1 Introduction: The first location problem? Early in the 17th century, Fermat (1601 - 1665) posed the following problem at the end of an essay on maxima and minima (Kuhn, 1967): "Let he who does not approve of my method attempt the solution of the following problem: Given three points in the plane, find a fourth point such that the sum of its distances to the three given points is a minimum." Donote the three given points by P 1 = (a 1 , b 1 ), P 2 = (a 2 , b 2 ), and P 3 = (a 3 , b 3 ), and let X = (x, y) be the fourth point to be found. The sum of the distances from X to the three given points is given by the function: f(X) = d(X,P 1 ) + d(X,P 2 ) + d(X,P 3 ), where d(X,P i ) = (x˚–˚a i ) 2 ˚+˚( y˚–˚b i ) 2 is the Euclidean distance for i = 1, 2, 3. The problem is to minimize f(X). (Figure 1) P 1 P 2 P 3 X Figure 1
38
Embed
Introduction: The first location problem? P1 P2 = (a2 …cecas.clemson.edu/~pmdrn/Dearing/location/totalcost.pdf2 The following development assumes the triangle ∆(P1, P2, P3) has
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
1
Introduction: The first location problem?
Early in the 17th century, Fermat (1601 - 1665) posed the following problem at the end of an
essay on maxima and minima (Kuhn, 1967):
"Let he who does not approve of my method attempt the solution of the following problem:
Given three points in the plane, find a fourth point such that the sum of its distances to the
three given points is a minimum."
Donote the three given points by P1 = (a1, b1), P2 = (a2, b2), and P3 = (a3, b3), and let X
= (x, y) be the fourth point to be found. The sum of the distances from X to the three given
points is given by the function:
f(X) = d(X,P1) + d(X,P2) + d(X,P3),
where d(X,Pi) = (x – ai)2 + (y – bi)2 is the Euclidean distance for i = 1, 2, 3.
The problem is to minimize f(X). (Figure 1)
P1
P2
P3X
Figure 1
2
The following development assumes the triangle ∆(P1, P2, P3) has all angles less than 120o. The
alternate case is considered subsquently.
Torricelli solved the problem before 1640. Circles circumscribing the equilateral triangles
constructed on the sides of and outside the triangle ∆(P1,P2,P3) intersect at the point X that
is sought, called the Torricelli point. (Figure 2)
Cavalieri, in "Exercitationes geometricae", 1647, shows that the lines from the Torricelli point to
the given points Pi subtend angles of 120o.
P1
P2
P3
X
Figure 2
3
Simpson, in "Doctrine and Application of Fluxions", 1750, proved that the three lines joining the
outside vertices of the equilateral triangles defined above to the opposite vertices of the given
triangle intersect at the Torricelli point. The three lines are called Simpson lines. (Figure 3)
P1
P2
P3
X
Sim
pson
line
Simpson line
Simpson line
Figure 3
Heinen, in 1834 proved that the lengths of the three Simpson lines are equal and equal to the
minimum sum of distances.
Observe that if one of the angles of ∆(P1, P2, P3) is greater than 120o, then the three Simpson
∆(P1, P2, P3). The case of having one angle greater than 120o is dealt with using properties
of mathematical programming.
Heinen also proved that for a triangle in which one angle is greater than or equal to 120o, the
vertex of this angle is the minimizing point.(Figure 4)
4
P1
P2
P3
X =
Figure 4
Fasbender, in 1846 proved that the perpendiculars to the Simpson lines through the three
given points are the sides of the largest equilateral triangle circumscribing these points, and
that the altitude of this triangle equals the minimum sum of distances. (Figure 5)
P1
P2
P3
XSimpson line
Sim
pson
line
Simpson line
Figure 5
5
Fasbender's results leads to a dual problem: Find the largest equilateral triangle circumscribing the
given triangle ∆(P1,P2,P3).
Restrict consideration to the case where the solution lies inside the triangle ∆(P1,P2,P3).
Lemma 1-1 : Let Z be any point interior to an equilateral triangle ∆(Z1,Z2,Z3) with altitude h.
Let hi be the length of the perpendicular from Z to the side opposite Zi. Then h = h1 + h2 +
h3.
