optimally locating facilities on a networklet’s consider objective functions: z minimize average...
TRANSCRIPT
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Optimally Locating Facilities Optimally Locating Facilities on a Networkon a Network
A Second Application of Transportation Network Analysis
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There Are There Are ThreeThree Things Things Important When Buying a House:Important When Buying a House:
LocationLocationLocation
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Examples:Examples:
LibrariesAmbulancesWarehousesFactoriesRestaurantsBanksTelephone centersMilitary facilities
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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
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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
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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
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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)
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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)
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TheoremTheorem
At least one k-median exists solely on the nodes of G.
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Proof by contradiction for Proof by contradiction for kk=1=1
p q
d(x,p) d(x,q)P
Q
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ProofProof
Start with 2 node problem, wp > wq.
J(x) =wpx + (1 −wp )(1− x)
= 2wpx + 1− x −wp = x(2wp −1) +1 −wp
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Proof Proof -- con'tcon't..
Draw this graphicallyAdd general multi-node networkAdd breakpoints
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1
2
3
4
5
2
22
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1
1
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1
2
3
4
5
2
22
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1
1
h1=0.1
h2=0.1
h3=0.0
h4=0.4
h5=0.4
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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
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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
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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
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Center ProblemsCenter Problems
G(N,A), undirectedLocate one facility so as to minimize
maximum distance to a node
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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
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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)[ ]
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Objective for the absolute center :
Find x* ∈ G such that m(x*) ≤m(x) for all x ∈ G.
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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: