2-edge connected subgraphswith bounded rings

B. FortzInstitut de Gestion et d’Administration, Louvain la Neuve, Belgique

A. R. MahjoubLIMOS, CNRS, Université Blaise Pascal, Clermont-Ferrand, France

S. T. McCormickFaculty of Commerce, Vancouver, Canada

P. PesneauLIMOS, CNRS, Université Blaise Pascal, Clermont-Ferrand, France

Pierre Pesneau 2

Outline

• Presentation of the problem

• Polyhedral study

• Branch&Cut algorithm

• Computational results

Pierre Pesneau 3

Network design

• Designing the network topology• Network survivability

2-edge connectivity

• Network performance in case of failure

bounded ring constraints

• Problem :

Find a 2-edge connected subgraph at minimum cost such that each edge belong to a cycle of length bounded by an integer K.

Pierre Pesneau 4

2-node connectivity(Fortz, Labbé, Maffioli 2000)

• Formulation in terms of edges and cycles• Valid inequalities and necessary and sufficient

conditions to be facet defining• Separation algorithms• Branch&Cut algorithm• Heuristics.

Pierre Pesneau 5

Cut inequalities

Let . Pose

Let

EF

V.W ,VW

.

F, e exF

otherwise0

if1)(

2))(( Wx W W\V

)(W

Pierre Pesneau 6

Cycle inequalities

e

T

• Let be a partition

such that ,• let ,

• let

• Every solution must verify :

p10 V , ,V ,V K p

p,VVe 0

.1

1

0

e,VV\ ET ii

p

i

0 x(e) x(T) Cycle configuration

1V

0V

2V

pV

Pierre Pesneau 7

Formulation

., 0,1

,1, 0

ion,configurat cycle ) ,(,0 )( )(

,,2 ))((

Eex(e)

Eex(e)

eTexTx

VWV WWx

cx Minimize

Pierre Pesneau 8

Cycle inequalities : facets

• Let G=(V,E) be a complete graph.• Let be a partition of V and• The associated cycle inequality is facet defining if

and only if :– ,– and

– for all

p, V, , VV 10 . ,VV e p0

Kp 10 V ,1KV

31 ii VV .1,,0 Ki

Pierre Pesneau 9

Cycle inequalities : separation

Let x be a solution.

The separation problem of cycle inequalities

for an edge e=st

The bounded (s,t)-path cut problem with B=K-1.

Pierre Pesneau 10

Bounded path cut problem

• Let G=(V,E) be a graph, s and t two nodes and B an integer.

• Bounded (s,t)-path cut : set of edges that cut all (s,t)-path of length

• Problem : Find a minimum cost bounded (s,t)-path cut (BPCP).

.B

Pierre Pesneau 11t

vtw ztw

' N'u 'v

'z

uvw

uvw

N u v z

suw szw

BPCP• If B=2 the problem is trivial.• If B=3 :

– The problem is polynomial.– It can be reduced to find a minimum cut in a particular

directed graph.s

t

N u v z

suw szw

vtw ztw

uvw

s

G 'G

Pierre Pesneau 12

BPCP

• If :Heuristic based on the Primal-Dual method :

While C is not a bounded (s,t)-path cut do

Find an (s,t)-path P of bounded length

Increase until an edge verifies

Improve C by removing useless edges of C

4B

;;0 Cy

Py Pf .

:f

QfQQ wy

;fCC

Pierre Pesneau 13

Cycle inequalities : separation

• Let G=(V,E) be a graph and e=st be an edge.• Calculate a bounded (s,t)-path cut C with B=K-1.• If x(C) < x(e), the we get a violated cycle

inequality and the associated partition is obtained by a breadth-first search from s in the graph G\C.

• We strengthen the partition by reducing to a single node.

KV

Pierre Pesneau 14

Cyclomatic inequalities

• Introduced by Fortz, Labbé (1999).• Let be a partition of V.• Every solution must verify :

pVV ,, 0

.KKp))VV((x p

1 ,, 0

Pierre Pesneau 15

Cyclomatic inequalities : separation

1st heuristic :Based on the separation of partition inequalities (Cunningham) :

Consider :

1)) ,, (( 0

KKpVVx p

.100with V

p

pVVx p )),,(( 0

We apply Barahona’s algorithm for the separation of the partition inequalities.

Pierre Pesneau 16

Cyclomatic inequalities : separation

2nd heuristic :Let G=(V,E) be a graph and x a solution

While and |V|>2 do

Find an edge e with the greatest value in x

Contract edge e in the graph G

If |V|>2 then we have a violated cyclomatic inequality and each element of the associated partition is given by the expansion of the nodes of the graph.

1

)(KpKEx

Pierre Pesneau 17

Cycle partition inequalities

• Let be a partition of V such that• Let T be the chords of the partition and C be the

other edges of the partition.• Every solution must verify :

p, V , V 0

.p)C(x)T(xK

pp 2 1

1

.Kp

Pierre Pesneau 18

Cycle partition inequalities : facets

• Let G=(V,E) be a complete graph.• Let be a partition of V.• The cycle partition inequality is facet defining if

and only if :– – there is at most one such that

– for

p, V , V 0

,Kp p...,,i 0 ,VV ii 11

2iV .p...,,i 0

Pierre Pesneau 19

Cycle partition inequalities : separation

Same idea than the separation of cyclomatic inequalities :

Let G=(V,E) be a graph and x be a solution

While |V|>K+1 do

Contract edge e with the greatest value in x

Search the order of the nodes of the final graph such that

is minimum.

If this value is <2K then, the expansion of the nodes of the final graph give a paretition inducing violated cycle partition inequality.

