ESA 2003 M. Pál. Universal Facility Location 1

Universal Facility Location

Mohammad Mahdian

MIT

Martin Pál Cornell University

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Metric Facility LocationF is a set of facilities.D is a set of clients.cij is the distance between any i and j in D

F.(assume cij satisfies triangle inequality)

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Problem Statement1) Allocate capacity ui at each facility i

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2) Assign every client to a facility (each facility i is assigned at most ui clients). Goal: Minimize the sum of allocation and assignment costs.

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The costCost of facility i is a function

fi(.) of demand allocated.

capacity

cost

fi(ui)

Assptn: fi(.) is nondecreasing.

Allocation cost cf(S): sum of

fi(ui) over all i in F.

Assignment cost cs(S): sum of distances from clients to assigned facilities.

Goal: Minimize cf(S) + cs(S).

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Special cases

demand

cost

Uncapacitated FL

cost

demandFL with soft capacities

demand

cost

Concave cost FL

cost

demand

FL with hard capacities

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Related work

Uncapacitated: primal-dual 1.52 [MYZ02] local search 3+ [CG99]

Soft capacities: primal-dual 2 [MYZ03] local search 4+ [AGK+01]

Concave cost: reduction to uncapacitated [HMM]

Hard capacities: local search 9+ [PTW01]

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Our result

First approximation algorithm for Universal FL

Generalizes local search algorithms for above special cases, with the same approx. guarantees

Our analysis: 8+ approximation (7.88 w. scaling)[CYZ] just claimed improved analysis: 7+

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Local Search

1) Start with any feasible solution.

2) Improve solution with “local operations”.

3) Stop when there are no remaining operations that lower the cost.

Well, almost… …we want to finish in poly-time:

Each operation is required to lower cost by a factor of /n.

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The add operationadd(s,) – Increase allocation at facility s by .

Reassign clients optimally.

Given u=(u1,u2 ,..., un), where do we send clients?

Solve a min cost flow.

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us

+1=2

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The pivot operation

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pivot(s,Δ): Pick a pivot point s and vector Δ.

For each facility i:

•Send -Δi units of demand from i to s (if Δi<0)

•Send +Δi units of demand from s to i (if Δi>0)s

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The pivot operationpivot(s,Δ): Pick a pivot point s and vector Δ.

For each facility i:

•Send -Δi units of demand from i to s (if Δi<0)

•Send +Δi units of demand from s to i (if Δi>0)

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Finding good operations

Want operations that decrease cost the most.

Plan: Try every s, find the best and Δ for add(s,) and pivot(s,Δ).

Easy to do for add: try all values for .

Too many possibilities for Δ to try them all. How to find the best pivot operation?

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How to find a good pivot?

Problem: find Δ to minimize cost(pivot(s,Δ)).

Consider facilities in any order.

Let p[d,i] be the min cost way of having d excess demand at the pivot from the first i facilities.

p[d,i+1] = min{ p[d-,i]+gi+1() }

p[0,n] is the optimum

Solution: dynamic programming, just like knapsack.

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Analysis

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Assignment cost

Thm: [Korupolu et al]

If no add(s,) op. improves a solution S,

cs(S) ≤ cs(SOPT) + cf(SOPT).

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Current sol S1 10

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Optimum SOPT

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Max(ui,uiOPT)

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Bounding the allocation cost

Bounding cf(S):

Local opt. cost(pivot(s,Δ)) ≥ 0 for any s, Δ.

Every operation gives a bound:

Σi fi(ui) ≤ Σi fi(ui + Δi) + flow cost

upper bound on cf(S) lower bound on cf(SOPT)

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First attempt

Take current S with some SOPT: let di = uiOPT - ui.

sources di < 0.

sinks di > 0.

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Reassignment costs may be high.

try pivot(s,d)

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Second attempt

Take current S with some SOPT: let di = uiOPT - ui.

di < 0.

di > 0.

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try swap(d): routing along min cost flow

Don’t have a swap operation.

sources

sinks

Flow is cheap: cost(y) ≤ cs(S) + cs(SOPT)

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What works

Take the swap(d) operation, let y be the flow.

Tile it with pivot operations so that:

•each source covered (at least) once

•each sink used at most k times (k=4)

•each pivot ships only along flow edges

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What works

Take the swap(d) operation, let y be the flow.

di < 0.

di > 0.

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Example: flow decomposed into two pivot op’s.

sources

sinks

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The flow y

Fact 1: cost(y) ≤ cs(S)+cs(SOPT)

Pf: use triangle inequality

Fact 2: y is min cost acyclic

Pf: augment along cycle to remove it

I.e. the flow is a tree (forest).

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Covering the tree

sources

sinks

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Subtrees of depth 2

sourcessinks

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Almost done..

Every source closed once

Every sink used ≤ 1+1=2 times(once as root, once as child)

Problem: flow cost

is nondominating if it sends ≥ ½ of its flow down.

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Nondominating sources

...

r

s1 s2 s3 sk

y(s1,r) ≤ y(s2,r) ≤ y(s3,r) ≤ ... ≤ y(sk,r)

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Summing up

Every facility opened 4 times

Flow at most tripled

cf(S) ≤ 7c(SOPT)

cs(S) ≤ 1c(SOPT)Assignment cost

Hence the algorithm is a (8+є)-approximation.

(7.88+є with scaling)

Total c(S) ≤ 8c(SOPT)