collection depots facility location problems in trees r. benkoczi, b. bhattacharya, a. tamir
DESCRIPTION
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. - PowerPoint PPT PresentationTRANSCRIPT
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.