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• 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)

– d(X, Pi) = |x - ai| + |y - bi|

• Straight line or Euclidean distance (L2 norm)

– d(X, Pi) =

• Tchebyshev distance (L norm)

– d(X, Pi) = max{|x - ai|, |y - bi|}

(x - a ) + (y - b )i2

i2

X = (x, y)

Pi = (ai, bi)

Pi = (ai, bi)

Pi = (ai, bi)

X = (x, y)

X = (x, y)

Classification of Planar Facility Location Problems

FacilityLocation

Single-Facility

Multi-Facility

Minisum

Minimax

Rectilinear

Euclidean

Tchebyshev

Rectilinear

Euclidean

Tchebyshev

Rectilinear

Euclidean

Tchebyshev

Rectilinear

Euclidean

Tchebyshev

Minisum

Minimax

# of facilities Objectives Distance measures

Planar Single-Facility Location Formulations

• Minisum Formulation :

Min f(x) =

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

For example, wi = ,

where ci : cost per hour of travel, ti : number of trips per month,

vi : average velocity.

• Minimax Formulation :

Min f(x) = Max {wi d(X, Pi)} Min z

s. t. wi d(X, Pi) z, i = 1, …, m

c i

t

vi

i

i = 1, …, m

w d X Pii

m

i1

,

Insights for the Minisum Problem with Euclidean Distance

• 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.

w5

w1

w2

w4

w3

P1

P2

P3

P4

P5

Weight proportional to wi

String

Hole

Horizontal pegboard

Minisum Location Problem with Rectilinear Distances

Min f(x, y) =

Note that f(x, y) = f1(x) + f2(y)

where f1(x) =

f2(y) =

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.

w [ | x a | | y b |i i ii =1

m

]

w | x a |i ii =1

m

w | y b |i ii =1

m

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)

a3a2a1

w3

w2

w1

w1 + w2 + w3

w1 + w2 - w3

w1 - w2 - w3

- w1 - w2 - w3

The slope changes sign

x

f1(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

m = 3

a1 = 10 a2 = 20 a3 = 40

w1 = 5 w2 = 6 w3 = 4

W = w1 + w2 + w3 = 15

W/2 = 7.5

w1 = 5 < 7.5

w1 + w2 = 11 > 7.5

Minimizing point : a2 = 20

• Problem Data :

• Solution :

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), Dual variables

s. t. x - r1+ s1 = a1, : y1 x

- r2+ s2 = a2, : y2

x - 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 r j 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

• Complementary slackness conditions :

0 < yj* < 2 wj x* = aj

1 2

0 y1 2 w1

a1

0 y2 2 w2

a2

a3

0 y3 2 w3

WW

a1 a2 a3

Example 1 : Dual Solution

• f1(x) = 5 |x - 10| + 6 |x - 20| + 4 |x - 40|

• W = 15

y1* = 10

y2* = 5

y3* = 0

0 < y2* < 12 x* = a2 = 20

1 2

0 y1 10

10

0 y2 12

20

40

0 y3 8

1515

Minisum Location Problem with Euclidean Distances

• Min f(x, y) =

• Colinear case : all the points are in a line

The problem reduces to minimizing f1(x), which is the rectilinear distance

problem.

(ai, bi)

The optimum location

is always in the convex

hull of

{(a1, b1), …, (am, bm)}

w [(x a ) (y b )i i2

i2

i =1

m

]1

2

(ai, bi)

Non-colinear Case

• The graph of is a cone (strictly convex function).

• f(x, y) = is strictly convex unless the

convex hull is a line segment.

(ai, bi)

contours

(ai, bi, 0)

[(x a ) (y b )i2

i2 ]

1

2

[(x a ) (y b )i2

i2 ]

1

2

y

x

yx

w [(x a ) (y b )i i2

i2

i =1

m

]1

2

Non-colinear Case (cont.)

• First order optimality conditions :

• Any point where the partial derivatives are zero is optimal.

