LP-Based Algorithms for Capacitated Facility Location Hyung-Chan An EPFL July 29, 2013 Joint work...

LP-Based Algorithms for Capacitated Facility Location

Hyung-Chan An

EPFL

July 29, 2013

Joint work with Mohit Singh and Ola Svensson

Capacitated facility location problem Given a metric cost c on

D: set of clients F: set of facilities

10

2

3

12



Ui: capacity of i∈F oi: opening cost of i∈F

5≤3

1≤2

20≤3




Want: Choose S⊆F to open

5≤3

1≤2

20≤3




Want: Choose S⊆F to open Assign every client to an open facility f : D→S

Capacities satisfied

5≤3

1≤2

20≤3




Want: Choose S⊆F to open Assign every client to an open facility f : D→S

Capacities satisfied Minimize Σi∈S oi + Σj∈D cj,f(j) = (20 + 5) + (2 + 2 + 2 +

5 + 2 + 3)

5

20

2

32

5

2 2

Successful special case Uncapacitated facility location problem

Ui = ∞ ∀i

NP-hard to approximate better than 1.463[Guha & Khuller 1999] [Sviridenko]

1.488-approximation algorithm [Li 2011]

Successful special case Uncapacitated facility location problem

Ui = ∞ ∀i

NP-hard to approximate better than 1.463[Guha & Khuller 1999] [Sviridenko]

1.488-approximation algorithm [Li 2011] Combining a linear program(LP)-rounding algorithm with a

primal-dual algorithm LP-rounding 3.16-approximation [Shmoys, Tardos, Aardal 1997], LP-rounding &

greedy 2.41-approximation [Guha & Khuller 1999], LP-rounding 1.74-approximation [Chudak & Shmoys 1999], primal-dual 3-approximation [Jain & Vazirani 2001], combining 1.73-approximation [Charikar & Guha 1999], primal-dual 1.61-approximation [Jain, Mahdian, Markakis, Saberi, Vazirani 2003], LP-rounding 1.59-approximation [Sviridenko 2002], primal-dual 1.52-approximation [Mahdian, Ye, Zhang 2006], combining 1.5-approximation algorithm [Byrka & Aardal 2010]

The Question

Can we use these LP-based techniques to solve the capacitated facility location problem?

The Question


All known approximation algorithms based on local search

[Bansal, Garg, Gupta 2012], [Pal, Tardos, Wexler 2001], [Korupolu, Plaxton, Rajaraman 2000]

5-approximation

The Question


Rich toolkit of algorithmic techniques

The Question


Rich toolkit of algorithmic techniques Per-instance performance guarantee Application to related problems

One of the ten Open Problems selected by the textbook of Williamson and Shmoys

The Question


Why is this hard? Standard LP relaxation fails to bound the

optimum within a reasonable factor Relaxed problems: uncapacitated problem,

capacities can be violated [Abrams, Meyerson, Munagala, Plotkin 2002], facilities can be opened multiple times [Shmoys, Tardos, Aardal 1997], [Jain, Vazirani 2001], opening costs are uniform [Levi, Shmoys, Swamy 2012]

The Question


Why is this hard? Standard LP relaxation fails to bound the

optimum within a reasonable factor No LP relaxation known that is algorithmically

amenable

Main result

There is a good LP: its optimum is within a constant factor of the true optimum.In particular, there is a poly-time algorithm that finds a solution whose cost is within a constant factor of the LP optimum.

Theorem

Our relaxation Standard LP rewritten

∀i∈F yi=1 if open, 0 if not ∀i∈F, j∈D xij=1 if j is assigned to i, 0

if not Consider a multicommodity flow

network: arc (j, i) of capacity xij

j∈D is a source of commodity j with demand 1

i∈F is a sink of commodity-oblivious capacity yi∙Ui

commodity-specific capacity yi∙1

≤3

≤2

≤3

y=1

y=0

y=1

All shown arcs are of capacity 1; others 0.


