Transcript Distance Measures

Facility Location
• Logistics Management
• Factors that Affect Location Decisions
• Distance Measures
• Classification of Planar Facility Location Problems
• Planar Single-Facility Location Problems
– Minisum Location Problem with Rectilinear Distances
– Minisum Location Problem with Euclidean Distances
– Minimax Location Problem with Rectilinear Distances
– Minimax Location Problem with Euclidean Distances
• Planar Multi-Facility Location Problems
– Minisum Location Problem with Rectilinear Distances
Logistics Management
•
Logistics Management can be defined as the management of the transportation and
distribution of goods. The term goods includes raw materials or subassemblies obtained
from suppliers as well as finished goods shipped from plants to warehouses or
customers.
•
Logistics management problems can be classified into three categories:
– Location Problems: involve determining the location of one or more new facilities
in one or more of several potential sites. The cost of locating each new facility at
each of the potential sites is assumed to be known. It is the fixed cost of locating a
new facility at a particular site plus the operating and transportation cost of serving
customers from this facility-site combination.
– Allocation Problems: assume that the number and location of facilities are known
a priori and attempt to determine how each customer is to be served. In other
words, given the demand for goods at each customer center, the production or
supply capacities at each facility, and the cost of serving each customer from each
facility, the allocation problem determines how much each facility is to supply to
each customer center.
– Location-Allocation Problems:involve determining not only how much each
customer is to receive from each facility but also the number of facilities along with
their locations and capacities.
Factors that Affect Location Decisions
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Proximity to source of raw materials.
Cost and availability of energy and utilities.
Cost, availability, skill, and productivity of labor.
Government regulations at the federal, state, county, and local levels.
Taxes at the federal, state, county, and local levels.
Insurance.
Construction costs and land price.
Government and political stability.
Exchange rate fluctuation.
Export and import regulations, duties, and tariffs.
Transportation system.
Technical expertise.
Environmental regulations at the federal, state, county and local levels.
Support services.
Community services - schools, hospitals, recreation, and so on.
Weather.
Proximity to customers.
Business climate.
Competition-related factors.
Distance Measures
• Rectilinear distance (L1 norm)
Pi = (ai, bi)
– d(X, Pi) = |x - ai| + |y - bi|
X = (x, y)
• Straight line or Euclidean distance (L2 norm)
– d(X, Pi) =
Pi = (ai, bi)
(x - a i ) 2 + (y - b i ) 2
• Tchebyshev distance (L norm)
X = (x, y)
Pi = (ai, bi)
– d(X, Pi) = max{|x - ai|, |y - bi|}
X = (x, y)
Classification of Planar Facility Location Problems
# of facilities
Objectives
Distance measures
Rectilinear
Minisum
Euclidean
Tchebyshev
SingleFacility
Rectilinear
Minimax
Euclidean
Tchebyshev
Facility
Location
Rectilinear
Minisum
Euclidean
Tchebyshev
MultiFacility
Rectilinear
Minimax
Euclidean
Tchebyshev
Planar Single-Facility Location
Formulations
• Minisum Formulation :
m