)()( 1

1 CxTxK

pp

Pierre Pesneau 20

Computational results

• Branch&Cut algorithm.• Tree manager : BCP (IBM).• Linear solver : CPLEX 7.1.• PC PIV 1,7 GHz, 1 Go RAM.• Random and real data.• • Complete graphs.• Time limit : 2 hours.

.103 K

Pierre Pesneau 21

Results : random instancesNode K Cut Cycle Metric Subset Cyclom. Cycle P Gap ro Gap

finCPU (s)

20 3 13.0 27.4 19.8 11.0 9.0 7.0 0.15 0.00 0.28

20 4 40.2 1000.4 496.8 118.2 56.8 28.8 3.46 0.00 54.28

20 5 52.0 2487.8 709.6 1068.6 81.8 28.0 4.36 0.00 276.30

20 6 61.6 1838.8 419.4 1656.0 62.8 20.6 4.47 0.00 173.64

20 10 38.8 484.2 124.4 1953.6 26.2 1.4 2.50 0.00 42.86

30 3 28.2 98.2 59.8 20.6 33.4 14.8 0.83 0.00 8.50

30 4 97.611029.

84768.2 646.4 457.8 125.0 5.99 1.97 7200.00

30 5 94.812905.

22462.0 2893.2 380.2 51.6 6.41 2.00 7200.00

30 6104.

412738.

01704.4 6560.8 207.4 18.0 7.86 3.85 7200.00

30 10121.

27326.4 647.8

23070.0

103.4 0.8 4.33 0.96 5504.88

40 3 53.4 311.0 175.8 40.0 192.4 53.6 1.31 0.00 362.92

40 4108.

87140.6 2824.2 453.0 658.4 109.0 5.98 3.12 7200.00

40 5120.

88141.8 1369.2 1540.6 395.2 41.0 8.21 5.54 7200.00

40 6118.

68776.6 932.6 3227.4 254.6 34.0 9.26 6.73 7200.00

40 10152.

08886.8 607.8

15942.2

127.0 0.5 4.32 2.04 7200.00

50 3 78.8 701.4 414.8 92.6 631.0 179.8 2.01 0.26 3522.34

50 4112.

24475.4 1592.6 253.6 902.6 83.6 7.21 5.54 7200.00

50 5111.

24982.6 787.0 928.4 497.4 45.0 9.34 7.55 7200.00

50 6112.

05420.2 527.4 1726.4 277.2 17.4 10.89 9.27 7200.00

50 10139.

47298.4 259.4

10923.2

102.6 1.6 6.23 4.58 7200.00

Pierre Pesneau 22

Results : real instancesNode K Cut Cycle Metric Subset Cyclom. Cycle P Gap ro Gap

finCPU (s)

12 3 2 4 6 2 1 1 0.00 0.00 0.01

12 4 10 22 22 12 9 0 0.52 0.00 0.13

12 5 14 28 28 43 4 0 1.77 0.00 0.29

12 6 8 8 17 38 5 0 0.72 0.00 0.15

12 10 4 0 20 0 6 0 0.83 0.00 0.09

17 3 13 43 41 15 11 6 0.51 0.00 0.52

17 4 18 159 111 40 16 3 1.73 0.00 2.81

17 5 28 97 42 103 16 2 2.21 0.00 1.47

17 6 9 7 11 26 6 2 0.00 0.00 0.08

17 10 2 0 0 0 2 0 0.00 0.00 0.03

30 3 30 91 50 18 49 17 0.58 0.00 9.96

30 4 110 11373 5768 816 443 102 4.59 1.12 7200.00

30 5 161 13161 2835 4185 405 76 4.72 0.98 7200.00

30 6 144 12874 2415 7987 264 24 5.50 1.92 7200.00

30 10 69 617 164 1668 27 0 1.41 0.00 58.40

52 3 87 584 358 73 837 108 1.31 0.00 2092.77

52 4 107 3656 1398 175 867 60 6.15 4.60 7200.00

52 5 130 3867 728 745 607 40 9.11 7.88 7200.00

52 6 128 4506 477 1400 308 8 8.09 6.72 7200.00

52 10 110 3552 118 7273 90 1 7.60 6.42 7200.00

Pierre Pesneau 23

Example of a solution

52 nodes

3K

35 minutes2047 constraints

Pierre Pesneau 24

Perspectives

• Solve bigger instances

(particularly for K=3,4 and 5).• Improve separation routines.• Find new classes of valid inequalities.

Pierre Pesneau 25

Subset inequalities

• Let T be an edge set such that G\T does not contain a feasible solution.

• We have :

• Separation : when we separate cycle inequalities, if two consecutive elements of the partition are reduced to a single node, then T induced a violated subset inequality.

.1)( Tx

Pierre Pesneau 26

Cut inequalities : facets

• Let G=(V,E) be a complete graph.• Let • The cut inequality is facet defining if and only if :

– either and

– or and

.VW,VW

2 4 W,K

32 3 ,W,K ,W\V 2

.,W\V 32

Pierre Pesneau 27

Cyclomatic inequalities : facets

• Let G=(V,E) be a complete graph.• Let be a partition of V.• The cyclomatic inequality is facet defining if and

only if and :– either and or and

for

– or for

pV,...,V 0

4K 2iV,p,...,i 0

32 3 ,V,K i .p,...,i 0

3iV 11 mod)1( Kp

Kp

Pierre Pesneau 28

Metric inequalities• Introduced by Fortz, Labbé, Maffioli (2000).• Let G=(V,E) be a graph and• Let be a set of node potential satisfying :

• Every solution must verify :

where

Separation : heuristic of Fortz et al. (2000).

.Eije Vkk

.Kji 1

eeEf

ff xxv

. allfor 1

1,0max,1min eEklf

Kv

ji

klf