∀i∈F yi=1 if open, 0 if not ∀i∈F, j∈D xij=1 if j is assigned to i,

0 if not Consider a multicommodity flow





≤3

≤2

≤3

y=1

y=0

y=1


Minimize Σi∈F oiyi + Σ i∈F, j∈D cijxij

subject to the flow network defined by (x, y) is feasible

xij, yi ∈ {0, 1}


∀i∈F yi=1 if open, 0 if not ∀i∈F, j∈D xij=1 if j is assigned to i,

0 if not Consider a multicommodity flow





≤3

≤2

≤3

y=1

y=0

y=1


Minimize Σi∈F oiyi + Σ i∈F, j∈D cijxij

subject to the flow network defined by (x, y) is feasible

xij, yi ∈ [0, 1]


∀i∈F yi=1 if open, 0 if not ∀i∈F, j∈D xij=1 if j is assigned to i, 0 if

not Consider a multicommodity flow network:

arc (j, i) of capacity xij




Instance with one client and one facility with capacity ≤10

≤3

≤2

≤3

y=1

y=0

y=1


Our relaxation Consider arbitrary partial assignment g : D↛F

Suppose that clients are assigned according to g

(x, y) should still give a feasible solution for the remaining clients: i.e., the flow network should still be feasible when

only the remaining clients have demand of 1 commodity-oblivious capacity of i is yi∙(Ui -|g-

1(i)|) In similar spirit as knapsack-cover inequalities

[Wolsey 1975] [Carr, Fleischer, Leung, Phillips 2000]

LP constraint: the flow network defined by (g, x, y) is feasible for all g Is this really a relaxation?

≤32

≤21

≤3

y=1

y=0

y=1


≤3

≤2

≤3

x

x

Our relaxation Is this really a relaxation? No. The only facility adjacent from

remaining clients has 1∙0 = 0 commodity-oblivious capacity

Introduce backward edges corresponding to g Now flows can be routed along

alternating paths

≤3

≤2

≤30

y=1

y=0

y=1


≤3

≤2

≤3

x

Our relaxationMinimize Σi∈F oiyi + Σ i∈F, j∈D cijxij

subject to MFN(g, x, y) is feasible ∀partial assignment g

where MFN(g, x, y) is a multicommodity flow network with

arc (j, i) of capacity xij, arc (i, j) of capacity 1 if g assigns j to i, j∈D is a source of commodity j with demand 1 if

not assigned by g, i∈F is a sink of

commodity-oblivious capacity yi∙(Ui -|g-1(i)|) and commodity-specific capacity yi∙1.

Our relaxationMinimize Σi∈F oiyi + Σ i∈F, j∈D cijxij


Automatically embraces Standard LP when g is the empty partial function Knapsack-cover inequalities

Can be separated with respect to any given g Algorithm uses feasibility of MFN(g, x, y) for a single g Invoke standard techniques [Carr, Fleischer, Leung, Phillips 2000], [Levi, Lodi, Sviridenko 2007]

LP-rounding without exact separation Relaxed separation oracle

Either finds a violated inequality or rounds the given point

Our rounding algorithm gives one It does not rely on optimality We can separate with respect to a given g

LP-rounding without exact separation Relaxed separation oracle

Either finds a violated inequality or rounds the given point

(Optimization) ellipsoid method Queries (exact) separation oracle with x*, which

either Finds an inequality violated by x* Determines x* is feasible – c(x) ≤ c(x*) added to

system Can run with a relaxed separation oracle

Finds an inequality violated by x* Rounds x* – c(x) ≤ c(x*) added to system

LP-rounding algorithm

LP-rounding algorithm I ← {i∈F | y*

i > ½}, S ← {i∈F | y*i ≤ ½}; open I

Can afford it Choose a partial assignment g : D↛I For each client j assigned by g,

assign j in the same way Remaining clients are to be assigned to S

Lemma We can choose a partial assignment g s.t. g is cheap MFN(g, x*, y*) has a feasible flow where all flow is

drained at S.

LP-rounding algorithm Remaining clients are to be assigned to S

MFN(g, x*, y*) remains feasible even when “restricted” to S MFN(∅, x*, y*) is feasible when restricted to remaining clients i.e., it is feasible for the standard LP Commodity-oblivious capacity of i∈S is yi∙Ui ≤ Ui / 2

Capacity constraints are not tight Can use Abrams et al.’s algorithm based on the standard LP

Finds 18-approx soln where capacities are violated by 2


drained at S.

LP-rounding algorithm Can use Abrams et al.’s algorithm based on the

standard LP Finds 18-approx soln where capacities are violated

by 2

A 288-approximation algorithm Obvious improvements, but perhaps not leading to

≤5 Determination of integrality gap remains openLemma We can choose a fractional partial assignment g

s.t. g is cheap MFN(g, x*, y*) has a feasible flow where at least half of

each commodity’s flow is drained at S.


drained at S.