Min f(x) =  w i  d X, Pi
i 1

where X = (x, y) : location of the new facility
Pi = (ai, bi) : location of the i-th existing facility, i = 1, …, m
wi : weight associated to the i-th existing facility
ti
For example, wi = c i ,
vi
where ci : cost per hour of travel,
ti : number of trips per month,
vi : average velocity.
• Minimax Formulation :
Min f(x) = Max
i = 1, …, m
{wi  d(X, Pi)}
 Min z
s. t. wi  d(X, Pi)  z, i = 1, …, m
Insights for the Minisum Problem with
Euclidean Distance
Hole
P2
P1
Horizontal
pegboard
P3
P5
String
P4
w5
w1
w2
w3
w4
Weight proportional
to wi
• Majority Theorem :
When one weight constitutes a majority of the total weight, an optimal new facility
location coincides with the existing facility which has the majority weight.
Minisum Location Problem with
Rectilinear Distances
m
Min f(x, y) =  w i  [|x  a i | |y  b i |]
i=1
Note that
f(x, y) = f1(x) + f2(y)
m
where
 w i |x  a i |
f1(x) = i=1
m
f2(y) =  w i |y  b i |
i=1
The cost of movement in the x direction is independent of the cost of
movement in the y direction, and viceversa.
Now, we look at the x direction.
f1(x) is convex  a local min is a global min.
Minisum Location Problem with
Rectilinear Distances (cont.)
• The coordinates of the existing facilities are sorted so that
a1  a2  a3  ….
• Now, we consider the case of m = 3.
Case x  a1 :
f1(x) = w1 |a1 - x| + w2 |a2 - x| + w3 |a3 - x|
= - (w1 + w2 + w3)x + w1 a1 + w2 a2 + w3 a3
= - W x + w1 a1 + w2 a2 + w3 a3,
where W = w1 + w2 + w3
Case a1  x  a2 :
f1(x) = w1 |a1 - x| + w2 |a2 - x| + w3 |a3 - x|
= (w1 - w2 - w3)x - w1 a1 + w2 a2 + w3 a3
= (- W + 2 w1) x - w1 a1 + w2 a2 + w3 a3
…
Objective Function f1(x)
f1(x)
- w1 - w2 - w3
The slope changes sign
w1 + w2 + w3
w1 - w2 - w3
w1 + w2 - w3
w1
w2
w3
a1
a2
a3
x
Minisum Location Problem with
Rectilinear Distances (cont.)
• Slopes of f1(x) :
M0 = - (w1 + w2 + w3) = - W
M1 = 2 w1 + M0
M2 = 2 w2 + M1
M3 = 2 w3 + M2 = w1 + w2 + w3 = W
• Median conditions :
f1(x) is minimized at the point where the slope changes from nonpositive
to nonnegative.
M1 = w1 - w2 - w3 < 0  2 w1 < (w1 + w2 + w3) = W
w1 < W/2
M2 = w1 + w2 - w3  0  2 (w1 + w2)  (w1 + w2 + w3) = W
(w1 + w2)  W/2
Example 1
• Problem Data :
m=3
a1 = 10
a2 = 20
a3 = 40
w1 = 5
w2 = 6
w3 = 4
• Solution :
W = w1 + w2 + w3 = 15
W/2 = 7.5
w1 = 5 < 7.5
w1 + w2 = 11 > 7.5

Minimizing point : a2 = 20
Linear Programming Formulation
Min f1(x) = w1 |a1 - x| + w2 |a2 - x| + w3 |a3 - x|

