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Collection Depots Facility Location Problems in Trees R. Benkoczi, B. Bhattacharya, A. Tamir. 陳冠伶 ‧ 王湘叡 ‧ 李佳霖 ‧ 張經略 Jun 12, 2007. Outline. By 陳冠伶. INTRODUCTION. C lient (demand service). Settings. F acility (service center). C ollection D epots. Cost of Service Trip. F. P 2. D.

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Collection depots facility location problems in trees r benkoczi b bhattacharya a tamir

Collection Depots Facility LocationProblems in TreesR. Benkoczi, B. Bhattacharya, A. Tamir

陳冠伶‧王湘叡‧李佳霖‧張經略

Jun 12, 2007



Introduction

By 陳冠伶

INTRODUCTION


Settings

Client (demand service)

Settings

Facility (service center)

Collection Depots


Cost of service trip
Cost of Service Trip

F

P2

D

P1

Service Cost

2(P1+P2)‧w(c)

C


Application 1
Application (1)

Express Transportation


Application 2
Application (2)

Garbage collection


Problem
Problem

  • IN: given a tree and

    • points of clients

    • points of collection depots

    • an integer k

  • OUT

    • Optimal placements of kfacilities

    • that minimizes some global function of the service cost for all clients.


Objective minimax
Objective – Minimax

  • Minimize the service cost of the most expensive client

C

C

D

D

D

C

D

F

D

C

C


Minimax center problems
Minimax – center problems

1-center

Minimize the maximum distance to the facility


Minimax center problems1
Minimax – center problems

k-center

Minimize the maximum distance to the closest facility


Objective minisum
Objective – Minisum

  • Minimize the total service cost

C

C

D

D

D

C

D

F

D

C

C


Minisum median problems
Minisum – median problems

1-median

Minimize the average distance to the facility


Minisum median problems1
Minisum – median problems

k-median

Minimize the average distance to the closest facility



Summary of results
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 problem

BY 王湘叡

1-CENTER PROBLEM


Prune and search
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


Observation
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
Where could the facility be?

  • A linear time algorithm could determine!

T1

T2

Ti

Tk


Initial tree
Initial tree

depot

client


Divide t i into s 1 and s 2
Divide T(i) into S1 and S2

  • Find the centroid and partition the tree into two parts

centroid

  • S1 > 1/3 |T(i)|

  • S2 > 1/3 |T(i)|


Find the x max
Find the Xmax

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

Xmax

f

S2

S1

foptmust be in S1


Special case

Xmax

Special case

  • Centroid is the optimal

Should be optimal

X’max


Partition the clients
Partition the clients

  • Compute all depot distance

  • Find the median δmed

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

δmed

  • S2


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

δ(f’)< δmed

f’

  • S1


Partition s 1 by med
Partition S1 by δmed

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

  • There are two cases, fopt is in ∪Tior not

f

T1

T2

T3


F opt is in t i
fopt is in ∪Ti

  • 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

δ(f’, u)

δ(f’, u)

fopt

f’


F opt is in not t i
fopt is in not ∪Ti

  • If fopt is not in red, consider K-,

    δ(K-)<δmed <δ(fopt)

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

  • Similar to previous

    case

  • Only fopt in ∪Ti is considered.

δ(fopt, u)

f’

fopt


Details on f opt is in t i
Details on fopt is in ∪Ti

  • 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))

δ(f’, u)

fopt

fopt


D f opt c d f opt c p f opt c t med
d(fopt, c)+d((fopt,c),p(fopt, c)) < tmed

  • consider tmed<tuv

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

δ(f’, u)

fopt

fopt


D f opt c d f opt c p f opt c t med1
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

δ(f’, u)

fopt

fopt


1 median problem

BY 李佳霖

1-MEDIAN PROBLEM


The 1 median problem
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)






Cost of subtree
Cost of Subtree

d

d2

v

dnew

d4

f

c2

d1

cj

d3

c1

c3

c4


Complexity
Complexity

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

  • Costs of the subtrees can be evaluated in constant timeonce 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.


Unrestricted k median problem

BY 張經略

UNRESTRICTEDK-MEDIAN PROBLEM


The objective
The objective

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


The property 1 4
The “自給自足” property (1/4)

  • We fixed an arbitrary optimal solution and explore its structure.


The property 2 4
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.

yv

Assumed (for contradiction) servicing facility for client C

xv

v

Tleft

Tright

client C


The property 3 4
The “自給自足” property (3/4)

Assumed (for contradiction) servicing facility for client C

yv

xv

v

client C

Tright

Tleft


The property 4 4
The “自給自足” property (4/4)

  • The blue part of the following graph is proven by symmetry.

yv

xv

v

Tleft

Tright


The intuition 1 2
The intuition… (1/2)

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

yv

xv

v

Tleft

Tright


The intuition 2 2
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
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.



Time complexity
Time complexity

  • Easily verified to be polynomial.


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