Z
Z1
Z2
Z3
h
h1
h2
h3
Figure 6
Proof : In Figure 6, the area of ∆(Z1, Z2, Z3) = area of ∆(Z, Z2, Z3) + area of ∆(Z1, Z, Z3) +
area of ∆(Z1, Z2, Z). Thus h2
3 =
h
3 (h1 + h2 + h3) and the lemma is proved.
6
Lemma 1-2 : (Weak duality) Given three points P1, P2, P3 in the plane and any equilateral
triangle with altitude h circumscribing ∆(P1, P2, P3), for any point X interior to the
equilateral triangle,
h ≤ d(X,P1) + d(X,P2) + d(X,P3).
Proof: For each i, construct a perpendicular from X to the side of the equilateral triangle
passing through Pi, and let hi be the length of the perpendicular. (Figure 7) Then hi ≤d(X,Pi) since the leg of a right triangle is no longer than the hypotenuse. Apply Lemma 1.
P1
P2
P3
X
2h
3h1h
d(X,P )1
d(X,P )3
d(X,P )2
Figure 7
7
Lemma 2 shows that the altitude of any circumscribing equilateral triangle is less than or equal the
sum of distances from the Pi to any point interior to ∆(P1, P2, P3).
In the proof of Lemma 2, observe that hi = d(X,Pi) iff the line segments X,Pi are
perpendicular to the sides of the equilateral triangle. This occurs when the line segments
X,Pi subtend angles of 120o at X.
Theorem 1-1: The three Simpson lines intersect at the point X. The angles formed at X are all
equal to 60o. (Figure 5)
Theorem 1 may be proved by elementary geometry.
The geometrical approach presented above is limited to 3 given points. For n > 3, and more
general functions of distance, mathematical programming techniques are used.
Exercise: Use straight edge and compass to construct the Torricelli point for the following three
points.
P12
P
3P
Exercise: Attempt to construct the Torricelli point when the triangle ∆(P1, P2, P3) has one angle
greater than 120o.
8
The generalized Fermat problem, or Steiner - Weber problem
Extend the Fermat problem by allowing more than three points, by allowing positive weights
associated with each point, and by allowing the points to be placed anywhere in the plane.
Given distinct points Pi = (ai, bi) in the plane, and positive weights wi, for i = 1, . . . n. The
problem is to find a point X = (x, y) that minimizes the weighted sum of Euclidean distances
from X to the given points. Let f(X) = ∑i=1
nwid(X,Pi) . The generalized Fermat problem is
to minimize f(X).
This problem is also called the "Weber" problem because the economists Alfred Weber considered
it in Uber den Standort der Industrien in 1909. In their book What is Mathematics (1941),
Courant and Robins call this the "Steiner" problem because of the contributions made by
Steiner. It is also called the "minisum" problem in the location literature.
Observe that since the distance function d(X,Pi) is a convex function of X, and the weighted sum
of convex functions is convex, then f(X) is a convex function of X.
If the n distinct points Pi are not collinear, f(X) is a strictly convex function of X.
If the n distinct points Pi are not collinear, f(X) has a unique minimum point.
If the n distinct points Pi are collinear, then f(X) is a piece-wise linear, convex function of
X on the line through the points Pi, and is strictly convex elsewhere. This case is handled
subsequently. The points are assumed to be non collinear for the remainder of this
development.
Necessary and sufficient conditions for a point X to be an optimal location are developed next.
These conditions lead to several properties of the optimal solution, and to an algorithm for
iterating to the optimal solution.
9
Consider the mechanical analog of the generalized Fermat problem. Holes are drilled through a
table top at positions corresponding to the given points Pi. For each i, a string is placed
through the hole corresponding to Pi, and a weight wi is attached to the string. The strings
are tied in a common knot on top of the table. Assuming no friction, and negligible weight to
each string, the equilibrium position of the knot is the minimum location for X. (Figure 8)
This model is also called Varignon's Frame, and was suggested by Georg Pick in his
Mathematical Appendix to Weber's book.
X1P
3P
2P
4P
5P
w1
w2
w3
w4
w5
Figure 8
Each string corresponds to a vector from X to Pi, given by (Pi – X). Normalizing the length
of each vector (Pi – X) and multiplying by the weight wi gives a vector wi (Pi – X)
d(X,Pi) of
magnitude of wi, which corresponds to the force generated by the weight wi acting along
each string from X to Pi.