Min z = w1 (r1+ s1) + w2 (r2+ s2) + w3 (r3+ s3),
s. t.
x - r 1 + s1
x
- r 2 + s2
x
Dual variables
= a1,
= a2,
: y1
: y2
- r3+ s3 = a3,
: y3
rj, sj  0, j = 1, 2, 3.
• Relationships among variables : aj - x = rj - sj , |aj - x| = rj + sj, rj, sj  0.
• If both rj, sj > 0, we can reduce each by j = min {rj, sj}.
• This maintains feasibility and reduces z
 In an optimal solution, at least one of the rj and sj is 0, i. e., rj  sj = 0.
Linear Programming Formulation (cont.)
• Dual Problem :
Max g = - a1y1 - a2 y2 - a3 y3 + (w1 a1 + w2 a2 + w3 a3)
s. t.
y1 + y2 + y3 = w1 + w2 + w3 = W
0  yj  2 wj,
j = 1, 2, 3
 Min a1y1 + a2 y2 + a3 y3
s. t.
y1 + y2 + y3 = W
0  yj  2 wj,
j = 1, 2, 3
0  y1  2 w1
a1
0  y2  2 w2
W
1
• Complementary slackness conditions :
0 < yj* < 2 wj  x* = aj
a2
a3
0  y3  2 w3
a1  a2  a3
W
2
Example 1 : Dual Solution
• f1(x) = 5 |x - 10| + 6 |x - 20| + 4 |x - 40|
• W = 15
0  y1  10
10
0  y2  12
15
1
20
40
0  y3  8
y1* = 10
y2* = 5
y3* = 0
0 < y2* < 12 
x* = a2 = 20
15
2
Minisum Location Problem with Euclidean
Distances
m
1
2 2
 w i  [(x  a i )  (y  b i ) ]
• Min f(x, y) = i=1
2
(ai, bi)
The optimum location
is always in the convex
hull of
{(a1, b1), …, (am, bm)}
• Colinear case : all the points are in a line
(ai, bi)
 The problem reduces to minimizing f1(x), which is the rectilinear distance
problem.
Non-colinear Case
1
2 2
• The graph of [(x  a i )  (y  bi ) ] is a cone (strictly convex function).
2
1
2 2
y
[(x  a i )  (y  bi ) ]
2
contours
(ai, bi)
x
(ai, bi, 0)
y
m
x
1
2 2
• f(x, y) =  w i  [(x  a i )  (y  bi ) ] is strictly convex unless the convex
i=1
hull is a line segment.
2
Non-colinear Case (cont.)
• First order optimality conditions :
 f(x, y)
0
x x
0

 f(x, y)
0
y y
(x0, y0) is optimal
0
• Any point where the partial derivatives are zero is optimal.
Let
 i (x, y) 
wi
m
and
1
2 2
[(x  a i )  (y  bi ) ]
2
 (x, y)    i (x, y)
i 1
Non-colinear Case (cont.)
 f(x, y) m
   i (x, y)  (x - a i ) = 0
i=1
x
 f(x, y) m
   i (x, y)  (y - b i ) = 0
i=1
y
m

x
 a i   i (x, y)
i=1
 (x, y)
m
y
 b i   i (x, y)
i=1
 (x, y)
No-colinear Case (cont.)
• If the optimal solution is in an exiting facility (ai, bi), then  i (x, y)   .
• A simple way to avoid the problem of division by zero is to “perturb”
the problem as follows :
m
Min f(x, y)   w i  [(x  a i )  (y  b i )   ]
2
2
1
2
i 1
f(x,y)
where  > 0 and small.
f(x,y) is flat near the optimum.
x
(x*,y*)
y
Weiszfeld’s Algorithm
1 m
Initialization : (i) (x , y )   (a i , bi )
m i 1
0
or
0
1 m
 w  (a i , bi ),
(ii) (x , y ) 
W i 1 i
0
0
where
m
Iterative step (k = 1, 2, …) :
xk =
 a i  i (x k-1 , y k-1 )
i=1
 (x k-1 , y k-1 )
m
yk =
Terminating conditions :
 b i  i (x k-1 , y k-1 )
i=1
 (x k-1 , y k-1 )
(i) (x k , y k ) - (x k-1 , y k-1 )  
or (ii) f(x k-1 , y k-1 ) - f(x k , y k )  
W = w1 + ... + w m
Example 2
Problem Data :
m=4
P1 = (0, 0)
w1 = 1
P2 = (0, 10)
w2 = 1
P3 = (5, 0)
w3 = 1
P4 = (12, 6)
w4 = 1
Solution :
x0 = (5+12)/4 = 4.25
y0 = (10+6)/4 = 4
k
(x, y)
f(x, y)
1
4.023, 3.116
24.808
2
3.949, 2.647
24.665
5
3.958, 2.124
24.600
10
3.995, 2.011
24.597
Optimum
4.000, 2.000
Minimax Location Problem with
Rectilinear Distances
• Possible example : locating an ambulance with the existing facilities being
the locations of possible accidents.
Hospital
Hospital
P1
h1
h3
P3
X
P2
P4
h4
Hospital :
h2
Hospital
Poss. Accident :
Ambulance :
Minimax Location Problem with
Rectilinear Distances (cont.)
• Notation
EF (existing facilities) locations : Pi = (ai, bi), i = 1, …, m
NF (new facility = ambulance) location : X = (x, y)
Travel distance from EF i to the nearest hospital = hi, i = 1, …, m
Travel distance from NF to EF i = |x - ai| + |y - bi|
• Formulation :
Min
where

