Optimally Locating Facilities Optimally Locating Facilities on a Networkon a Network

A Second Application of Transportation Network Analysis

There Are There Are ThreeThree Things Things Important When Buying a House:Important When Buying a House:

LocationLocationLocation

Examples:Examples:

LibrariesAmbulancesWarehousesFactoriesRestaurantsBanksTelephone centersMilitary facilities

Let’s Consider Objective Let’s Consider Objective Functions:Functions:

Minimize average travel time or distanceMinimize worst case (maximum) travel time or distanceMinimize fraction of population greater than 10 minutes from a facilityMaximize minimum travel time

Classic Location ProblemsClassic Location Problems

Median Problems– Minimize average travel distance (time)– Sometimes called Minisum

Center Problems– Minimize maximum distance to (from) a

facilityRequirements Problems– Allocate to achieve some objective

Median ProblemMedian Problem

Nodal weights hj, representing fraction of customers from node jSum of hj's equals one.We have an undirected network G(N,A)Objective: locate k facilities on G such that mean travel distance to a closest facility is minimized

G(N,A), | N |= n

hj ≥ 0 hjj=1

n

∑ =1

Xk = {x1, x2 ,..., xk}; x j ∈ Gd(Xk, j)≡min. distance between any one o points xi ∈ Xk and the node j ∈ N.

d(Xk, j)≡MINxi ∈ Xk

d(xi, j)

J(X k) ≡ hjd(Xkj=1

n

∑ , j ) = mean travel time to (from) closest facility

Find Xk* ∈ G such that for all Xk ∈ G,

J(Xk* ) ≤ J(Xk )

Xk* is a k - median of G(N,A) with a given h = (h1,h2, ...,hn)

TheoremTheorem

At least one k-median exists solely on the nodes of G.

Proof by contradiction for Proof by contradiction for kk=1=1

p q

d(x,p) d(x,q)P

Q

ProofProof

Start with 2 node problem, wp > wq.

J(x) =wpx + (1 −wp )(1− x)

= 2wpx + 1− x −wp = x(2wp −1) +1 −wp

Proof Proof -- con'tcon't..

Draw this graphicallyAdd general multi-node networkAdd breakpoints

1

2

3

4

5

2

22

2

1

1

1

2

3

4

5

2

22

2

1

1

h1=0.1

h2=0.1

h3=0.0

h4=0.4

h5=0.4

1

2

3

4

5

2

22

2

1

1

h1=0.1

h2=0.1

h3=0.0

h4=0.4

h5=0.4

1

2

3

4

5

2

22

2

1

1

h1=0.1

h2=0.1

h3=0.0

h4=0.4

h5=0.4

Summary for 1Summary for 1--MedianMedian

Step 1: Find travel distance matrix DStep 2: Find (hj,d(i,j))Step 3: Find row sums of (hj,d(i,j))– Take the minimum

Center ProblemsCenter Problems

G(N,A), undirectedLocate one facility so as to minimize

maximum distance to a node

1

2

3

4

5

2

22

2

1

1

h1=0.1

h2=0.1

h3=0.0

h4=0.4

h5=0.4

Recall:

0 2 3 2 4 2 0 2 3 3D= 3 2 0 1 1 2 3 1 0 2 4 3 1 2 0

Nodes 2, 3 and 4 all vertex centers

Center ProblemCenter Problem

No nodal solution in general for the absolute center, only for the vertex centerLet x ∈ G(N,A)Define the "maximum distance function":

m(x) ≡MAXj ∈ N

d(x, j)[ ]

Objective for the absolute center :

Find x* ∈ G such that m(x*) ≤m(x) for all x ∈ G.

Single Absolute Center Single Absolute Center Algorithm:Algorithm:

1. Find all local centers (one for each link)2. Choose a local center with smallest m(xl). That is an absolute center x* of G.

Optimally Locating Facilities on a NetworkThere Are Three Things Important When Buying a House:Examples:Let’s Consider Objective Functions:Classic Location ProblemsMedian ProblemTheoremProof by contradiction for k=1ProofProof - con't.Summary for 1-MedianCenter ProblemsCenter ProblemSingle Absolute Center Algorithm: