Collection Depots Facility LocationProblems in Trees

R. Benkoczi, B. Bhattacharya, A. Tamir

  陳冠伶‧王湘叡‧李佳霖‧張經略

Jun 12, 2007

Outline

INTRODUCTIONBy 陳冠伶

Settings

Collection DepotsCollection Depots

Facility (service center)Facility (service center)

Client (demand service)

Client (demand service)

Cost of Service Trip

F

C

D

P1

P1

P2

P2

2(P1+P2)‧w(c)

Service CostService Cost

Application (1)

Express Transportation

Express Transportation

Application (2)

Garbage collectionGarbage collection

Problem

• IN: given a tree and• points of clients• points of collection depots• an integer k

• OUT• Optimal placements of k facilities• that minimizes some global function of the service

cost for all clients.

Objective – Minimax

• Minimize the service cost of the most expensive client

F

C

DD

C

DD

DD

DD

CC

C

DD

1-center

Minimize the maximum distance to the facility

Minimax – center problems

k-center

Minimize the maximum distance to the closest facility

Minimax – center problems

Objective – Minisum

• Minimize the total service cost

F

C

DD

C

DD

DD

DD

CC

C

DD

Minisum – median problems

1-median

Minimize the average distance to the facility

Minisum – median problems

k-median

Minimize the average distance to the closest facility

Classifications

Summary of Results

• Unrestricted 1-center problem• O(n)

• Unrestricted median problems• 1-median: O(nlogn)• k-median: O(kn3)

• Restricted k-median problem• NP-complete• Facility setup costs are not identical

1-CENTER PROBLEMBY 王湘叡

Prune and Search

• Every iteration, eliminate a fraction of impossible instances.

• Binary Search• T(n)=T(n/2)+1• T(n)=O(lg n)

• How about ( ) ((1 ) ) , 0 1T n T c n n c

2 1( ) (1 ) (1 )T n n c n c n n

c

Observation

• c(f)=max min r(f, vi)

• Service cost is non-decreasing when the facility goes away from the client.

Where could the facility be?

• A linear time algorithm could determine!

T1

T2

Ti Tk

Initial tree

clientclient

depotdepot

Divide T(i) into S1 and S2

• Find the centroid and partition the tree into two parts

centroidcentroid

S1 > 1/3 |T(i)| S2 > 1/3 |T(i)|

Find the Xmax

• Find the client Xmax with the largest service cost from the centroid.

S1 S2

XmaxXmax f

fopt must be in S1fopt must be in S1

Special case

• Centroid is the optimal

Xmax X’max

Should be optimalShould be optimal

Partition the clients

• Compute all depot distance

• Find the median δmed

• Separate all clients into two sets, K+ (red) and K- (blue)

S2 δmed δmed

• Consider f’ in S1, that depot distance δ(f’)< δmed

δ(f’)< δmed δ(f’)< δmed

S1

f’f’

Partition S1 by δmed

• Find all f’, they form trees T1, T2, …,Tn

• There are two cases, fopt is in T∪ i or not

T1

T2

T3

f

fopt is in T∪ i

• If fopt in red, consider K+, δ(fopt)<δmed<δ(K+)

• For a facility F’ in S1 and a client in S2, δ(fopt, u) is in S1

foptf’

δ(f’, u)δ(f’, u)δ(f’, u)δ(f’, u)

fopt is in not T∪ i

• If fopt is not in red, consider K-,

δ(K-)<δmed <δ(fopt)

• For a facility F’ in S and a client in S2, δ(fopt, u) is in S2

• Similar to previous case• Only fopt in T∪ i is

considered. fopt

f’δ(fopt, u)δ(fopt, u)

• Arbitrarily paired clients in K+

• For each pair (u, v), Compute tuv s.t. w(v)(tuv+d(c,v))=w(u).(tuv+d(c,u))

• Compare tmed and

d(fopt, c)+d((fopt,c),p(fopt, c))

foptfopt

δ(f’, u)δ(f’, u)

Details on fopt is in T∪ i

foptfopt

δ(f’, u)δ(f’, u)

d(fopt, c)+d((fopt,c),p(fopt, c)) < tmed

• consider tmed<tuv

• d(fopt, c)+d((fopt,c),p(fopt, c))<tmed<tuv

foptfopt

δ(f’, u)δ(f’, u)

d(fopt, c)+d((fopt,c),p(fopt, c)) > tmed

• consider tmed>tuv

• d(fopt, c)+d((fopt,c),p(fopt, c))>tmed>tuv

• ¼ K+ can be removed

1-MEDIAN PROBLEMBY 李佳霖

The 1-median Problem• Find a placement for facility to minimize the cost

of all tours.• i.e. minimize the sum of weighted distances of the

facility to client, then to optimal depot, and return to facility.

• For the path of a facility to a client, the closest depot can be found efficiently.

• Brute Force: Ο(n2)• Using Spine decomposition and pre-sorting: Ο(nlogn)

The Spine Decomposition

r0

33 5

23

Construct Search Tree

r0r0

Search Tree of SD

r0

Super-path of Search Tree

r0

f

Cost of Subtree

c2

dnewdnew

cjcj

| | | |

1 1 1

2 [ ( ) ( , ) ( ) ( , ) ( ) ( ) ( , ( , ))]v vT Tj

i i v i new i i ii i i j

w c d v c w T d f v w c d w c d d p v c

f

v

c3c4

c1

d d2

d4

d3

d1

Complexity• Construction for the SD has time complexity Ο(n) and space complexity Ο(n)

• Costs of the subtrees can be evaluated in constant time once j is determined.• If we use binary search with dnew, we spend Ο(logn)

time for every subtree. So Ο(log2n).• Use the sequential search in sorted order. So Ο(logn).

• The 1-median collection depots problem in tree can be sloved in Ο(nlogn) time and Ο(n) space.

UNRESTRICTEDK-MEDIAN PROBLEM

BY 張經略

The objective

• To minimize the sum of facility opening costs plus service costs for servicing the clients.

The “自給自足” property (1/4)• We fixed an arbitrary optimal solution and

explore its structure.

The “自給自足” property (2/4)• Consider an arbitrary vertex v.

• xv: minimize the trip cost of serving v

• yv:be a closest facility to v.

client C

v

xv

Assumed (for contradiction) servicing facility for client C

yv

Tleft Tright

The “自給自足” property (3/4)

client C

v

xv

Assumed (for contradiction) servicing facility for client C

yv

Tleft

Tright

The “自給自足” property (4/4)• The blue part of the following graph is proven

by symmetry.

v

xv

yv

Tleft

Tright

The intuition… (1/2)

• The total cost can be partitioned into four categories: the red, yellow, blue cost and v.

v

xv

yv

Tleft

Tright

The intuition… (2/2)

• The optimal solution has to be a combination of optimal substructures• You have to be “optimal” in the red (to minimize

the red cost) and the yellow (to minimize the yellow cost).

• This almost leads to Dynamic Programming already!

The technical things

• Due to some complications, the final Dynamic Programming is much more complicated…

• But the proof requires no special technique beyond the “自給自足”  property.

• The challenge is to devise the “right” recurrences to carry out the aforementioned intuitive approach.

Simple intuition, complicated recurrences… take a look

Time complexity

• Easily verified to be polynomial.