Let

and

ii

i2

i2

(x , y)w

[(x a ) (y b )

]1

2

f(x , y)

x0

f(x , y)

y0

x

y

0

0

(x0, y0) is optimal

(x , y) (x , y)i

i

m

1

Non-colinear Case (cont.)

f(x , y)

x(x , y) (x - a )

f(x , y)

y(x , y) (y - b )

ii =1

m

i

ii =1

m

i

xa (x , y)

(x , y)

yb (x , y)

(x , y)

i ii =1

m

i ii =1

m

= 0

= 0

No-colinear Case (cont.)

• If the optimal solution is in an exiting facility (a i, bi), then .

• A simple way to avoid the problem of division by zero is to “perturb” the

problem as follows :

where > 0 and small.

f(x,y) is flat near the optimum.

M in f(x , y) w [(x a ) (y b ) ]i i2

i2

1

2

i 1

m

x

y

f(x,y)

(x*,y*)

i (x , y)

Weiszfeld’s Algorithm

Initialization :

Iterative step (k = 1, 2, …) :

Terminating conditions :

(i) (x , y )1

m(a , b )

(ii) (x , y )1

Ww (a , b ) , w h ere W = w + . . . + w

0 0i i

i 1

m

0 0i i i

i 1

m

1 m

(i) (x , y ) - (x , y )

(ii) f(x , y ) - f(x , y )

k k k -1 k -1

k -1 k -1 k k

x = a (x , y )

(x , y )

y = b (x , y )

(x , y )

ki i

k -1 k -1

i =1

m

k -1 k -1

ki i

k -1 k -1

i =1

m

k -1 k -1

or

or

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

1

2

5

10

Optimum

(x, y)

4.023, 3.116

3.949, 2.647

3.958, 2.124

3.995, 2.011

4.000, 2.000

f(x, y)

24.808

24.665

24.600

24.597

Minimax Location Problem with Rectilinear Distances

• Possible example : locating an ambulance with the existing facilities

being the locations of possible accidents.

XP3

P4

P2

P1

h3

h4h2

h1

Hospital

Hospital

Hospital

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 g(x, y)

where g(x, y) = max {|x - ai| + |y - bi| + hi : i = 1, …, m}

Min z

s. t. |x - ai| + |y - bi| + hi z, i = 1, …, m

Minimax Location Problem with Rectilinear Distances (cont.)

• We want to make the inequalities linear

|x - ai| + |y - bi| z - hi

x - ai + y - bi z - hi (1)

ai - x + bi - y z - hi (2)

ai - x + y - bi z - hi (3)

x - ai + bi - y z - hi (4)

Make the

intersection as small

as possible with the

largest diamond as

small as possible.

(2) (4)

(3) (1)

Minimax Location Problem with Rectilinear Distances (cont.)

Min z

s. t. 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

- x + y + z - ai + bi + hi, i = 1, ..., m

Min z

s. t. x + y - z c1 where c1 = Min {ai + bi - hi}

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

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 = 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

z (c2 - c1)/2 (lower bound)

z (c4 - c3)/2 (lower bound)

Minimax Location Problem with Euclidean Distances

• Examples : helicopter in an emergency unit,

radio transmitter

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, i = 1, …, m

min z

s. t. (x - ai)2 + (y - bi)2 z , i = 1, …, m

(ai, bi)(x, y)

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 :

BD

Defining points :

ABD

Defining points :

ABD

Defining points :

AD

A

B

Defining points :

BCD

Defining points :

ACD

Defining points :

ABD

Defining points :

AD

A

B

A

Defining points :

CD

Defining points :

BD

Figure 1. Alternative B Figure 2. Alternative C

B

C

C

Planar Multi-Facility Location Problems

Old Facility :

New Facility :