Simplifying assumption For each client j, all its

incident edges in the support have equal costs

Support of x* := { (i, j) | x*ij > 0

}

≤2

≤1

≤1

≤2

≤2

x-values shownon edges

1/21/2

1/21/2

1/31/31/3

1

1

y=1

y=1

y=1

y=1/2

y=1/2


drained at S.


Simplifying assumption For each client j, all its

incident edges in the support have equal costs

Support of x* := { (i, j) | x*ij > 0

}

≤2

≤1

≤1

≤2

≤2


drained at S.



i > ½}, S ← {i∈F | y*i ≤

½}; open I Find a maximum bipartite matching

g on the support of x*

j∈D is matched at most once i∈I is matched up to Ui times i∈S is not matched

≤2

≤1

≤1

≤2

≤2

I

S


i > ½}, S ← {i∈F | y*i ≤




Now we observe MFN(g, x*, y*) There exists no path from a remaining

client to a facility in I that is “undermatched”

≤2

≤1

≤1

≤2

≤2

I

S

≤2

≤1

≤1

≤2

≤2

I

S


i > ½}, S ← {i∈F | y*i ≤ ½};

open I Find a maximum bipartite matching g

on the support of x*




“Fully matched” facility has zero capacity All flow drained at S

≤2

≤1

≤1

≤2

≤2

I

S

≤21

≤10

≤10

≤2

≤2

I

S

xxx

Lifting the assumption

Lifting the assumption Problem

g may become expensive compared to c(x*)

Solution Find a fractional partial assignment g ≤ x*




Σ i∈I gij ≤ 1 ∀j∈D

Σ j∈D gij ≤ Ui ∀i∈I

gij = 0 ∀i∈S




1/2

1/31/31/3

1

1

S

s t

11111

112




But we defined MFN(g, x*, y*) only for integral g…

(Extending) our relaxationMinimize Σi∈F oiyi + Σ i∈F, j∈D cijxij







subject to MFN(g, x, y) is feasible ∀fractional part. asgn. g








arc (j, i) of capacity xij, arc (i, j) of capacity gij, j∈D is a source of commodity j with demand 1 if






arc (j, i) of capacity xij, arc (i, j) of capacity gij, j∈D is a source of commodity j with demand

dj := 1 - Σ i∈I gij, i∈F is a sink of







commodity-oblivious capacity yi∙(Ui - Σ j∈D gij) and commodity-specific capacity yi∙1.






commodity-oblivious capacity yi∙(Ui - Σ j∈D gij) and commodity-specific capacity yi∙dj.

Lifting the assumptionLemma We can choose a fractional partial assignment g s.t.

g ≤ x*

MFN(g, x*, y*) has a feasible flow where at least half of each commodity’s flow is drained at S.


g ≤ x*


Hoping too much:

y=.01

≤1

≤1y=.99

.01

.99

.01dj = 0.0001.01

.99

.99


g ≤ 2x*


What’s the goal? Goal is not in making MFN(g, x*, y*) feasible – the

ellipsoid method guarantees this Will find an assignment that leaves on clients

“too much” demand to be served by remaining capacities in I

Lifting the assumption Two different type of “paths”

In MFN(g, x*, y*) In the residual graph of the partial assignment

Previous argument A path from a client with positive demand to a facility in I with positive capacity corresponds to a path from an undermatched client to an undermatched facility and therefore does not exist.


i > ½}, S ← {i∈F | y*i ≤






≤2

≤1

≤1

≤2

≤2

I

S

≤2

≤1

≤1

≤2

≤2

I

S

Lifting the assumption Two different type of “paths”

In MFN(g, x*, y*) In the residual graph of the partial assignment

≤1

≤1

≤1

≤1

All weights on edges 1/2

x* g

≤1

≤1

Residual Graph

≤1

≤1

MFN(g, x*, y*)

Lifting the assumption Consider a maximum fractional matching g

and its residual graph

s t



R: reachable from an undermatched client

N: not reachable

s t

DR

DN

IR

IN

S




N: not reachable

Consider MFN(g, x*, y*)

s t

DR

DN

IR

IN

S

DR

DN

IR

IN

S

Lifting the assumption No facility in IR is undermatched

a


N: not reachable

s t

DR

DN

IR

IN

S

DR

DN

IR

IN

S

Lifting the assumption No facility in IR is a sink

s t

DR

DN

IR

IN

S

DR

DN

IR

IN

S

possible sinks


Every client with positive demand is in DR

s t

DR

DN

IR

IN

S

DR

DN

IR

IN

S

possible sinkspossible sources


Every client with positive demand is in DR

Consider j ∈ DR and i ∈ IN (j, i) is saturated: gij = 2x*ij

s t

DR

DN

IR

IN

S

DR

DN

IR

IN

S


jj

i i

×

Lifting the assumption Consider j ∈ DR and i ∈ IN (j, i) is saturated: gij = 2x*ij