Min
s. t.
g(x, y)
g(x, y) = max {|x - ai| + |y - bi| + hi : i = 1, …, m}
z
|x - ai| + |y - bi| + hi  z,
i = 1, …, m
Minimax Location Problem with
Rectilinear Distances (cont.)
Make the
intersection as small
as possible with the
largest diamond as
small as possible.
• We want to make the inequalities linear
(3)
(2)
(1)
(4)
|x - ai| + |y - bi|  z - hi

x - ai + y - bi  z - hi
ai - x + bi - y  z - hi
ai - x + y - bi  z - hi
x - ai + b i - y  z - h i
(1)
(2)
(3)
(4)
Minimax Location Problem with
Rectilinear Distances (cont.)
Min
z
x + y - z  ai + bi - hi,
s. t.
i = 1, ..., m
x + y + z  ai + bi + hi, i = 1, ..., m
- x + y - z  - ai + bi - hi, i = 1, ..., m
- x + y + z  - ai + bi + hi, i = 1, ..., m

Min
s. t.
z
x + y - z  c1
where
c1 = Min {ai + bi - hi}
i = 1, ..., m
x + y + z  c2
c2 = Max {ai + bi + hi}
- x + y - z  c3
c3 = Min {- ai + bi - hi}
- x + y + z  c4
c4 = Max {- ai + bi + hi}
i = 1, ..., m
i = 1, ..., m
i = 1, ..., m
Minimax Location Problem with
Rectilinear Distances (cont.)
Min
z
s. t.
- x - y + z  - c1
x + y + z  c2
x - y + z  - c3
- x + y + z  c4
c5 = Max {c2 - c1, c4 - c3}
 z  (c2 - c1)/2 (lower bound)
 z  (c4 - c3)/2 (lower bound)
z = c5/2
Optimal solution :
(x1, y1, z1) = 1/2 (c1 - c3, c1 + c3 + c5, c5)
(x2, y2, z2) = 1/2 (c2 - c4, c2 + c4 - c5, c5)
The line segment joining (x1, y1) and (x2, y2) is the set of optimal NF
locations
Minimax Location Problem with Euclidean
Distances
• Examples :
helicopter in an emergency unit,
radio transmitter
(x, y)
(ai, bi)
EF : (ai, bi), i = 1, …, m
NF : (x, y)


