Disconnecting Networks via Node Deletions

Transcription

1 1 / 27 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

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

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

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

5 4 / 27 Introduction Motivation and Contributions The MaxNum and MinMaxC on general graphs: N P-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 duals, and integrate the two stages 3 Linearize the monolithic MIP, and solve it to optimality

6 4 / 27 Introduction Motivation and Contributions The MaxNum and MinMaxC on general graphs: N P-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 duals, 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.

7 5 / 27 Exact MIP Interdiction Models Master Problem (MaxNum) max s.t. ( ) η(x, y) 1 nx (1 x i ) (1a) n i=1 X (1 x i ) B (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)

8 6 / 27 Exact MIP Interdiction Models Maximizing the Number of Components (MaxNum) MaxNum Subproblem: Solving η(x, y) Formulate on a directed transformation network eg(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 η(x, y) = min X i N z i (2a) s.t.: e V paths from node 0 to every active node i (2b) f 0ik + 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)

9 6 / 27 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:

10 7 / 27 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

11 8 / 27 Exact MIP Interdiction Models Minimizing the Largest Component Size (MinMaxC) MinMaxC The master problem is similar to MaxNum except an obj modification: ( ) min η (x, y) + 1 nx (1 x i) : (1b) (1e), (2) n 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 i=1

12 9 / 27 Exact MIP Interdiction Models Minimizing the Largest Component Size (MinMaxC) MinMaxC: A Monolithic Model ( ) min λ + 1 nx (1 x i ) (3a) n i=1 s.t. (1b) (1e), and σ kk = 1 k N λ X 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.

13 10 / 27 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 arcs) and five 30-node (having 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

14 11 / 27 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

15 12 / 27 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 )

16 12 / 27 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

17 12 / 27 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.

18 13 / 27 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 ( ) Illustration of ( ) when an open set is present. Note that here because the open set itself is not counted in. Illustration of ( is present. ( ) ) when no open set

19 14 / 27 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 s.t. n i = n v + p v v S i v S i Every open set O v becomes a new component after merging.

20 14 / 27 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 s.t. n i = v S i n v + v S i p v Every open set O v becomes a new component after merging. When p i = 1 (not deleted): f i (1, n i ) = min f v (p v, n v ) v S i s.t. n i = v S i n v All open sets O v will merge with O i to form a larger-cardinality open set at i.

21 14 / 27 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 s.t. n i = v S i n v + v S i p v Every open set O v becomes a new component after merging. When p i = 1 (not deleted): f i (1, n i ) = min f v (p v, n v ) v S i s.t. n i = v S i n v All open sets O v will merge with O i to form a larger-cardinality open set at i. Calculate f i (p i, n i ) by sequentially merging one subtree at a time Since n i n, computing f i is O(n 2 ), for all i V. Total complexity: O(n 3 ) for solving MaxNum on trees.

22 15 / 27 Just Solve the MIP... O(n 3 log n) DP algorithms for MinMaxC f i (o i, m i ): the fewest number of deletions on subtree T i, given an open set of size o i exists on i a maximum component size of m i (excluding O i ) However, since both o i and m i n, merging requires O(n 5 ) steps

23 15 / 27 Just Solve the MIP... O(n 3 log n) DP algorithms for MinMaxC f i (o i, m i ): the fewest number of deletions on subtree T i, given an open set of size o i exists on i a maximum component size of m i (excluding O i ) However, since both o i and m i n, merging requires O(n 5 ) steps Define f i (o i, τ) instead: the fewest number of deletions on subtree T i, given that no component has a larger size than τ (a fixed target) it generates an open set of size o i where o i τ f l (1, τ) = 0 at every leaf node l V for any τ 1.

24 15 / 27 Just Solve the MIP... O(n 3 log n) DP algorithms for MinMaxC f i (o i, m i ): the fewest number of deletions on subtree T i, given an open set of size o i exists on i a maximum component size of m i (excluding O i ) However, since both o i and m i n, merging requires O(n 5 ) steps Define f i (o i, τ) instead: the fewest number of deletions on subtree T i, given that no component has a larger size than τ (a fixed target) it generates an open set of size o i where o i τ f l (1, τ) = 0 at every leaf node l V for any τ 1. Employ a binary-search scaling scheme over τ; update f i (o i, τ) for all i V for a given τ

25 16 / 27 Just Solve the MIP... Update f i (o i, τ) given f v (o v, τ), v S i When o i = 0 (subtree root i is deleted): X f i(0, τ) = min f v(o v, τ) + 1. v S i The largest component size is automatically not more than τ.