Set gij ← 0

Increases dj by Σ i∈IN 2x*ij, which is twice the total

capacity from j to IN

s t

DR

DN

IR

IN

S

DR

DN

IR

IN

S


jj

Lifting the assumption For every j ∈ DR, its demand is at least twice

the total capacity of arcs from j to IN in MFN(g, x*, y*)

s t

DR

DN

IR

IN

S

DR

DN

IR

IN

S


jj

Lifting the assumption For every j ∈ DR, its demand is at least twice the

total capacity of arcs from j to IN in MFN(g, x*, y*)

Compare to: For every j ∈ DR, there does not exist a path

from j to IN in MFN(g, x*, y*)

s t

DR

DN

IR

IN

S

DR

DN

IR

IN

S


jj

Lifting the assumption Say: arcs from j to IN is labeled as commodity j

s t

DR

DN

IR

IN

S

DR

DN

IR

IN

S


Lifting the assumption Say: arcs from j to IN is labeled as commodity j Want: there exists a feasible flow where, for

every j ∈ DR, every arc from j to IN is used by commodity j

s t

DR

DN

IR

IN

S

DR

DN

IR

IN

S



Recall gi’j’ ← 0 for j’ ∈ DR and i’ ∈ IN

s t

DR

DN

IR

IN

S

DR

DN

IR

IN

S


×


Recall gi’j’ ← 0 for j’ ∈ DR and i’ ∈ IN Suppose gi’j’ > 0 for some j’ ∈ DN and i’ ∈ IR

s t

DR

DN

IR

IN

S

DR

DN

IR

IN

S


j’×

i’


Recall gi’j’ ← 0 for j’ ∈ DR and i’ ∈ IN Suppose gi’j’ > 0 for some j’ ∈ DN and i’ ∈ IR

s t

DR

DN

IR

IN

S

DR

DN

IR

IN

S


×

Lifting the assumption Say: arcs from j to IN is labeled as commodity j Recall gi’j’ ← 0 for j’ ∈ DR and i’ ∈ IN Suppose gi’j’ > 0 for some j’ ∈ DN and i’ ∈ IR Any flow path ending at IN contains an arc from

DR to IN

s t

DR

DN

IR

IN

S

DR

DN

IR

IN

S


Lifting the assumption Any flow path ending at IN contains an arc from

DR to IN

s t

DR

DN

IR

IN

S

DR

DN

IR

IN

S


j

Lifting the assumption Any flow path ending at IN contains an arc from DR to IN Suppose j ∈ DR drains the highest fraction of its flow at

IN and this fraction is >1/2 Want: there exists a feasible flow where, for every j ∈

DR, every arc from j to IN is used by commodity j

s t

DR

DN

IR

IN

S

DR

DN

IR

IN

S


j

Want: there exists a feasible flow where, for every j ∈ DR, every arc from j to IN is used by commodity j Any flow path ending at IN contains an arc from DR to IN Suppose j ∈ DR drains the highest fraction(>½) of its flow

at IN There is a flow path of j that “steals” an arc of k ∈ DR


s t

DR

DN

IR

IN

S

DR

DN

IR

IN

S


j

k

Want: there exists a feasible flow where, for every j ∈ DR, every arc from j to IN is used by commodity j k ∈ DR recovered its stolen arc


s t

DR

DN

IR

IN

S

DR

DN

IR

IN

S


j

k

Main result

There is a good LP: its optimum is within a constant factor of the true optimum.In particular, there is a poly-time algorithm that finds a solution whose cost is within a constant factor of the LP optimum.

Theorem

The Question

Can we use LP-based techniques to solve the capacitated facility location problem? Yes!

Can we use other LP-based techniques with our new relaxation?

Can we apply these results to related problems? Can we improve our integrality gap bounds?

Thank you.

LP-Based Algorithms for Capacitated Facility Location Hyung-Chan An EPFL July 29, 2013 Joint work...

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