10
The resultant of all the forces along the strings at a point X is given as follows.
If X ≠ Pi the resultant force R(X) at X is given by:
R(X) = ∑i=1
n
w i (P i – X )d(X,Pi)
provided X ≠ Pi for all i.
If X = Pj for some j = 1, . . . n, define Rj as the vector sum of all forces from Pj to Pi
for i ≠ j, but disregarding the force wj at Pj :
Rj = ∑i=1 i≠j
n
w i (Pi – Pj)d(Pj,Pi)
.
Define the resultant force R(Pj) at Pj to incorporate the force wj as follows:
R(Pj) = max( |Rj| – wj, 0 )Rj|Rj|
.
Thus, if wj ≥ |Rj| , then R(Pj) = 0, else, there is a resultant vector in the direction of Rj
with magnitude |Rj| – wj.
Theorem 1-2 : The point X minimizes f(X) iff R(X) = 0.
Proof : If X is not a vertex, then the convexity and differentiability of f imply that the first order
condition ∇ f(X) = R(X) = 0 is necessary and sufficient for X to be a minimum.
If X = Pj for some j, consider the change in f along the direction Z from Pj to Pj + tZ
where |Z| = 1 and t ≥ 0.
Direct calculation shows that ddt f(Pj + tZ) = wj – RjZ for t = 0.
Hence the direction of greatest decrease is Z = Rj|Rj|
. Thus Pj is a local minimum iff
wj – RjRj|Rj|
≥ 0, i.e., if the directional derivative is nonnegative in the direction of greatest
decrease, then the directional derivative is nonnegative in all directions. However, wj – RjRj|Rj|
= wj – RjRj|Rj|
= wj – |Rj| ≥ 0 implies R(Pj) = 0.
11
Example: Suppose n = 3, all wi = 1, and the triangle ∆(P1, P2, P3) has no angle greater than
120o. A zero resultant force R(X) at a point X ≠ Pi implies that the normalized vectors(P i – X )d(X,Pi)
subtend angles of 120o.
P12P
3P
X
If the triangle ∆(P1, P2, P3) has one angle greater than 120o, say at P1, the resultant vector
R1 = (P2 – P1)d(P2,P1) +
(P3 – P1)d(P3,P1) has magnitude < 1 = w1, and P1 is optimal.
P1
2P3P R1
12
Example: Suppose n = 4, all wi are equal, and all four points Pi are extreme points to the
convex hull of the Pi, i.e., none of the given points Pj is interior to the triangle formed by the
remaining three points. Then the lines connecting "opposite" points, see Figure 9, intersect at
a point X interior to the convex hull of the four points.
The conditions of optimality state that X is the optimal solution since R(X) = 0, i.e., the
normalized resultant forces from X to each Pi "cancel out" and the resultant force is zero.
X1
P
3P
2P
4P
If one of the given four points, say Pj, is interior to the triangle formed by the remaining three
points, then X = Pj is the optimal solution, since |Rj| < 1 = wj.
1P
3P
2P
4P
R1
13
Corollary 1-1 : (Majority Property) If wj ≥ ∑i=1 i≠j
nwi for some j, then Pj is the optimal location.
Proof : Observe that |Rj| = | ∑i=1 i≠j
n
w i P i – P jd(Pj,Pi)
| ≤ ∑i=1 i≠j
n
w i |Pi – Pj|d(Pj,Pi)
= ∑i=1 i≠j
nw i .
If wj ≥ ∑i=1 i≠j
nwi ≥ |Rj| , then R(Pj) = 0, and Pj is the optimal solution.
The following stronger result is proven in Kuhn and Kuenne (1962) and Hamacher (1995).
Theorem 1-3 : Pj is the optimal location if and only if
wj2 ≥ ( ∑i=1 i≠j
n
w i ai – ajd(Pj,Pi)
)2 + ( ∑i=1 i≠j
n
w i b i – bjd(Pj,Pi)
)2 .
Theorem 1-4 : If the point X minimizes f(X), then X is in the convex hull of the points Pi.