X2

X1

P4

P3

P2

P1

v12

w24

w23

w12

w11

Minisum Multi-Facility Location Problem with Rectilinear Distances

ij

n

1j

m

1iji

nkj1kjjk axwxxv

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) = ij

n

1j

m

1iji

nkj1kjjk bywyyv

Example 3• Problem Data : n = 2 (NF) m = 3 (EF)

v = [vjk] = w = [wji] =

x = [xj] = (x1, x2) a = [aj] = (10, 20, 40)

Min f1(x1, x2) = 2 |x1 - x2| + 2 |x1 - 10| + |x1 - 20| + 4 |x2 - 20| + 5 |x2 - 40|

Min f1(x1, x2) = 2 (p12 + q12) + 2 (r11 + s11) + (r12 + s12) + 4 (r21 + s21) + 5 (r23 + s23)

s. t. x1 - x2 - p12 + q12 = 0

x1 - r11 + s11 = 10

x1 - r12 + s12 = 20

x2 - r21 + s21 = 10

x2 - r23 + s23 = 40

• 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

0 2

0 0

2 1 0

4 0 5

Example 3 (Dual Problem)

Max (- 10 u11 - 20 u12 - 10 u21 - 40 u23) + (102 + 201 + 104 + 405)

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 v w w v

2 1

7

1 2

ji jk i ji jk

0

4 0 5

0 2

2 0

5

Equivalent Network Flow Problem

After drawing the network, the solution can be usually obtained by inspection.

N1

N2

E3

E1

E2 N31

4

8 12

Cap =

Cap =

Cap =

u11 4

u12 2

z12 4

u21 8

u23 10

(0)

(0)

(0)

(0)

(0)

12(20)

(10)

(40)

5

7

Equivalent Network Flow Problem (cont.)

Complementary slackness conditions :

1. 0 < zjk* xk* xj*

zjk* < 2 vjk xj* xk*

In particular,

0 < zjk* < 2 vjk xj* = xk*

2. 0 < uji* ai xj*

uji* < 2 wji xj* ai

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) w1 = 3 P2 = (7, 6) w2 = 8

P3 = (4, 2) w3 = 2 P4 = (16, 3) w4 = 10 W =

Computation of x*:

a3 = 4 w3 = 2 w3 = 2

a1 = 5 w1 = 3 w3 + w1 = 5

a2 = 7 w2 = 8 w3 + w1 + w2 = 13 x* =

a4 = 16 w4 = 10

Computation of y*:

b3 = 2 w3 = 2 w3 = 2

b4 = 3 w4 = 10 w3 + w4 = 12 y* =

b2 = 6 w2 = 8

b1 = 10 w1 = 3

Example 5

Find the optimal location of an ambulance

with respect to four (known) possible

accident locations which coordinates are

P1=(6,11), P2=(12,5), P3=(14,7), and

P4=(10,16). The objective is to minimize

the maximum distance from the ambulance

location to an accident location and from

the accident location to its closest hospital.

The distances from the accident locations to

their closest hospitals are h1=10, h2=16,

h3=14, and h4=11. Assume that distances

are rectilinear. If multiple optima exist, find

all optimal solutions.

(5, 4) 6 8 10 12 14

(6, 11)

(10, 16)

(12, 5)

(10, 7)

(14, 7)

(12, 9)

h4 = 11

h2 = 16

h3 = 14

h1 = 10

16

14

12

10

8

6

Example 5 Solution

m = 4 P1 = (6, 11) h1 = 10 P2 = (12, 5) h2 = 16

P3 = (14, 7) h3 = 14 P4 = (10, 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:

z* =

Set of optimal solutions: line segment defined by the following end points:

(x1*, y1*) = (c1 - c3, c1 + c3 + c5) = ( - , + + ) = ( , )

(x2*, y2*) = (c2 - c4, c2 + c4 - c5) = ( - , + - ) = ( , )

25c

1

21

2

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

P1

P2

P4

C1

P5

P2

P3

P1

Figure 2

P3

P1

P2

P4

P5

C2

P2P1

P4

Figure 3

P3

P1

P2

P4

P5

C3

P4

P5P1