max (1a) n i = 1 X ( 1 − x i ) ≤ B s.t. (1b) i ∈V x i + x j − 1 ≤ y ij ∀ ( i , j ) ∈ E (1c) x i ∈ { 0 , 1 } ∀ i ∈ V (1d) 0 ≤ y ij ≤ 1 ∀ ( i , j ) ∈ E , (1e) Undirected graph G ( V , E ) , where V = { 1 , . . . , n } and E ⊂ V × V η ( x , y ) : Subproblem objective, e.g., number of components for MaxNum x i ∈ { 0 , 1 } : x i = 1 if node i is not deleted, and x i = 0 if i is deleted y ij ∈ { 0 , 1 } : y ij = 1 if edge ( i , j ) exists, and y ij = 0 otherwise ( y ij = x i x j ) B : Given node deletion budget (positive integer) 5 / 27

Node Deletion and Node Disconnection Exact MIP Interdiction Models Maximizing the Number of Components (MaxNum) MaxNum Subproblem: Solving η ( x , y ) Formulate on a directed transformation network e G ( N , A ) Design a dummy node 0 and a unit cost for constructing arc ( 0 , i ) , ∀ i ∈ V GOAL : To flow | e V| paths from 0 to every active node i ∈ e V Decision Variables: z i : = 1 if ( 0 , i ) is constructed and = 0 otherwise; f ijk : Flow on arc ( i , j ) with respect to path 0– k X η ( x , y ) = min z i (2a) i ∈N | e V| paths from node 0 to every active node i s.t.: (2b) − f 0 ik + z i ≥ 0 ∀ i , k ∈ N (2c) − f ijk ≥ − y ij ∀ ( i , j ) ∈ A , k ∈ N (2d) z i ∈ { 0 , 1 } , f ijk ≥ 0 . (2e) 6 / 27

Node Deletion and Node Disconnection Exact MIP Interdiction Models Maximizing the Number of Components (MaxNum) MaxNum Subproblem: Solving η ( x , y ) A transformed directed graph and a feasible solution illustration: 6 / 27

Node Deletion and Node Disconnection Exact MIP Interdiction Models Maximizing the Number of Components (MaxNum) Solving MaxNum Good News: ) Fix ( x , y ) at binary values, and a subproblem LP gives the convex hull in terms of variables z . Solution Scheme: Replace η ( x , y ) in the master problem by the subproblem LP dual Linearize bilinear terms of “ x × duals" and “ y × duals" by using McCormick inequalities (since both x and y are binary-valued). Monolithically solve MaxNum in a “max{max} = max" framework 7 / 27

Node Deletion and Node Disconnection Exact MIP Interdiction Models Minimizing the Largest Component Size (MinMaxC) MinMaxC The master problem is similar to MaxNum except an obj modification: ( ) X n η ′ ( x , y ) + 1 ( 1 − x i ) : (1b)–(1e) min , (2) n i = 1 where η ′ ( x , y ) represents the largest component size for a given ( x , y ) . Subproblem Notation: σ ik ∈ { 0 , 1 } : = 1 if nodes i and k belong to the same component σ kk = 1 , ∀ k ∈ N λ = η ′ ( x , y ) represents the largest component size 8 / 27

Node Deletion and Node Disconnection Exact MIP Interdiction Models Minimizing the Largest Component Size (MinMaxC) MinMaxC: A Monolithic Model ( n ) λ + 1 X min ( 1 − x i ) (3a) n i = 1 (1b)–(1e), and σ kk = 1 ∀ k ∈ N s.t. X λ ≥ σ ik ∀ k ∈ N (3b) i ∈N σ jk − σ ik ≥ y ij − 1 ∀ ( i , j ) ∈ A , k ∈ N (3c) σ ik ∈ { 0 , 1 } ∀ i , k ∈ N . (3d) (3b) enforces λ to be the largest component size (3c) pushes σ jk = 1 if σ ik = 1 and y ij = 1. That is, nodes j and k are in the same component, if nodes i and k are in the same component and j is connected to i (3) yields the convex hull even with (3d) being linear. 9 / 27