Proof : If the point X is a vertex Pi, then it is trivially in the convex hull. Otherwise, the
condition R(X) = 0 yields
∑i=1
n
wiPi
d(X,Pi) = ∑
i=1
n
wiX
d(X,Pi)
or X = ∑i=1
n
wiPi
d(X,Pi) / ∑
i=1
n
wi
d(X,Pi) (*)
Define λi = wi
d(X,Pi) / ∑
i=1
n
wi
d(X,Pi) so that X = ∑
i=1
n
λiPi , and ∑i=1
n
λi = 1, which is a
convex combination of the points Pi.
Corollary 1-2 : If the point X minimizes f(X) and is not a vertex, then X is not on the
boundary of the convex hull.
Proof : This follows because the λi > 0 for each i.
14
Iterative solution methods for the weighted Euclidean distance minisum problem
An iterative method is suggested by expression (*). This method attempts to find an X that
satisfies the conditions of optimality.
Xk+1 = ∑i=1
n
wiPi
d(Xk,Pi) / ∑
i=1
n
wi
d(Xk,Pi) .
A suitable starting point is given by X0 = ∑i=1
n
wiPi / ∑i=1
nwi which is the center of gravity of
the Pi.
This iterative scheme is equivalent to the gradient method with specified step length (Kuhn,
1967). Note that –R(X) is the gradient of f(X) when it exists. This method was first
suggested by E. Weiszfeld (1937). Convergence is proved by Kuhn (1973).
Another approach taken by Love and Morris (1975) approximates the nondifferentiable Euclidean
distance function by a differentiable function called the hyperbolic approximation:
dε(X, Pi) = [(x – ai)2 + (y – bi)2 + ε ]1/2 for ε a small positive real number.
The generalized Fermat problem is given by
min fε(X) = ∑i=1
nwidε(X,Pi) .
Love and Morris show give convergence results and computational experience that show
fε(X) → f(X) as ε → 0.
15
Generalized Fashbender duality
A dual to the generalized Fermat problem is developed that generalizes the geometric dual
developed for the case with n = 3 and all weights equal to 1.
Given distinct points Pi = (ai, bi) in the plane, and positive weights wi, for i = 1, . . . n.
Let Ui = (ui, vi) denote n two-dimensional vectors. The dual to the generalized Fermat
problem asks for the vectors Ui which maximize
g(U1, U2, . . . , Un) = ∑i=1
n
Ui.Pi
s.t. ∑i=1
n Ui = 0 (4)
|Ui| ≤ wi for i = 1, . . . , n. (5)
Theorem 1-5 : (weak duality) For any X and any set of Ui, satisfying (4) and (5),
g(U1, U2, . . . , Un) ≤ f(X).
Proof : g(U1, U2, . . . , Un) = ∑i=1
n
Ui.Pi = ∑i=1
n
Ui.Pi – ∑i=1
n Ui X = ∑
i=1
n U i(P i – X ) .
≤ ∑i=1
n |Ui| |Pi – X| (by the Schwartz inequality, Ui(Pi – X) ≤ |Ui| |Pi–X| )
≤ ∑i=1
nwid(X,Pi) = f(X).
16
Theorem 1-6 : (Strong duality) Given any X that solves the generalized Fermat Problem, there
exists feasible Ui such that g(U1, U2, . . . , Un) = f(X).
Proof : Case 1: X ≠ Pi for all i = 1, . . . , n.
Set Ui = wi
d(X , Pi) (Pi – X) for i = 1, . . . , n.
Then ∑i=1
n Ui = R(X) = 0, and |Ui| =
wid(X , Pi)
|Pi – X| = wi, for i = 1, . . .,
n.
Thus the Ui are feasible, and the inequalities in the proof of the weak duality Theorem
become equations so that g(U1, U2, . . . , Un) = f(X).
Case 2: X = Pj for some j.
Set Ui = wi
d(Pj, Pi) (Pi – Pj) for i ≠ j, and define Uj = – ∑
i=1 i≠j
nUi .
Then ∑i=1
n Ui = 0, and |Ui| = wi, for i ≠ j as above.
Observe that Uj = – Rj and R(Pj) = 0 imply wj ≥ |Rj| = |Uj|.
Thus the Ui are feasible, and the inequalities in the proof of the weak duality Theorem
become equations. The jth term contributes nothing to the sum since