Collection Depots Facility Location Problems in Trees R. Benkoczi, B. Bhattacharya, A. Tamir
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Transcript of Collection Depots Facility Location Problems in Trees R. Benkoczi, B. Bhattacharya, A. Tamir
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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
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Outline
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INTRODUCTIONBy 陳冠伶
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Settings
Collection DepotsCollection Depots
Facility (service center)Facility (service center)
Client (demand service)
Client (demand service)
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Cost of Service Trip
F
C
D
P1
P1
P2
P2
2(P1+P2)‧w(c)
Service CostService Cost
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Application (1)
Express Transportation
Express Transportation
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Application (2)
Garbage collectionGarbage collection
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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.
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Objective – Minimax
• Minimize the service cost of the most expensive client
F
C
DD
C
DD
DD
DD
CC
C
DD
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1-center
Minimize the maximum distance to the facility
Minimax – center problems
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k-center
Minimize the maximum distance to the closest facility
Minimax – center problems
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Objective – Minisum
• Minimize the total service cost
F
C
DD
C
DD
DD
DD
CC
C
DD
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Minisum – median problems
1-median
Minimize the average distance to the facility
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Minisum – median problems
k-median
Minimize the average distance to the closest facility
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Classifications
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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
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1-CENTER PROBLEMBY 王湘叡
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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
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Observation
• c(f)=max min r(f, vi)
• Service cost is non-decreasing when the facility goes away from the client.
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Where could the facility be?
• A linear time algorithm could determine!
T1
T2
Ti Tk
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Initial tree
clientclient
depotdepot
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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)|
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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
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Special case
• Centroid is the optimal
Xmax X’max
Should be optimalShould be optimal
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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
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• Consider f’ in S1, that depot distance δ(f’)< δmed
δ(f’)< δmed δ(f’)< δmed
S1
f’f’
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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
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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)
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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)
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• 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
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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
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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
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1-MEDIAN PROBLEMBY 李佳霖
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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)
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The Spine Decomposition
r0
33 5
23
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Construct Search Tree
r0r0
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Search Tree of SD
r0
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Super-path of Search Tree
r0
f
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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
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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.
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UNRESTRICTEDK-MEDIAN PROBLEM
BY 張經略
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The objective
• To minimize the sum of facility opening costs plus service costs for servicing the clients.
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The “自給自足” property (1/4)• We fixed an arbitrary optimal solution and
explore its structure.
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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
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The “自給自足” property (3/4)
client C
v
xv
Assumed (for contradiction) servicing facility for client C
yv
Tleft
Tright
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The “自給自足” property (4/4)• The blue part of the following graph is proven
by symmetry.
v
xv
yv
Tleft
Tright
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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
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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!
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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.
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Simple intuition, complicated recurrences… take a look
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Time complexity
• Easily verified to be polynomial.