min
g(x, y)
where
g(x, y) = max {[(x - ai)2 + (y - bi)2]1/2, i = 1, …, m}
min
z
s. t.
[(x - ai)2 + (y - bi)2]1/2  z,
min
z
s. t.
(x - ai)2 + (y - bi)2  z ,
i = 1, …, m
i = 1, …, m
Elzinga-Hearn Algorithm (1971)
Step 1. Choose any two points and go to Step 2.
Step 2. Find the minimum covering circle for the chosen points*. Discard from the set
of chosen points those points not defining the minimum covering circle, and go
to Step 3.
Step 3. If the constructed circle contains all the points, then the center of the circle is a
minimax location, so stop. Otherwise, choose some point outside the circle,
add it to the set of points defining the circle, and go to Step 2.
* Find the minimum covering circle for the chosen points :
A.
If there are two points, let the two points define the diameter of the circle.
B.
If there are three points defining a right or obtuse triangle, let the two points
opposite to the right or obtuse angle define the diameter of the circle.
Otherwise, construct a circle through the three points (see Figure 1).
C.
If there are four points, construct a circle using as defining points those
indicated in Figure 2.
Elzinga-Hearn Algorithm (cont.)
Defining points :
Defining points :
Defining points :
ABD
BCD
CD
Defining points :
C
BD
A
Defining points :
A
Defining points :
B
BD
ACD
A
Defining points :
Defining points :
B
B
C
AD
Defining points :
ABD
Defining points :
AD
ABD
Figure 1. Alternative B
Figure 2. Alternative C
Planar Multi-Facility Location Problems
P1
P4
w24
w11
X2
X1
v12
w23
P2
w12
P3
Old Facility :
New Facility :
Minisum Multi-Facility Location Problem
with Rectilinear Distances
Location of new facilities:
Xj = (xj, yj), j = 1, …, n.
Location of existing facilities:
Pi = (ai, bi), i = 1, …, m.
Weight between new facilities j and k: vjk,
where k > j.
Weight between new facility j and existing facility i: wji.
Problem formulation:
Min f((x1,y1), …, (xn, yn)) = f1(x1, …, xn) + f2(y1, …, yn)
where
f1(x1, …, xn) =
f2(y1, …, yn) =
n
m
v
jk
x j  x k   w ji x j  a i
v
jk
y j  y k   w ji y j  bi
1 j k  n
1 j k  n
j1 i 1
n m
j1 i 1
Example 3
•
Problem Data : n = 2 (NF)
m = 3 (EF)
 0 2
v = [vjk] = 

 0 0
 2 1 0
w = [wji] = 

 4 0 5
x = [xj] = (x1, x2)
a = [aj] = (10, 20, 40)
Min f1(x1, x2) = 2 |x1 - x2|
+ 2 |x1 - 10|
+ |x1 - 20|
Min f1(x1, x2) = 2 (p12 + q12) + 2 (r11 + s11) + (r12 + s12)
s. t.
+ 4 |x2 - 20|

+ 4 (r21 + s21) + 5 (r23 + s23)
x1 - x2 - p12 + q12
x1
=0
- r11 + s11
x1
= 10
- r12 + s12
x2
= 20
- r21 + s21
x2
•
+ 5 |x2 - 40|
= 10
- r23 + s23
Relationships among variables :
x1 - x2 = p12 - q12,
|x1 - x2| = p12 + q12,
p12, q12  0
xi - aj = rij - sij,
|xi - aj| = rij + sij,
rij, sij  0
= 40
Example 3 (Dual Problem)
Max (- 10 u11 - 20 u12 - 10 u21 - 40 u23) + (102 + 201 + 104 + 405)
 Min
10 u11
+ 20 u12
+ 10 u21
+ 40 u23
s. t. z12 + u11 + u12
=5
- z12
+ u21
+ u23
=7
0  z12  4
0  u11  4
0  u12  2
0  u21  8
0  u23  10
w ji
 2 1 0


 4 0 5

wi 
v jk
 0 2


  2 0

w v
ji
 5
 
 7
12
jk
Equivalent Network Flow Problem
E1
(10)
u11  4
(0)
4
5
N1
8
12
(0)
(0)
1
u12  2
Cap = 
(20)
z12  4
u21  8
Cap = 
E2
N3
(0)
7
N2
u23  10
(40)
Cap = 
(0)
E3
After drawing the network, the solution can be usually obtained by inspection.
12
Equivalent Network Flow Problem (cont.)
Complementary slackness conditions :
1.

xk*  xj*
zjk* < 2 vjk 
xj*  xk*
0 < zjk*
In particular,
0 < zjk* < 2 vjk  xj* = xk*
2.