26 16 / 27 Just Solve the MIP... Update f i (o i, τ) given f v (o v, τ), v S i When o i = 0 (subtree root i is deleted): X f i(0, τ) = min f v(o v, τ) + 1. v S i The largest component size is automatically not more than τ. When o i > 0 (not deleted): X f i (o i, τ) = min f v(o v, τ) v S i s.t. o i = X o v + 1 τ. v S i

27 16 / 27 Just Solve the MIP... Update f i (o i, τ) given f v (o v, τ), v S i When o i = 0 (subtree root i is deleted): X f i(0, τ) = min f v(o v, τ) + 1. v S i The largest component size is automatically not more than τ. When o i > 0 (not deleted): X f i (o i, τ) = min f v(o v, τ) v S i s.t. o i = X o v + 1 τ. v S i Initial: Upper bound UB = n B; Lower bound LB = 1; τ = n B+1 2 Step 1: Solve MinMaxC for a current τ (O(n 3 ) steps) Step 2: Update τ: If LB < UB, update τ = (UB + LB)/2 ; go to Step 1 (O(log n) iterations) Total complexity: O(n 3 log n) for solving MinMaxC on trees.

28 17 / 27 Just Solve the MIP... k-hole graphs A hole of a graph: a set of nodes v 1,..., v m such that an edge exists between v i and v j (i < j) if and only if i = j 1 or i = 1 and j = m. M 1,..., M k : the k holes in a graph, where nodes {v 1,..., v q } compose the union of the nodes in these holes Transform a k-hole graph into a weighted hole tree

29 18 / 27 Just Solve the MIP... MaxNum and MinMaxC on k-hole-graphs Case 0 (no node is deleted in any hole) Every M j is a hole-node with size M j Yield a tree structure with weighted hole-nodes Use the same DP recursions as before, but prohibit deletions of hole-nodes

30 18 / 27 Just Solve the MIP... MaxNum and MinMaxC on k-hole-graphs Case 0 (no node is deleted in any hole) Every M j is a hole-node with size M j Yield a tree structure with weighted hole-nodes Use the same DP recursions as before, but prohibit deletions of hole-nodes Case i (delete node v i and obtain a p-hole-graph such that p < k) Recursively solve on a resulting p-hole-graph Γ(k) = the complexity on k-hole-graphs, we have that Γ(k) = O(n Γ(k 1)) Base case: 0-hole-graph (i.e., a tree)

31 18 / 27 Just Solve the MIP... MaxNum and MinMaxC on k-hole-graphs Case 0 (no node is deleted in any hole) Every M j is a hole-node with size M j Yield a tree structure with weighted hole-nodes Use the same DP recursions as before, but prohibit deletions of hole-nodes Case i (delete node v i and obtain a p-hole-graph such that p < k) Recursively solve on a resulting p-hole-graph Γ(k) = the complexity on k-hole-graphs, we have that Γ(k) = O(n Γ(k 1)) Base case: 0-hole-graph (i.e., a tree) Complexities on k-hole-graph: O(n 3+k ) for MaxNum, and O(n 3+k log n) for MinMaxC.

32 19 / 27 Just Solve the MIP... Incorporate DP Solutions into the MIP Framework Idea 1: Optimal DP solutions obtained on k-hole subgraphs of G provide bounds for the real subproblem objectives. However... Our computational results show: Bounds are generally not very tight, but tighter on smaller G instances (i.e., 20-node as opposed to 30- and 40-node graphs) Idea 2: Employ a graph-partition strategy, solve the DP on each partition, and generate valid inequalities for MIPs.

33 20 / 27 Valid Inequalities from Partitions Reformulating the MIP Notation (MaxNum for instance): Partition graph G into m subgraphs G 1,..., G m k i : the number of holes in each subgraph G i, i = 1,..., m Execute DP on each k i -hole subgraph G i for a budget B η i (B i ): maxnum obtained on G i for deletion budgets B i = 0,..., B (variables) g i (B i ): Piecewise-linear concave envelope function of η i (B i ) such that η i (B i ) g i (B i ) for all B i = 0,..., B.

