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 張經略 K-median problem 李佳霖 1-median problem 王湘叡 1-center problem 陳冠伶 introduction

INTRODUCTION By 陳冠伶

Settings C ollection D epots F acility (service center) C lient (demand service)

Cost of Service Trip F F C C D D P1P1 P1P1 P2P2 P2P2 2(P 1 +P 2 ) ‧ w(c) Service Cost

Application (1) Express Transportation

Application (2) Garbage 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 F C C D D C C D D D D D D C C C C C C D D

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 F C C D D C C D D D D D D C C C C C C D D

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(kn 3 ) Restricted k-median problem NP-complete Facility setup costs are not identical

1-CENTER PROBLEM BY 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

Observation c(f)=max min r(f, v i ) Service cost is non-decreasing when the facility goes away from the client.

Where could the facility be? A linear time algorithm could determine! T1T1 T2T2 TiTi TkTk

Initial tree client depot

Divide T(i) into S 1 and S 2 Find the centroid and partition the tree into two parts centroid S 1 > 1/3 |T(i)|S 2 > 1/3 |T(i)|

Find the X max Find the client X max with the largest service cost from the centroid. S1 S2 X max f f f opt must be in S 1

Special case Centroid is the optimal X max X’ max Should be optimal

Partition the clients Compute all depot distance Find the median δ med Separate all clients into two sets, K+ (red) and K- (blue) S2S2 δ med

Consider f’ in S 1, that depot distance δ(f’)< δ med δ(f’)< δ med S1S1 f’

Partition S 1 by δ med Find all f’, they form trees T 1, T 2, …,T n There are two cases, f opt is in ∪ T i or not T1T1 T2T2 T3T3 f f

f opt is in ∪ T i If f opt in red, consider K +, δ (f opt )< δ med < δ (K + ) For a facility F’ in S 1 and a client in S 2, δ (f opt, u) is in S 1 f opt f’ δ (f’, u)

f opt is in not ∪ T i If f opt is not in red, consider K -, δ (K - )< δ med < δ (f opt ) For a facility F’ in S and a client in S 2, δ (f opt, u) is in S 2 Similar to previous case Only f opt in ∪ T i is considered. f opt f’ δ (f opt, u)

Arbitrarily paired clients in K+ For each pair (u, v), Compute t uv s.t. w(v)(t uv +d(c,v))=w(u).(t uv +d(c,u)) Compare t med and d(f opt, c)+d((f opt,c),p(f opt, c)) f opt δ (f’, u) Details on f opt is in ∪ T i

f opt δ (f’, u) d(f opt, c)+d((f opt,c),p(f opt, c)) < t med consider t med <t uv d(f opt, c)+d((f opt,c),p(f opt, c))<t med <t uv

f opt δ (f’, u) d(f opt, c)+d((f opt,c),p(f opt, c)) > t med consider t med >t uv d(f opt, c)+d((f opt,c),p(f opt, c))>t med >t uv ¼ K + can be removed

1-MEDIAN PROBLEM BY 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(n 2 ) Using Spine decomposition and pre-sorting: Ο(nlogn)

The Spine Decomposition r0r

Construct Search Tree r0r0 r0r0

Search Tree of SD r0r0

Super-path of Search Tree r0r0 r0r0 f f

Cost of Subtree c2c2 c2c2 d new cjcj cjcj f f v v c3c3 c3c3 c4c4 c4c4 c1c1 c1c1 d d d2d2 d2d2 d4d4 d4d4 d3d3 d3d3 d1d1 d1d1

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 d new, we spend Ο(logn) time for every subtree. So Ο(log 2 n). 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

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. x v : minimize the trip cost of serving v y v :be a closest facility to v. client C v xvxv Assumed (for contradiction) servicing facility for client C yvyv T left T right

The “ 自給自足 ” property (3/4) client C v xvxv Assumed (for contradiction) servicing facility for client C yvyv T left T right

The “ 自給自足 ” property (4/4) The blue part of the following graph is proven by symmetry. v xvxv yvyv T left T right

The intuition… (1/2) The total cost can be partitioned into four categories: the red, yellow, blue cost and v. v xvxv yvyv T left T right

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.