Robust Assignments with Vulnerable Nodes David Adjiashvili Viktor Bindewald Dennis Michaels Technische Universit¨ at Dortmund 9-13, January 2017, Aussois This work is supported by DFG through RTG 1855 “Discrete Optimization of Technical Systems under Uncertainty”
1. Problem definition and applications 2. Hardness and algorithmic results Approximation for the weighted version Approximation for the unweighted version 3. Robust assignments with vulnerable edges Hardness and algorithmic results The simplest non-trivial case
Problem definition and applications
Motivation × 3/34
Robust Assignment Problem Certain problem • given G = ( U ˙ ∪ W , E ), costs c ∈ R W ≥ 0 • nodes in U are called jobs and those in W machines • a set of machines X ⊆ W is called an assignment if all jobs can be performed using machines in X • i.e. the induced graph G [ U ˙ ∪ X ] contains a U -perfect matching (a set of non-adjacent edges covering U ) • objective: choose a minimum-cost assignment 4/34
Robust Assignment Problem Robust version • G = ( U ˙ ∪ W , E ), costs c ∈ R W ≥ 0 , | U | < | W | • Scenario set F = { f 1 , . . . , f k } ⊆ W (vulnerable machines) • If scenario f i emerges, then machine f i is deleted from G min c ( X ) s.t. ∀ f ∈ F : X \ { f } contains an assignment ( RAP ) X ⊆ W 5/34
Robust Assignment Problem Interpreted as a game: 1. the adversary announces a list of machines he wants to sabotage 2. the decision maker buys a set of machines 3. the adversary destroys one machine from the list 4. the decision maker selects an assignment from the remaining machines he bought and performs the jobs 6/34
Example (unit weights) 1 1 1 1 2 2 2 a 3 3 a 3 3 b 4 4 4 4 b c 5 5 c 5 6 6 6 6 F = { 1 , 2 , 4 , 6 } optimal solution 7/34
Classification and potential areas of application • RAP is a redundancy-based robust problem (example: finding k -connected spanning subgraphs) • more precisely, it is a bulk-robust problem ([Adjiashvili, Stiller, Zenklusen ’15]) The concept of bulk-robustness models long-term decision problems where “infrastructure” must be fixed in advance. In our setting, • staff scheduling / rostering • hiring / subcontracting • reservation systems • manufacturing process flexibility 8/34
Hardness and algorithmic results
How hard is the problem? Theorem Set Cover can be restated equivalently as RAP , preserving the cost. 9/34
Set Cover Problem Recap: • Given a tuple ( { 1 , . . . , n } , C ), where C = { S 1 , S 2 , . . . , S l } with S j ⊆ { 1 , . . . , n } . • Find a minimum-cardinality sub-collection C ′ ⊆ C such that � S ∈C ′ S = { 1 , . . . , n } . Theorem [Feige ’98] For any d < 1, there is no d · log n –approximation for the Set Cover problem, unless NP ⊆ DTIME ( n log log n ). 10/34
Hardness of RAP Set Cover → RAP ( { 1 , . . . , n } , { S 1 , S 2 , . . . , S l } ) → bipartite graph on 2 n + l nodes w 1 u 1 w S 1 c w = 1 w 2 u 2 w S 2 c w = 1 w S l c w = 1 w n u n W 2 = F W 1 U 11/34
Asymptotically tight algorithm for RAP Theorem RAP admits a (log | U | + 2)-approximation. Approach: Select a cheap matching and complete it to a solution. 12/34
(log | U | + 2) -Approximation for RAP Require: Feas. instance: G = ( U ˙ ∪ W , E ), c ∈ R W ≥ 0 , F ⊆ W 1: for each w ∈ W : for each e ∈ δ ( w ) : d e := c w 2: M ← minimum-cost U -perfect matching w.r.t. d 3: W M ← V ( M ) ∩ W 4: U F ← { jobs matched to a vulnerable machine in M } 5: for each w ∈ W \ W M do R w ← { u ∈ U F | ∃ M -alt. u - w -path in G [ V ( M )] + w } 6: 7: S ← apx. sln. to SC instance ( U F , { R w | w ∈ W \ W M } ) ∪{ w ∈ W \ W M | R w ∈ S} 8: return W M ˙ 13/34
(log | U | + 2) -Approximation for RAP Proof: • c ( ALG ) = c ( W M ) + c ( S ) • c ( W M ) < c ( OPT ) • c ( S ) ≤ (log | U F | + 1) · c ( OPT (Set Cover)) ([Chvatal ’79]) • c ( OPT (Set Cover)) ≤ c ( OPT ) and | U F | ≤ | U | � 14/34
Approximation for minimum-cardinality RAP We now focus on the variant of RAP with unit weights, i.e. c w = 1, ∀ w ∈ W ( card-RAP ). Theorem card-RAP with F = W admits a 1 . 75-approximation algorithm. Approach: Greedily augment a U -perfect matching. 15/34
1 . 75 -Approximation for card-RAP Require: Feas. card-RAP -instance: G = ( U ˙ ∪ W , E ), F = W . 1: M ← any U -perfect matching in G 2: X ← V ( M ) ∩ W (machines from matching M ) 3: for each machine node w ∈ W \ X do R w ← { u ∈ U | ∃ M -altern. u - w -path in G [ V ( M )]+ w } 4: 5: U F ( X ) ← U 6: while | U F ( X ) | > 0 do w ← arg max {| U F ( X ) ∩ R w ) | | w ∈ W \ X } 7: U F ( X ) ← U F ( X ) \ R w 8: X ← X ∪ { w } 9: 10: return X 16/34