34 Valid Inequalities from Partitions Reformulating the MIP Notation (MaxNum for instance): Partition graph G into m subgraphs G 1,..., G m k i : the number of holes in each subgraph G i, i = 1,..., m Execute DP on each k i -hole subgraph G i for a budget B η i (B i ): maxnum obtained on G i for deletion budgets B i = 0,..., B (variables) g i (B i ): Piecewise-linear concave envelope function of η i (B i ) such that η i (B i ) g i (B i ) for all B i = 0,..., B. Append the following valid inequalities into the MaxNum MIP: η mx η i 0 (4a) i=1 η i g i (B i ) 0 i = 1,..., m (4b) B i = X j V i (1 x j ) i = 1,..., m. (4c) 20 / 27

35 21 / 27 Valid Inequalities from Partitions Example 1: Solving MaxNum Given a 20-node graph G and B = 10, solving the 1 st partition G 1 (10-node):

36 21 / 27 Valid Inequalities from Partitions Example 1: Solving MaxNum Given a 20-node graph G and B = 10, solving the 2 nd partition G 2 (10-node):

37 21 / 27 Valid Inequalities from Partitions Example 1: Solving MaxNum Inequalities (4a) and (4c) are: η η 1 + η 2, B 1 = X i G 1 (1 x i), B 2 = X i G 2 (1 x i). (5) Associated with the three-segment g 1(B 1), for G 1, we generate (4b) as η 1 2B 1 + 1, η 1 B 1 + 4, η (6) Similarly, corresponding to each segment of g 2(B 2), for G 2, (4b) become η 2 (4/3)B 2 + 1, η 2 B 2 + 3, η (7)

38 22 / 27 Valid Inequalities from Partitions Example 2: Solving MinMaxC g i(b i) is the convex envelop of η i (B i), and signs in (4a) and (4b) are flipped.

39 22 / 27 Valid Inequalities from Partitions Example 2: Solving MinMaxC g i(b i) is the convex envelop of η i (B i), and signs in (4a) and (4b) are flipped.

40 22 / 27 Valid Inequalities from Partitions Example 2: Solving MinMaxC Inequalities (4a) and (4c) are: η η 1 + η 2, B 1 = X i G 1 (1 x i), B 2 = X i G 2 (1 x i). (8) The following two sets of inequalities are generated to describe g i(b i), for i = 1, 2: η 3B , η 2B 1 + 9, η B 1 + 6, η 0.5B 1 + 4, η 1 (9) η 1.5B , η B 2 + 8, η 1 (10)

41 23 / 27 CPU Time Comparison CPU Times for 20-node Instances Using 2-Partition Instance Prob. B = 4 B = 8 Orig. 2-Partition Orig. 2-Partition MaxNum [34.52] 5.94 [16.85] MinMaxC [1.90] MaxNum MinMaxC MaxNum [33.84] MinMaxC [19.29] MaxNum [88.48] MinMaxC MaxNum [34.65] MinMaxC

42 24 / 27 CPU Time Comparison CPU Times for 30-node Instances Using 3-Partition Instance Prob. B = 4 B = 8 Orig. 3-Partition Orig. 3-Partition MaxNum MinMaxC [204.12] MaxNum [218.07] MinMaxC [334.29] MaxNum MinMaxC [121.11] MaxNum MinMaxC [351.84] [96.36] MaxNum MinMaxC

43 25 / 27 CPU Time Comparison 40-node Instances Using 2- and 4-Partition None 40-node instances can be solved within a one-hour time limit. Thus, we report gaps (%) reported by CPLEX instead Instance Prob. B = 4 B = 8 Orig. 2-Partition 4-Partition Orig. 2-Partition 4-Partition MaxNum % 58.12% % 87.82% 48.07% 87.79% MinMaxC 27.82% 11.11% 27.85% 62.47% 32.82% [62.49%] MaxNum % % % 84.68% 33.78% 74.97% MinMaxC 26.19% 6.95% 20.42% 58.52% 21.10% [58.68%] MaxNum % 44.99% % 85.94% 85.92% [88.85%] MinMaxC 25.92% 25.77% 25.85% 58.17% 57.09% 47.38% MaxNum % 59.95% [128.20%] 95.47% 49.98% 86.80% MinMaxC 27.93% 27.93% 27.87% 61.52% 47.38% 47.50% MaxNum % 44.99% % 84.15% 53.76% [100.18%] MinMaxC 26.25% 26.21% 26.20% 59.17% 44.65% 51.80%

44 26 / 27 Summary and Future Research Future Research Vary partition patterns, and test the computational efficacy of different valid inequalities Dynamically update partitions within a branch-and-bound (B&B) tree The locally valid inequalities may lead to a quicker termination and more effective fathoming rules for the B&B algorithm

45 27 / 27 Summary and Future Research Thank you Questions?...