the Robust Network Loading Problem with Dynamic Routing Sara Mattia - PowerPoint PPT Presentation

the Robust Network Loading Problem with Dynamic Routing Sara Mattia DIS - Dipartimento di Informatica e Sistemistica “Antonio Ruberti” Università degli Studi di Roma “La Sapienza” mattia@dis.uniroma1.it

outline the problem problem ■ description NL ■ mathematical model RNL previous works example the branch-and-cut algorithm algorithm ■ separation routines conclusions ■ branch-and-cut heuristic conclusions ■ preliminary computational results ■ future work Sara Mattia - RNL Aussois 2010 – 2 / 26

the Network Loading Problem 2 d 12 = 3 . 2 2 2 d 15 = 5 problem 5 d 24 = 0 . 7 NL 1 3 1 d 34 = 1 . 1 2 RNL d 35 = 5 . 3 3 previous works 7 2 d 45 = 5 . 3 example d j = 0 otherwise 5 4 1 algorithm conclusions given ■ a graph G ( V, E ) with per-unit edge costs c : E → R + ■ a traffic matrix: set of point-to-point traffic demands (commodities) problem compute minimum cost integer capacities such that all the demands can be routed simultaneously on the network Sara Mattia - RNL Aussois 2010 – 3 / 26

the Robust Network Loading Problem 2 d 12 = 3 . 2 d 13 = 0 . 7 2 2 d 15 = 5 d 14 = 2 . 1 problem 5 d 24 = 0 . 7 d 23 = 0 . 7 NL 1 3 1 d 34 = 1 . 1 . . . d 24 = 1 . 9 2 RNL d 35 = 5 . 3 d 34 = 1 . 8 3 previous works 7 2 d 45 = 5 . 3 d 35 = 2 example d j = 0 d j = 0 5 4 1 algorithm conclusions given ■ a graph G ( V, E ) with per-unit edge costs c : E → R + ■ a set of traffic matrices D to be served non simultaneously problem compute minimum cost integer capacities such that every d ∈ D can be supported Sara Mattia - RNL Aussois 2010 – 4 / 26

the hose model the demand set D problem ■ D explicitly given (list of matrices) NL ■ D implicitly described (polyhedral representation [ 1 ] ) RNL previous works example the hose polyhedron feasible demands must respect bounds on node traffic [ 2 ][ 3 ] algorithm conclusions ■ symmetric: a single bound on the sum of the incoming and outgoing traffic ■ asymmetric: two bounds, one for the incoming traffic and one for the outgoing traffic [ 1 ] Ben-Ameur, Kerivin (2005) [ 2 ] Duffield, Goyal, Greenberg, Mishra, Ramakrishnan van der Merwe (1999) [ 3 ] Fingerhut, Suri, T urner (1997) Sara Mattia - RNL Aussois 2010 – 5 / 26

flows flows problem ■ unsplittable: each commodity must be routed on a single NL path RNL ■ splittable: the flow for a commodity can be splitted along previous works several paths example algorithm conclusions 2 2 s s 1 3 1 3 5 4 t 5 4 t Sara Mattia - RNL Aussois 2010 – 6 / 26

routing routing scheme problem ■ static: the routing must be the same for all d ∈ D NL ■ dynamic: we can choose a (possibly) different routing for RNL every matrix previous works example algorithm conclusions d p ∈ D d 1 ∈ D 2 2 1 3 1 3 . . . . . . 5 4 5 4 Sara Mattia - RNL Aussois 2010 – 7 / 26

related problems this talk exact approach for RNL with splittable flows and dynamic problem routing NL RNL previous works related problems example ■ Robust Network Design Problem ( RND ), RNL where algorithm capacities can be fractional conclusions ■ Virtual Private Network Problem ( VPN ), RND with unsplittable flows and static routing Sara Mattia - RNL Aussois 2010 – 8 / 26

the literature this talk exact approach for RNL with splittable flows and dynamic problem routing NL RNL previous works previous works example ■ many approximation results for RND and VPN [ 4 ][ 5 ][ 6 ] algorithm ■ branch-and-cut-and-price for VPN [ 7 ] conclusions ■ branch-and-cut for RNL with static routing [ 8 ] ■ no previous exact approaches for RNL with dynamic routing [ 4 ] Goyal, Olver, Shepherd (2009) [ 5 ] Chekuri, Oriolo, Scutellà, Shepherd (2007) [ 6 ] Eisenbrand, Grandoni, Oriolo, Skutella (2005) [ 7 ] Altın, Amaldi, Belotti, Pınar (2007) [ 8 ] Altın, Yaman, Pınar (2009) Sara Mattia - RNL Aussois 2010 – 9 / 26

an example static vs dynamic problem NL i: 2 (a) (b) (c) 2 2 2 RNL o: 0 previous works i: 0 3 1 2 example 4 1 o: 2 i: 1 algorithm 1 3 1 3 1 3 o: 0 2 1 2 conclusions 3 1 1 4 1 1 i: 2 4 4 4 16 17 o: 0 for RND the gap between the optimal dynamic solution and the optimal static solution is O ( log n ) [ 4 ] . [ 4 ] Goyal, Olver, Shepherd (2009) Sara Mattia - RNL Aussois 2010 – 10 / 26