1 . 75 -Approximation for card-RAP Proof idea: • an iteration is called good, if | U F ( X ) | is decreased by at least 2 • g : ⇔ total decrease of | U F ( X ) | obtained by good iterations • | ALG | ≤ 2 n − g 2 • | OPT | ≥ max { n , 2( n − g ) } � 18/34
Hardness of approximation for card-RAP Theorem There is no PTAS for card-RAP , unless P = NP . This is also true for F = W . Proof idea • Vertex Cover in subcubic graphs ( VC 3 ) is APX -hard ([Alimonti, Kann ’97]) • Restate VC 3 in terms of a set cover problem, • use an extended version of the prior presented reduction � 19/34
Robust assignments with vulnerable edges
Edge-Robust Assignment Problem Certain problem • Given G = ( U ˙ ∪ W , E ) with | U | = | W | , costs c ∈ R E ≥ 0 • Find a minimum-cost perfect matching M ⊆ E of G Robust version ( E-RAP ) • Scenario set F = { f 1 , . . . , f k } ⊆ E (vulnerable edges) • If scenario f i emerges, then edge f i is deleted from G min c ( X ) s.t. ∀ f ∈ F : X \ { f } contains a perf. matching X ⊆ E 20/34
Example (unit weights) a x a x y y b b c z c z w w d d F = {{ a , z } , { d , z } , { d , w }} optimal solution 21/34
How hard is the problem? Theorem Set Cover can be restated equivalently as E-RAP , preserving the cost. For details see ”Robust Assignments via Ear Decompositions and Randomized Rounding” [Adjiashvili, B., Michaels ’16] Theorem E-RAP admits a randomized O (log n )-approximation algorithm. Approach: Obtain a matching-covered subgraph via randomized rounding. 22/34
Technical ingredients I Definition A graph G = ( V , E ) is called matching-covered if each of its edges is part of some perfect matching of G Why is it useful? Up to technicalities, inclusion-wise minimal feasible solutions to E-RAP are matching-covered subgraphs and vice versa. 23/34
Technical ingredients II We use the following IP formulation for E-RAP : (Given G = ( U ˙ ∪ W , E ) , F = E , c ∈ R E ≥ 0 ) c ⊤ y min x − f s.t. ∈ P MATCH ( G ) , for each f ∈ F , x − f = 0 , for each f ∈ F , f x − f , ≥ for each f ∈ F , y x − f { 0 , 1 } E , ∈ for each f ∈ F , ∈ { 0 , 1 } E y 24/34
O (log n ) -Approximation for RAP Require: G = ( U ˙ ∪ W , E ), F = E and c ∈ R E ≥ 0 . 1: ( x , y ) ← optimal solution of the LP relaxation 2: X ← ∅ 3: while X is infeasible do Select f ∈ F such that X \ { f } has no perf. matching 4: Decompose x − f into p. matchings: x − f = � k i =1 λ i χ M − f 5: i Select ¯ M ∈ { M − f | i ∈ [ k ] } with probability λ i 6: i Augment X using edges from ¯ M connecting distinct con- 7: nected components in ( V [ G ] , X ) 8: return X Can be adopted to F � E . 25/34
O (1) -Approximation for minimum-cardinal. E-RAP We now focus on the variant of E-RAP with unit weights, i.e. c e = 1, ∀ e ∈ E ( card-E-RAP ). Theorem card-E-RAP admits a O (1)-approximation algorithm. Approach: Exploit ear decompositions of matching-covered graphs. 27/34
Bipartite ear decomposition Definition P 0 1 2 Consider a bipartite graph H and a subgraph H ′ ⊆ H . An odd ear of H relative to H ′ is an odd path P in H connecting nodes u , v in H ′ P 2 3 4 such that only u and v are in H ′ . Bipartite Ear Decomposition If H can be constructed by starting with an edge via odd ear additions, then H P 1 5 6 has a Bipartite Ear Decomposition. We write H = P 0 + P 1 + · · · + P r . 28/34
A characterization of bipartite m.-c. graphs Theorem [Lovasz, Plummer ’86] A bipartite graph H is matching-covered if and only if H has a Bipartite Ear Decomposition. Observation Removing ears of length one from a Bipartite Ear Decomposition yields a matching-covered graph. 29/34
O (1) –Approximation algorithm for card-E-RAP Require: G = ( U ˙ ∪ W , E ) and F ⊆ E 1: Remove all dispensable edges from G 2: Compute an ear decomposition G = P 0 + P 1 + · · · + P r � � { E ( P i ): | E ( P i ) | > 1 , i = 1 , . . . , r } � 3: X ← E ( P 0 ) ∪ 4: return X 1 2 • if F = E , then the approximation guarantee 3 4 is 1 . 5 • if F ⊆ E , it is 3 30/34 5 6
Hardness of approximation for card-E-RAP Theorem There is no PTAS for card-E-RAP , unless P = NP . This is also true for F = E . 31/34
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