Node Deletion and Node Disconnection MIP Bounds and Inequalities Just Solve the MIP... How efficient the Monolithic MIP models are? Experimental Tests: CPLEX 11.0 & C++; a Dell PowerEdge 2600 UNIX machine with two 3 . 2 GHz processors; a one-hour time limit Five 20-node (having 40 - 60 arcs) and five 30-node (having 100-200 arcs) graph instances with varied B -values Result Observations: CPU time: 10s-100s for most 20-node instances; 100s-800s for 30-node instances CPU time ↑ as B ↑ at the begining, and then CPU time ↓ as B continue to ↑ above a threshold of approximately 0 . 25 |V| 10 / 27

Node Deletion and Node Disconnection MIP Bounds and Inequalities Just Solve the MIP... On the other hand... Given a tree T ( V , E ) , a DP algorithm can solve: O ( n 3 ) ⇒ MaxNum on trees O ( n 3 log n ) ⇒ MinMaxC on trees Extend the results to k -hole-graph for some k : O ( n 3 + k ) ⇒ MaxNum O ( n 3 + k log n ) ⇒ MinMaxC 11 / 27

Node Deletion and Node Disconnection MIP Bounds and Inequalities Just Solve the MIP... DP Algorithms for Specially-Structured Graphs For an undirected tree T ( V , E ) , r : root node T i : subtree rooted at node i ( T = T r ) 12 / 27

Node Deletion and Node Disconnection MIP Bounds and Inequalities Just Solve the MIP... DP Algorithms for Specially-Structured Graphs For an undirected tree T ( V , E ) , r : root node T i : subtree rooted at node i ( T = T r ) Key Concept: Open set O i : All nodes in the same component to which subroot i belongs, and o i = | O i | If i is deleted, then O i is empty and o i = 0 12 / 27

Node Deletion and Node Disconnection MIP Bounds and Inequalities Just Solve the MIP... DP Algorithms for Specially-Structured Graphs For an undirected tree T ( V , E ) , r : root node T i : subtree rooted at node i ( T = T r ) Key Concept: Open set O i : All nodes in the same component to which subroot i belongs, and o i = | O i | If i is deleted, then O i is empty and o i = 0 Incumbent Initial Step: There exists an optimal solution to all MaxNum and MinMaxC instances on tree graphs in which NO leaf node is deleted. 12 / 27

Node Deletion and Node Disconnection MIP Bounds and Inequalities Just Solve the MIP... O ( n 3 ) DP algorithms for MaxNum f i ( p i , n i ) : the fewest number of deletions required on subtree T i , given that p i : = 0 if subtree root i is deleted, and = 1 otherwise n i : Number of components created, not including O i Note: f l ( 1 , 0 ) = 0 at every leaf node l ∈ V ( ) ( ) ( ) when an open set is ( ) when no open set Illustration of Illustration of present. Note that here because the is present. open set itself is not counted in . 13 / 27

Node Deletion and Node Disconnection MIP Bounds and Inequalities Just Solve the MIP... Update f i ( p i , n i ) given f v ( p v , n v ) , ∀ v ∈ S i When p i = 0 (subtree root i is deleted): � f i ( 0 , n i ) = min f v ( p v , n v ) + 1 v ∈ S i � � n i = n v + s.t. p v v ∈ S i v ∈ S i Every open set O v becomes a new component after merging. 14 / 27

Node Deletion and Node Disconnection MIP Bounds and Inequalities Just Solve the MIP... Update f i ( p i , n i ) given f v ( p v , n v ) , ∀ v ∈ S i When p i = 0 (subtree root i is When p i = 1 (not deleted): deleted): � f i ( 1 , n i ) = min f v ( p v , n v ) � f i ( 0 , n i ) = min f v ( p v , n v ) + 1 v ∈ S i v ∈ S i � n i = s.t. n v � � n i = n v + s.t. p v v ∈ S i v ∈ S i v ∈ S i All open sets O v will merge with O i to form a larger-cardinality open Every open set O v becomes a new set at i . component after merging. 14 / 27