ai  xj*
uji* < 2 wji 
xj*  ai
0 < uji*
In particular,
0 < uji* < 2 wji  xj* = ai
In Example 3,
0 < z12* < 2 v12  x1* = x2*,
0 < u12 = 2 w12  x2* = a1 = 10.
If the network is not connected, then the problem decomposes into independent
problems, one for each component.
Example 4
Four hospitals located within a city are cooperating to establish a centralized blood-bank facility that
will serve the hospitals. The new facility is to be located such that the (total) distance traveled is
minimized. The hospitals are located at the following coordinates: P1=(5,10), P2=(7,6), P3=(4,2), and
P4=(16,3). The number of deliveries to be made per week between the blood-bank facility and each
hospital is estimated to be 3, 8, 2, and 10, respectively. Assuming rectilinear travel, determine the
optimum location.
m=4
P1 = (5, 10)
P3 = (4, 2)
Computation of x*:
a3 = 4
a1 = 5
a2 = 7
a4 = 16
Computation of y*:
b3 = 2
b4 = 3
b2 = 6
b1 = 10
w1 = 3
w3 = 2
P2 = (7, 6)
P4 = (16, 3)
w2 = 8
w4 = 10
w3 = 2
w1 = 3
w2 = 8
w4 = 10
w3 = 2
w3 + w1 = 5
w3 + w1 + w2 = 13

x* =
w3 = 2
w4 = 10
w2 = 8
w1 = 3
w3 = 2
w3 + w4 = 12

y* =
W=
Example 5
(10, 16)
16
h4 = 11
Find the optimal location of an ambulance
with respect to four (known) possible
14
accident locations which coordinates are
P1=(6,11), P2=(12,5), P3=(14,7), and
P4=(10,16). The objective is to minimize
12
(6, 11)
the maximum distance from the ambulance
location to an accident location and from
10
h1 = 10
(12, 9)
the accident location to its closest hospital.
The distances from the accident locations
8
to their closest hospitals are h1=10, h2=16,
(14, 7)
h3=14, and h4=11. Assume that distances
are rectilinear. If multiple optima exist,
(10, 7)
6
h3 = 14
(12, 5)
find all optimal solutions.
h2 = 16
(5, 4)
6
8
10
12
14
Example 5 Solution
m=4
P1 = (6, 11)
P3 = (14, 7)
h1 = 10
h3 = 14
P2 = (12, 5)
P4 = (10, 16)
h2 = 16
h4 = 11
c1 = min {ai + bi - hi} = min {6+11-10, 12+5-16, 14+7-14, 10+16-11} =
c2 = max {ai + bi + hi} = max {6+11+10, 12+5+16, 14+7+14, 10+16+11} =
c3 = min {-ai + bi - hi} = min {-6+11-10, -12+5-16, -14+7-14, -10+16-11} =
c4 = max {-ai + bi + hi} = max {-6+11+10, -12+5+16, -14+7+14, -10+16+11} =
c5 = max (c2 - c1, c4 - c3}= max {
-
,
-
}=
Optimal objective value:
c
z* = 5 
2
Set of optimal solutions: line segment defined by the following end points:
1
(c1 - c3, c1 + c3 + c5) =
2
1
(x2*, y2*) =
(c2 - c4, c2 + c4 - c5) =
2
(x1*, y1*) =
1
(
2
1
(
2
-
,
+
+
-
,
+
-
)=(
)=(
,
,
)
)
Example 6
Given five existing facilities located at points P1, P2, P3, P4, and P5 as shown below, determine the
optimum location for a new facility which will minimize the maximum distance to the existing
facilities. Assume that distances are Euclidean.
Elzinga-Hearn algorithm:
Figure 1 : Initial set of points = {P1, P2, P3}; center = C1.
Figure 2 : 2nd set of points = {P1, P2, P4}; center = C2.
Figure 3 : 3rd set of points = {P2, P4, P5}; center = C3 (optimal location).
Figure 1
P3 
P5
P2
P1
C1
P1 
P2
P3
P4
Figure 2
P4
P5
P2
P1
C2
P3
P2
P1
P4
Figure 3
P4
P5
P2
P1
C3
P3
P1
P5
P4