the flow formulation � min c e  e problem e ∈ E algorithm � ( ƒ kd − ƒ kd j ) = − d k  ∈ V, k ∈ K, d ∈ D ( 1 ) formulation j j ∈ N (  ) B&C �� � separation ( ƒ kd + ƒ kd mx j ) ≤  e e = ( , j ) ∈ E ( 2 ) heuristic j d ∈ D k ∈ K conclusions ƒ ≥ 0  e ∈ Z + notation ■  e , capacity installed on e ∈ E ■ ƒ kd flow for commodity k and demand d on e = ( , j ) j Sara Mattia - RNL Aussois 2010 – 11 / 26

the flow formulation � min c e  e problem e ∈ E algorithm � ( ƒ kd − ƒ kd j ) = − d k  ∈ V, k ∈ K, d ∈ D ( 1 ) formulation j j ∈ N (  ) B&C �� � separation ( ƒ kd + ƒ kd mx j ) ≤  e e = ( , j ) ∈ E ( 2 ) heuristic j d ∈ D k ∈ K conclusions ƒ ≥ 0  e ∈ Z + remark it is non-compact, while there exists a compact formulation for static routing [ 7 ][ 8 ] [ 7 ] Altın, Amaldi, Belotti, Pınar (2007) [ 8 ] Altın, Yaman, Pınar (2009) Sara Mattia - RNL Aussois 2010 – 12 / 26

the flow formulation � min c e  e problem e ∈ E algorithm � ( ƒ kd − ƒ kd j ) = − d k  ∈ V, k ∈ K, d ∈ D ( 1 ) formulation j j ∈ N (  ) B&C �� � separation ( ƒ kd + ƒ kd mx j ) ≤  e e = ( , j ) ∈ E ( 2 ) heuristic j d ∈ D k ∈ K conclusions ƒ ≥ 0  e ∈ Z + question is there a compact formulation for RNL with dynamic routing? Sara Mattia - RNL Aussois 2010 – 13 / 26

the capacity formulation � min c e  e problem e ∈ E algorithm �� � ℓ μ formulation � � μ e  e ≥ mx k d k μ ≥ 0 ( 3 ) B&C d ∈ D e ∈ E k ∈ K  ∈ V separation  e ∈ Z + heuristic conclusions remarks ■ due to metric inequalities [ 9 ] , it is non-compact (even for NL ) ■ there is a non-metric capacity formulation for RNL with static routing [ 8 ] [ 8 ] Altın, Yaman, Pınar (2009) [ 9 ] Onaga, Kakusho (1971), Iri (1971) Sara Mattia - RNL Aussois 2010 – 14 / 26

the capacity formulation � min c e  e problem e ∈ E algorithm �� � ℓ μ formulation � � μ e  e ≥ mx k d k μ ≥ 0 ( 3 ) B&C d ∈ D e ∈ E k ∈ K  ∈ V separation  e ∈ Z + heuristic conclusions question is there a non-metric formulation for RNL with dynamic routing? Sara Mattia - RNL Aussois 2010 – 15 / 26

the algorithm polyhedral properties problem ■ all properties that are valid for NL(G,d) are also valid for RNL(G,D) algorithm formulation ■ all facet-defining inequalities are tight metrics B&C separation � μ e  e ≥ R μ μ ∈ Met G heuristic e ∈ E conclusions μ T y : y ∈ RNL ( G, D ) � � R μ = min the algorithm the formulation is non-compact, we have to use branch-and-cut ■ separating tight metric inequalites is NP -hard ■ no algorithm, not even heuristic, is known Sara Mattia - RNL Aussois 2010 – 16 / 26

separation separation strategy as a first step, look for violated metric inequalites problem algorithm separation of metric inequalites formulation given ¯  , find either an inequality B&C separation � � � heuristic ℓ μ � � μ e  e ≥ mx k d k μ ∈ Met G d ∈ D conclusions e ∈ E k ∈ K  ∈ V violated by ¯  , or conclude that none exists remarks ■ for NL separating metric inequalites is easy, the separation problem can be formulated as an LP ■ for RNL separating metric inequalites is difficult ■ how difficult? Sara Mattia - RNL Aussois 2010 – 17 / 26

separation separation problem for metric inequalities it can be formulated as bilevel programming problem. For problem other bilevel separation problems see [ 10 ] algorithm formulation ℓ μ � � � �  e μ e − β ¯  e μ e − ¯ min min k d k B&C e ∈ E e ∈ E k ∈ K  ∈ V separation ℓ μ kj ≤ ℓ μ ℓ μ kj ≤ ℓ μ k + μ e k ∈ K, e ∈ E k + μ e k ∈ K, e ∈ E heuristic � � conclusions μ e = 1 μ e = 1 e ∈ E e ∈ E μ ≥ 0 , ℓ free � � d k + d t ≤ b   ∈ V ℓ μ � � t ∈ V k ∈ K β = mx k d k μ, d ≥ 0 , d  = 0 ℓ free k ∈ K  ∈ V � � ( φ  ) d k + d t ≤ b   ∈ V k ∈ K t ∈ V d ≥ 0 , d  = 0 Lodi, Ralphs (2009) [ 10 ] Sara Mattia - RNL Aussois 2010 – 18 / 26